8SRK1HV LIBRARY U8WER9TV ol CALISOP-'IA METALLURGICAL CALCULATIONS % Qraw-MlBock &, 7m PUBLISHERS OF BOOKS F O R^ Coal Age * Electric Railway Journal Electrical World ^ Engineering News-Record American Machinist v Jngenieria Internacional Engineering g Mining Journal ^ Po we r Chemical 6 Metallurgical Engineering Electrical Merchandising METALLURGICAL CALCULATIONS BY JOSEPH W. RICHARDS, A.C., Ph.D. Professor of Metallurgy in Lehigh University, Secretary (and Past-President) of the American Electrochemical Society, V ice-President, American Institute of Mining Engineers, Mem- ber of the U. S. Naval Consulting Board, Author of t( Aluminium, Its Metallurgy, etc. 1 ' ONE VOLUME EDITION PART I Introduction, Chemical and Thermal Principles Problems in Combustion, and Radiation and Conduction of Heat PART II Applications to the Metallurgy of Iron and Steel PART III Applications to Other Metals (Non-ferrous Metals) THIRD IMPRESSION McGRAW-HILL BOOK COMPANY, INC, NEW YORK: 239 WEST 39TH STREET LONDON: 6 & 8 BOUVERIE ST., E. C. 4 1918 COPYRIGHT, 1910, 1915, 1917, 19x8, BY THE McGRAw-HiLL BOOK COMPANY, INC. COPYRIGHT, 1906, 1907, 1908, BY THE McGRAw PUBLISHING COMPANY. THK MAPLE PRESS YORK PA PREFACE While issuing Parts I, II and III separately, for the use of students, it is thought that most persons, particularly practical metallurgists, will prefer to have the whole work in one volume. The writer has endeavored to correct all the mistakes which crept into former editions, and to add such new and useful phys- ical and chemical data as have appeared to date. Some new features are the tables of thermochemical constants (which will assist in predicting undetermined heats of formation), the esti- mation of many latent heats of fusion and vaporization of the elements (which data can be used pro tern, as first approxima- tions), and the evaluation of vapor tension formulas for the ele- ments in both liquid and solid states (likewise to be used as first approximations). My thanks are due to Mr. Leonard Buck and Mr. Y. Takikawa, students in metallurgy, for considerable assistance given in calculating and completing the new tabular data. The author is grateful to many friends whose kindly criticisms have pointed out mistakes or suggested improvements. He ap- preciates the careful translations which have been made into Italian by Sr. Remo Catani, into Russian by Engineer Koshkin, into German by Prof. Neuman and Engineer Brodal, and regrets that the breaking out of the great war has interrupted transla- tions already begun into French and Spanish. These inter- national recognitions are simply proofs that the day of Quanti- tative Metallurgy is dawning all over the world, and that the application of Metallurgical Calculations is the path to Metallurgical Efficiency. JOSEPH W. RICHARDS. LEHIGH UNIVERSITY, January 15, 1918. CONTENTS. PAGE INTRODUCTION. Scope of the Treatise iii-xi CHAPTER I. THE CHEMICAL EQUATION , . 1-12 Atomic Weights 1 Relative Weights 2 Relative Volumes of Gases 2 Exact Weights and Exact Volumes 3 Weights and Volumes of Gases 4 Corrections for Temperature 5 Corrections for Pressure 7 Corrections for Temperature and Pressure 7 Problem 1. Combustion of Coal 8 Problem 2. Combustion of Natural Gas 10 Problem 3. Oxidation in a Bessemer Converter.. 11 CHAPTER II. THE APPLICATIONS OP THERMOCHEMISTRY. 13-40 Thermochemical Nomenclature , 13 Thermochemical Data 18-40 Oxides ;. 18 Hydrates 19 Sulfides 21 Selenides, Tellurides 22 Arsenides, Antimonides, Phosphides, Nitrides ... 23 Hydrides, Hydrocarbons 24 Carbides 25 Silicides, Fluorides 26 Chlorides 27 Carbonates 28 Bi-Carbonates 29 Nitrates 30 Silicates 31 Sulphates ' 32 Bi-sulphates, Phosphates, Arsenates, Tungstates. 33 Borates, Molybdates, Titanates 34 Manganates, Aluminates, Cyanides 35 Cyanates, Metallo-cyanides 36 vii Viii CONTENTS. Amalgams, Alloys 37 Thermochemical Constants of Bases and Acids. 38 CHAPTER III. THE USE OP THE THERMOCHEMICAL DATA . 41-60 Simple Combinations Complex Combinations 41 Double Decompositions . . . . 43 Calorific Power of Fuels 46 Problem 4. Combustion of Natural Gas 47 Dulong's Formula 48 The Theoretical Temperature of Combustion 50 Specific Heats of Gases Produced 51 Combustion with Heated Fuel or Air 52 Problem 5. Calorific Intensity of Natural Gas . 54 The Eldred Process of Combustion 55 Temperatures in the "Thermit Process" 58 CHAPTER IV. THE THERMOCHEMISTRY OF HIGH TEMPERA- TURES 61-117 Reduction of Zinc Oxide by Carbon 63 General Remarks 68 Reduction of Iron Oxide by Hydrogen 69 Specific Heats of the Elements 70 Latent Heats of Fusion of the Elements 71 Latent Heats of Vaporization of the Elements 73 Thermophysics of the Elements 74 Tables of Thermophysics of the Elements 75-103 Efficiency of Furnaces 104 Problem 6. Efficiency of a Rockwell Furnace. . 106 Problem 7. Efficiency of an Amalgam Retort. . 107 Problem 8. Efficiency of a Zinc Distilling Fur- nace 108 Thermophysics of Alloys 109 Tables of Thermophysics of Alloys 109-113 Problem 9. Efficiency of a Steel Melting Fur- nace 114 Problem 10. Efficiency of a Siemen's Furnace. . 114 Problem 11. Efficiency of a Foundry Air-fur- nace 114 Problem 12. Efficiency of a Foundry Cupola. . 115 CHAPTER V. THERMOPHYSICS OF CHEMICAL COMPOUNDS 118-144 Oxides. . 118 CONTENTS. ix Problem 13. Efficiency of Alumina Melting Fur- nace 127 Problem 14. Reduction of Iron Oxides by CO gas. 128 Problem 15. Reduction of Iron Oxides by Car- bon 129 Chlorides 131 Bromides, Iodides 133 Fluorides, Sulfides 134 Compound Sulfides 135 Arsenides, Antimonides, Selenides, Tellurides 136 Cyanides, Hyposulfites, Sulfates 136 Carbonates, Nitrates 137 Borates, Phosphates, Chromates, Arsenates 138 Chlorates, Aluminates, Titanates, etc 139 Silicates 139 Miscellaneous Materials 142 Slags 144 CHAPTER VI. ARTIFICIAL FURNACE GAS 145-184 Simple Producers 146 Problem 16. Investigation of a Simple Pro- ducer 149 Problem 17. Drying of Moist Producer Gas. . . 154 Mixed Gas Producers 158 Problem 18. Investigation of a Morgan Pro- ducer 163 Mond Gas 169 Problem 19. Calculations on a Mond Producer . 169 Reactions on Pre-heating Mond Gas 175 Water Gas 177 Problem 20. On a Dellwick-Fleischer Plant. ... 179 Problem 21. Blast Required for D-F. Plant ... 182 CHAPTER VII. CHIMNEY DRAFT AND FORCED DRAFT . . 185-199 Chimney Draft 185 Available Head 191 Problem 22. Design of Puddling Furnace Chim- ney 192 Problem 23. Raising Steam by Waste Heat ... 194 X CONTENTS. CHAPTER VIII. CONDUCTION AND RADIATION OF HEAT. 186-200 Principles of Heat Conduction 200 Table of Heat Conductivities of Metals 203 Principles of Heat Transfer 206 Problem 24. Cooling of Hot Air in a Tube 208 Table of Heat Conductivities of Insulating Ma- terials 211 Radiation .'. 213 Table of Radiating Capacity (Emissivity) 215 APPENDIX 217-231 Problem 25. Combustion of Hydrocarbons 217 Problem 26. Calorific Values of Coal, Oil and Gas.. 219 Problem 27. Calorific Values of Various Gases 219 Problem 28. Combustion of Coal under a Boiler. . . 220 Problem 29. Combustion of Anthracite Coal 221 Problem 30. Air Required for Powdered Coal 221 Problem 31. Efficiency Test on a Boiler 222 Problem 32. Temperature of Powdered-coal Flame. 223 Problem 33. Temperature at Tuyeres in Blast Fur- nace 223 Problem 34. Efficiency of Powdered Coal Firing, . . 224 Problem 35. Calculation of Chimney Draft 224 Problem 36. Calorimeter Test of Temperature 224 Problem 37. Calorimeter Test for Specific Heat 225 Problem 38. Economy of a Feed- water Heater 225 Problem 39. Working of Gas-engine 226 Problem 40. Efficiency of a Gas Producer 226 Problem 41. Temperature of Producer-gas Flame. . 227 Problem 42. Drying Kiln for Peat 227 Problem 43. Power from Waste Coke-oven Gases. . 228 Problem 44. Efficiency of By-product Coke Ovens. 229 Problem 45. Decomposition of Steam in a Producer . 229 Problem 46. Chimney for a Gas Furnace 230 Problem 47. Loss of Heat through Sides of Chimney 231 CONTENTS TO PART II. PAGE CHAPTER I. BALANCE SHEET OF THE BLAST FURNACE. .235-249 Fuel 236 Ore 239 Flux... 243 Blast 244 Problem 51. Balance of a Swedish Furnace. . . 245 CHAPTER II. CALCULATION OF A FURNACE CHARGE .. 250-267 Flux and Slag 251 Problem 52. Calculation of Flux needed 259 Comparison of Values of Fuels 261 Comparison of Values of Fluxes 263 Comparison of Values of Ores 265 Problem 53. Cost of Pig Iron calculated 267 CHAPTER III. UTILIZATION OF FUEL IN THE BLAST FUR- NACE 268-280 Problem 54. Generation of Heat in Blast Fur- nace 268 Problem 55. Efficiency of Hot Blast Stoves 270 Problem 56. Power from Blast Furnace Gases. 272 Gruners Ideal Working 274 Minimum Carbon necessary in the Furnace 277 CHAPTER IV. HEAT BALANCE SHEET OF BLAST FUR- NACE 281-304 Heat received and developed 281 Combustion of Carbon. Illustration 281 Sensible Heat in Hot Blast. Illustration 284 Formation of Pig Iron 285 Formation of Slag 286 xi xii CONTENTS TO PART II. PAGE Heat absorbed and disbursed 287 Heat in Waste Gases 287 Heat in Slag 288 Heat in Pig Iron 289 Heat conducted away and radiated 289 Heat in Cooling Water 290 Heat for drying and dehydrating charges 290 Heat for decomposing carbonates 291 Heat for reduction of iron oxides 292 Heat for reduction of other oxides 292 Decomposition of Moisture of Blast 293 Problem 57. Complete Heat Balance Sheet 294 CHAPTER V. RATIONALE OF HOT-BLAST AND DRY-BLAST 305-312 Problem 58. (for practice) Heat Balance Sheet. 305 Hot Blast. Temperatures obtained 308 Dried Blast. Increased Temperatures obtained 309 Illustration: Temperatures in a specific case. ... 310 Table of temperatures under various conditions. 312 CHAPTER VI. PRODUCTION, HEATING AND DRYING OP BLAST 313-332 Production of blast. Work required 313 Problem 59. Work of a blowing engine 315 Measurement of pressure of blast. Indicator cards . . 317 Heating of the blast 318 Problem 60. Efficiency of an iron-pipe stove. . . 320 Problem 61. Efficiency of fire-brick stoves 322 Drying air blast 325 Illustration: Increased efficiency of blowing en- gines t 326 Illustration: Refrigeration required 326 Problem 62. Efficiency of drying apparatus. . . 327 Illustrations: Calculation of saturation tempera- ture 331 Problem 63. Cooling compressed air by river water 331 CHAPTER VII. THE BESSEMER PROCESS. 333-350 Air required 333 Problem 64. Maximum and minimum blast. . 334 CONTENTS TO PART II. xiii PAGE Air received 336 Problem 65. Volume efficiency of blowing plant 336 Blast pressure 340 Illustration: Back pressure in converter 341 Problem 66. Distribution of blast pressure .... 343 Flux and Slag 346 Illustration: Lime needed in a basic blow 347 Recarburization 348 Problem 67. Recarburization balance sheet 349 CHAPTER VIII. THERMO-CHEMISTRY OF THE BESSEMER PROCESS. 351-367 Elements consumed 351 Heat balance sheet items 353 Heat developed by oxidation 356 Heat of formation of slag 357 Illustration: Heat of formation of basic slag 358 Heat in escaping gases 360 Heat conducted to the air 362 Heat radiated 363 Problem 68. Complete heat balance sheet 363 CHAPTER IX. THE TEMPERATURE INCREMENT IN THE BESSEMER CONVERTER 368-380 Combustion of Silicon 369 Combustion of Manganese 371 Combustion of Iron 372 Combustion of Titanium 373 Combustion of Aluminum, Nickel, Chromium 374 Combustion of Carbon 375 Combustion of Phosphorus 377 Resume' 379 CHAPTER X. THE OPEN-HEARTH FURNACE 381-395 Gas producers 381 Flues to furnace; regenerators 382 Problem 69. Sizes of regenerators 384 Valves and ports 388 Problem 70. Areas of gas and air ports 389 Laboratory of furnace 391 XIV CONTENTS TO PART II. PAGE Problem 71. Efficiency of laboratory 393 Chimney flues and chimney, miscellaneous 395 CHAPTER XI. THERMAL EFFICIENCY OF OPEN-HEARTH FURNACES 396-428 Heat in warm charges ; in gas used ' 397 Heat in air used 398 Heat of combustion; of oxidation of the bath 399 Heat of formation of slag; heat in melted steel 399 Reduction of iron ore; decomposition of carbonates . . 400 Heat in slag 401 Loss by imperfect combustion ; in chimney gases .... 402 Loss by conduction and radiation 403 Problem 72. Heat balance sheet. . . 403 Problem 73. Thermal efficiencies 411 Problem 74. New Siemens' furnace 418 Problem 75. The Monell process 424 CHAPTER XII. THE ELECTROMETALLURGY OF IRON AND STEEL > -.429-451 Electrothermal reduction of iron ores 429 Problem 76. Production of pig-iron electrically . 430 Problem 77. Production of nickelif erous pig . . . 435 Production of Steel 440 Problem 78. The induction electric furnace. . . 442 Problem 79. The electric " pig and ore " process 443 Problem 80. Electrolytic refining; Burgess' process 446 APPENDIX: PROBLEMS FOR PRACTICE 452-465 Problem 81. Diameter of tuyeres for a blast fur- nace 452 Problem 82. Blast furnace slag calculation 452 Problem 83. Carbon burned before the tuyeres 453 Problem 84. Calculation of blast furnace charge. . . 454 Problem 85. Slag calculation for various slags 454 Problem 86. Weight of flux needed 455 Problem 87. Volume of blast, and cooling of same by expansion 456 Problem 88. Output of a blast-furnace 456 CONTENTS TO PART II. XV PAGE Problem 89. Efficiency of the blowing engines 456 Problem 90. Power producible from blast-furnace gases ^ 457 Problem 91. Proportion of gases required by hot- blast stoves 457 Problem 92. Complete balance sheet of a blast fur- nace 458 Problem 93. Working of a foundry cupola 460 Problem 94. The Monell process in an open-hearth furnace 460 Problem 95. Volume of blast required by a Besse- mer converter 461 Problem 96. Volume of blast received by a Besse- mer converter 461 Problem 97. Loss of weight of a Bessemer charge. 462 Problem 98. Balance sheet of a complete Bessemer blow 462 Problem 99. Lime needed in a basic Bessemer blow 464 Problem 100. Power of blowing engines required for a converter 465 INDEX . , 466 CONTENTS TO PART III. PAGE CHAPTER I. THE METALLURGY OF COPPER 469-577 Roasting and Smelting of Copper Ores - 469 The Roasting Operation 475 Problems 101-104. Pyritic Smelting 483 Fundamental Principles 485 Theoretical Temperature at the Focus 486 Use of Auxiliary Coke 488 Rate of Smelting 489 Problems 105, 106. Smelting of Copper ores 499 Reverberatory Smelting 501 Problems 107, 108. "Bringing Forward" of Copper Matte 514 The Welsh Process 514 Problem 109. "Blister Roasting" or Roasting-Smelting 523 " Bessemerizing" Copper Matte 525 Theoretical Temperature Rise 527 Problem 110. The Electrometallurgy of Copper 534 Electrolytic Processes 536 Treatment of Ores by Electrolysis 536 Problem 111. Electrolytic Treatment of Matte 539 Use of Matte as Anode 539 Problem 112. Use of Matte as Cathode 545 Problem 113. Extraction from Solutions 548 (1) Use of Iron Anodes 548 xvii xviil CONTENTS TO PART III. PAGE Problems 114, 115. (2) Use of Insoluble Anodes 551 Problems 116, 117, 118 Electrolytic Refining of Impure Copper 558 Energy Absorbed by Electrolyte , 559 Problem 119. Energy Lost in Contacts 562 Problem 120. Energy Lost in Conductors 567 Problem 121. Electric Smelting 572 Problem 122. CHAPTER II. THE METALLURGY OF LEAD 578-604 The Volatility of Lead 583 Roasting of Lead Ores 586 Problem 123. Reduction of Roasted Ore 591 Problem 124. The Electrometallurgy of Lead 598 Problems 125, 126. CHAPTER III. THE METALLURGY OF SILVER AND GOLD. 605-619 Electrolytic Refining of Silver Bullion 605 Problems 127, 128. The Volatilization of Silver and Gold 614 CHAPTER IV. THE METALLURGY OF ZINC 620-657 Roasting of Sphalerite 620 Problems 129, 130. Reduction of Zinc Oxide 627 Thermochemical Considerations 628 Heating up the Charge 628 Problem 131. Distillation of the Charge 632 Problems 132, 133, 134. Electric Smelting of Zinc Ores 637 Zinc Vapor 643 CONTENTS TO PART III. xix PAGE Problem 135. Condensation of Zinc and Mercury 649 Metallic Mist or Fume Vapor 656 CHAPTER V. METALLURGY OF ALUMINIUM 658-661 Electrolytic furnace reduction of alumina 658 INDEX. . 663 METALLURGICAL CALCULATIONS. INTRODUCTION. The making of calculations respecting the quantitative work- ing of any process, furnace or piece of apparatus used in metal- lurgical operations is of the greatest importance for estimating the real efficiency of the process, for determining avenues of waste and possible lines of improvement, and for obtaining the best possible comprehension of the real principles of operation involved. The possibility of making such calculations respecting any process, furnace or apparatus depends on skill in collecting such accessary data as can be obtained by observation or measure- ment, the insight or intuition to see the further use which can be made of aid data when once obtained, and, finally on the possession of a working knowledge of the fundamental chem- ical, physical and mechanical principles involved in the calcu- lations. The highest desideratum, all in all, however, is a plain analytical, common sense mind, capable of clear, logical thinking It is the writer's conviction that no study of details, or even observation of plants in actual operation, can supply the insight into metallurgical processes and principles, such as is gained by these, calculations, in addition to the high grade of mental training involved. SCOPE OF THE TREATISE Discussion of the chemical equation. Weights and volumes of gases. Correction of gas volumes for temperature and pressure. Combustion of commercial fuels. Heat of chemical combination; of combustion. Theoretical flame temperatures: With pure oxygen. With ordinary air. xxi xxii INTRODUCTION. With diluted air: Farley's system. With hot air and cold gas. With hot air and hot gas. Effect of excess air. Calculation of furnace efficiencies. Chimney draft. Water gas. Producer gas: Efficiency, effect of drying. Mixed gas: Use of steam in producers. Increased efficiency. Maximum steam permissible. Transmission of heat through metals, brick, etc. Regenerative gas furnace: Proportioning of gas and air regenerators. Efficiency of regenerators. Heat balance sheet. Theoretical temperatures under different conditions. Gas engines: Calculation of temperature in cylinder. Efficiency; balance sheet. Cupolas: Amount of blast required. Efficiency of running. Blast furnaces: Balance sheet of materials. Calculation of blast received. Efficiency of blowing engines. Power and dimensions of blowing engines. Carbon consumed at tuyeres. Effect of atmospheric changes. Effect of the moisture in the blast. Calculation of the temperature. Effect of hot-blast. Heat balance sheet of the furnace. Power producible from the waste gases. Hot-blast stoves: Theory of iron-pipe and fire-brick stoves. Efficiency. Bessemer Converters: Blast required and time of operation Balance sheet of materials. INTRODUCTION. xxiii Heating efficiency of various ingredients of bath. Heat balance sheet. Theoretical rise in temperature. Conversion of copper matte. Open-hearth Furnaces: Pig and ore process; calculation of charge. Heat evolved or absorbed in bath reactions. Efficiency of furnaces; of furnaces and producers. Heat balance sheet. Pyritic smelting. Electric furnaces: Working temperatures. Heat balance sheet. Efficiency, Electrolytic furnaces: Absorption of heat in chemical decompositions. Equilibrium of temperature attained. Ampere and energy efficiency. Electrolytic refining: Calculation of plant and output Power requirements; temperature of baths. Ampere and energy efficiency. Condensation of metallic vapors: Principles involved. Application to condensation of zinc and mercury. CHAPTER I. THE CHEMICAL EQUATION. The calculation of the quantitative side of many metallur- gical processes depends upon the correct understanding of chemical equations. Every chemical equation is capable of giving three most important sets of data concerning the pro- cess which it represents; its shows the relative weights of the reacting substances, their relative volumes, when in the gas- eous state, and the surplus or deficit of energy involved in the reaction, when the heats of formation of the substances con- cerned are known. ATOMIC WEIGHTS. These are the basis of all quantitative chemical calculations. For metallurgical purposes we may use them in round num- bers as: Hydrogen H 1 Lithium Li 7 Beryllium Be 9 Boron B 11 Carbon C 12 Nitrogen . .N 14 Oxygen O 16 Fluorine F 19 Sodium Na 23 Magnesium Mg 24 Aluminium Al 27 Silicon Si 28 Phosphorus P 31 Sulphur S 32 Chlorine Cl 35.5 Potassium ". K 39 Calcium.. . .Ca 40 Arsenic As 75 Selenium Se 79 Bromine. Br 80 Strontium Sr 87 Zirconium Zr 90 Columbium Cb 94 Molybdenum Mo 96 Palladium Pd 106 Silver Ag 108 Cadmium Cd 112 Tin Sn 118 Antimony Sb 120 Tellurium Te 126 Iodine I 127 Barium Ba 137 Tantalum Ta 183 Tungsten W 184 2 METALLURGICAL CALCULATIONS. Titanium Ti 48 Tridium Ir 193 Vanadium V 51 Platinum Pt 195 Chromium Cr 52 Gold Au 197 Manganese Mn 55 Mercury Hg 200 Iron . . .Fe 56 Thallium Tl 204 Nickel ..Ni 58.5 Lead Pb 207 Cobalt Co 59 Bismuth Bi 208 Copper Cu 63.6 Thorium Th 232 Zinc Zn 65 Uranium. U 238 RELATIVE WEIGHTS. Writing any chemical equation between elements or their compounds, the relative weights concerned in the reaction are obtained directly from using these atomic weights, which are, themselves, of course, only relative. E.g., the slagging of iron in Bessemerizing copper matte: 2FeS + 3O 2 + 2SiO 2 = 2(FeO.Si0 2 ) + 2S0 2 176+ 96+ 120= 264 +128 These relative weights may be called kilograms or tons, pounds, ounces or grains; whatever units of weight we may be working in. In most metallurgical work we use kilograms or pounds as the convenient weight units. RELATIVE VOLUMES OF GASES. Where gases are involved, the relative number of molecules of the gaseous substance concerned in the reaction stands for the relative volume of that gas concerned in the reaction. It is usual and convenient to designate these relative volumes by Roman numerals, placed above the formulae. The following are some examples: Complete combustion of carbon: i i C + O 2 = CO 2 Incomplete combustion of carbon: i 2C + O Combustion of marsh gas: i n 2C + O 2 = 2CO i n i ii CH 4 + 2O 2 = CO 2 + 2H,0 THE CHEMICAL EQUATION. 3 Production of water gas: i i i C + H 2 O = CO + H 2 In each case above, the volume of a solid or liquid cannot be stated, but the relative volumes of all the gases taking part in a reaction are derived simply from the number of molecules of each gas concerned. These relative volumes may be called so many cubic meters or liters, or cubic feet, or whatever measure is wanted or being used. In most metallurgical calculations it is convenient to use cubic meters or cubic feet. EXACT WEIGHTS AND EXACT VOLUMES. If we specify or fix the weights used, as, for instance, so many kilograms of each substance as the numbers representing the relative weights, then we can, by using one constant factor, convert all the relative volumes into the real or absolute vol- umes corresponding to the weights used. If, for instance, we take the equation of the production of water gas: i i i C + H 2 = CO + H 2 12+ 18 = 28 + 2 With the relative weights written beneath and the relative volumes above, then if we fix the weights as kilograms, the relative volumes can be converted into actual volumes in cubic meters by multiplying by 22.22. A cubic meter of hydrogen gas (under standard conditions) weighs 0.09 kilogram, and thence 2 kilograms will have a volume of 2-7-0.09 = 22.22 cubic meters. But the relative volumes show that the CO and H 2 O gas are the same in volume as the hydrogen, and it, there- fore, follows that each Roman I stands for 22.22 cubic meters of gas, if the weights underneath are called kilograms. The consideration of these relations is very advantageous, because, by means of this factor (22.22) we pass at once from the weight of a gas to its volume; each molecule or molecular weight of a gas, in kilograms (or, briefly, each kilogram-mole- cule), represents 22.22 cubic meters of that gas. The conversion from weight to volume is quite as simple using English measures; and, by a strange coincidence, the same factor can be used as in the metric system. The coin- 4 METALLURGICAL CALCULATIONS. cidence alluded to is the fact (which the writer, as far as he can discover, was the first to notice) that there happens to be the same numerical relation between an ounce (av.) and a kilo- gram, as there is between a cubic foot and a cubic meter; in short, there are 35.26 ounces (av.) in a kilogram, and 35.31 cubic feet in a cubic meter. The difference is only one-seventh of one per cent., which can be ignored, and we can therefore say that if the relative weights in an equation are called ounces (av.), each molecule of gas in the equation represents 22.22 cubic feet. Example. The production of acetylene from calcium carbide : i CaC 2 + H 2 O = CaO + C 2 H 2 64 + 18 = 56 + 26 Interpreting by weights, and calling the relative weights ounces, we can call the I molecule of C 2 H 2 gas 22.22 cubic feet, so that, theoretically, 64 ounces of pure carbide, acting on 18 ounces of water, produce 56 ounces of lime and 26 ounces of C 2 H 2 gas, the volume of which is 22.22 cubic feet. WEIGHTS AND VOLUMES OF GASES. The weight of one cubic meter of dry air, under standard con- ditions (at Centigrade and at a pressure of 760 millimeters of mercury), is 1.293 kilograms. The composition of air is: By Weight. By Volume. Oxygen 3 21 Nitrogen ,. 10 80 or, in percentages, Oxygen 23.1 20.8 Nitrogen 76.9 79.2 While these may not represent the absolutely accurate aver- age composition of dry air, yet the variations are such that the above simple ratios, 3 to 10 and 21 to 80, are close enough for all practical purposes in metallurgy. The weight of one cubic foot of dry air is 1.293 ounces (av.). The weight of one cubic meter of hydrogen gas, at standard conditions, is 0.09 kilogram (1 cubic foot, 0.09 ounces). The formula of hydrogen gas is H 2 , is molecular weight 2; and since the densities of all gases are found experimentally to THE CHEMICAL EQUATION. 5 be proportional to their molecular weights, it follows that the density of any gas referred to hydrogen is expressed numer- ically by one-half its molecular weight. But the weight of a cubic meter of gas is the weight of a cubic meter of hydrogen multiplied by the density of the gas referred to hydrogen; thus, is obtained the weight of a cubic meter of any gas whose formula is known. Examples follow: Molecular Density Referred Weight of 1 Formula. Weight. to Hydrogen. Cubic Meter. Hydrogen , H 2 2 1 0.09 kilos. Water vapor H 2 O 18 9 0.81 " Nitrogen N 2 28 14 1.26 " Oxygen O 2 32 16 1.44 " Carbon monoxide. . .CO 28 14 1.26 " Carbon dioxide CO 2 44 22 1.98 " Marsh gas CH 4 16 8 0.72 " Etc., etc. In the case of water vapor, a particular explanation is neces- sary. It cannot exist under standard conditions, but condenses to liquid at 100 C. if under 760 millimeters pressure. It does exist at lower temperatures than 100, but only under partial pressures of fractions of an atmosphere; thus, at a pressure of 1-50 atmosphere (when it forms 1-50 of a mixture of gases) it can exist un condensed at ordinary temperatures (15 C. or 60 P.). The above weight for a cubic meter of water vapor (0.81 kilos, per cubic meter at standard conditions) is, there- fore, only a hypothetical value, but it is extremely useful, be- cause it enables us to calculate, by the principles to be ex- plained further on, the weight of a cubic meter of water vapor under any conditions of temperature and pressure at which it is possible for it to exist. CORRECTIONS FOR TEMPERATURE. The volumes of all permanent gases increase uniformly for uniform increase of temperature, so that, starting with a given volume at C., it is found that their volume increases 1/273 for every degree Centigrade rise in temperature. Thus, at 273 C., the volume is just double the volume at 0. Stating this fact in another way, we may say that the gas acts as if it would have no volume at 273 C., and would increase uniformly 6 METALLURGICAL CALCULATIONS. in volume from this point up to all measurable temperatures, the increment being, for each degree, 1-273 of the volume which the gas has at C. A still briefer statement is that the volume of a gas is proportional to its temperature above 273 C., or to its absolute temperature the latter being its tem- perature in C + 273. The converse of these principles is, that the density of a gas; that is, the weight of a unit volume, varies inversely as its absolute temperature. In Fahrenheit degrees, we can say that a gas expands 1-490 (1-273x5-9) for every degree rise above 32 C. ; or that the volume is proportional to the absolute temperatures, i.e., to the F. -32 + 490 ( = F. + 458). These principles are in constant use in metallurgical calcula- tions. Thus, one kilogram of coal will need 8 cubic meters of air to burn it, at and 760 millimeters pressure. What volume will that be at 30 C. and the same pressure? Since 30 C. is 30 + 273 = 303 absolute, the two temperatures will be 273 and 303, and the 303 Volume at 30 C. = volume at C.X It is always to be recommended to make such calculations in the above form; that is, to first put down the known volume, and then to multiply it by a fraction, the numerator and de- nominator of which are the two absolute temperatures. A moment's reflection will show which way the fraction must be written; if the new volume must be greater than the old, the value of the fraction must be greater than unity, the higher temperature must be in the numerator; if the fraction were inverted, we know that the result would be less than the start- ing volume instead of greater, which would be wrong. Taking an example in Fahrenheit degrees: What is the vol- ume under standard conditions of 175 cubic feet of gas meas- ured at 90 F. and standard pressure (29.93 inches of mer- cury)? Since 32 F. is 490 absolute and 90 F. is 548 abso- lute, and the new volume must be less than the starting vol- ume, we have Volume at 32 F, = 175x = 156.5 cubic feet THE CHEMICAL EQUATION, 7 CORRECTIONS FOR PRESSURE. The principle is that the volumes of a gas are inversely as the pressure upon it, so that doubling the pressure halves the volume, etc. Since the practical problems almost always pre- sent the pressure as two numbers, all that is necessary is to multiply the original volume by a fraction whose numerator and denominator are the two pressures concerned, and ar- ranged with the numerator the larger or the smaller of the two numbers, according as to whether the final volume should be greater or less than the starting one. Putting the solution in this manner avoids the primary school method of making a proportion, which is so apt to be expressed upside down, and absolutely avoids error with the minimum exercise of brain power. Examples. What is the volume of 100 cubic meters of any gas, if the pressure is changed to 700 millimeters? 760 Answer: 100 X z = 108.6 cubic meters. What is the volume at standard pressure of 150 cubic feet of gas measured at 28.50 inches of mercury? oo en Answer: 150 X^-^ = 142.8 cubic feet. Z\y . y5 CORRECTIONS FOR TEMPERATURE AND PRESSURE. These can be both allowed for, by simply correcting first for one, and then for the other. Actually, the simplest statement is to put down the original volume, then to multiply it by one fraction, which corrects for temperature, and again by another fraction correcting for pressure, thinking out carefully for each fraction the proper way of expressing it, i.e., whether it should increase or decrease the volume. Examples. What does 100 cubic meters of air at standard conditions become at 50 C. and 780 millimeters pressure? en i 070 Solution: 100X I* X~ = 115.3 cubic meters. to(J What is the weight of one cubic meter of hydrogen at 1000 8 METALLURGICAL CALCULATIONS. C. and 250 millimeters pressure, its weight at standard condi- tions being 0.09 kilograms? 273 250 Solution: 0.09X 1000 + 2 73 X 76Q = - 00637 kil S rams - What weight of oxygen is in 1500 cubic feet of dry air at 100 F. and at 28.50 inches of mercury? (Refer to weight of air at standard conditions, and percentage composition.) o Solution: l^gSX XXx 1 X 1500 = 374 ounces. lo OOo Z\).\)o What is the weight of 50 cubic meters of water vapor at a temperature of 30 C. and a pressure of 31.6 millimeters? 070 01 fi Solution: 0.81X^X^~X50 = 1.517 kilograms. 070 qi a or 50X^X^^X0.81 = 1.517 kilograms. oOo /OU The first expression calculates the weight of a cubic meter of water vapor at the assumed conditions, and multiplies by 50; the second calculates the hypothetical volume of the 50 cubic meters if reduced to standard conditions, and multiplies by the hypothetical weight of a cubic meter at those condi' tions. Problems Illustrating Preceding Principles. Problem 1. A bituminous coal contains on analysis: Carbon ............... 73.60 Moisture ............. 0.60 Hydrogen ............ 5.30 Ash ................. 8.05 Nitrogen ............. 1.70 Sulphur .............. 0.75 100.00 Oxygen .............. 10.00 It is powdered and blown into a cement kiln by a blast of air. Required: 1. The volume of dry air, at 80 F. and 29 inches barometric pressure theoretically required for the perfect com- bustion of one pound of the coal. 2. The volume of the products of combustion, using no excess THE CHEMICAL EQUATION. 9 of air, at 550 F. and 29 inches barometer, and their percentage, composition. Solution: The reactions of the combustion are: C + O 2 = CO 2 S + O 2 = SO 2 12 32 44 32 32 64 2FP + O 2 = 2H 2 O 4 32 36 Requirement (1): The oxygen required for burning one pound of coal is: Oxygen for carbon ........ = 0.7360x32/12 = 1.9(53 pounds. Oxygen for hydrogen ____ -= 0.0530x32/4 = 0.424 Oxygen for sulphur ...... = 0.0075X32/32 = 0.0075 " Total required ......................... ...... 2.3945 " Oxygen in coal .............................. 0.1000 " Oxygen to be supplied ....................... 2.2945 " Nitrogen accompanying ....................... 7.6483 Air necessary ....... . ........................ 9.9428 = 159.08 ounces (av.). Volume of air necessary (standard conditions) no = 123.03 cubic feet. Volume of air necessary at 80 F. and 29 inches barometer = ?^ = 139.4 cubic feet. (1) Requirement (2): Pounds. The weight of CO 2 formed is ........... 0.7360+1.963 = 2.699 The weight of H 2 O formed is ........... 0.0530 + 0.424 = 0.477 The weight of moisture is .......... .... 0.006 The weight of SO 2 formed is ............ 0.0075 + 0.0075 = 0.015 The weight of nitrogen altogether is 7.6483 + 0.0170 = 7.6653 pounds. .Converting these weights into ounces, and dividing each by the weight of a cubic foot of each gas in ounces, we have the volume of these theoretical products -at standard conditions : 10 METALLURGICAL CALCULATIONS. Volume CO 2 - 2.699x16+198 = 43.184+ 1.98 = 21.80 cubic feet. Volume H 2 O - 483 X 16 + 0.81 = 7728 + 0.81. = . 954 " Volume SO 2 = 0015x16 + 2.88 = 02400 + 2.88 - 0.08 " Volume N 2 - 7665X16 + 1.26 = 122.645 + 1.26= 9734 " Total volume at standard conditions. .... = 128.76 Volume at 550 F. and 29 inches barometer = - 2m " " . The percentage composition by volume follows from the above volumes as: CO 2 ........... 17.0 per cent. N 2 .......... 75.5 per cent. H 2 ....... ... 7.4 rr SO 2 ........... 0.1 Problem 2. Natural gas in the Pittsburg district contains: Marsh gas ................. CH 4 60.70 per cent. Hydrogen, ............... H 2 2903 Ethane ................... C 2 H 6 7.92 Olefiant gas ............. . .C 2 H< 0.98 Oxygen .................. .O 2 0.78 Carbonous oxide ............ CO 0.58 Required: (1) The volume of air necessary to burn it. (2) The volume of the products of combustion. Reactions : CH' + 2O ? = CO 2 + 2H 2 O 2H ? f O 2 = 2H 2 O 2C 2 H 6 + 7O- = 4CO ? + 6H 2 O C 2 H 4 + 3O 2 = 2CO ? + 2H 2 O 2CO -f O 2 = 2CO 2 Solution : Oxygen required for CH< ..... = 0.6070X 2 = L2140 parts. Oxygen required for H 2 ...... = 0.2903 X i =-- 1451 " THE CHEMICAL EQUATION. 11 Oxygen required for C 2 H 6 . . . . = 0.0792x7/2 = 0.2772 parts. Oxygen required for C 2 H 4 ____ = 0.0098 X 3 = 0.0294 " Oxygen required for CO ...... = 0.0058 X i = 0.0029 " 1.6686 " Deduct oxygen already present ................ 0.0078 Leaves oxygen to be supplied ................ 1.6608 Corresponding to air ................ ,' "I - = 7.985 " (1) U . Volumes of products of combustion: CO 2 H 2 O N 2 From CH 4 ........... 0.6070 1.2140 From H 2 ......... .. .. 0.2903 From C 2 H 6 ........... 0.1584 0.2376 From C 2 H 4 ........... 0.0196 0.0196 From CO ............ 0.0058 From air.. 6.3242 Total products . . . . 7908 1 . 761 5 6 . 3242 (2) The above solution is entirely in relative volumes, which may be all considered cubic feet or cubic meters, and are true for equal conditions of temperature and pressure. Problem 3. A Bessemer converter contains 10 metric tons of pig iron of the following composition: Carbon 3 . 00 per cent. Manganese 0.50 Silicon 1.50 Iron... 95.00 On being blown, one-third the carbon burns to CO 2 , the rest to CO; 5 per cent, of iron is oxidized, and no free oxygen escapes from the converter. Blast is assumed to be dry. Requirements : (1) What weight of oxygen is needed during the blow. (2) How many cubic meters of air, at standard conditions, will be needed. (3) What will be the average composition of the gases. 12 METALLURGICAL CALCULATIONS Reactions : C + O 2 = CO 1 Si-fO' - SiO 2 12 32 44 28 32 60 2C + O 2 = 2CO 2Fe + O* = 2FeO 24 32 56 112 32 144 2Mn + O 2 = 2MnO 110 32 142 Oxygen needed: C to CO 2 . . . . 100 kilos. X 32/12 = 266.7 kilos. CtoCO ..... 200 ' X32/24 =2667 " Mn to MnO. . 50 " X 32/1 10 = 14.5 " Si to SiO 2 . .150 " X 32/28 = 171.4 " Fe to FeO. . .500 " X32/112 --= 142.8 " Total ............................... 862.1 " (1) Nitrogen accompanying this ......... =2873.7 Air needed ........................... 3735 8 " 3735 8 Volume of air ...... = ~f~293' = 2889 ' 3 cublc meters ( 2 ) Volume of products of combustion: ) 7 CO 2 = 1004266.7 = 366.7 kilos = -~ = 185.2 cu. m. CO 200 + 266.7 --= 466.7 kilos = = 370.4 cu. m. 1 .zo N 2 . , 2280.7 " Total volume ............................. 2836.3 " Percentage composition by volume: CO 2 .......................... 6.5 per cent. CO .......................... 13.1 N 2 . -.80.4 CHAPTER II. THE APPLICATIONS OF THERMOCHEMISTRY. The ordinary interpretation of the chemical equation by weight gives us the quantitative relations governing the re- actions of substances upon each other, when the reaction pro- ceeds to a finish. Unfortunately, much chemical instruction, as given in our elementary schools, and even in some of the higher ones, stops with the consideration of the weight re- lations and does not proceed to those equally important rela- tions, the energetics of chemical reactions. In most of the reactions which occur in practical metallurgy, the quantity of the combustible used, or, more broadly, the amount of energy in the form of heat or electrical energy, necessary for pro- ducing the reactions desired, is the controlling factor regulat- ing the practicability or impracticability, the commercial success or failure, of the process. The relative values of fuels, the manufacture and utilization of gas, the principles of the regenerative furnace, the Bessemer process, electric reduction, and a host of metallurgical pro- cesses, depend essentially on the realization and utilization of chemical energy, and the only way to become conversant with the amounts of energy involved or evolved in these opera- tions is to understand thoroughly the thermochemistry of the reactions concerned. THERMOCHEMICAL NOMENCLATURE The heat evolved when compounds are formed from the elements (in a few cases heat is absorbed) is determined ex- perimentally by the use of the calorimeter. This branch is sometimes called "chemical calorimetry ; " it is practically a department of experimental physics. The data obtained give the heat evolved for the total change from the components at room temperature to the resulting products, at room teir- 13 14 METALLURGICAL CALCULATIONS. perature (or very near to it), and may be expressed per unit of weight of either component or of the substance formed. Thus, if carbon is burned in a calorimeter to carbonic acid gas, CO 2 , the heat evolved may be reported or expressed as 8100 gram calories per gram of carbon burnt. Or, 3037 gram calories per gram of oxygen used. Or, 2209 gram calories per gram of CO 2 formed. Of these three methods of expressing the results, the first is the more often used, especially by the physicist. The chemist, however, finds it often more convenient and logical to express these heats of combination per formula weight of the substances combining and of product formed. E.g., In the case of CO 2 , which contains 12 parts of car- bon and 32 of oxygen in 44 of the gas, the chemist would write, (C, O 2 ) = 97,200, meaning thereby that when 12 grams of carbon is burnt by 32 grams of oxygen, forming 44 grams of carbon dioxide, there is evolved 97,200 gram-calories. These are the laboratory units; for practical purposes, we call the weights kilograms and the heat units kilogram -calories (gram-calories are abbreviated to " cal. "; kilogram -calories to " Cal."). Ostwald, in his thermo- chemical tables writes (C, O 2 ) = 9721 K, where K represents a unit 100 times as large as a gram-calorie, if weights are taken as being grams. Berthelot writes (C, O 2 ) = 97.2, where the heat units are kilogram-calories, if the weights concerned are taken as grams. Both these methods of expression are liable to cause confusion ; the writer prefers to follow the older thermo- chemists (Hess, Naumann) and to use the larger number, e.g., 97,200, which then means gram-calories, if weights are taken in grams (laboratory units), and kilogram-calories, if weights are taken in kilograms (practical units). If it is desired to work in " British Thermal Units " (1 B. T. U. is the heat needed to raise one pound of water one degree Fahrenheit), the weights represented by the formula may be called pounds, and then the expression is as follows: (C, O 2 ) = [97,200X9/5] = 174,960 B. T. U. The factor 9/5 is simply the relation of 1 C. to 1 F. ; the APPLICATIONS OF THERMOCHEMISTRY. 15 reasoning is, of course, that if the combustion of 12 kilograms of carbon evolves 97,200 kilogram calories, or would heat 97,200 kilograms of water 1 C., that the combustion of 12 pounds of carbon would heat 97,200 pounds of water 1 C., or 174,960 pounds 1 F. From the equation as thus expressed and interpreted, the heat of combination per pound of carbon burnt, or of oxygen used, or of product formed, may be found in B. T. U. by dividing 174,960 by 12, 32 or 44, respectively. Since it is very inconvenient, as well as unscientific, to have the two unrelated heat units, with their resulting double sets of experimental data, I strongly recommend the use of the metric data and metric Centigrade heat unit. It is, however, sometimes convenient, when all the data of a problem are given in English weights, to use as the unit of heat the " pound 1 C.," or the heat required to raise the temperature of one pound of water 1 C. This may be called the " pound cal.," as distinguished from the B. T. U. The advantage of using it is that all the experimental data of the metric system units are at once transferable to the English weights. E.g., (C, O 2 ) = 97,200 pound cal., if the weights concerned in the formula (12, 32, 44) are called pounds. The thermochemist gives all his experimental data in the form above explained, and we will now give all the important thermochemical data known which are useful in metallurgical calculations, the data being for the reactions beginning and ending at 15 C. (60 F.): INTRODUCTION TO THERMOCHEMICAL TABLES. In the thermochemical tables following, the heats of formation or combination are given first for the molecular weights repre- sented by the formulas. What these molecular weights are, can be found by adding up the atomic weights of the atoms repre- sented in the formula. The atomic weight table on pages 1 and 2 supplies the necessary data. The formula given represents, thermochemically, the union of the constituent parts separated by commas, the whole reaction being enclosed in parentheses, to distinguish it from the ordinary chemical formula. Thus, while (Ca, 0) represents the union of 40 parts of calcium with 16 parts of oxygen to form 56 parts of calcium oxide, and (Ca, Si, Oa) represents the combination of 40 parts of calcium with 16 METALLURGICAL CALCULATIONS. 28 parts of silicon and 48 parts of oxygen to form 116 parts of calcium silicate, (CaO, SiO 2 ) represents the combination of 56 parts of lime with 60 parts of silica to form 116 parts of calcium silicate. In order to facilitate the use of these data in metallurgical calculations, the succeeding columns give the heat evolution per unit weight of the metal or base involved, of the acid element or acid radical involved, and of the compound formed. This will save much calculation, for while the heat evolution per formula weight is of prime use in discussing the heat energy of chemical reactions, the heats of combination per unit weight of constituents or of the product are most convenient in ordinary calculations. The column headed "To Dilute Solution" gives the total heat evolution inclusive of the heat of solution in a large excess of water. The differences between corresponding columns under this heading and the heading "Anhydrous," are the heat of solu- tion, per formula weight, per unit of base, of acid, or of compound. The order in which the elements are arranged in each table is the order of their heats of combination with unit weight of the acid element or radical, which expresses the real order of affinity of the metals in each case. The order is given for the "An- hydrous" condition, because this is more frequently concerned in metallurgy than "To Dilute Solution." Theoretically, the order of arrangement "To Dilute Solution" is the more uniform and logical, besides being invariable, but it is less useful in metal- lurgy, except where dilute solutions are concerned. In order that an estimate may be made of heats of formation so far undetermined, there is given at the end of the tables a list of the " Thermochemical Constants of the Elements" per chemical equivalent or per unit of valence with which they enter into combination. These are based on the known law that the heats of formation of salts to dilute solution are additive functions, being simply the sum of the thermochemical combining power of the basic element and that of the acid element or radical. Assuming arbitrarily that hydrogen has zero for its thermo- chemical constant, the constants for those elements can be cal- culated for which any thermochemical data, even a single heat of formation, exist. Thus, referring to the table, on page 38, 57,200 for Na means that 23 parts of sodium going into combina- APPLICATIONS OF THERMOCHEMISTRY. 17 tion contributes 57,200 calories towards the heat of formation of the compound formed, irrespective of what it is combining with. Similarly, 39,400 for Cl means that 35.5 parts of chlorine going into combination contributes 39,400 calories towards the heat of formation of any chloride, irrespective of the element it is combining with. The sum of these, 96,600, is the heat of formation of (Na, Cl) to dilute solution. By means of the table of constants given, simple addition gives the heat of formation of a large number of salts, from their elements (taken from their usual physical condition at room temperature) per chemical equivalent of base and acid, or of compound, involved. This is not per formula weight, except for monovalent elements, but is per chemical equivalent weight; it must be multiplied by the valency of the base in the formula, that is, by the number of chemical equivalents represented by the formula, to get the heat of formation per formula weight. Thus, ;HjAl 40,100 plus Cl 39,400 gives 79,500 as the heat of formation of J^Al C1 3 ; from which we have 238,500 as the heat of formation of Al CU, per formula weight. While this table has considerable uses, it does not give the heats of formation of anhydrous compounds; its values must be cor- rected by the heats of solution, where known, to give the values for the anhydrous condition. Yet, these are frequently correc- tions of a minor order, and leave the table highly useful for getting first approximations to undetermined thermochemical data. Names of some elements are introduced whose thermo- chemical constants are unknown, being placed, at a guess, in their most probable position in the table. All quantities are positive, except these indicated negative. The + exponent after a symbol represents one positive valence, the exponent one negative valence. A more detailed explanation of the matter can be found in an article by the writer in the Trans. American Electrochemical Society, 1903, Vol. iv, page 142. 18 METALLURGICAL CALCULATIONS. HEAT OF FORMATION OF OXIDES Formula Anhydrous To Dilute Solution Molecu- lar Per Unit Weight of Molecu- lar Per Unit Weight of Metal Oxygen Oxide Metal Oxygen Oxide (Th, 2 ) 326,000 1,304 10,188 1,235 (Mis, OO 1 . 472,000 1,655 9,850 1,417 (La 2 , Os) 444,700 1,602 9,265 1,364 (Nd !f 3 ) 435,100 1,506 9,006 1,291 (Mg, 0) 143,400 5,975 8,963 3,585 148,800 6,200 9,300 3,720 (Pr, 00 412,400 1,467 8,592 1,297 (Li 2 , 135,800 9,700 8,488 4,527 167,000 11,929 10,437 5,567 (Ba.O) 133,400 974 8,338 872 161,500 1,179 10,938 1,051 (Ca. 0) 131,500 3,288 8,219 2,348 149,600 3,740 9,350 2,671 (Sr, 0) 131,200 1,508 8,200 1,274 158,400 1,821 9,900 1,538 (Ah. OO 392,600 7,270 8,179 3,849 (V., Oi) 353,200 3,463 7,358 2,355 (Ce, 00 ' 224,600 1,603 7,019 1,306 (Ti, 00 , 218,400 4,550 6.825 2,730 (U 3 , 00 845,200 1,184 6,603 1,004 (V, 0) 104,300 2,045 6,519 1,557 (U, 00 303,900 1,277 6,341 1,063 (Na 2 , 0) 100,400 2,183 6,275 1,620 155,900 3,389 9,744 2,515 (Kt. 0) 98,200 1,259 6,138 1,045 165,200 2,118 10,325 1,758 (Si, OO 196,000 7,000 6,125 3,267 (Mn, O) 90,900 1,653 5,681 1,280 (B, OO 272,600 12,391 5,679 3,894 279,900 12,723 5,831 3,999 (Vt, Os) 447,000 4,324 5,588 2,423 (Zr, 0.) 177,500 1,972 5,547 1,455 (Zn, 0) 84,800 1,305 5,300 1,058 (Mm, OO 328,000 1.98S 5,125 1,434 (Cra, 0.) 243,900 2,345 5,081 1,605 (Li., 0*) 152,650 10,904 4,833 3,318 159,840 11,417 4,995 3,474 (P. 6 ) 365,300 5,895 4,566 2,572 400,900 6,466 5,013 2,823 (Ba, 0.) 145,500 1,062 4,547 861 (Mo, O) 142,800 1,488 4,463 1,116 (Sn, 0). 70,700 599 4,419 527 (Sn, 00 141,300 1,197 4,416 942 (CO, 0) (gas) 68,040 2,430 4,253 1,546 73,940 2,641 4,621 1,680 [ solid 70,400 35,200 4,400 3,911 (Hi, 0) 1 liquid 69,000 34,500 4,313 3,833 I gas 58,060 29,030 3,629 3,226 (Pe,, 4 ) 270,800 1,612 4,231 1,167 (Cd, 0) 66,300 592 4,144 518 (Pe, 0) 65,700 1,173 4,106 913 (W. 00 131,400 714 4,106 620 (W, 00 196,300 1,067 4,089 846 (Fes, 00 195,600 1,746 4,075 1,223 (Co 0) 64,100 1,086 4,006 855 (Mn, OO 125,300 2,278 3,916 1,440 Mi = Mischmetal a mixture of rare earth metals sold commercially, containing cerium, thorium, yttrium, etc. APPLICATIONS OF THERMOCHEMISTRY. 19 HEAT OF FORMATION OF OXIDES. (Continued] Formula Anhydrous To Dilute Solution ifdte*- lor Per Unit Weight of Molecu- lar Per Unit Weight of Meal Oxygen Oxide Metal Oxygen Oxide (Ni, 0) 61,500 1.051 3,844 826 (Na z , 2 ) 119,800 2,604 3,744 1,536 (Mo, Oa) 175,000 1,823 3,646 1,215 (Sb, 3 ) 166,900 695 3,479 580 (Sbi, O) 209,800 874 3,278 690 (Ast, O 3 ) 156,400 1,043 3.258 790 148,900 993 3,102 752 (Pb, 0) 50,800 245 3,175 228 (C, Oi) (gas) 97,200 8,100 3.038 2,209 103,100 8,592 3,222 2,343 (Cr, Oi) 140,000 2,692 2,917 1,400 (Bit. O 3 ) 139,200 335 2,900 300 (Sb, Os) 231,200 963 2,890 723 (As. Oa) 219,400 1,463 2,743 954 225,400 1,503 2,818 980 (CU2, 0) 43,800 344 2,738 306 (Tl, 0) 42,800 105 2,675 101 39,700 97 .2,481 94 (Te, Oj) 78,300 624 2,447 497 (Cu, 0) 37,700 593 2,356 474 (S, Oi) 69,260 2,196 2,196 1,098 77,600 2;425 2,425 1,213 (Pb, O z ) 63,400 306 1,981 265 (S, 3 ) (gas) 91,900 2,872 1,915 1,149 141,000 4,406 2,938 1.763 (Tit, Oi) 87,600 215 1,825 192 (C, 0) (gas) 29,160 2,430 1,823 1,041 (H, Oi) 47,300 23,650 1,478 1.391 (Hgi, O) 22,200 56 1,388 53 (Hg, 0) 21.500 108 1,344 100 (Pd, 0) 21,000 198 1,313 172 (Pt, 0) 17,000 87 1,063 81 (Ag. 0) 7,000 32 438 30 (Am. Oi) -11,500 -29 -240 -26 HEAT OF FORMATION OF HYDRATES A. From the Elements Formula and Reaction Anhydrous To Dilute Solution Per Formula Weight Per Unit Weight of Metal Per Formula Weight Per Unit Weight of Metal (Li, 0, H) 112,200 16,030 118,000 16,860 (Mg, 2 , H 2 ) 217,800 9,075 (Sr, 2 , H 2 ) 217,300 2,498 227,400 2,614 (Ca, 0,, H,) 215,600 5,390 219,500 5,488 (K, 0, H) 104,600 2,682 117,100 3,003 (Na, 0, H) 102,700 4,465 112,450 4,889 (Al, 0,, H 3 ) 301,300 11,160 (N, H, 0, H) 88,800 90,000 5,294 (Zn, Oi, HO 151,100 2,325 solid 70,400 (H, 0, H) liquid 69,000 gas 58,060 (Tl, O, H) 57,400 260 54,300 246 (Bi, 0,, H,) 171,700 825 20 METALLURGICAL CALCULATIONS. HEAT OP FORMATION OF HYDRATES Continued A. From the Elements Formula Anhydrous To Dilute Solution and Reaction Per Formula Weight Per Unit Weight of Metal Per Formula Weight Per Unit Weight of Metal (Te, 4f H 2 ) 166,740 1,323 (Te, O lf H 2 ) 145,600 1,156 (Se,0 3 , HO 124,500 1,576 (Se, 4 , HO 128,220 1,623 145,020 1,836 (Tl, 3 , HO 78,700 386 B. From Metal, Oxygen and Water Formula and Reaction Anhydrous To Dilute Solution Per Formula Weight Per Unit Weight of Metal Per Unit Weight of Oxygen Per Formula. Weight Per Unit Weight of Metal Per Unit Weight of Oxygen (Li 2 , 0, H 2 0) 155,400 11,100 9,710 167,000 11,930 10,440 (Mg, 0, H 2 0) 148,800 6,200 9,300 (Sr, O, H 2 O 148,300 1,705 9,280 158,400 1,822 9,900 (Ca, 0, H 2 0) 146,600 3,665 9,162 150,500 3,763 9,406 (K 2 , 0, H 2 0) 140,200 1,797 8,762 165,200 2,118 10,325 (Na 2 , 0, H 2 0) 136,400 2,965 8,525 155,900 3,389 9,744 (A1 2 , 3 , 3H 2 0) 395,600 7,326 8,242 (Zn, 0, H 2 0) 82,100 1,263 5,131 (T1 2 , 0, H 2 0) 45,800 112 2,863 39,600 97 2,475 (Bi 2 , 3 , 3H 2 0) 136,400 328 2,842 (Te, 2 , H 2 0) 76,600 6,079 2,394 (Te, 3 , H 2 0) 97,740 776 2,036 (Se, 2 , H 2 0) 55,500 703 1,735 (Se, O 3 , H 2 O 59,220 750 1,234 76,020 962 1,583 (T1 2 , O 3 , 3H 2 O) 86,400 212 1,800 C. From Metallic Oxide and H Z Anhydrous To Dilute Solution Formula and Per Per Unit Weight of Per Per Unit Weight of Reaction Formula Weight Metallic Oxide H 2 Hydrate Formula Weight Metallic Oxide Com- bined mo Dis- solved Hydrate (Li, O, H 2 O) 19,600 653 1,089 408 31,200 1,040 1,733 650 (MgO, H 2 0) 5,400 135 300 93 (SrO, H 2 O) 17,100 166 950 141 27,200 264 1,511 225 (CaO, HO) 15,100 270 839 204 19,000 339 1,056 257 (K 2 0, H 2 0) 42,000 447 2,333 375 67,000 713 3,722 598 (Na 2 0, H 2 0) 36,000 581 2,000 450 55,500 895 3,083 694 (N H4) 2 0, H'O (A1 Z 3 , 3H 2 0) 3,000 30 56 19 (ZnO, H 2 O) -2,700 -33 -150 -27 (ThO, HzO) 3,000 7 167 7 -3,200 -7 -178 -7 (BizOa, 3H 2 0) -2,800 *T -52 -6 (TltOi, 3H'0) -1,200 -3 -22 -2 APPLICATIONS OF THERMOCHEMISTRY. 21 HEAT OF FORMATION OF SULFIDES Formula Anhydrous To Dilute Solution of Per Formula Weight Per Unit Weight of Per Formula Weight Per Unit Weight of Metal Sulfur Sulfide Metal Sulfur Sulfide (Li 2 S) 115,400 8,243 3,606 2,509 \- L ' 1 2j *-V (K,, S) 103,500 1,327 3,234 941 113,500 1,455 3,547 10,319 (Ba, S) 102,900 751 3,216 609 109,800 802 3,431 474 (Sr, S) 99,300 1,141 3,103 835 106,700 1,225 3,334 897 (Na, f S) 89,300 1,941 2,791 1,145 104,300 2,267 3,259 1,337 (Nd 2 , S,) 285,900 993 2,977 744 (Ca t S) 94,300 2,358 2,947 1,310 100,600 2,515 3,144 1,397 (Mg, S) 79,400 3,308 2,481 1,418 (Mn, S) 45,600 829 1,425 524 (Zn, S) 43,000 662 1,344 443 (A1 2 , SO 126,400 2,341 1,316 843 (N, H 5 , S) 40,000 2,105 1,250 776 36,700 1,931 1,147 720 (Cd, S) 34,400 307 1,075 239 (K, SO 59,300 1,521 927 576 59,700 1,531 933 579 (Na, S,) 49,500 2,152 773 556 54,400 2,365 850 611 (B 2 , S 3 ) 75,800 3,445 790 642 (Fe, S) 24,000 428 750 273 (Co, S) 21,900 371 685 241 (T1 2 , S) 21,600 106 675 92 (Cu, f S) 20,300 160 634 127 (Pb, S) 20,200 98 631 85 (Si, SO 40,000 1,429 625 435 (Ni, S) 19,500 333 609 215 (Sb,, SO 34,400 143 358 102 (Hg f S) 10,600 53 331 46 (Cu, S) 10,100 159 316 106 (H 2 , S) (gas) 4,800 2,400 150 141 9,500 4,750 297 2Y9 (Ag 2 , S) 3,000 14 94 12 (C, SO (gas) -25,400 -2,117 -397 -334 (liquid) -19,000 -1,583 -297 -250 (1,3) 22 METALLURGICAL CALCULATIONS HEAT OF FORMATION OF SELENIDES Anhydrous To Dilute Solution Formula Per Per Unit Weight of Per Per Unit Weight of Weight Metal Selen- ium Selen- ide Wright Metal Selen- ium Selen- ide (Liz, Se) 83,000 5,929 1,051 892 93,700 6,693 1,186 1,008 (K 2 , Se) 79,600 1,021 1,008 507 87,900 1,127 1,113 560 (Ba, Se) 69.900 510 885 324 (Sr, Se) 67,600 777 856 408 (Na 2 . Se) 60,900 1,324 720 487 78,600 1,709 995 629 (Ca> Se) 58,000 1,450 727 487 (Zn, Se) 30,300 466 384 210 (Cd, Se) 23,700 212 300 124 (Mn, Se) 22,400 407 284 167 (N, Hs, Se) 17,800 937 225 184 12,800 674 162 136 (Cu, Se) 17,300 272 219 114 (Pb, Se) 17,000 82 215 59 (Pe, Se) 15,200 262 ,192 111 (Ni, Se) 14,700 251 186 107 (Cd, Se) 13,900 236 176 101 (Th, Se) 13,400 33 170 21 (Gu, Se) 8,000 63 101 39 (Hg, Se) 6,300 32 80 23 (Ag 2 , Se) 2,000 93 25 7 (H 2 , Se) (gas) -25,100 -12,550 ^-318 -310 - 15,800 -7,900 -200 -195 (N, Se) -42,300 -3,021 -534 ,'-455 HEAT OF FORMATION OF TELLURIDES Anhydrous Formula Per Formula Weight Per Unit Weight of Metal Tellurium Telluride (Zn, Te) 31,000 477 246 162 (Cd, Te) 16,600 148 132 70 (Co, Te) 13,000 220 103 70 (Fe, Te) 12,000 214 95 66 (Ni, Te) 11,600 198 92 63 (T1 2 , Te) 10,600 26 84 20 (Cu,, Te) 8,200 64 65 32 (Pb, Te) 6,200 30 49 19 (H 2 , Te) (gas) -34,900 -17,450 -277 -272 APPLICATIONS OF THERMOCHEMISTRY. 23 HEAT OF FORMATION OP ARSENIDES Formula Per Formula Weight Per Unit Weight of Metal Arsenic Arsenide (H 3 , As) (gas) -44,200 -14,733 -589 -567 HEAT OF FORMATION OF ANTIMONIDES Formula Per Formula Weight Per Unit Weight of Metal Antimony Antimonide (H, f Sb) (gas) -86,800 -28,933 -723 -706 HEAT OF FORMATION OF PHOSPHIDES Per Per Unit Weight of Formula Formula Weight Metal Phosphorus Phosphide (Mn 3 , P,) 70,900 430 1,144 312 (H 3 , P) (gas) 4,900 1,633 158 144 (Fe, P) HEAT OF FORMATION OF NITRIDES Anhydrous To Dilute Solution Formula Per Per Unit Weight of Per Per Unit Weight of Weight Metal Nitrogen Nitride Weight Base Acid Com- pound (Bus. N 2 ) 149,400 363 5,336 340 (Mga, N 2 ) 119,700 1,662 4,275 1,197 (Cas, N 2 ) 112,200 935 4,007 758 (Li,, N) 49,500 2,357 3,536 1,414 (Al, N) 45,450 1,683 3,246 1,109 (Li, N) 18',750 2,679 1,339 893 (Hs, N) (gas) 12,200 4.067 871 718 21,000 7,000 1,500 1,235 (liquid) 16,600 5,533 1,186 976 (P, N,) 81,500 876 1,164 500 24 METALLURGICAL CALCULATIONS. HEAT OF FORMATION OF HYDRIDES Formula Per Formula Per Unit Weight of Per rTOtflttlCL Per Unit Weight of Weight Metal Hydro- gen Hy- dride Weight Base Acid Com- pound (Ca, HO* 46,200 1,155 23,100 1,100 (Li, H) 21,600 3,086 21,600 2,700 (Sr, HO 38.400 441 19,200 431 (Ba, H) 37,500 274 18,750 271 (Ptis, H) 16,950 6 16,950 6 (Ptio, H) 14,200 7 14,200 7 (Pu, He) 53,400 144 8,900 .141 (Pdu, H) 4,600 3 4,600 3 (P, Hi) (gas) +4,900 158 + 1,633 144 (Si, H4) (gas) -6,700 -239 -1,675 -209 (N4, HO -19,000 -339 -4,750 -317 -26,100 -466 -6,525 -435 (As, H 3 ) -44,200 -589 -14,730 -567 (Sb, HJ) -86,800 -723 -28,930 -706 HEAT OF FORMATION OF HYDROCARBONS (From Amorphous Carbon, to Gas, Unless Otherwise Specified} Formula and Reaction Name State Per Formula Weight Per Unit Weight of Carbon Hydro- gen Com- pound (C, HO Methane (Marsh gas) Gas +21,700 + 1,808 +5,425 + 1,356 (C 2 , HJ) Acetylene Gas -52,300 -2,179 -26,150 -2,016 (C 2 , HO Ethylene (olefiant gas) Gas -8,700 -363 -2,175 -311 (C,, He) Ethane (ethylene hydride) Gas +29,100 + 1,213 +4,850 +970 (Ca, HO Allylene Gas -44,000 -1,222 -11,000 -1,100 (C 3 , HO Propylene Gas -700 -22 -133 -19 (Cj. HO Propane (propylene hydride) Gas +39,200 + 1,089 +4.900 +891 (Ca, HO Benzene Gas +7,200 + 100 + 1,200 +92 Liquid + 14,400 +200 + 2,400 + 185 (Cr, HO Toluol Liquid + 21,900 +261 + 2,738 +238 (Cio, HO Naphthalene Liquid + 9,100 +76 + 1,138 + 71 Solid + 13,700 + 114 + 1,713 + 107 (Cio, Hie) Turpentine Gas +23,800 + 197 + 1,488 + 175 Liquid + 33,200 +277 +2,075 +244 (Ci4. Hio) Anthracene Solid -2,600 -16 -260 -15 (C, Hi,'0) Methyl alcohol (wood alcohol) Gas +56,000 4,667 14,000 1,750 Liquid + 65,050 5,420 16,260 2.033 (Ct. He, 0) Ethyl alcohol (grain alcohol) Gas + 64,200 2,675 10,700 1.396 Liquid +74,900 3,120 12,480 1,628 (C., He, 0) Acetone Gas + 63,150 1,754 10,525 1,089 Liquid + 71,300 1,980 11,880 1,229 APPLICATIONS OF THERMOCHEMISTRY. 25 HEAT OF COMBUSTION OF HYDROCARBONS Formula Name State Per Formula Weight Calor- ies per M3 Calor- ies per Kg B.T.U. B.T.U. per /M per Ib. Calorime- ter Value Practical Value Water formed not condensed CH4 CiHj C,H4 CiHe C 3 H4 CaHe C 3 Hs CeH 6 CyHs CioHs CioH 16 CuHio CHiO C 2 HO CaH.O Methane Acetylene Ethylene Ethane Allylene Propylene Propane Benzene Toluol Naphthalene Turpentine Anthracene Methyl alcohol Ethyl alcohol Acetone Gas Gas Gas Gas Gas Gas Gas Gas Liquid Liquid Liquid Solid Gas Liquid Solid Gas Liquid Gas Liquid Gas Liquid 213,500 315,700 341,100 372,300 473.600 499.300 528,400 783,000 775,800 934,500 1,238,900 1,234,300 1,500,200 1,490,800 1,708,400 179,200 170,150 337,200 326,500 435,450 427,300 191,620 304,760 319,220 339,480 451,720 466,480 484,640 750,180 742,980 890,740 1,195,150 1,190,550 1,412,680 1,403,280 1,653,700 157,320 148,270 304,380 293,680 402,630 394,480 8,623 13,714 14,365 15,277 20,327 20,992 21,812 33,758 63,570 11,976 11,722 11,400 11,315 11,295 11,105 11,015 9,618 9,525 9,682 9,337 9,300 10,385 10,318 9,290 970 1,543 1,616 1,719 2,287 2,362 2,464 3,798 21,555 21,000 20,520 20,365 20,150 19,995 19,825 17,312 17,145 17,430 16,805 16,740 18,695 18,570 16,545 8,854 8,340 11,910 11,490 12,495 12,240 7,152 7,079 4,919 4,633 786 13,700 6,617 6,384 1,541 18,120 6,942 6,801 2,038 HEAT OF FORMATION OF CARBIDES Formula Per Formula Weight Per Unit Weight of Metal Carbon Carbide (A1 4 , C.) 232,000 2,148 6,445 1,815 (Si, C) 26,520 947 2,210 663 (Mn,, C) 9,900 60 825 57 (Fe 3 , C) 8,460 50 705 47 (Ca, CO -7,250 181 302 113 (Na, C) -4,400 191 367 126 (Li, C) -5,750 821 479 303 (Ag, C) -43,575 403 3,631 363 (N 2 , CO -73,000 (gas) -2,607 -3,042 -1,404 28 METALLURGICAL CALCULATIONS. HEAT OF FORMATION OF CARBONATES A. From the Elements Formula Anhydrous To Dilute Solution Per Formula ' Weight Per Unit Weight of Metal Per Formula Weight Per Unit Weight of Metal (Ba, C, 0.) 286,300 2,090 (K 2 , C, 3 ) 282,100 3,617 288,600 3,700 (Sr, C, 3 ) 281,400 3,234 (Ca, C, 3 ) 273,850 6,846 (Na 2 , C, 3 ) 273,700 5,950 279,300 6,072 (Mg, C, 3 ) 269,900 11,245 (Mn, C, 3 ) 210,300 3,824 (N, H 6 , C, 3 ) 208,600 10,980 202,300 10,640 (Zn, C, 3 ) 197,500 3,038 (Pe, C, 3 ) 187,800 3,354 (Cd, C, 3 ) 183,200 1,636 (Pb, C, 3 ) 170,000 821 (Cu, C, 3 ) 146,100 2,301 (Ag 2 , C, 3 ) 123,800 593 B. From Metal, Oxygen and C0 2 Anhydrous To Dilute Solution Formula and Per Per Unit Weight of Per Per Unit Weight of Weight Metal Oxygen Weight Metal Oxygen (Ba, 0, C0 2 ) 189,100 1,380 (K 2 , O, CO 2 ) 184,900 2,370 191,400 2,454 (Sr, 0, C0 2 ) 184,200 2,117 (Ca, 0, C0 2 ) 176,650 4,416 (Na 2 , O, CO 2 ) 176,500 3,837 182,100 3,959 (Mg, 0, C0 2 ) 172,700 7,195 (Mn, O, CO 2 ) 113,100 2,056 (N, H 6 , O, CO 2 ) 111,400 5,863 105,100 5,532 (Zn, O, CO 2 ) 100,300 1,543 (Fe, 0, C0 2 ) 90,600 1,618 (Cd, 0, C0 2 ) 86,000 768 (Pb, O, CO 2 ) 72,800 352 (Cu, 0, C0 2 ) 48,900 770 (A g2 , 0, C0 2 ) 26,600 123 APPLICATIONS OF THERMOCHEMISTRY. C. From Metallic Oxide and CO Z 29 Anhydrous To Dilute Solution Formula Per Unit Weight of Per Unit Weight of Reaction Per Formula Per Formula Weight Metallic Oxide CO 2 Car- bonate Weight Metallic Oxide COt Car- bonate (BaO, CO 2 ) 55,700 364 1,266 282 (K 2 O, CO 2 ) 86,700 922 1,970 628 93,200 992 2,118 675 (SrO, CO*) 53,000 515 1,205 361 (CaO, CO 2 ) 45,150 806 1,026 451 (Na 2 O, CO 2 ) 76,100 1,219 1,718 713 81,700 1,318 1,857 771 (MgO, CO 2 ) 29.300 733 666 349 (MnO, CO 2 ) 22,200 313 505 193 (NHs, O, CO 2 ) 22,600 646 514 286 16,300 466 370 206 (ZnO, CO 2 ) 15,500 191 352 124 (FeO, CO 2 ) 24,900 346 566 215 (CdO, CO 2 ) 19,700 154 448 115 (PbO, CO 2 ) 22,000 99 500 83 (CuO, CO 2 ) 11,200 140 255 91 (Ag 2 O, CO 2 ) 19,600 82 445 71 HEAT OF FORMATION OF BICARBONATES A. From the Elements Anhydrous To Dilute Solution Formula and Reaction Per Formula Weight Per Unit Weight of Metal Per Formula Weight Per Unit Weight of Metal (K, H, C, Os) 233,300 5,987 228,000 5,846 (Na, H , C, Os) 227,000 9,870 222,700 9,683 (N, H H, C, Os) 205,300 199,000 B. From Metal, Oxygen, H Z and CO Z Formula and Reaction Anhydrous To Dilute Solution Per For- mula Weight Per Unit Weight of Per Formula Weight Per Unit Weight of Metal CO 2 Metal C0 2 Com' Pound (K 2 , O, H 2 0, 2C0 2 ) 203,200 2,605 2,309 1,016 192,600 2,469 2,189 963 (Na 2 , O, H 2 O, 2C0 2 ) 190,600 4,143 2,166 1,135 182,000 3,957 2,036 1.083 C. From Metallic Oxide, H Z and CO Z Formula Anhydrous TOiOO^t^OOiOCOCO-Tttt^.-io-<*i^-t>-o5O(Noicoooo5iO'*iosr^ o CO COC^i^T-t^Hi-HOt^-C^^CO 1-4 -H i-H i-H 99 00 00 S flirtriortdC^^ !-! 2 21 co ^ * ^4i*J4i*J12*i4i^ . 90,5000600500,50035 O5GO^i- 278 400 987 1 210 K44. (3SrO, As 2 Os) 148,000 567 643 302 (3CaO, As 2 O 6 ) 118,900 708 502 299 (3MgO, As 2 Os) 63,000 525 274 151 (3Ba 2 O, As 2 O 5 ) 9,600 21 42 14 HEAT OF FORMATION OF BORATES. A. From the Elements Formula and Reaction Anhydrous To Dilute Solution Per Formula Weight Per Unit Weight of Metal Per Formula Weight Per Unit Weight of Metal (Na, B4, OT) 748,100 16,263 758,300 16,485 B. From Metallic Oxide and B 2 3 Formula and Reaction Anhydrous To Dilute Solution Per For- mula Weight Per Unit Weight of Per For- mula Weight Per Unit Weight of Metallic Oxide B 2 0a Borale Metallic Oxide BzOt Borate (Na 2 O, 2B 2 Oa) 102,500 1,653 732 507 112,700 1,818 805 558 HEAT OF FORMATION OF MOLYBDATES. Anhydrous A. From the Elements B. From Metallic Oxide and MoOa Formula and Reaction Per Formula Weight Per Unit Weight of Metal Formula and Reaction Per Formula Weight Per Unit Weight of Metallic Oxide MoOs Molyb- date (Naz, Mo, 4 ) 363,800 7,909 (Na 2 O, MoOa) 88,400 1,426 612 429 HEAT OF FORMATION OF TITANATES. Anhydrous A. From the Elements B. From Metallic Oxide and TiO 2 Formula and Reaction Per Formula Weight Per Unit Weight of Metal Formula and Reaction Per Formula Weight Per Unit Weight of Metallic Oxide Ti0 2 Titan- ate (Naz, Ti, Os) 388,500 8,446 (Na 2 O, TiO 2 ) 69,700 1,124 871 491 APPLICATIONS OF THERMOCHEMISTRY. 35 HEAT OF FORMATION OF MANGANATES AND PER-MANGANATES Anhydrous A. From the. Elements B. From Metallic Oxide and Mn 2 O7 Formula and Reaction Per Formula Weight Per Unit Weight of Metal Formula and Reaction Per Formula Weight Per Unit Weight of Metallic Oxide MniQi MniOi Com- pound (Nai, Mn, O 4 ) 269,400 5,857 (NazO. Mn0 3 ) (K, Mn, 04) 200,050 5,129 (K 2 0, MnzOy) HEAT OF FORMATION OF ALUMINATES A nhydrous A. From the Elements B. From Metallic Oxide and AhOa Formula and Reaction Per Formula Weight Per Unit Weight of Metal Formula and Reaction Per Formula Weight Per Unit Weight of Metallic Oxide AhOi Alumin- ate (Na, Al, O 2 ) 533,000 11,587 (Na 2 O, A1 2 3 ) 40,000 645 394 244 (Ca, Ah, 4 ) 524,550 13,114 (CaO, A1 2 O 3 ) 450 8 4 3 (Ca 2 , Ah. Os) 658,900 8,236 (2CaO, AhOO 3,300 29 32 15 (Caa, Ah, Oe) 780,050 6,500 (3CaO, AhOi) -7,050 -42 -69 -26 HEAT OF FORMATION OF CYANIDES Anhydrous Formula and Reaction Per Formula Weight Per Unit Weight of Per Formula Weight Per Unit Weight of Metal CN Cyanide Metal CN Cyan- ide (Ca, C*. N 2 ) 41,650 1,041 801 453 (K, C, N) 33,450 858 1,287 515 30,250 776 1,163 465 (Na. C, N) 25,950 1,128 998 530 25,450 1,107 979 519 (Sr, C 2 , N 2 ) 47,000 540 904 338 (Ba, C 2) NO 48,300 353 929 256 (Ca, Ct, NO 41,650 1,041 801 453 (Mg, C 2 , N 2 ) 34,000 1,417 654 447 (Ni, C. Ni) -23,400 -400 -450 -212 (Zn, C 2 , NO -24,550 -378 -472 -210 (Per, Cis, NiO -256,700 -655 -549 -298 (Cd, C 2 , NO -31,850 -284 -612 -194 (Cu, C, N) -20,375 -320 -784 -227 (Pd, Cx N) -49,250 -465 -947 -312 (gas) (H, C, N) -27,150 -27,150 -1,044 -1,006 -21,050 -21,050 -810 -781 (Hg, C 2 , NO -59,150 -296 -1,137 -235 (Ag, C, N) -34,000 -315 -1,308 -254 To Dilute Solution 36 METALLURGICAL CALCULATIONS. HEAT OF FORMATION OF CYANATES A. From the Elements Formula and Reaction Anhydrous To Dilute Solution Per Formula Weight Per Unit Weight of Per Formula Weight Per Unit Weight of Metal Cyanate Metal Cyanate (K, C. N. 0) 105,850 2,714 1,307 100,650 2,581 1,242 (Na, C, N, 0) 105,050 4,567 1,616 100,250 4,359 1,542 (Ag, C, N, O) 26,450 245 176 (Hg, Ci. Ni, Oi) -62,900 -315 -221 B. From Cyanide and Oxygen Anhydrous To Dilute Solution Formula and Per Per Unit Weight of Per Per Unit Weight of Reaction Formula Formula Weight Cyan- ide Oxygen Cyanate Weight Cyan- tde Oxygen Cyanate (KCN, 0) 72,400 1,115 4,525 894 67,200 1,034 4,200 830 (NaCN, O) 79,100 1,614 4,944 1,217 74,300 1,516 4.644 1,143 (AgCN. 0) 60,450 451 3,781 403 (HgCjN,, 2 ) -3,750 -15 -117 -13 HEAT OF FORMATION OF METALLO-CYANIDES A. From the Elements Anhydrous To Dilute Solution Formula and Reaction Per Per Unit Weight of Per Per Unit Weight of Weight Metal CN Cyanide Weight Metal CN Cyanide (K 4 . Pe, Ce, Ne) 157,300 1,007 1,007 428 145,300 931 931 395 (H 4 , Fe, Ce. Ne) - 102,000 -25,500 -652 -472 -101,500 -25,375 -650 -470 (Ki, Pe, Ce, Ne) 129,600 1,108 831 394 100,800 862 647 307 (Hi. Pe, Ce, Ne) -127,400 -42,467 817 593 (K.'Ag, C. Ni) 13,700 351 264 69 5,350 137 103 27 B. From the Constituent Cyanides Formula and Reaction Anhydrous To Dilute Solution Per Formula Weight Per Unit Weight of Per Formula Weight Per Unit Weight of Cyan- ide I Cyan- ide II Com- pound Cyan- ide I Cyan- ide II Com- pound (KCN. AgCN) -86,100 -1.325 -642 -433 -94.450 - 1,452 -704 -474 APPLICATIONS OF THERMOCHEMISTRY. 37 HEAT OF FORMATION OF AMALGAMS Formula and Reaction To Formula Given Per Formula Weight Per Unit Weight of Mercury Metal Amalgam (Hg, Na) (Hg 2 , Na) (Hg 4 , Na) (Hg, Na) (Hg,, Na) (Hg 2 , K) (Hg 4t K) (Hg 9 , K) (Hg 12 , K) (Hg,, K) (Hg,, Ag) (Hg,, Au) (Hg,., Zn) \ 8.7% Zn f (Hgo.94, Zn \ 25.9% Zn / 10,300 17,800 20,000 21,900 19,000 20,000 29,700 33,000 34,600 25,600 2,470 2,580 -1,345 -703 52 45 25 18 448 774 870 952 826 513 762 846 887 656 23 13 -21 -11 46 42 24 18 50 37 18 14 46 35 18 14 -2 -3.8 -1.8 -2.8 HEAT OF FORMATION OF ALLOYS Formula or Composition To Formula or Composition Given Percentage Composition Per Formula Given Per Unit Weight of Metal I Metal 11 Alloy Metal I Metal II Per Cent. Per Cent. (Cu, Znj) 10,143 160 78 52 Cu 32.8 Zn 67.2 (Cu, Zn) 5,783 91 89 45 Cu 49.4 Zn 50.6 (Cus, Al) 26,910 141 997 124 Cu 87.6 Al 12.4 (Cu2. Al) 21,278 167 788 138 Cu 82.5 Al 17.5 (Cua, Ah) 17,395 91 322 71 Cu 77.9 Al 22 . 1 (Cu. Al) 1,887 30 70 21 Cu 70.2 Al 29.8 (Cu2, All) 10,196 80 126 49 Cu 61.1 Al 38.9 (Cu, Ah) 31,900 502 591 271 Cu 54.0 Al 46.0 (Nat. K) (liquid) -2,930 -64 -75 -34 Na 54.1 K 45.9 (Na, K 2 ) (liquid) 1,940 84 25 19 Na 22.77 K 77.23 (Na, Ka) (liquid) 1,160 50 10 8 Na 16.43 K 83.57 (Pbu, Zn) 23,188 6 357 5.8 Pb 98.4 Zn 1.6 (Pb, Zn) 952 4.6 15 + 3.5 Pb 76.1 Zn 23.9 (Pbi.2, Bi) - 1,782 -6.9 -8.6 -3.8 Pb 55.6 Bi 44.4 (Sne.i, Zn) -4,789 -6.6 -74 -6.1 Sn 91.7 Zn 8.3 (Pb, Snu.r) -8,034 -39 -4.3 -3.9 Pb 10.0 Sn 90.0 (Pb, Sm.s) -1,028 -5.0 -3.1 -1.9 Pb 38.2 Sn 61.8 (Pbi.i, Sn) 450 1.0 3.5 0.8 Pb 79.0 Sn 21.0 (Pbio.8, Sn) 2,594 1.2 22 1.1 Pb 95.0 Sn 5.0 (Pbt. Sn) 5,914 1.0 50 1.0 Pb 98.0 Sn 2.0 (Mg, Zm) 24,900 1,038 192 162 Mg 15.6 Zn 84.4 (Mg4, Ah) 164,800 1,717 204 93 Mg 54.2 Al 45.8 (Mg, Cd) 17,700 738 158 130 Mg 17.6 Cd 82.4 (Na, Cd.) 30,800 1,339 137 125 Na 9.3 Cd 90.7 (Na, Cd) 60,600 2,635 108 104 Na 3.9 Cd 96.1 (Ca, Zm) 55.600 1,390 214 185 Ca 13.3 Zn 86.7 (Ca, Zmo) 199,100 4,978 306 29 Ca 5.8 Zn 94.2 (Cui, Cdi) 47,700 376 142 103 Cu 27.4 Cd 72.6 38 METALLURGICAL CALCULATIONS. THERMOCHEMICAL CONSTANTS OF BASES TO DILUTE SOLUTION. (Per Chemical Equivalent of Base.} 'For explanation of nature and use of these tables, see page 16. Caesium Rubidium Lithium Potassium Beryllium (Glucinum) Barium Thorium Strontium Sodium Lanthanum Neodymium Calcium Praesodymium Magnesium Aluminium Ammonium (from the elements) Vanadium Cerium Titanium Uranium Vanadium Vanadium Silicon Boron Vanadium Zirconium Manganese Zinc Chromium Phosphorus Lithium (per-salts) Barium (per-salts) Iron (ferrous) Tungsten (Wolfram) Tungsten (Wolfram) Cadmium Cobalt . Manganese Sodium (per-salts) Nickel Antimony Arsenic Sulfur Cs+ Rb+ Li+ K+ Na (N, H 4 ) +81,240 76,265 62,900 61,900 60,350 59,950 58,700 57,200 55,600 54,400 54,300 40,100 33,400 24,900 17,200 10,900 9,000 8,200 7,700 2,800 APPLICATIONS OF THERMOCHEMISTRY. 39 Iron (ferric) Tin (stannous) Tin (stannic) Lead Carbon Chromium Bismuth Antimony Arsenic Copper (cuprous) Hydrogen Tellurium Tellurium Sulfur Selenium Selenium Tellurium Copper (cupric) Lead (per-salts) Thallium Carbon Hydrogen (per-salts) Mercury (mercurous) Palladium Mercury (mercuric) Platinum Silver Gold (auric) Gold (aurous) 27,30 1,900 Cu+ H+ Te+ Ag Au Au 400 -900 -1,300 -7,900 14,250 19,450 25,300 -30,300 THERMOCHEMICAL CONSTANTS OF ACIDS TO DILUTE SOLUTION (Per Chemical Equivalent of Acid Element or Radical] Fluorine Chlorine Bromine Oxygen lodine Sulfur Selenium Tellurium Hydrate Sulf -hydrate Elements F~ Cl~ Br~ I" Acid Radicals (from the elements) (O, H)~ (S, H)~ 52,900 39,400 27,500 20,700 13,200 -5,100 -17,900 55,200 3,400 40 METALLURGICAL CALCULATIONS. Selen-hydrate Hypochlorite Chlorate Per-chlorate Hypobromite Bromate lodate Per-iodate Hypo-sulfite Sulfite Bi-sulfite Pyro-sulfite Sulfate Bi-sulfate Per-sulfate Di-thionate Tri-thionate Tetra-thionate Penta-thionate Selenite Selenate Hypo-nitrite Nitrite Nitrate Phosphate Mono-H- Phosphate Di-H-Phosphate Arsenite Arsenate Mono-H- Arsenate Di-H-Arsenate Cyanide Cyanate Sulfo-cyanate Ferro-cyanide Fern-cyanide Carbonate Bi-carbonate Formate Acetate Oxalate (Se, H)- 19,100 (Cl, CO- 27,500 CCI, oo- 21,900 (Cl, 4 )- 39,400 (Br, 0)- 22,800 (Br, 3 )- 6,700 (i, oo- 60,470 (i, oo- 48,070 ^(S 2 , 3 )~ 71,750 M(S, 3 )~ 75,100 (H, S, 3 )- 149,400 K(S 2 , oo 115,200 ^(S, 4 )~ 107,000 (H, S, 4 )- 211,100 H(S 2 , 00" 158,100 ^(S 2 , 0.) 138,350 H(S >f o.) 136,500 K(S 4 , 6 )~ 130,600 H(Si, o) 133,100 H(Se, 0,) 60,050 H(Se, 4 )~ 72,800 (N, 0)- -3,800 (N, 2 )- 27,000 (N, 0,)- 48,800 M(P, 4 ) 99,300 H(H, P, 4 )" 152,750 (H 2 , P, 4 )- 307,700 H(As 2 , 4 )~ 102,150 H(As, 4 ) 70,200 H(H, As, 00 108,050 (H 2 , As, 4 )- 217,200 (C, N)- -34,900 (C, N, 0)- ' 37,100 (C, N, S)- -18.100 K(Pe, C., N 6 ) -25,600 H(Fe, C 6 , NO -52,800 H(C, 0,) 82,450 (H, C, 3 )- 169,100 (C, H, 00- 104,600 (C 2 , H,, 2 )- 120,500 H(C fc 00" 99,800 CHAPTER III. THE USE OF THE THERMOCHEMICAL DATA. SIMPLE COMBINATIONS. If the problem is the simple calculation of how much heat is evolved in the combination of a given weight of one element with another, the factors needed, the heat evolved per unit weight of substance combining, are obtained by a simple divi- ion from the thermochemical data given. Thus, suppose the question to be the total heat evolved in Problem 3, in the oxidation in a Bessemer converter of 100 kilos, of carbon to carbonic oxide. 200 kilos, of carbon to carbonous oxide. 50 kilos, of manganese to MnO. 150 kilos, of silicon to SiO 2 . 500 kilos, of iron to FeO. The solution would be Q7 200 100X is = 100X8100 = 810,000 Calories. 200 X ??ii2 = 200X2430 = 486,000 LZ on QDO 50 X J* - 50X1653= 82,650 oo 150X 18 o' OQ = 150X7000 = 1,050,000 500 X 'ft = 500X1173 = 586,500 Total 3,015,150 COMPLEX COMBINATIONS. If the problem is a step more complex, that is, includes the combination of compounds with each other to form a more 41 42 METALLURGICAL CALCULATIONS. complex compound, the molecular weights are still our guide, together with the thermochemical data given. If, for in- stance, the question above solved is complicated by the further requirement, to add the heat evolved by the formation of the slag, that is, of the MnO and FeO with SiO 2 to form silicate, we may calculate the heat of union of FeO and MnO with SiO 2 as follows: (Mn, Si, O 3 ) = 276,300 Calories. But (Mn, O) = 90,900 and (Si, O 2 ) = 180,000 therefore (MnO, SiO 2 ) = 5,400 or, per kilo, of MnO = - : - = 76 Calories. 71 Similarly (Fe, Si, O 3 ) = 254,600 Calories. (Fe, O) = 65,700 " (Si, O 2 ) = 180,000 " (FeO, SiO 2 ) = 8,900 per kilo, of FeO = ~ = 124 Calories. {& Therefore the additional heat of formation of the slag may be Wt. MnO = 64.5 kilos. X 76 = 4,902 Calories. Wt. FeO = 642.8 " X124 = 79,707 Sum = 84,609 Similar principles of calculation apply to all the oxygen- containing salts. Thus, if from the heat of formation of any sulphate we subtract the heat of formation of SO 3 , and also of the metallic oxide present, the residue is the heat of combina- tion of the metallic oxide with SO 3 , the weights involved being always those represented by the formulae. Thus, calling MO any metallic oxide, we may express the principle as follows: (MO, SiO 2 ) = (M, Si, O 3 ) (M, O) (Si, O 2 ) (MO, SO 3 ) = (M, S, O 3 ) (M, O) (S, O 3 ) (MO, CO 2 ) = (M, C, O 3 ) (M, 0) (C, O 2 ) (3MO, P 2 O 5 ) = (M 3 , P 2 , O 8 ) 3(M, 0) (P 2 , O 5 ) etc, etc. THERMOCHEM1CA L DATA. 43 Example. What is the heat required to calcine limestone? (CaO, CO 2 ) = (Ca, C, O 3 ) (Ca, O) (C, O 2 ) = 273,850 131,50097,200 45,150 Calories. This is the heat required to split up CaCO 3 (100 parts) into CaO (56) and CO 2 (44) ; the heat required is therefore 45,150-i- 100 = 451.5 Calories per kilo, of CaCO 3 decomposed. 45,150-^44=1026. " " " CO 2 driven off. 45,150^-56 = 806. " " " CaO remaining. And either of these quantities may be. used, according to convenience in working the problem. DOUBLE DECOMPOSITIONS. If the thermochemical problem involves the simultaneous decomposition of one or more substances and formation of one or more others, then the chemical equation should be written, and a thermochemical interpretation given of all the energy involved in the passage from starting compounds to products. Every chemical equation can be thus interpreted, if the heats of formation of all the compounds represented in it are known. We obtain the net energy of the reaction by assuming that all the substances used are resolved into their elements, and that all the products are formed from their elements; the first item is therefore to add together the heats of formation of all the substances used, starting with their elements, and by changing the algebraic sign of this sum we have the heat necessary to decompose the substances used into their elements; the second item is similarly found by adding together the heats of forma- tion of all the substances formed; the difference between these two items is the net energy of the reaction. In making these summations, regard must, of course, be paid to the number of molecules of each substance concerned, as shown in the re- action (because the tabulated data are the heats of formation of one molecule only) and to the algebraic sign of the heats of formation. Examples. (a) What heat is evolved when dry ferric oxide is reduced by aluminium, in the Goldschmidt " Thermite " process? 44 METALLURGICAL CALCULATIONS. Fe 2 O 3 + 2Al = Al 2 O 3 + 2Fe 195,600 + 392,600 + 197,000 Calories. (b) Write the equation showing the reaction of metallic sodium on water in excess: 2H 2 + 2Na = 2NaOH + H 2 + aq. (excess water) 2(69,000) 138,000 + 2(112,500) + 225,000 + 87,000 Calories. (c) What heat is evolved in the reaction of water* on cal- cium carbide to form acetylene gas? CaC 2 +2H 2 O = Ca0 2 H 2 + C 2 H 2 (6250) 138,000 131,750 + 215,600 +(54,750) + 160,850 + 29,100 The last case leaves out the very small heat of solution of the calcium hydrate which would normally go into solution. If such a large excess of water was used that all the calcium hydrate formed could dissolve, the heat of formation of the hydrate in dilute solution, 219,500 Calories, would be used, and the final result would be 33,000 Calories. The case is also very instructive, because it contains two compounds which are endothermic, that is, their heat of formation is negative, and therefore, as in the case of CaC 2 , heat is given out when it decomposes, while in the case of C 2 H 2 heat is absorbed when it is formed. (d) What are the heats of combustion of the gaseous hydro- carbons which form constitutents of common fuels, expressed per cubic meter of gas burning, and with the water formed assumed remaining as vapor, uncondensed? To calculate these we write the equations of combustion, with water as gas, and find the sum total of heat evolved; since every molecule of gas burning represents 22.22 m 3 of gas, we can, by a simple division, obtain the value sought. THERMOCHEMICAL DATA. 45 Methane: CO 2 + 2H 2 O + 97,200 + 2(58,060) 22,250 + 191,070 Calories. = 8,598 per m 8 CH. Ethane: 2C 2 H 6 + 70 2 = 4CO 2 + 6H 2 O 2(26,650) | +4(97,200) + 6(58,060) + 683,860 Calories. =15,387 perm 3 C 2 H 8 . Propane: C 3 H 8 + 50 2 = 3CO 2 + 4H 2 O 33,850 +3(97,200) +4(58,060) + 489,990 Calories. = 22,050 per m 3 C 3 H 8 . By applying these principles to all the hydrocarbons whose heat of formation has been given in the tables, we obtain the following useful table, water being considered as uncondensed and cold: Molecular Heat of Combustion . Heat of Combustion of Hydrocarbon. Calories. 1 m 3 (Calories) 1 ft 3 (B.T.U.) Methane (gas) .......... 191,070 8,598 966 Ethane (gas) ........... 341 ,930 15,387 1728 Propane (gas) .......... 489,990 22,050 2477 Ethylene (gas) .......... 321 ,770 14,480 1627 Propylene (gas) ......... 471 ,830 21,232 .2385 Toluene (liquid) ........ 906,990 ...... ____ B (liquid ....... 758,130 ...... (gas .......... 765,330 34,440 3869 - ,. j liquid ..... 1,428,930 ...... ____ Turpentine | ....... AT (solid ..... 1,223,690 Naphthaline j Anthracine (solid) ...... 1,690,150 ...... ____ Acetylene (gas) ......... 307,210 13,825 1555 TUT xt, 1 1 LI (liquid.. 148,270 Methyl alcohol | ^ ^^ 7,050 799 _,,, j liquid... 295,330 Ethyl alcohol j gas ^^ 13 7M 1M4 46 METALLURGICAL CALCULATIONS. Molecular Heat of Combustion. Heat of Combustion of . (liquid 396,130 (gas 403,630 18,163 2040 (To the above we will add data for three other common gaseous fuels.) Carbonous oxide (gas) 68,040 3,062 344 Hydrogen (gas) 58,060 2,613 293 . 5 Hydrogen sulphide 122,520 5,513 619 CALORIFIC POWER OF FUELS. By using the principles explained we can calculate the cal- orific power of any combustible, with the aid of one or two simple assumptions. Regarding the water formed, we will con- sider in all ordinary metallurgical calculations that it remains uncondensed, thus putting it on the same footing as the other products of combustion. This amounts virtually to assuming that the latent heat of condensation of the vapor formed has not been generated in the furnace, an assumption which is quite justified if, as is almost always the case, the water formed inevitably escapes as vapor. If, in any special case, the pro- ducts of combustion are in reality cooled down so far that the water does condense, then it would be proper to assume the higher calorific powers for the combustion of the hydrogen- containing materials, and thus a more accurate heat-balance of the furnace operation could be constructed. An instance of the latter might be the case of the partial combustion of a fuel in a gas producer, where the fuel gas is afterwards cooled be- fore using, in order to condense from it ammonia water, etc. (Mond's producer). In this case, it is possible that some water formed by the partial combustion would be afterwards condensed, and to obtain a correct heat-balance-sheet of the producer it would be necessary to assume that this heat had also been generated. Such cases occur very rarely in metal- lurgical practice, and it is therefore recommended to always use the low r er calorific values, assuming water vapor uncon- densed and its latent heat of condensation not to have been generated unless the conditions of the problem point plainly to the opposite course as correct. It is no more correct to charge against a furnace the latent heat of condensation of the THERMOCHEMICAL DATA. 47 water vapor formed, than it would be to charge against it the latent heat of condensation of the carbonic oxide formed, if the conditions of practice are such that it is impossible to utilize in any useful manner either one of them. Those who persist in assuming that the latent heat of vaporization of the water formed by combustion is generated in and chargeable against the furnace, and then, of necessity, charge the same quantity against the heat carried off in the products of combustion, are in the great majority of cases adding to both sides of the balance sheet a quantity which cannot possibly be utilized, and are therefore distorting all the factors of heat generation and distribution. By thus abnormally increasing the sensible heat in the waste gases, they abnormally decrease the proportionate value of other items of heat distribution and utilization. I have dwelt on this matter at length, because it is of importance in furnaces using fuel, such as gas, rich in hydrogen; where, in some cases, the counting of the latent heat of condensation of the water vapor formed in the escaping gases, would perhaps double the apparent chimney loss, and give distorted values to the whole heat balance sheet. What we are after, in every case, are the real facts and figures as to the working of a furnace or operation, and we must therefore judge the operation by a theoretically perfect standard, it is true, but not by a theo- retically impossible standard; i.e., our standard must be the theoretically possible one under the practical conditions neces- sarily prevailing. A radical reform is very much needed in just this respect, in the treatment of the latent heat of vapor- ization of the water formed by writers on metallurgical pro- cesses and problems of combustion. Problem 4. A natural gas from Kokomo, Ind., contained by analysis: CH 4 94.16, H 2 1.42, C 2 H 4 0.30, CO 2 0.27, CO 0.55, O 2 0.32, N 2 2.80, H 2 S 0.18 per cent. Required : (1) What is its metallurgical calorific power per cubic meter and per cubic foot? (2) Comparing it with coal, having a calorific power of 8,000 Calories, how much gas gives the same generation of heat as a metric ton of coal? 48 METALLURGICAL CALCULATIONS. (3) If the natural gas costs $0.15 per 1,000 cubic feet, at what price per metric ton would the coal furnish heating power at the same cost for fuel? Solution. Requirement (1) Heat of combustion of 1 cubic meter: CH 4 0.9416X 8,598 = 8095.9 Calories. H 2 0.0142X 2,613 = 37.1 C 2 H 4 0.0030 X 14,480 = 43.4 CO 0.0055X 3,062 = 16.8 H 2 S0.0018X 5,513 = 9.9 Total - 8203.1 In British thermal units per cubic foot we have 8,203.1X3.967 ^ 35.314 = 8,203.1X0.11233 = 921.5 B. T. u. per ft 3 (1) Requirement (2): 8000X1000 8,203.1 = 34,450 ft 3 . If the ton of coal be taken as 2240 pounds, instead of the metric ton of 2204 pounds, the equivalent volume is |rr = 3 5j013 ft s B Requirement (3): 34 450 T6oo~ x0 ' 15 = $5 - 16 * P er metric ton - or, per ton of 2240 pounds, 2240 S5.16JX 2204 = $5 ' 25 per long ton. DULONG'S FORMULA. When a solid or liquid carbonaceous fuel is to be burned, its calorific power may either be determined directly in a calori- meter (which is the best way, if carefully done), or may be calculated from its analysis. In case it is determined calori- THERMOCHEMICAL DATA. 49 metrically, the weight of water produced per unit of fuel should be determined either in the same or by a separate experi- ment, and the latent heat of vaporization of this water sub- tracted from the calorimetric value of the fuel, in order to get its metallurgical, or practical, calorific power. Example. A coal gave 9215 calories per gram in the calori- meter, and its products of combustion gave up 0.45 grams of condensed water. What is its practical calorific power? Taking the products cold, the heat of condensation per gram of water vapor may be taken as 606.5 calories (Regnault), and we therefore have obtained in the calorimeter 0.45X606.5 = 273 Calories more than if the products were cold and water uncondensed. The practical calorific power is therefore 9215 - 273 = 8942 Calories per kilogram. Dulong's simple formula, or method of calculation, is the state- ment that the calorific power of a fuel can be calculated from the amount of carbon, hydrogen, and oxygen it contains by assuming all the carbon free to burn, the oxygen to be all combined with hydrogen in the proportions of O 2 to 2H 2 (32 to 4), and that the rest of the hydrogen is free to burn. Example. Taking the bituminous coal of Problem 1, con- taining carbon 73.60, hydrogen 5.30, nitrogen 1.70, sulphur 0.75, oxygen 10.00, moisture 0.60, ash 8.05, its calculated calorific power is per kilogram: Carbon 0.7360 X 8100 = 5962 Calories. Hydrogen 0.0530 0.0125 0.0405 X 34,500 = 1397 " (water liquid) Sulphur 0.0075 X 2164= 17 Total 7376 This calculation is made, however, on the assumption that all the water in the products is condensed to liquid, i.e., not only the water formed by combustion of the free hydrogen, but also the already-formed H 2 O containing the oxygen of the coal, as well as the moisture present to start with. To obtain 60 METALLURGICAL CALCULATIONS. the practical calorific power by calculation we must subtract from the above-generated heat the latent heat of all the vapor of water in the products, as follows: Moisture in coal . 0060 kilos. Moisture formed by combined hydrogen . . . 1 125 " Moisture formed by free hydrogen 0.3645 " Total. 0.4830 " Latent heat of vaporization = 606.5x0.4830 = 293 Calories. Practical calorific power, water all as vapor = 7083 Calories. The values thus calculated are found to agree satisfactorily with the laboratory and the practical calorific powers of the fuels. THE THEORETICAL TEMPERATURE OF COMBUSTION. If we start with a cold fuel and cold air the maximum tem- perature which can be obtained by the combustion is simply that temperature to which the heat generated can raise the products of combustion, assuming that all that heat resides primarily in the products as sensible heat. This is then a prob- lem in physics, in which we have a known amount of heat available, and the question is to what temperature it will raise the products of combustion. The problem is at once soluble when we know the mean specific heats of the products of combustion and their respective quantities. Until a few years ago these specific heats at high tempera- tures were unknown, and very false results were obtained by assuming constant specific heats from ordinary temperatures up, and values were thus calculated which were known to be hundreds, and in some cases, thousands, of degrees too high. The most satisfactory values for the specific heats of the fixed gases and water vapor and carbonic oxide are those deter- mined by Mallard and Le Chatelier, and their proper use has entirely solved the question of theoretical temperatures of com- bustion, and removed a positively disgraceful discrepancy between theory and practice. The specific heats of these gases increase with temperature, so that the actual specific heat of a cubic meter (measured at standard conditions) is, at the tem- perature t, THERMOCHEMICAL DATA. 51 for N 2 , O 2 , H 2 , CO S = 0.303 + 0.0000541 " < CO 2 S = 0.37 +0.00044t 11 " (vapor) H 2 O S = 0.34 + 0.00030t For calculating the quantity of heat needed to raise the gas from O to t, however, we need the mean specific heats, Sm, between O and t. These are, of course, the above constants plus half the increase, viz.: for N 2 , O 2 , H 2 , CO Sm - 0.303 + 0.000027t CO 2 Sm = 0.37 + 0.00022t (vapor) H 2 O Sm = 0.34 +0.00015t and the quantity of heat needed to raise 1 cubic meter (at standard conditions) from O to t is for N 2 O 2 H 2 , CO Q (o-t) = 0.303t + 0.000027t 2 CO 2 Q (o-t) = 0.37t + 0.00022t 2 H 2 O O (o-t) = 0.34t + 0.0001 5t 2 or, finally, the quantity of heat needed to raise 1 cubic meter (measured at standard conditions) from t to t' is for N 2 , O 2 , H 2 , CO Q (t'-t) = 0.303 (t'-t) +0.000027 (t /2 -t 2 ) CO 2 Q (t'-t) = 0.37 (t '-t) + 0.00022 (t /2 -t 2 ) H 2 Q (t'-t) = 0.34 (t'-t) +0.00015 (t' 2 -t 2 ) And the mean specific heat, Sm, between any two tempera- tures is for N 2 O 2 , H 2 , CO Sm (t'-t) = 0.303 + 0.000027 (t' + t) CO 2 Sm (t'-t) = 0.37 +0.00022 (t' + t) H 2 O Sm (t'-t) = 0.34 +0.00015 (t' + t) Examples : (1) What is the temperature of the hottest part of the oxy- hydrogen blowpipe flame? According to the equation 2H 2 + O 2 = 2H 2 O, the hydrogen gas forms an equal volume of water vapor. (Equal numbers of molecules.) The heat of combustion of one cubic meter of hydrogen is 2613 Calories. The question is, therefore: "To 52 ME 1 A LLURGIOA L CA UCULA TIONS. what temperature will 2613 Calories raise one cubic meter of water vapor? " The answer is, using the data above, Q ( -t) = 0.34t + 0.00015t 2 = 2613 Whence t = 3191 (2) What is the maximum temperature of the hydrogen flame burning in dry air? The heat evolved is, as before, 2613 Calories, and there is formed also one cubic meter of water vapor, but the products will contain also the nitrogen which accompanied the 0.5 cubic meter of oxygen necessary for combustion, viz.: ' 0.5 U . ^Uo = 1.9 cubic meters, and this is heated as well as the water vapor. Therefore, Q ( -t) = (0.34t + 0.00015t 2 ) + 1.9 (0.303t + 0.000027t 2 ) = 2613 Whence 0.916t + 0.0002013t 2 = 2613 And t = 2010 (3) What is the temperature of the air-hydrogen blowpipe, if 25 per cent, excess of air is used, above that required ? All the data are the same as above, except that to the pro- ducts will' be added 0.25 (0.5 + 1.9) =0.6 cubic meters of unused air, which has the same specific heat as nitrogen, and therefore the equation becomes Q (o-t) = (0.34t + 0.00015t 2 )+2.5 (0.303t + 0.000027t 2 ) = 2613 Whence t = 1764 This calculation brings out very clearly the uselessness and ineffectiveness of using more air than is theoretically neces- sary; any excess simply passes unused into the products of combustion, and thus reduces their maximum possible tem- perature. The obtaining of the maximum possible temperature depends upon accurately proportioning the supply of oxygen or air to the quantity of gas burned ; an excess or a deficiency will result in a lower temperature. COMBUSTION WITH HEATED FUEL OR AIR. If the fuel itself or the air which burns it is preheated, the sensible heat in either one or in both is simply added to the heat generated by the combustion, to give the total amount of TEERMOCHEMICAL DATA. 53 heat which must be present as sensible heat in the products of combustion. The effect is exactly the same as if the heat developed by combustion had been increased by the sensible heat in the fuel or air used. What is the calorific intensity (theoretical maximum tem- perature) obtained by burning carbonous oxide gas? (a) Cold, with cold air; (b) cold, with hot air at 700 C.; (c) hot, with hot air, both at 700 C. (a) Take one cubic meter of carbonous oxide. Calorific power 3,062 Calories; product one cubic meter of carbonic oxide, and 1.904 cubic meters of nitrogen. Let t be the temperature attained ; then Heat in the 1 m 3 carbonic oxide == 0.37t + 0.00022t 2 1.904m 3 nitrogen gas = 0.577t + O.Q000514t 2 11 products = 3,062 Cal's. = 0.947t + 0.0002714t 2 Whence t = 2050 (b) If the 2.404 cubic meters of air needed are preheated to 700 C., they will bring in as sensible heat Q (o-700) = 2.404 [0.303 (700) +0.000027 (700) 2 = 552 Calories. The total heat in the products will be 3062 + 552 = 3614 Calories, and therefore 0.947t + 0.000271 4t 2 = 3614 Whence t = 2189 (c) If the gas itself is also preheated it brings in Q (o-700) = 0.303 (700) +0.000027 (700) 2 = 225 Calories. The total heat in the products will be 3614 + 225 = 3839 Calories, and therefore 0.947t + 0.0002714t 2 = 3839 Whence t = 2284 Heating both gas and air to 700 before they burn thus raises the theoretical temperature from 2050 to 2284 or 234. Problem 5. Statement. The natural gas of Kokomo, Ind., contains by analysis: Methane (marsh gas) 94.16, ethylene (olefiant gas) 54 METALLURGICAL CALCULATIONS. 0.30, hydrogen 1.42, carbonous oxide 0.55, carbonic oxide 0.27, oxygen 0.32, nitrogen 2.80, hydrogen sulphide 0.18, in per- centages, by volume. Required: (1) The maximum flame temperature, if burned cold with the theoretical amount of cold, dry air necessary. (2) The calorific intensity, if burned cold, with the requisite air preheated to 1000 C. (3) The calorific intensity, if burned cold, with 25 per cent, more air than theoretically necessary, preheated to 1000. Solution : [The practical calorific power of this gas has been already calculated in Problem 4 as 8203 Calories per cubic meter. The gas itself is always burned cold, because, if preheated, it decomposes and deposits carbon in the regenerators.] The products of combustion of the gas, say per cubic meter, must first be found, using the formula? for combustion. Requirement (1): 1 Cubic Meter of Gas. Oxygen Needed. m 3 CO 2 Products. H 2 O CH 4 .9416 1 .8832 .9416 1 .8832 C 2 H 4 .0030 .0090 .0060 .0060 H 2 .0142 .0071 0.0142 CO .0055 .00275 .0055 .... CO 2 .0027 .0027 .... O 2 0.0032 0.0032 N 2 .0280 .... .... .... H 2 S .0018 .0027 .0018 SO 2 N 2 0.0280 0.0018 1.90155 0.9558 1.9052 0.0018 0.0280 Air required - 9.14 (carrying in N 2 ) = 7-238_ Total N 2 77266" The heat generated will exist as sensible heat in the CO 2 , H 2 O, SO 2 and N 2 of the products. The mean specific heat of SO 2 per cubic meter is 0.444 at ordinary temperatures; what it is at high temperature has not been determined; we will give it the same index of increase as the analogous gas CO 2 , THERMOCHEMICA L DATA. 55 and take for it Sm (o-t) = 0.444 + 0.00027t, or Q (o-t) = 0.444t + 0.00027t 2 . At the temperature attained by combustion, t, the products will contain the following amounts of heat: N 2 = 7.266 (0.303t + 0.000027t 2 ) H 2 O = 1.9052 (0.34t +0.000l5t 2 ) CO 2 = 0.9558 (0.37t + 0.00027t 2 )* SO 2 = 0.0018 (0.444t + 0.00027t 2 ) Total = 3.2044t + 0.00074057t 2 = 8203 Calories. From which t = 1806 (1) Requirement (2): Heat in 1 m 3 of air at 1000 = 0.303 (1000) +0.000027 (1000) 2 . = 330 Calories. " 9.14m 3 " 1000 = 3016 " Therefore : 3.2044t + 0.00074057t 2 = 8203 + 3016 Whence t = 2288 Requirement (3): Excess air used = 9.14x0.25.= 2.285m 3 . Heat in 11.425m 3 at 1000 = 11.425x330 = 3770 Calories, The heat capacity of the excess air must be added to the heat in the products at temperature t, viz.: Excess air = 2.285 (0.303t + 0.000027t 2 ) making the total heat in the products (adding to previous ex- pression) : 3.8968t + 0.00080227t 2 = 8203 + 3770 Whence t = 2134 THE ELDRED PROCESS OF COMBUSTION. A means of regulating the temperature of the flame has been proposed by Eldred, and described by Mr. Carlton Ellis in the December, 1904, issue of El. Chem. Ind. The proposition is simply to mix with the air used for combustion a certain * These and some succeeding problems were worked out using this specific heat. Later the author has adopted a slightly different value for CO 2 , viz., (0.37 + 0.00022t), which he now considers more nearly correct. 56 METALLURGICAL CALCULATIONS. proportion of the products of combustion themselves. The principle is easily understood when the requisite calculations of the theoretical flame temperatures are made. When solid fuel is burnt the temperature is often too high locally, and re- sults in burning out grate bars or overheating the brick work of the fire-place, or overheating locally the material which is mixed with the fuel. If the air is diluted with products of combustion the initial theoretical temperature is lowered, and the above evils may be obviated. Using solid fuel, the heat in the fuel before the air actually burns it must be added to the heat generated by combustion to get the actual temperature in the hottest part of the fire. Examples: (1) What will be the highest temperature in a charcoal fire fed by air, assuming complete combustion without excess of air? Assuming the charcoal to be pure carbon, and to be heated to the maximum temperature t before it burns (by the com- bustion of the preceding part), the heat available is: Heat of combustion of 1 kilo of carbon 8100 Cal's. Sensible heat in the carbon at t. . . .0.5t 120 " Total heat available to raise the temperature. . .7980 + 0.5t Products of combustion CO 2 1.85 m 3 ..N 2 7.04 " Heat of product at t CO 2 1.85 (0.37t +0.00022t 2 ) N 2 7.04 (Q.303t + O.OQ0027t 2 ) Sum 2.81t + 0.00060t 2 therefore 2.81t + 0.00060t 2 = 7980+0.5t whence t = 2199 (2) What will be the temperature in the same case, if the air used is diluted with an equal volume of the products of combustion ? The heat available is the same as before, 7980 + 0.5t; but since the mixed air for combustion contains only half as much oxygen per cubic meter, the products will be exactly doubled in amount for a unit weight of carbon burnt, and we therefore have directly 2 (2.810t + 0.00060t 2 ) = 79SO + 0.5t whence t = 1213 THERMOCHEMICA L DA TA . 57 It is therefore evident that the maximum temperature of the hot gases at their moment of formation is nearly halved by the procedure stated. (3) Taking the cases cited by Mr. Ellis, where the products contained originally 15 per cent, oxygen and 6 per cent, carbon dioxide, and after mixing the air used with half its volume of the chimney gases, 9 per cent, oxygen and 12 per cent, carbon dioxide (the gas-air mixture entering containing 15 per cent. oxygen and 6 per cent, carbon dioxide), what are the maximum temperattires obtained in the two cases? Case 1 : The heat available is as before ; the products of com- bustion are CO 2 1.85 m 3 , O 2 4.62 m 3 , N 2 24.36 m 3 , and their sensible heat at temperature t Heat in CO 2 = 1.85 (0.37t +0.00027t 2 ) = 28.98 (0.303t + 0.000027t 2 ) Sum 9.46t + 0.00128t 2 therefore 9.46t + 0.001280t 2 = 7980 + 0.5t whence t = 800 Case 2: The products, per kilogram of carbon burnt, will be CO 2 3.70 m 3 , O 2 2.77 m 3 , N 2 24.36 m 3 , and we have Heat in CO 2 = 3.70 (0.37t +0.00027t 2 ) O 2 + N 2 = 27.13 (0.303t + 0.000027t 2 ) Sum = " 9.09t + 0.00173t 2 therefore 9.09t + 0.00173t 2 = 7980 + 0.5t whence t = 764 Conclusions: The calculations regarding the Eldred process bring out what was not stated in the printed description of the method, viz.: that if a small excess, or no excess of oxygen escapes, to the chimney, the temperature of the flame will be greatly reduced by the dilution of the air used, because more gas-air mixture will be required per unit of fuel burnt; but if there is any large amount of unused oxygen escaping in the first instance, the dilution practiced will scarcely affect the temperature of the flame at all, because it makes very little difference whether the fuel is heating up unused oxygen or carbon dioxide dilutant substituted for some of it, as long 58 METALLURGICAL CALCULATIONS. is there is more than enough oxygen to burn the fuel. The fact of the specific heat of carbon dioxide being greater than that of oxygen, is the only reason for the small calculated dif- ference of 36, in cases 1 and 2. It is true that the dilution practiced will result in less heat being lost in the waste gases, in the Example (3), but the same economy could be obtained by simply using less excess of air in the first instance. TEMPERATURES IN THE " THERMIT " PROCESS. The Goldschmidt porcess of reducing metallic oxides by powdered aluminium, igniting the cold mixture, is only a spe- cial case of our general rule, as far as concerns the calcula- tion of the theoretical temperatures obtained. In any case, the total heat available is the surplus evolved in the chemical reaction, and the temperature sought is that to which this quantity of heat will raise the products of the reduction. The products are alumina and the reduced metal. The heat in the latter, in the melted state, is well known in many cases; it is most clearly expressed by the sum of the heat in such melted metal just at its melting point (which is easily determined calorimetrically, and is well known for many metals), plus the heat in the melted metal from the melting point to its final temperature, which is equal to t, minus the melting tem- perature, multiplied by the specific heat in the melted condition These data are known for many metals; for some they may have to be assumed from some general laws correlating these values. The heat in melted alumina has not been determined calorimetrically, to my knowledge. Its sensible heat solid is 0.2081t + 0.0000876t 2 (determination made in author's labora- tory), which, evaluated for the probable melting point, 2200 C., would give the heat in it at that temperature 881.8 Calories; the latent heat of fusion of molecular weight is very probably 2.1 T, where T is the absolute temperature of the melting point, making the latent heat of fusion per kilogram 2.1 (2200 + 273) -^ 102 = 5193-^102 = 50.9 Calories. The specific heat in the melted state is probably equal to the specific heat of the solid at the melting point, viz.: 0.2081+0.0001752 (2200) = 0.5935. We, therefore, have for the heat in melted alumina at temperature t THERMOCHEMICAL DATA. 59 Heat in solid alumina to the melting point 881 .8 Calories Latent heat of fusion 50.9 Heat in liquid alumina to its setting point 0.5935 (t-2200) Total. 932. 7 + 0.5935 (t-2200) or, fora molecular weight, 102 kilograms: 95,135 + 60.54 (t-2200). Examples: (1) If black cupric oxide is reduced by powdered aluminium what is the temperature attained? The react ion "is 3CuO + 2Al = Al 2 O 3 + 3Cu 3 (37,700) ' + 392,600 + 279,500 The products must therefore be raised to such a tempera- ture that they contain 279,500 Calories. The heat in molecular weight of alumina at temperature t has already been found; that in copper at t is, for one kilogram (using the author's determinations) : Heat in melted copper at setting point =162 Calories. Heat in melted copper above 1065 = 0.1318 (t 1065) Calories. Total = 162 + 0.1318 (t 1065). Per atomic weight (63.6 kilos.) = 10303 + 8.3825 (11065) From these data there follows the equation: 95,135 + 60.54 (t 2200) + 3[10,303 + 8.3825 (t 1065)] = 279,500 whence t = 3670 A little further calculating will show that approximately one- third of all the heat generated exists in the copper, and two- thirds in the melted alumina. (2) If pure ferric oxide is reduced by the " Thermit " process, what is the temperature of the resulting iron and melted alumina ? The reaction and the heat evolved have been already given in the preceding discussion of these calculations (page 44.) They show that per molecular weight of alumina formed there are two atomic weights (112 kilos.) of iron formed, and there is disposable altogether 197,000 Calories. The heat in a kilo- 60 METALLURGICAL CALCULATIONS. gram of pure iron at its melting point (1600) is 300 Calories, the latent heat of fusion approximately 69 Calories, and the specific heat in the melted condition 0.25. The total heat in a kilogram of melted iron is therefore 369 + 0.25 (t 1600) Cal- ories, or per atomic weight = 20,664 + 14 (t 1600) Calories. We, therefore, can write the equation: 95,135 + 60.54 (t 2200) + 2[20,664 + 14 (t 1600)] = 197,000 whence t = 2694 A similar calculation made for the reduction of MnO by the theoretical amount of aluminium, shows a reduction tem- perature less than the melting point of alumina. This would mean that the melting down of the mass to a fused slag of pure alumina could not take place. What happens in the re- duction of manganese is that an excess of manganous oxide is used, whereby all the aluminium is consumed, and none at all gets into the reduced manganese, and, furthermore, the excess of manganous oxide unites with the alumina to form a slag of manganous aluminate, which is fusible at the temperature attained. Without the latter arrangement no fused slag could result. Similar calculations, made with silicon as the reducing agent, show similar difficulties regarding the theoretical tem- peratures attainable, when iron or manganese are reduced. By using an excess of the oxides of these metals, however, calculation shows that temperatures sufficient to fuse the metals and the manganous silicate, or ferrous silicate produced, ought to be obtained. CHAPTER TV. THE THERMOCHEMISTRY OF HIGH TEMPERATURES The problem is: Knowing the heat evolved (or absorbed) in the formation of a compound, or in a double reaction, start- ing with the reacting materials cold, and ending with the pro- ducts cold, what is the heat evolved (or absorbed) in either of the following cases? (a) Starting with the reagents cold and ending with the products hot. (b) Starting with the reagents hot and ending with the products hot. (c) Starting with the reagents hot and ending with the pro- ducts cold. Of these three cases (b) is the most general form of the prob- lem, and occurs frequently in metallurgical practice, particu- larly in electrometallurgy; (a) is a more limited form, and requires less data for its calculation, while it is very frequently the desideratum in discussing the thermochemistry or heat re- quirements of a metallurgical process; (c) is derivable at once if the data exist for calculating (b), and is of such rare occur- rence in practice that we can dispense with its lengthy dis- cussion. The thermochemical data already given and described in pre- ceding sections enable us to calculate the heat of any chemical reaction starting with cold reagents and ending with the pro- ducts cold. For instance: (Zn, O) = 84,800 Calories means that if we take 65 kilograms of solid zinc, at room tempera- ture, and 16 kilograms of oxygen, as gas at room temperature, ignite them, and after the reaction cool the 81 kilograms of zinc oxide formed down to the same starting temperature, there will be developed a total of 84,800 Calories. Similarly, using the datum (C, 0) = 29,160 Calories, we can construct 61 60 METALLURGICAL CALCULATIONS. gram of pure iron at its melting point (1600) is 300 Calories, the latent heat of fusion approximately 69 Calories, and the specific heat in the melted condition 0.25. The total heat in a kilogram of melted iron is therefore 369 + 0.25 (t 1600) Cal- ories, or per atomic weight = 20,664+14 (t 1600) Calories. We, therefore, can write the equation: 95,135 + 60.54 (t 2200) + 2[20,664+14 (t 1600)] = 197,000 whence t = 2694 A similar calculation made for the reduction of MnO by the theoretical amount of aluminium, shows a reduction tem- perature less than the melting point of alumina. This would mean that the melting down of the mass to a fused slag of pure alumina could not take place. What happens in the re- duction of manganese is that an excess of manganous oxide is used, whereby all the aluminium is consumed, and none at all gets into the reduced manganese, and, furthermore, the excess of manganous oxide unites with the alumina to form a slag of manganous aluminate, which is fusible at the temperature attained. Without the latter arrangement no fused slag could result. Similar calculations, made with silicon as the reducing agent, show similar difficulties regarding the theoretical tem- peratures attainable, when iron or manganese are reduced. By using an excess of the oxides of these metals, however, calculation shows that temperatures sufficient to fuse the metals and the manganous silicate, or ferrous silicate produced, ought to be obtained. CHAPTER TV. THE THERMOCHEMISTRY OF HIGH TEMPERATURES. The problem is: Knowing the heat evolved (or absorbed) in the formation of a compound, or in a double reaction, start- ing with the reacting materials cold, and ending with the pro- ducts cold, what is the heat evolved (or absorbed) in either of the following cases? (a) Starting with the reagents cold and ending with the products hot. (b) Starting with the reagents hot and ending with the products hot. (c) Starting with the reagents hot and ending with the pro- ducts cold. Of these three cases (b) is the most general form of the prob- lem, and occurs frequently in metallurgical practice, particu- larly in electrometallurgy; (a) is a more limited form, and requires less data for its calculation, while it is very frequently the desideratum in discussing the thermochemistry or heat re- quirements of a metallurgical process; (c) is derivable at once if the data exist for calculating (6), and is of such rare occur- rence in practice that we can dispense with its lengthy dis- cussion. The thermochemical data already given and described in pre- ceding sections enable us to calculate the heat of any chemical reaction starting with cold reagents and ending with the pro- ducts cold. For instance: (Zn, O) = 84,800 Calories means that if we take 65 kilograms of solid zinc, at room tempera- ture, and 16 kilograms of oxygen, as gas at room temperature, ignite them, and after the reaction cool the 81 kilograms of zinc oxide formed down to the same starting temperature, there will be developed a total of 84,800 Calories. Similarly, using the datum (C, O) = 29,160 Calories, we can construct 61 62 METALLURGICAL CALCULATIONS. and interpret the reduction reaction, starting with cold ma- terials and finally ending with cold materials, as follows: ZnO + C = Zn + CO 84,800 I +29,160 55,640 This reaction, as interpreted, stands for none of the above (a), (b) or (c) ; in fact, it represents only a calorimetric determination in the laboratory, and does not correspond to either of the three cases actually taking place in practice, that is, it is not directly applicable to practical conditions, without being modified by the conditions actually arising in practice. Case (a): If we, in practice, start with the reagents cold, and the products pass away from the furnace hot, at some de- termined temperature t, the total heat energy necessary to cause this transformation is calculable in two ways. The first way is to follow the course of the reaction, and to say that the total heat absorbed is that necessary to heat the reacting bodies to the temperature t, plus the heat of the chemical re- action assumed as starting and finishing at that temperature. The first of these items can be obtained if we know the sensi- ble heat in the reacting bodies up to the temperature t ; it is a question of specific heats of the reacting bodies up to t (in- cluding any physical changes of state occurring in them be- tween o and t); the second item requires a knowledge of the heat of formation of all the compounds involved, starting with their ingredients at t, and ending with the products at t. But the latter item is the general question of the heat of a reaction starting with the ingredients hot and ending with the pro- ducts hot; it is the most general case, which we have desig- nated as Case (6), and which will be discussed later. This way of solving Case (a), therefore, really includes the solu- tion of Case (b), and we will defer its consideration for the present. The second method of solving Case (a), and one which does not involve the more general solution, is to take the heat of the reaction, starting with the ingredients cold and ending with the products cold the ordinary heat of the re- action from ordinary thermochemical data and to add to this THERMOCHEMISTRY OF HIGH TEMPERATURES. 63 the amount of heat which would be required to raise the pro- ducts from zero to the temperature t. This, of course, does not actually represent the sequence in which the reactions take place in practice, but it accurately represents the heat involved or evolved in passing from the cold reagents to the hot pro- ducts, and is, therefore, exactly the practical quantity which we are endeavoring to find. Moreover, it involves a knowledge of only the specific heats of the products, and not that of the ingredients or substances reacting. Illustration: Starting with a mixture of zinc oxide and car- bon in the proportions Zn O and C, at ordinary temperature, and shoveling them cold into a retort, how much heat is ab- sorbed in converting them into Zn vapor and CO gas, issuing from the retort at 1300 C.? If we start to calculate this quantity, by first finding the heat necessary to heat Zn O and C up to 1300, that is obtained by multiplying the weights of each by their mean specific heats from to 1300, and then by 1300, as follows: Calories. 81 kilos. ZnOX*[0.1212 (1300) +0.0000315 (1300) 2 ] = 17,075 12 " C X*[0.5 (1300) 120] = 6,360 Sum = 23,435 To this must then be added the heat of the reaction ZnO + C = Zn + CO, starting with the reacting bodies at 1300, and ending with the products at the same temperature. This can only be found by solving the general Case (b) for this par- ticular reaction, which we will find involves a knowledge of the heat required to raise Zn, O, C. ZnO and CO from to t. Anticipating such a solution, we may say that the heat of the reaction starting and ending at 1300 is 80,166 Calories, making the sum total of energy required 103,601 Calories. The solution is usually much simpler if we take the second method, and add to the heat of the reaction, starting and ending at zero ( 55,640 Calories), the heat required to raise 65 kilos, of zinc and 28 kilos, (22.22 cubic meters) of carbonous *Heat in 1 kilo of carbon, for temperature above 1,000, 0.5t - 120 (deduction from Weber's results) ; zinc oxide 0.1212 t + 0-0000315 1 2 (de- termination by the author). 64 METALLURGICAL CALCULATIONS. oxide, from their ordinary condition at zero to their normal condition at 1300. The calculation for the CO gas is simply: 22.22X[0.303 (1300)4-0.000027 (1300) 2 ] = 9,766 Calories. For the zinc, the calculation is more complicated: Heat in solid zinc to the melting point (420) : 65X[0.09058 (420) +0.000044 (420) 2 ] = 2,977 Calories. Latent heat of fusion 65X22.61 =1,470 " Heat in melted zinc, 420 to boiling point (930) : 65 X [0.0958 + 0.000088 (420)]X (930420) =4,228 " Latent heat of vaporization (Trouton's rule) 23 X (930 + 273) = 27,670 u Heat in zinc vapor (monatomic) 5X (1300 930) = 1,850 Calories. Sum = 38,195 " The total sensible heat required is, therefore, 9,766 + 38,195 = 47,961 Calories, which, added to the 55,640 absorbed in the chemical reaction, if it started and ended at zero, makes a total heat requirement of 103,601 Calories for the practical carrying out of this reaction, starting with the reagents cold and ending with the hot products at 1300. [To be absolutely accurate, regard should be paid in the above case to the fact that the above calculations are based on the substances being all at atmospheric pressure, while in the mixture of Zn vapor and CO gas each is under only 0.5 atmospheric tension. Since each of these represents a molecular weight, the outer work which has been included in the calcu- lations is 2X2T = 2X2 (1300 + 273) = 6292 Calories, whereas it should really be only half that much, or 3146 Calories. The corrected heat required is, therefore, 103,6013,146 = 100,455 Calories, or 1,545 Calories per kilogram of zinc. This datum is exactly the net heat requirement on which calculations of the net electrical energy required to produce zinc from its oxide, or calculations of the net efficiency of an ordinary zinc furnace, would be based.] THERMOCHEMISTRY OF HIGH TEMPERATURES. 65 Case (6): To calculate the heat of a chemical reaction start- ing and finishing at any temperature t, two methods are avail- able; The most general solution, and that easiest to understand, is to calculate for each compound involved the heat of its formation at the temperature t, that is, the heat evolved if the elements start at t and the product is cooled to t. Having these heats of formation at t, they are used in the equation in just the same manner as the heats of formation at zero are ordinarily used in obtaining the heat of the reaction start- ing and ending at zero. The calculations are based on this general principle: The heat evolved when the cold elements unite to form the hot product at temperature t equals the heat of union at zero, minus the heat necessary to raise the product from zero to t ; if to this difference we add the heat which would be necessary to heat the uncombined elements from zero to t, the sum is the desired heat of formation at t. Illustration: The heat of formation of ZnO at zero is 84,800, starting with cold Zn and O and ending with cold ZnO. If we started with cold Zn and O and ended with hot ZnO, say at 1300, the heat evolved altogether would be less than 84,800 by the sensible heat in the 81 kilograms of ZnO at 1300, which has already been calculated (see previous illustration) to be 17,075 Calories. The difference is 67,725 Calories, and repre- sents the transformation from cold Zn and O to hot ZnO. If the Zn and O were heated to 1300 before combining, they would contain as sensible heat the following quantities: Heat in 65 kilograms of Zn (0 to 1300) already calculated 38,195 Calories Heat in 16 kilograms of O = 11.11 m 3 X [0.303 (1300) +0.000027 (1300) 2 ] = 4,884 Sum 43,079 And the heat evolved in passing from the hot reagents to the hot product at 1300 must be 67,725 plus this sensible heat, or 110,804 Calories. We can express this datum as follows: (Zn, O) 1300 = 110,804, which means that when the zinc and oxygen taken in their normal state at 1300 combine to form ZnO at 1300, 110,804 Calories are evolved. 66 METALLURGICAL CALCULATIONS. A similar calculation for CO is as follows: (C, O) = 29, 160 Calories Sensible heat in CO (0 to 1300) already cal- culated = 9,766 Heat evolved when cold C and O form CO at 1300 = 19,394 Sensible heat in C (0 to 1300) = 6360 " O ( " ) = 4884 11,244 Calories Heat evolved hot C and O to hot CO = Sum = 30,638 Or (C, O) 1300 = 30,638 Uniting the two data found, for the heats of formation of ZnO and CO at 1300, in the equation of reduction, we have ZnO + C = Zn + CO at 1300 = -110,804 | +30,638 80,166 The actual reduction is thus seen to absorb 24,526 Calories more at 1300 than at zero, an increase of over 42 per cent. If to this heat of reaction at 1300 we add the heat neces- sary to raise the ZnO and C to 1300 (already calculated under Case (a) as 23,435 Calories), we will have a total heat require- ment of 103,601 Calories, which would be required practically if we started with cold ZnO and C and ended with the hot Zn and CO. This agrees, as it indeed must, with the sum of the heat absorbed in the reaction at zero, 55,640, increased by the sensible heat in Zn and CO at 1300, already found to be 47,961, or a total of 103,601. Another method of calculating the heat of the reaction ZnO + C = Zn + CO at 1300, without using the heat of forma- tion of ZnO and CO at 1300, is based on the following gen- eral principle: If from the heat of any reaction, starting and ending cold, there be subtracted the heat necessary to raise the products from to t, the difference is the heat of the trans- formation from cold reagents to hot products; if to this be added the heat which would be contained in the reagents if they were heated to t, the sum is the heat of transformation from heated reagents to heated products, all at the tempera- ture. THERMOCHEMISTRY OF HIGH TEMPERATURES. 67 Illustration: The heat of the reaction we have been study- ing, starting and ending cold, is 55,640 Calories Sensible heat in Zn at 1300 = 38,195 " CO " = 9,766 47,961 Difference 103,601 Sensible heat in ZnO at 1300 = 17,075 C = 6,360 23,435 " Sum 80,166 The above is the simplest way of calculating the heat of any chemical reaction at any desired temperature, since it involves the knowledge of the heat capacities of only those substances which occur individually in the reaction, and not that of the elements of which the compounds present are composed. The above calculation, for instance, involves the heat capacities of ZnO, C, Zn, and CO, but not that of oxygen, which does not occur free in the reaction. Case (c) : This hardly needs discussion, because if we have the data for calculating Case (b), this case can be easily worked. If to the heat of the reaction at t we add the heat given out by the products in cooling from t to 0, the sum is the total heat evolved in passing from the hot reagents to the cold pro- ducts. This solution of the problem presupposes, however, the solution of Case (b) and requires the maximum amount of data, but it has the advantage of following and representing the logical course of the reaction. A simpler solution is to add to the heat of the cold reaction at zero, the heat necessary to heat the ingredients to t, and the sum will be the quantity required. Illustration : Taking the same case as before : . Heat of reaction at 1300 = 80,166 Sensible heat in products at 1300 = , 47,961 Sum = 32,205 or, Sensible heat in reagents at 1300 = 23,435 Heat of reaction at zero = 55,640 Sum = 32,205 The reasoning involved in the above is simply that, starting with the reagents hot and ending with the products -cold, the 68 METALLURGICAL CALCULATIONS. heat evolution must be the same whether we suppose the sys- tem to pass along the one path or the other. GENERAL REMARKS. It will be evident from this enunciation of principles and the illustrations adduced, that for many practical purposes the calculation according to Case (a) will suffice for finding the net energy involved in many metallurgical operations; it gives the sum total of energy necessary to be supplied to pass from the cold reagents to the hot products as they come from the furnace, but the two items of which this sum is composed do not actually represent the two items of the division of this total in the furnace, i.e., into heat necessary to heat the re- agents to the reacting temperature and heat actually absorbed as the reaction takes place. The latter item can only be calcu- lated by one of the two methods explained under Case (b) t and then the former item can be obtained by difference, or by direct use of the heat capacities of the reacting substances. It is, of course, obvious that this whole subject of calcu- lating the thermochemistry of high temperatures (as distin- guished from the ordinary zero thermochemistry} necessitates the use of all available data regarding specific heats in the solid, liquid and gaseous states of both elements and com- pounds, and also in many cases of their latent heats of fusion and vaporization. When, however, the necessary physical data are known, or can be assumed with approximate accuracy, the way is open to make many calculations of the greatest value in practical metallurgy and chemistry. The exact heats of forma- tion of chemical compounds at a certain temperature, which may be, and often are, very different from these values at zero, and, having frequently different relations to each other, will explain in many cases many hitherto little understood and ap- parently contradictory reactions. The calculations also enable us to understand many reactions taking place only at high tem- peratures, to calculate whether they are really exothermic or endothermic at the temperatures at which they occur, and to compare these data with the data as to the heat of such re- actions obtained by studying the chemical equilibrium of such reactions and deducing the heat of the reaction from the rate of displacement of the chemical equilibrium. One such ex- THERMOCHEMISTRY OF HIGH TEMPERATURES. 69 ample may suffice to suggest the large field here open to the scientific metallurgist. Example: G. Preuner (Zeitschrift fur Physikalische Chemie* March 15, 1904, p. 385) reports a long and careful investigation of the equilibrium of the reaction Fe 3 O 4 + 4H 2 ;=.' 3Fe + 4H 2 in which he finds the equilibrium constant for different tem- peratures, and calculates from its value at 960, by the use of Van't Kofi's formula, that the heat value of the reduction at that temperature is 11,900 Calories, and noticing the great difference between this value and the value derived from the ordinary heats of formation (270,800 + 4(58,060) = 38,560), he concludes that the Van't Hoff formula gives wrong re- sults when applied to this class of reactions. Now, the facts are that Preuner's observations were good, Van't Kofi's for- mula is correct, and applies, but the thermochemical value of the reaction, 38,560 Calories is the correct value only for the reaction beginning and ending at ordinary temperature. Our thermochemical principles enable us to calculate the heat of the reaction at 960, as follows: Heat of the reaction beginning and ending at zero = 38,560 Calories Heat in products at 960: 3Fe = 3(56) X [0.218(960) 39] Pionchon = 28,560 4H 2 = 4(22.22) X [0.34(960) +0.00015. (960) 2 ] = 41,300 69,860 Calories Heat in reagents at 960: Fe 3 O 4 = 232X[0.1447 (960) +0.0001878 (960) 2 ]* = 72,384 4H 2 = 4(22.22) X[0.303(960) +0.000027 (960) 2 ] = 28,075 100,459 " The heat of the reaction at 960 is, therefore, according to our method of Case (6), 38,560 69,860 + 100,459 = 7,961 * Approximate determination of heat in Fe 3 4 , made recently in author's laboratory, Q== 0.1447t + 0,0001878t 2 . 70 METALLURGICAL CALCULATIONS. Calories. It thus appears that Premier's dilemma was mostly caused by his thinking that the thermochemical value of the reaction at zero should be a constant for any temperature: a proper thermochemical calculation removes the dilemma. Since the whole treatment of the subject of the thermo- chemistry of high temperatures requires a knowledge of data concerning the heat capacity of elements and compounds in the solid, liquid and gaseous states, as well as of their, latent heats of fusion and vaporization, the next instalment of these cal- culations will supply these data as far as they are known, and discuss a number of applications of these principles to various metallurgical processes. In order to apply the principles explained in the preceding discussion, two sets of data are necessary; first, the thermo- chemical data, such as are ordinarily obtained by laboratory experiments at laboratory temperatures; second, physical data concerning the specific heats and latent heats of fusion, and volatilization of elements and compounds. The first have been given, at least for all important compounds met with in metallurgy, in a previous place (p. 18), the latter will now be discussed and the data presented as far as they have been determined. SPECIFIC HEATS OF THE ELEMENTS. Dulong and Petit's law announces the fact that the specific heat of atomic weight of a solid metal is nearly constant, the value varying between 6 and 7, and averaging 6.4. This gen- eralization was made chiefly upon the basis of the specific heats of the metals, as determined in the range 100 C. to 10 or 15 C., such as in Regnault's accurate experiments. About the only notable exceptions to this rule are carbon, boron and silicon, and it has been naively remarked by more modern physicists that these exceptions to the rule disappear if we find the spe- cific heat of these three elements at high temperatures, that, for instance, the specific heat of carbon above 1000 C. is 0.5. making its atomic specific heat 0.5X12 = 6, and therefore the exceptions to the rule are all accounted for. Now, the rule is an important one, and has done good service, but the exceptions just noted and their behavior at high temperatures really prove that the rule must be made more general, or else THERMOCHEMISTRY OF HIGH TEMPERATURES.. 71 abandoned altogether. The fact is, that the specific heats of almost all the solid elements increase with the temperature at a rate equal to an increase of about 0.04 per cent, of their value for each degree centigrade, so that the atomic specific heat of the majority of the elements, which is about 6.4 at ordinary temperatures, becomes about 40 per cent, greater at 1000 C. for such elements as are not melted at that tempera- ture. Therefore, while we may say that at ordinary tem- peratures the specific heat of atomic weight of a solid element is 6.4, and its specific heat per unit of weight is 6.4, divided by the atomic weight, yet it will be more accurate, if actual determinations have not been made, to assume that the actual specific heat increases 0.04 per cent, for every degree rise in temperature, and mean specific heat to zero half that fast. The specific heat in the liquid state has not been determined for many elements. It is in general higher than the specific heat of the solid at ordinary temperatures; in fact, it appears to be more nearly equal to the specific heat of the solid element just before fusion, and may be so assumed if no determina- tions have been made. It is found, furthermore, not to change perceptibly with rise of temperature, so that it may be as- sumed constant. The specific heat of the gaseous elements has been deter- mined only for those which are gaseous at low temperatures. For the metals which, as far as known, have monatomic vapors, i.e., vapors in which the atoms exist alone and uncoupled with each other, the specific heat of atomic weight, occupying 22.22 cubic meters at standard conditions, should be theoretically 5.0 Calories at constant pressure, or 0.225 per cubic meter. We may thus estimate the specific heat of metallic vapors which have not been determined. LATENT HEATS OF FUSION OF THE ELEMENTS. The passage from the solid to the liquid state is in all cases accompanied by an absorption of heat, which in amount varies from one or two Calories up to 100 Calories per unit of weight. This quantity has been most frequently determined by finding calorimetrically how much heat is given out by unit weight of the melted element just at its setting point, in cooling to 72 METALLURGICAL CALCULATIONS. ordinary temperatures, and subtracting from this the heat in unit weight of the solid substance at the melting point, a e determined most accurately by exterpolating the value of mean specific heat of the solid up to the melting point. In this manner the latent heat of fusion for a number of elements has been directly determined. If a crucible full of melted metal is allowed to cool, the temperature falls regularly until the melting point is reached, and then stays constant, or nearly so, for some time, while the metal is setting. A comparison of the rate of cooling before and after setting, with the length of time during which the temperature was constant, gives the relative value of the latent heat of fusion in terms of the specific heat of the melted metal and of the solid metal near to the melting point. While the latent heat of fusion per unit weight shows no perceptible regularities, it is found that as soon as the latent heat of fusion is expressed per atomic weight of the element (analogous to specific heat of atomic weight) that notable regularities appear. The elements with high melting points have high atomic heats of fusion, and vice versa, so that if the elements are arranged in the order of their melting points their latent heats of fusion per atomic weight of each are in the same order, and very nearly proportionately so. If, for instance, a chart or diagram is made, using the melting points as abscissas, and latent heats of fusion of atomic weights as ordinates, the latter will lie very nearly in a straight line. Numerically, if the melting points be expressed in degrees of absolute temperature (centigrade temperatures plus 273), the latent heats of fusion of atomic weights average about 2.1 times the temperature of the melting point. This rule may be used to predict an undetermined latent heat of fusion. In addition to the above general rule, another one bearing on the same question was also discovered and applied by the writer. (See Journal of the Franklin Institute, May, 1897.) According to this observation, the continued product of the latent heat of fusion of atomic weight and the coefficient of expansion and the cube root of the atomic volume (atomic weight divided by specific gravity) is a constant. If the co- efficient of linear expansion between and 100 C. is used. THERMOCHEMISTRY OF HIGH TEMPERATURES. 73 the constant is 0.095, or if the actual linear expansion of unit length from to 100 C. is used, the constant is 9.5. This rule, applied to all elements whose latent heat of fusion is known, gives satisfactory agreements, and enables us therefore to predict the latent heat of fusion of nearly a dozen other elements for which the coefficient of expansion is known. LATENT HEATS OF VAPORIZATION OF THE ELEMENTS. This datum has been determined for but a very few elements. Some of the metalloids, like sulphur, phosphorous and arsenic, are known to become complex vapors immediately above their boiling point, corresponding to such formulae as S 6 , P 4 , As 4 ; the metals, as far as they have been tested, pass into monatomic vapors, such as Na, K, Hg, Zn and Cd, in which each atom represents a molecule. In the latter cases the following gen- eralization may be made: The latent heat of vaporization of atomic weight is proportional to the absolute temperatures of the boiling point at atmospheric pressure, and is numerically equal to about twenty-three times that temperature (twenty- one times, if the outer work of overcoming the atmospheric pressure be not included). From this rule it is possible to estimate the amount of heat necessary to vaporize any metal whose boiling point under atmospheric pressure is known. Examples'. The boiling point of carbon under atmospheric pressure is 3,700 C., and if its vapor is monatomic, the latent heat of vaporization is, for an atomic weight of carbon (C = 12), 23X (3700 + 273) = 92,080 Calories, equal to. 7,673 Calories per kilogram of carbon. If the vapor is diatomic, and its for- mula C 2 , then the above latent heat is for twenty-four parts of carbon, and for one part by weight is 3,837 Calories. Other considerations, from thermochemistry, make the latter value the more probable one. The boiling point of cadmium is 772 C., and its vapor is known to be monatomic, what is its latent heat of vaporiza- tion? The atomic weight being LI 2, the latent heat of vapor- ization of this quantity is 23 X (772 + 273) = 24,035 Calories, which is 215 Calories per kilogram. In making such calculations it must be strictly observed that the boiling point under atmospheric pressure is to be used, and not any temperature at which vapors may appear at partial 74 METALLURGICAL CALCULATIONS. tensions which may be only small fractions of atmospheric pressure. THERMOPHYSICS OF THE ELEMENTS. Having laid down the laws and the empirical rules which appear to govern these phenomena, we will now discuss the data for the common elements, giving both what is known and what may be assumed as probably true wherever actual deter- minations have not been made. The elements will be taken up in the order of their atomic weights, the only scientifically logical order. In all cases, the actually measured mean specific heats, Sm, will be given. In the case of gases, these will be the mean specific heats under constant pressure ; if they are desired under constant volume the amount of outer work must be calculated in Calories (two Calories per degree for a molecular weight of a gas, 0.09 Calories per degree for 1 cubic meter, and 0.09^ weight of 1 cubic meter for a kilogram of gas), and subtracted from the specific heat at constant pressure. The specific heat of 1 cubic foot is, in pound Calories, the specific heat per cubic meter divided by 35.32 and multiplied by 2.204, or, in brief, multiplied by 0.0624; in British thermal units it will be the same as in pound Calories, since t is then Fahrenheit degrees. If, from the data given, it is desired to find the mean specific heat between any two temperatures, t and t', instead of the mean specific heat from t to 0, as given directly by the for- mula, it need only be observed that Sm from t' to t is obtain- able by finding Sm from to (t' + t). If, for instance, Sm (o to t) = 0.303 + 0.000027t, then Sm (t to t') = 0.303 + 0.000027 (t'+t). Furthermore, if the actual specific heat at any temperature t is desired, it is equal to the mean specific heat from to 2t; e.g., in the above cases, S (at any tempera- ture t) = 0.303 + 0.000027 (2t) = 0.303 + 0.000054 (t). Temperatures will be always given and represented in centi- grade degrees, except where the specific heat is given in British thermal units, in which case / represents, of course, Fahrenheit degrees, and will also be printed in italics, /, to further dis- tinguish it from t in centigrade degrees. Absolute tempera- tures, if used, will be designated as T, and are equal to t + 273. THERMOCHEMISTRY OF HIGH TEMPERATURES. 75 The vapor tensions of all the elements are expressed in the ^ form log p = +B, where p is used in millimeters of mercury, T is the corresponding temperature in absolute measure (K), and A and B are constants for the elements in question, which can be found if merely two values of p and T are known. The latent heat of vaporization of molecular volume is, thermodynamically, 4.57A. B is found nearly constant for all elements, averaging 7.9 to 8.4 in the case of the liquid element and 8.85 for the solid element. If only one value of p and T are known, then B is assumed either 7.9 for elements of low boiling point or 8.4 for elements of high boiling point. A little study of the question will show that these values correspond to using 23 or 25.1 for Trouton's constant (KT) expressing the latent heat of vapori- zation of molecular volume at p = 760 mm. (A/T = Trouton's constant/4.57). The vapor tension at the melting point is cal- culated by putting T = M. P., at which point the liquid and solid have the same vapor tension. This gives one value of p and T A' in the analogous formula for the solid state, (log. p= -~r+B'), where A' can be calculated from A and the latent heat of fusion, and B' thus becomes known. The vapor tension down to C. is then calculated from this formula. HYDROGEN. From the experiments of Mallard and LeChatelier we deduce : Sm (0-t) 1 kilogram (up to 2000 C.) =3.370+0.0003t Calories. 1 pound (up to 2000 C.) =3.370+0.0003t pound Calories. 1 pound (up to 3600 F.) =3.370+0.00017* B. T. U. 1 cu. meter (up to 2000 C.) = 0.303 +0.000027t Calories. 1. cu. foot (up to 2000 C.) =0.0189+0.0000017t pound Calories. 1 cu. foot (up to 3600 F.) =0.0189+0.0000009* B. T. U. For higher temperatures, such as electric furnace heats be- 76 METALLURGICAL CALCULATIONS. tween 2000 and 4000 C. (3600 to 7200 P.), Berthelot and Vielle have made experiments which give us : Sm (0 t) 1 kilogram 1 pound pound = 2.75 +0.0008t Calories. = 2.75 +0.0008t pound Cal. 1 pound = 2.75 +0.00044* B. T. U. 1 cubic meter = 0.2575 +0.000072t Calories. 1 cubic foot = 0.0161 +0.0000045t pound Cal. 1 cubic foot = 0.0161+0.0000025* B. T. U. LITHIUM. Sm (0 t) S (at M. P. = 179) Q (solid, at M. P.) L. H. Fusion Q (liquid, at M. P.) S (liquid) Q (liquid at B. P. 76 o = 1450) L. H. Vaporization (1450) Q (vapor; 1450) S (vapor) per cubic meter per kilo Vapor tension, liquid : log p 0.7895+0.0024t+ 0.0000063t 2 (Bernini). 2.255. 254 Cal. 136 Cal. (calculated by 2.1 Trule). 173 Cal. (calculated by second rule) 390 Cal. (calculated). 2.255 (assumed). 3483 Cal. (calculated). 5660 Cal. (Trouton's rule, K = 23). 9140 Cal. (calculated). 0.225 (assumed). 0.714 (calculated). ' -^+7.92 at M. P. solid : at C. logp = - (B = 7.92, assumed). 4.9XlO- 12 mm. (calculated). ^2+8.38. = 6.0X10- 25 mm. (calculated). BERYLLIUM (GLUCINUM). Sm (0-t) = 0.38+0.0004t (Nilson & Pettersson). S(atM.P. = 1430) = 1.52. Q (solid, at M. P.) = 1358 Cal. THERMOCHEMISTRY OF HIGH TEMPERATURES. 77 L. H. Fusion , = 400 cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 1758 Cal. S (liquid) =1.52 (assumed). BORON. Sm (0 t) (t under 233) =0.22 + 0.00035t (Weber). S (solid, at 500) = 0.57 (by extension of above formula). S (solid, above 500) = 0.57 (assumed to remain constant) . * 88 Sm (0 t) (t over 500) = 0.57 ' Q (solid, at M. P. = 2500) = 1337 Cal. (calculated). L. H. Fusion = 265 Cal. (calculated by 2.1 T rule, and molecule B 2 ). Q (liquid, at M. P.) =1602 Cal (calculated). S (liquid) = 0.57 (assumed). Q (liquid at B. P. 76 o = 3700) = 2285 Cal. L. H. Vaporization (3700) = 4155 Cal. (Trouton's rule, K = 23, and molecule B 2 ). Q (vapor, 3700) = 6440 Cal. S (vapor) per cubic meter = 0.315 (assumed for B 2 ). per kilo = 0.318 (calculated). Vapor tension, liquid : log p = ^ |-7.92 (B = 7.92 assumed), at M. P. =5 mm. solid: logp= -?M?9+8.36. at C. = 3.9X10- 70 mm. (calculated). CARBON. {Amorphous and Graphitic.) Sm (0 1) (up to 250) = 0.1567+0.00036t. (0t) (250 1000) = 0.2142 + 0.000166t. S at 1000 = 0.528 (Violle); 0.546 (Weber). Q at 1000 = 380 Cal. Sm (0-t) (t over 1000) = 0.4430 +0.0000425t-^^ (Violle). *This figure would give atomic specific heat = 6.27, and we assume this constancy by analogy with similar behavior of carbon. 78 METALLURGICAL CALCULATIONS. Q (0 t) (t over 1000) = 0.4430t+0.0000425t 2 - 106.5. Q (solid) (atB. P. 76 o = 3700) = 2114 Cal. L. H. Sublimation (3700) = 4140 Cal. (Richard's rule, K = 2.1, and Trouton's rule, K = 23;forC 2 ), Q (vapor) (3700) = 6254. S (solid, at M. P. = 4400) = 0.777 (Violle's equation). S (vapor) per cubic meter = 0.315 (assumed). per kilo = 0.292 (calculated for C 2 ). Q (solid) (at M. P. 4400) = 2666 Cal. (by Violle's equation). L. H. Fusion (4400) = 409 Cal. (by 2.1 T rule for C 2 ). = 396 Cal. (by second rule, for C 2 ). Q (liquid) (4400) = 3066 Cal. (calculated). S (liquid) = 0.777 (assumed). Q (vapor) (4400) = 6806 Cal. (calculated). 24 Vapor tension, solid: log p = -^jr+9.0 (B' = 9.0, assumed) . at M. P. = 6310 mm. = 8.3 atmospheres. r -A 1 22,195 liquid: log p = -- ^ -- 1-8.5. at 0C. = 1 X 1C- 80 mm. (calculated). NITROGEN. Sm (0 t) 1 kilogram (up to 2000 C.) = 0.2405 +0.0000214t Cal. 1 pound (up to 2000 C.) = 0.2405+0. 00002 14t pound Cal. 1 pound (up to 3600 F.) = 0.2405+0.0000119* B. T. U. 1 cubic meter (up to 2000 C.) = 0.303 +0.000027t Cal. 1 cu. foot (up to 2000 C.) = 0.0189 +0.0000017t pound Cal. 1 cu. foot (up to 3600 F.) = 0.0189+0.0000009* B. T. U. For temperatures between 2000 and 4000 C. the following are the values of the mean specific heats to zero : Sm (0 t) 1 kilogram = 0.2044 +0.000057t Cal. 1 pound = 0.2044+0.000057t pound Cal. THERMOCHEMISTRY OF HIGH TEMPERATURES. 79 1 pound = 0.2044+0.000032* B. T. U. 1 cubic meter = 0.2575 +0.000072t Cal. 1 cubic foot = 0.1601 +0.0000045t pound Cal. 1 cubic foot = 0.0161+0.0000025* B. T. U. OXYGEN. Sm (0 t) 1 kilogram (up to 2000 C.) = 0.2104+0.0000187t Cal. 1 pound (up to 2000 C.) = 0.2104+0.0000187t pound Cal. 1 pound (up to 3600 F.) = 0.2104+0.0000104* B. T. U. 1 cu. meter (up to 2000 C.) = 0.303 +0.000027t Cal. 1 cu. foot (up to 2000 C.) = 0.0189+0.0000017t pound Cal. 1 cu. foot (up to 3600 F.) = 0.0189+0.0000009* B. T. U. Sm (0 t) 1 cu. meter (2000 -4000C) =0.2575 +0.000072t Cal. 1 cu. foot (2000- 4000 C.) = 0.0161 +0.0000045t pound Cal. leu. foot (3600 -7200 F.) = 0.0161+0.0000025* B. T. U. 1 kilogram (2000-4000 C) = 0.1788+0.00005t Cal. 1 pound (2000 -4000 C.) = 0.1788+0.00005t pound Cal. 1 pound (3600-7200 F.) = 0. 1788+0. 00003 1 B. T. U. SODIUM. Sm (0 t) = 0.2932+0.00019t (Bernini). Q (solid, at M. P. = 98) = 30.6 Cal. (by formula). L. H. Fusion = 27.2 (Rengade). Q (liquid, at M. P.) = 57.8 Cal. S (liquid, 98 100) = 0.333 (Bernini). Q (liquid at B. P. 760 = 877) = 317 Cal. (calculated). L. H. Vaporization (877) = 905 Cal. (from vapor tension constants) . Q (vapor, at 877) = 1222 Cal. S (vapor) for cubic meter = 0.225 (assumed), per kilo = 0.218 (calculated). 80 METALLURGICAL CALCULATIONS. Vapor tension, liquid: logp = -- ^-+6.85 (from Gebhardt's data) at M. P. = 3.55 X10~ 6 mm. (calculated). 4700 solid: log.p =- -4jF+7.30. at C. = 8.1 X 10~ 9 mm. (calculated). MAGNESIUM. Sm (0 1) = 0.2372+0.000093t+ 0.0000000685t 2 (Stucker). Q (solid, at M. P. = 650) = 212 Cal. (by formula). L. H. Fusion = 70 Cal. (Roos). Q (liquid, at M. P.) = 282 Cal. S (liquid) = 0.445 (assumed; same as solid at M. P.). Q (liquid, at B. P. 76 o = 1120) = 491 Cal. (calculated). L. H. Vaporization (1120) = 1443 Cal. (Trouton's rule, K = 25.2). Q (vapor, at 1120) = 1934 Cal. . S (vapor) per cubic meter = 0.225 (assumed). kilo = 0.208 (calculated). Vapor tension, liquid : log p = -- ~-+8.4 (constants from Trouton's rule). at M. P. =1.15 mm. (calculated). solid : log p = - + 8 - 8L at C. = 5.25 X10~ 20 mm. (calculated). ALUMINIUM. Sm(0 1) = 0.2220+0.00005t (Richards). Q (solid, at M. P. = 657) =*= 167.4 Cal. (from formula). L. H. Fusion = 90.9 Cal. (Richards). Q (liquid, at M. P.) = 258.3 Cal. (Richards). S (liquid) = 0.308 (Pionchon). Q (liquid, at B. P. 76 o = 2200) = 733.5 Cal. (calculated). L. H. Vaporization (2200) = 2305 Cal. (Trouton's rule, K = 25.2). THERMOCHEMISTRY OF HIGH TEMPERATURES. 81 Q (vapor, at 2200) = 3039 Cal. S (vapor) per cubic meter = 0.225 (assumed). kilo = 0.185 (calculated). i Q Vapor tension, liquid : log p = -- '^ -- h8.4 (constants from Trouton's rule). at M. P. = 5.25 X10~ 7 mm. (calculated). 14 1QO solid : log p = -^^+8.98. at C. = 1.0X10- 43 mm. (calculated). SILICON. Sm(0 t)(up to 234) = 0.17+0.00007t (Weber). Q (solid, at M. P. = 1430) = 386 Cal. (by extension of for- mula). L. H. Fusion = 128 Cal. (by 2.1 T rule). = 108 Cal. (by second rule). Q (liquid, at 1430) = 494 Cal. S (liquid) = 0.37 (assumed, same as solid at M. P.). Q (liquid, at B. P. 760 = 2300) = 816 Cal. calculated). L. H. Vaporization (2300) = 1153 Cal. (Trouton's rule, K = 25.2, vapor = Si 2 ). Q (vapor, at 2300) = 1969 Cal. S (vapor) per cubic meter = 0.315 (assumed, for 812). per kilo = 0.125 (calculated). 14 200 Vapor tension, liquid : log p = -- ^ -- 1-8.4 (constants from Trouton's rule). at M. P. = 1.32 mm. (calculated). solid: logp= - at C. = 8X 10~ 47 mm. (calculated). PHOSPHORUS. Sm(-20to7) ,= 0.1788. Q (solid, at M. P. = 44) = 8 Cal. L. H. Fusion (44) = 5 Cal. (Person). Q (liquid, at 44) = 13 Cal. S (liquid, 44 to 98) = 0.205. Q (liquid at B. P. 76 o = 287) = 63 Cal. 82 METALLURGICAL CALCULATIONS. L. H. Vaporization (287) Q (vapor, at 287) S (vapor) per cubic meter per kilo = 130 Cal. = 193 Cal. = 0.405 (assumed, for P 3 ). = 0.097 (calculated). Vapor Tension, liquid : log p = at M. P. solid : at C. 2160 +7.0. T = 1.63 mm. logp= _?26C> +7 .34. T = 0.115 mm. SULFUR. Sm(15 97) Q(solid, at M. P. = 113) L. H. Fusion Q (liquid, at M. P.) Sm(0t) (liquid, 114 445) Q (liquid at B. P. 76 o = 444.5) L. H. Vaporization (444.5) Q (vapor, at 444.5) S (vapor) per cubic meter per cubic meter per cubic meter per kilo Vapor tension, liquid : log p at M. P. solid : log p atOC. 0.18 (Regnault). 20 Cal. 9.4 (Person). 29.4 Cal. 0.338+0.000187t- 0.00000058t 2 . 126 Cal. 72 Cal. 198 Cal. 0.855 for S 8 , close to 445. 0.674 for S 6 , up to 500. 0.315 for S 2 , over 800. 0.074 for S 8 , close to 445. 0.078 for S 6 , up to 500. 0.109 for S 3 , over 800. -^+8.1. 0.028 mm. 4140 T lXlO- 6 mm. CHLORINE. Sm (gas) per cubic meter per kilo = 0.40 (13 202) (Regnault). = 0.1241 (Regnault). THERMOCHEMISTRY OF HIGH TEMPERATURES. 83 POTASSIUM. Sm(0 1) = 0.1858+0.00008t (Bernini). Q (solid, at M. P. = 60) = 11.4 Cal. L. H. Fusion = 13.6 Cal. (Bernini). Q (liquid, at M. P.) = 25.0 Cal. Sm (liquid) (0 1) = 0.1422+0.00067t (Rengade). Q (liquid, at B. P. 760 = 757) = 181.5 Cal. L. H. Vaporization (757) = 607 Cal. (Trouton's rule, K = 23). Q (vapor, at 757) = 789 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.128 (calculated). Vapor Tension, liquid : log p = jp +7.92 (constants from -^ Trouton's rule), at M. P. = 2.25 X10~ 8 mm. solid : logp = - at 0C. = 7.lXlO- 12 mm. CALCIUM. Sm(0 100) = 0.1704 (Bunsen). Q (solid, at M. P. = 800) = 136.3 Cal. L. H. Fusion = 56.3 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 192.6. TITANIUM. S (solid, at 0) = 0.0978 (Nilson and Pettersson) . Sm (0 1) = 0.0978+0.0000035t (assum- ing atomic specific heat = 10 at M. P.). Q (solid, at M. P. = 1795) = 187 Cal. (from above formula). L. H. Fusion = 90 Cal. (from 2.1 T rule). Q (liquid, at M. P.) = 277 Cal. S (liquid) = 0.2084 (assumed, same as solid at M. P.). Q (liquid, at B. P. 76 o = 2475) = 319 Cal. 84 METALLURGICAL CALCULATIONS. L. H. Vaporization (2475) = 143 Cal. (from Trouton's rule, K = 25). Q (vapor, at 2475) = 462 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.104 (calculated). Vapor tension, liquid : log p = -- ^ -- \- 8.4. at M. P. = 11.75 mm. solid : log p = -- ^p? + 8 . 85> atOC. = 8.3XlO- 49 mm. VANADIUM. Sm (0 1) = 0.105+0.00003t. Q (solid, at M. P. = 1710) = 267 Cal. L. H. Fusion = 82 Cal. (by 2.1 T rule). Q (liquid, at M. P.) , = 349 Cal. S (liquid) = 0.208 (assumed, same as solid at M. P.). CHROMIUM. Sm (0 1) (to 600) = 0.1039+0.00000008t 2 (Adler). Q (solid, at M. P. = 1489) = 352 Cal. L. H. Fusion = 71 Cal. (calculated by 2.1 T (rule). Q (liquid, at M. P.) = 423 Cal. S (liquid) =0.24 (estimated). Q (liquid at B. P. 76 o = 2200) = 594 Cal. L. H. Vaporization (at 2200) = 1197 Cal. (Trouton's rule, K = 25.1). Q (vapor, at 2200) = 1791 Cal. S (vapor), per cubic meter = 0.225 (assumed). per kilo = 0.096 1 o Vapor tension, liquid : log p. = -- ^ -- h 8. 4 (const ants from Trouton's rule, K = 25.1). at M. P. = 4.47 mm. 14 solid : log p. = - -- + 8.86. at C. = 7.8 X 10~ 45 mm. THERMOCHEMISTRY OF HIGH TEMPERATURES. 85 MANGANESE. Sm (0 1) = 0.1088+0.000103t. Q (solid, at M. P. = 1207) = 281 Cal. L. H. Fusion = 43 Cal. (calculated by second rule.) Q (liquid, at M. P.) = 324 Cal. S (liquid) = 0.357 (assumed from S solid at M. P.). Q (liquid at B. P. 76 o = 1900) = 572 Cal. L. H. Vaporization (1900) = 995 Cal. (Trouton's rule, K = 25.1). Q (vapor, at 1900) = 1567 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo. = 0.091 (calculated). 11 995 Vapor tension, liquid : log p = ^ h 8.4 (Constants from Trouton's rule, K = 25.1). at M. P. = 1.98 mm. Hg. solid : log p. = - i^P + 8 .85. at C. = 3.2X10- 38 mm. Hg. IRON. Sm(0 1) up to 660 = 0.1 1012+0. 000025t+ 0.0000000547t 2 (Pionchon; also Oberhoffer). Q at 660 = 96.1 Cal. Q at 750 = 125.6 Cal. L.H. Change of state (730) =5.3 Cal. (Pionchon); 5.0 (Leschtschenko) . S above 750 = 0.1675 (approximately con- stant.) Q (0 1) t = 750 to 1500 = 0.1675t - 51 Cal. L. H. Change of state (900) = 6.0 Cal. (Pionchon); 6.1 (Leschtschenko) . Q (solid, at M. P. = 1535) = 256 Cal. L. H. Fusion = 66 Cal. (calculated by 2.1 T rule). = 69 Cal. (calculated by second rule). (4.3% C.) = 59 Cal. (Schmidt). 86 METALLURGICAL CALCULATIONS. Q (liquid, at 1535) = 322 Cal. S (liquid) = 0.20 (estimated). Q (liquid at B. P. 760 = 2450) = 505 Cal. L. H. Vaporization (2450) = 1224 Cal. (Trouton's rule, K = 25.1). S (vapor) per cubic meter = 0.225 (assumed). per kilo. = 0.089 (calculated). Vapor tension, liquid : log p. = ^ -- h 8.4 (Constants from Trouton's rule, K = 25.1). at M. P. = 1.20 mm. Hg. solid: log p. = ~ .. = at C. = 6.9X10- 50 mm. NICKEL. Sm (0 1) up to 230 = 0. 10836 +0.00002233t (Pionchon) . L. H. Change of State = 4.64 Cal., 230 to 400, (Pionchon) . = 3.1.1 Cal. 363 (Leschtschenko) . Sm (0 1) t = 440 to 1050 = 0.099 +0.00003375t + ^ t (Pionchon) . Q (solid, at M. P. = 1450) = 221 Cal. L. H. Fusion = 68 Cal. (calculated by second rule). Q (liquid, at 1450) = 289 Cal. S (liquid) = 0.197 (estimated). Q (liquid, at B. P. 76 o = 2150) = 427 Cal. L. H. Vaporization (2150) = 1042 Cal. (Trouton's rule, K= 25.1). S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.086 (calculated). 13 375 Vapor tension, liquid : log p. = -- ^ -- f- 8.4 (Constants from Trouton's rule, K = 25.1). at M. P. = 4.36 mm. Hg. solid : log p. = - p- + 8.9. at C. = 5.25 X 10~ 44 mm. Hg. THERMOCHEMISTRY OF HIGH TEMPERATURES. 87 COBALT. Sm (0 1) up to 900 = 0.105844-0.00002287t-f 0.000000022t 2 (Pionchon). Sm (0 t)t over 900 = 0.124+0.00004t - ^ (Pionchon). Q (solid, at M. P. = 1490) = 259 Cal. (extension of above formula) . L. H. Fusion = 68 Cal. (calculated by second rule). Q (liquid, at 1490) = 327 Cal. S (liquid) = 0.243 (assumed, same as solid at M. P.). COPPER. Sm (0 t) = 0.0939+0.00001778t (Frazier and Richards). Q (solid, at M. P. = 1083) = 118.7 Cal. L. H. Fusion = 43.3 Cal. (Richards). Q (liquid, at M. P.) = 162.0 Cal. (Frazier & Richards) . S (liquid) = 0.156 (Glaser). Q (liquid, at B. P. 760 = 2310) = 353 Cal. L. H. Vaporization (2310) = 1087 Cal. (calculated, from va- por tension curve). Q (vapor, at 2310) = 1440 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.079 (calculated). Vapor tension, liquid : log p. = -- ^ -- h 8.75 (from Green- wood's data). at M. P. = 3.5XlO- 3 mm. Hg. solid : log p. = - + 9.17. at C. = 2.6 X10~ 49 mm. Hg. ZINC. Sm (0 1) = 0.0906 +0.000044t. Q (solid, at M. P. = 419) = 45.2 Cal. 88 METALLURGICAL CALCULATIONS. L. H. Fusion Q (liquid, at M. P.) S (liquid) Q (liquid, at B. P. 760 = 930) L. H. Vaporization (930) -4 Q (vapor at 930) S (vapor) per cubic meter per kilo Vapor tension, liquid : log p = at M. P. solid : log p. at C. 22.6 Cal. (Richards). 67.8 Cal. 0.179 (Glaser). 159 Cal. 446 Cal. (calculated from vapor tension curve). 605 Cal. 0.225 (assumed). 0.077 (calculated). /O> pr h 8.17 (from various < observations). 9.3XlO- 2 mm. Hg. = 1.03XlO- 16 mm. Hg. Sm (12 to 23) Q (solid, at M. P. = 30) L. H. Fusion Q (liquid, at M. P.) S (liquid, 30 119) Sm (21 to 65) amorphous crystalline Q (solid, at S. P. 76 o = 616) L. H. Sublimation (616) - Q (vapor, at 616) S (vapor) per cubic meter per kilo Q solid, at M. P. = 827) L. H. Fusion (827) Q (liquid at 827) L. H. Vaporization, 827) Q (vapor, at 827) GALLIUM. = 0.079 (Berthelot). = 2.4 Cal. = 19.3 Cal. (Berthelot). = 21.7 Cal. = 0.084 (Berthelot). ARSENIC. = 0.0758 (Bettendorf and Wullner). = 0.083 (Bettendorf and Wullner). = 47 Cal. = 114 Cal. (calculated from vapor tension curve, for As 3 ) . = 161 Cal. = 0.405 (assumed for Asa). = 0.040 (calculated). = 63 Cal/ = 10 Cal. = 73 Cal. = 104 Cal. = 177 Cal. under 10,000 mm. pressure. THERMOCHEMISTRY OF HIGH TEMPERATURES. 89 Vapor tension, solid : log p. = jp- + 9.10 (from experi- mental data), at M. P. = 10,000 mm. Hg. liquid : log p. = h 8.65 (from experi- mental data.) at C. = 2.8 X10- 21 mm. Hg. SELENIUM. Sm (60 200) = 0.084 (Bettendorf and Wullner). Q (solid, at M. P. = 217) = 18 Cal. L. H. Fusion =4 Cal. Q (liquid at M. P.) = 22 Cal. S (liquid) = 0.12 (assumed). Q (liquid, at B. P, 760 = 690) = 79 Cal. L. H. Vaporization (690) = 97 Cal. (calculated from vapor tension curve, for Sea). Q (vapor, at 690) = 176 Cal. S (vapor) per cubic meter = 0.405 (assumed for Sea), per kilo = 0.038 (calculated). Vapor tension, liquid : log p = ^ f-8.10 (from experi- mental data), at M. P. = 0.71 mm. Hg. solid : log. p = f-8.56 atO C. = 2.lXlO- 11 mm.Hg. BROMINE. Q (solid, at M. P. = -7) = -16.9 Cal. L. H. Fusion ( - 7) = 16.2 Cal. (Regnault). Q (liquid, at M. P.) = -0.7 Cal. Sm (liquid, -6 to +58) = 0.105+O.OOllt Q (liquid, at B. P. 76 o = 58) = 10 Cal. L. H. Vaporization (58) = 43.7 Cal. (Berthelot and Ogier). Q (vapor, at 58) = 53.7 Cal. S (vapor) per cubic meter =0.40 per kilo = 0.0555 (Regnault). 1 A^? f\ Vapor tension, liquid : log p = +7.8 (from experimental data). at M. P. = 45 mm. Hg. at C. = 66 mm. Hg. 90 METALLURGICAL CALCULATIONS. RUBIDIUM. Sm (20 to 35) = 0.0792+O.OOOOlt (Deusz). Q (solid, at M. P. = 38) = 3 Cal. L. H. Fusion - 6 Cal. (E. Duess). Q (liquid, at 38) = 9 Cal. S (liquid) = 0.080 (assumed). Q (liquid, at B.P.76Q = 696) = 62 Cal. L. H. Vaporization (696) = 261 Cal. (from Trouton's rule; K = 23). Q (vapor, at 696) = 323 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.059 (calculated). 4890 Vapor tension, liquid : log. p = +7.90 (constants from Trouton's rule; K = 23). at M. P. = 1.5 X10- 8 mm. Hg. solid : logp = ~r~ +8.28. at C. = 6.7 X 10~ 11 mm. Hg. STRONTIUM. Sm (0 100) = 0.0735 (assumed by Dulong and Petit's Law). Q (solid, at M. P. = 800) = 59 Cal. L. H. Fusion = 26 Cal. (calculated by 2.1 T rule). Q (liquid, at 800) = 85 Cal. S (liquid) = 0.092 (assumed, from at. sp. (heat = 8.0). ZIRCONIUM. Sm (0 100) = 0.0662 (Mixter and Dana) . Q (solid, at M. P. = 2350) = 155 Cal. L. H. Fusion = 61 Cal. (calculated by 2.1 T rule). Q (liquid, at 2350) = 216 Cal. S (liquid) = 0.11 (assumed, from at. sp. heat = 10). COLUMBIUM (NIOBIUM). Sm (0100) = 0.068 (assumed, from Dulong Q (solid, at M. P. = 1950) = 133 Cal. and Petit's Law). THERMOCHEMISTRY OF HIGH TEMPERATURES. 91 L. H. Fusion = 50 Cal. (calculated by 2.1 T rule). Q (liquid, at 1950) = 183 Cal. S (liquid) = 0.107 (assumed from at. sp. heat = 10). MOLYBDENUM. Sm (0 1) = 0.0655 +0.00002t (Defacqz and (Guichard). Q (solid, at M. P. = 2500) = 289 Cal. L. H. Fusion = 61 Cal. (calculated by 2.1 T rule). Q (liquid, at 2500) = 350 Cal. S (liquid) = 0.165 (assumed, from extension of formula for S (solid) . Q(liquidatB. P. 760 = 3600) = 522 Cal. L. H. Vaporization (3600) = 1015 Cal. (calculated from Trouton's rule; K = 25). Q (vapor, at 3600) = 1537 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.052 (calculated). 21 380 Vapor tension, liquid : log p = ^ 1-8.4 (constants from Trouton's rule, K = 25). at M. P. = 4.9 mm. Hg. solid : log p = -^j^+8.85. at C. = 7.6 X 10~ 75 mm. Hg. RUTHENIUM. Sm (0 100) = 0.0611 (Bunsen). Sm (0 1) = 0.0601 -f-O.OOOOlt (function of t assumed). Q (solid, at M. P.= 1950) = 155 Cal. (extension of above formula) . L. H. Fusion = 46 Cal. (calculated by second rule). Q (liquid, at M. P.) = 201 Cal. S (liquid) = 0.099 (assumed same as solid at M. P.). Q (liquid at B. P. 76 o = 2520) = 257 Cal. L. H. Vaporization = 689 Cal. (calculated from Trouton's rule, K = 25). 92 METALLURGICAL CALCULATIONS. Q (vapor, at 2520) = 946 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.049 (calculated). Vapor tension, liquid : log p = ^ (-8.4 (constants from Trouton's rule; K = 25). at M. P. = 28.8 mm. Hg. solid: logp= - ^^+8.86. at C. = 4X10- 52 mm. Hg. RHODIUM. Sm (10 97) = 0.0580 (Regnault). Sm (0 1) = 0.0574 +0.00001 1 (function of t assumed). Q (solid, at M. P. = 1970) = 152 Cal. L. H. Fusion --= 53 Cal. (calculated by second rule). Q (liquid, at M. P.) = 205 Cal. S (liquid) = 0.097 (assumed same as solid at M. P. Q (liquid at B. P. 76 o = 2500) = 256 Cal. L. H. Vaporization (2500) = 677 Cal. (calculated from Trouton's rule, K = 25). Q (vapor, at 2500) = 933 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.049 (calculated). Vapor tension, liquid : log p = ^ f-8.4. at M. P. = 38 mm. Hg. solid: logp= -^P+8.94. at = 3.2 X 10~ 52 mm. Hg. PALLADIUM. Sm (0 1) = 0.0582+O.OOOOlt (Violle). Q (solid, at M. P. = 1549) = 110 Cal. (Violle). L. H. Fusion = 36 Cal. (Violle). Q (liquid, at M. P.) = 146 Cal. (Violle). S (liquid) = 0.089 (assumed same as solid at M. P.). THERMOCHEMISTRY OF HIGH TEMPERATURES. 93 Q (liquid, atB. P. 76 o = 2535) = 238 Cal. L. H. Vaporization (2535) = 667 Cal. (calculated from Trou- ton's rule; K = 25). Q (vapor, at 2535) = 905 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.047 (calculated). Vapor tension, liquid : log p = -- ^ -- h8.4 (constants from Trouton's rule, K = 25.1). at M. P. = 7.8 X 10- 1 mm. Hg. solid : log p = - ^i2 + 8>85 at = 1 .4 X 10- 60 mm. Hg. SILVER. Sm (0 1) up to 400 = 0.0555 +0.00000943t (Pionchon) . over 400 = 0.05758+0.0000044t+ 0.000000006t 2 (Pionchon). Q (solid, at M. P. = 962) = 64.8 Cal. L. H. Fusion = 24.35 Cal. (Pionchon). Q (liquid, at M. P.) = 89.15 Cal. (Pionchon), S (liquid, 962 to 1020) = 0.0748 (Pionchon). Q (liquid at B. P. 760 = 2040) = 170 Cal. L. H. Vaporization (2040) = 490 Cal. (calculated from vapor tension data) . Q (vapor, at 2040) = 660 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.046 (calculated). 1 1 Vapor tension, liquid : log p. = -- Jp -- f- 7.9 (from experi- mental data). at M. P. = 3.2 X10- 2 mm. Hg. 19 solid : log p = - ^ + 8.36. at C. = 5.6 X10~ 37 mm. Hg. CADMIUM. Sm (0 1) = 0.0546+0.000012t (Naccari). Q (solid, at M. P. = 321) = 18.8 Cal. L. H. Fusion = 13.0 Cal. (Person). 94 METALLURGICAL CALCULATIONS. w Q (liquid, at M. P.) = 31.8 (Person). S (liquid) = 0.0623 (calculated same as solid, at M. P.). Q (liquid, at B. P. 760 = 778) = 60.3 Cal. L. H. Vaporization (778) = 251 Cal. (calculated from vapor tension curve) . Q (vapor, at 778) = 311 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo. = 0.045 (calculated). *~t ~[ Vapor tension, liquid : log p = \- 8.74 (from experi- mental data), at M. P. = 2.6XlO- 2 mm. 6470 solid : log p = - ^-^ + 9.28. at C. = 3.8 X 10- 15 mm. Hg. TIN. Sm (0 1) = 0.0560+0.000044t (Bede, combined with Regnault.). Q (solid, at M. P. = 232) = 14.34 Cal. L. H. Fusion = 13.82 Cal. (Richards). Q (liquid, at M. P.) = 26.16 Cal. (Richards). Sm (0 t)(t = 232 to 1000) = 0.06129- 0.00001047t+ 0.00000001035t 2 + ^^ b (Pionchion) . S (liquid) at t = 232 to 1000 = 0. 06 129-0. 00002094t+ 0.00000003 105t 2 (Pionchon). Q (liquid, at B. P. 760 = 2250) = 157 Cal. L. H. Vaporization (2250) = 538 Cal. (calculated from Trouton's rule, K = 25.1). Q (vapor, at 2250) = 695 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.042 (calculated). 13 930 Vapor tension, liquid : log p = ^ 1-8.4 (Constants from Trouton's rule, K = 25.1). at M. P. = 6.3XlO- 20 mm. Hg. solid: logp = -^|?? + 9.1. at C. = 5.6 X 10~ 44 mm. Hg. THERMOCHEMISTRY OF HIGH TEMPERATURES. 95 ANTIMONY. Sm (0 1) = 0.04864+0.0000084t (Naccari) . Q (solid, at M. P. = 630) = 34 Cal. L. H. Fusion . = 40.3 Cal. (Richards). Q (liquid, at M. P. ) = 74.3 Cal. (Richards). S (liquid, 632 to 830) = 0.0605 (Richards). Q (liquid, at B. P. 76 o = 1500) = 127 Cal. L. H. Vaporization (1500) = 124 Cal. (calculated from Trouton's rule for Sbs, K = 25.1). Q (vapor, at 1500) = 251 Cal. S (vapor) per cubic meter = 0.405 (assumed, for vapor = Sb 3 ). per kilo = 0.025 (calculated for Sb 8 ). 9790 Vapor tension, liquid : log p = \- 8.4 (constants from Trouton's rule, K = 25.1). at M. P. = 3.55XlO- 3 mm. Hg solid : log p = - ^0 + 9.5 8 . at C. = 6 . 5 X 10~ 31 mm. Hg. IODINE. Sm (9 98) - = 0.05412 (Regnault). Q (solid, at M. P. = 114) = 6.2 Cal. L. H. Fusion = 11.7 Cal. (Person). Q (liquid, at M. P.) = 17.9 Cal. S (liquid) = 0.0676 (assumed). Q (liquid, at B. P. 760 = 185) = 22.7 Cal. L. H. Vaporization (185) = 23.95 Cal. (Fabre and Silber- mann) . Q (vapor, at 185) = 46.65 Cal. per kg. S (vapor) per kilo (206 to 377) = 0.034 (Strecker). per cubic meter = 0.389 (calculated). percu. meter ( > 1200) = 0.225 (assumed for I gas). per kilo = 0.039 (calculated). L. H. Change of State (I 2 = 21) = 145 Cal. per kilo (calculated from vapor tension curve). = 1658 Cal. per cubic meter 12. 96 METALLURGICAL CALCULATIONS. Q (vapor, at 1200 = I vapor) = 226 Cal. per kilo. 2390 Vapor tension, liquid : log p = -- [-8.10 (from experi- mental data). at M. P. = 89 mm. Hg. solid : log p = -p + 10.43. atO = 2XlO- 2 mm. Hg. i TELLURIUM. Sm (0 455) = 0.0613 (Richards). Sm (0 1) = 0.0495+0.000026t (Richards). Q (solid, at M. P. = 455) = 27.3 Cal. L. H. Fusion = 19.0 Cal. (Richards). Q (liquid, at M. P.) = 46.3 Cal. (Richards). S (liquid) = 0.0731 (calculated same as solid at M. P.) Q (liquid at B. P. 760 = 1390) = 115 Cal. L. H. Vaporization (1390) = 166 Cal. (calculated by Trou- ton's rule for Te2*, K = 25.1). Q (vapor, at 1390) = 281 Cal. S (vapor) per cubic meter = 0.315 (assumed for Te2) per kilo = 0.028 (calc. for Te 2 ). Vapor tension, liquid : log p = -- ^+8.4 (constants from Trouton's rule, K = 25.1). at M. P. = 6.3X10~ 5 mm. Hg. solid: logp= -i|?0+9.84. at C. = 8.1 X 10~ 32 mm. Hg. CAESIUM. Sm (0 26) = 0.0482 (Eckhardt and Graefe). Q (solid, at M. P. = 29) =1.4 Cal. L. H. Fusion = 4.8 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 6.2 Cal. L. H. Vaporization (670) = 163 Cal. (calculated by Trou- ton's rule; K = 23). THERMOCHEMISTRY OF HIGH TEMPERATURES. 97 S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.038 (calculated). 4760 Vapor tension, liquid : log p = -- ^-+7.92 (constants from Trouton's rule; K = 23). at M. P. = 1.2XlO- 8 mm. Hg. 4QOO solid : log p = at C. = 2.5 X 10- 10 mm. Hg. BARIUM. Sm (0 100) = 0.05 (Mendeleeff). Q (solid, at M. P. = 850) = 42.5 Cal. L. H. Fusion = 17.0 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 59.5 Cal. CERIUM. Sm (0 100) = 0.0448 (Hillebrand) . Q (solid, at M. P. = 623) = 27.9 Cal. L. H. Fusion = 13.4 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 41.3 Cal. S (liquid) = 0.07 (assumed). TANTALUM. Sm. (0 100) = 0.0354 (assumed, from at. sp. heat = 6.4). Sm (0 1) = 0.0338+0. 000016t (function of t assumed). Q (solid, at M. P. = 2900) = 233 Cal. L. H. Fusion = 37 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 270 Cal. TUNGSTEN. Sm (0 1) (to 423) = 0.033 +0.00001 It (Defacqz and Geuchard). Q (solid at M. P. = 3350) = 234 Cal. L. H. Fusion = 40 Cal. (calculated by 2.1 T rule). 98 METALLURGICAL CALCULATIONS. Q (liquid, at M. P.) = 274 Cal. S (liquid) = 0.10 (assumed, same as solid at M. P.). Q (liquid, at B. P. 76 o = 5164) = 455 Cal. L. H. Vaporization (5164) = 743 Cal. (B. P. according to Langmiur) . Q (vapor, at 5164) = 1198 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo - 0.027 (calculated). 30 000 Vapor tension, liquid : log p = -- ^ -- h8.4 (constants from Trouton's rule, K = 25.1). at M. P. = 1.3 mm. Hg. solid : log p = - .. at C. = 7.4 X 10- 108 mm. Hg. OSMIUM. Sm (19 98) = 0.03113 (Regnault). Sm (0 1) = 0.0305+0.000006t (function of t assumed). Q (solid, at M. P. = 2200) = 96 Cal. L. H. Fusion = 36 Cal. (calculated by second rule). Q (liquid, at M. P.) = 132 Cal. S (liquid) = 0.057 (assumed from sp. ht. solid at M. P.). Q (liquid, B. P. 760 = 2600) = 155 Cal. L. H. Vaporization (2600) = 379 Cal. (calculated from Trouton's rule, K = 25.1). Q (vapor, at 2600) = 534 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.026 (calculated). Vapor tension, liquid : log p = -- ^ -- [-8.4 (constants from Trouton's rule; K = 25.1). at M. P. = 97.7 mm. Hg. solid: logp= -M8_+8.81. at C. = 9.6 X10~ 54 mm. Hg. THERMOCHEMISTRY OF HIGH TEMPERATURES. 99 IRIDIUM. Sm (0 1) = 0.0317+0.000006t (Violle). Q (solid, at M. P. = 2360) = 108 Cal. L. H. Fusion = 28 Cal. (calculated by second rule). Q (liquid, at M. P.) = 136 Cal. S (liquid) = 0.060 (assumed, same as solid at M. P.). Q (liquid, at B. P. 76 o = 2550) = 147 Cal. L. H. Vaporization (2550) = 368 Cal. (calculated from Trouton's rule; K = 25.1). Q (vapor, at 2550) = 515 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.020 (calculated). Vapor tension, liquid : log p = -- 7p -- f-8.4 (constants from Trouton's rule; K = 25.1). at M. P. = 302 mm. Hg. solid : log p = - at C. = 5.6 X 10~ 53 mm. Hg. PLATINUM. Sm (0 1) = 0.0317+0.000006t (Violle). Q (solid, at M. P. = 1755) = 75.2 Cal. (Violle). L. H. Fusion = 27.2 Cal. (Violle). Q (liquid at M. P.) = 102.4 Cal. (Violle). S (liquid) = 0.053 (assumed, same as solid at M. P.). Q (liquid, at B. P. 76 o = 2450) = 139 Cal. L. H. Vaporization (2450) = 351 Cal. (calculated from Trouton's rule; K = 25.1). Q (vapor, at 2450) = 490 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.026 (calculated). Vapor tension, liquid : log p = -- 7p -- h8.4 (constants from Trouton's rule; K = 25.1). at M. P. = 9.55 mm. Hg. solid: logp = -i^_+9.0. at C. = 3.0 X 10~ 51 mm. Hg. 100 METALLURGICAL CALCULATIONS. S (0 600) Sm (0 1)600 to 1064 Q (solid, at M. P. = 1064) L. H. Fusion Q (liquid, at M. P.) S (liquid) Q (liquid, at B. P. 760 = 2530) L. H. Vaporization (2530) Q (vapor, at 2530) S (vapor) per cubic meter per kilo GOLD. = 0.0316 constant (Violle). Qfi = 0.0289+0.0000045t+ j (Violle). = 34.63 Cal. (from Violle's equa- tion) . = 16.30 Cal. = 50.93 Cal. (Roberts-Austen). = 0.0358 (assumed, same as solid at M. P.). = 103.4 Cal. = 358 Cal. (calculated from Trouton's rule; K = 25.1). = 461 Cal. = 0.225 (assumed). = 0.025 (calculated). Vapor tension, liquid : log p = 15,470 +8.4 (constants from at M. P. solid : at C. Trouton's rule, K = 25.1). = 6.8XlO- 4 mm. Hg. log p = - 16,175 +8.93. = 4.8XlO- 51 mm. Hg. MERCURY. = 0.0319 (Regnault). = -4.1 Cal. = 2.84 Cal. (Person). = -1.30 Cal. = 0.0333 (Pettersson). = 0.03337 -0.0000027t+ 0.00000000055t 2 (Naccari). Q (liquid, at B. P. 760 = 357) = 11.7 Cal. S (solid) at -59 Q (solid) at M.P. -39 L. H. Fusion (-39) Q (liquid, at M. P.) Sm (liquid, -36 to 0) Sm (0 1) up to 250 L. H. Vaporization (357) (357) (0) L.H.V. at critical temperature (1139) = 67.8 Cal. (Kurbatoff). = 66.8 Cal. = 77.14 Cal. 0.0 Cal. (Calc. thermo- dy n ami call y, from vapor ten- sion curve). 1139 is calcu- lated critical temperature. THERMOCHEMISTRY OF HIGH TEMPERATURES. 101 Q (vapor, at 357) = 79.5 Cal. (vapor, at 1139 = critical temp.) = 100 Cal. S (vapor) per cubic meter. = 0.225 (assumed), per kilo = 0.025 (calculated). Q Q Q r\ f* Vapor tension, liquid : log p = ^r~ + 1.75 log. T 0.00221T+4.6544 (from vapor tensions at 357, 260 and 140) . at M. P. = 5.13X10- 6 mm. Hg. solid : log p = -?^+8.26 (8.26 assumed). THALLIUM. Sm (17 -100) = 0.03355 (Regnault). Sm (0 1) = 0.030+0.00002t. Q (solid, at M. P. = 303) = 10.9 Cal. (calculated from for- mula) . L. H. Fusion = 7.2 Cal. (Robertson). Q (liquid, at M. P.) = 18.1 Cal. S (liquid) = 0.042 (assumed, from solid at M. P.). Q (liquid, at B. P. 760 = 1700) = 77 Cal. L. H. Vaporization (1700) =221 Cal. (calculated by Trouton's rule; K = 23). Q (vapor, at 1700) = 298 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.025 (calculated). 9 940 Vapor tension, liquid : log. p = ^p h 7.92 (constants from Trouton's rule; K = 23). at M. P. = 4.7XlO- 10 mm. Hg. r A 1 10,260 , _ solid: logp = ^ h-8.5. at C. = 8.3 X 10- 30 mm. Hg. LEAD. Sm (0 1) = 0.02925 + 0.000019t(Bede, com- bined with Regnault). Q (solid, at M. P. = 327) = 11.6 Cal. (Le Verrier). L. H. Fusion = 6.0 Cal. (Richards). Q (liquid, at M. P.) = 17.6 Cal. (Person). S (liquid, 335 to 430) = 0.0402 (Person). 102 METALLURGICAL CALCULATIONS. Q (liquid, at B. P. 76 o = 1580) = 68 Cal. L. H. Vaporization = 209 Cal. (calculated from vapor tension curve). Q (vapor, at 1580) = 277 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.024 (calculated). Vapor tension, liquid: log. p = h 8.0 calculated from Greenwood's data), at M. P. = 1.5XlO- 8 mm. Hg. 9744 solid: logp = ^-+8.41. at C. = 4.9 X10- 28 mm. Hg. BISMUTH. Sm (0 1) = 0.0285+0.00002t (Bede, com- bined with Regnault). Q (solid, at M. P. = 269) = 9.0 Cal. L. H. Fusion = 12.0 Cal. Q (liquid, at M. P.) = 21.0 Cal. (Person). Sm (liquid, 280 to 360) = O.Q363 (Person). Q (liquid, at B. P. 760 = 1435) = 63 Cal. L. H. Vaporization (1435) = 208 Cal. (calculated from vapor tension curve). Q (vapor, at 1435) = 271 Cal. S (vapor) per cubic meter = 0.225 (assumed), per kilo = 0.024 (calculated). 9460 Vapor tension, liquid : log p = [-7.45 (from experimen- tal data). at M. P. = 1 X 10- 10 mm. Hg. r , 10,000 . AK solid : log p = i=p h 8.45. at C. = 6.6 X 10~ 29 mm. Hg. RADIUM. Sm = 0.027 (from Dulong and Petit 's Law). Sm (0-t) = 0.027 +0.000015t (coefficient of t assumed). Q (solid, at M. P. = 700) = 26.3 Cal. L. H. Fusion = 9.0 Cal. (calculated by 2.1 T rule). THERMOCHEMISTRY OF HIGH TEMPERATURES. 103 Q (liquid, at M.P.) = 35.3 Cal. S (liquid) = 0.048 (assumed, same as solid at M. P.). S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.022 (calculated). THORIUM. Sm (0-100) = 0.0276 (Nilson). Sm (0-t) = 0.0270+0.000008t (coefficient of t assumed). Q (solid, at M. P. = 1690) = 68.5 Cal. L. H. Fusion = 18.0 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 86.5 Cal. S (liquid) = 0.054 (assumed, same as solid at M. P.). S (vapor) per cubic meter = 0.225 (assumed) per kilo = 0.022 (calculated). URANIUM. Sm (0-98) = 0.028 (Blumcke). Sm (0-t) = 0.0275 +0.000007t (coefficient of t assumed). Q (solid, at M. P. = 2000) = 83 Cal. L. H. Fusion = 20 Cal. (calculated by 2.1 T rule). Q (liquid, at M. P.) = 103 Cal. S (liquid) = 0.055 (assumed, same as solid at M. P.). Q (liquid, at B. P. 760 = 2900) = 152 Cal. L. H. Vaporization (2900) = 333 Cal. (calculated from Trouton'srule; K = 25.1). Q (vapor, at 2900) = 485 Cal. S (vapor) per cubic meter = 0.225 (assumed). per kilo = 0.021 (calculated). Vapor tension, liquid : log p = -- ^ -- h 8.4 (constants from Trouton's rule; K = 25.1). at M. P. = 5.5 X 10~ 6 mm. Hg. solid: logp= -- atO = lX10- 57 mm. Hg, 104 METALLURGICAL CALCULATIONS. in several of the preceding instalments we have given the heats of formation of alloys and compounds and the thermo- physics of the elements. Before passing to the thermophysics of alloys and compounds and problems involving their use, we will consider a few simple cases of the application of data so far given. Such include operations in which metals are melted or volatilized, or amalgams retorted. A few words may be in order, to clear the ground, regarding what is to be regarded as the efficiency of a furnace. Efficiency of Furnaces. Under this term we must distinguish a generic sense and a specific sense, the first referring to furnaces in which the object is to maintain a certain temperature for a certain time with the minimum consumption of fuel, the second, in which the object is to perform a certain thermal operation with the small- est consumption of fuel. In the first case, one furnace may be compared with another, and thus comparative efficiencies cal- culated; in the second case real or absolute efficiencies can be also calculated. A few examples will illustrate this difference, which is an essential difference as far as making calculations is concerned. Cases of Specific Efficiency: Whenever it is desired to melt a metal for the purpose of casting it, a certain definite amount of heat must be imparted to the metal, and the ratio between this efficiently utilized heat and the heating power of the fuel consumed, is the efficiency of the furnace. If the furnace is electric the theoretical heat value of the electric energy used is the divisor. If, in addition to the heat required to raise the substances to the desired temperature, there is also heat absorbed in chemical reactions, this amount can be added in as usefully applied heat, and the sum of this and the heat in the final products be regarded as the total efficiently applied heat. If a blast furnace takes iron ore and furnishes us melted pig iron, the sum of the heat absorbed in the chemical decom- position of the iron oxide and the sensible heat in the melted pig iron is the efficiently applied heat, because it is the neces- sary theoretical minimum required ; all other items are more THERMOCHEMISTRY OF HIGH TEMPERATURES. 105 or less susceptible of reduction, but these are necessary items and, therefore, measure the net efficiency. If my dwelling re- quires 200 cubic feet of hot air per minute at 150 F. to keep it at 65 F., while the outside air is at F., the ratio of the heat required to warm the 200 cubic feet of air from F. to 150 F., to the calorific power of the fuel used per minute, measures the specific efficiency of the "heater;" the question whether this amount of hot air keeps the temperature of the rooms at 65 F. is a question of the general efficiency of the construction of the house. Cases of Generic Efficiency. Such are those in which prac- tically all the heat generated eventually leaves the furnace by radiation or conduction, or useless heat in waste gases; this is the case when a certain temperature has to be continuously maintained for a given time, and where the time element is the controlling one, and not any definite amount of thermal work is to be done. Examples are numerous: An annealing furnace, where steel castings, let us say, are to be kept at a red heat for two days, or a brick kiln, where several days slow burning are required, or a puddling furnace, where the melted iron must be held one to two hours to oxidize its impurities. In all these cases we may say that one furnace keeps its con- tents at the right heat for the right time with so much fuel, another does the same work with. 10 or 25 per cent, less fuel, and is, therefore, 10 or 25 per cent, more efficient; but we can- hot, in the nature of the case, speak of the absolute or specific efficiency of the furnace, because there is no definite term, expressible in calories, to compare with the thermal power of the fuel. In many cases the two efficiencies are mixed in the same process or operation, and then the calculation of absolute or specific efficiency can be made for that portion of the operation wherein a certain definite amount of thermal work is done. Thus, in an annealing kiln, 50 tons of castings may be brought up to annealing heat in 24 hours, starting cold, and the heat absorbed by the castings compared with the calorific power of the coal burnt during this period, is a measure of the real effi- ciency of this part of the operation. During the rest of the operation, while the castings are simply kept at annealing heat, 106 METALLURGICAL CALCULATIONS* there can be no calculation of the absolute or specific efficiency of the furnace, because one of the terms necessary for the comparison has disappeared, in that part of the process we can only speak of relative efficiency compared to some other furnace doing a similar operation. It goes, almost without saying, that we can, of course, apply the conception of efficiency in its relative or general sense to the whole operation or to any part of it Problem 6. The Rockwell Engineering Co. state in their current adver- tisements that their regenerative oil-burning furnace melts 100 pounds of copper with the consumption of less than 1.5 gallons of oil. Assume that 1.5 gallons of oil is used, and that the copper is heated from 25 C. to melted metal 100 C. above its melting point. Required: The "efficiency" of the furnace; i.e., its specific efficiency as calculated from the net heat utilized. Solution: One gallon of fuel oil averages in weight 7.5 pounds, and its calorific power 11,000 Calories per kilogram, or 11,000 pound Calories per pound. The calorific power of the fuel used in melting 100 pounds of copper is therefore: Heat generated 11,000X7.5X1.5 = 82,500X1.5 <= 123,750 pound. Calories. The heat imparted to the copper is as follows, taking the data from Article V. of these calculations: (p. 68.) Heat in 1 Ib. melted copper at melting point = 162 Ib. Cal. Heat in 1 Ib. solid copper at 25 C. = 2 Heat required to just melt the copper = 160 " Heat to superheat liquid copper 100 C. 0.133X100= 13 " Total heat expended on each pound of copper = 173 Heat usefully applied per 100 pounds = 17,300 1 7 onn Net efficiency of furnace = ^3750 ' 14 = 14 % It is proper to remark that although this efficiency appears low, yet it is considerably greater than is attained in simple melting holes or wind furnaces, and yet the calculations show THERMOCHEMISTRY OF HIGH TEMPERATURES. 107 what a large margin for improvement and greater efficiency exists in even some of the best and relatively most efficient metallurgical furnaces. Problem 7. In the distillation of silver amalgam in iron retorts, 1000 kilos, of amalgam, containing 200 kilos, of silver, is retorted with the consumption of 550 kilos, of wood, the mercury vapor passes off at an average temperature of 450 C., and the silver is raised towards the end of the operation to 800 C., in order to expel the last of the mercury. Assume the calorific power of the wood 3000 Calories. Required: The net efficiency of the furnace. Solution: The heating power of the wood is 550X3000 = 1,650,000 Calories. The heat utilized is that absorbed in separating the silver from the mercury plus the sensible heat in the mercury vapor at 450, plus the sensible heat in the silver at 800. These are calculated as follows: Heat to decompose amalgam = 2470 Calories per 108 kilos, of silver = 200 X (2470 - 108) = 200X22.9 = 4,580 Calories. Heat in silver at 800, using Pionchon's formula, is 800 [0.05758 + 0.0000044 (800) +0.000000006 (800) 2 ]X200 = 10,390 Calories. Heat in 800 kilos, of mercury vapor at 450 is (a) heat to boiling point (Naccari) 357 [0.033370.00000275 (357) + 0.0000000667 (357) 2 ] X 800 . = 11 ,677 Cal. (b) heat to vaporize 72.5X800 = 58,000 " (c) heat in vapor at 450 = 0.025 X (450 357) X 800 = 1,880 " Total = 71,557 " Heat usefully applied : In decomposing amalgam = 4,580 " In mercury vapor, as sensible heat = 71 ,557 " In silver, as sensible heat = 10,390 " Total = 86,527 86 527 Efficiency of furnace = 1 ' nnr . = 0.052 = 5.2%. 1,DOU,UUU 108 METALLURGICAL CALCULATIONS. Problem 8. In a zinc works, impure zinc is refined by redistillation in fire-clay retorts, a bank of retorts distilled 970 kilos, of zinc with the expenditure of 912 kilos, of small anthracite coal. Assume that the zinc vapors pass out of the muffles at the boiling point (930). Required: (1) The net efficiency of the furnace. (2) The electrical power which would be required, in horse- power-hours, to do the same work, assuming the heating effi- ciency of the electric furnace is 75 per cent. Solution: (1) The small anthracite may be assumed to have a calorific power of 7850 Calories; therefore, the total heat which should be developed is 7850X912 = 7,159,200 Calories. The heat in 1 kilo, of zinc in the state of vapor at its boiling point can be calculated from the thermophysical data supplied for zinc as: (a) In solid zinc to melting point (420) ........... 45.20 Cal. (6) Latent heat of fusion ........................ 22.61 " (c) Heat in melted zinc to boiling point (930) ____ 65.05 " (d) Latent heat of vaporization ................... 425.00 " Total 557.86 " Heat required for 970 kilos. = 541,124 " 541 124 Efficiency of furnace = 2QQ = 0.075 = 7.5%. (2) One electric horse-power-hour = 644 . Cal. Efficiently applied heat = 644 X. 75 = 483.0 " 489 995 Electric horse-power-hours required = r~- = 1013 E.H.P hours 4oo One metric ton of zinc requires = 1044 E.H.P. hours. Cost of power, at $20.00 per E. H. P. year = TQQ X 1044 = 0.00228X1044 = $2.38. 912 This cost of electric power would replace the use of ' ^ metric tons of small anthracite, equal to a cost of $2.53 for THERMOCHEMISTRY OF HIGH TEMPERATURES. 109 electric power sufficient to replace a metric ton of coal for this purpose. Many other examples could be given of the technical use of the thermophysical data concerning the elements, but the problems given illustrate the methods of calculation. When one is acquainted with some of the ordinary metal- lurgical operations, such as melting and distilling the metals, it is surprising to notice how little is known or thought of the efficiency or lack of efficiency of the furnaces used. One man melts 100 pounds of metal by the use of 150 pounds of coal, he builds a new furnace and does it more cheaply by using 100 pounds of coke, which is certainly relatively more efficient; but it is seldom that the operator knows that in one case he is getting probably only 7 per cent, efficiency from his fuel and in the other case only 10 per cent. It is the knowledge of these absolute efficiencies which tells the practical man just what he is accomplishing, and shows him how much room there still remains for improvement. THERMOPHYSICS OF ALLOYS. There does not exist, in technical literature, much data of this nature concerning alloys. There is here a wide and inter- esting field for metallurgical research, whose cultivation would yield results both of high practical and high theoretical inter- est, and yet it is comparatively untouched. What is wanted is complete data concerning the specific heat of solid and liquid alloy, and latent heat of fusion. These, combined with the determination of the heat evolved in the alloying, would fur- nish a sound basis for a practical theory of alloys, besides enabling workers with these alloys to control the efficiency of their furnaces and, in general, to know with scientific exactness what they are accomplishing. ALLOYS OF TIN AND LEAD. Per Cent, of Tin. Sm. Latent Heat of Fusion. 4.8 (Pb 16 Sn) 5.5 (Mazotto) 10.2 (Pb 5 Sn) 8.0 at 307 (Spring) 12.5 (Pb 4 Sn) 8.3 at 292 (Spring) 110 METALLURGICAL CALCULATIONS. Sm. Per Cent, of Tin. Sm. Latent Heat of Fusion. 16.0 (Pb 3 Sn) 9.1 at 289 (Spring) 22.2 (Pb 2 Sn) 9.5 at 270 (Spring) 7.9 (Mazotto) 36.3 (PbSn) 0.04073 (12*-99) 11.6 at 241 (Spring) (Regnault) 9.4 (Mazotto) 50.0 = total heat to in 1 kilo. melted metal 18.0 from 202 (Ledebur) 53.3 (PbSn 2 ) 0.04507 (10 99) 10.5 at 197 (Mazotto) (Regnault) 63.1 (PbSn 3 ) 15.5 at 179 (Spring) 69.5 (PbSn 4 ) 17.0 at 188 (Spring) 83.0 = total heat to in 1 kilo. melted metal 21.5 from 205 (Ledebur) 90.1 (PbSn 16 ) 12.9 (Mazotto) ALLOYS OF TIN AND BISMUTH. Per Cent, of Tin. Sm. Latent Heat of Fusion. 6.7 (Bi 8 Sn) 11.4 Cal. (Mazotto) 22.1 (Bi 2 Sn) . . . 11.2 Cal. (Mazotto) 36.2 (BiSn) 0.0400 (20 99) 11.6 Cal. (Mazotto) (Regnault) 53.1 (BiSn 2 ) solid, 0.0450 11.6 (Cal. Mazotto) (Regnault) liquid, 0.0454 (146 275) Person 69.1 (BiSn 4 ) 11.1 Cal. at 140 (Mazotto) 82.7 (BiSn 8 ) 12.6 Cal. (Mazotto) 90.1 (BiSn 16 ) 12.8 Cal. (Mazotto) ALLOYS OF TIN AND ZINC. Per Cent, of Tin. Sm. Latent Heat of Fusion. 78.4 (ZnSn 2 ) . . . .. 23.5 Cal. (Mazotto) 92.7 (ZnSn 7 ) .- 16.2 Cal. at 197 (Mazotto) 95.6 (ZnSn 12 ) 16.3 Cal. (Mazotto) 97.3 (ZnSn 20 ) 15.1 Cal. (Mazotto) THERMOCHEMISTRY OF HIGH TEMPERATURES. Ill ALLOYS OF TIN AND COPPER. Bell metal (20 per cent, tin) Sm (14 98) = 0.0862 (Reg- nault). Bronze (15 per cent. tin). Total heat in melted metal (to 0) = 130 Cal. (Ledebur). If strongly superheated = 143.5 Cal. (Ledebur). ALLOYS OF TIN AND ANTIMONY. Britannia metal (90 per cent, tin ) requires to melt it, starting cold, 28.0 Calories per kilogram (melting point 236); with 82 per cent, of tin, 25.7 Calories; melting point 205 (Ledebur). ALLOYS OF TIN, BISMUTH AND ANTIMONY. BiSn 2 Sb (bismuth 34.3, tin 41.9, antimony 23.8 per cent.). Sm (15 100) = 0.0462 (Regnault). ALLOY OF TIN, BISMUTH, ANTIMONY AND ZINC. BiSn 2 SbZn 2 (bismuth 29.8, tin 34.0, antimony 17.3, zinc 18.9 per cent.). Sm (15 100) = 0.0566 (Regnault). ALLOYS OF LEAD AND BISMUTH. Per Cent, of Lead. Sm. Latent Heat of Fusion. 11.1 (PbBi 8 ) 10.2 (Mazotto) 33.2 (PbBi 2 ) 6.4 (Mazotto) 39.9 (Pb 2 Bi 3 ) solid, 0.03165 (16 99) Person, liquid, 0.03500 (144 358) Person 42.7 (Pb 3 Bi 4 ) 4.7 at 127 (Mazotto) 49.9 (PbBi) 4.0 (Mazotto) 66.6 (Pb 2 Bi) 3.6 (Mazotto) 88.8 (Pb'Bi) 4.9 (Mazotto) ALLOYS OF LEAD AND ANTIMONY. With 63.0 per cent of lead, Sm (10 98) = 0.0388 (Regnault) With 82.0 per cent, of lead, heat in 1 kilo, melted metal = 15.6 Calories (Ledebur). With 90.0 per cent, of lead, total heat in 1 kilo, melted metal = 13.8 Calories (Ledebur). 112 METALLURGICAL CALCULATIONS. ALLOYS OF LEAD, TIN AND BISMUTH. D'Arcet's Alloy, containing 32.5 lead, 18.5 tin, 48.7 bismuth Sm solid (5 65) = 0.0372 (Mazotto) Sm solid (12 50) = 0.049 (Person). Sm solid (14 80) = 0.060 (Person). Sm liquid (107 136) = 0.047 (Person). Sm liquid (120 150) = 0.0399 (Mazotto). Sm liquid (136 300) = 0.0360 (Person). Latent heat of fusion = 5.96 Cal. at 96 (Person). = 5.77 Cal. at 99 (Mazotto). Rose's Alloy, containing 24.0 lead, 27.3 tin, 48.7 bismuth: Sm solid (5 65) = 0.375 (Mazotto). Sm fluid (119 338) = 0.0422 (Person). Latent heat of fusion = 6.85 Cal. at 99 (Mazotto). Fusible Alloy, containing 31.8 lead, 36.2 tin, 32.0 bismuth: Sm solid (18 52) = 0.0423 (Person). Sm solid (11 98) = 0.0448 (Regnault). Sm fluid (143 330) = 0.0460 (Person). Latent heat of fusion = 7.63 Cal. at 145 (Person). Wood's Alloy, containing 25.8 lead, 14.7 tin, 52.4 bismuth: 7 cadmium: Sm solid (5 50) = 0.0352 (Mazotto). Sm fluid (100 150) = 0.0426 (Mazotto). Latent heat of fusion = 7.78 Cal. at 75 (Mazotto). Lipowitz's Alloy, containing 25.0 lead, 14.2 tin, 50.7 bismuth, 10.1 cadmium: Sm solid (5 50) = 0.0345 (Mazotto). Sm fluid (100 150) = 0.0426 (Mazotto). Latent heat of fusion = 8.40 Cal. at 75 (Mazotto). ALLOYS OF COPPER AND ZINC. Red Brass S at = 0.0899 (Lorenz) S at 50 = 0.0924 (Lorenz) (Copper 85%) S at 75 = 0.0940 (Lorenz) Yellow Brass Sat = 0.0883 (Lorenz) at 50 = 0.0922 (Lorenz) (Copper 65%) at 175= 0.0927 (Lorenz) Heat in 1 kilo, of melted, somewhat superheated, brass =* 130 Calories (Ledebur). THERMOCHEMISTRY OF HIGH TEMPERATURES. 113 ALLOYS OF COPPER, ZINC AND NICKEL. German Silver: (74Cu. 20Zn. 6Ni) Sm(0 t) = 0.0941 +0.0000053t (Tomlinson) ALLOYS OF COPPER AND ALUMINIUM. Copper 88.7% Sm (20 100) = 0.10432 (Luginin) ALLOYS OF SILVER AND PLATINUM. Silver 66.7% Sm (0 t) = 0. 04726 + 0.0000 138t (Tomlinson). ALLOYS OF MERCURY AND TIN. HgSn (37.1% Sn) Sm (30 15) = 0.04083 (Schiiz) (25 15) = 0.04218 (Schiiz) ( 22 99) = 0.07294 (Regnault) HgSn (54.1% Sn) Sm ( 25 99) = 0.06591 (Regnault) HgSn 5 (74.7% Sn) Sm (16 15) = 0.05039 (Schiiz) ALLOYS OF MERCURY AND LEAD. Pb Hg (50.9% Pb) Sm (69 20) = 0.03458 (Schuz) Sm ( 23 99) = 0.03827 (Regnault) Pb 2 Hg (67.4% Pb) Sm (72 20) = 0.03348 (Schiiz) ALLOYS OF CADMIUM AND TIN. CdSn 2 (67.8% Sn) Sm (77 20) = 0.05537 (Schuz) whence we have Sm ( t) = 0.0557 + 0.00000366t (Schuz) ALLOYS OF IRON AND CARBON. Soft Steel (0.15% carbon) Sm (20 98) = 0.1165 (Regnault). Hard Steel (1.00% carbon) Sm (20 98) = 0.1175 (Regnault). Total heat in 1 kilo melted steel at 1350 = 300 Calories (Ledebur). Cast Iron (4.0% carbon) Sm (01200) = 0.175. Sm (0 1) = 0.12-f 0.000046t Total heat in 1 kilo, melted at 1200 = 245 Calories (Ledebur). Total heat in 1 kilo, coming from blast furnace = 250 to 325 Calories (Akermann). 114 METALLURGICAL CALCULATIONS. Problem 9. A steel-melting crucible contains 110 pounds of steel, which is melted in a wind furnace with the use of 150 pounds of coke. Assume the coke to be 90 per cent, fixed carbon, and the steel to be superheated 100 C. above its melting point. Req^lired: The net efficiency of the furnace. Solution: The calorific power of the coke may be assumed as 90 per cent, that of pure carbon, and therefore: = 1 50 X (8100X0.90) = 150X7290 = 1,093,500 Ib. Cal. Heat in steel at melting point: 110X300 (Ledebur) == 33,000 Heat to superheat 100: 110X100X0.15 (assumed) = 1,650 Total 34,650 Ib. Cal. Efficiency of furnace - yj - 0.032 = 3.2% Problem 10. A Siemen's regenerative furnace holds eighteen steel cruci- bles, each containing 100 pounds of steeL Assume that the efficiency of utilization of the heat for melting the steel is 5 per cent., and that the furnace is fed by natural gas, having a calorific power of 512-pound Calories per cubic foot. Required: The number of cubic feet of natural gas required per furnace heat of eighteen crucibles = 1800 pounds of cast steel. Solution: Heat in steel = 1800x315 = 567,000 Ib. Cal. Heating power of gas required = ~- =11,340,000 Ib, Cal. Cubic feet of gas required = - ^ -- = 22,150 cubic feet. 1 Ju Gas required per ton of steel, 2000 Ibs. = 24,610 cubic feet. Cost of gas, at $0.08 per 1000 cubic feet = $1.97. Problem 11. In a malleable-casting foundry the pig iron is melted in a reverbatory air furnace, 3000 kilos, being melted in two hours THERMOCHEMISTRY OF HIGH TEMPERATURES. 115 by the combustion of 1200 kilos, of bituminous coal, having a calorific power of 8500 Calories. Required: The melting efficiency of the furnace. Solution'. Calorific power of coal used: 1200X8500 = 10,200,000 Calories. Heat in melted iron at foundry heat: 3000X250 (Ledebur) = 750,000 Calories. Efficiency of furnace = = 0.0735 = 7.35% Problem 12. In an iron foundry cupola 14 metric tons of pig iron are melted in one hour, using 1.5 tons of coke (90 per cent, car- bon). The gases passing away contain by volume CO 13 per cent., CO 2 13 per cent., nitrogen 74 per cent., and leave the cupola at 500 C. The body of the cupola is 1.5 meters in diameter outside and 4 meters high. Required: (1) The net melting efficiency of the cupola. (2) The proportion of the calorific power of the coke lost. (a) By the sensible heat .of the hot gases escaping. (b) By the imperfect combustion of the coke. (c) By radiation from bottom and walls of the cupola. (3) The amount of heat in Calories radiated, on an average, from each square meter of outside surface per minute. Solution : (1) Calorific power of the coke; 1500X0.90X8100. = 10,935,000 Calories. Heat in melted iron: 14,000X250 (Ledebur) = 3,500,000 Calories. Efficiency of melting = - 0.32 = 32% (2 a) Weight of carbon escaping = 1500X0.90 = 1350 kilos. 116 METALLURGICAL CALCULATIONS. Volume of CO and CO 2 escaping = ^ - 2500 m* (Because 1 m 3 of either gas carries 0.54 kilos. C.) Volume of escaping gas = n 13 + 13 Volume of nitrogen (by difference) = 7115m 8 Sensible heat of nitrogen and CO (71 15 + 1250) X [0.303 (500) -f- 0.000027 (500) 2 ] - 1,323,760 Cal. Sensible heat of CO 2 1250 X [0.37 (500) + 0.00022 (500) 2 ] - 300.000 " Total sensible heat in gases 1,623,760 " Proportion of calorific power of the fuel thus lost: 1 ' 623 ' 760 --01485= 1485. Required: (I) The thermal value of the reaction at 900 130 METALLURGICAL CALCULATIONS. (2) The temperature of the resultant products after the reac- tion. Solution : (1) Value of the reaction at zero =- 4(195,600) Cal. + 3(97,200) + 6(29,160) Add: 315,840 Heat in 4Fe 2 O 3 at 900 = + 180,800 " Heat in 9C at 900 = 9(12) [0.2142(900) + 0.000166(900 2 )] = + 35,340 " Subtract: Heat in 4Fe 2 at 900 = 70,340 " Heat in 3CO 2 at 900 = 34,080 " Heat in 6CO at 900 = 6 (22.22) [0.303 (900) + 0.000027 (900 2 )] = 39,275 " Algebraic sum = 243,395 " Absorbed per molecule of Fe 2 O 3 = 60,850 " This reaction is therefore strongly endothermic, and therefore tends to check itself constantly by the resultant cooling effect. (2) If the Fe 2 3 and C were at 900 to start with, the pro- ducts would be below 900 after the reaction, since they would have to furnish the deficit of heat. The products would give out in cooling from 900 to t. 4Fe 2 = 4(112) [0.218(900 t)] SCO 2 = 66.66 [0.37(900 t) +0.00022 (900 2 t 2 )] 6CO = 133.33 [0.303(900 1)+ 0.000027 (900 2 t 2 )] Sum = 0.0183t 2 -162.73t + 161,250,= 243,395 Cal. Whence t = 537. This result is, of course, due to the hypothetical assumption that this endothermic reaction would go on until completed without checking itself. The impossible result obtained means that if this single reaction really goes on, it will soon check itself on account of the decrease of temperature. There are a few other compounds than those previously mentioned whose thermophysics has been investigated, but their number is in reality infinitesimal compared with the num- ber of those which have not been touched. There are many THERMOPHYSICS OF CHEMICAL COMPOUNDS. 131 compounds of great importance in metallurgy whose specific heats are not known, to say nothing of their latent heats of fusion, etc. The wide introduction of the electric furnace has rendered desirable the latent heats of vaporization and specific heats at high temperatures, but these are altogether lacking; we can only estimate their values. It is to be hoped that many metallurgical laboratories may be incited to take up this very neglected field, and thus bring forward numerical data which would be of the greatest theoretical and practical value. One example of such work, which deserves special commendation, is the quite recent publication of Prof. J. H. L. Vogt, of Chris- tiania, on the " Silikatschmelzlosungen," i.e., on " Melted Silicate Solutions," in which determinations of the melting points and latent heats of fusion of many simple and complex silicates are given for the first time ; we venture to predict that the data and conclusions of Prof. Vogt will be of great value to the physicist, chemist, metallurgist, and geologist, in both a practical as well as a theoretical sense. In collating the remaining available data, it is rather difficult to decide as to wha.t has immediate metallurgical interest and what has not. Electrometallurgy, in particular, is busying itself with the treatment and decomposition of so many com- pounds heretofore considered outside of the metallurgist's sphere of interest, that almost all of the common chemical compounds now have either a present or a prospective interest to the metallurgist. THERMOPHYSICS OF CHLORIDES Substance HC1 (gas) Tem- per a tur e 22-214 L. H. Specific Heat Fusion 1867 L. H. Va port- A uthority zalion Regnault HC1 (liquid) -83 0.3067(lm 3 ) 97 . 5 Elliott & RbCl (fused) 16-45 112 Mclntosh . Kopp LiCl (fused) 13-97 2821 Regnault KC1 KC1 14-99 77-2 0.1730 86.0 Regnault Plato NaCl 15-98 2140 Regnault NaCl 804 123 . 5 Plato NH 4 C1 23-100 0.3908 . Neumann 132 METALLURGICAL CALCULATIONS. Substance Tem- perature Specific Heat L. H. L. H. Fusion Vapor 'i- Authority zation NH 4 C1 350 709 Marignac BaCl 2 (fused) 14-98 0.0896 Regnault BaCl 2 (fused) 959 27 . 8 Plato SrCl 2 13-98 0.1199 Regnault SrCl 2 872 25.6 Plato CaCl 2 (fused) 23-99 0.1642 Regnault CaCl 2 (fused) 774 54.6 Plato MgCU (fused) 24-100 0.1946 Regnault A1C1 3 -22-15 0.188 Band TiCl 4 (solid) 13-99 0.1881 Regnault TiCl 4 (gas) 163-271 0.1290 Regnault SiCl 4 (liquid) 10-15 0.1904 Regnault SiCl 4 (gas) 90-234 0.1322 Regnault SiCl 4 (gas) 90-234 1.0113(lm 3 ) Regnault SiCl 4 (liquid) ? 37.3 Ogier CC1 4 (liquid) 20 0.207 Timofejew CC1 4 (liquid) 52 . Regnault MnCl 2 15-98 (?) 0.1425 Regnault MnCl 2 (liquid) ? 49. 4(?) Ogier. ZnCl 2 (fused) 21-99 0.1362 Regnault. / 0525 T1C1 (solid) TJC1 (at M. P.) 0-t 427 I +0.000007t 0.0585 Russell, Goodwin & T1C1 (solid) 427 15.6 T1C1 (liquid) 427-530 0.0590 CrCl 2 ? 0.1430 Kopp PC1 3 (solid) 11-98 0.2092 Regnault PC1 3 (liquid) 78.5 51.42 Andrews PCI, (gas) 111-246 0.135 Regnault AsCl, (liquid) 14-98 0.1760 Regnault AsCls (liquid) 69 . 74 Regnault AsCl 3 (gas) 159-268 0.1122 Regnault SnCl 2 (fused) 20-99 0.1016 Regnault SnCl 4 (liquid) 10-15 0.1904 Regnault SnCl 4 (liquid) 112 46.84 Regnault SnCl 4 (gas) 149-273 0.0939 Regnault HgCl (solid) 7-99 0.0521 Regnault HgCl 2 (solid) 13-98 0.0689 Regnault CuCl (solid) 17-98 0.1383 Regnault PbCl 2 (solid) PbCU (at M. P 0-t ) 498 / 0.0630 I +0.00002U 0.0839 Lindner, Goodwin & PbCl 2 (solid) 498 18.5 PbCl 2 (liquid) > 498 0.1035 Ehrhardt AgCl (fused) 0-t f 0.0897 I +0.0000125t f Regnault I Goodwin, THERMOPHYSICS OF CHEMICAL COMPOUNDS. 133 Substance Tem- peratur L. H. Specific Heat Fusion L. H. Vapor i- Authority zation AgCl (at M. P.) 455 0.1011 Robertson AgCl (solid) 455 30.5 & AgCl (liquid) 455-533 0.129 Kalmus THERMOPHYSICS OF BROMIDES HBr (gas) 11-100 0.0820 Strecker HBr (gas) 11-100 0.2989 (perm*) Strecker HBr (liquid) -70 48.68 Estreigher & Schnerr KBr (fused) 16-98 0.1132 Regnault NaBr 0-96 0.1170 Koref AlBr 3 93 10.47 Kablukow SbBr 3 95 9.76 Telloczko AsBr 3 31 8.93 Telloczko SnBr 4 30 6.26 Telloczko TIBr 460 ....... 12.7 Goodwin & Kalmus PbBr 2 (fused) 0-t f 0.0526 \+0.000006t Goodwin & Kalmus PbBr 2 (at M. P.) 488 0.0585 from formula PbBr 2 (at M. P.) 488 9.9 G. & K. PbBr 2 (at M. P.) 488 12.34 Ehrhardt PbBr 2 (liquid) 488-587 0.0780 G. & K. 07^6 Regnault, AgBr (fused) 0-t +0.0000025t Goodwin & Kalmus. AgBr (at M. P.) 430 0.0757 Calculated AgBr (at M. P.) 430 : 12.6 G & K. AgBr (liquid) 430-563 0.0760 G. & K. THERMOPHYSICS OF IODIDES HI (gas) 21-100 0.0550 ..... Strecker HI (gas) 21-100 0.3168 (perm 3 ) Strecker HI (liquid) -37 / Elliott & o5.00 < -_ T , , I Mclntosh KI (solid) 13-48 0.0766 Nernst Nal (solid) 16-99 0.0868 Regnault Cul (solid) 18-99 0.0687 Regnault Hg,I (red) 0-100 0.0406 Guinchan' Hg 2 I (yellow) 0-247 0.0446 . . .^. . Guinchant Hg 2 I (liquid) 250-327 0.0554 Guinchant Hgl (solid) 17-99 0.0395 Regnault HgI 2 (solid) 18-99 0.0420 Regnault Hglj (at M. P.) 250 9.79 Guinchant 134 METALLURGICAL CALCULATIONS. Substance Tem- perature Specific Heat L. H. Fusion L. H. Vapori- zation Authority PbI 2 (solid) 0-375 0.0430 Ehrhardt PbI 2 (at M. P.) 375 11 . 50 Ehrhardt PbI 2 (liquid) > 375 0.0645 Ehrhardt Agl (solid) 15-264 0.0577 Bellati & Romanese PbI 2 .AgI 10-124 0.0476 Bellati & Romanese THERMOPHYSICS OF FLUORIDES HP (liquid) . ? .... 360 Guntz KP -76-0 0.1930 Koref KP (M. P.) 860 ....... 108 Plato NaP 15-53 0.2675 Band 3NaP.AlF 3 16-99 0.2522 Oeberg (Cryolite) CaF 2 15-99 0.2154 Regnault A1F 3 15-53 0.2294 Band A1F 3 .7H 2 O 15-53 0.342 Band PbF 2 0-34 0.0722 Schottky THERMOPHYSICS OF SULFIDES H 2 S (gas) 20-206 . 2451 Regnault H 2 S (gas) 20-206 0.3750(lm 3 ) Regnault H 2 S (gas) 0-t (Im3)0.34 Richards +0.00015t CS 2 (liquid) 14-29 0.2468 Person CS 2 (gas) 86-190 0. 1596 . . . Regnault CS 2 (gas) 86-190 0.5458(lm 3 ) Regnault CS 2 (liquid) 46 83, 8 Wirtz MnS 10-100 0.1392 Sella ZnS 15-98 0. 1230 Regnault CdS 26 0.0908 Russell FeS 17-98 0. 1357 Regnault Fe 7 S 8 20-100 0.1602 .., Regnault (Pyrrhotite) FeS 2 19-98 0. 1301 Regnault FeS 2 565 (Gives off S) Richards CoS 15-98 0. 1251 Regnault NiS 15-98 . 1281 Regnault MoS 2 20-100 0.1233 Regnault (Molybdenite) Cu 2 S 9-97 0. 1212 Regnault (Chalcocite) 103 Transformation Point / Bellati & 190 0.1454 , \Lussana THERMOPHYSICS OF CHEMICAL COMPOUNDS. 135 Tem- L. H. L. H. Substance perature Specific Heat Fusion Vapori- Authority zation CuS 25 0.1243 Russell (Covellite) AsS 20-100 0.1111 Naumann (Realgar) As 2 S 3 20-100 0.1132 Naumann (Orpiment) (Orpiment) (gas) 0-t (Im 3 )0.34 Richards +0.00015t Sb 2 S 3 23-99 0.0840 Regnault (Stibnite) SnS. 13-98 0.0837 Regnault SnS 2 12-95 0.1193 ..... Regnault PbS 16-98 0'.0509 Regnault PbS (liquid) 1050 ....... 50 Richards (104 to 0) Bi 2 S 3 (fused) 11-99 0.0600 Regnault HgS 14-98 0.0512 Regnault Ag 2 S 7-98 0.0746 Regnault A g2 S 0-t 0.0685 Bellati & +0.00005t Lussana THERMOPHYSICS OF COMPOUND SULFIDES 3Cu 2 S.Pe 2 S 3 10-100 0.1177 Sella (Bornite) Cu 2 S.Fe 2 S 3 14-98 0.1310 Kopp (Chalcopyrite) (Chalcopyrite) 720 (Gives off S) Richards Cu Matte 0-t 0.2110 Landis (47% Cu) -0.0000366t Cu Matte 1000 30 Landis (47% Cu) Cu Matte (liquid) 1000 (204 Landis toO) 4Cu 2 S.Sb 2 S 3 10-100 0.0987 ..... Sella (Tetrahedrite) 2PbS.Cu 2 S.Sb 2 S 3 10-100 0.0730 ..... Sella (Bournonite) FeS 2 .PeAs 2 10-100 0.1030 ..... Sella (Arsenopyrite) CoS 2 .CoAs 2 15-99 0.0991 ...'.. Sella (Cobaltite) 3Ag 2 S.As 2 S 3 10-100 0.0807 ..... Sella 3A g2 S.Sb 2 S 3 10-100 0.0757 Sella 136 METALLURGICAL CALCULATIONS. THERMOPHYSICS OF SELENIDES AND TELLURIDES Substance Tern- L. H. L. H. perature Specific Heat Fusion Vapori- A uthority zation Cu 2 Se 20-200 0.1047 j Bellati & Ag 2 Se 37-187 0.0684 / Lussana THERMOPHYSICS OF ARSENIDES AND ANTIMONIDES PeAs 2 10-100 0.0864 Sella (Lollingite) CoAs 2 10-100 0.0830 Sella (Smaltite) Cu 3 As 10-100 0.0949 Sella (Domeykite) Ag 3 Sb 10-100 0.0558 Sella (Dyscrasite) THERMOPHYSICS OF CYANIDES K 4 Fe(CN) 6 1-46 0.2142 Nernst K 2 Zn(CN) 2 14-46 0.241 Kopp Hg(CN) 2 (cryst) 11-46 0.100 Kopp THERMOPHYSICS OF HYPOSULFITES K 2 S 2 0, 20-100 0.197 Pape Na 2 S 2 O 8 25-100 0.221 Pape BaS 2 O 3 17-100 0.163 Pape PbS 2 8 15-100 0.092 Pape THERM.OPHYSICS OF SULFATES H 2 SO 4 10.5 26.0 Bronsted H 2 SO 4 5-22 0.332 Cattanes H 2 SO 4 326 122.1 Person K 2 S0 4 15-98 0.1901 Regnault KHSO 4 19-51 0.2440 Kopp Na 2 SO 4 17-98 0.2312 Regnault Na 2 SO 4 .10H 2 O 31 51.2 Cohen BaSO 4 10-98 0.1128 Regnault SrS0 4 21-99 0.1428 Regnault CaSO 4 13-98 0.1965 Regnault MgS0 4 25-100 0.2250 Pape A1 2 K 2 (SO 4 ) 3 .24H 2 O 15-52 0.3490 Band (NH 4 ) 2 SO 4 .12H 2 O 13-45 0.3500 Kopp ZnSO 4 22-100 0.1740 Pape MnSO 4 21-100 0.1820 Pape Cr 3 K 2 (S0 4 ) 4 .24H 2 19-51 0.3240 Kopp THERMOPHYSICS OF CHEMICAL COMPOUNDS. 137 Substance Specific Heat Authority FeSO 4 .7H 2 O 9-16 0.3460 .; Kopp CoSO 4 .7H 2 O 15-80 0.3430 Kopp NiS0 4 15-100 0.1840 Pape CuSO 4 23-100 0.1840 Pape PbSO 4 20-99 0.0872 Regnault Hg 2 S0 4 0-34 0.0624 Schottky THERMOPHYSICS OF NITRATES HNO 3 -47 9.54 ..... Berthelot HNO 3 86 .. 115.1 Berthelot LiNOa 169-250 0.387 Goodwin & Kalmus LiNOa 250 88.5 Goodwin & Kalmus LiNO 3 (liquid) 250-302 0.390 Goodwin & Kalmus KN0 3 13-98 0.239 Regnault KN0 3 334 48.9 Person KNO 3 (liquid) 350-435 0.332 Person NaNO 3 14-98 0.278 Regnault NaNO 3 306 64.9 Person NaNO 3 (liquid) 320-430 0.413 Person K 2 Na(NO 3 ) 2 15-100 0.235 Person NH 4 NO 3 14-31 0.455 Kopp BaNO 3 13-98 0.1523 Regnault SrNO 3 17-47 0.181 Kopp CaNO 3 .4H 2 O 42 33.5 Pickering AgNO 3 16-99 0.1435 Regnault AgNO, 209 17.6 ..... Guinchant AgNO 3 (liquid) 208-281 0.187 Guinchant PbN0 3 17-100 0.1173 . Neumann THERMOPHYSICS OF CARBONATES Pb 2 CO 3 (fused) 18-47 0.123 ..... Kopp K 2 CO 3 23-99 0.2162 Regnault Na 2 CO 3 16-98 0.2728 Regnault BaCO 3 11-99 0.1104 Regnault SrCO 3 8-98 0.1475 Regnault CaCO 3 (calcite) 20-100 0.2086 Regnault CaCO 3 (argonite) 18-99 0.2085 Regnault CaCO 3 (marble) 23-98 0.2099 Regnault CaMg(CO 3 ) 2 20-100 0.2179 Regnault (Dolomite) Mg 7 Fe 2 (C0 3 ) 9 17-100 0.2270 Neumann (Brown magnesite) 138 METALLURGICAL CALCULATIONS. Substance pZaTure Sp^fic Heat Fusion L. H Vapor i- Authority zation ZnCO 3 o-t I - 1407 1 l+o.oooitj Lindner CuCO 3 .Cu(OH) 2 15-99 0.1763 Oeberg (Malachite) FeCO 3 9-98 0.1935 Regnault PbCOs 16-47 0.0791 Kopp THERMOPHYSICS OF CHROMATES K 2 CrO 4 19-98 0.1851 Regnault K 2 Cr 2 7 0-t 0.1803 +0.00008t ( Regnault, & j Goodwin & I Kalmus K 2 Cr 2 O 7 397 29.8 Goodwin & K. K 2 Cr 2 O 7 (liquid) 397-484 0.335 Goodwin &K. Na 2 CrO 4 23 39.2 Berthelot FeCrO 4 19-50 0.1590 Kopp (Chromite) PbCrO 4 19-50 0.0900 Kopp THERMOPHYSICS OF BORATES K 2 B 2 4 16-98 0.2048 Regnault K 2 B 4 7 18-99 0.2198 .*.... Regnault Na 2 B 2 O 4 17-97 0.2571 Regnault Na 2 B 4 O 7 16-98 0.2382 Regnault PbB 2 O 4 15-98 0.0905 Regnault PbB 4 7 16-98 0.1141 Regnault THERMOPHYSICS OF PHOSPHATES H 8 P0 4 18 25.71 Thomsen K 4 P 2 7 . 17-98 0.1910 Regnault Na 4 P 2 07 17-98 0.2283 Regnault CaP 2 O 6 15-98 0.1992 Regnault 3Ca 3 P 2 O 8 .CaF 2 15-99 0.1903 Oeberg (Apatite) Pb 2 P 2 O 7 (fused) 11-98 0.0821 Regnault Pb 3 P 2 8 11-98 0.0798 Regnault Ag 3 P0 4 19-50 0.0898 Kopp THERMOPHYSICS OF ARSENATES Pb 8 As 2 8 (fused) 13-97 0.0728 Regnault KaAs a O 6 (fused) 17-99 0.1563 Regnault THERMOPHYSICS OF CHEMICAL COMPOUNDS. 139 THERMOPHYSICS OF CHLORATES AND PER-CHLORATES Substance Tem- perature Specific Heat 7 f7 Vapor i- Authority KC1O 3 (fused) 16-98 0.2096 Regnault KC1O 4 14-45 0.190 Kopp NaClOa (fused) 184-255 . 320 Goodwin & Kalmus NaClO 3 (fused) 255 49 . 6 Goodwin & Kalmus NaClO 3 (liquid) 255-299 0.325 Goodwin & Kalmus THERMOPHYSICS OF ALUMINATES MgAl 2 O 4 15-47 0.1940 Kopp (Spinel) BeAl 2 O 4 0-100 0.2004 Nilson & (Chrysoberyl) Petersson THERMOPHYSICS OF TITANATES FeTiOa 15-50 0.177 Kopp THERMOPHYSICS OF MOLYBDATES PbMoO 4 15-50 0.083 Kopp (Wulfenite) THERMOPHYSICS OF TUNGSTATES CaWO 4 15-50 0.097 Kopp (Scheelite) Fe(Mn)WO 4 15-50 0.098 Kopp (Wolframite) THERMOPHYSICS OF MANGANATES KMnO 4 15-50 0.179 ..... Kopp THERMOPHYSICS OF SILICATES Mg 2 SiO 4 0-100 0.2200 Vogt (Olivin) 140 METALLURGICAL CALCULATIONS. Substance Tem- perature Specific Heat L. H. Fusion L. H. Vapor i- Authority zation Mg 2 SiO 4 1400 130 Vogt (Olivin) Mg 2 SiO 4 (liquid) 1400 [520 Vogt toO] MgSiO 3 0-t 0.1974 Vogt (Enstatite) +0.000086t MgSiO 3 1300 125 Vogt (Enstatite) MgSiO 3 (liquid) 1300 [403 Vogt toO] CaSiO 3 0-t 0.1690 Vogt (Wollastonite) +0.0001t CaSiO 3 1250 100 Vogt (Wollastonite) CaSiO 3 (liquid) 1250 [360 ..... Vogt toO] CaMgSi 2 O 6 0-t 0.186 Vogt +0.00008t CaMgSi 2 O 6 1225 100 Vogt CaMgSi 2 O 6 (liquid) 1225 [344 Vog toO] Ca,MgSi 4 Ou 0-t 0.179 Vog (Malacolite) +0.00007t Ca 3 MgSi 4 Oi 2 1200 94 Vogt Ca 3 MgSi 4 Oi 2 (liquid ) 1200 [319 Vogt toO] H 4 Al 2 Si 2 O 9 20-98 0.2243 Ulrich (Kaolin) Al 2 Si(F)O 6 12-100 0.1997 Joly (Topaz) KAlSisOs 20-100 0.1877 Oeber (Orthoclase) KAlSi 3 O 8 1200 100 Vogt (Orthoclase) KAlSi 3 O 8 20-100 0.197 Bogajaw- ( Micro cline) lensky KAlSi 3 O 8 1170 83 Vogt (Microcline) CaAl 2 Si 2 O 8 0-t 0.1790 Vogt (Anorthite) +O.OQ01t CaAl 2 Si 2 O 8 1220 100 ..... Vogt (Anorthite) CaAl 2 Si 2 O 8 (liquid) 1220 [358 Vogt toO] THERMOPHYSICS OF CHEMICAL COMPOUNDS. 141 Substance Tent' perature Specific Heat ZrSiO 4 15-100 0.1456 (Zircon) BeAl 2 Si 2 8 12-100 0.2066 (Beryl) Fe 3 Al 2 Si 3 Oi2 16-100 0.1758 (Iron Garnet) Al 2 SiO 6 0-100 0.1684 (Audalusite) Asbestos 20-98 0.1947 Serpentine 16-98 0.2586 Talc (soapstone) 20-98 0.2092 Potash mica 20-98 0.2080 Sodium mica 20-98 0.2085 Magnesia mica 20-98 0.2061 Oligoclase 20-98 0.2048 Spodumene 20-100 0.2176 Labradorite 20-98 0.1949 Hypersthene 20-98 0.1914 Augite 20-98 0.1931 Hornblende 20-98 0.1952 Granite 20-524 0.2290 Basalt 0-t 0.1937 -f-0.000086t Basalt ? Lava (Etna) 0-500 0.1820 +0.000155t Lava (Etna) 500-800 0.270 Pumice 10-100 0.240 Ca, K, Glass 14-99 0.198 Ca, K, Glass 0-300 0.190 Flint Glass 10-50 0.117 Flint Glass 18-100 0.082 Crown Glass 10-50 0.161 Mirror Glass 10-50 0.186 Thermometer glass 0.1869 (French, hard) Jena Glass 18-99 0.2182 Thermometer glass 19-100 0.1988 (Normal, German) Porcelain 0-t 0.2826 -0.0000264t Cement clinker 28-40 0.186 Portland cement 28-30 0.271 Humus (soil) 20-98 0.443 Quartz sand 20-98 0.1910 L. H. Fusion L. H. Vapori- Authority zation Regnault 130 Joly Oeberg Lindner Ulrich Oeberg Ulrich Ulrich Ulrich Ulrich Ulrich Schulz Ulrich Ulrich Ulrich Ulrich Bajrtoli Bartoli Tamman Bartoli Bartoli Dunn Regnault Dulong& Petit H. Meyer Winkelmann H. Meyer H. Meyer Zonboff Winkelmann Winkelmann Harker Hart! Haiti Ulrich Ulrich 142 METALLURGICAL CALCULATIONS. Substance Sandstone Slag Enamel slag Bessemer slag Tem- perature 0-100 C 14-99 15-99 14-99 Specific Heat 0.174 0.1888 0.1865 0.1691 L. H. Fusion L. H. Vapor i- Authority zation Hecht Oeberg Oeberg Oeberg MISCELLANEOUS MATERIALS Paraffin (solid) 0-t Paraffin (liquid) 52-63 Beeswax (solid) 26-42 Beeswax (liquid) 65-100 Soft Para Rubber 0-100 Vulcanite 20-100 Anthracite 0-12 Oak Wood 0.532 +0 rf 0012t 0.706 0.82 0.50 0.48 0.331 0.312 0.57 Fir Wood 0.65 Charcoal 20 Red Brick 0.22 Ether 25 Ether 35 0.54 Alcohol 20-26 Alcohol (liquid) 78 0.58 Benzol (liquid) 10 Benzol (solid) 5 0.40 Benzol (liquid) 80 Glycerin 0.58 Glycerin 13 Machine oil 0.40 Petroleum 50 Turpentine Turpentine 159 0.42 Stearic acid 64 Ammonia (gas) 0-t Ammonia (solution) 18 Ammonia (liquid) 16 0.38 +0.00016t 1.00 Methane (CH 4 ) 0-t Ethelyne (C 3 H 4 ) 0-t 0.38 +0.00022t 0.46 +0.0003t 42.5 47.6 Battelli Battelli Person Person Gee & Terry A. M. Mayer Hecht , Regnault 90 Brix Bose 206 Schall Pickering 30 J. Meyer 94 Tyrer Berthelot 74 Brix . . . . Bruner . LeChatelier Thomsen 297 Regnault LeChatelier For silicates in general, the specific heat is found to agree fairly well with what would be calculated from their percent- THERMOPHYSICS OF CHEMICAL COMPOUNDS. H3 age composition, giving each oxide constituent its proper specific heat. Those oxides whose specific heat at are known are as follows: SiO 2 = 0.1833 (Richards) A1 2 3 = 0.2081 (Richards). Fe 2 O 3 = 0.1456 (Richards). MgO = 0.2420 (Regnault). CaO = 0.1779 (calculated). Those which are not known give good results if assumed to be as follows: FeO = 0.1460 (Vogt). MnO = 0.1511 (Vogt). K 2 O = 0.1390 (Vogt). Na 2 O = 0.2250 (Vogt). Li 2 O = 0.4430 (Vogt). Using these data, S at is calculated from the percentage composition of the silicate. For S at higher temperatures it can be assumed with considerable approximation to the truth, that S increases 0.078 per cent, for each degree, and so, calling S the specific heat at zero, we would have S = S (l+0.00078t) Sm = S (l+0.00039t) Q (0 1) = SmXt Besides the above generalizations, which enable one to cal- culate the heat in the solid silicate at the melting point (when the latter is known), Vogt has shown that if the heat neces- sary to heat the silicate from 273 to the melting point is calculated, the latent heat of fusion may be taken as being 20 to 25 per cent, of this quantity, say an avejage of 22.5 per cent, and thus the heat required for fusion may be approximately calculated. Akerman has determined for many metallurgical silicate slags the total heat contained per kilogram of melted slag at the melting point. These quantities vary from 347 to 530 Calories, depending principally on the elevation of the melting point of the slag. A brief tabulation of Akerman 's results are as follows, arranged according to the amount of heat in the just-melted slag: 144 METALLURGICAL CALCULATIONS. Calories. % 347.. I SiO 2 59 1 350 L 39 63 58 58 53 360 41 38 39 37 66 59 48 40 380 34 31 46 58 58 62 38 400 25 44 60 65 41 37 21 43 %CaO %A1 2 3 36 5 42 19 35 2 35 7 37 5 37 10 42 17 47 15 43 19 40 23 32 2 38 3 42 10 48 12 48 18 37 32 37 17 32 10 27 15 37 1 52 10 34 41 33 23 20 20 35 52 7 53 10 32 47 30 27 The above includes most varieties of acid and basic iron blast-furnace slags. . Data for other slags made in the metal- lurgy of iron and of other metals, are almost altogether lack- ing. A wide field is here open for metallurgical experiments; data thus obtained would be immediately useful in practical calculations. CHAPTER VI. ARTIFICIAL FURNACE GAS. There are many different forms of producers for making artifi- cial furnace gas. For the purposes of making calculations upon them they may be conveniently divided into four classes, as follows : 1. Simple producers, those which use ordinary fuels, such as wood, peat, lignite, bituminous coal or anthracite, and in which no water or water vapor is introduced other than the water in the fuel itself and the normal moisture of the air used. 2. Mixed gas producers, in which water vapor or steam is introduced with the air for combustion, in such amount as to be entirely decomposed in passing through the fuel. 3. Mond gas producers, in which, for a special purpose, more steam is introduced than can be decomposed in the producer, thus producing very wet gas. 4. Water gas producers, in which air and steam alone are alternately fed to the producer, the former for heating up, the latter for producing water gas. As far as the calculations are concerned, the essential dif- ference between these classes is the varying amount of water vapor or steam introduced under the fuel bed while producing the gas, from all air in Class 1 to all steam in Class 4. The calculations which it is of immediate interest to make, and the results of which are of immediate value to the metallurgist, are those concerned with the volume of gas produced per unit of fuel, its calorific power compared to that of the fuel from which it is produced, the items of the heat losses during the operation of transforming the solid fuel into gaseous fuel, the function of steam in the producer, the limits up to which the use of steam is permissible, the increase of efficiency of the gas by subsequently drying it, the advantages as to final efficiency which are gained by gasifying the fuel over burning solid fuel directly. 145 146 METALLURGICAL CALCULATIONS. For information as to the construction and operation of gas producers, reference may be made to treatises such as Sexton's or Wyer's " Producer Gas," Groves and Thorp's " Chem- ical Technology, Vol. V., Fuels." Trade pamphlets and cata- logues, such as those of R. D. Wood and Co. on " Gas Pro- ducers," of the De la Vergne Machine Co. on " Koerting Gas Engines and Gas Producers," etc., contain a great deal of exact and useful information, and may be usually had for the asking. The monograph of Jiiptner and Toldt on " Genera - toren und Martinofen " (Felix, Leipzig, 1900), is concerned wholly with calorimetric calculations concerning the produc- tion of gas and its utilization in regenerative gas furnaces. 1. SIMPLE PRODUCERS. In these a deep bed of fuel is burnt by air or fan blast, in- troducing no more moisture than happens to be in the atmo- sphere at the time being. The fuel fed into the producer is first dried by the hot gases, then is heated and distilled or coked, and finally is oxidized by the incoming air. The residue is the ash of the coal, which is ground out at the bottom or drops through the grate, containing more or less unburnt fixed car- bon. Great loss of efficiency sometimes occurs from the ashes being rich in carbon. The escaping gases issue at tempera- tures of 300 up to 1000 C., carrying much sensible heat out of the producer. Calculations as to the amount of gas produced per unit of fuel consumed are to be based entirely on the carbon. The gas mtist be carefully analyzed, so that it can be calculated from this analysis how much carbon, in weight, is contained in a given volume of gas. (An alternative method is to take a carefully measured volume of the gas, mix it with excess of oxygen, and explode in a gas burette, determining the amount of carbon dioxide formed, and from that calculate the weight of carbon in the volume of gas taken.) Knowing this, the rest is simple; the carbon in unit weight of fuel minus the carbon lost in the ashes by poor combustion gives the weight of carbon gasified ; this divided by the weight of carbon in unit volume of gas produced, gives the volume of the latter per unit weight of fuel. Illustration', A fuel used in a gas producer contains 12 per ARTIFICIAL FURNACE GAS. 147 cent, of ask and 72 per cent, of carbon. The ashes made con- tain 20 per cent, of unburnt carbon; the gas produced contains by analysis and calculation 0.162 ounce of carbon per cubic foot of gas, measured at 60 F. and 29.8 inches barometric pressure. What volume of gas, measured at above conditions, is being produced per ton of 2240 pounds of coal used? Solution: The distinction between ash and ashes must be noted; the former is the analytical expression for the amount of inorganic material left after complete combustion during the chemical analysis; the latter term means the waste matter pro- duced industrially, and consists, if weighed dry, of the true ash, plus any unburnt carbon. The calculations are therefore as follows : T . Ash in 2240 pounds of coal = 2240 X 0. 12 = 268 . 8 Ashes corresponding = 268.8-^0.80 = 336.0 (the ashes are 80 per cent, ash) Carbon in the ashes = 67 . 2 Carbon in the coal = 2240X0.72 = 1612.8 Carbon going into the gas (gasified) = 1545.6 Carbon in 1 cubic foot of gas = 0.162-7-16 0.010125 Volume of gas produced per 2240 pounds of coal = 1545.6 -=-0.010125 = 152,652 cu. ft It will be noted that these calculations absolutely require the percentage of total carbon in the fuel, as determined by chemical analysis. This is not a difficult analysis, as it consists in burning the carbon in a heated tube in a stream of oxygen or air free from carbon dioxide; the products of combustion are dried and then passed through caustic potash solution to absorb CO 2 gas, the weight of which is obtained by the in- creased weight of the potash bulb, and the total carbon thus obtained. The fixed carbon and volatile matter of the coal, as determined by the ordinary proximate analysis, cannot be used in this calculation, since while all the fixed carbon is carbon, the volatile matter is of variable composition, containing such varying proportions of carbon that no fixed percentage of the latter in it can be assumed without considerable possible error. The calorific power of the gas, per cubic meter or cubic foot, can be calculated from its analysis, using the calorific powers of the combustible constituents as already given in our 148 METALLURGICAL CALCULATIONS. tables. This, multiplied by the volume of gas produced per unit of coal, gives the calorific power of the gas as compared with that of the coal from which it is made. The difference is the heat loss in the operation of producing the gas, including loss by unburnt carbon in the ashes. In fact, we may state that the heat balance is based on the following equations: Heating power of the coal, per unit Heating power of the gas per unit of coal = Calorific losses in conversion. The latter item is composed of: Loss by unburnt carbon in the ashes. Sensible heat of the hot gases issuing. Heat conducted to the ground. Heat radiated to the air. These items may be modified as follows: If the air used is hotter than the normal outside temperature, its sensible heat above this datum should be added to the heating power of the coal, because it increases the total available heat. If the ashes are removed hot, and not allowed to be completely cooled by the incoming air, their sensible heat should be included in the calorific losses during conversion. If the air used is moist, its moisture will be decomposed to hydrogen and oxygen, but the heat absorbed in doing this is exactly represented by the calorific power of this increased amount of hydrogen in the gases, and the heat absorbed is not lost but really represents so much saved as available calorific power of the gases. This item must, therefore, not be counted as one of the heat losses during the operation, as those losses have been defined by us. If the fuel is wet, considerable heat is required to evaporate the mois- ture in it, but this heat is not to be reckoned as one of the losses in conversion, if we have taken as the heating power of the coal the practical metallurgical value; that is, its value assuming all the water in its products of combustion to re- main as vapor and none to condense. If this value has been so taken, the heat required to vaporize the moisture in the coal will have already been allowed for. Similarly, it may take a little heat energy to break up a bituminous coal so as to expel its volatile matter, but this should not be reckoned in as ARTIFICIAL FURNACE GAS. 149 a heat loss in the producer, because a little reflection will show that, whatever this amount may be, it has been properly al- lowed for in the determination or calculation of the total calo- rific power of the fuel. Jiiptner and Toldt call this the " gas- ifying heat," and use it in all their calculations, but it is doubt- ful whether it really amounts to an appreciable quantity, for one thing, and even if it does it should not be reckoned as a heat loss in the producer. Problem 16. Jiiptner and Toldt ran a gas producer with lignite of the following composition: (Generatoren, p. 49). Carbon 69.83 per cent. Hydrogen 4.33 Nitrogen 0.50 Oxygen 12.38 Moisture 7.25 Ash 5.71 Of this coal, 3214 kilograms was used in 8 hours, 50 minutes, producing gas which contained, analyzed dry, by volume- Carbon dioxide, CO 2 . 5. 21 per cent. Carbon monoxide, CO 23.99 Oxygen, O 2 0.63 Methane, CH 4 0.25 Hydrogen, H 2 10.64 Nitrogen, N 2 59.28 The ashes produced weighed 22.23 kilograms per 100 kilo- grams of coal used, and contained 68.76 per cent, of unburned carbon. The calorific power of the coal, determined in the calorimetric bomb (in compressed oxygen, moisture resulting condensed) was 6949 Calories per kilogr. Temperature of hot gases, 282 C.; temperature of air used, 9 C.; humidity, 62 per cent.; barometer, 712 millimeters of mercury. Required : 1. The volume of gas, measured at and 760 mm. pressure (and assumed dry), produced per metric ton (1000 kilos. = 2204 pounds) of fuel used. 150 METALLURGICAL CALCULATIONS. 2. The calorific power of the coal, per kilogram, with mois- ture formed by its combustion assumed uncondensed. 3. The proportion of the calorific power of the coal develop- able by burning the gas produced from it. 4. The loss of heat in conversion. 5. The loss of heat by unburnt carbon in the ashes. 6. The loss of heat as sensible heat in the gases. 7. The loss of heat by radiation and conduction, expressed: (a) Per unit of coal burnt. (b) Per minute. 8. The volume of air required by the producer, at the con- ditions of the atmosphere, per kilogram of coal burnt. Solution : (1) The gas contains, per cubic meter at standard condi- tions : Carbon in CO 2 0.0521 X 0.54 kilos. Carbon in CO 0.2399x0.54 " Carbon in CH 4 0.0025 X 0.54 " Total 0.2945X0.54 = 0.1590 kilos. The carbon gasified from 1000 kilograms of coal is: Carbon in coal = 698.3 kilos. Carbon in ashes 222.3X0.6876 = 152.8 " Carbon gasified = 545.5 " Therefore, C/1K C Gas (dry) produced = ^ "^ = 3430 cubic meters. (1) (2) The calorific power of the coal as given must be dimin- ished by the heat required to vaporize all the moisture formed by its combustion, leaving such moisture as theoretical mois- ture at C. There will be formed per kilogram of coal: From moisture of coal = 0.0725 kilos. From hydrogen 0.0433x9 = 0.3897 " Total = 0.4622 " To evaporate this to theoretical moisture at zero (thus ARTIFICIAL FURNACE GAS. 151 putting the water vapor on the same footing as the other pro- ducts of combustion, CO 2 and N 2 ) requires: 0.4622X606.5 (Regnatdt), = 280 Calories, leaving as the metallurgical or practical calorific power 6949280 = 6669 Calories, (2) (3) The calorific power of each cubic meter of gas (meas- ured dry at standard conditions) is CO = 0. 2399 m 3 X 3,062 = 734 . 6 Calories CH 4 = 0.0025 m 3 X 8,598 = 21.5 H 2 = 0.1064 m 3 X 2,613 = 278.0 Total = 1034.1 Calorific power of gas from 1 kilogram of coal: 1034.1X3.43 = 3547.0 Calories. which equals F- = 53.2 per cent. (3) (4) The loss of calorific power in conversion is 100 53.2 = 46.8 per cent, of the calorific power of the coal, or per kilo- gram of coal: 66693547 = 3122 Calories. (4) (5) Carbon in ashes = 0.1528X8100 = 1237.7 Cal. = 18.6 per cent. (5) (6) The gases produced carry off, per cubic meter measured dry, the following amounts of heat: Volume X mean specific heat (0 282) = heat capacity per 1. CO 2 0.0521X0.432 = 0.0225 CH 4 0.0025X0.428 = 0.0011 CO 0.9454X0.311 = 0.2940 N 2 Sum = 0.3176 Heat carried out = 0.3176X282 = 89.56 Calories. Per kilogram of coal = 89.56x3.43 = 307 Calories. 307 Proportion of calorific power = ^^ = 4.6 per cent. 152 METALLURGICAL CALCULATIONS. The above result is, however, subject to a small correction, because some of the moisture in the coal goes undecomposed into the gases, and is not represented in the analysis of the dried gas. The amount of this moisture can be obtained with sufficient accuracy by finding how much moisture would be obtained by burning the dried gas from 1 kilogram of coal, and comparing this with the moisture which would be ob- tained from 1 kilogram of coal itself; the difference must rep- resent the moisture accompanying the gas as water vapor, and which has not been included in the above computation. Burning 1 cubic meter of gas, the H 2 vapor is: From CH 4 0.0025X2 = 0.0050 cubic meters. FromH 2 0.1064X1 = 0.1064 " 0.1114 " Per kilo, of coal = 0.1114X3.43 = 0.3821 " But the weight of water vapor from burning 1 kilogram of coal has already been found to be (2) 0.4622 kilograms, the volume of which is 0.4622 -r- 0.81 = 0.5706 cubic meters. Leaving, therefore, 0.57060.3821 = 0.1885 cubic meters of water vapor as such accompanying the 3.43 cubic meters of (dried) gas from 1 kilogram of coal. This would take out 0.1885X0.382X282 = 20.3 Calories. Thus increasing the sensible heat in the gases to 307 + 20.3 = 327.3 Calories = 4.9 per cent. (6) [In reality, a still further correction should be made; viz.: to add in the moisture in the air used, because it would also reappear as moisture on final combustion of the gases. Its amount is found from the amount of air used, which, if 62 per cent, saturated at 9 would carry moisture having 0.62 X 8.6 mm. (if saturated) = 5.3 millimeters tension, which repre- sents, barometer being 712 mm., 0.7 per cent, of the volume of the air used, or practically 0.9 per cent, of the volume of nitrogen in the air. Since the nitrogen in the gas represents almost entirely the nitrogen in the air used, the moisture to be accounted for from the air amounts to 0.5928X0.009 = 0.0053 ARTIFICIAL FURNACE GAS. 153 cubic meters per cubic meter of gas, or = 0.0182 cubic meters per kilogram of coal burnt. This correction is altogether too small to affect the results in this case, but should be taken into account whenever the air used is warm and moist.] (7) The calorific loss in conversion was 3122 Calories. Of this we have accounted for: Lost by unburnt carbon in ashes 1237.7 Calories. Sensible heat of gases- (including moisture) 327.3 Total 1565.0 Loss by radiation and conduction = 1557.0 " (a) Per 8 hours 5 minutes there is burnt 3214 kilograms of fuel, making the loss of heat by radiation and conduction per minute = 1557X3214 530 = 9,442 Calories. (b) (8) At the conditions given, each cubic meter of moist air used contained 0.7 per cent, of its volume of moisture, making its percentage composition by volume: Water vapor 0.70 per cent. j Oxygen 20.65 Air ( Nitrogen 78.65 The volume of gas produced per kilogram of coal is 3.43 cubic meters, of which 59.28 per cent, is nitrogen, equal to 2.0333 cubic meters, and weighing 2.0333x1.26 = 2.562 kilograms. Of this 0.0050 kilograms came from the coal itself, leaving 2.512 kilograms to come from the air, or 2.512 -i- 1.26 = 1.9921 cubic meters. This would correspond to 1.9921 -=-0.7865 = 2.5329 cubic meters of moist air if measured at standard con- ditions. At 9 and 712 mm. pressure the real volume of moist air used at prevailing conditions, per kilogram of coal burnt, is 7fif) ^ = 2.80 cubic meters. (8) It must not be thought that the conditions of working in the above producer represent good practice; they are very poor practice as far as concerns the utilization of the fuel. Many producers make gas having 75 to 90 per cent, of the 154 METALLURGICAL CALCULATIONS. calorific power of the coal from which it is made, so that the losses by unburnt carbon and radiation and conduction in this case must be regarded as highly abnormal and very poor prac- tice. The writer chose this example for calculating, because of the carefulness with which Juptner and Toldt had collected the necessary data, and because it illustrated so well the principles to be employed in similar calculations. Problem 17. A gas producer run in Sweden uses saw-dust of the following composition : Water 27.0 per cent. Ash 0.5 Carbon 37.0 Hydrogen 4.4 Oxygen . 30.6 Nitrogen 0.5 Assume that it is run by dry air and that 0.5 per cent, of ashes are made. The gas formed, dried before analysis, con- tains, by volume: Carbon dioxide, CO 2 , . 6.0 per cent. Carbon monoxide, CO 29.8 Ethylene, C 2 H 4 0.3 Methane, CH 4 6.9 Hydrogen, H 2 6.5 Nitrogen, N 2 50.5 The gas actually produced is -partly dried before use by having its temperature reduced by cold water, in a surface condenser, to 29 C., in order to increase its calorific intensity of combustion. Required: (1) The proportion of the moisture in the moist gas which is condensed out. (2) The calorific intensity of the moist gas, if burned pre- heated to 800 C. by the theoretical quantity of air preheated also to 800 C. ARTIFICIAL FURNACE GAS. 155 (3) The calorific intensity of the dried gas, burnt under exactly similar conditions. Solution : (1) It is first necessary to find the weight or volume of water vapor accompanying the gas before condensation, next that accompanying it after passing the condenser. The first can be calculated best on the basis of the hydrogen present in the fuel and in the (dried) gas made from it; the difference is the hydrogen of the moisture removed before analysis, i.e., the hydrogen of the moisture in the wet gas. The first step is to find the volume of gas (dry) produced per unit of fuel, as follows: Carbon in 1 kilo, of fuel = 0.370 kilos. Carbon in 1 m 3 of gas = CO 2 + CO + CH 4 + 2C 2 H 4 (0.060 + 0.298 + 0.069 + 0.006) X 0.54 = 0.2338 " Dry gas per kilo, of fuel = ' - = 1.5825 m 3 U . Zooo The next step is to calculate the water which would be formed by the combustion of 1 kilogram of fuel: Water present in fuel 0. 270 kilos. Water produced by hydrogen Q. 396 " Total = 0.666 " Volume at standard conditions = ' = 0.8222 m e . o 1 ( ) From this we subtract the moisture which would be pro- duced by the combustion of the 1.5825 cubic meters of dry gas, obtained as follows: Water from ethylene = 0.003 X 2 X 1.5825 m 3 Water from methane = 0.069 X2X 1.5825 Water from hydrogen = 0.065X1 X 1.5825 = 0.209 XI. 5825 = 0.3307m 8 Difference = water vapor to 1.5825 m 3 of dried gas = 0.82220.3307 = 0.4915m 3 = 0.4915-5-1.5825 = 0.3106 m 3 per 1 m 8 of dried gas, as analyzed 156 METALLURGICAL CALCULATIONS. This is to be compared with the amount of moisture ac- companying the same quantity of (dried) gas as it escapes from the condenser. This is obtained directly from the fact that the gas escaping will be saturated with moisture at 29 C., that the latter will, therefore, have a tension of 30 milli- meters (tables), and that, assuming the barometer normal (760 mm.), the partial tensions of moisture and gas proper are as 30 to 76030, or as 30 to 730. Since their respective volumes (if both were measured separately at normal pres- sures) are in the same proportion, it follows that each cubic meter of (dry) gas is accompanied by 30 -r- 730 = 0.0411 cubic meters of uncondensed moisture. The respective quantities of moisture accompanying 1 cubic meter of dry, uncondensable gas, are, therefore, 0.3106 before cooling and 0.0411 after cooling, showing that 13.2 per cent, of all the moisture escapes condensation, and that, therefore, 86.8 per cent, of moisture is condensed. (1) (2) The wet gas has a calorific power, calculating on the basis of 1 cubic meter of dried gas analyzed: CO 0.298 X 3,062 = 912.5 Calories. C 2 H 4 0.003 X 14,480 = 43.4 CH 4 0.069 X 8,598 = 593.3 H 2 0.065 X 2,613 = 169.8 Total = 1719.0 There is added to this available heat, when burned, the sensible heat in the gas itself, at 800 C., and also that of the necessary air, also at 800. Heat in gas: CO,H 2 ,N 2 = 0.868 m 3 X 0.3246 = 0.2817 Cals. per 1 CO 2 = 0.060 m 3 X 0.5460 = 0.0328 CH 4 = 0.069 m 3 X 0.4485 = 0.0309 C 2 H 4 = 0.003 m 3 X0.50 = 0.0015 H 2 = 0.3106 m 3 X 0.460 =0.1429 Calorific Capacity = 0.4898 Total sensible heat = 0.4898X800 = 391.8 Calories. ARTIFICIAL FURNACE GAS. 157 The air required theoretically is: For CO 0.298 m 3 = 0.1490 m 3 oxygen For C 2 H 4 0.003 m 3 = 0.0090 m 3 For CH 4 0.069 m 3 = 0.1380 m 3 For H 2 0.065 m 3 = 0.0325 m 3 Sum = 0.3285 m 3 = 1.58m 3 air Heat in this at 800 = 1.58X0.3246X800 410.3 Calories. Sum total of heat going into the products: Developed by combustion 1719.0 Cals. Sensible heat in gas 391.8 " Sensible heat in air 410.3 " Total 2520.0 " The products of the combustion are CO 2 , H 2 O and N 2 , as follows: CO 2 = 0.060 (in gas) +0.298 (from CO) +0.006 (from C 2 H 4 )+ 0.069 (from CH 4 ) = 0.433m 3 . H 2 = 0.311 (with gas) +0.006 (from C 2 H 4 ) +0.138 (from CH 4 )+ 0.065 (from H 2 ) = 0.520m 3 . N 2 = 0.505 (in gas) +[1.58 0.33 = 1.25] (from air) = 1.755 m 3 . Since the 2520.0 Calories remains as sensible heat in the above products, at some temperature t, we have Heat capacity of the CO 2 = 0.433 (0.37 +0.00022t) Heat capacity of the H 2 O = 0.520 (0.34 +0.00015t) Heat capacity of the N 2 = 1.755 (0.303 + 0.000027t) Heat capacity of the products = 0.8688 + 0.00022065t and the calorific intensity t must be 2520.0 t " 0.8688 + 0.00022065t whence t = 1942. (2) (3) When the dried gas is burned under similar conditions, the only difference is that 0.2695 m 3 of water vapor are absent from the gas and from the products, having been condensed. 158 METALLURGICAL CALCULATIONS. This reduces the available heat by the sensible heat in this much water vapor at 800, viz. : 0.2695X0.460X800 = 99.2 Calories, and decreases the calorific capacity of the products by 0.2695 (0.34 + 0.00015t). Our equation, therefore, becomes = 2360.2 " 0.7772 + 0.00018023t whence t - 2056 The increased efficiency of the dried gas for obtaining high temperatures is too evident to need further comment, the difference being in round numbers 150 C., equal to 270 F. f in favor of the dried gas. 2. MIXED GAS PRODUCERS. This Ciass of producers are those most commonly used. -In them a moderate amount of steam or vapor of water passes with the air into the fire, and is decomposed, producing carbon monoxide and hydrogen gases by the reaction: i i i H 2 O + C = CO + H 2 18 12 28 2 which may be read as follows: One volume of steam forms one volume of carbon monoxide and one volume of hydrogen; or eighteen parts by weight of water vapor act upon twelve parts of solid carbon, producing twenty-eight parts of carbon monoxide and two parts of hydrogen. If we speak of kilo- grams as the above weights, then we can call each " volume " spoken of 22.22 cubic meters: or if we call the weights ounces avoirdupois, each " volume " represents 22.22 cubic feet. The water vapor is admitted either automatically, as in the old Siemen's type of producer, where water was run into the ash pit to be evaporated by the heat radiated from the grate or by hot ashes falling into it, or as in the modern water-seal ARTIFICIAL FURNACE GAS. 159 bottom producer, where the ashes rest upon water in a large pan, and so are continually kept soaked by capillary action; or, finally, steam is positively blown under the grate, either as a simple steam jet, or, more economically, by using it in a steam blower, so as to have it produce by injector action an air blast sufficient to run the producer. In the latter case the proportions of air and steam may be regulated with precision, and the blast action produced makes the production and de- livery of the gas practically independent of chimney dralt. The use of steam also rots or disintegrates the ashes, preventing or breaking up masses of clinker, and so facilitating the re- moval of the ashes. Steam or water vapor cools down the fire in the producer so that it runs cooler; at the same time gas is produced which is rich in hydrogen, and, therefore, of higher calorific power. This saves unnecessary waste of heat in the producer, and in- creases the efficiency of the gas in the furnace in which it is burnt. The scientific reasons for these facts are to be found in a consideration of the thermochemistry of the reaction by which steam is decomposed. H 2 O + C = CO + H 2 69,000 +29,160 = 39,840 Calories. This would be the deficit in decomposing 18 kilograms of water if it starts in the liquid state. If, however, it is used as steam at 100 C., each kilogram contains 637 Calories of sensi- ble heat, making 18X637 = 11,466 Calories altogether, leav- ing the deficit 28,374 Calories, or the deficit = 1,576 Calories per kilogram of steam decomposed. = 2,364 Calories per kilogram of carbon thus burnt. In making this calculation it might be objected that the steam used is often at three or four atmospheres pressure, and its temperature, therefore, over 100 C. ; but it must not be overlooked that this steam expands suddenly to atmospheric tension, and that in so doing it cools itself to an amount roughly proportional to its excess pressure so that the expanded steam at atmospheric pressure is usually close to 100 C. When mixed with air, the temperature of the mixture is usually below 100, some 40 to 50 C., and in this condition some of 160 METALLURGICAL CALCULATIONS. the steam is possibly condensed to fog, but it must not be for- gotten that the heat of condensation thus given out has been absorbed in raising the temperature of the admixed air, and therefore goes as sensible heat into the bed of burning fuel. For the purpose of calculation, we will, therefore, be very nearly right in assuming that we are dealing in each case with steam at 100 C., requiring the above calculated deficits to be made up, in order for the decomposition to proceed. It will be evident that any heat thus used in the producer must be sensible heat of the hot carbon, which, if not so used, would be lost as sensible heat in the gases produced or radi- ated and conducted away from the producer. The basic pro- cess of the running of the producer is the burning of carbon ttf carbon monoxide, liberating 2,430 Calories per kilogram of carbon, which is 2,4304-8,100 = 30 per cent, exactly of the calorific power of the carbon. This heat, if no water vapor gets into the producer, is lost as sensible heat of the hot gases and by radiation and conduction, and is thus largely a dead loss. In Problem 16, for instance, these items figured out 28.25 per cent, of the calorific power of the coal used. Now, the facts are, that while small producers need that much heat to keep them up to working temperature, large producers need very much less, and run far too hot if no steam is admitted to check the rise of temperature. In the largest, producers the sensible heat in the gases, plus the losses by radiation and conduction, does not exceed 10_per cent, of the calorific power of the fuel. Out of the 30 per cent, of the calorific power of the carbon inevitably generated, only some 10 per cent, is, therefore, needed to supply the losses in a large producer, leaving 20 per cent, applicable to decomposing steam. We therefore have: Per 1 Kilo, of Carbon Gasified. Heat generated 2,430 Calories. Necessary loss in a large producer 810 Useful for decomposing steam 1,620 Required to decompose 1 kilo, steam at 212 F. (69,000 11,466) -5- 18= 3,196 Maximum steam decomposable 1,620-7-3,196 = 0.507 kilo. Calculation, therefore, shows that about one-half of a unit weight of steam is the maximum which can be used per unit ARTIFICIAL FURNACE GAS. 161 weight of carbon burnt to carbon monoxide, consistent with keeping the producer at proper working temperature. This would be equal to 0.088 kilos, of steam per kilo, of air used. 0.088 pounds of steam per pound of air used. 0.114 kilos, of steam per cubic meter of air used. 0.007 pounds of steam per cubic foot of air used. The above discussion is on the assumption that the bed of fuel in the producer is thick enough, and its temperature al- ways high enough, to burn all the carbon to carbon monoxide. Such is only an ideal condition, for the irregular charging, descent and working of the fuel always allow of some carbon dioxide being produced, and the best regular gas usually con- tains 1 to 5 per cent, of dioxide, representing some 3 to 20 per cent, of the total carbon oxidized in the producer. Assuming that on an average 10 per cent, of the carbon oxidized inevitably forms carbon dioxide, we can calculate under these more usual conditions how much steam can be. used, because for each kilogram of carbon oxidized, 0.1 kilo, then gives us 8,100 Calories per kilogram instead of 2,430 in the producer, a surplus of 0.1 X (8,100 2,430) = 567 Calories. over the conditions when only monoxide is formed. Instead of the 1,620 Calories available for decomposing steam we will now have 2,187 Calories, which will decompose o -I o'-r = 0.684 kilo, of steam. which reckoned on the air used would be 0.108 kilos, of steam per kilo, of air used. 0.108 pounds of steam per pound of air used. 0.140 kilos, of steam per cubic meter of air used. 0.009 pounds of steam per cubic foot of air used. The above proportions are those which cannot practically be exceeded, if producing gas as low as is usually possible in carbon dioxide. [Working the producer comparatively cold, with excess of 162 METALLURGICAL CALCULATIONS. steam, much larger proportions of carbon dioxide are formed and correspondingly larger proportions of steam are decom- posed, but this manner of working is abnormal, is not unusual, and will be discussed under the next heading, treating of Mond gas.] Making gas containing any given proportion of CO 2 to CO, by volume, the ratio thus obtained is identical with the pro- portionate weight of carbon oxidized to CO 2 and CO in the producer, and by the application of the principles just described it can be calculated how much steam can be used per unit of carbon oxidized, making proper allowance for losses by radia- tion, etc. Illustration'. A Siemen's producer with chimney draft pro- duced gas containing, by volume, 4. 3 per cent. CO 2 and 25.6 per cent. CO. How much steam could be used per pound of air used, assuming 50 per cent, of the heat generated by the oxidation of the carbon to be needed to run the producer? Solution : 4 3 Per cent, of carbon burnt to CO 2 = = 14.4 per cent. 4 . o -J~ do . b Heat generated by C to CO 2 = 0.144X8,100 =1,166 Ib. Cal. Heat generated by C to CO = 0.856X2,430 = 2,080 Heat generated per kilo, of C = 3,246 Heat lost by radiation, etc. (50 per cent.) = 1,623 Heat available for decomposing steam = 1,623 1 623 Steam decomposable = ' = 0.508 pounds. O j A V/O The above is expressed per kilogram of carbon oxidized, but the same proportion is true per pound. The air required per pound of carbon oxidized is found from the oxygen required to form CO and CO 2 : Oxygen for C to CO 2 = 0.144x8-3 = 0.384 Ibs. Oxygen for C to CO =0.856X4-3 = 1.141 " Total = 1.525 ". fi na Air = 6 - 608 " Volume of air = - 81.8 cu. feet. ARTIFICIAL FURNACE GAS. 163 Steam per cubic foot of air = ' = 0.006 Ibs. ol . o Steam per pound of air = ' = 0.077 " Air required per pound of steam = 13.0 It should easily be seen that whatever heat is absorbed in the producer in decomposing steam, is entirely recovered when the hydrogen thus produced is burnt and steam is reproduced. If then, in any case, it is possible to absorb in the producer, in the decomposition of steam, an amount of heat equal to say, 20 per cent, of the total calorific power of the fuel, then that 20 per cent, is regained and capable of being utilized when the hydrogen so produced is burnt in the furnace. In other words, 20 per cent, less of the calorific power of the fuel will be lost in the process of conversion into gas in the producer, and 20 per cent, more will be obtained in the burning of the gas when it is used. The great advantages of using steam judiciously are thus clearly evident. Problem 18. R. W. Hunt Co. report the following tests made of the running of a Morgan continuous gas producer. Coal used, " New Kentucky " Illinois coal, run of mine. Composition: Fixed carbon 50.87 per cent. Volatile matter 37.32 Moisture 5.08 Ash , 6.73 100.00 The ultimate composition was: Total carbon 69.72 Hydrogen 5.60 Nitrogen 2.00 Total sulphur 0.94 Oxygen 11.00 Moisture 5.08 Inorganic residue (less sulphur) 6.66 The ash, on combustion, contains 1.12 per cent, of its weight 164 METALLURGICAL CALCULATIONS. of sulphur (as FeS); the ashes obtained from the producer contain 4.66 per cent, of unburnt carbon. The gas produced, dried, contained by volume: Carbon monoxide (CO) 24.5 per cent. Marsh gas (CH 4 ) 3.6 Ethylene (C 2 H 4 ) 3.2 Carbon dioxide (CO 2 ) 3.7 Hydrogen (H 2 ) 17.8 Oxygen (O 2 ) 0.4 Nitrogen (N 2 ) (by difference) 46.8 [The moisture and sulphur compounds in the gas not having been determined, we can calculate the former, and, for the purposes of calculation, are justified in assuming the sulphur present in the gas as H 2 S, and in subtracting it from the hydro- gen. We will also assume that the moisture in the coal goes unchanged into the gases as moisture, and that all the steam used is decomposed. The 0.94 per cent, of sulphur in the coal will furnish 0.86 per cent, to the gases, because 6.73 X 0.0112 = 0.08 per cent, will go into the ash as ferrous sulphide. The 0.86 pound of sulphur would produce 0.91 pounds of H 2 S, equal in volume to 9.56 cubic feet per 100 pounds of coal used, or (since we will see later that 53.83 cubic feet of gas are pro- duced per pound of coal) there will be 0.2 per cent, of H 2 S in the gases, leaving 17.6 per cent, of hydrogen.] On the basis of above data and assumptions: Required: (1) The volume of gas produced per pound of fuel used in the producer. (2) The weight of steam used per 100 cubic feet of air blown in, assuming the air dry. (3) The proportion of the total heat generated in the pro- ducer which is utilized in decomposing steam. (4) The percentage of increased economy thus obtained reckoned on the calorific power of the fuel. (5) The efficiency lost by unburnt carbon in the ashes. (6) The efficiency of the gas, burnt cold, compared with the coal from which it is made. Solution: (1) Per pound of coal burnt, there remains in the ashes 0.0673 X (0.0466 -i- 0.9534) =0.0033 pounds of un- burnt carbon, leaving 0.06972 0.0033 = 0.6939 pounds gasified. ARTIFICIAL FURNACE GAS. 165 One cubic foot of gas, at 32 F., contains the following weight of carbon: C in CO = 0.245X0.54 ounces CinCH 4 = 0.036X0.54 " CinC 2 H 4 = 0.032X1.08 " CinCO 2 = 0.037X0.54 * Total = 0.382X0.54 " = 0.20628 ounces av = 0.01289 pounds av. Gas produced, measured dry, per pound of coal, at 32 F.: 0.6939 0.01289 = 53.83 cubic feet (2) If the moisture of the coal is assumed to pass unchanged into the gas, as moisture, then all the hydrogen in the dry gas, in any form or combination, must have come either from hydrogen in the coal or in the air blast. The hydrogen in the coal is given as 5.60 per cent. The hydrogen in the gas is calculated as follows, per cubic foot: H 2 in H 2 = 0.176X0.09 ounces. H 2 in H 2 S = 0.002X0.09 " H 2 in CH 4 = 0.036X0.18 " H 2 inC 2 H 4 = 0.032X0.18 " Total = 0.314X0.09 " = 0.02826 ounces av. = 0.00177 pounds av. Hydrogen in gas from 1 pound of coal: 0.00177X53.83 = 0.0953 pounds. Hydrogen from decomposition of steam: 0.09530.0560 = 0.0393 pounds. Weight of steam decomposed per pound of coal used: 0.0393X9 = 0.3537 pounds. To express this weight relatively to the air blown in, we must calculate the air used per pound of coal, as follows: 166 METALLURGICAL CALCULATIONS. Nitrogen in gas per cubic foot 0.468X(14X.09) = 0.5897 oz. av. = 0.036856 Ibs. av, Per pound of coal = 0.036856X53.83 = 1.9840 Subtract N 2 in 1 pound of coal = 0.0200 Leaves N 2 from air = 1.9640 " 1 S Weight of air = 1.9640Xy^ = 2.5532 Volume of air = 2.5532x16^-1.293 = 31.61 cu. ft. Steam used per 100 cubic feet of air blown in: 0.3537 31.61 X100 = 1.119 pounds. (2) (3) The heat utilized in decomposing steam has been found to be 3,196-pound Calories per pound of steam at 212 F. We therefore, have the heat so used per pound of fuel used: 0.3537X3,196 = 1,130-pound Calories. This quantity must now be compared with the total heat gen- erated in the producer, and the latter quantity can be deter- minated in two ways: (1) We may subtract from the total calorific power of the coal the calorific power of the gas pro- duced and of the unburnt carbon in the ashes; the difference must be the net heat generated in the producer, i.e., the total heat generated minus that absorbed in decomposing steam. The total heat generated is the net heat thus calculated plus the heat absorbed in decomposing steam. (2) We may calcu- late the heat of formation of the CO and CO 2 in the gas, and assume that as the total heat generated in the producer. This method is not so accurate as (1). The total calorific power of the coal is given as practically 7,747-pound Calories per pound, water formed being con- densed, which would be decreased by the latent heat of vapor- ization, if the latter is assumed un condensed. The deduction is 606.5 X [(0.056X9) +0.0508] = 337-pound Calories, leaving 7,410-pound Calories as the practical metallurgical calorific power of the fuel. ARTIFICIAL FURNACE GAS. 167 Calorific power of the gas (dried) per cubic foot: CO 0.245 X 3,062 = 750.2 ounce Cal. CH 4 0.036 X 8,598 = 309.5 C 2 H 4 0.032X14,480 = 463.4 H 2 0.176X 2,613 = 459.9 H 2 S 0.002X 5,513 = 11.0 Total = 1994.0 = 124.6 pound Cal. Calorific power of gas per pound of coal: 124.6X53.83- 6,707 pound Cal. Calorific power of carbon in ashes: 0.0033X8,100= 27 Sum = 6,734 Calorific power of 1 pound coal = 7,410 Net heat lost in conversion = 676 Used in decomposing steam = 1,130 Gross heat generated in producer = 1,806 Proportion of this utilized in decomposing steam: = 0.626 = 62.6 per cent. (3) (4) We can state this result in another way, by saying that 1,806-7-7,410 =-- 24.40 per cent, of the calorific power of the fuel is generated in the producer, of which 1,130 -h 7, 410 = 15.25 per cent, is utilized to decompose steam, and 9.15 per cent, is lost by radiation, conduction and sensible heat in the gases. The calorific power of the gases represents 6,707 -r- 7,410 = 90.50 per cent, of the calorific power of the coal of which 15.25 per cent., however, is clear gain from the employ- ment of steam. Reckoning on the total calorific power of the coal, the increased economy from the use of steam is 15.2 per cent. (4) (5) The loss of calorific power by the unburnt carbon in the ashes is 27-pound Calories, or, on the whole heat available, = 7^0 =0.36 per cent. 168 METALLURGICAL CALCULATIONS. This loss is exceptionally low, and may be very profitably compared with the analogous loss of 18.6 per cent, occurring in the case discussed in Problem 16. (6) This has already been calculated as 90.50 per cent. ,4-LU on the assumption that the gases are burnt cold. If they are burnt hot, say issuing from the producer at 1200 F. (649 C.), and are burnt when at 1,000 F. (548 C.) their sensible heat at 1,000 F. will be added to their efficient heating power, and can be calculated with exactness, using the principle ex- plained and used in requirement (6) of Problem 14. Under such conditions the total efficiency of the producer, reckoned on the calorific power of the coal used, approximates 95 per cent. To be fair to everybody concerned, however, we must de- duct from this the coal required to be burnt to raise the steam used. There is a little over one-third pound of steam used per pound of coal used in the producer. This requires in ordinary boiler practice ~o"X-^- = 7^7 pound of coal, or about 4 per cent, of the weight of fuel burnt in the producer. The results of the previous calculation must, therefore, be dimin- ished in this proportion, if the steam has to be raised by burning coal under boilers. Under these conditions (6) becomes: 90.50-^1.04 = 87 per cent, efficiency. (6) If the steam can be obtained from waste gases of a blast furnace, or from the hot producer gases themselves (in case they are going to be burnt cold), then no such deductions need be made. It is very evident, however, that if 9.15 per cent, greater efficiency is gained by burning 4 per cent, more coal to raise the steam, that the net gain would be only 5.15 per cent. Even under these conditions it pays to use steam, be- cause of the greater calorific intensity of the richer gas, the usefulness of the steam for supplying air blast, and the rotting of the clinkers therewith obtained. ARTIFICIAL FURNACE GAS. 169 3. MOND GAS. In the Mond producer, an excess of steam is introduced with the heated air used for combustion; the producer is thus run much colder than the ordinary producer, and far more carbon dioxide is present in the gas. In fact, the gas, compared with ordinary producer gas, is very high in carbon dioxide (10 to 20 per cent.), very high in hydrogen (20 to 30 per cent.), very low in carbon monoxide (10 to 15 per cent.), low in nitrogen (40 to 50 per cent.), and carries an extraordinary amount of undecomposed moisture. The fuel used is low-grade bitumin- ous slack, and the object of using so much steam is to keep the temperature in the producer so low that a maximum amount of the nitrogen in the coal is evolved as ammonia. The gas is cooled to ordinary temperature by contact with water spray, so that all but a small amount of moisture is condensed, the ammonia is removed by dilute sulphuric acid, and the cold nearly dry gas is then used for gas engines or in furnaces. The calorific power of the gas is not low, because the high proportion of hydrogen compensates for the low carbon mon- oxide, while the great heat evolved by the large formation of carbon dioxide has been mostly absorbed in decomposing steam, and is, therefore, potentially present in the gas in the form of hydrogen. Problem 19. Bituminous slack coal used in Mond producers for gener- ating gas for a gas-engine power plant contained: Moisture 8.60 per cent. Carbon.. . ..62.69 Hydrogen 4.57 Oxygen . . .10.89 Nitrogen 1.40 Ash 10.42 Calorific power, determined in a bomb calorimeter, water condensed, 6,786 Calories per unit of dried fuel. Ashes pro- duced 268 pounds per ton (2,240 pounds) of moist slack used; contains 12 per cent, of carbon. Air used for running is heated to 300 C. by the waste heat of producer gases, and carries in 2J tons of water as steam (at 170 METALLURGICAL CALCULATIONS. same temperature) for every ton of fuel burnt. The steam is generated by a tubular boiler run by the escape gases from the gas engines, but is heated from 100 C. to 300 C. by the waste heat of the producer gases. The latter escape from the producer at 350 C. Composition of waste gases passing out of condensers at 15 C.: Carbon monoxide (CO) 11.0 per cent Hydrogen (H 2 ) 27.5 Marsh gas (CH 4 ) 2.0 Carbonic oxide (CO 2 ) 16.5 Nitrogen (N 2 ) 41.3 Water vapor (H 2 0) 1.7 " 100.0 . Assume all the nitrogen of the fuel to form ammonia gas NH 3 . Required: (1) The calorific power of the Mond gas. (2) The volume of gas produced per ton of fuel used. (3) The efficiency of the producer. , (4) The weight of steam which is decomposed in the pro- ducer. (5) The proportion of the calorific power of the fuel saved to the gas by the decomposition of steam. (6) The proportion of the heat generated in the producer which is saved to the gas by the decomposition of steam. (7) The proportion of the calorific power of the coal lost from the producer by radiation and conduction. Solution: (1) One cubic foot of the gas at C. would gen- erate the following amounts of heat (water un condensed). CO = 0.110 cubic feet X 3062 = 336.8 ounce Calories. H 2 = 0.275 cubic feet X 2613 = 718.6 CH 4 = 0.020 cubic feet X 8598 = 172.0 Total =1227.4 = 76.7 pound = 138.1 B. T. U. Per cubic meter = 1227.4 kilo. ARTIFICIAL FURNACE GAS. 171 If measured at 15 C. (60 F.) the above values will be re- duced by the factor 273 -r (273 + 15), and become Per cubic foot = 72.7 pound Calories. Per cubic meter = 1163 kilo. " (1) (2) Carbon in 1 cubic foot of gaa at C.: In CO 0.110X0.54 ounces In CH 4 0.020X0.54 " In CO 2 0.165X0.54 " 0.295X0.54 " = 0.1593 ounces = 0.009956 pounds Carbon going into gas per pound of fuel burnt: Carbon in fuel 0.6269 pounds Carbon in ashes ~-X0.12 = 0.0144 Carbon in gas = 0.6125 " Volume of gas (at C.) per pound of fuel used: 0.6125 = 61.52 cubic feet 0.009956 Per ton of 2240 pounds = 137,805 cubic feet At 15 C. (60 F.) = 137,805 X 27 ;!* 15 == 145,375 " " (2) (3) The calorific power of the dried fuel is given as 6,786 Calories per pound, water condensed. Since one pound of wet fuel contains 1008.60 = 91.40 per cent, dried fuel, the calorific power of 1 pound of wet fuel, moisture condensed, is 6,786X0.9140 = 6,202 pound Calories. But, 1 pound of wet fuel would produce, on combustion, 0.0860 + 9 (0.0457) = 0.4973 pounds moisture, which, remaining vaporized at 15, would retain 0.4973X596 = 296 Calories, leaving the net metallurgical calorific power as 6,202296 = 5,906 Calories. 172 METALLURGICAL CALCULATIONS. The 61.52 cubic feet of gas produced per pound of moist fuel will have a calorific power, burnt cold, of 61.52X76.7 = 4,719 Calories, making the efficiency of the producer, on fuel consumed in it, 4 71Q g^- 0.799 = 79.9 per cent. (3) The above figure is true only on the assumption that the steam used is obtained from waste heat, and therefore does not require the combustion of extra fuel (4) The gas produced contains, per cubic foot, the following amount of hydrogen, free and as CH 4 : As H 2 0.275 cubic feet X 0.09 =0.02475 ounces. As CH 4 0.040 cubic feet X 0.09 = 0.00360 Sum = 0.02835 = 0.001 772 pounds Per pound moist fuel = 0.001772X61.52 = 0.1090 Present as ammonia gas 0.0140 X (3 H- 14) = 0.0030 Total (not including H as water) = 0.1120 Hydrogen in 1 pound of coal = 0.0457 " Hydrogen from decomposition of steam = 0.0663 " Water decomposed in the producer = 0.5967 "(4) Since the steam introduced weighs 2.5 pounds for every pound of fuel burnt, we see that only 0.2387 = 23.87 per cent. j . of the steam introduced is decomposed. [In the writer's opinion this unused 76.85 per cent, can only pass in and pass out carrying out sensible heat, and it seems a very wasteful method of keeping down the temperature in the producer. From the standpoint of regarding the 2 pounds of steam as a mere absorber of sensible heat, it could probably be replaced by some of the gases of combustion from the gas engine or open-hearth furnace. The products of combustion of the above gas would contain approximately Nitrogen 68.7 per cent. Carbon dioxide 14.7 Water vapor 16.6 ARTIFICIAL FURNACE GAS. 173 And if the J pound of water vapor decomposed in the pro- ducer were thus supplied, it would bring in, per pound of coal burnt, Nitrogen ......................... 41.2 cubic feet. Carbon dioxide ................. 8.8 Water vapor ................... 9.9 And if the CO 2 thus introduced were reduced to CO, as it prob- ably would be, the gases would receive from this source Nitrogen ....................... 41.2 cubic feet. Carbon monoxide ............... 27.5 Hydrogen ...................... 9.9 thus producing gas quite up to standard as regards combus- tibles, while the heat absorbed in the reduction of CO 2 to CO would cool the fire down quite as effectually as the extra steam now used.] (5) The steam decomposed, 0.5967 pounds per pound of fuel burnt, may be assumed to be at 100 C. on entering the fire, and to therefore absorb 3,196-pound Calories per pound of steam decomposed '. The heat absorbed in the producer, and thus transferred into potential calorific power is, therefore 3196X0.5967 = 1907 pound Calories, which expressed in per cent, of the calorific power of the fuel is = 0.323 = 32.3 per cent (5) (6) The heat generated in the producer equals the calorific power of the coal minus the heat lost by carbon in the ashes, minus the calorific power of the gases, plus the heat absorbed in decomposing steam. The loss by carbon in the ashes is X0.12X8100 = 117 Calories. 2240 The heat generated in the producer is, therefore, 5906117 1719 + 1850 = 2920 Calories, equal to 2920 -r- 5906 = 49.4 per cent, of the calorific power of 174 METALLURGICAL CALCULATIONS. the coal. Of this, 1907 Calories is absorbed in decomposing 1907 steam, which is TTQ = 65.3 per cent, of the total heat gener- ated. (6) (7) There are 2922 Calories generated in the producer, of which 1907 are absorbed in decomposing steam, leaving 1015 Calories to supply radiation and conduction and as sensible heat in the hot gases, to which must be added the sensible heat in the hot air and steam used at 300 C. The air used per pound of coal is found from the nitrogen in the gases: Nitrogen in 1 cubic foot of producer gas = 0.413 cubic foot Nitrogen in 61.52 cubic foot of producer gas = 25.40 Air used = 25.4-0.792 =32.08 Steam used = (2.5X16)^0.81 = 49.40 Heat in steam and air at 300 C. (from 15 C.): 32.08X0.3116X285 = 2850 ounce Calories 49.40X0.3872X285=5450 Sum = 8300 = 519 pound Calories Total heat radiated, conducted and in hot gases: 1015 + 519 = 1534 Calories. The heat in the hot gases is as follows, per cubic foot of gas produced : CO 0.110X0.313X335] H 2 0.275X0.313X335 [ = 83.7 N 2 0.413X0.313X335 J CO 2 0.165X0.450X335 = 24.9 CH 4 0.020 X 0.460 X 335 = 3. 1 Sum = 111.7 ounce Calories = 7.0 pound Calories. Per pound of coal burnt : 7.0X61.52 = 430 pound Calories ARTIFICIAL FURNACE GAS 175 To this must be added the heat 'in undecomposed water vapor in the gases, as follows: Steam used per pound of coal = 2.50 pounds Steam decomposed per pound of coal = 0.60 Steam remaining in gases = 1.90 Moisture in coal = 0.086 Total in gases = 1.986 Volume = (1.986X16)-*- 0.81 = 39.2 cubic feet Heat contained in this at 350 C. : 39.2X0.395X335 = 5187 ounce Calories = 324 pound Calories Total heat in producer gases: 430 + 324 = 754 Calories Heat lost by radiation and conduction: 1589754 = 827 Calories Proportion of calorific power of coal thus lost: = 0.140 = 14.0 per cent. (7) When Mond gas is heated in the regenerator of an open- hearth furnace it changes considerably in composition, as is shown by the following analyses made by Mr. J. H. Darby, on gas dried before analysis: Before After Regenerator. Regenerator. Carbonic acid gas, CO 2 17.8 10.5 Carbon monoxide, CO 10.5 21.6 Ethylene, C 2 H 4 0.7 0.4 Methane, CH 4 2.6 2.0 Hydrogen, H 2 24.8 17.7 Nitrogen, N 2 43.6 47.8 100.0 100.0 The above changes are very interesting, and their discussion profitable. Before heating, the gas burns with a non-luminous 176 METALLURGICAL CALCULATIONS. flame; after heating, it burns with a brilliant white flame. Let us inquire: (1) What chemical change occurs during the heating? (2) What are the relative volumes of the gas before and after heating (excluding water)? (3) What change in the calorific power is produced by the heating ? (1) An inspection of the analyses shows undoubtedly that at a high temperature the CO 2 cannot hold all its oxygen in the presence of such a large amount of hydrogen, and that the following reaction must occur: i i i i C0 2 + H 2 = CO + H 2 O The figures given do not check exactly, but assuming that the above reaction takes place, we can find to what extent, by expressing the composition of the gas after heating for the same amount of nitrogen as was in the gas before heating, since this gas is unchanged: Before After Loss Heating. Heating. or Gain. CO 2 .................... 17.8 9.6 8.2 CO ..................... 10.5 19.7 +9.2 H 2 ................ .'....24.8 16.1 8.7 N 2 ..................... 43.6 43.6 0.0 Since, according to the reaction written, the volume of CO 2 reduced to CO will be equal to the volume of H 2 thus con- sumed, and will produce an equal volume of CO; the above table proves, within the probable limits of error, that the reaction written actually takes place. The separation of luminous carbon is probably due to the splitting up of C 2 H 4 . The relative volumes of the heated and unheated gases will be inversely as the percentage of nitrogen in each (since this gas is unchanged), viz.: as 47.8 to 43.6, or as 100 to 91.2. The contraction, 8.8 parts, would again correspond almost exactly to the amount of water formed in the assumed reaction, which would be equal to the hydrogen ^so used, thus giving another check on the validity of the reaction assumed to take place. ARTIFICIAL FURNACE GAS. 177 The calorific power of 1 cubic foot of original gas is: CO 0.105X 3,062= 321.5 ounce Calories C 2 H 4 0.007X14,480 = 101.4 " CH 4 0.026 X 8,598 = 223.5 " H 2 0.248 X 2,613 = 648.0 " Sum = 1294.4 " = 80.9 pound " Per 100 cubic feet = 8090 " The calorific power of 1 cubic foot of heated gas is: CO 0.2I6X 3,062= 661.4 ounce Calories C 2 H 4 0.004X14,480 - 57.9 " CH 4 0.020 X 8,598 = 172.0 " H 2 0.177X 2,613 = 462.5 " Sum = 1353.8 " = 84.6 pound For 91.2 cubic feet the calorific power would be 84.6X91.2 = 7715 pound Calories. The net conclusion is that the heated gas (aside from its sensible heat) gives less heat by combustion in the furnace per unit of coal gasified in the producers than the correspond- ing quantity of unheated gas; but the difference is only some 4 per cent. On the other hand, the calorific power of the heated gas per cubic foot is some 5 per cent, greater than that of the unheated gas, and, therefore, its calorific intensity will have been increased by the heating quite aside from the question of its temperature being higher, and therefore its sensible heat greater. 4. WATER GAS. This gas is made intermittently, by first burning part of the fuel until the fire is very hot, and then introducing steam, cool or superheated, into the fire. The reaction producing the gas is: i ii C + H 2 = CO + H 2 And if the reaction is complete, and only carbon is present as fuel, the gas produced is theoretically composed of equal parts by volume of CO and H 2 , the volume of each of these being equal to that of the steam used (if measured at the same tem- perature and pressure). 1/8 METALLURGICAL CALCULATIONS. Deviations from this ideal composition are caused in practice by the occurrence of undecomposed steam in the gas, also of CO 2 and N 2 , which come from the residual gas formed during the heating up of the fire, some of which will get into the first portion of the water-gas produced, and also hydrocarbons, tar and ammonia from the distillation of the fuel. A typical analysis of water-gas, given by Mr. W. E. Case, is: Hydrogen (H 2 ) 48.0 Carbon monoxide (CO) 38.0 Methane (CH 4 ) . 2.0 Carbon dioxide (CO 2 ) 6.0 Nitrogen (N 2 ) 5.5 Oxygen (O 2 ) 0.5 The production of water-gas is more expensive than that of other artificial producer gases, because of the large amount of steam necessarily required, the heat lost during the heating up, and the essentially intermittent character of the operation. Its essential advantages are its very high proportion of com- bustibles, averaging 90 per cent, and the consequent high calorific intensity which it is capable of producing. (Uncar- buretted, it is well known that water-gas is a valuable domestic fuel, and illuminant when used in mantle burners; carburetted, it forms our principal illuminating gas, and as such is manu- factured on an immense scale. We will treat here only of its metallurgical uses.) Discussing the manufacture of the gas, the first step is the heating up of the fuel. This is accomplished by blowing air through it. At this point we can distinguish two systems. The older one is that of blowing in air under moderate pres- sure, which passes through the fire at moderate velocity, and produces a fair grade of ordinary producer gas. This gas is either wasted, or burned in furnaces requiring such quality of gas, or burned in regenerators where heat is stored up to be utilized in the next stage in superheating steam. The newer system is to blow in air at high pressure, such that a large percentage of carbon dioxide remains in the gases, thus nearly completely burning what carbon is oxidized, and storing up a corresponding quantity of heat in the remaining carbon. The incombustible gases produced are passed through a recuperator ARTIFICIAL FURNACE GAS. 179 or regenerator, where their sensible heat is partly communi- cated to the steam used, so as to superheat it when forming water-gas. This system consumes the minimum of carbon in " heating up " the fuel, saves time in this unproductive period, and allows the producer to be run longer " on steam." In practice the fuel is probably raised to an average temperature of 1500 C. during the heating up. We can calculate the efficiency of this heating up, that is, the proportion of the calorific power of the carbon burnt which is stored up in the remaining fuel, if we know how much fuel is in the producer, how much air is blown through, the com- position of the gases produced, and the average rise in tem- perature of the fuel. Since this operation is, however, only supplementary to the real formation of water-gas by the de- composition of steam, we will first make calculations upon the cooling down period, during which water-gas is made. During the use of steam the reaction absorbs heat, and the producer rapidly cools. Steam is passed through until the temperature of the fuel is 800 C., and must then be stopped, because between 800 and 600 the reaction is mostly C + 2H 2 = CO 2 + 2H 2 resulting in a rapid increase of CO 2 in the gas. The heat to decompose steam is furnished partly by the oxidation of carbon to CO, and partly by the sensible heat in the carbon itself. Since the carbon, from temperatures of about 1000 C. up has a specific heat of 0.5, we can easily calculate how much steam can be decomposed before the temperature falls to 800. Problem 20. A Dellwick- Fleischer water-gas producer contains 3 tons of coke (90 per cent, carbon), heated up to 1500 C. Steam, heated to 300 C., is passed through for 8 minutes, until the temperature of the gas escaping is 700 C., or 800 at the fuel bed. The composition of the gas is (Prof. V. B. Lewes): Hydrogen 50.0 Carbon monoxide 40 . Carbon dioxide 5.0 Oxygen 1.0 Nitrogen 4.0 180 METALLURGICAL CALCULATIONS. Assume 9000 Calories per minute lost by radiation and con- duction. Required: (1) The amount of steam used in the 8 minutes (2) The volume of gas produced in the 8 minutes. (1) The amount of steam which could have been used is limited by the available heat to decompose it. The latter is furnished by (a) Oxidation of carbon to monoxide. (6). Oxidation of carbon to dioxide. (c) Sensible heat of the carbon and ash of fuel. (d) Sensible heat of the steam used. While the items of heat absorption and loss are: (e) Decomposition of the steam. (/) Sensible heat in the gases. (g) Radiation and conduction. The simplest way to arrive at a solution is to make a few necessary assumptions, and to then let X represent the weight of steam used. The assumptions are that the average tem- perature of the fuel bed falls to 1000 C., that the average temperature of the escaping gases is (1500 + 700) -=-2 = 1100 C., that the average specific heat of carbon in the range 1000 1500 is 0.5, and of the ash of the fuel 0.25. Then, casting up a heat balance sheet in terms of X, we can finally arrive at an expression for the heat which was available during the 8 minutes for decomposing steam, and thus at the weight of steam decom- posed, and (making this equal to X) at a solution. (a) The analysis of the gas shows that eight times as much carbon was burnt to monoxide as to dioxide, making the equa- tion of combustion: 9C+10H 2 O = 8CO + C0 2 +10H 2 By weight, 8X12 = 96 parts of carbon was burnt to CO per 10 X 18= 180 parts of steam used, or 0.533 parts per one part of steam. The heat generated in forming monoxide is, therefore : 0.533XXX2430 = 1296 X Calories. (6) Only one-eighth as much carbon burns to dioxide, giving, therefore, as the heat evolved: (0.533H-8)XXX8100 = 540 X Calories. ARTIFICIAL FURNACE GAS. 181 (c) 3000 kilos, of fuel, representing 2700 kilos, of carbon and 300 kilos, of ash, cool from 1500 to 1000: 2700X0.50X500 = 675,000 Calories 300X0.25X500 = 37,500 Sum = 712,500 (d) X-J-0.81 will be the volume of the steam used at normal conditions, which brings in considered only as vapor, at 300: (X * 0.81) X 0.385X300 = 142.6 X Calories. The sum total of (a) + (b) + (c) + (d) gives the total available heat, viz.: 1978.6 X-f 712,500 Calories. (e) The steam used requires for its decomposition, con- sidered theoretically as cold steam, producing cold products: (X ^ 9) X 29,042 = 3227 X Calories (/) The gas being 50 per cent, hydrogen, and the latter being equal to the volume of steam used, the volume of gas must be 2X (X-:- 0.81), which multiplied by the mean specific heat of gas of this composition between and 1100 per cubic meter, and by the temperature, will give the heat thus carried out of the producer: 2 (X-r-0.81)X0.347XllOO = 942.5 X Calories (g) 9000X8 = 72,000 Calories. The sum total of (e) + (f) -f (g) gives the total heat distribu- tion, viz.: 4.169.5 X + 72,000 Calories. Since the heat available equals the heat distributed: 1.978.6 X-f 712,500 = 4,169.5 X 4- 72,000 or X = 292 kilograms (1) (2) The volume of this steam, at assumed standard condi- tions, would be 292 H- 0.81 = 360 cubic meters, and of gas, since it is 50 per cent, hydrogen, 360X2 = 720 cubic meters. 182 METALLURGICAL CALCULATIONS. The above figures represent the maximum attainable pro- duction, on the assumption that sufficient steam-generating power is available to furnish the steam, and that the fuel in the producer is of small size and the bed so uniform that the production of gas is regular all over it. In practice, figures con- siderably below this are attained, but it is always well to know the possible maximum which is attainable. Problem 21. In a Dellwick-Fleischer water-gas producer the heating up is accompliched in 2 minutes by blast from a Root blower, fur- nishing air through a 9.5 in h pipe at a total water-gauge pressure of 19 inches of. water, temperature of air 15 C. The gases escaping from the producer analyze: Carbon dioxide 17.9 per cent. Carbon monoxide 1.8 Nitrogen 78.6 " Oxygen 1.7 Temperature of waste gases 900 C. ; heat lost by radiation and conduction 9000 Calories per minute; assume producer to contain 3000 kilos, of fuel, consisting of 90 per cent, carbon and 10 per cent. ash. Required: (1) The average rise in temperature of the fuel bed. (2) The proportion of the heat generated which is thus stored up as useful heat for producing water-gas. (3) Assuming that the production of water-gas lasts 8 min- utes, during which 2500 Calories are absorbed from the fuel bed per kilogram of steam used, what should be the steam supply in kilograms per minute? (4) The ratio between the volume of air supplied during the blowing up period and the weight of steam used in the gas- making period. Solution: (1) We imist first find the amount of air fur- nished by the blower. To do this, we calculate the pressure head in terms of air at 15 C., instead of water, and then apply the well-known formula V V 2g. h. "^ater is 772 times as heavy as air at C., and, therefore, 1^ inches of water pres- sure would represent 19X772+12 = 1222 fee' of air pressure (that is, a column of fluid as light as air, 1222 feet high). But ARTIFICIAL FURNACE GAS. 183 air at 15 is lighter still than air at 0, in the ratio 273 to 288. so that the air pressure measured in terms of air at 15 will be = 1290 feet The velocity of the air supplied will therefore be, in feet per second : _ V =V 64X1290 = 287 feet per second, and the volume delivered, per minute, at 15 C.: 287 X 60 X (0.7854 X 9.5 X ( .5 -5- 144) cubic feet = 17,220X0.492 = 8472 cubic feet, which in terms of air at C., would be 97*3 847 X Sss = 8050 cubic feet Zoo = 228 cubic meters The volume of waste gases produced in the 2 minutes can be found from the relative percentages of nitrogen in the air (79. 2^ and in the gases (78.6), as follows: 79 9 228X2X;r~-^ = 459.5 cubic meters /o. o Containing, therefore, from its analysis: Carbon dioxide .............. 82.25 cubic meters Carbon monoxide ............ 8.27 " " Oxygen ..................... 7.81 " Nitrogen .................... 361.15 " 459.48 The carbon burnt to CO 2 and CO will be: C to CO 2 82.25X0.54 - 44.42 kilos. CtoCO 8.27X0.54= 4.47 " Sum = 48.89 " And the heat thus generated: C to CO 2 - 44.42X8100 = 359,800 Calories CtoCO = 4.47X2430 =_JA860 Sum = 370,660 184 METALLURGICAL CALCULATIONS. To find the amount of this heat left in the producer at the end of the 2 minutes blowing up, we must subtract the 2X 9000 = 18,000 Calories lost by radiation and conduction, and then, in addition, the heat carried out by the hot gases, at an average temperature of 900 C., which latter will be: CO, O 2 , N 2 377.25X0.327X885 = 109,150 Calories CO 2 82.25X0.571X885 = 41,565 Sum = 150,715 * * Heat left in the fuel bed: 370,660168,715 = 201,945 Calories heat capacity of the fuel bed per 1 C.: 3000X0.9 kilos, carbon X0.5 = 1370 Calories 3QOOX0.1 kilos, ash X0.25 = 75 Sum = 1445 Average rise in temperature of the fuel bed. uo-c. a) (2) The useful heat thus stored up in the fuel bed amounts to the following proportion of the total heat generated during the blowing up: = 0.545 = 54.5 per cent. (2) (3) Steam which can be decomposed in the gas-producing period: = 80.8 kilograms = 10.1 kilograms per minute (3) (4) Air supplied in 2 minutes = 456 cubic meters. Steam used in 8 minutes = 80.8 kilograms. 80 8 Ratio = -TVTT = 0.177 kilos, steam per 1 m 8 air (4) 4ob = 0.177 ounce steam per 1 ft 3 air 1.1 pounds steam per 100 ft 3 air CHAPTER VII. CHIMNEY DRAFT AND FORCED DRAFT. In all problems concerning combustion, we must furnish the air needed for combustion either by suction or by pressure. The original and almost universal method is by chimney draft; the more positive and reliable method is forced draft. Often the two are combined with very satisfactory results. The waste heat from any metallurgical process or furnace is generally considerable. Most furnaces must be kept above a red heat, and the gases pass directly out of the furnace into the chimney. In such cases the chimney is indicated as the proper source of draft, because it utilizes, although very in- efficiently, the ascensive force of the hot gases, and thus works by otherwise wasted energy. In other cases it is practicable to pass the gases through boilers before they go to the chim- ney, and thus to raise large amounts of steam. The gases are then cooled down so far that they enter the chimney too cold to furnish all the draft needed; in such cases a small fraction of the steam generated will run a steam engine or steam tur- bine, and run a fan capable of furnishing all the draft needed. In this manner considerable steam is available for other pur- poses, and great economy is effected. CHIMNEY DRAFT. The principles involved are not obscure or complicated. The total pull, or suction, which a chimney can produce, as- suming it to be filled with hot air, is simply due to the ascen- sive force of the hot air inside, and the measure of this is the difference of weight of the chimney full of hot gases and what it would be if filled with cold air of the temperature outside. Illustration: A chimney is 6 feet square inside and 100 feet high, uniform, with the gases inside at an average tempera - ,ture of 500 F., and specific gravity (air = 1) of 1.06. The 185 186 METALLURGICAL CALCULATIONS. air outside is at 80 F. What is the ascensive force of the hot gas inside, in total pounds, in ounces per square inch and inches of water gauge ? The volume of the space in the chimney chimney volume is 100X6X6 = 3600 cubic feet. This volume, filled with air at 32 F., would weigh 3600X1.293 = 4654.8 oz. av. = 290.9 pounds. And, filled with gas at 500 F., 491 290.9X1.06X 500 _ 32 + 491 = 157.9 pounds. If filled with outside air, at 80 F., the weight would be 4Q1 We, therefore, see that the hot gases in the chimney, weigh- ing 157.9 pounds, displace 265.0 pounds of cold air, and the tendency of the former to rise upwards in this ocean of air must be 265.0157.9 = 107.1 pounds. To put it in another way, if a piston fitted into the chimney at the bottom, and could move without friction, the piston would have to be loaded with 107.1 pounds to keep it from moving up the chimney. The total upward pull of the chimney is therefore 107.1 pounds. Since this would be exerted on a piston 6X6 = 36 square feet in area, the pull or suction per square foot, in pounds, is 107.1-7-36 = 2.98 pounds, and in ounces per square inch. (2.98-5-144) X 16 = 0.331 ounce per square inch. If the pull or suction is measured on a gauge, as by water pressure, the pressure, of a 1-foot column of water at ordi- nary temperatures is 1000 ounces per square foot, or 1 inch of water is ( 1 000 -T- 144) -7- 12 = 0.597 ounces per square inch. The total pull of the chimney is therefore equivalent to ' = 0.57 inch of water gauge. U . U4J7 CHIMNEY DRAFT AND FORCED DRAFT. 187 By exactly similar methods of calculation the theoretical total suction of a chimney of any given height and temperature of gases inside and of air outside may be obtained. The suction expressed in ounces per square inch, or in water gauge is, of course, independent of the cross-sectional area of the chimney; it depends only on its height and on the temperatures inside and outside. The above calculated total suction (allowing nothing for friction, etc.) is called the total head of the chimney, and is usually expressed in terms of cold air (at C.) instead of in water. Cold air is a fluid, and water is 772 times as heavy as it; therefore, a gauge pressure or hydrostatic head of 0.57 inch of water is the same as 0.57X772 = 440 inches of air. = 36.5 feet of air. What this head really represents is clearly seen from the above calculation. Its value is to be obtained directly from the height of the chimney, temperature inside and out and specific gravity of the chimney gases (air = 1) by the fol- lowing relations, in which h = total head in feet of air at 32 F. h = total head in meters of air at C. t = temperature in the chimney, F. t = temperature in the chimney, C. t' = temperature of outside air, F. t' = temperature of outside air, C. D = Specific gravity of chimney gases, air = 1. H = Height of chimney in feet. H = Height of chimney in meters. coef = coefficient of gaseous expansion, F coef = Coefficient of gaseous expansion, C = -^=^ r(i l Ql +coef (t f 32)] [1+coef (t 32) r(l D + coef (t DtT| le ~ (l + at')(l+crt) J 188 METALLURGICAL CALCULATIONS. The author is not fond of using formulas whenever their use can be avoided. The above formulas express in the sim- plest mathematical form the principles which have been so far explained and used in the calculations, but it is strongly urged that the formulas be kept " for exhibition purposes only," and that when any specific case is to be worked it be attacked from the standpoint of the principles involved, as explained in the case worked. In other words, if one understands properly and thoroughly the basic principles, he has no need of the formula; if one does not understand the principles, the formula had better be kept forever in " innocuous desuetude." The total head, obtained as above, is the theoretical head. It is like the pressure on the piston of a locomotive the total available force for all purposes. Just as the pressure on the locomotive piston is used up in friction in the engine and in moving the engine itself, and the residue is the available pull on the draw-bar which moves the train, so the total head of the chimney is partly used up in friction in the chimney itself, partly in giving velocity to the gases as they pass out of the chimney, and the residue is the available head which draws or pulls the gases through fire-grates, furnaces and flues up to the base of the chimney. If the chimney could be momen- tarily completely closed at the bottom, except for the gauge opening, and the gas inside be brought to rest, the gauge would show the total head ; as soon as dampers are opened connecting the flues, gas moves up the chimney, and the gauge pressure is lessened by the head required to move the gases and that absorbed in the friction in the chimney. Head Represented in Velocity of Issuing Gases. This item always exists .when the chimney is working, and depends only on the velocity of the gases as they escape and their tempera- ture. The hydraulic head necessary to give any fluid a velocity V is simply the same as the height which a falling body must fall in order to acquire that same velocity; i.e.'. in which expression g is the constant acceleration of gravity, 9.8 meters or 32.2 feet, and the velocity is in meters or feet per second. If we know, therefore, the velocity of the gases CHIMNEY DRAFT AND FORCED DRAFT. 189 issuing from the chimney, or can calculate or assume it, we can get h. In practice the velocity does not vary within very wide limits. In small house chimneys it may not exceed 3 feet per second, in boiler chimneys 6 to 12 feet per second, in fur- nace chimneys 12 to 20 feet per second. The temperatures of these issuing gases is, moreover C F In small chimneys ........ 100 to 200 200 350 In boiler chimneys ........ 100 to 300 200 550 In furnace chimneys ...... 300 to 1000 550 18QO If the chimney in question has, therefore, a known velocity of exit of its gases, h can be calculated; but it must not be forgotten that h will be in terms of the kind of gases which is escaping; i.e., of hot gas, and to subtract it from or com- pare it with h we must reduce it to its equivalent head in terms of cold gas. This is merely a matter of taking into account the specific gravities or relative densities of hot gas and cold air, which are inversely proportional to their absolute temperatures; that is, if D represents the relative density of air and chimney gases at the same temperature: V, vel. V, V h hx i+t" Illustration: Assuming the actual velocity of the gases is- suing from a furnace chimney to be 15 feet per second, and their temperature 500 F., density 1.06 (air = 1), what will be the head represented by the velocity of these gases in terms of cold air at 32 F. ? The head represented, in terms of hot gases at 500 F., is In terms of air at 500 F. is 3.5X1.06 = 3.71 feet. And in terms of air at 32 F., 4Q1 190 METALLURGICAL CALCULATIONS. Out of the total head which this chimney produces (say 36.5 feet), 1.9 feet is represented by the velocity of the issuing gases, or 5.2 per cent, of the whole, leaving 34.6 feet to repre- sent loss by friction in the chimney and the available head. We will proceed to discuss the loss of head due to friction in the chimney. Head Lost in Friction in the Chimney. This varies with the smoothness or roughness of the walls, and has been deter- mined experimentally for air moving with different velocities. ' The manner of expressing the friction loss is, to put it as a function of the head necessary to give the gases their actual velocity, assuming there were no friction. Thus, supposing as in the preceding paragraph, the actual velocity of the hot gases is 15 feet per second, and the head (in terms of cold air) necessary to give that velocity, not considering friction, is 1.9 feet, then the head lost in friction in getting up this velocity will be h friction = i.gx-^K. d That is, it will be proportional to H, the height of the chimney, inversely as d, the diameter or side, if square, and to a co- efficient K, determined by experiment. The latter varies, ac- cording to Grashof s experiments, between 0.05 for a smooth interior to 0.12 for a rough one, and averages 0.08. Illustrations'. Assuming the height of the chimney, 100 feet, its section to be 6 feet square, the coefficient of friction K = 0.08, and the head represented by the net velocity of the hot gases in the chimney to be 1.9 feet of cold air, what is the head lost in friction in the chimney? The ratio of height to side is 100-7-6 = 16.67, which mul- tiplied by K gives 1.33 as the value of the function containing these three terms. This means that 1.33 times as much head has been lost in friction as is represented by the net actual velocity of the gases as they pass up the chimney. Therefore, h friction = 1.9X1.33 = 2.5 feet cold air. Another way of looking at this, which is sometimes useful in considering the height of a chimney, is to say h riction = 0.025 H. CHIMNEY DRAFT AND FORCED DRAFT. 191 Or, that in this case, the head lost in friction amounts numer- ically to one-fortieth the height of the chimney. If we subtract the head lost in friction plus that represented in the net velocity of the gases, from the total gross head, the residue is that available for doing work external to the chimney. In the specific case of the preceding illustrations we have h = total head = 36.5 feet = 100 per cent, h velocity = velocity hea'd = 1.9 " = ~~5 k friction = friction in chimney = 2.5 " = 7 h available = available head =32.1 " = 88 Available Head of a Chimney. This is the part of the total head which remains after subtracting the head lost in friction in the chimney and that represented by the velocity of the issuing gases. In the specific cases considered in the above illustrations, the net available head amounted to 88 per cent, of the whole theoretical head. If we assume limiting con- ditions as found in practice, we can find the limiting values of this proportion. Calling the cases I and II, those with mini- mum and maximum absorption of head in the chimney itself, we have Case I. Case II. Temperature of issuing gases 100 C. 1000 C. Velocity of issuing gases per second 1 meter 7 meters Ratio H to d 10 50 Coefficient K 0.05 0.12 Specific gravity of gases (air = 1) 1.00 1.06 Head as velocity of gases (meters of air). .0.04m. 0.56m. Head as velocity of gases (feet of air) 0.13 1.87 Head absorbed in friction (meters of air) . .0.02 3.36 Head absorbed in friction (feet of air) 0.07 11.2 Head used up in chimney (meters) 0.06 to 3.92 Head used up in chimney (feet) . 0.20 to 13.0 Water gauge pressure thus lost, m.m 0.1 to 5.0 Water gauge pressure thus lost, inches 0.003 to 0.2 The available head, will, therefore, be the theoretical total head minus a loss in the chimney itself, which may amount to a maximum of 3.9 meters or 13 feet, representing an ab- sorption of water gauge pressure up to 5 millimeters, or 0.2 192 METALLURGICAL CALCULATIONS. inch at a maximum. Under ordinary conditions half these quantities would be a rather high chimney loss. In most conditions which confront the metallurgist, the question is to determine how high a chimney should be built in order to supply a certain available draft determined by practice to be necessary. For instance, to burn a certain amount of coal per hour on any grate requires a certain amount of draft. This amount is increased if the draft is increased, and vice versa. In boilers, 18 pounds of coal burned per square foot of grate surface per hour is highly economical practice, and requires a draft of 0.4 inch to 0.8 inch of water gauge, according to the kind of coal 'burned. In furnaces where the amount of coal burned is greater per hour there will be usually a correspondingly greater temperature in the chimney. To calculate the height of chimney required it is necessary to assume only the temperature in the chimney > the available draft re- quired and an average chimney loss. Problem 22. It is desired to design a chimney for a puddling furnace, the grate of which is 4 feet by 6 feet, and which shall burn 30 pounds of bituminous coal per hour per square foot of grate surface. Temperature of gases entering the chimney 1200 C., at the top probably 1000 C. Specific gravity of gases 1.03 (air = 1). Draft required 0.6 inch of water gauge. Outside temperature 30 C. Solution: We can assume that since the gases will be at an average temperature of 1100 C. in the chimney, their ve- locity will be high, and that at least 0.1 inch of water gauge pressure will be absorbed by the chimney itself. This makes a total requirement of 0.7 inches of water for total head, or h = 0.7X772^12 = 45 feet of cold air. Or an unbalanced pressure or ascensive force of 45X1.293'-*- 16 = 3.64 pounds per square foot. Considering the air outside the chimney, its weight at 30 C. is equal, per cubic foot, 1.293 X -16 = 0.073 pounds. oUo CHIMNEY DRAFT AND FORCED DRAFT. 193 The gases inside the chimney weigh, per cubic foot, 273 1.293Xl.03X n()Q + 273 ^16 = 0.0166 pounds. The height of the chimney being called H and its cross- section S, the volume is HxS, and the weight of hot gas inside it is (HXS)X 0.0166 pounds. And of an equal volume of cold air outside (H X S) X 0.073 pounds, giving a total ascensive force of (H X S) X 0.0564 pounds. But there is needed a total ascensive force of SX3.64 pounds, in order to give the pull of 3.64 pounds per square foot, and, therefore, of necessity, HXSX 0.0564 = SX3.64, from which Concerning the cross-section of this chimney, it would not be safe to make it less in diameter than one-fiftieth of its height, because of lack of stability; in fact, one -twenty-fifth would be better practice. This consideration would make its in- ternal diameter 2 ft. 7 inches, area 5.2 square feet. Another way of arriving at a diameter is to calculate the volume of the hot gases which must pass up the chimney and assume for them some maximum velocity in the chimney, such as, let us say, 6 meters (20 feet) per second, and so get the minimum area necessary for filling this condition as follows: Coal burnt per hour 4 X 6 X 30 = 720 pounds. Air theoretically necessary, assuming aver- age bituminous coal (see Prob. 1) = 123 X 720 = 88,560 cubic feet. 194 METALLURGICAL CALCULATIONS. Products of combustion at standard condi- tions = 129X720 = 92,880 cubic teet Volume chimney gases at 1100 C. = . 467 , 100 -. Volume per second = 130 Area of chimney, if maximum velocity is 20 feet per second = 6 . 5 sq. feet Diameter, if round = 2 ft. 10 in. This chimney would do its work better, and there would be much less loss in friction, if the internal diameter were made 25 per cent, greater than the above calculated minimum, say, there- fore, 3 feet 6 inches, making the area nearly 50 per cent, greater and cutting down the velocity in the chimney to 13.5 feet per second. Problem 23. In the case of the puddling furnace of Problem 22, assume that the hot gases, instead of going directly into the chimney, are passed through the flues of a boiler placed above the fur- nace, and thence pass into the chimney at a point 15 feet higher than before. Assume chimney 3 feet 6 inches internal diameter, 64.5 feet high above the furnace flue, and that the gases now passing into it 15 feet higher up are at 350 C., and cool to 250 C. at the top of the chimney. The boiler flues in- troduce additional frictional resistance equal to 0.1 inch of water. The boiler raises steam at a net efficiency of 45 per cent., the steam engine utilizes the steam at a mechanical efficiency of 20 per cent., and a centrifugal fan supplies the forced draft needed at a mechanical efficiency of 25 per cent. Required: (1) The total head of the chimney, when the furnace discharged directly into it, and the average tempera- ture of the gases in it was 1100 C., and specific gravity 1.03 (air = 1). (2) The head absorbed as velocity of the outgoing gases, their temperature being 1000 C. (3) The head lost in friction in the chimney, in this case. (4) The head which was available to run the puddling furnace. (5) The total head of the chimney with the gases entering 15 feet above former flue, and average temperature 300 C. CHIMNEY DRAFT AND FORCED DRAFT 195 (6) The head absorbed in this case as velocity of outgoing gases, their temperature being 250 C. (7) The head lost in friction in the chimney in this case. (8) The available head to draw gases into the chimney. (9) The deficit of head which must be made up by forced blast under the grate of puddling furnace. (10) The horse-power absorbed by the fan which furnishes this blast. (11) The horse-power furnished by the engine using the steam from the boiler. (12) The excess of power which is thus saved and available for other purposes. Solution : (1) Volume of gases in chimney: 64.5X3.5X3.5X0.7854 = 620.5 cu. ft. Weight at 32 F. (0 C.): 620.5 X (1.293 4- 16) X 1.03 = 51.65 Ibs. Weight if temperature is 1100 C.: 070 Weight of equal volume of air outside at 30 C.: 070 620. 5X (1.293-;- 16) X on , * = 45.18 Ibs oU -T Z< o Difference of weight = ascensive force, 45.1810.27 = 34.91 Ibs. Ascensive force per square foot, 34.91 -v- 9.62 = 3.63 Ibs. Total head in terms of cold air at C., 3.63 -T- (1.293 H- 16) = 44.9ft. (1) In terms of water gauge pressure, 44.9X12-772 = 0.685 ins. (1) 196 METALLURGICAL CALCULATIONS. (2) Volume of gases per hour at C., (Prob. 22) = 92,880 cu. ft. Volume at 1000 C., = 92,880 X 10 1: 273 = 433,100 cu. ft. Velocity per second, 433, 100 -h (3600) -9, 62 = 12.50ft. Head necessary to give this velocity, in terms of hot gases, at 1000 = (12.50) 2 -^-64.3 (2g) = 2.43ft. In terms of gases at C., In terms of air at C., 0.52X1.03 = 0.55ft. (2) In terms of water gauge pressure, 0.55 X 12 -J- 772 = 0.008 in. (2) (3) Assuming K, the coefficient of friction, 0.08, then h friction = __ H 2g 273 +t d This is only an abbreviated form of the operations done under (2), adding the terms which account for the height, diameter and friction. Now, the velocity per second: i inn -4- 97*} V = 92,880 x"Jl -H 3600-^9.62 = 13.5ft. Head necessary to give this velocity in terms of air at 0, 070 a3.5)'H-64.3Xg ?gTn55 Xl.03 = 0.58 ft. Proportion of this velocity head lost in friction = K - ^XO.08 - 1.47 ' a 3.5 CHIMNEY DRAFT AND FORCED DRAFT. 197 / Head lost in friction in chimney, 0.58X1.47 = 0.85ft. (3) In terms of water gauge pressure, 0.85X12*772 = 0.013 in. (3) (4) Cold Air. Water Gauge. Total head ....................... 44.90 feet 0.685 inch Absorbed in velocity of gases. ...... 0.55 " 0.008 " Absorbed in friction in chimney .... 0.85 " 0.013 " Available for the furnace ........... 43.50 " "O664 " (4) (5) Volume of chimney gases, (64.5 15) X 3.5X3.5X0.7854 = 475.7 cu. ft. Weight at 300 C., specific gravity 1.03 (air = 1). 07Q 475 . 7 X (1 . 293 * 16) X 1 03 X 273 + 300 = 18.86 Ibs. Weight of equal volume of outside air at 30 C., 273 475. 7X (1.293* 16) X 273 30 = 34.68 Ibs. Ascensive force of air per square foot, (34.68 18.86) -9.62 = 1.64 Ibs. Total head in terms of cold air, 1.64* (1.293-5-16) = 20.3 ft. (5) In terms of water gauge pressure, 20.3X12*772 = 0.32 in. (5) (6) Velocity of issuing gases, per second, at 250 C., OCQl 070 92,880 X I"' -5-3,600*9.62 = 5.14ft Head as velocity in terms of cold air at C., (5. 14) 2 * 64 3X^X1.03 = 0.22ft, In terms of water gauge pressure = 0.003 in. (7) Average velocity of gases in chimney at 300 C., onn I 970 92,880 * 3,600 X- 1T *9.62.= 5.63ft. AIO 198 METALLURGICAL CALCULATIONS. \ Head lost in friction in terms of cold air, 27S 4Q 5 (5.63) 2 -^54.3X 3QO | 273 Xl.03X-^yX0.08 = 0.27ft. (7) In terms of water gauge pressure = 0.004 in. (7) (8) Cold Air. Water Gauge. Total head .................... . . .20.30 feet 0.320 inch Absorbed in velocity of gases ....... 0.22 " 0.003 " Absorbed in friction in chimney ---- 0.27 " 0.004 " Available to draw gases in, ........ 19.81 " 0.313 " (8) (9) Available head needed for puddling (4)... ......................... 43.50 " 0.664 " Available head needed for boiler. . . . 6.43 " 0.100 " Total head needed for both ........ 49.93 " - 0.764 " Available head from chimney (8). . .19:81 " 0.313 " Deficit, to be supplied by blast ..... 30.12 " 0.451 " (10) The 0.451 inches of water gauge equals (0.451 -5-1 2) X 62.5 = 2.35 Ibs. per sq. ft. The volume of air to be supplied is, at 30 C., 88,560 (Prob. 22)^60X^|^^ = 1,638 cu. ft. per min. Net work dpne by the fan, 1,638X2.35 = 3,850 ft. Ibs. per min Gross power needed by the fan, 3,850-7-0.25 (efficiency) = 15,400 ft. Ib. per min. Horse-power needed to drive the fan, 15,400-^-33,000 = 0.47 H. P. (10) (11) The boiler receives the gases at 1200 C. and discharges them at 350, and 92.880 cubic feet of gases (measured at standard conditions) pass through per hour. The composition of these gases is not given, but from the specific gravity we might conclude that they contain on an average 10 per cent, of carbon dioxide, since if they contained the maximum amount CHIMNEY DRAFT AND FORCED DRAFT. 199 of that gas (about 20 per cent.) their specific gravity would be 1.06 (air=l). Assuming them, therefore, to contain CO 2 10 per cent. H 2 O . 10 " " CO, N 2 , O 2 80 their heat capacity per degree per cubic foot would be, between 350 and 1200. CO 2 0.1X[0.37 +0.00022 (350 + 1200)] = 0.0711 oz. cal. " H 2 O 0. IX [0.34 +0.00015 (350 + 1200)] = 0.0573 " CO, N 2 , O 2 0.8X[0.303 + 0.000027 (350 + 1200)] = 0.2759 " Sum = 0.4043 Heat given up per cubic foot, 0.4043 X (1200 350) = 343.7 oz. cal. Heat given up by gases per hour to boiler, 92,880X343.7 = 31,922,850 oz. cal. 1,995,000 Ib. cal. Heat in the steam produced per hour, 1,995,000X0.45 (efficiency) = 897,750 Ib. cal. Heat equivalent of mechanical energy of steam engine per hour, 897,750X0.20 (efficiency) = 179,950 Ib. cal. Heat equivalent of 1 hp hour = 635 kg. cal. = 1,400 Ib. cal. Horse-power generated by the engine, 179,950-5-1,400 = 128 H. P. (11) (12) Net available power after supplying fan, 1280.5 = 127.5 H. P. (12) CHAPTER VIII. CONDUCTION AND RADIATION OF HEAT. These two factors are of the greatest practical importance to the metallurgist, yet they are also the subjects of all others upon which the practical metallurgist is usually the most poorly informed. It often happens that a furnace is built of twice the capacity of its predecessor or of its neighbors, and the manager unexpectedly and most agreeably discovers that it keeps up a more uniform heat and requires considerably less than twice the amount of fuel to keep it running. Two reasons were operative; first, the walls were made thicker to support the heavier structure, but that made them also poorer conduc- tors of heat; second, the capacity increased as the cube of a linear dimension, while the radiating surface increased as its square, so that radiation was, therefore, less than twice the primary amount from the furnace of double capacity. Many practical metallurgists have learned the practical results, but are entirely ignorant of the principles upon which the results are obtained. While it is well to be successful, it is better to be intelligently so. PRINCIPLES OF HEAT CONDUCTION. It is well known that the ability of metals to conduct heat is very nearly the same as their ability to conduct electricity. The order of metals in these two series is almost identical. Further, the specific conductance for heat is closely analogous to specific conductance for electricity. Just as we say that the electrical resistance of a conductor is proportional to its length and inversely as its cross-section, so we can make the same statement concerning heat resistance. Just as we can add electrical resistances when the bodies are in series, and must add electrical conductances when they are in parallel, so we 200 CONDUCTION AND RADIATION OF HEAT. 201 can add heat resistances when the bodies are in line, and must add heat conductances when they are hooked up together in parallel. The unit of electrical resistance is an ohm, and a substance has unit resistivity when a cube of it, 1 centimeter on a side, has that resistance. In such a case a drop of potential of 1 volt from one side to the opposite one sends through the cube 1 coulomb of electricity per second. Since conductivity is merely the reciprocal quality, actually and mathematically, to resistivity, the unit of conductivity is defined in exactly the same way, and the conductivity of any substance may be ex- pressed as so many reciprocal ohms. If we read in the preceding paragraph, thermal resistance for electrical resistance, degrees temperature for volts, and gram calories for coulombs, we have defined the corresponding unit of thermal resistivity. A substance has unit capacity for conducting heat when a cube 1 centimeter on a side trans- mits 1 gram calorie of heat per second, with a drop of tem- perature from one surface to the other of 1 C. Many investi- gators have adopted slightly different units, such as 1-kilo- gram Calorie per cubic meter per hour per degree difference; but such are only simple multiples or fractions of the centi- meter-gram-second unit above denned. Illustration'. If the thermal conductivity of copper is 0.92 units, what would be the amount of heat passing per hour through a sheet 1 millimeter thick and 1 meter square, with a constant difference of 1 C. between the two sides? Solution: The 0.92 units means that a column of copper 1 centimeter long and 1 square centimeter in cross-section, hav- ing a difference of temperature at its two surfaces of 1 C., would allow 0.92 gram-calories of heat to flow through it per second. For a sheet one-tenth as thick, 10,000 times the area and during 1 hour, the heat passing per 1 C. difference will be 0.92XyX 10 ^ X360Q = 331,200,000 cal. = 331, 200 Cal. Such calculations as the above are, of course, very useful for comparing different thicknesses of plates of different material, as to their relative heat-carrying capacity. With some slight 202 METALLURGICAL CALCULATIONS. modifications they are applicable to the actual conveyance of heat through such walls, from a fluid on one side to a fluid on the other. This introduces an idea analogous to transfer or contact resistance in electrolytic conduction. Thus, if we call R the thermal specific resistance (resistivity) in C. G. S. units, of the material of a partition having a thickness of d centi- meters, and an area of S sqtiare centimeters, then the thermal resistance of the body of the partition will be R X d and its thermal conductance -- =k R X d Besides this resistance in the body of the material there is transfer resistance at its two sides, or at its inner and outer surfaces, which depends on the materials which communicate heat and receive heat, and their velocity. These resistances may be expressed as so many units per square centimeter, the unit being of exactly the same nature as R, except that it connotes no thickness but merely a surface effect. Calling R 1 the specific resistance of transfer from the inner fluid to the inner surface of the material, and R 2 the outside specific trans- fer resistance to the fluid outside, we can consider all three resistances as being in series, and the total thermal resistance to transfer from the inside fluid to the outside fluid to be R X d R 1 R 2 or the thermal conductance of the system (RX<*)-fR' The following table gives the thermal resistivity of various materials, in C. G. S. gram-Calorie units, also the thermal conductivity in units which are reciprocals of the resistance units: CONDUCTION AND RADIATION OF HEAT. 203 Material. Conductivity (in reciprocal resistance units . ) Silver (at 0) 1.10 Copper (0 30). . '. 0.92 Copper, commercial 0.82 Copper, phosphorized 0. 72 Magnesium . 38 Aluminium (0) 0.34 Aluminium (100) 0.36 Zinc (15) 0.30 Brass, yellow (0) 0.20 Brass, yellow (100) 0. 25 Brass, red (0). 0.25 Brass, red (100) 0.28 Cadmium (0). , 0.20 Tin (15) 0.15 Iron, wrought (0), 0.21 Iron, wrought (100). 0. 16 Iron, wrought (200). C. 14 Iron, steel, soft 0.11 Iron, steel, hard 0.06 German silver (0) 0.07 German silver (100) 0.09 Lead (0) 084 Lead (100) 0.076 Antimony (0) ". . . . 0.044 Mercury (0)., 0.015 Mercury (50) 0.019 Mercury (100) 0.024 Bismuth (0) 0.018 Wood's alloy (7) 0.032 Allov, 1 Sn. 99 Bi. . 0.008 Resistivity (inC.G S. gram-cal. units.} 0.91 09 22 39 63 2 94 2.75 33 00 00 00 3.57 5.00 6.67 4.76 6.25 7.14 9.01 16.67 14.28 11.11 11.90 13.16 22.73 66.67 52.63 41.67 55.55 31.25 125.00 The last alloy, bismuth, with only 1 per cent, of tin, is -re- markable for its extremely low heat conductivity, less than one hundredth as good as copper. It has also the lowest electrical conductivity of any alloy, about one two hundredth that of copper. These peculiar properties ought to make it of use for some particular purposes. The above data enables one to calculate the rate at which heat 204 METALLURGICAL CALCULATIONS. will pass through a metallic partition or wall when the tempera- ture of its two surfaces is known. This is a very useful calcu- lation, for in many cases the temperature inside a partition or pipe is known, or can be easily determined, and when the tem- perature of the- outside surface is determined, the calculation can be made. The temperature of the outside of a partition or pipe can be found by several methods: one is to lay a very flat bulb thermometer (made especially for this purpose) against it; another is to put the junction of a thermocouple against it, covering the couple with a little putty or clay; another is to take small pieces of metals, or alloys of known melting points, and see which melts against the hot metal. The only uncer- tainty then in the calculation is the question as to what differ- ence in the conductivity may be caused by the higher tempera- ture, The conductivities in most cases decrease as tempera- ture rises, but in others increase. There is here a large field for metallurgical experiment, in determining the heat con- ductivities of metals, alloys and fire-resisting materials at high temperature. Illustration'. An iron pipe, 6 centimeters in diameter out- side and 5 centimeters inside, is filled with water at 10 C., and surrounded by hot gases at 198 C. From the rise in tempera- ture of the water it is known that for each square centimeter of heating surface 0.084-gram calories of heat pass per second. Assuming the thermal conductivity of the iron (k) = 0.14, what is the difference of temperature of the two surfaces, inside and outside, of the pipe?- Solution: The thickness of the walls is (65)^2 = 0.5 centimeter. If the walls were 1 centimeter thick, 1 difference would transmit 0.14 calorie; but being only half that thick, 1 difference would transmit 0.28-gram calorie. The actual dif- ference of temperature of the two surfaces must then be Of course, such calculations can be turned around, and if the temperature of the inside and outside surfaces is known, the heat being transmitted can be calculated; or if the temperature of these two surfaces is known and the heat being transmitted is measured, the thermal conductivity of the partition can be CONDUCTION AND RADIATION OF HEAT. 205 reckoned; or, again, if the temperature of the two surfaces is known, and also the thermal conductivity of the partition, and the temperature of either fluid on either side, the thermal re- sistance of the transfer from either of the fluids to the sur- face of the partition can be calculated. Illustration', In the previous illustration the temperature of the hot gases was 198, while that of the pipe in contact with them was practically 10. What was the thermal resistivity of the transfer from gas to metal? Solution-. The thermal resistivity is the reciprocal of the thermal conductivity, and in the case of this transfer resistance is the reciprocal of the number of gram calories which will be transferred to 1 square centimeter of pipe surface from the hot gases per second per each degree centigrade of difference of temperature causing the flow. This is a surface or skin resistance, and, therefore, no linear dimension representing thickness enters into the calculation. There is 0.084 calorie per second being transferred to each square centimeter of pipe, with a difference of temperature acting as propelling force of 19810 = 188. The thermal conductivity of this transfer, k, is, therefore, 0.084 -M 88 - 0.00045 And the thermal resistivity, R, the reciprocal, viz.: 2222. Since the thermal resistivity of the iron wall is l-r-0.28 = 3.57 units, it follows that the contact surface offers 2222^-3.57 = 622 times as much resistance to the flow of heat as the metal itself. In this specific instance, we can conclude that the transfer of heat from the gases to the pipe is the principal item which conditions the flow of heat, the passage of the heat through the wall of the pipe itself taking place 622 times as readily. The valuable practical conclusion is that the thermal re- sistance of the walls of the pipe is insignificant as compared with the whole thermal resistance, and the thickening or thin- ning of the walls of the pipe, or the substitution of copper for iron, because of its greater thermal conductivity, is prac- tically unnecessary for thermal considerations, since such can be expected to make practically no change in the thermal re- sistance of the whole system. If the pipe under discussion, however, acquires during use a 206 METALLURGICAL CALCULATIONS. layer of scale deposited from the water, then the thermal re- sistance of the pipe or of the system is materially affected. A study of the thermal resistivity of various boiler scales would be a very useful and practical subject, but has not been done, as far as the writer is aware. Assuming a deposit, 0.5 centimeter thick, of material having the thermal conductivity of plaster of paris, for which k = 0.0013, the specific resistance of this ma- terial is 0.14-J-0.0013 = 108 times that of iron, and therefore 0.5 centimeter of this represents 54 centimeters thickness of iron. The thermal resistances in this case, neglecting that of transfer from the water to the pipe or scale, are as follows: Resistance of transfer, gases to pipe = 2222.0 = 85 per cent. Resistance of 0.5 c.m. iron = . ' . = 3.6 =0 " 014 Resistance of 0.5 c.m. scale = \f = 384.4 = 15 U . UUlo Total = 2610.0 It thus appears that a deposit of scale from the water to a thickness of 5 millimeters increases the total thermal resistance of the system some 18 per cent, of its original amount, and would cut down the efficiency of this part of the heating surface of a boiler by this amount. These considerations are not only of vital importance to the steam boiler engineer, but they are the essential principles which condition the efficiency of steam-heating apparatus, feed-water heaters, hot-air stoves and ovens, air-cooling of parts of furnaces and the efficiency of water jackets. PRINCIPLES OF HEAT TRANSFER. We have already had to speak of the transfer of heat from fluids to solids, or vice versa, and in one specific case we de- duced the value 2222 for the transfer resistivity from hot gases to the surface of iron pipe, meaning thereby that for each degree of temperature difference between the gases and out- side of the pipe 0.00045 gram calorie passed per second through each square centimeter of contact surface. A consideration of the transfer of heat through such contact surfaces, from gases or liquids to solids and vice versa, has shown that the CONDUCTION AND RADIATION OF HEAT. 207 transfer resistivity varies with the solid and with the fluid con- cerned, but much more with the latter than with the former, and is very largely dependent upon the circulation of the fluid, that is, upon the rate at which it is renewed, and there- fore upon its velocity. The conductivity or resistivity of such a transfer must, therefore, contain a term which includes the velocity of the fluid. Various tests by physicists have shown the specific conductance (or conductivity of transfer) to vary approximately as the square root of the velocity of the fluid. From metal to air or similar gases, the mean velocity of flow being expressed in centimeters per second, and the other units being square centimeters and gram calories, the transfer resistivity is approximately 36,000 K = - r=. 2 + Vv and the transfer conductivity of the contact k = 0.000028 From hot water to metal the relations are similar, but the conductivity is much better Experiments show values as follows : k = 0.000028 (300 + 180\/v) 36,000 K. = 300 + lSO\/v Illustration: In the preceding case of the iron pipe, calculate the difference of temperature of the water in the pipe and the inner surface of the pipe, assuming the water to be passing through at a velocity of 4 centimeters per second. Using the above given formula, the heat transfer per 1 difference would be 0.000028 (300 + 180\/4)' = 0.0185 calories, and the difference to transfer 0.084 calories per second will be -46 0.0185 The inner surface of the iron pipe will be, therefore, con- tinuously 4.6 higher than the water, and, therefore, at 14.6; 208 METALLURGICAL CALCULATIONS. the outer surface will be continuously 0.3 higher, or prac- tically at 15. Illustration: A steam radiator, surface at about 100 C., caused a current of hot air to rise having a velocity of about 10 centimeters per second, which was insufficient to keep the room warm. An electric fan was set to blow air against the radiator, which it did with a velocity of about 300 c.m. per second, and keeping the room comfortably warm. What were the relative quantities of heat taken from the radiator in the two cases? The relative thermal conductivities of transfer were 2 + VT(j : 2 + X/300 or 5 : 16 Showing over three times as much heat taken away per unit of time in the second instance. This illustration proves the great efficiency which the metal- lurgist may attain in air cooling of exposed surfaces, by blow- ing the air against them instead of merely allowing it to be drawn away by its ascensive force. Problem 24. Dry air of the volume of 33,000 cubic meters per hour passes through an iron pipe exposed to the air, 30 meters long, 1.5 meters inside diameter, thickness of walls 1 centimeter, and lined inside with 5 centimeters of fire-brick. Assume the hot air entering at 1,000, the outside air to be at 3, the coefficient of internal transfer 0.000028 (2 + Vv), the conductivity of the fire-bricks 0.0014, of the iron 0.14, of external transfer 0.000028 (2 + x/v7, the ve l citv f tne w i n< l against the outside 10 kilometers per hour. Required: The temperature of the hot air leaving the tube. Solution: The mean temperature in the tube is the factor which conditions the mean velocity in the tube, and the rate of flow of heat towards the outside. If, therefore, we let t repre- sent that mean temperature, the solution can be stated in the simplest terms. We then have: Volume of air per hour, at 0, = 33,000 cubic m. Volume of air per second, at 0, = 9.167 CONDUCTION AND RADIATION OF HEAT. 209 t + 273 Volume of air per second, at t, = 9.167 _ Z7o Cross-section of inside of tube (1.5 0.1) 2 X0.7854 = 1.54 sq. m. Velocity of air at t, in pipe = 5.95 (1 + at) m. p. s. = 595 (1 + at) c.m. p. s. Coefficient of internal transfer = 0.000028 Coefficient of conductance of 5 c.m. of fire-brick lining = 0.0014 -T- 5 = 0.00028 Coefficient of conductance of 1 c.m. of iron = 0.14-^1.= 0.14 Coefficient of external conductance (v = 278 c.m. per sec.) = 0.000028 (2 + X/278)" = 0.000524 Total fall of temperature of air = 2 (1000 t) End temperature of the air = 1000 2 (1000 t) = 2 t 1000 Mean specific heat of air per cubic meter between 1000 and end temperature = 0.303 + 0.000027 (2t) Heat given out by the air per hour = 33,000X2(1000 t)X (0.303 + 0.000027 (2t) ) = 19,998,00016,434 1 3.564 t a This heat is the quantity transferred through the pipe in kilogram Calories. The outside surface of the pipe may be taken as the conducting surface, because the larger part of the resistance to the flow of heat takes place there. If we desired to be more exact, the mean area of iron, fire-bricks and the inner surface could be each used separately. The outside diameter being 150 centimeters, and the length 30 meters, the outer surface is 1.52X3.1416X30=142.3 square meters = 1,423,000 square centimeters, the total driving force is the inside temperature minus that outside, or t 3, and the total 210 METALLURGICAL CALCULATIONS. thermal resistance is the sum of the four thermal resistances in series, that is, the sum of thermal resistance gas to fire-bricks = (14- a t) thermal resistance of fire-brick lining = 1 0.00028 thermal resistance of iron shell = 1 0.14 thermal resistance of iron to air = 1 0.000524 The reciprocal of the sum of these four resistances is the thermal conductance of the system per square centimeter, which, multiplied by the conducting surface, 1,423,000 square c.m., and by the total difference of temperature, t 3, gives the heat transferred per second in gram calories. We, therefore, have the final equality expressed as the heat given up by the air, per second, equals the heat transmitted to the outside air per second; i.e., (19,998,000 16.434 1 3.564 t z ) =- 0.000028 (2 + V595 (l + t) 0.00028 0.14 0.000524 XI, 423,000 X (t 3) Whence t = 965.5 And the temperature of the air at the end of the tube is 2t 1000 = 931 (1) It would be interesting to compare the temperatures of the inside and outer surfaces of the tube. The air inside is at a mean temperature of 965, the heat transmitted per second is 0.1574-gram calories per square centimeter of outer surface, CONDUCTION AND RADIATION OF HEAT. 211 whose thermal conductivity is 0.000515, and therefore, the out- side surface must be 0.1574 0.000515 hotter than the surrounding air; that is, must be at 309 C. The extra loss of heat by radiation from this outside surface would be small at that low temperature, but has not been allowed for in the working of the problem. For such metallurgical problems we need the data as to the thermal conductivity or resistivity of ordinary furnace ma- terials. These have been determined in but few instances, and in most cases not at the high temperatures at which they are practically used. A whole series of physico-metallurgical ex- periments is needed just upon this point. The following are probably nearly all that have been determined and the values published. The unit k is the C. G. S.-gram calories unit, the same as used for the metals. k. Ice (datum useful in refrigerating plants, where pipes become coated with ice, as in Gayley's method of drying blast) 0.00500 Snow 0.00050 Glass (10 15) 0.00150 Water 0.00120 Quartz sand (18 98) 0.00060 Carborundum sand (18 98) 0.00050 Silicate enamel (20 98) . .' 0.00040 (Explains the small conductance of enameled iron ware.) Fire-brick, dust (20 98) 0.00028 Retort graphite dust (20 100) . 00040 (Datum useful where articles are packed in this poorly conducting material.) Lime (20 98) 0.00029 (Datum would be highly useful for oxyhydrogen platinum furnaces, if it were only known at high tem- peratures.) Magnesia brick, dust (20 100) . 00050 Magnesia calcined, Grecian, granular (20 100) 0.00045 Masonry 0.0036 to 0.0058 Water, uncirculated 0.0012 to 0.0016 212 METALLURGICAL CALCULATIONS. Magnesia calcined, Styrian, granular (20 100) 0.00034 Magnesia calcined, light, porous (20 100) 0.00016 Infusorial earth (Kieselguhr) (17 98) 0.00013 Infusorial earth (0 650) 0.00038 Clinker, in small grains (0 700) 0.00110 Coarse ordinary brick dust (0 100) 0.00039 Chalk (0 100) 0.00028 Wood ashes (0 100). . . : 0.00017 Powdered charcoal (0 100) 0.00022 Powdered coke (0 100) 0.00044 Gas retort carbon, solid (0 100) .0.01477 Cement (0 700) f 0.00017 Alumina bricks (0 700) 0.00204 Magnesia bricks (0 1300) 0.00620 Fire-bricks (0 1300) 0.00310 Fire-bricks (0 500) 0.00140 Marble, white (0) 0.0017 Pumice 0.0006 Plaster of paris 0.0013 Felt 0.000087 Paper .0.00040 Cotton , 0.000040 Wool 0.000035 Slate 0.00081 Lava 0.00008 Pumice 0.00060 Cork 0.00072 Pine wood ' 0.00047 Oak wood 0.00060 Rubber 0.00047 A study of the above table will in many cases show the metallurgist how conduction of heat can be checked, and to what degree. The substance chosen must be able to stand the temperature without being destroyed; but it is in many cases possible to use one kind of material inside, where the heat is greatest, and a material of much poorer conductivity outside, where the heat will not destroy it. Such compound linings or coverings may be very advantageous. Infusorial earth is one of the very best insulators for moderate temperature; above CONDUCTION AND RADIATION OF HEAT. 213 a bright red it loses efficiency greatly, and is then hardly better than powdered fire-brick. Mr. Irving Langmuir has recently conducted important ex- periments on convection and radiation of heat, recorded in a paper before the American Electrochemical Society (Trans- actions (1913) 23, 299-332). His experimental results for convection alone, from horizontal and vertical surfaces, to still air at 27 C., are as follows, first in calories per second and next in watts: GRAM-CALORIES PER SEC. PER SQ. CM. TO STILL AIR AT 27 C. 100 200 300 400 500 Downward from - from Upward from - Surface Surface - Surface 0.002 0.005 0.011 0.014 0.029 0.032 0.024 0.050 0.055 0.035 0.074 0.081 0.047 0.099 0.108 WATTS PER SQ. CM. TO STILL AIR AT 27 C. 100 200 300 400 500 Downward from - from Upward from Surface Surface - Surface 0.010 0.020 0.046 0.060 0.120 0.132 0.100 0.210 0.233 0.146 0.308 0.341 0.196 0.416 0.452 These figures are for convection alone, and do not include radiation. They are different from the formula given on page 207, but are probably more reliable. If data are required for over 500, the above values may be plotted and the curves extrapolated. If the air is in motion, Mr. Langmuir found that the heat loss by convection could be expressed as -\/ times the loss to \ oo still air, where V is the velocity of air current in centimeters per second. Therefore, the velocity being known, evaluate the above expression and multiply the tabulated values by this factor. RADIATION. A body placed in a vacuum, with no ponderable substance in contact with it, radiates heat to its surroundings. All experi- ments so far made confirm the accuracy of Stefan's law, that every hot body radiates energy in proportion to the fourth power of its absolute temperature, and that the transfer of heat by radiation is therefore proportional to the difference 214 METALLURGICAL CALCULATIONS. between the fourth powers of the absolute temperatures of the hot body and its surroundings, respectively. Mr. Langmuir uses Stefan's law of radiation expressed in the following form Q R = E 5.9 X 10- 12 (T 2 4 TV) watts per sq. cm. = E 1.41X10- 12 (T 2 4 TV) cal. per sec. per sq. cm. These formulae express the fact that the radiation from a body at absolute temperature T 2 to one at 7\, is proportional to the difference between the fourth powers of the temperatures in question, and that for each unit of such difference the radiation loss from the hotter body is 1.41X1Q- 12 calories per square centimeter per second, if the body has unfi;y radiating power (black body), or Exl.41 XlO~ 12 calories if its coefficient ot radiating power (emissivity) is E. If t 1 .c heat flow is expressed as energy flow, 1 . 41 calories per second =5.9 watts. For convenience sake, to avoid large numbers, the absolute temperatures may be expressed on a scale of 1000, in which case 10~ 12 disappears, and the formula becomes: = EX1.41 viooo/ watts 1000/ UOOO cal. per sec. In the above form, this formula is very convenient for cal- culating radiation losses from any body at absolute temperature T 2 to its surroundings at r b provided the emissivity of the hot body, E, is known. Dull lampblack has an emissivity nearly 1 . 00, but all other materials have emissivities less than unity, the polished metals having values as low as 0.02. Unfortu- nately, total energy emissivity is not quite independent of temperature, and most of the determinations have been made for low temperatures only. In the following table, some emissivities have been calculated from the older experiments on radiation, and some are Mr. Langmuir 's own values. CONDUCTION AND RADIATION OF HEAT. 215 Radiating Capacity (Emissivity) Theoretical black body. Lampblack (At all tempe (50) 0.95 - (50) 0.68 (50) 0.64 (50) 0.56 (50) 0.52 (50) 0.17 (1200) 0.28 (1600) 0.28 (500) 0.85 (440) 0.91 (50) 0.56 (50) 0.47 (50 1 0.11 (50) 0.04 (500) 0.98 (50) 0.99 (50) 0.60 (50) 0.65 (50) 0.90 (50) 0.77 (50) 0.39 (50) 0.03 (1075) 0.16 (240) 0.156 (50) 0.67 (50) 0.04 (50) 0.02 (50) 0.03 (50) 0.33 (50) 0.035 (1160) 0.14 (50) 0.07 (50) 0.50 (50) 0.50 (50) 0.11 (50) 0.04 (50) 0.05 (50) 0.35 (800) 0.68 (50) 0.04 (50) 0.07 (50) 0.72 (3250) 0.39 (500) 0.10 (50) 0.06 (50) 0.06 (50) 0.10 (50) 0.49 (50) 0.25 (300) 0.73 (50) 0.60 (50) 0.60 (50) 0.60 ratures) = l.( - 0.98 (300) 0.67 (1750) 0.28 (1000) 0.88 (468) 0.85 (1000) 0.99 (720) 0.60 (300) 0.82 (300) 0.70 (300) 0.26 (1000) 0.09 (1175) 0.15 (360) 0.19 (127) 0.45 (300) 0.03 (500) 0.07 (126) 0.37 (300) 0.061 (1430) 0.16 (300) 0.30 (300) 0.41 (500) 0.19 (500?) 0.06 (300) 0.40 (1000) 0.75 (300) 0.14 (1000) 0.72 (Langmuir) . (500) 0.09 (500) 0.09 (500) 0.40 (500) 0.50 (400) 0.79 )0 (500) 0.53 (Thwing) (1200) 0.89 (945 C ) 0.60 (400^) 0.92 (500) 0.73 (500) 0.23 (1275) 0.13 (500) 0.04 (500) 0.08 (1695) 0.175 (500) 0.38 (1000) 0.37 (1000) 0.17 (500) 0.48 (1075) 0.80 (500) 0.23 (1170) 0.61 (Burgess) (500) 0.95 (1000) 0.70 (1000) 0.20 (1290) 0.1 5 Thwing (1000) 0.124 (Burgess) (710) 0.62 (1300) 0.88 Smoothblack paint Black paper Cast iron, rusted Cast iron, new Cast iron, polished Soft steel, liquid Iron oxide scale Iron rust Russian sheet iron Ordinary sheet iron Leaded sheet iron Tinned sheet iron Hematite Chromite Cuprous oxide Graphite, mat Copper, oxidized Copper, calorized Copper, polished Copper, liquid Aluminium polished. . . . Aluminium paint Brass, polished Gold, polished Gold, enamel Platinum, polished Silvered paper Monel metal, bright. . . . Monel metal, oxidized. . Zinc, bright.. Nickel, bright Nickel oxide Tin bright Magnesium. Silicon Alumina Magnesia Lime.. Glass Porcelain. White fire clay. Building stone . Plaster Wood. APPENDIX TO PART I. Problem 25. (1) Write the equations showing the relative weights and relative volumes (of gases) concerned in the combustion of Methane (marsh gas) C H 4 Acetylene C 2 H 2 Ethylene (olefiant gas) C 2 H 4 Methylene C 2 H 6 Allylene C 3 H 4 Propylene C 3 H 8 Propylene hydride C 3 H 8 Benzine C 6 H 8 Turpentine (liquid) C 10 H 18 Napthaline (liquid) C 10 H 8 (2) The molecular heats of combustion of the above gases to CO 2 and liquid H 2 O are given by Berthelot as: for C H 4 213,500 Calories " C 2 H 2 315,700 " C 2 H 4 341,100 C 2 H e 372,300 * " C 3 H 4 473,600 " C 3 H 6 499,300 " " C 3 H 8 528,400 " C 8 H 6 784,100 C io H ie (liquid ) 1,490,800 C io H s (liquid) 1,241,800 * (a) Calculate the molecular heats of combustion to CO 2 and H 2 O vapor. (b) Calculate the molecular heats of formation of the hydro- carbons themselves, using (C, O 2 ) 97,200 and (IPO) 69,000 liquid or 58,060 gas. (c) Calculate the heat of combustion, to CO 2 and H 2 vapor, of one cubic meter of each gas, and of one cubic foot in B. T. units. 217 218 METALLURGICAL CALCULATIONS. Answers (1): i ii i ii CH 4 + 20 2 = CO 2 + 2H 2 O 16 64 44 36 ii V IV ii 2C 2 H 2 + 50 2 = 4CO 2 + 2H 2 O 52 160 176 36 i in ii ii C 2 H 4 + 3O 2 = 2CO 2 + 2H 2 O 28 96 88 36 ii VII IV VI 2C 2 H 6 + 7O 2 = 4CO 2 + 6H 2 O 60 224 176 108 i IV in ii C 3 H 4 + 40 2 PP 3C0 2 + 2H 2 40 128 132 36 ii IX VI VI 2C 3 H 6 + 9O 2 = 6CO 2 + 6H 2 O 84 288 264 108 i V in IV C 3 H* + 5O 2 = 3C0 2 + 4H 2 O 44 160 132 72 i XIV X VIII C 10 H 16 + 14O 2 = 10CO 2 + 8H 2 O 136 448 440 144 i XII X IV C 10 H 8 + 12O 2 = 10C0 2 + 4H 2 O 128 384 440 72 (2) Molecular Heat Molecular Heat Heat of Combustion of of Formation of Combustion 1 m 3 (Cal.) if*. 3 (C Amorphous.) (to H*O Vapor.) B.T.U. C H 4 + 21,700 + 191,620 + 8,623 + 970 C 2 H 2 52,300 + 304,760 + 13,714 + 1,543 C 2 H 4 8,700 + 319,220 + 14,365 + 1,616 C 2 H 6 + 29,100 + 339,480 + 15,277 + 1,719 C 3 H 4 44,000 + 451,720 + 20,327 + 2,287 C 3 H 6 700 + 466,480 + 20,992 + 2,362 C 3 H 8 + 39,200 + 486,640 + 21,899 + 2,464 C 6 H 6 + 6,100 + 751,280 + 33,808 + 3,803 C io H ie (liquid) + 33,200 + 1,403,280 .... .... C 10 H 16 (gas) + 23,800 + 1,412,680 + 63,570 + 7,152 C 10 H 8 (liquid) + 6,200 + 1,198,040 .... .... APPENDIX. 219 Problem 26. The following typical analyses are from Poole's book on 11 The Calorific Power of Fuels ": Bituminous Coal. Fuel Oil . Natural Gas.^ "Carnegie, Pa." "Lima, O." "Findlay, 0." Carbon 77.20 80.2 H 2 1.64 Hydrogen 5.10 17.1 CH 4 93.35 Oxygen 7.22 1.3 C 2 H 4 0.35 Nitrogen 1.68 1.4 CO 2 0.25 Sulphur 1.42 CO 0.41 Moisture 1.45 O 2 0.39 Ash 5.93 N 2 3.41 H 2 S 0.20 Required: (1) The practical, metallurgical calorific powers of (a) The coal, per kilogram or pound 7,447 (b) The oil, per kilogram or pound 11,393 (c) The gas, per cubic meter 8,143 kg. Cal. (d) The gas, per 1000 cubic feet 509,000 Ib. Cal. (2) Compare the calorific values of 1 ton (2240 Ibs.) of coal, 1 ton (2000 Ibs.) of oil, and 1000 cubic feet of natural gas. 32.8:44.8:1. Problem 27. The composition of some commercial gases used as fuels is given by Wyer (Treatise on Producer-Gas and Gas Producers, p. 50), as in following table: H 2 CH 4 C 2 H 4 N 2 CO O 2 CO 2 Natural gas (Pitts- burg) 3.0 92.0 3.0 2.0 Oil gas 32.0 48.0 16.5 3.0 0.5 Coal gas, from re- torts 46.0 40.0 5.0 2.0 6.0 0.5 0.5 Coke-oven gas 50.0 36.0 4.0 2.0 6.0 0.5 1.5 Carburetted water- gas. . 40.0 25.0 8.5 4.0 19.0 0.5 3.0 Water gas 48.0 2.0 5.5 38.0 0.5 6.0 Producer gas (hard coal, using steam) 20.0 49.5 25.0 0.5 5.0 Producer gas (soft coal) 10.0 3.0 0.5 58.0 23.0 0.5 5.0 220 METALLURGICAL CALCULATIONS. Requirements: For each of the above gases calculate (1) The volume of air theoretically necessary to burn it. (2) The volume of the products of combustion. (3) The calorific power (a) per cubic meter, in kilogram Calories. (b) per cubic foot, in pound Calories. (c) per cubic foot, in British Thermal Units. (4) The theoretical temperature, if burned cold with cold air (calorific intensity). 7?/>c*,/7/c' Volume Volume J\VWU*. of air to of prod- Calorific Power. Calori- burnl ucts per Kg. Cal. Lb. Cal. B.T.U. fie In- volume. 1 volume per m*. per ft*. per ffi. tensity. Natural gas (Pittsburg) 9 . 35 10.33 8423 526 948 1852 Oil gas : 7.74 8.58 7350 460 828 1915 Coal gas, from retorts . . 5 . 79 6 . 50 5550 347 625 1896 Coke-oven gas 5 . 36 6 . 08 5159 322 579 1892 Carburetted water-gas.. 5. 06 5.77 5008 313 563 1914 Water gas 2.24 2.81 2590 162 292 1928 Producer gas (with steam) 1.08 1.86 1290 81 146 1676 Producer gas (ordinary) 1.13 1.98 1300 82 148 1555 Problem 28. An anthracite coal containing on analysis Carbon 89 per cent. Hydrogen 3 Oxygen 1 Ash 7 is burned under a boiler, and the ashes produced weigh, dry, 10 per cent, of the weight of the coal used. The chimney gases, dried and then analyzed, contain CO 2 15.3 per cent., O 2 3.5 per cent., N 2 81.2 per cent., and 25.9 grams of water is obtained for each cubic meter of dry gas collected. Required: (1) The volume of chimney gas (measured dry) produced per kilo, of coal burned 10.41 m 3 . (2) The volume of air used, measured dry at normal condi- tions, per kilo, of coal burned 10.67 m s . (3) The weight of dry air used 13.80 kg. (4) The volume of dry air theoretically necessary for the complete combustion of one kilo, of coal 8.72 m 3 . APPENDIX. 221 (5) The excess of air used, in percentage of that theoretically necessary for the perfect combustion of the coal 22.36%. (6) The volume of chimney gas, at standard conditions, per kilo, of coal burned 10.74 m 3 . Problem 29. An anthracite coal contains Carbon 89 per cent. Hydrogen 3 " Oxygen 1 Ash 7 It is burned on a grate, using 15 per cent, more air than is theoretically needed for its perfect combustion. The ashes weigh 10 kilos, per 100 kilos, of coal burned. Required: (1) The volume of air (assumed dry and at standard conditions) used per kilo, of coal burned 9.71 m 3 . (2) The number of cubic feet of air per pound of coal 155.4. (3) The percentage composition (by volume, of course) of the chimney gas, assuming it contains no soot or unburned gas. N 2 77.9, CO 2 16.1, O 2 2.6, H 2 O 3.4. (4) The percentage composition of the same, if first dried and then analyzed N 2 80.6, CO 2 16.7, O 2 2.7. (5) The number of grams of moisture carried per cubic meter of dried gas measured 28.4. (6) The number of grains per cubic foot 12.4. (7) The volume of the chimney gases, at standard conditions, (water assumed uncondensed) per kilo, of coal burned 9.85 m 3 . (8) The number of cubic feet of products per pound of coal 157.6. (9) The volume of the products in cubic meters at 350 C. and 700 m.m. pressure 24.4. (10) The volume of the products in cubic feet at 600 F. and 29 inches pressure 349.6. Problem 30. A bituminous coal contains Carbon 75 per cent. Hydrogen 5 " Oxygen 10 Ash.. .10 222 METALLURGICAL CALCULATIONS. It is powdered and injected into a cement kiln with 25 pet cent, more air than is theoretically necessary for its complete combustion. The kiln calcines 200 barrels of cement daily, using 125 pounds of coal per barrel of cement. Assume outside air at C. Hot gases enter stack at 819 C. Omit from cal- culation the water vapor and carbonic acid gas expelled from the charge. Required: (1) The volume of air per minute which the blower or fan must furnish. 2,672 ft 3 . (2) The volume of the hot products of combustion per min- ute, as they pass into the stack. 11,044 ft 3 . Problem 31. A bituminous coal (as of Problem 30) contains Carbon 75 per cent. Hydrogen 5 Oxygen..., 10 Ash 10 It is burned under a boiler with 25 per cent, more air than is theoretically necessary for its complete combustion. As- sume combustion perfect, and no unburnt carbon to remain in the ashes'. The wet steam produced contains 3 per cent, of water of priming, and 8.82 pounds of water is evaporated per pound of "coal burnt. Steam pressure 6 atmospheres effective pressure. Outside air C., feed -water 10 C., stack gases 310 C. Required: (1) The calorific power of the coal, by calculation, water formed considered uncondensed, in pound calories per pound of coal, and British Thermal Units per pound. 7096; 12773. (2) The percentage of the calorific power of the coal repre- sented by the heat in the dry steam produced 78.0. (3) ditto in the water of priming 0.6. (4) ditto in the sensible heat of the chimney gases 14.5. (5) ditto lost by radiation and conduction 6.9. (6) What increase in the net efficiency (requirement 2) would be obtained by using a feed water heater, which ab- stracted 60 per cent, of the sensible heat in the chimney gases, all other conditions remaining constant? 8.6%. APPENDIX. 223 Problem 32. A bituminous coal contains (as in Problems 30 and 31): Carbon 75 per cent. Hydrogen , 5 Oxygen ' 10 Ash 10 It is blown into a revolving cylindrical furnace with a high pressure blast which supplies only 14.9 percent, of the air neces- sary for its complete combustion, producing near the burner a highly luminous flame surrounded by a cylindrical sheath of auxiliary air drawn in by the injector action of the fuel-air jet. Assuming that in the body of the jet the hydrogen only of the coal is consumed, the luminosity being due to un con- sumed particles of carbon, and ash; that the mean specific heat 120 of carbon is 0.5 , of the ash 0.25 t Required: (1) The theoretical temperature of the interior of the jet of burning fuel 1140C. (2) The temperature if the supply of air in the fuel jet is increased to that theoretically necessary for perfect combustion of the whole fuel . 1945 C. Problem 33. Calculate the maximum temperature obtainable in the region of the tuyers of a blast-furnace. (1) Using cold, dry air, 1683 C. (2) Using dry air, heated to 700 C. 2272 C. (3) Using moist air, heated to 700 C., the amount of moisture present being such as air at 37 C. can carry when saturated with moisture. 1947 C. (4) Using cold, moist air of above composition. 1367 C. Problem 34. Powdered coal having the following composition is burnt in a cement kiln: Carbon 73.60 per cent. Hydrogen 5.30 Nitrogen 1.70 Sulphur. . . . 0.75 224 METALLURGICAL CALCULATIONS. Oxygen 10.00 per cent. Ash 8.05 Moisture 0.60 The finely-ground coal is burnt by cold air, at 20 C., and 760 m.m. pressure. The gases resulting, together with carbon dioxide gas from the charge, pass into the stack at 820 C. The analysis of the flue gases, by volume, dried before analysis, is: Carbon dioxide 25.9 per cent. Oxygen 3.1 Carbon monoxide 0.2 . " Sulphur dioxide not determined. Nitrogen difference. Of the carbon dioxide in the gases assume 40 per cent, of it to come from the carbonates in the charge being treated, and not from the coal. Assume no water in the furnace charge. The air used is saturated with moisture, at 20 C., tension of the moisture 22 m.m. of mercury. Required: (1) The theoretical calorific power of the coal 7090. (2) The theoretical temperature of the hottest part of the flame 1593 C. (3) The proportion of the calorific power of the fuel carried out by the hot gases 50.4%. (4) The percentage excess of air admitted above that theoret- ically required 16.4%. Problem 35. Calculate the draft of a chimney, in inches of water gauge and feet of cold air, if measured at its base, its height being 120 feet, inside diameter, (round) 6 feet, temperature of gases inside at bottom 300 C., at top 200 C., specific gravity (air = 1) 1.03, and taking in 200 cubic feet (measured at and 760 m.m.) of products of combustion per second. Section inside uniform top to bottom, sides fairly smooth, assume K = 0.04. Outside temperature C. 0.91 in.; 59.25 ft. Problem 36. A copper cylinder weighing 22.092 grams was placed in a small iron box on the end of a rod, and held several minutes in the hot blast main of a blast-furnace. On removal, and drop- APPENDIX. 225 ping instantly into a calorimeter, containing 301.3 grams of water, the temperature rose 4. 183 C. in three minutes. From previous experiments with this calorimeter, it was known that in three minutes it would abstract from the water 30 gram calories for each 1 observed rise of temperature above starting. The mean specific heat of copper between and t being 0.09393 + 0.00001778t, what was the temperature of the hot- blast? [N.B. Calculate what would theoretically have been the temperature of the water of the calorimeter if no heat had been lost to the calorimeter, and work out using this as the final temperature of the water.] 618 C. Problem 37. A piece of tin-stone (Cassiterite, SnO 2 ) from Bolivia was tested to obtain its specific heat. It was heated to two tem- peratures determined by a Le Chatelier thermo-electric pyro- meter, and dropped into a calorimeter. Corrections for calori- meter losses were made as explained in the Journal of the Frank- lin Institute, August, 1901. Data of the two tests were as follows : Experiment 1. Experiment 2. Weight of tin-stone used 12.765 gms. 12.765 gms. Temperature of same 476 C. 1018 C. Weight of water in calorimeter. . .299.4 gms. 300.7 gms. Temperature of same, starting 16.527 18.705 Temperature of same, 3 minutes. . 18. 444 22. 985 Water value of calorimeter per 3', for each 1 rise 30 cal. 30 cal. Required: A formula of the form S m = a + /h, for the particu- lar piece of Cassiterite used. 0.1050 +0.000006 It Problem 38. The chimney gases from a boiler enter the stack at a tem- perature of 400 C. Their composition is: Carbon dioxide 15.0 per cent. Oxygen 5.9 Nitrogen 79.1 What percentage of the total calorific power of the coke burnt will be saved by using a feed-water heater which reduces the 226 METALLURGICAL CALCULATIONS. temperature of these gases to 200 C. and sends 75 per cent. of the heat thus abstracted into the boiler with the feed-water? 7.92%. Problem 39. A blast-furnace gas contains, by volume: Carbon monoxide 23 per cent. Carbon dioxide 12 Hydrogen 2 Methane 2 Water vapor '. 3 Nitrogen . .58 Its temperature is 20 C., and it is taken to a gas engine, mixed with the theoretical amount of dry air needed for perfect combustion, and compressed to 4 atmospheres tension (total pressure) before being ignited. Neglect the heating of the gas-air mixture by compression, i.e., assume the temperature of the compressed mixture 20 C. before ignited. Assume that at the instant of ignition the volume occupied by the gases remains constant, so that the specific heat at constant volume applies. (Specific heat of 1 cubic meter at constant volume equals specific heat at constant pressure 0.09.) Required: (1) The calorific power of the gas per cubic meter, at constant pressure, and at constant volume. 928.5; 925.5. (2) The theoretical temperature at the moment after ignition has taken place. 1592 C. (3) The maximum pressure exerted. 22 atmospheres. (4) The volume of gas needed per horse-power hour, at a nechanico-thermal efficiency of 30 per cent. 2.81 m 3 , at 20 C. Problem 40. Bituminous coal containing carbon 78 per cent., hydrogen 5, oxygen 8, ash 8, water 1, is used in a gas producer. Assume the calorific power of the coal (water formed uncondensed) as 7480 Calories; ashes formed 12 per cent. Gas formed leaves producer at 600 C. ; composition : Carbon monoxide 35 per cent. Carbon dioxide 5 Methane 5 Hydrogen 5 Nitrogen 50 " APPENDIX. 227 Required: (1) The"vblume of gas obtained per kilo, of coal burnt. 3.045 m 3 (2) The calorific power of the gas per cubic meter. 1633 Calories. (3) The proportion of the calorific power of the fuel obtain- able on burning the gas. 66.5 per cent. (4) The proportion of the calorific power of the fuel which has been sacrificed in making the gas, assuming it burnt cold. 33.5 per cent. Problem 41. Producer gas of the following composition: Carbon monoxide 28 per cent. Carbon dioxide 4 Hydrogen 4 Methane 2 " Water vapor 1 " Nitrogen 61 is burned with 10 per cent, more air than theoretically re- quired, both 'air and gas being preheated to 1000 C. Required: The theoretical maximum temperature of the flame. 2100 C. Problem 42. Kiln -dried peat from Livonia contained by analysis: Carbon 49.70 per cent. Hydrogen 5.33 Oxygen ' 30.76 Nitrogen 1.01 Ash 13.23 The wet peat, as taken from the ground, carried 75 per cent. of water, when air-dried 20 per cent., and when kiln -dried none, as per analysis. The air-dried peat is dried in a kiln by means of a current of hot air, which enters the kiln, and comes in contact with the freshly-charged peat, at a temperature of 150 C., while it leaves the kiln, near the discharge end, at 50 C. Outside tem- perature C., outside air dry. The kiln loses by radiation 10 per cent, of the sensible heat of the hot air coming into it, the rest represents the sensible heat of the warm moist air and 228 METALLURGICAL CALCULATIONS. dried peat, issuing at 50 C., and the heat necessary to evaporate the moisture. The air required is heated in a stove where kiln-dried peat is burned, 75 per cent, of the heat generated being transferred to the air. Mean specific heat of kiln-dried peat 0.25. Required: (1) The practical calorific powers of wet peat, air dried peat and kiln dried peat. 607; 3278; 4249. (2) The volume of air, at standard conditions, needed for drying one metric ton of air-dried peat. 5,122 m 3 . (3) The percentage degree of saturation, with moisture, of the issuing air. 35.4. (4) The amount of kiln-dried peat required to be burned in the stove per ton of air-dried peat put through the kiln, and the percentage of the total fuel necessary for this purpose. 70.5 kg. 9.25%. Problem 43. A set of four coke ovens produce 10,000 cubic feet of gas per hour, measured at 60 F., and having the composition H 2 64.3 per cent., CO 20.7, CH< 5.4, C 2 H< 0.5, C 6 H 0.5, CO 2 2.0, O 2 1.0, N 2 5.6 per cent. Temperature leaving the ovens 2900 F. For half of each hour the whole gas is burned by the theoretical amount of cold air, as it passes into recuperators, whence the products at 1900 F. pass under steam boilers where their temperature is reduced to 500 before passing to the stack. For the second half of each hour the gases pass through the recuperators unmixed with air, are there heated to 1900 F., and at that temperature pass under boilers where they meet with the theoretical quantity of air needed for combustion, are burned, and the products pass to the stack at 500 F, Required: (1) The horse-power of the boilers during the first 30 minutes, calling 1 horse-power the ability to evaporate 34J pounds of water per hour at 212 F., and assuming the boilers to produce steam representing 50 per cent, of all the heat re- ceived by them and generated within them (net efficiency 50 per cent.). 23.4 H. P. (2) The same, for the second 30 minutes. 57.0 H. P. Problem 44. A plant of by-product coke ovens uses bituminous coal con- taining 76 per cent, of carbon, and having a calorific power APPENDIX. 229 of 9000, produces coke containing an average of 86 per cent, of carbon, and having a calorific power of 7000,. while the by- product tar produced contains 20 per cent, of carbon. The coke weighs 70 per cent, and the tar 5 per cent, of the weight of the coal used. The average analysis of the gases for the month of January, 1904 (samples dried before analysis) was: Carbon dioxide 3.00 per cent. Oxygen 0.50 Carbon monoxide 5.10 Marsh gas, CH 4 35.00 fC 2 H 4 2.13 Illuminants { C 3 H 8 1.06 [C 6 H 8 1.06 Hydrogen 40.00 Nitrogen. . '. 12.15 Required: (1) The volume of by-product gases, at standard barometric pressure and at 60 F., produced per ton (2000 pounds) of coal used. 16,294 cubic feet. (2) The proportion of the calorific power of the coal repre- sented by the calorific powers of the coke and the gases. 54.4%; 27.4%. (3) Using half the gases produced in gas-engines, at an effi- ciency of conversion into power of 25 per cent., how many horse-power-hours could be thus generated per pound of coal coked? 0.22 H. P. hours. Problem 45. A gas producer uses coal which analyzed: total carbon 75.68 per cent., oxygen 12.70 per cent., hydrogen 4.50 per cent., ash 7.12 per cent. The ashes produced contain 21.07 per cent, of unburnt carbon. The gas carried 30.4 grams of moisture to each cubic meter of dried gas collected, and the latter shows on analysis: Carbon dioxide 5.7 per cent. Carbon monoxide 22.0 Methane (CH 4 ) 2.6 Ethylene (C 2 H 4 ) 0.6 Hydrogen 10.5 Oxygen 0.4 Nitrogen 58.2 230 METALLURGICAL CALCULATIONS. A steam blower furnishes blast, forcing in the outside air which carries 15 grams of moisture per cubic meter, measured dry at 20 C. Required: (1) The volume of gas (dry) produced per kilo, of coal used. 4.337 m 3 . (2) The weight of steam used by the blower per kilo, of coal used. 0.2691 kg. (3) The weight and volume of air blown in, per kilo, of steam used by the blower. 15.5 kg. ; 12.95 m 3 . (4) The percentage of the steam blown in which is not de- composed in the producer, assuming the moisture in the gas tc represent steam used and not decomposed (assumption not strictly accurate). 49.3%. (5) The mechanical efficiency of the blower, assuming it uses steam at 4 atmospheres effective pressure, and produces 10 centimeters of water gauge pressure in the ash pit of the producer. 4.67%. Problem 46. Assume the gas of Problem 42 to be produced by the com- bustion of 1000 kilos, of coal per hour in the producers, and to be burned in a furnace with such excess of air (carrying at 20 C. 15 grams of moisture per cubic meter measured dry) that the chimney gas, analyzed dry, contains: Carbon dioxide 12.7 per cent. Nitrogen 80.6 Oxygen 6.7 These products of combustion enter the chimney at 500 C. and leave it at 350 C., their velocity entering at the base is 4 meters per second. Outside air 20. The efficient draft is 2.5 centimeters of water gauge, measured at the base; assume this 90 per cent, of the total head of the chimney. Required: (1) The diameter of the chimney, assuming it round, and its height, assuming its section uniform. 1.52 meters; 41.3 meters. (2) The work done by the chimney, expressed in horse- power. 2.6 (3) The energy efficiency of the chimney, i.e., the ratio of the mechanical work it performs to the mechanical equivalent of the heat which it receives. 0.079 per cent. APPENDIX. 231 Problem 47. Assume the chimney of Problem 46 to be built of fire-brick of an average thickness of 60 centimeters; that the gases passing through per hour are 1586 cubic meters of nitrogen and air, 315 cubic meters of carbon dioxide and 234 cubic meters of water vapor; the temperature at the base is 500 C., at the top 350; the velocity of the hot gases at the base, 4 meters per second; diameter 1.52 meters, height 41.3 meters. Assume further the coefficient of transfer of heat from gas to brick and brick to air to be tt _ ..J; (in C. G. S. units) , and the do, DUD velocity of the wind outside to be 20 kilometers per hour. Required'. The coefficient of conductivity of the fire-brick in C. G. S. units. 0.31 PART II. IRON AND STEEL. 233 [AFTER I. BALANCE SHEET OF THE BLAST FURNACE. As the most important factor in the production of the most important metal, the blast furnace is the most important fur- nace or piece of metallurgical apparatus in the world. It is, therefore, proper that we should commence a series of articles on the application of metallurgical principles and calculations to the metallurgy of iron, by a discussion of the blast furnace; and since this discussion, to be complete, must include a wide range of topics, we will commence with the simplest, viz.: the balance sheet of materials, entering and leaving the furnace. Later we can discuss the balance sheet of heat entering in, developed within and leaving the furnace, the reactions taking place in the furnace, the action of hot and of dried blast, the calculation of the proper constituents of the charge, the tem- peratures attained before the tuyeres, unused combustible energy of the gases, efficiency of the hot-blast stoves, and other interesting and practically valuable factors in the running of the furnace. The blast furnace may be regarded from several points of view; we will mention two. First, it may be regarded as a huge gas producer, run by hot, forced blast, in which the in- combustible portions of the contents are melted down (with a little unburnt carbon) to liquid metal and slag, and are run out beneath, while the gaseous products pass upwards through 50 to 100 feet of burden, and escape above. The escaping gases are primarily of N the composition of producer gas, with some of its carbonous oxide changed to CO 2 by the oxygen abstracted from the" burden, with some CO 2 added from the decomposition of the carbonates of the charge, and with the usual increment of moisture from the charge and volatile matter (if any) from the distillation of the fuel. From this point of view, the blast furnace is a huge gas producer, giving a rather inferior quality of combustible gas in very large quantities, and incidentally 235 236 METALLURGICAL CALCULATIONS. reducing to metal and slag the burden of iron ore and flux (limestone) which is put in with the fuel. The treatment of the furnace as a metallurgical problem may then proceed as the discussion of a gas producer, witli the composition of the gas produced somewhat modified by the amount of oxygen given up to the gas by the reducible portions of the charge of the furnace. The other viewpoint is to regard the furnace as primarily an apparatus for deoxidizing or reducing iron ore, for which purpose the ore is charged with sufficient carbonaceous fuel to do two things, viz.: to abstract all the oxygen from the reducible metallic oxides, and to furnish enough heat, or high enough temperature, to melt down to superheated liquids the pig iron and slag (combinations of irreducible metallic oxides) formed. In this view, the fuel must supply the reducing energy and the melting-down or smelting requirements; the first by acting upon the metallic oxides at a red to a white heat and abstracting their oxygen; the second, by being burned at the foot of the furnace by hot air blast, and there generating the heat and higher temperatures necessary for the smelting down of the already reduced materials. MATERIALS CHARGED AND DISCHARGED. The materials put into a blast furnace may all be classed under four heads: Fuel ......... ] Iron ore ..... \ Charged at the throat. Fluxes ....... j Blast ............. Blown in at the tuyeres. The materials discharged from the furnace may be classed under four heads also: iron ...... Tapped from the crucible. I Passing out at the top. We will discuss the resolution of each of the four materials charged into the four avenues of escape. FUEL. The fuel used is sometimes charcoal, but in the great ma- jority of cases coke, with perhaps some raw bituminous coal BALANCE SHEET OF BLAST FURNACE. 237 or anthracite coal, or in a few cases all raw bituminous coal. The composition of these fuels consists of moisture, volatile matter, fixed carbon, sulphur and ash consisting of silica, lime, iron, alumina, alkalies, etc. The moisture is driven off near the top, and goes into the gases as moisture. The volatile matter is expelled near to the top; almost all of it goes unchanged into the gases, but part of the hydrocarbons thus expelled may be decomposed and deposit fixed carbon on the iron oxides, etc., surrounding them. This carbon, however, will take up oxygen from the charge lower down in the furnace, and thus eventually pass into the gases as CO or CO 2 . We can, therefore, assume without error that all the volatile constituents of the fuel pass into the gases, but cannot be certain in exactly what state of combination, except as regards the moisture. It will be quite exact if we know the ultimate composition of the volatile matters of the coal, as so much carbon, hydrogen, oxygen, nitrogen, sulphur, etc., to charge them thus entirely to the gases. The fixed carbon all finds its way ultimately into the dust or the gases, either as CO, CO 2 , CH 4 or HCN, or alkali cyanides, excepting the amount represented by the carbon in the pig iron. Subtracting the carbon in the pig iron from the total fixed carbon in the fuel, the difference can safely be put down as entering the gases or being in the dust carried away by the gases. The sulphur in the fuel has a more varied history. When it is partly present in the form of iron pyrites, some may go into the gases as sulphur vapor, and eventually be burned to SO 2 when the gases are burned; another part may be oxidized in the furnace itself to SO 2 , and as such appear in the gases; the rest, along with organic sulphur, passes either into the slag or the pig iron. Sulphur passing into the slag seems to do so as calcium sulphide, CaS, formed by some such reaction as CaO + C + FeS = CaS + CO + Fe or, if the sulphur was present in the fuel as gypsum, CaS0 4 + 4C = CaS + 4CO The amount of sulphur going into the iron depends really upon 238 METALLURGICAL CALCULATIONS. the opportunity for it to go into the slag. If the temperature of the furnace at the tuyeres is very high, and especially if the slag is low in silica, sulphur will keep out of the iron and go into the slag, to the extent of ten or twenty times as much being in the slag as in the iron ; but if the temperature is low and the slag rich in silica the reverse may be the case. A high temperature and a high percentage of lime in the slag are the blast furnace manager's means of keeping down the sulphur in the iron, although high magnesia or high alumina are also efficacious. In casting up the balance sheet it can be assumed that when using coke or charcoal all the sulphur of the fuel goes either into the slag or the iron, and knowing from the analysis of the pig iron made how much goes into it, the rest can be calculated as going into the slag as CaS. If raw coal is used, it is uncertain how much sulphur goes into the gases, and an exact analysis of either the slag or gases, for sul- phur, in addition to that of the pig iron, would be necessary to fix its distribution. The ash of the fuel counts in with the other incombustible ingredients of the charge. Some of the silica in it may be reduced to silicon, and some of the CaO to Ca, to form CaS; while most of the iron will pass into the pig iron. It is in most cases uncertain whether the silicon in the pig iron comes at all from the fuel ash, so it is usual to assume it as coming from the silica of the ore only; as to the iron, it is best to assume it all reduced to the metallic state, as is probably always the case. Besides all these avenues of escape for the constituents of the fuel, it is sometimes necessary to take into account the possibility of some of it, in fine particles, being carried out of the furnace bodily with the outgoing gases. If the amount of this in the dust is determined, it must be subtracted in toto from the fuel charged, and then the remainder distributed as just discussed. Illustration. A blast furnace is charged, per 1,000 kilos, of pig iron produced, with 925 kilos, of coke, containing by analysis: Fixed carbon, 86 per cent.; volatile carbon, 2; hydrogen, 1; oxygen, 0.5; nitrogen, 0.5; sulphur, 1.0; iron, 2; silica, 5; lime, 1; moisture, 1. The pig iron contains 3.5 per cent, of carbon and 0.1 per cent, of sulphur. The dust carries 15 kilos, of dry BALANCE SHEET OF BLAST FURNACE. 239 coke per metric ton of pig iron. Required the distribution of the coke in the furnace per ton of pig iron made: Charges. Pig Iron. Slag. Gases. Dust. Coke 15.0 VAJKC . . . V^ KA Na 2 = X " Na 2 uZ Illustration. An iron ore contributes to the slag 350 pounds of silica, 12 of FeO and 60 of MnO. It is desired to flux it by using limestone containing 38.1 per cent, lime, 13.6 magnesia, 3.4 silica, and 44.9 carbonic oxide (CO 2 ). How much flux must be used to produce a ratio of silica to summated lime = 0.8? Solution. Letting x be the pounds of limestone used, the ingredients of the slag will be SiO 2 = 350 + 0.034 x CaO = 0.381* MgO = 0.136 x FeO = 12 MnO = 60 256 METALLURGICAL CALCULATIONS. The lime equivalents of the MgO, FeO and MnO are 56 CaO equivalent of MgO = 0.136 *X = 0.1904 x FeO - 12 X^ = 9.33 MnO =60 X^ = 47.32 CaO = 0.381 xX 1 = 0.381 * Summated CaO = 0.5714 # + 56.65 The ratio of silica to summated lime is therefore: 350 + 0.034* = U.o 56.65 + 0.5714* Whence * = 720 pounds. It is easily seen that this method of solution, calling * the weight of flux used, and then getting expressions for the weight of each ingredient in the slag and the weight of the whole slag, is a very general solution which is applicable to any kind of assumed composition to which it is desired that the slag shall conform. (3) If alumina is present in the slag-forming constituents of the ore it also may be reckoned with in several ways. It may be reckoned as so much by weight, and added in as such to the silica or the lime, or it may be calculated to its silica or its lime equivalent, and added into the summated silica or the summated lime. Here we touch on a question which has agitated blast furnace managers and theorists for a generation: Should the alumina be reckoned with the bases or with the acids; summated as silica or as lime? It would be presump- tuous to set forth a dictum on a subject which has been so long and so ably discussed by some of the best iron metal- lurgists, but we will assume, as somewhere near the truth, that as far as the elimination of sulphur in the slag is concerned, CALCULATION OF THE CHARGE. 257 alumina acts in slags low in silica as though it were lime, not in the proportions of its lime equivalent (1 AS \ j02 X weight of aluminaj but rather in about the proportions of its simple weight. As far as fusibility is concerned, in high silica slags alumina increases the fusibility up to a certain point, above which it decreases it. It acts in these, therefore, like lime, and may be classed with the bases. In low silica slags-, below 45 per cent., alumina acts like silica when considerable is present, and like lime when less is present; for instance, in a low lime, high alumina slag, alu- mina and silica may be substituted one for the other within wide limits, without materially affecting the fusibility of the slag; in a high lime, high alumina slag, alumina and lime may be substituted for each other within wide limits without sen- sibly changing the fusibility. To state the matter as succinctly as possibe, in slags low in silica (30 to 35 per cent.), alumina reinforces the bases in the elimination of sulphur; in regard to fusibility, it acts like silica in a slag low in both silica and lime, and like lime in all other blast furnace slags. Illustration. An iron ore carries 10 per cent, of its weight of silica and 6 per cent, of alumina. The lime stone on hand con- tains 37.3 per cent, lime, 13.3 magnesia, 3.3 silica, 44 carbonic oxide (CO 2 ), and 2.1 per cent, alumina. How much flux is required per 1,000 parts of ore to make (a) a slag with 49 per cent, of silica plus alumina; (b) a slag with 33 per cent, silica; (c) a slag with summated silica = the summated lime? Solution. (a) Weight of SiO 2 in slag = 100 + 0.033 x Weight of A1 2 3 in slag = 60 + 0.021 x Weight of CaO in slag = 0.373 x Weight of MgO in slag = 0.133 x Total weight of slag = 160 + 0.560 x therefore, 160 + 0.054* = 0.49 (160 + 0.560 x) whence *= 370 258 METALLURGICAL CALCULATIONS. (b) 100 + 0.033 x = 0.33 (160 + 0.560 x) whence x = 313 (c) Silica = 100 +0.033* 180 Silica equivalent of alumina = (60 + 0.021 x) Summated silica =153 +0.0515* Lime 0.373 x eft Lime equivalent of magnesia = (0.133 x) 4U Summated lime = 0.5592* therefore 153 + 0.0515 * = 0.5592 * whence * = 300 Returning to the slag resulting from the proportions of flux chosen in Problem 51, and containing: SiO 2 33.74 per cent. FeO 1.54 per cent. APO 3 5.43 " MnO 4.21 CaO 46.23 " K 2 O 1.54 " MgO 7.11 " CaS 0.20 " We see that its percentage of silica is low, therefore it is adapted to produce pig iron low in sulphur; its percentage of alumina is low, and therefore its presence increases the fusibility of the slag, which would otherwise be rather deficient, because of the high lime and somewhat considerable amount of other bases. In such a slag, alumina would be summated with the bases. oo T A The ratio of silica to bases is _' = 0.516 ob.Ub The ratio of silica to lime plus magnesia is ' = 0.632 oo . oo The summated lime = 46.23 + CaO equivalent of APO 3 = ^ (5.43) = 8.94 CALCULATION OF THE CHARGE. 259 cc + CaO equivalent of MgO = (7.11) = 9.95 4U + CaO equivalent of FeO = (1-54) = 1.20 H-CaO equivalent of MnO = ~ (4.21) = 3.32 + CaO equivalent of K 2 O = -^- (1.54) = 0.92 70.56 33 74 Ratio silica to summated lime = ^-^ = 0.478 70.56 Problem 52. In a blast furnace charge, consisting of 1530.2 pounds of ore and 682 pounds of charcoal, per 1,000 pounds of pig iron made, it is known from the balance sheet of above materials that they will contribute to the slag the following slag-forming ingre- dients. (See balance sheet, Problem 51) : SiO 2 70.9 pounds. FeO 3.2 pounds. APO 3 11.6 " MnO 9.3 CaO 39.9 " K 2 O 3.4 MgO 15.5 ." CaS 0.4 The limestone at hand contains : CaO 53.74 per cent. APO 3 0.32 per cent. MgO 0.17 " Fe 2 O 3 0.18 " SiO 2 3.14 " CO 2 42.42 " Required. The weight of limestone to be used to make: (1) A slag containing 33.74 per cent, of silica. (2) A slag in which the ratio of silica to bases is 0.516. (3) A slag in which the ratio silica to summated lime "s 0.478. Solution. This problem embodies the conditions which con- 260 METALLURGICAL CALCULATIONS. front the metallurgist when desiring to calculate the flux needed by any given furnace, and we have assumed certain working ratios to be aimed at in the slag, in order to elucidate the method of solution. SiO 2 CONSTII From On and Fuel. 70.9 ruENTs OF SLAG. ? From Flux. 0.0314* 0.0032* 0.5374* 0.0017* |g (0.0018*) Total. 70.9 + 0.0314* 11.6 + 0.0032* 39.9 + 0.5374* 15.5 + 0.0017* 3.2 + 0.0016* 93 APO 3 11.6 CaO .... 39.9 MgO.. 15.5 FeO 3.2 MnO 9 3 K 2 3 4 34 . . . . CaS . . 4 0.4 154.2 0.5753* 154.2 + 0.5753.* (1) To make a slag with 33.74 per cent, of silica we Must have 70.9 + 0.0314* = 0.3374 (154.2 + 0.5753*) whence * = 116 pounds. (2) To make a slag with ratio of silica to bases 0.516 we must have 70.9 + 0.0314* = 0.516 (82.9 + 0.5439*) whence * = 116 pounds. (3) To make a slag with ratio of silica to summated lime 0.478, we must first summate the lime as follows: Lime = 39.9 + 0.5374* Lime equiv. of APO 3 = (11.6 + 0.0032*) = 19.1 + 0.0053* MgO = -j- (15.5 + 0.0017*) = 21.7 + 0.0024* KC FeO = -|j( 3.2 + 0.0016*) = 2.5 + 0.0012* CALCULATION OF THE CHARGE. 261 MnO = -|^ ( 9.3) = 7.3 K 2 = -^ ( 3.4) = 2.0 Summated Ihne = 92.5 + 0.5463 x therefore 70.9 + 0.0314* = 0.478 (92.5 + 0.5463*) whence * = 116 pounds. COMPARISON OF FUELS, FLUXES AND ORES. By properly utilizing the preceding principles, it is possible to compare different varieties of fuels, fluxes or ores with each other, and thus to determine their relative values to the fur- nace, as far as can be inferred from their chemical composi- tion. (Some of the following methods are from a paper by Mr. F. W. Gordon, Trans. Am. Institute Mining Eng., 1892, p. 61.) COMPARISON OF FUELS. If different qualities of fuel are available it is possible to calculate which is the most advantageous to use in the furnace. The fixed carbon only is efficient for the furnace, and not all of that, because the ash of the fuel needs to be fluxed to slag, and a certain amount of the fixed carbon will need to be burned simply to melt this slag. Then the cost of the limestone to be used to flux this ash must be counted in, and finally part of the labor costs of running the furnace must' be charged against the slag. We can thus calculate the total ^charges against the fuel to supply one part of available carbon, which is the best basis upon which to compare different fuels. Illustration. Two varieties of coke are available for a blast furnace, analyzing respectively: No. 1. No. 2. Fixed carbon 84 per cent. 90 per cent. Volatile matter 2 1 Moisture 5 " 3 " Ash 9 " 6 " And costing respectively $4.50 and $5.50 per ton. The ash of the fuels analyzes respectively: 262 METALLURGICAL CALCULATIONS. No. 1. No. 2. Silica 55 per cent. 25 per cent. Alumina 25 " 5 " Lime 15 " 50 " Magnesia 5 " 10 " Ferric oxide " 10 " They are to be fluxed with the limestone of preceding prob- lem, assumed to cost $1.00 per ton, and to make a slag carrying 40 per cent, of silica and alumina together. Assume an average of 0.228 parts of fixed carbon necessary to melt down 1 part of slag, and that the manufacturing costs borne by the slag amount to $1.00 per ton. What are the relative values of the two fuels in this furnace? Solution. The amounts of flux needed to 100 parts of each fuel burned will be found as follows, letting x be the amount of flux used: Slag No. 1. Slag No. 2. Silica 4.95 + 0.0314* 1.50 + 0.0314* Alumina 2.25 + 0.0032* 0.30 + 0.0032* Lime 1.35 + 0.5374* 3.00 + 0.5374* Magnesia 0.45 + 0.0017 * 0.60 + 0.0017 * Ferrous oxide.. 0.0016* 0.54 + 0.0016* Total weights 9.00 + 0.5753 * 5.94 + 0.5753 * Therefore, in case No. 1: 7.20 + 0.0346* = 0.40 (9.00 + 0.5753*) whence * = 18.4 And in case No. 2: 1.80 + 0.0346* = 0.40 (5.94+0.5753*) whence * = 2.9 The negative value in case No. 2 simply means that the ash of fuel No. 2 is more basic than the slag, and, therefore, requires no limestone, but itself acts as a basic flux. Per ton of fuel burned, there would be required respectively 0.184 and 0.029 tons of limestone, and the weights of slag would be 0.196 tons and 0.0427 tons. (Substituting values of * in the total weights of slags.) The weights of fixed carbon necessary to smelt these weights of slag would be: CALCULATION OF THE CHARGE. 263 No. 1. 0.196 X0.228 = 0.0447 tons. No. 2. 0.0427X0.228 = 0.0097 " Leaving as available fixed carbon for the furnace in each case: No. 1. 0.84- 0.0447 = 0.7953 tons. No. 2. 0.90- 0.0097 = 0.8903 " From these figures the cost of 1 ton of available carbon fur- nished by each fuel, adding in manufacturing cost chargeable against the slag, is No. 1. Cost of coke, $4.50 ^ 0.7953 = $5 . 658 Cost of limestone, $1.00X0.184^0.7953 = 0.231 Costs against slag, $1.00X0. 196 -r- 0.7953 = 0.246 $6.135 No. 2. ^Cost of coke, $5.50-^0.8985 = $6.233 Cost of limestone, $1.00 X (- 0.092) + 0.8903 = -0. 101 Cost against slag, $1.00X0.0065X0.8903 = 0.007 $6.139 The two fuels, at the prices given, are therefore of almost exactly the same value to the furnace. The solution along the lines shown is general, for any de- sired composition of slag, or for use with any given limestone, and is a valuable means of comparing the values of different fuels. The cost of a ton of puie, available carbon, when fur- nished by any given fuel, is an item which is useful when com- paring the relative values of different fluxes or ores with each other. COMPARISON OF FLUXES. If different qualities of flux are available it is very desirable to be able to calculate which is the most economical to use in the furnace. Any acid ingredients in the flux diminish very sharply its efficient fluxing power, because they must first be satisfied from the bases present in the same proportions as acid to bases in the final slag. The slag thus formed from the im- purities requires further to be melted, and other costs are properly chargeable against it. The best comparison is finally 264 METALLURGICAL CALCULATIONS. obtained by calculating for each flux available the cost from it of pure net lime, or net summated lime, analogous to the cal- culation for pure net carbon in the case of fuels. Any ordinary condition may be imposed upon the slag which the furnace is to produce. Illustration. There are available for a furnace two qualities of limestone, containing respectively: CaO 53.74 per cent. 47.80 per cent. MgO 0.17 " 4.61 " SiO 2 3.14 " 5.12 " APO 3 0.32 "' 3.36 Fe 2 3 0.18 " 1.10 " CO 2 42.42 " 37.55 " The first costs $1.00 per ton, the second $0.80. Assume them smelted with fuel furnishing pure available fixed carbon at $6.135 per ton ; that 0.228 tons of pure fixed carbon is needed to smelt 1 ton of slag; that manufacturing costs against slag are $1.00 per ton, and that the slag to be made in the furnace must have summated silica equal to summated lime. Compare the relative values of the two fluxes. Solution. We will direct our calculations towards finding the net cost of 1 ton of pure available summated lime from each of the two limestones. The summated lime and silica in each flux are: No. 1. No. 2. Summated lime. 0.5411 0.5502 Summated silica . . . . . 0342 . 0808 Excess of summated lime . 5069 . 4694 The weights of slag formed from the impurities present in each limestone will be: No. 1. 'No. 2. CaO (difference between amount present and excess of summated lime found) . . . 0305 . 0086 MgO 0.0017 0.0461 FeO 0.0016 0.0099 SiO 2 0.0314 0.0512 APO 3 .. ..0.0032 0.0336 Totals.. ..0.0684 0.1494 CALCULATION OF THE CHARGE 265 The cost of 1 ton of pure available lime from each of thes<5 fluxes will therefore be, adding in costs chargeable against the slags formed by the impurities present: No. 1. Cost of limestone, $1.00^0.5069 = $1.973 Cost of carbon for melting slag, $6.135X0.228X0.0684 -^ 0.5069 = 0.188 Costs of running, chargeable against slag, $1.00 X 0.0684 -T- 0.5069 =0.135 No. 2. Cost of limestone, $0.80-^0.4694 Cost of carbon for melting slag, $6.135X0.228X0.1494 ^0.4694 = 0.465 Cost of running, chargeable against slag, $1.00X0.1494 4-0.4694 = 0.318 $2.487 The conclusion is that the poorer limestone, at $0.20 per ton less cost, is in reality costing $0.191 per ton more for pure available lime, or is in reality 8.3 per cent, dearer than the first, instead of being 20 per cent, cheaper. The method of calculation here described is quite general for any compositions of limestone or other flux, and for any as- sumed conditions which the slag must conform to. COMPARISON OF ORES. As the more complicated, we come to the comparison of various ores which may be at the iron master's disposal. Here a similar method of procedure is advisable. It can be calcu- lated first, for a unit weight of ore, how much pure lime would be required to flux its impurities, how much pure carbon would be required to melt the slag thus formed, and to the costs of each of these would be added the handling of the slag. Each of these can be also expressed per unit of pure oxide of iron in the ore ; and if to their sum we add the cost of ore necessary to furnish unit weight of pure oxide of iron we obtain the total costs per unit weight of pure iron oxide (Fe 2 O 3 ). This is the basis on which different ores may then be compared. It must not be forgotten that one ore may, because of higher sulphur 266 METALLURGICAL CALCULATIONS. content, require the production of a more basic slag, so that the amounts of flux required and slag formed will be influenced by the condition necessary to impose on the slag in each case. Illustration. The iron ore briquettes (of Problem 51) con- tained Fe 2 O 3 , 85.93 per cent.; FeO, 3.96; SiO 2 , 5.50; MnO, 0.63; A1 2 O 3 , 0.76; CaO, 2.23; MgO, 0.97 per cent. If these cost $4.40 per ton, and are smelted in a furnace making slag with ratio of silica to bases 0.516, and assuming 0.3 per cent, of the iron, 82.7 per cent, of the silica, and 96.6 per cent, of the manga- nese to go into the slag, what is the cost per ton of pure Fe 2 O 3 from this source, charging pure lime for fluxing at $2.296 per ton, pure carbon for smelting slag at $6.135 per ton, and re- quiring 0.228 tons of carbon for one of slag, adding also manu- facturing costs at $1.00 per ton of slag? Solution. The slag-forming ingredients from 1 ton of ore briquettes are: FeO 0.003 x X 55 |7 o.8593X ~ + o.0396x = 0.0024 tons MnO 0.966X0.0063 =0.0061 CaO = 0.0223 MgO = 0.0097 A1 2 O 3 = 0.0076 SiO 2 = 0.0455 and the bases to satisfy the silica present must be 0.0455 -v- 0.516 , =0.0882 But, sum of bases already present = 0.0482 therefore, pure lime to be added = 0.0417 and total weight of slag = 0.1337 Efficient Fe 2 O 3 in 1 ton of ore = 0.8593 plus Fe 2 O 3 equivalent of reduced FeO Ifin = ~ (0.0396- 0.0024)= 0.0413 Total available Fe 2 O 3 = 0.9006 u METALLURGICAL CALCULATIONS. 267 Cost of 1 ton pure available Fe 2 O 3 from these briquettes: Cost of ore, $4.40 -*- 0.9006 = $4 . 885 Cost of pure lime, $2.296x0.0417-^0.9025 = 0.106 Cost of carbon for melting slag, $6.135X0.228X0.1337 -0.9025 = 0.207 Costs chargeable against slag, $1.00X0.1337-=- 0.9025 = 0.148 Total = $5.346 A similar calculation is possible with any ore of any given composition, and making any assumed quality of slag. The costs of 1 ton of pure ferric oxide thus calculated will give the relative costs of the iron obtained from these different sources, and therefore indicate the relative values of the different ores to the blast furnace manager. Problem 53. Assume a blast furnace manager to use the ore of preceding illustration, furnishing pure Fe 2 O 3 at a net cost of $5.336 per ton, and to use with it fuel furnishing pure carbon at $6. 135 per ton, there being required for reduction and melting the iron produced and furnishing it with carbon, 0.66 tons of pure available carbon per ton of pig iron produced, and the pig iron containing 96.656 per cent, of iron. The running costs of the furnace are $3.00 per ton of pig iron produced (costs against slag not included) . Required: The cost of the pig iron per ton. Solution: Fe 2 3 required 160 -5- 112 X 0.96656 = 1 . 3808 tons. Cost of the ore, $5.336 X 1.3808 = $7.368 Cost of fuel, $6.135X0.66 = 4.049 Cost of manufacturing (share against pig iron) = 2 . 000 Total cost = $13.417 CHAPTER III. UTILIZATION OF FUEL IN THE BLAST FURNACE. The blast furnace, in its simplest terms, may be regarded as a huge gas producer, producing in the region of the tuyeres pure producer gas from fixed carbon and heated air; the gas thus produced is partly oxidized in its ascent through the furnace by the oxygen abstracted from the charge (which latter item is almost a constant quantity per unit of pig iron made), and has added to it carbon dioxide from the carbonates of the charge. But, after all, the unoxidized and combustible ingredients of the gas escaping represent a large part, in fact, often the largest part, of the total calorific power of the fuel. Problem 54. A blast furnace uses 2,240 pounds of coke, containing 90 per cent, fixed carbon and 350 pounds of limestone, containing 10 per cent, of carbon (as carbonic acid, CO 2 ) to produce a ton of pig iron containing 4 per cent, of carbon. The gases contain 24 per cent, of carbonous oxide, CO, 12 per cent, of carbonic oxide, CO 2 , 2 per cent, of hydrogen, 2 per cent, of methane, and 60 per cent, of nitrogen. Required. (1) The volume of gas, as analyzed, produced per ton of pig iron made. (2) The calorific power of the gas. (3) The proportion of the calorific power of the coke which has been generated in the furnace. Solution. (1) The carbon going into the gases will be that in the coke, less that in the pig iron, plus that in the carbonates of the charge. Carbon in coke = 2,240X0.90 = 2,016 pounds Carbon in carbonates = 350X0.10= 35 Carbon charged = 2,051 Carbon in pig iron = 2,240X0.04 = 89.6 " Carbon going into the gases = 1,961.4 " 268 UTILIZATION OF FUEL IN BLAST FURNACE. 269 Carbon in 1 cubic foot of gas: In CO 0.24X0.54 In CO 2 0.12X0.54 In CH 4 0.02X0.54 Total 0.38X0.54 = 0.2052 ounces av. = 0.012825 pounds Gas produced per ton of pig iron = ' ' = 152,935 cu. ft.(l) (2) Calorific power of 1 cubic foot of gas: CO 0.24X3,062 = 734.9 oz. cal. H 2 0.02X2,613 = 52.3 " CH 4 0.02X8,598 = 172.0 " Sum = 959.2 = 59.95 pound cal. per 152,935 cubic feet = 9,168,450 pound cal. (2) (3) The calorific power of the coke considering it to contain simply 90 per cent, of fixed carbon, would be 8,100X0.90 = 7,290 pound cal. per pound. The presence of CH 4 in the gases points, however, to there being probably some available hydrogen in it, which would in- crease its calorific power somewhat. A closer approximation to the calorific power of the coke could, therefore, be obtained by assuming at least as much available hydrogen in it as would correspond to the hydrogen in the CH 4 in the gas. CH 4 in 152,935 cu. ft. of gas 3,059 cu. ft. Weight of this = 3,059 X (0.09 X 8) = 2,202 oz. av. 137.7 Ibs. Hydrogen = 137.7X (4-J-16) 34.4 Ibs. Available hydrogen in coke = 34.4 -s- 2,240 1.54 percent. Calorific power 0.0154X29,030 447 Ib. cal. per Ib. Total calorific power of the coke 7,737 " " Calorific power of coke used per ton 7,737X2,240 = 17,330,880 Ib. cal. Calorific power of the gases = 9,168,450 " Calorific power generated = 8,162,430 " = 47.1 per cent. 270 METALLURGICAL CALCULATIONS. This figure, however, practically over-charges the furnace little bit, because the pig iron with its 4 per cent, of carbon really takes out of the furnace some unburnt fuel, whose heat of combustion may be utilized outside the furnace as in the Bessemer converter. The furnace does not generate heat from this, representing: 2,240X0.04X8,100 = 725,760 Ib. cal. leaving as actually generated in the furnace 8,162,430- 725,760 = 7,436,670 Ib. cal. = 42.9 per cent. (3) Such an average blast furnace cannot, therefore, be accused of generating within it over some 43 per cent, of the calorific power of the fuel put into it, while the heat rejected as poten- tial energy of combustion of the waste gases amounts to more than half the calorific power of the fuel. Fifty years or more ago, when these waste gases were allowed to burn truly to waste the blast furnace was indeed a devourer of fuel, but matters have been improved by the utilization of the waste gases to heat the blast, and thus one of the largest "leaks" of heat from the furnace has been patched up to some extent, although yet far from satisfactorily. Problem 55. Assume that in problem 54, one-third of the gases produced are burnt in hot-blast stoves, preheating the air blown in at the tuyeres, and that the blast is thus preheated to 450 C. Required. (1) The amount of blast blown in per ton of pig iron made. (2) The heat in the blast. (3) The efficiency of the hot-blast stoves. (4) The increased efficiency of the blast furnace plant as a whole in generating the calorific power of the fuel, when thus provided with this hot-blast apparatus. Solution. (1) Volume of (dry) gases per ton = 152,935 cu. ft. Nitrogen present in these (60%) = 91,761 Air containing this = ^~ = 115,860 " UTILIZATION OF FUEL IN BLAST FURNACE. 271 This is the volume of the blast per ton of pig iron produced, assuming no nitrogen to come from the coke used, and the blast to be dry. If the blast were moist, and its hydrometric condition known, the volume of moist blast could be calculated. (2) Assuming the blast dry, it is heated to 450 C., requiring 1 15,860 X [0.303 + 0.000027 (450)]X450 = 16,430,975 oz. cal. = 1,026,936 Ib. cal. (3) The hot-blast stoves receive one-third of all the gas pro- duced, having, therefore, a calorific power of 9,168,450-^3 = 3,056,150 Ib. cal. Efficiency of the stoves = o''n = - 336 = 33 - 6 P er cent - 6, (Job, 150 /3\ (4) The blast furnace was primarily rejecting unused 57.1 per cent, of the calorific power of the fuel, 4.2 per cent., how- ever, as a necessary loss, to supply the carbon in the pig iron, but 52.9 per cent, as combustible power of unburnt waste gases. If one-third of these gases are completely burnt in hot-blast stoves, then the combined plant furnace plus stoves is utilizing 17.6 per cent, more of the calorific power of the fuel than before, or 42.9 + 17.6 = 60.5 per cent., and, therefore, re- jecting undeveloped 39.5 per cent, of the calorific power. (4) [The net effect of the use of the hot-blast stove, upon the heat generation in the furnace, is practically to put back into the furnace, as sensible heat, 1-3X33.6 = 11.2 per cent, of the calorific power of the waste gases, equal, therefore, to 0.112X 52.9 = 5.9 per cent, of the total calorific power of the fuel. This renders available, for the working of the furnace, 42.9 + 5.9 = 48.8 per cent, of the calorific power of the fuel, an in- 5 9 crease of available heat for reducing and smelting of 777-7: = 4 L . y 13.7 per cent, of the former available quantity.] The practical conclusion is that a blast fuinace generates in itself not much over 40 per cent, of the calorific power of the fuel used, and rejects nearly 60 per cent. ; by using part of the waste gases to heat the blast, however, some of this re- jected heat, to an amount representing net 5 to 10 per cent, of the calorific power of the fuel used, is returned to and injected 272 METALLURGICAL CALCULATIONS. bodily into the furnace, thus rendering available for the pur- poses of running the furnace some 50 per cent, of the calorific power of the fuel as a maximum. The efficiency with which the furnace applies this 50 per cent, usefully to the objects of reducing and smelting, is another question for investigation. Problem 56. Assume that at the furnace of Problems 54 and 55 the two- thirds of the waste gases are burnt under boilers, raising steam which *runs the blowing engines, hoists and pumps, and pro- viding 10 effective horse-power for each ton of pig iron made per day in the furnace. Required. (1) The efficiency of development of the calorific power of the fuel in the plant (furnace, stoves, boilers, engines) regarded as a whole. (2) The thermo-mechanical efficiency of the boiler and engine plant. (3) The power which could be generated if gas engines, at 25 per cent, thermo-mechanical efficiency, were used in their stead. Solution. (1) Since the stoves completely burn one-third of the waste gases, and the boilers the other two-thirds, all the combustible power of the waste gases is developed in the com- bined plant, and the only part of the calorific power of the fuel which is unused is the 4.9 per cent, represented by the carbon necessarily entering into the composition of the pig iron. The plant as a whole, therefore, develops or generates 95.1 per cent, of the calorific power of the fuel used. (2) For each ton of pig iron, the heat developed under the boilers will be two-thirds of the calorific power of the gases, or 9,168,450X2-3 = 6,112,300 Ib. cal. There is generated thereby 10 effective horse-power days, equal to 10X33,000X60X24 = 475,200,000 ft. Ibs But 1 Ib. cal. = 425X3.2808 = 1394.3 " Therefore, thermal equivalent of , - 475,200,000 work done = ' ' -* 340,800 Ib. cai. . o UTILIZATION OF FUEL IN BLAST FURNACE. 273 Thermo-mechanical efficiency of boiler and engine plant: 340,800 6,112,300 (3) Gas engines, at 25 per cent, thermo-mechanical efficiency, would give power representing per ton of pig iron produced : 6,112,300X 0.25 = 1,528 075 Ib. cal. Equal to 1,528,075X1394.3 = 2,130,645,900 ft. Ibs. 2,130,645,900 , - ^ 33,000X60X24 " A quicker solution is: 10 X^ = 44.9 " " (3) Leaving net surplus power per ton of pig iron produced per day = 34.9 horse-power. The preceding problems have elucidated the question of the small proportion of the calorific power of the fuel which is generated in a blast furnace, showing it to be, in usual practice only 40 to, at most, 50 per cent, of the calorific power. The discussion has not explained " why," but a further consideration will throw light on this question also. The proportion of the calorific power of a fuel which is gen- erated in a blast furnace is solely a question of how much of it is burned to carbonic oxide, CO 2 , and how much to carbonous oxide, CO? If all the carbon were burned to CO 2 , practically all the calorific power of the fuel would be generated; if all 2 430 were burned to CO, only ~r^= 0.30 = 30 per cent, of the o,IUU heating power of the carbon would be generated. If a blast furnace was filled with nothing but coke, and air blown in as usual at the tuyeres, carbon would be burned in the furnace only to CO, and but 30 per cent, of its calorific power be gen- erated and available for the needs of the furnace. The entire gain over this percentage is due to the oxidation of CO to CO 2 by the oxygen abstracted from the solid charges, that is, by the act of reduction. In Problem 54 we calculated that under ordinary conditions, between 40 and 50 per cent, of the calorific 274 METALLURGICAL CALCULATIONS. power of the fuel is generated in the furnace; the excess of this above 30 per cent, is due to the oxidation of CO to CO 2 during the reduction of the metallic oxides in the charge. From this standpoint it is advisable to strive to perform the greatest possible proportion of the reduction in the furnace by CO gas, because in this case the total generation of heat in the furnace per unit of fuel charged will tend towards a maximum. Since no carbon can be burned to CO 2 at the tuyeres, it follows that, from the standpoint of the generation of the maximum quantity of heat in the furnace, from a given weight of fuel, Gruner was right in formulating his dictum of the ideal working, of a blast furnace, viz.: GRUNER'S " IDEAL WORKING." All the carbon burnt in the furnace should be first oxidized at the tuyeres to CO, and all reduction of oxides above the tuyeres should be caused by CO, which thus becomes CO 2 . This dictum is not in Gr liner's own words, but expresses their sense, and from the point of view of the present discussion, it is the correct principle upon which to obtain the maximum gen- eration of heat in the furnace from a given weight of fuel. It practically directs us to generate at the tuyeres 30 per cent, of the calorific power of the carbon oxidized in the furnace, and the rest that can be obtained from the carbon is to be generated during the reduction of the charge. If we apply this principle to the furnace and data of Problem 54, we should first observe that the carbon oxidized in the furnace is: Carbon in coke charged 2016 . pounds Carbon in pig iron produced 89 . 6 . " Carbon oxidized in furnace 1926. 6 Requiring, if all oxidized to CO at the tuyeres 4 1926. 6 X-o- Ibs. oxygen = 2569. pounds. u But, Problem 52 shows us that there was actually blown into this furnace 115,860 cubic feet of blast, containing, there- fore, 115,860X0.208X(1.44-*-16) = 2,169 Ibs. oxygen, UTILIZATION OF FUEL IN BLAST FURNACE. 275 capable of oxidizing to CO at the tuyeres 2,169X0.75 = 1,627 Ibs. carbon. Proportion of carbon gasified burnt at the tuyeres, 1,627 1926.6 0.844 = 84.4 per cent. It is, therefore, true of the furnace under discussion, that if Gruner's ideal working be called standard, this furnace attains to 84.4 per cent, of that ideal; and it is also true that this fur- nace generates from the carbon burnt at the tuyeres 84.4 per cent, of the amount of heat which could have been generated if Gruner's ideal working had been attained. It is always possible to find out for any given blast furnace, by similar calculations, how much carbon is burned at the tuyeres, and how much is burned above the tuyeres, and thus to determine how closely the furnace running approximates to Gruner's ideal working. This proportion or percentage will not necessarily express how efficiently the furnace is running, as regards fuel used per unit of iron made, but it will tell what proportion of the calorific power of the fuel used is being gen- erated at the tuyeres, and in possibly nine cases out of ten this proportion indicates the general efficiency of the furnace as regards fuel consumption. It will be next profitable to inquire when and under what conditions Gruner's ideal working does not correspond to maximum fuel economy, and why it usually does. The answer is not difficult to understand: if all the carbon gasified in the furnace is burned to CO at the tuyeres, 30 percent of the total calorific power of the carbon burned is there developed, which is more than half of all the heat generated from carbon in the furnace. To this must also be added the sensible heat in hot blast, which may amount (as in Problem 52) to some 5.9 per cent, of the calorific power of the carbon, making, therefore, a total of 35 per cent, of the calorific energy of the fuel gen- erated at the tuyeres out of a total of about 50 per cent, de- veloped in the furnace. If, however, the blast be heated to a very high temperature, or particularly if it be dried, or if the ore and fuel are extra pure, so that a smaller quantity of heat is needed to melt down slag at the tuyeres, then there may not 276 METALLURGICAL CALCULATIONS. be needed at the tuyeres the generation by combustion of so much heat as Griiner's ideal working would require and cause to be produced, and to burn at the tuyeres all the carbon oxi- dized in the furnace would be wasteful of fuel. In this, case although less heat would be generated per unit of fuel, by burning some of it above the tuyeres, yet economy in fuel con- sumption as a whole would be attained, because of the better distribution of the heat which was generated from a smaller total quantity of fuel. Illustration. In Problem 55 we assumed that the furnace ran with blast heated to 450 C., and that this hot blast, burn- ing at the tuyeres 84.4 per cent, of all the carbon gasified in the furnace, smelted down pig iron and slag satisfactorily and kept the. tuyere region at proper temperature. If the temperature of this blast were raised to 900 C., how much greater pro- portion of heat would be available in the tuyere region? We have already calculated that the 115,860 cubic feet of blast used per ton of iron made, brought in at 450 C., 1,026,936 pound calories of heat, equal to 5.9 per cent, of the calorific power of the fuel put into the furnace. If the temperature were 900 C., the heat brought in would be 115,860 X [0.303 + 0.000027 (900)] X 900 = 37,920,978 oz. cal. = 2,370,061 Ib. cal. which equals = 0.137 - 13.7 per cent. of the calorific power of the fuel, a gain of 7.8 per cent, added to the heat available in the tuyere region. This causes a very great increase in the smelting-down power of the furnace, enabling the same work per ton of ore smelted to be done with much less consumption of fuel in the tuyere region. An idea of this increased smelting-power may be obtained from the following comparison of heat available for smelting purposes in the tuyere region in the two cases just discussed: CASE 1. Heat developed by oxidation of carbon per ton of iron made, 1,627 lbs.X2,430 = 3,953,610 Ib. cal. Sensible heat in blast at 450 C. = 1,026,936 Total heat available = 4,980,546 UTILIZATION OF FUEL IN BLAST FURNACE. 277 CASE 2. Heat developed by same quantity of air burning same carbon = 3,953,610 " Sensible heat in blast at 900 C. = 2,370,061 Total heat available = 6,323,671 It is, therefore, seen that the heat generated and available at the tuyeres is increased 1,343,125 calories, amounting to 27 per cent, of the amount disposable in Case 1. It follows, there- fore, that the smelting down power has been increased 27 per cent., and that, if the 4,980,546 calories were sufficient for satisfactorily smelting down the iron and slag in the first in- stance, that the extra heat of Case 2 can all be utilized for smelting down 27 per cent, extra burden. We can, there- fore, charge 27 per cent, more burden per unit weight of coke in Case 2, because we have the requisite smelting down power at the region of the tuyeres, which amounts to saying that we can charge 22 per cent, less coke for a given weight of pig iron made. In actual practice, as the amount of burden is increased and the temperature of the blast increased, the change causes more and more of the carbon to be oxidized above the tuyeres, and a smaller proportion to be oxidized at the tuyeres, thus ob- taining less service in the furnace from oxidation of carbon as a whole, but compensating for this by the extra heat in the hot blast. Or, looking at it in another way, we may say that the same heat could be made available in the region of the tuyeres, when using hot blast, by the combustion there of a smaller quantity of carbon; therefore, we can burn more of it above the tuyeres and yet work more economically on the whole, than we were working in the first instance, with the colder blast. MINIMUM CARBON NECESSARY IN THE FURNACE. Many writers have assumed that in the reduction of iron, oxide, such as Fe 2 O 3 , the reaction of its reduction by carbonous oxide, CO, is expressed as follows: Fe 2 O 3 + SCO = 2Fe + 3CO 2 If this were true, there would need to be burnt to CO at the 278 METALLURGICAL CALCULATIONS. tuyeres only 3C, or 36 parts of carbon, to ensure the reduction of Fe 2 O 3 , representing 112 parts of iron. The reaction does not, however, progress as shown, because CO 2 acts oxidizingly on Fe to such a degree that when ICO 2 is present in the gas for ICO left unused, the reduction practically stops, even though the gases are moving slowly through the warm ore. The real reaction of reduction by CO gas is therefore more nearly represented by Fe 2 O 3 + 6CO = 2Fe + 3CO + 3CO 2 , which shows that 112 parts of iron would require at least 72 parts of carbon to be oxidized at the tuyeres to CO, in order to produce the gas necessary for its reduction. The presence of some CO 2 in the furnace coming directly from carbonates in the charge would neutralize still more of the reducing power of the CO gas, and cause still more of it to be theoretically re- quired for reduction. The minimum amount of carbon neces- sary to be charged in the furnace will be that necessary to furnish fixed carbon enough for this reducing gas and for the carbon in the pig iron. This would be, per 100 parts of pig iron, containing say 93 iron and 3 carbon, and using coke con- taining 90 per cent, of fixed carbon: Carbon burnt at tuyeres = 72X93-^112 =59.8 Carbon in pig iron . =3.0 Total fixed carbon necessary = 62 . 8 Total coke to supply this = 62.8 4- 0.9 =69.8 It results from these calculations that if " Griiner's ideal working " of a blast furnace were carried out to the practical extent of reducing all the charge by carbonous oxide, CO, and oxidizing no carbon at all directly above the tuyeres, that about 63 parts of fixed carbon would be required per 100 of pig iron made, requiring from 70 to 80 parts of fuel, according to its richness in fixed carbon (90 to 80 per cent.). In prac- tice, as is well known, more than this is commonly used, be- cause of the larger proportion of unused CO in the gases than above assumed: and less than this has been regularly used, showing that economy of fuel can be attained without adher- ing to " Griiner's ideal working," in fact, by transgressing it as far as one dares. UTILIZATION OF FUEL IN BLAST FURNACE. 279 The principle involved can be best grasped by a calculation of the amount of carbon which would be required by the fur- nace, supposing all the heat necessary for melting down the charge were supplied by electrical means, thus dispensing, for the purposes of this supposition, with the necessity of blast and the consequent necessity of oxidizing any carbon by any other agent than the oxygen given up by the ore. - In this case the gases resulting would be, let us assume, of the same com- position as before, that is, containing equal volumes of CO and CO 2 , and since this oxygen is abstracted altogether from the ore, the reaction is Fe 2 3 + 2C = 2Fe + CO + C0 2 This would represent the utilization of carbon in an electrically- heated furnace, and would require per 100 of pig iron made, assuming it 3 per cent, carbon and 93 iron : Carbon for reduction 24 X 93 -^ 1 12 =19.9 Carbon in pig iron = 3.0 Total fixed carbon necessary = 22 . 9 Or only a little over one-third as much as the minimum re- quired when the smelting down is done by blast. Aside from electrical furnace practice, however, this discus- sion proves that whatever fixed carbon burns or oxidizes above the region of the tuyeres, in a blast furnace, absorbs oxygen from the charges with three times the efficiency of carbon first burnt at the tuyeres. Every pound of oxygen abstracted from the charges by solid carbon requires the use or intervention of only one-third as much carbon as that which is abstracted by CO gas; or, each pound of carbon abstracting oxygen directly from the charge takes from it three times as much oxygen as a pound of carbon first burnt to CO at the tuyeres possibly can. The ordinary furnace produces at the tuyeres, in order to get heat enough to melt down the charges, more CO gas than is needed to abstract all the oxygen from the charges; under these conditions it is uneconomical to oxidize any carbon at all above the tuyeres. The exceptional furnace, because of pure ores, small amount of slag, pure fuel, high temperature of blast, or dry blast, gets heat enough at the tuyeres to melt 280 METALLURGICAL CALCULATIONS. down the charges without producing enough CO gas to reduce all the charges; under these conditions more or less reduction is effected by solid carbon, and with the greatest economy in quantity of carbon required in the furnace. These are the conditions under which, having passed the turning point, the greater economy of fuel is attained the farther away one can get from " Gruner's ideal working." CHAPTER IV. THE HEAT BALANCE SHEET OF THE BLAST FURNACE. Twenty-eight years ago, Sir Lothian Bell first constructed a satisfactory heat -balance sheet for a blast furnace. His ob- servations were largely, and his experience altogether, confined to the reduction of the argillaceous siderite ores of the Cleve- land district, England, and although he made numerous at- tempts to draw general conclusions from the data at hand applicable to iron smelting in general, yet many of his deduc- tions remain true only for the particular ores and manner of working characteristic of the Cleveland district. No treatment of this subject, however, can be based other- wise than upon Bell's researches, following the lines laid down, in his " Principles of the Manufacture of Iron and Steel." HEAT RECEIVED AND DEVELOPED. The items on this side of the balance sheet are: (1) Combustion of carbon to carbonous oxide (CO). (2) Combustion of carbon to carbonic oxide (CO 2 ). (3) Sensible heat of the hot blast. (4) Heat of formation of the pig iron from its constituents. (5) Heat of formation of slag from its oxide constituents. (1) and (2) Combustion of carbon in the furnace. There is but one satisfactory way to determine with exactness the amounts under this heading. From the balance sheet, the total amount of carbon passing into the gases is obtained; from the analysis of the gases, the weight of carbon per unit volume of gases is calculated; the first divided by the second gives the volume of gases per unit weight of pig iron produced. The amount of CO and CO 2 in these gases is then obtained by use of the gas analysis, and if from the total CO and CO 2 in the gases there be subtracted the CO and CO 2 contributed as such by the solid charges, the difference is the CO and CO 2 which have been formed in the furnace. The heat evolved in the formation of these quantities can then be calculated. 281 282 METALLURGICAL CALCULATIONS. Illustration. In Problem 51 it was calculated that per 1,000 kilos, of pig iron produced, 534.09 kg. of carbon went into the gases; also that the analysis of the gases showed 0.20736 kg. of carbon in each cubic meter of gas. The quotient indicated, therefore, 2575.6 cubic meters of gas produced per ton of pig iron. From the analysis of the gases there was in this volume, 2575.6X0.231 = 595.0 m 3 of CO 2575.6X0.148 = 381.2 m 3 of CO 2 whose weights were 595.0X1.26 = 749.7kg. CO 381.2X1.98 = 754.8kg. CO 2 The balance sheet shows, however, 49.1 kg. of CO 2 contained in the limestone flux used, which can be assumed as entering the gases bodily. Subtracting this we have 705.7 kg. of CO 2 formed in the furnace, and 749.7 kg. of CO, containing respectively 705. 7 X = 192.5 kg. of C in CO 2 749.7X^1 = 321.3 kg. of C in CO The heat generated in the furnace by the oxidation of carbon is, therefore, 192.5X8100 = 1,559,250 Calories 321.3X2430 = 780,760 " 2,340,010 If this carbon could have been entirely burnt to CO 2 , there would have been generated 513.8X8100 = 4,161,780 Calories Showing that only 56 per cent, of the calorific power of the carbon was developed in the furnace; the other 46 per cent, exists as potential calorific power in the waste gases, and part THE HEAT BALANCE SHEET. 283 of it is really put back into the furnace as sensible heat in the hot blast. There is a little doubt as to how to consider the CH 4 in the gases; that is, whether the heat of its formation should be reckoned in as developed in the furnace. This would be (C,H 4 ) = 22,250, or 1,854 Calories per kg. of carbon contained therein. Its presence in the gas probably results largely from the dis- tillation of the fuel at a high temperature, and the heat re- quired to disunite the CH 4 from the solid fuel is probably as great as is represented by its heat of formation from carbon and hydrogen. The item is, therefore, a doubtful one, and as far as we know, we may be coming about as near to the truth by omitting it altogether as by counting it in. If we wished to add it in the illustration just given the calculation would be: Volume of CH 4 = 2575.6 X 0.005 = 12.88 m 3 Weight of C = 12.88X0.54 = 6.9 kg. Heat of formation 6.9 X 1,854 = 12,793 cal. It should be emphasized that in this calculated heat of oxi- dation of carbon in the furnace, no account has been taken whatever of where in the furnace this heat is generated. Above all, the mistake should not be made of supposing that the 780,760 Calories produced by formation of CO represents the heat generation at the region of the tuyeres ; nothing could be further from the truth. A great deal of carbon is burnt to CO at the tuyeres, and some above the tuyeres, but a goodly pro- portion of this CO oxidizes by abstracting oxygen from the charge and becomes CO 2 . It would not be incorrect, however, to divide the heat of oxidation of carbon in the furnace into two parts, viz: to assume all the carbon as first forming CO, and part of this CO afterwards forming CO 2 , corresponding to the amount of the latter formed in the furnace. If this were done, we would have 513.8 kg. C to CO = 513.8X2430 = 1,248,535 Cal. 449.2 kg. CO to CO 2 = 449.2 X 2430 = 1,091,475 " 2,340,010 This analysis gives us the information that of the total heat generated by the oxidation of carbon in the furnace, some- 284 METALLURGICAL CALCULATIONS. where about one-half is generated by its burning to CO, and the other half by the further oxidation of CO to CO 2 ; and we also know that the larger part of the former takes place at the tuyeres, and all of the latter takes place during the reduction of the charges in the upper part of the furnace. If we know, however, or have calculated the amount of the blast received by the furnace, or, more properly speaking, the amount of oxygen in the blast, then the heat generated by oxidation of carbon at the tuyeres becomes known. In the previous illustration, taken from Problem 51, we can also take from the same problem the weight of oxygen in the blast, 557.7 kilos. This would burn 557.7X0.75 = 418.3 kg. of carbon to CO at the tuyers, generating there 418.3X2430 = 1,016,410 Calories, or 44 per cent, of all the heat generated by oxidation of carbon in the furnace, leaving 1,323,610 Calories as generated above the tuyeres by the agency of the oxygen of the charges. These figures tell us just where and how the principal items of heat were generated in this particular furnace, and the similar cal- culation may be made for any blast furnace for which we have the necessary data. (3) Sensible heat in the hot blast. To calculate this item, we need to know the weight or volume of the different con- stituents of the blast and their temperature. The question at once arises, as to what base line of temperature shall be chosen. It is most -convenient to choose C., since that is not over 15 from the average temperature in the largest iron produc- ing countries. However, any other prevailing temperature may be taken as the base line, involving merely a little more calculation, since our specific heats are reckoned from C. The temperature ought, moreover, to be taken as near to the tuyeres as possible, to properly take into account the effect of cooling of the blast in the bustle and feeder pipes, from radiation and expansion. The blast consists of air p oper and moisture, the former with a mean specific heat between and t C of 0.303 + 0.000027t, in kilogram Calories per cubic meter, or in ounce calories per cubic foot, the latter with a similar mean specific heat of 0.34+ O.OOOlSt. Since the mois- ture at times amounts to as much as 5 per cent, of the blast, it should be calculated separately. THE HEAT BALANCE SHEET. 285 Illustration: With the outside air at 30 C., and saturated with moisture (raining), calculate the heat carried into a blast furnace by blast carrying in 1859.1 kilos, of nitrogen, the tem- perature of the heated blast being 600 C. Barometer 720 millimeters of mercury. Temperature base line C. Solution: One cubic meter of the moist blast, as taken into the blowing cylinders, carries all the moisture it can hold, the tension of which is therefore 31.5 millimeters. The tension of the air proper present is therefore 720-31.5 = 688.5 milli- meters, and each cubic meter of moist air carries 31 5 ' = 0.0438 cubic meter of moisture, and oo fr ' = 0.9562 cubic meter of air proper. 7 20 Whatever the temperature of the blast, the moisture and air proper will be in this same proportion whenever its tempera- ture is over 30 C. If the temperature were C. the moisture would be mostly condensed, but for the purposes of calculating the heat brought in we may assume the moist air to be at C., with its moisture uncondensed. That volume of blast which would be 1 cubic meter at and 760 mm. pressure, would, therefore, bring in, at 600 C., the following quantity of heat : H 2 0.0438X[0.34 +0.00015 (600)] X600 = 11.3 Calories Air 0.9562 X [0.303 + 0.000027 (600)] X 600 = 179.7 " Total 191.0 Since the nitrogen present in this is 0.9562 X 1.293 X-^ = 0.9511 kg. lo the heat brought in per 1859.1 kg. of nitrogen is 373 > 344 Calories. An amount equal to over one-third of all the heat generated by combustion of carbon at the tuyeres. (4) Heat of formation of pig iron from its constituents. The 286 METALLURGICAL CALCULATIONS. pig iron contains several per cent., some 5 to 10 altogether, of carbon, silicon, manganese, phosphorus, sulphur and other elements. The energy of their combination with the iron is a somewhat indefinite quantity, and in no case can be consid- able. Berthelot states the energy of combination of carbon with iron as (Fe 3 ,C) = 8,460, which would be 705 Calories per kilogram of carbon, and another investigator (Ponthiere) states the heat of combination of phosphorus with iron to be zero. In the present state of uncertainty it is hardly allowable to add in any other than the heat of combination of the car- bon in the iron, and leave out that of the other elements. (5) Heat of formation of the slag from its constituent oxides, Here we touch upon a qur'itity of more than insignificant proportions, yet which is not yet quantitatively known with satisfactory accuracy. The main constituents of the slag are SiO 2 , APO 3 , CaO, MgO and CaS, which are provided by clay, limestone and iron sulphide. If we allow, on the other side of the balance sheet, for the heat necessary to de-hydrate clay, drive carbonic acid off carbonates, and break up iron sul- phide and enough CaO to furnish Ca for CaS, we are then en- titled, on the other hand, to place in the heat evolution column the heat of combination of aluminum silicate with lime and magnesia, the heat of formation of CaS and its heat of so- lution in the silicate slag. The heat of formation of CaS is 94,300 Calories, or 2,947 Calories per kilogram of sulphur; its heat of combination with a silicate slag is unknown. The heat of combination of lime with aluminum silicate has been determined only for the proportions 3CaO to APSi 2 O 7 , that is, for 168 parts of CaO uniting with 222 parts of aluminum sili- cate. This has been determined in Le Chatelier's laboratory as (3CaO, Al 2 Si 2 O 7 ) = 33,500 Calories which is 200 Calories per unit of CaO combining, or 150 Calories per unit weight of A1 2 O 3 + SiO 2 . The calculation would be made on the basis of the amount of lime (plus lime equivalent of magnesia present), if it were present in a smaller ratio than 168 to 222 of silica and alumina, and on the basis of the silica and alumina, if their ratio to the summated lime were less than 222 to 168. It is probable that in the near future these quantities will be known more accurately. One item of heat received by the furnace has not been men- THE HEAT BALANCE SHEET. 287 tioned, because of its usual absence, viz.: heat in hot charges. Very rarely roasted ore comes hot to the furnace, in which case its sensible heat must be counted in, else the thermal sheet of the furnace will be that much out of balance. HEAT ABSORPTION AND DISBURSEMENT. The items on this side of the balance sheet are: (1) Sensible heat in waste gases, including water vapor only as vapor. (2) Sensible heat in outflowing slag. (3) Sensible heat in outflowing pig iron. (4) Heat conducted to the ground. (5) Heat conducted and radiated to the air. (6) Heat abstracted by cooling water, tuyeres, etc. (7) Heat for de-hydrating the charges. (8) Heat for vaporizing water from charges. (9) Heat absorbed by decomposition of carbonates. (10) Heat absorbed in reduction of iron oxides. (11) Heat .absorbed in reduction of other metallic oxides. (12) Heat absorbed by decomposition of moisture of the blast. (1) Sensible heat in waste gases. The amount of these gases is known only from the carbon contained in unit volume, by analysis, and the known weight of carbon entering and leav- ing the furnace. If there is much fine coke carried over by the blast, allowance must be made for the carbon in it, be- cause this would not be represented in the gas analysis. The analysis of completely dried gas is that usually obtained, be- cause if the gas is measured without drying, an uncertain amount of moisture is condensed, and, therefore, it is usual to dry before measuring and analyzing. The amount of mois- ture in the gases is either assumed as that driven off from the charges, as shown by the balance sheet, or else is determined directly by drawing the gases through a calcium chloride tube or other dessicating apparatus. Several tests should be made to get a fair average, because much more will be in the gases immediately after charging than immediately before. Dust must be excluded from the drying tube by filtering the gases through dry asbestos. The average temperature of the gases should be known over a considerable period, a thermo-couple 288 METALLURGICAL CALCULATIONS. in the down-comer gives this most accurately and more uni- formly than if inserted above the stock line in the furnace. The weight of moisture per unit volume of dry gas is then converted into volume at standard conditions by dividing by 0.81 (Icubic meter = 0.81 kg; 1 cubic foot = 0.81 ounce avoirdup.). The sensible heat of the gases is then calculated, using C. as the base line, and the proper mean specific heats of the gases per unit of volume. The water vapor will here be considered simply as a gas, and its sensible heat above water vapor at only calculated. This leaves the latent heat of vaporization of this water to be considered as a separate item (606.5 Calories), that is, as heat absorbed by reactions in the furnace, thus putting it on exactly the same footing as the CO 2 in the gases which has been expelled from carbonates in the furnace. By so proceeding much uncertainty as to the heat in the water vapor is avoided. If the amount of flue dust is considerable its quantity should be ascertained, and the heat in it also calculated and added in to the heat in the moist gases. Its specific heat may be ap- proximated as so much carbon, iron oxide and silica, the pro- portions of each of these present being known. (2) Sensible heat in outflowing slag. The weight of slag produced is seldom taken directly, but can be reckoned up with all needful accuracy from the balance sheet of materials entering and leaving. Its temperature and specific heat, solid and liquid, melting point and latent heat of fusion, are un- fortunately almost always unknown factors. The one datum which is needful, however, is the total heat in a unit weight of liquid slag as it flows from the furnace, and this is not a diffi- cult quantity to obtain. A rough calorimeter with a reliable thermometer and containing a carefully weighed quantity of water, may be easily constructed. Some liquid slag is run directly into it, and by observing the rise of temperature and afterwards filtering out, drying and weighing the granulated slag, a satisfactory determination can be arrived at. This is corrected to C. by using an approximate specific heat, say 0.20, for the range of final calorimeter temperature to zero. In this connection it is important to note that the calorimetric determinations of Akerman on blast-furnace slags, give the heat in the just-melted slag, whereas slags flowing out of a THE HEAT BALANCE SHEET. 289 furnace are considerably, some 200 to 500 C., above their melting point, and therefore contain some 50 to 150 Calories more heat than that given by Akerman for a slag of similar com- position. Since Akerman 's values run from 350 to 400 Calories, the actual heat in the outflowing slag may be between 400 and 550 Calories. Akerman himself states that an average of twenty-seven Swedish furnaces gave 530 Calories as the actual heat in unit weight of outflowing slag, and Bell uses 550 in most of his calculations on Cleveland (England) furnaces. (3) Heat in outflowing pig iron. The heat in just-melted pig iron is evidently too small a quantity to use in this con- nection. The heat in the outflowing pig iron at 200 to 500 above its melting point will be 50 to 100 Calories greater. The former quantity is about 245 Calories; the latter will be 300 to 350. Bell takes 330 for Cleveland furnaces; Akerman states 250 to 325 for Swedish furnaces. We may conclude then, to use 300 Calories for a coke furnace running cool, and up to 350 Calories for a very hot furnace. (4) Heat conducted to the ground. This is a very uncertain quantity. It varies with the kind of ground, and is more nearly a constant per day than per unit of pig iron produced. It is, therefore, expressed per unit of iron produced, larger for small furnaces run slowly than for large furnaces run fast. It is less when running rich ores and greater with poor ores, other things being equal. As nearly as can be assumed we would put this item as lying between 60 and 200 Calories per unit of pig iron made. Bell uses 169 on one Cleveland furnace, but it is certainly less than 100 in some charcoal furnaces using pure ores and fuel, and consequently with a small heat re- quirement. (5) Heat conducted and radiated to the air. This item is likewise more nearly a constant quantity per day for a given furnace, and is therefore less per unit of iron produced the faster the furnace is run. It may vary between 60 and 250 Calories per unit of pig iron, the former in furnaces of low heat requirement per unit of iron produced, the latter in those of high heat requirement. If the amount were calculated it would figure out as a time function, and would require the temperature of the outside shell, that of the air, the velocity of the wind, and the total outside surface, in order to calculate 290 METALLURGICAL CALCULATIONS. by the principles of heat radiation and conduction, the amount radiated per day. No one. has done this yet for any one fur- nace, and, in brief, items (4) and (5) of this schedule are usually grouped together and determined simply by difference, their sum aggregating from 100 to 500 Calories per unit of pig iron, averaging 100 to 150 for charcoal furnaces of low heat requirement, 200 to 450 for Cleveland furnaces (Bell), and 300 to 500 for large, modern furnaces with thin walls and great height. (6) Heat abstracted by cooling water. In the old-fashioned heavy masonry, cold blast furnace, this item was zero. With the advent of hot blast, the water needed for cooling the tuyeres entered as a heat abstracting factor. It is greater the harder a furnace is blown, but does not increase proportionately with the output. The heat lost by tuyere-cooling water may be 50 to 100 Calories per unit of pig iron made. That for cooling of bosh plates and the outside of the crucible in modern fur- naces, may vary all the way up to 200 Calories. These two items are very large in a modern furnace, but are necessary expenditures of heat energy in order to preserve the lines of the furnace during fast running. For any particular furnace they may be determined with all needful accuracy by mea- suring the amount of water pumped or used for these purposes and its temperature before and after using. (7) and (8) Drying and de-hydrating charges. Water goes into the furnace as moisture and as combined water of the charges. To convert the moisture, such as is evaporated by a current of moderately warm air, into vapor requires 606.5 Calories per unit of water. This allows merely for its vapor- ization in the furnace, and not for any sensible heat which it may carry out of the furnace at the temperature of the waste gases. This latter item is properly considered in with the sensible heat of the waste gases. The common practice of saying that it takes 637 Calories to evaporate the moisture of the charges is wrong, because this amount would convert water at to vapor at 100, and, therefore, would include part of what is properly the sensible heat of the waste gases. On the other hand, it is equally wrong to say that this heat of vapor- ization should be counted in as sensible heat in the hot gases; it would be just as logical or, rather, equally illogical, to count the latent heat of vaporization of CO 2 as sensible heat in the gases. THE HEAT BALANCE SHEET. 291 To drive off water of hydration from hydrated minerals in the charge requires an additional amount of chemically-ab- sorbed heat. As far as is known, this is small for the water driven off hydrated iron oxides, so small as to be a doubtful quantity and safely left out ; but if it comes from clay the large amount of 611 Calories is absorbed in merely separating it from its chemical combination (2H 2 O, APSi 2 O 7 ) = 22,000 Calories, which would require 611 + 607 = 1,218 Calories to put into the state of vapor each unit weight of water entering the fur- nace chemically combined in clay. (This does not concern the ordinary moisture in moist clay, expelled at 100 C., but only the combined water in the dry clay). Where much clay occurs in the ores this quantity becomes important, and its amount explains some of the difficulties met in working clayey charges, particularly since a large part of this chemically com- bined water is expelled only at a red heat, and, therefore, cools greatly the hotter zones of the furnace. (9) Decomposition of carbonates. Raw limestone, or dolo- mite, is the usual flux of the blast furnace, and its carbonic acid is evolved at temperatures between 600 and 800. Whether some of this is subsequently decomposed by contact with carbon and reduced to CO, is immaterial to the balance sheet, because more than enough CO 2 escapes from the furnace to represent the CO 2 of the flux, and we charge the furnace only with the formation of the CO and CO 2 actually found in the gases, less the CO 2 from fluxes. Bell charges up the heat absorbed also in the assumed reaction CO 2 + C = 2CO, but this is an error, because it is doubtful how much of the CO 2 is thus decomposed, and the question, in its last analysis, is one of heat distribution in the furnace, and does not concern the totals of heat absorbed or evolved. We can, therefore, omit the item of decomposition of this CO 2 (as likewise, and for analogous reasons, the heat evolved in carbon deposition in the upper part of the furnace 2CO = C + CO 2 ), and need consider only the heat required to expel the CO 2 from car- bonates. This is : (CaO, CO 2 ) = 45,150 Calories = 1,026 Calories per kg. CO 2 (MgO, CO 2 ) = 29,300 " = 666 (MnO, CO 2 ) = 22,200 " = 500 " (FeO, CO 2 ) = 24,900 " = 566 " (ZnO, CO 2 ) = 15,500 " = 352 " 292 METALLURGICAL CALCULATIONS. By using the above figures, in connection with the known composition of ore and fluxes, the heat required to decompose carbonates can be correctly calculated. (10) Reduction of iron oxides. The heats of formation of the various oxides of iron are: (Fe,O) = 65,700 Calories = 1,173 Calories per kg. iron (Fe 3 ,O 4 ) = 270,800 " = 1,671 (Fe 2 ,O 3 ) = 195,600 " = 1,746 And, therefore, just these quanities of heat are required per unit weight of iron reduced from these compounds. If the ore is a carbonate the heat absorbed in driving of CO 2 from FeCO 3 can be first allowed for, and then the heat required for reduction of the FeO calculated on the weight of the re- duced iron. If some FeO goes into the slag it will be as FeO, and if the ore was Fe 2 3 , or Fe 3 4 , the reduction of unit weight of iron from the state of Fe 2 3 , or Fe 3 O 4 , to the state of FeO, absorbs respectively 573 or 498 Calories, as may be readily deduced from the heats of formation of the three oxides con- cerned. If FeS is present its heat of formation is (Fe, S) = 24,000 Calories = 429 Calories per kg. Fe. If the iron is charged partly as silicate, such as mill or tap cinder, an additional amount of heat will be required for re- duction, equal to that needed to separate the iron oxides from their combination with silica. The heat of formation of the bi-silicate slag only has been determined. (FeO, SiO 2 ) = 8,900 Calories = 148 Calories per kg. SiO 2 . And since the cinders concerned contain relatively more iron than this, we can best make allowance for the heat required to set free the silica. It is necessary, therefore, to take the amount of silica in the iron cinder charged, and allow, as necessary for its decomposition into FeO and SiO 2 , 148 Calories for each unit weight of SiO 2 contained. (11) Reduction of non-ferrous oxides. Silicon is usually present in pig iron, its reduction from silica requiring: (Si, O 2 ) = 180,000 Calories = 6,413 Calories per kg. Si. There is a little doubt about this (Berthelot's) figure; more THE HEAT BALANCE SHEET 293 recent determinations, not yet published, point rather to 196,000 and 7,000 Calories respectively. Manganese is often present in the ores as Mn 2 O 3 , Mn 3 O 4 , or MnO 2 , and going partly into the slag as MnO. The heat ab- sorbed in reduction to manganese is: (Mn, O) = 90,900 Calories = 1,653 Calories per kg.Mn (Mn 3 , O 4 ) = 328,000 " = 1,988 " (Mn, O 2 ) = 125,300 " = 2,278 " For the MnO produced and going into the slag, the reduc- tion from Mn 3 O 4 , or MnO 2 , per unit weight of contained man- ganese, absorbs 335 or 625 Calories respectively. Sulphur generally comes from the reduction of FeS, re- quiring 667 Calories per kg. of sulphur; but care must be taken not to allow for this heat twice, since if reckoned once under the head of iron reduction it must not be reckoned on the balance sheet a second time under sulphur. One reduction of FeS liberates both constituents. Phosphorus may be reduced in large quantity in making basic iron. It probably comes mostly from calcium phosphate, in which case we must reckon on not only the heat of oxidation of phosphoric oxide but also its heat of combination with lime: (P 2 , O 5 ) = 365,300 Cal. = 5,892 Cal. per kg. of P. (3CaO, P 2 O 5 ) = 159,400 " = 2,410 " " contained. Making a total heat requirement of 8,302 Calories to separate unit weight of phosphorus from phosphate of lime, and leave free lime. In a furnace making a pig iron with several per cent, of phosphorus, this item becomes quite large. Calcium occurs in the slag as CaS, its reduction from lime requiring (Ca, O) = 130,500 Calories = 3,263 Calories per kg. Ca. Other elements than the above rarely occur in pig iron in notable quantity. If rare ones occur, their heat of reduction can be sought in thermochemical tables. Those of tungsten, titanium, molybdendum and chromium are, however, not at present known. (12) Decomposition of moisture in the blast. This is to be 294 METALLURGICAL CALCULATIONS. counted as vapor of water, the heat required to decompose, which is: (H 2 , O) vapor = 58,060 Cal. = 3,226 Cal. per kg. H 2 O. = 29,030 " " H 2 . It is not correct to allow here for the sensible heat in this water vapor coming in with the hot blast, because that heat should go on the other side of the balance sheet as heat de- livered to the furnace. Neither is it correct to subtract the heat of combination of the oxygen of this moisture with car- bon to form CO at the tuyeres; because, although that com- bination actually does take place, yet the heat thereby evolved properly belongs also on the other side of the balance sheet, and is there properly taken care of as part of the heat of oxida- tion of carbon in the furnace. Problem 57. (Data partly from paper by the author, Transactions Ameri- can Institute of Mining Engineers, 1905). A blast furnace running on Lake Superior ore has the fol- lowing charges, per 100 of pig iron produced: Hematite ore: 177.6; composition H 2 O 10.0 per cent. SiO 2 10.0 A1 2 O 3 - 3.5 Fe 2 3 76.5 Limestone: 44.4; composition SiO 2 5.0 MgO 4.8 CaO -47.6 CO 2 42.6 Coke: 95.8;composition SiO 2 - 5.3 CaO 5.3 H 2 O - 1.0 C 88.0 Pig iron produced: 100; composition Si 1.0 C 4.0 Fe 95.0 Gases produced : composition dry : CO 2 13.0 "v\. CO 22.3 N 2 64.7 THE HEAT BALANCE SHEET. 295 Blast used: Contained 5.66 grains of moisture per cubic foot of dry air, at 24 C. = 75 F. Charges in pounds: Coke, 10,200; ore, 20,000; stone 5,000. Product per day: Pig iron, 358 tons = 801,920 pounds. Coke used per day: 768,626 pounds. Temperature of blast 720 F. = 382 C. Temperature of waste gases = 538 F. = 281 C. Displacement of blowing engines = 40,000 ft. 3 per min. Heat in one unit of pig iron = 325 Calories. Heat in one unit of slag = 525 Calories. Cooling water, per day, heated 50 C. = 300,000 gallons. Required; (1) The volume of gases per 100 kg. of pig iron (2) A balance sheet of materials entering and leaving the furnace, per 100 unit of pig iron. (3) The volume and weight of blast per 100 kg. of pig iron. (4) The efficiency of the blowing plant. (5) The heat balance sheet of the furnace. (6) The proportion of the fixed carbon of the fuel burnt at the tuyeres. (7) The proportion of the whole heat generated at the tuyeres. (8) The proportion of the iron reduced in the furnace which is reduced by solid carbon from FeO. (9) The theoretical maximum of temperature at the tuyeres. (10) The theoretical maximum if all the moisture were removed from the blast. Solution: (1) To find the volume of the gases : Carbon in the coke used 95.8X0.88 = 84.3 kg. Carbon m the limestone 44.4X0.426 X - = 5.2 " Total carbon entering furnace =89.5 Carbon in 100 of pig iron = 4.0 Carbon entering gases =85.5 Carbonous and carbonic oxides in gases = 35.3 per cent Carbon in 1 m 3 of dried gas = 0.54X0.353 = 0.19062 kg. 85 5 Dry gas per 100 kg. of pig iron = () 19Q62 = 448.5 m 3 . (1) 296 METALLURGICAL CALCULATIONS. Dry gas per 100 oz. of pig iron = 448.5 ft. 3 Dry gas per 100 pounds of pig iron = 448.5 X16 =7176.0" Dry gas per 2240 pounds of pig iron = 7176.0 X22.4 = 160,742 " 160,742X358 (tons) Dry gas per minute = ' 6Q><24 = 39,962" (2) BALANCE SHEET OF MATERIALS, PER 100 OF PIG-IRON. CO {^ Charges Fe 2 O 3 135.6 Pig Iron Slag Fe 95 . Gases o 40 7 1-* o H 2 O 17.8 H 2 O 7 8 6 SiO 2 17.8 Si 1.0 SiO 2 15.7 O 1.1 ^ SiO 2 2.2 SiO 2 2.2 s CaO 21.1 CaO 21.1 q MgO 2 . 1 MgO 2 . 1 s CO 2 19 CO 2 19 C 84 3* C 40 c 80 3 % SiO 2 5 . 3 SiO 2 5.3 H CaO 5 . 3 CaO 5.3 fe H 2 O 9 H 2 O 9 (N O 2 96 . 4 o 96.4 s N 2 321 3 N 2 321 3 s H 2 O 4.5 1 H 5 5 [o 4.0 Total 740.0 100.0 58.0 582.0 The charges are calculated simply from the weights of ore, flux and fuel used, and their percentage composition. The blast is calculated as given in solution of requirement (3). The 1.0 of silicon in the pig iron requires 1.0 X (60-5-28) = 2.1 parts of SiO 2 to furnish it, leaving 15.7 of unreduced SiO 2 to go into the slag, and 1.1 of oxygen to the gases. (3) The balance sheet shows that the solid charges furnish to the gases 41.8 of O, 190 of CO 2 and 18.7 of H 2 0. The H 2 O goes as such into the gases, and therefore its oxygen is not present in the sample of dry gas analyzed. The oxygen in CO 2 is 19.0 X (32 -r- 44) = 13.8 kg., which added to the 41.8 gives 55.6 of oxygen getting into the gases as CO or CO 2 , from the solid charges. THE HEAT BALANCE SHEET. 297 The oxygen in the CO and CO 2 of the gases is to be calcu- lated from the oxygen in unit volume of gas and the total volume of gas produced. The oxygen in 1 cubic meter of gas will be: O in CO = 0.223X (o.09 X y \ X ^| = 0.16056kg. (0.09 X f ) roo 0.09X -^1 X = 0.18720 0.34776 ' A quicker solution is to note that CO 2 represents O 2 , its o/m volume of oxygen, and CO represents O, or half its volume of oxygen, and since 1 cubic meter of oxygen weighs 0.09X (32-* 2) = 1.44 kilograms, the weight of oxygen required is (n 22*3 \ -^~ + 0.13J Xl.44 = 0.34776 kg. The total oxygen in 448.5 m 3 of gases is therefore 448.5X0.34776 = 156 kg subtracting that furnished by the solid charges = 55.6 " there remains oxygen furnished by blast =100.4 " If the blast were perfectly dry air, its nitrogen would be 100.4 X (10 *3) = 334.7 " and the weight of dry blast = 435.1 " and its volume at C. 435.1-5-1.293 = 336:5 in 3 The blast, however, is not dry, and the 100.4 kilos, of oxygen includes that brought in by the moisture. The moisture weighs 5.66 grains per cubic foot of measured dried air; it is, there- fore, simplest to calculate the oxygen entering the furnace per unit volume of air proper entering. Since 5.66 grains = 5.66 ^437.5 = 0.01294 ounces av., the calculation can be made in ounces per cubic foot or kilograms per cubic meter in identical figures as follows: Oxygen in 0.01294 parts of water = 0.01294 X-f- =0.0115 y Oxygen in 1 volume of air proper, at 24 C. o 070 - 1 - 283 * 13* 278 + 24 =' 2743 Sum = 0.2858 298 METALLURGICAL CALCULATIONS. The blast received, per 100.4 kg. of oxygen thus received, is therefore, measured dry at 24 C: ' 351.8 cubic meters, n (J . and in cubic feet, per 100.4 pounds of oxygen: 100.4X16 0.2858 = 5628.8 cubic feet. The volume of the moist air containing this will be the volume of assumed dried air plus the volume of the water vapor. The latter is, per unit volume of dried air: 0. 01294 -f- (o.OQXy X 27 \^ 24 ) =0.01738 and the volume of moist air received, at 24, is, therefore, 351.8X1.01738 = 357.9 cubic meters. Or 5628.8X1.01738 = 5726.6 cubic feet. The weights of water, oxygen and nitrogen received per 100 of pig iron are (as already given on the balance sheet): H 2 0.01294X351.8= 4.52kg. O 2 0.2743 X351.8 = 96.4 " N 2 0.9143 X351.8 = 321.3 " And the volumes of these, considered theoretically at C. and standard pressure, per 100 kg. of pig iron made: H 2 O vapor 4 . 52 -r- . 81 = 5.6 cubic meters. Air 417.7 -7-1.293 = 322.8 " Sum = 328.4 There are several other modifications of this solution which will suggest themselves to anyone who studies out the question, but while half a dozen ways may be equally correct, that one is logically to be preferred which is most easily understood and is the shortest. One solution, however, based on volume relations, is well to know. Water vapor, H 2 O, represents half THE HEAT BALANCE SHEET. 299 its volume of contained oxygen, while air has 0.208 oxygen. The cubic foot of dried air at 24 C. was accompanied by 070 j_ 04 0.01294^-0. SIX *JI = 0.0174 cubic foot of moisture, which was removed in making the test. The oxygen per cubic foot of dry blast was, therefore, at 24 C. : as H 2 = 0.0174 -v- 2 = 0.0087 cubic foot. O 2 as air = 1.0000X0.208 = 0.2080 " 0.2167 " Weight of this oxygen : 070 And volume of dry blast, at 24, per 100 of pig iron: 100.4 0.2868 350.1 cubic meters. The difference of about 0.4 per cent between this and the pre- viously-calculated result is due to the figures not being carried out to a sufficient number of decimal places. (4) The efficiency of the blowing plant is found by calcu- lating the volume of air and moisture received by the furnace per minute, calculated to 24 C. = 75 F., and comparing this with the piston displacement of 40,000 cubic feet per minute. Volume of moist air received, at 24 C., per 100 Ibs. of pig iron made = 5726.6 ft. 3 Volume per day = 5726.6X8019.2 = 45,922,750 ft. 3 Volume per minute = 45,922,750-^1440 = 31,891 " o-j CQf) Efficiency of blowing plant - =79.7 per cent. (4) 4U,UUU (5) The heat balance sheet is based for most of its data upon the balance sheet of materials, the calculations already made and the additional data given in the statement. The balance sheet shows 80.3 kilos, of carbon oxidized in the furnace. Of this, the amount oxidized to CO and remain- 300 METALLURGICAL CALCULATIONS. ing as such in the gases is obtained from the amount of CO in the gases, as follows: C in CO = 448.5X0.223X0.54 = 54.0 kg. C oxidized to CO 2 is therefore 80.3- 54 = 26.3 kg. The heat in hot blast is calculated from its volume assumed to be at C., and already calculated, viz.; 322.8 cubic meters of air proper and 5.6 cubic meters of water vapor, with mean specific heats per cubic meter between and 382 C. of 0.313 and 0.40 respectively. The heat of formation of the pig iron may be taken from the amount of carbon in the iron, as 705 Calories per kilogram of carbon, and that of silicon omitted from consideration. The heat of formation of the slag, since it contains 29.5 of silica and alumina to 28.5 of lime and magnesia, may be taken as 150 Calories per unit of silica and alumina. The total heat received and generated, and to be accounted for, in the furnace, is therefore, per 100 kg. of pig iron: Carbon to CO 54.0X2430 = 131,220 Calories. Carbon to CO 2 26.3X8100 = 213,030 In H 2 O vapor 5.6X0.40 X382 \ on QQK In air proper 322.8X0.313X382 / Solution of C in iron 4x705 = 2,820 Formation of slag 29.5X150 = 4,425 -t Total = 390,880 The items of heat distribution are 325 Calories in each kilo- gram of pig iron, 525 Calories in each kilogram of slag, heat in the waste gases at 281, heat in cooling water, lost by radia- tion and conduction (by difference), evaporation of the mois- ture in charges, expulsion of CO 2 from carbonates, decom- position of moisture of blast, reduction of silicon and iron. Reduction of Fe 95 kg.X 1,746 = 165,870 Calories. Reduction of Si 1 " X 7,000 = 7,000 Expulsion of CO 2 from CaCO 3 16.7 " XI. 026 Expulsion of CO 2 from MgCO 3 2.3 " X 666 Evaporation of H 2 O 18.7 " X606.5 - 11,342 THE HEAT BALANCE SHEET. 301 X281 = 43,836 Heat in waste gases: Nand CO 400 m 3 X 0.3106 CO 2 58.3 m 3 X 0.446 H 2 O23.1 m 3 X 0.382 Decomposition of moisture of blast: H 2 O 4.5 kg. X (29,040 -r9) = 14,511 Heat in slag 58 kg. X525 = 30,450 Heat in pig iron 100X325 = 32,500 Heat in cooling water (300,000 gallons per diem) 300,000X8.3 (Ibs.) X 50 -r- 8019.2 = 15,525 Loss by radiation and conduction (differ- ence) = 51,180 Total = 390,880 " (5) (6) The proportion of the fixed carbon burned at the tuyeres is obtained by calculating the weight of carbon which the oxygen entering at the tuyeres could oxidize to CO, as follows: Oxygen entering at the tuyeres = 100.4 kg. . Carbon burned to CO = 100.4X^1 = 75.3" lo Fixed carbon charged = 84.3 ~" Proportion burned at tuyeres = 89.3 per cent. A more just proportion is that between the carbon burned at the tuyeres and the total fixed carbon oxidized, because the fixed carbon which carbonizes the iron cannot be oxidized under any circumstances. This proportion in this furnace is: indicating a very fair approximation to Gruner's ideal work- ing. If we make the further allowance, that the silicon in the pig iron is necessarily reduced by solid carbon, and that there- fore the solid carbon needed to reduce the 1 kilogram of silicon [IX (24^-28)] is in no case available for combustion at the tuyeres, we have the approach to Gruner's ideal working meas- ured by the ratio 80.3-0.9 302 METALLURGICAL CALCULATIONS. in spite of which, however, the furnace is not doing very good work. (7) The proportion of the heat requirement generated or available at or about the tuyeres is determined as follows: Combustion of C to CO = 75.3X2430 = 182,979 Calories. Heat in hot blast = 39,385 Format on of pig iron = 2,820 Formation of slag = 4,425 ". Total = 229,609 Against which must be charged heat required to decompose moisture of blast = 14,511 Leaving net heat available = 215,098 " which is 215,098^-390,880 = 55 per cent, of the total heat generated and available in the furnace. Another way in which it is sometimes put, is that the oxida- tion of carbon to CO or CO 2 furnishes a certain amount of heat to the furnace (346,250 Calories in this case), of which a certain amount is generated by combustion at the tuyeres (182,979 Calories), making the ratio thus considered, in this case, 53 per cent., which is almost the same figures as above, but not so significant, since it is illogical to consider the heat brought in by the hot blast as not being available heat for doing work in the tuyere region. (8) The proportion of iron reduced by solid carbon is found by finding how much carbon is used up in that reduction. Fixed carbon charged =84.3 kg. Fixed carbon carbonizing the pig iron = 4.0 " Fixed carbon oxidized in the furnace =80.3 Fixed carbon oxidized by the blast = 75.3 Fixed carbon oxidized above the tuyeres = 5.0 " Carbon needed to reduce 1 kg. silicon = 0.9 " Carbon used for reducing FeO = 4.1 P/> Amount of Fe, thus reduced = l.lX-r = 19.1 " Proportion of the total Fe, thus reduced - ~ =20 per cent. (8) THE HEAT BALANCE SHEET. 303 (9) The maximum temperature of the gases in the region of the tuyeres is that temperature to which the heat there available will raise the products of combustion. This question is best resolved by simply considering the combustion of 1 kilogram of carbon, evolving 2430 calories, while the heat in the hot carbon itself just before it is burnt, and that in the hot blast required, will also exist as sensible heat in the pro- ducts, the whole diminished by the heat necessary to decom- pose the water vapor blown in. Since, per 75.3 kg. of carbon burned at the tuyeres there enter 5.6 m 3 of water vapor and 322.8 m 3 of air proper, meas- ured at 0, the volume of blast per kilogram of carbon oxidized at the tuyeres is H 2 O 5.6-V-75.3 = 0.0738m 3 = 0.0598 kg. Air 322.8^75.3 = 4.2869m 3 The products of the combustion are, per kg. of carbon burned: CO 22.22-^12 = 1.8519 cubic meters. N 2 321.3* 1.26-v- 75.3 =3.3865 " H 2 equal to H 2 O decomposed = 0.0738 " Total = 5.3122 The heat available to raise their temperature is: Heat of combustion of 1 kg. carbon = 2430 Calories. Heat in hot blast = 39,385^-75.3 = 523 Heat in hot carbon at t (or very nearly) = 0.5t-feo " Less heat absorbed in decomposing steam 14,511-*- 75.3 = 193 Net heat available in gaseous products =2640 + 0.5t Cal Calorific capacity of gaseous products = 5.3122 (0.303t + 0.000027t 2 ) Therefore 5.3122 (0.303t + 0.000027t 2 ) = 2640 + 0.5t Whence t = 1910 (9) This represents the absolute maximum of temperature which the gaseous products formed at the tuyeres can possess. (10) If all the moisture were removed from the blast, the heat available would be: 304 METALLURGICAL CALCULATIONS. By combustion of 1 kg. carbon = 2430 Calories. Heat in 4.4685 m 3 of dry air at 382 C. = 4.4685X0.313X382 = 574 Heat in 1 kg. of carbon at t = 0.5t- 120" Net heat available in gaseous products = 0.5t + 2884 Cal. Calorific capacity of gaseous products = 5.3976 (0.303t + 0.000027t 2 ) Therefore 5.3976 (0.303t + 0.000027t 2 ) = 2884 + 0.5t Whence t = 2018 (10) It is to be noted that this is 108 C. = 194 F. higher than with moist blast ; and while the slag and iron in passing through this zone of high temperature will not reach this maximum temperature, yet they would be heated approximately 100 C. higher when using dry blast, if all other conditions were kept constant. N. B. In working this problem, the metric measurements and English measures have been purposely used interchange- ably, in order that readers may understand better that if weights are taken in pounds and the heat unit is the pound Calorie (1 C.) , the same numbers represent a solution in either system. When volumes are concerned, cubic feet and ounces, or ounce calories, have practically the same numerical expression as cubic meters and kilograms or kilogram calories. CHAPTER V. THE RATIONALE OF HOT-BLAST AND DRY-BLAST. Problem 58 (for practice). The blast furnace of Problem 57 had its blast dried before using, to the extent of leaving on an average 1.75 grains of moisture per cubic foot of air proper, at 5 C., in the air pumped. The composition of the ore, limestone, and coke used was unchanged, also that of the pig iron made. The weights charged per 100 of pig iron were: Ore 177.6, flux 44.4, fuel 77.0, and the blast used calculates out oxygen 76.5, nitrogen 255.0, moisture 1.0. Analysis of gases: CO 19.9, CO 2 16, N 2 64.1 per cent. Product per day 447 tons. Temperature of gases 191 C., of blast 465 C. Piston displacement (air at - 5 C.) 34,000 cubic feet per minute. Assume heat in pig iron and slag same as before, in cooling water 20 per cent, greater per day. Required. (1) The volume of gases per 100 kg. of pig iron made. Answer. Measured dry, 355.9 m 3 . (2) A balance sheet of materials entering and leaving the furnace, per 100 units of pig iron. (See table on page 281.) (3) The volume of blast per 100 kg. of pig iron. Answer. 252.9 m 3 at- 5 C. (4) The efficiency of the blowing plant. Answer. 82.3 percent. (5) The heat balance sheet of the furnace, per 100 of pig iron. Developed. Carbon to CO 92,950 Calories. ^Carbon to CO 2 206,955 "" HeaV;.in hot blast 37,850 Soluti^a; of carbon in iron 2,820 " Formation of slag 4,260 Total 344,835 305 306 METALLURGICAL CALCULATIONS. Distributed. Reduction of iron 165,870 Calorics. Reduction of silicon 7,000 Expulsion of carbonic oxide (CO 2 ) 18,666 Evaporation of moisture of charges 11,342 Heat in waste gases 23,799 " Decomposition of moisture of blast 3,225 Heat in slag . 29,820 Heat in pig iron 32,500 Heat in cooling water 14,922 Lost by radiation and conduction (diff.) . 37,791 344,835 (6) Compare the heat items which are substantially different for the furnace run by moist and dried blast. Moist Blast. Dried Blast. Combustion of C to CO 131,220 92,950 Heat in waste gases 43,836 23,799 Decomposing moisture in blast 14,511 3,225 Loss by radiation and conduction 51,180 37,791 It may be noticed, that using moist blast too much carbon was burnt to CO at the tuyeres; the chief item of economy with dried blast is the ability to get along with less. The smaller total amount of gases, particularly nitrogen, accounts for the lower temperature of the waste gases, with dried blast. The direct saving by reason of decomposition of the moisture is the smallest item of economy. The reduced losses by radia- tion and conduction are mainly because of the faster rate of running, these losses being nearly constant per day. The ratio of these losses is found to be 0.74, whereas, the inverse ratio of the pig iron productions per day was 0.79. (7) Calculate the carbon burnt at the tuyeres, the propor- tion of the carbon available thus consumed. Compare these items with those of the furnace on moist blast. Moist Blast. Dried Blast. Carbon burnt at tuyeres 75.3 58 . 05 Total fixed carbon charged 84 . 3 67 . 8 Proportion burnt at tuyeres 89 . 3 85 . 6 Fixed carbon really available 79 . 4 62.9 Proportion burnt at tuyeres 94 . 8 92 , 3 RATIONALE OF HOT-BLAST AND DRY-BLAST 307 (In some charcoal furnaces of low heat requirement, i.e., with pure ores and fuel, as little as 37 parts of carbon is burned at the tuyeres per 100 of pig iron produced, which represents, moreover, only 70 to 75 per cent, of the available fixed carbon in these furnaces.) Charges. Pig Iron. Slag. Gases. CO fc Fe 2 O 3 135.7 H 2 O 17 8 Fe 95.0 40.7 H 2 O 17.8 SiO 2 17.8 A1 2 O 3 6 3 Si 1.0 SiO 2 15.7 A1 2 O 3 6 . 3 1.1 SiO 2 9 2 SiO 2 2 2 3 CaO 21 1 CaO 21 . 1 % MgO 2 1 MeO 2 1 E CO 2 19.0 CO 2 19.0 C 67.8 C 4.0 C 63.8 fc SiO 2 4 2 SiO 2 4 . 2 v CaO 4 2 CaO 4 2 Z H 2 O 0.8 H 2 O 0.8 U5 O 2 76 . 5 O 76.5 n N 2 255 N 255.0 f H 2 O 1 f H 0.1 5 \ O 0.9 Totals 631.5 100.0 55.8 475.7 (8) The proportion of the whole heat requirement available in the region of the tuyeres. Answer. 53 per cent. (9) The proportion of the iron in the furnace which is re- duced by solid carbon from FeO. Answer. 23.8 per cent. (10) The theoretical maximum of temperature at the tuyeres. Answer. 1965 C. 308 METALLURGICAL CALCULATIONS. HOT BLAST. For several centuries blast furnaces were run by charcoal as fuel and with cold blast. How great the variations in tem- perature of the cold blast may be, can be inferred from the experience of a furnace manager in the Urals, Russia, who noted temperatures of the air nearly 40 C. in the summer and 60 C. in the winter. Assuming an average temperature of C. for unheated blast, burning charcoal to CO, the theo- retical maximum of temperature before the tuyeres can be calculated as follows: Heat generated by combustion 2430 Calories Heat in hot carbon being burnt 0.5t 120 Volume of CO and N 2 formed 5.3795 cubic meters 2310 + 0.5t Temperature == 5 . 3944 (o. 3 03 + Q.Q00027t) Whence t = 1678 This does not mean that the pig iron and slag will be car- ried to this temperature, any more than if a locomotive could run alone 90 miles an hour it could therefore pull a train of cars that fast. The hot gases, CO and N 2 , are at the start at this temperature, and as they ascend and come in contact with the descending iron and slag, these are raised to temperatures approximating towards, but always lower than, the above. In fact, the temperature -to which the iron and slag is raised depends on the relative quantities of iron and slag to fuel burnt, and the speed with which the furnace is worked. If the blast is heated, its sensible heat is simply added to the numerator of the above expression. We can easily find out how much sensible heat the 4.4685 cubic meters of air brings in, at any temperature desired [Q = 4.4685 (0.303t + 0.000027t 2 )], and then solve the quadratic anew. We thus find Temp, of Blast. Heat in Blast. Theoretical Temp. C. - Calories 1678 C. 100 " 137 " 1762 " 200 " 276 " 1845 " 300 " 417 " 1929 " 400 " 561 " 2012 " 500 " 707 " 2096 " 600 " 856 " 2180 " RATIONALE OF HOT-BLAST AND DRY-BLAST 309 Temp, of Blast. Heat in Blast. Theoretical Temp. 700 C. 1007 Calories. 2265 C. 800 " 1160 " 2350 " 900 " 1316 " 2435 " 1000 " 1475 " 2520 " The heating of the blast thus raises the maximum tempera- ture available some 85 C. for each 100 C. to which the blast is heated. It not only increases the temperature available, but also the number of heat units, thus increasing both the quan- tity and the intensity of the heating in the tuyere region. Of these two items of increase, that of the intensity factor is the most important, since it regulates the rapidity of transfer of heat to the charge and the efficiency and speed of the smelting action of the furnace. DRIED BLAST. Each kilogram of water vapor decomposed absorbs 29040 * 9 = 3227 Calories, which would not be needed if the 0.67 kilo, of carbon thus employed was oxidized by air instead of by moisture. Per kilogram of carbon oxidized by water vapor, there is absorbed 58,080 -=-12 = 4,840 Calories, while this kilo- gram of carbon can only furnish 2430 Calories in becoming CO, leaving a net absorption of 2410 Calories per kilogram of carbon thus burned, against which, however, can be credited the heat in the kilo, of carbon burned and the sensible heat in the water vapor itself; the former is 0.5t 120, and the latter can be calculated from the volume of the water vapor, 1.8519 cubic meters. We thus have, per kilogram of carbon thus oxidized : Temperature of Water Vapor. Heat in Moisture. Heat in Products. 100 66 Calories 0.5t- 2460 Calories 200 137 " 0.5t-2393 300 214 0.51-2316 400 296 " 0.5t-2234 500 384 " 0.51-2146 600 478 " 0.5t-2042 700 577 " 0.51-1953 800 682 " 0.51-1848 900 792 " 0.51-1738 1000 907 " 0.51-1623 310 METALLURGICAL CALCULATIONS. Since the heat in the carbon burnt (0.5t 120) can never equal numerically 1623, it is seen that under no circumstances can the water vapor do anything but diminish the quantity of heat available at the tuyeres, while the products of its decom- position CO and H 2 increase the volume of the products and so diminish still further the intensity of temperature attainable. The best way to determine the amount of moisture in the' air is to draw it through a tube containing concentrated sulphuric acid, measure the quantity of dry air drawn through, and weigh the amount of moisture caught by the tube. This gives the weight of moisture accompanying unit volume of dry air (not weight of moisture in unit volume of moist air). De- terminations by the wet and dry bulb thermometers, the whirled psychrometer, humidity gauges, etc., are none of them so re- liable as the above described method, which is direct, simple and accurate. The results may be obtained in grains per cubic foot of dry air or milligrams per liter. The best way to express them, for further use, is in ounces avoirdupois per cubic foot, or kilograms per cubic meter. The first is obtained "by dividing the number of grains by 437.5, the second by divid- ing the milligrams per liter by 1000. The numbers thus ob- tained are practically identical in the two systems. The theoretical temperatures attainable with moist blast of varying degrees of moisture and heated to various tempera- tures are obtained by applying the previously explained prin- ciples. We have already calculated the temperature obtained by burning carbon with dry air of various temperatures. We have also calculated a table of the deficit of heat available pro- duced by the entrance of 1.5 kilos, of water vapor (which would oxidize 1 kilo, of carbon and produce 1.8519 cubic meters of CO and H 2 ). We are, therefore, prepared to calculate a table of the maximum temperature attainable using blast of any degree of humidity heated to any practical temperature. Before giving the table we will run through the details of one calcula- tion, to make clear the method employed. Illustration. What is the theoretical maximum tempera- ture using air which carries normally 10 grams of moisture per cubic meter of dry air, reduced to standard conditions (i.e. ,10 grams per 1.293 kilograms of dry air), and heated to 500 C.? It takes 3.5275 cubic meters of dry air at standard condi- RATIONALE OF HOT-BLAST AND DRY-BLAST 311 tions to burn 1 kilogram of carbon. If dry, and at 500, there is 2430 Calories generated by combustion, 707 Calories in the dry air used, 0.5t 120 Calories in the hot carbon, and the total heat thus at hand raises the 5.3944 cubic meters of pro- ducts to the temperature of 2096 C., as is determined by solving the equation 2430 + 707+ (0.5t- 120) 5.3944 (0.303 + 0.000027t) = 2096 As modified by the moisture, the 4.4685 cubic meters of dry air would be accompanied by 44.685 grams of moisture, or 0.044685 kg., which would oxidize two-thirds of its weight of carbon, or 0.02979 kg. of carbon, which would contribute to the heat available 0.02979 (0.5t-2146) Calories, making a con- tribution of 0.015t 64 Calories to the numerator of above equation. The products of combustion will be increased by H 2 and CO equal in volume to twice the volume of the moisture, or 2 X 0.044685 H- 0.81 = 0.1102 cubic meters, the mean specific heat of which goes into the denominator. We then have = 3017 + 0.5t + 0.015t-64 " (5.3944 + 0.1102) (0.303 + 0.000027t) " Another method of solution, not using the previously calcu- lated tables but based entirely on first principles, is to base the whole calculation on the use of 1 cubic meter of dry air, with its accompanying moisture, as follows: Oxygen present in 1 m 3 dry air = 1,293X3/13 = 0.2984 kg. Oxygen present in the moisture = 0.010X8/9 = 0.0089 " Total = 0.3073 " Carbon consumed = 0.3073X0.75 = 0.2305 " Volume of moisture = 0.010 H- 0.81 = 0.0123 m 3 Volume of oxygen in dry air = 0.2078 ' Volume of products from dry air = 1.0000 + 0.2078 = 1.2078 " Volume of products from moisture = 2X0.0123 = 0.0246 " Total volume of products = 1.2324 " Heat of combustion of carbon = 0.2305X2430 = 560 Cal. Heat in carbon at t = 0.2305X (0.5t- 120) = 0.1152t- 28 " 312 METALLURGICAL CALCULATIONS. Heat in dry air at 500 = IX [0.303 + 0.000027 (500)] 500 158 Cal. Heat in moisture at 500 = 0.0123 [0.34 + 0.00015 (500)] 3 Heat absorbed in decomposing moisture = 0.0123 X2614 = -32 " Whence results the equation 0.1152t + 661 0.2324 (0.303 + 0.000027t) By applying either of the two methods of calculation ex- plained, the temperatures in the following table are obtained: THEORETICAL TEMPERATURES BEFORE THE TUYERES j Grams per cubic meter of dry air reduced to C. MOISTURE, j Grains per cubic f oot of dry air re( i uce( } t o C. Grams 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 Grains. 2.19 4.38 6.57 8.75 10.94 13.13 15.32 17.50 Blast. 40 1678 1647 1615 1573 1536 1507 1471 1443 100 1723 1692 1666 1627 1587 1548 1526 1496 200 1807 1776 1751 1712 1673 1636 1612 1584 300 1892 1861 1837 1800 1760 1725 1700 1673 400 1976 1945 1921 1885 1846 1813 1786 1760 500 2061 2030 2007 1970 1933 1902 1874 1848 600 2146 2115 2093 2055 2020 1991 1962 1936 700 2232 2201 2178 2144 2108 2080 2050 2025 800 2318 2287 2264 2227 2195 2169 2138 2114 900 2404 2373 2351 2313 2282 2257 2226 2203 1000 2490 2459 2437 2399 2369 2345 2314 2292 The calculations show that the temperature before the tuyeres may vary as much as 235 C. = 423 F., from change in the moisture of the air, from dryness to warm air saturated with moisture. CHAPTER VI. PRODUCTION, HEATING AND DRYING OF AIR BLAST. This is a subject intimately related to the running of iron blast furnaces, and incidentally related more or less to all classes of metallurgical operations. The principles involved are those of physics mechanical and thermal and when once thoroughly understood can be used for the most various classes of metallurgical problems. PRODUCTION OF BLAST. There are two different principles upon which air is com- pressed, represented by the fan and the piston machines. The first is constant in its operation, the second intermittent, the first draws in and expels a continuous current of air, the second draws in a given quantity of air, compresses it and expels it. Measuring the work done by the difference in static conditions of the compressed and of the uncompressed air, the work actu- ally done is a fixed and calculable quantity, independent of what type of machine performs it. During the compression, heat is generated, and the mechanical work done includes the mechanical equivalent of the heat thus generated. This is allowed for in the well-known formula for adiabatic com- pression : in which f = the ratio of the specific heat of air at constant pres- sure to that at constant volume = 1.408. V = the volume of the uncompressed air. p = the tension of the uncompressed air. pi = the tension of the compressed air. If we use the value of f = 1.408, the coefficient r be- r 1 comes 3.45, and - - =0.29. If we then use the other di- r 313 314 METALLURGICAL CALCULATIONS. mensions in feet and pounds, the resulting work done will be in foot-pounds; if we use meters and kilograms, in kilogram- meters. The element of time does not enter into this equation, the work actually done is the same for compressing a given amount of air, and is independent of the time. If I lift a kilo- gram 1 meter high, the same amount of work is done whether I lift it in 1 minute or in 1 second, the rate of doing work would be, however, sixty times as great in the last instance as in the first, but the actual amount of work accomplished is the same in each case. If, therefore, we use in the formula the volume of air entering the compressor per minute, the result will give the work done per minute; if we use the volume per second we get the work done per second. If we wish to transpose the work done into horse-power, we bear in mind that a horse- power is understood in English units as 33,000 foot-pounds of work done per minute, and in the metric system as 75 kilogram- meters of work done per second. In applying the formula we may further notice that ex- Po presses the ratio of compression; that is, by how many times the tension of the original air is increased. If air at one at- mosphere tension (such as the air we usually have to breathe) is compressed to two atmospheres tension, the ratio of com- pression is 2; if air entered a machine at two atmospheres ten- sion and was therein compressed to four atmospheres, the ratio of compression would be likewise 2, and the work done (for a given quantity of air) the same as before. The effective pressure, as shown on a gauge, is the difference between these two tensions: it is not pj. It is highly important that this point be held clearly in mind. The tension of the uncompressed air is its barometric pressure, as measured against a vacuum; the tension of the compressed air is likewise its pressure as measured against a vacuum; the effective pressure of the com- pressed air is the difference between these two tensions; the tension of the compressed air is, therefore, (by transposition) equal to the barometric pressure of the uncompressed air plus the effective pressure of the compressed air, i.e., plus the gauge pressure. Never use the effective pressure of the compressed air as p l5 but always add to it the barometric pressure of the uncompressed air p for the correct value of p t . PRODUCING HEATING AND DRYING BLAST. 315 Regarding the volume of uncompressed air, V is its volume measured at the pressure P , i.e., the actual volume of uncom- pressed air at its actual tension. If more convenient, however, we may use the volume of this air measured at standard baro- metric pressure, if we multiply it by p , the standard pres- sure. Thus, if we know the volume of uncompressed air meas- ured at one atmosphere standard tension, we may use for the p immediately following it the standard pressure 10.334 (kilos, per square meter), or 2,120 (pounds per square foot). By the law of the reciprocal nature of volume and pressure, the pro- duct thus obtained is the same as would be found by multi- plying the volume of the same air at any other pressure by that pressure (V p = Vj p x = V 2 p 2 , etc.). Whatever pres- sure we use in getting the product V p , however, we must use only the actual tension of the uncompressed air as the denominator in getting the ratio of compression (p!-v-p ). Problem 59. A blast furnace requires 2,615 cubic meters of air, measured at 5 C., for every metric ton of pig iron made. Assuming outside barometric pressure 735 millimeters of mercury column, the efficient pressure of the compressed blast to be 1.0 kilo- gram per square centimeter (as measured in a blast regulating reservoir) , a mechanical efficiency of the blowing engine of 90 per cent., of delivery of blast 82.3 per cent., and output 447 tons per day: Required : (1) The ratio of compression of the blast. (2) The piston displacement per ton of pig iron made. (3) The work of the blowing cylinders, per ton of iron made. (4) The gross or indicated horse-power of the steam cylinders. Solution : (1) The effective pressure being 1.0 kilogram per square 1 0000 centimeter, represents 0004^ 760 = 735 millimeters of mer- cury. But the uncompressed air is at 735 mm. barometric tension, therefore, the total tension of the compressed air, p lf is 1470 mm. of mercury, and the Ratio of compression = =2.0 ^1) 7oo 316 METALLURGICAL CALCULATIONS. (2) Piston displacement per metric ton of pig iron: g-g23 = 3165 cubic meters. (2) = 111,560 cubic feet. (3) The volume displaced must be the basis of calculating the power required, since subsequent losses of air (100 82.3 = 17.7 per cent.) do not affect the work done in the blowing cylinder. This volume is 3,165 cubic meters, representing that volume of uncompressed air at 5 C. and 735 mm. pressure. The initial temperature of the air does not affect the work to be done, and we can either use its volume at 735 mm. tension or correct this to 760 mm. tension, just as we chose. If we consider the volume at 735 mm. tension, then the product of volume and pressure is Vp = 3,165 X (lO,334 x ~\ = 31,755,000 If we first change the volume to standard pressure, we have (70K \ 3,165 X ^-j X 10,344 = 31,755,000 The work done in compression, in the blowing cylinder is, therefore, W = 3.45X3,073X10,334X[2- 29 - 1] = 109,554,750 X[l. 2225 - 1] = 24,326,800 kilogrammeters. (3) (4) The steam in the steam cylinders does more mechanical work than would be represented by the compression of air in the blowing cylinders. In this case, we assume 90 per cent, mechanical efficiency, or 10 per cent, mechanical loss. The gross work of the steam is, therefore, 24,326,800^-0.90 = 27,029,800 kilogrammeters, or, in horse-power, 27,029,900X447 1440X4500 = 1,865 horse-power. This amounts to 4.17 h.p. per metric ton of pig iron made per day. PRODUCING, HEATING AND DRYING BLAST. 317 MEASUREMENT OF PRESSURE. In making calculations of the work done in furnishing blast, as in the preceding problem, it is important to note how the pressure of the compressed air is measured. The real pressure is measured correctly by a pressure gauge only where the air is comparatively at rest, as on a pressure-equalizing reservoir. If the pipe communicating with a gauge is connected with a blast main, in which the velocity of the air is considerable, the pressure recorded will vary greatly with the direction of the end of this tube relative to the direction of the air current. The total pressure of the air in motion is the pressure which it exerts against the sides of the main, plus the pressure which has been used in giving it velocity. If the pressure gauge tube is cut off at right angles to the flow of air in the main, the pressure recorded is even less than the actual static pres- sure of the air against the sides of the main (because of suction effect), and does not include at all the pressure represented by the velocity. The only way to measure correctly in a main the total pressure which has been impressed upon the blast is to bend the end of the pressure tube until it is parallel with the axis of the main and faces the current of air. The gauge then records the static pressure, plus the velocity head, and gives the proper pressure to use in calculating the work done by the blowing engine. However, it is still preferable to put the gauge on a blast reservoir, if there is one, where the air is nearly at rest and velocity head approximates zero for the reason that the velocity of air passing through a tube is greatest in the center and least against the walls, and it is diffi- cult to place the pressure tt:be so as to measure the mean ve- locity. An "approximation to the mean value is found by placing the end of the pressure tube not in the center but at one-third of the radius of the main from the center. INDICATOR CARDS. If the blowing engine cylinder can be tested by pressure indicator cards, then a different method of obtaining the work of producing the blast is furnished. The integration of the diagram gives the mean pressure on the piston during the stroke. Expressing this in kilograms per square meter or pounds per square foot, and multiplying by the area of the 318 METALLURGICAL CALCULATIONS. piston and the length of the cylinder, we get the work done per stroke; again multiplying by the number of strokes per minute we have the work done per minute, from which is readily obtained the horse-power. These operations are usually com- bined in the formula: Work = PXLXAXN If we observe that LxAXN is the piston displacement per minute, the formula becomes: Work = P X Piston displacement per minute, in which, it must not be forgotten, P represents mean pres- sure on the piston during the stroke, and is not to be con- founded with gauge pressure of the compressed air. Such a formula is, therefore, totally inapplicable to finding the work done by a fan or rotary blower, in which only the final pressure of the compressed air is known. The mistake of so applying it is often made. When only the final pressure of the compressed air is known, the formula for adiabatic compression is the only correct one to use. It may not be useless to call attention to the fact that when using the formula for adiabatic compression, the raising of the ratio of compression to the 0.29 power has to be done by logar- ithms : If a table of logarithms is not at hand, a satisfactory ap- proximation may be made by taking the cube root of the ratio, since HEATING OF THE BLAST. Air blast is commonly heated either continuously, by direct transmission of heat through metal or earthenware pipes, or discontinuously by the heating up of fire-brick surfaces, which are subsequently cooled by the blast. It is not within the prov- ince of these calculations to enter into a description of the various types of such hot-blast stoves, but to indicate the principles upon which the metallurgist can base calculations as to the efficiency of such stoves, and thus be prepared to find PRODUCING, HEATING AND DRYING BLAST. 319 out which do the best work, and wherein lie the principal ad- vantages or disadvantages of each type. The efficiency of a hot-blast stove is measured by the ratio of the heat imparted to the blast to that contained in the air and fuel used and generated by combustion. It is a furnace whose useful effect is the heat imparted to the air, and all other items of heat distribution are more or less necessary losses. The problem is simplest when the stove is a recuperator (con- tinuous type), using solid fuel. Then the item of heat genera- tion is perfectly definite, since the amount of fuel used in a given time can be definitely determined. When the hot-blast stove receives gaseous fuel, however, from a blast furnace, the amount of gas received by the stoves is usually a very uncertain quantity, since only part of all the gas produced by the furnace is used by the stoves, and the question of what fraction is thus used is difficult to determine. In such cases, the usual method of comparing the sizes of the pipes leading gas to the stoves and those carrying gas to boilers, etc., affords but a very un- certain determination, since the draft and consequent velocity of the gas may be very different in the two sets of pipes. In such cases, knowing the total volume of gas produced by the furnace, not only the relative sizes of the hot-blast stove pipes should be noted, but also the relative velocities of the gas currents in the two sets of pipes. Another method is to de- termine the quantity of chimney gas passing away from the stoves into the chimney flues, by measuring the size of the chimney flue, temperature of the gases and average velocity; given in addition the chemical analyses of the chimney gas and of the furnace gas, the quantity of the latter being used can be calculated, using the carbon in the two gases as the basis of calculation. Problem 60. Statement: Air for drying peat in a kiln is heated from C. to 150 C. by an iron pipe stove, the latter consuming dried peat, whose ultimate analysis is: Carbon 49.70 per cent. Hydrogen . 5.33 " Oxgyen 30.76 " Nitrogen 1.01 " Ash.. . 13.23 " 320 METALLURGICAL CALCULATIONS. The calorific power of this peat is 4249, and by the com- bustion of 92.5 kilograms, 5122 cubic meters of air is heated to the required temperature. The products of combustion pass away from the stove at 200 C., and contain (analyzed dry) 14.8 per cent, of CO 2 , and no CO or CH 4 or other gas containing carbon. Required: (1) The heat generated in the stove. (2) The heat in the hot air, and the net efficiency of the stove. (3) The volume and composition of the products of com- bustion in the stove. (4) The heat carried out in the chimney of the stove. (5) The heat lost by radiation and combustion. (6) The excess of air used in burning the peat. * (1) Heat generated in the stove: 4249X92.5 = 393,033 Calories (1) (2) Heat imparted to the air: [0.303 + 0.000027 (150)] X 150 X 5122 = 235,907 Net efficiency = 60.0 per cent. (2) (3) Volume of products of combustion: Weight of carbon in fuel burnt 92.5 X 0.4970 = 45.97 kilos. Volume of CO 2 thus produced- 45.97 -7-0.54 = 85.13 cub. meters Volume of (dry) gas produced 85.13 -i-0.148 = 575.20 " Volume of N 2 and O 2 in products 575.2-85.13 = 490.07 " Volume of water vapor in products 92.5 X 0.0533 X 9 -=- 0.81 = 54.78 " Percentage composition of products of combustion: Moist. Dried. CO 2 13.5 14.8 N 2 andO 2 77.8 85.2 H 2 0.. . 8.7 PRODUCING, HEATING AND DRYING BLAST. 321 To separate the nitrogen and oxygen would necessitate con- siderable calculation, and is not required just here, because the two gases have the same heat capacity per cubic meter, and therefore can thermally be reckoned together: (4) Heat in the chimney gases at 200: CO 2 85.13 m 3 X0.4140 = 35.24 N 2 and O 2 490.07 m 3 X 0.3084 = 151.14 H 2 O 54.78 m 3 X 0.3700 = 20.27 206.65X200 = 41,330 Cal. Proportion thus lost = 10.5 P. C. (4) (5) Loss by radiation and conduction 100 - (60.0 + 10.5) = 29.5 P. C. (5) (6) The separation of N 2 and O 2 in the products is not easy. It is best based on the consideration that part of these con- sists of the nitrogen of the air which was necessary for com- bustion (plus the nitrogen of the fuel) and of the excess air. The first can be calculated, and thus the latter obtained by difference, and then the percentage of air in excess determined. Oxygen necessary for combustion: 00 C to CO 2 = 45.97X7= = 122.59 kilos, iz Hto H 2 = 4.93X8 = 39.44 " 162.03 " Oxygen present in peat 92.5X0.3076 = 28.45 " Oxygen to be supplied = 133.58 Nitrogen accompanying = 445.27 Air necessary = 578.85 = 447.7 m 3 Nitrogen from coal, 92.5X0.0101 = 0.93 kilos. Total nitrogen from coal and necessary air = 446.2 Volume = 446.2-r-l.26 =354.1 m 3 Total N 2 and O 2 in products = 490.1 " Therefore, excess air = 136.0 " Percentage of excess air ^-^ = 0.303 = 30.3 P. C. (6) 322 METALLURGICAL CALCULATIONS. Problem 61. A bla$t furnace has three hot-blast stoves, two of which are always on gas and one on blast. Per long ton of pig iron pro- duced there issues from the furnace (reduced to standard tem- perature and pressure) : Nitrogen 81,763 cubic feet Carbon monoxide 25,383 Carbon di-oxide 20,409 " Water vapor 8,230 For the same quantity of pig iron made, 92,330 cubic feet (standard conditions) of air is heated in the stoves from 5 C. to 465 C. The hot gases reach the stoves at 175 C., are there burned by 10 per cent, excess of air at 0, and the chimney gases resulting (assume perfect combustion) pass out of the stoves at 120 C. Required: (1) The net efficiency of the stoves, assuming they receive 25, 30, 35, 40, 45 or 50 per cent, of the gas produced by the furnace. (2) Assuming that each stove radiates and loses to the ground one-third as much heat as the blast furnace itself (the heat balance sheet of the furnace showed 846,518 pound Cal- ories thus lost per long ton of pig iron produced), what pro- portion of the whole gas goes to the hot-blast stoves, and what is the net efficiency of the stoves? Solution : (1) We will first calculate the efficiency of the stoves, as- suming that they received all the gas produced, from which datum can then be obtained the efficiency, supposing any as- sumed fraction of the gas goes to the stoves. Heat in the blast ( -5 to 465) : 92,330 X [0.303 + 0.000027 (465 -5)] X [465 - ( -5)] = 92,330 XO. 31542X470 = 13,687,683 ounce cal. = 855,480 pound Cal. Sensible heat in the hot gases (0 to 175) : N 2 and CO 107,146X0.3077 32,969 CO 2 20,409X0.4085 8,337 H 2 O 8,230X0.3763 3,097 Heat capacity per 1 44,403 ounce cal. 2,775 pound CaL Heat capacity per 175.. 2,775X175 = .485,625 " PRODUCING, HEATING AND DRYING BLAST. 323 Heat generated by combustion: CO to CO 2 25,383X3062 = 77,722,746 ounce cal. 4,857,672 pound Cal. Total heat available = 5,343,297 " If the gas were all used in the stoves the efficiency of the latter would be only - 855,480 5,343,297 = 0.160 = 16.0 per cent. With smaller proportions of the whole gas used the efficiency of the stoves calculates out as follows: Using 50 per cent, of the gas Efficiency 32.0 per cent. "45 " " " 36.0 " "40 " " " 40.0 " "35 " " 45.7 " "30 " " " 53.3 "25 " " ....... " 64.0 " 16 " " " 100.0 " (1) All that the above analysis tells us is, that certainly over 16 per cent, of the gas produced by the furnace iriust be used by the stoves ; but if we can deduce any probable value for the percentage of the gas actually used, such as by measuring the several gas mains and the velocities of the gas in each, we can then reckon out the value of the efficiency of the stoves, with about the same degree of probability. Blast furnaces use from 33 to 60 per cent, of the gas they produce in the stoves. If we assume 50 per cent., in this case, the efficiency of the stoves would be 32 per cent. (2) There is another way of solving the problem, which is to either measure, calculate or assume the heat lost by radia- tion and conduction from the stoves; adding to this the heat going out in the products of combustion, and the net heat in hot blast, the sum is the total heat which has been brought into and generated within the stoves. But, all the gas would bring in and generate in the stoves 5,343,297 Calories; we can, therefore, find easily what proportion of all the gas is being used in the stoves. In requirement (2) we are told to as- sume that the three stoves lost by radiation and conduction 324 METALLURGICAL CALCULATIONS. 846,518 pound Calories per long ton of pig iron made, an amount equal to that similarly lost by the furnace itself. The heat in the chimney gases, at 120, can be thus calculated: Oxygen required ( CO j = 12,692 cubic feet Air required = 61,017 " Excess of air used = 6,102 " Nitrogen of required air = 48,325 " " Nitrogen already in gas = 81,763 " " Nitrogen in these two items = 130,088 " Chimney products: CO 3 = 45,792 " H 2 O = 8,230 " N 2 = 130,088 " Excess air = 6,102 " Heat in chimney gases : CO 2 45,792X0.3964 = 18,152 ounce cal. H 2 O 8,230X0.3580 = 2,946 " N 2 and air 136,190X0.3064 = 41,729 " Heat capacity per 1 = 62,827 " = 3,927 pound Cal. Heat capacity per 120 = 471,122 " If all the furnace gas were burnt in the stoves, 471,122 pound Calories would go up the stove chimneys. But, since only a part of the gas is so used, only a fraction of this amount of heat is lost to the chimneys. If we call X the proportion of the gases used, then 471,122 X will represent the chimney loss, and the total heat requirements of the stoves will be: Heat in air blast = 855,480 Ib. Cal. Heat in chimney gases = 471,122 X " Radiation and conduction = 846,518 Total = 1,701,998+471,122 Xlb. Cal. The proportion of the total gases needed to supply this, X. is 1,701,998 + 471, 122 X 5,343,297 whence X = 0.349 = 34.9 per cent. (2) PRODUCING, HEATING AND DRYING BLAST. 325 and the net efficiency of the stoves is 855,480 855,480 5,343,297 X . 349 1,864,812 = 45.9 per cent. . (2) Arithmetically, the finding of X can be simplified, perhaps, by considering that the chimney loss represents in any case 471,122-^-5,343,297 = 0.088 = 8.8 per cent, of the total heat received by the stoves, leaving 91.2 per cent, to be applied to heating the blast and for radiation and conduction loss. The last two items, however, must amount to 1,701,998 Calories, and, therefore, the total heat requirement of the stoves is 1 ' 866 ' 200 P und requrng iggO= 0.349 . 34.9 per cent. (2) of all the gas produced by the furnace. In the above solution the only uncertain factor in the calcu- lation is the radiation and conduction loss from the stoves, and this uncertainty does not largely affect the reliability of the result obtained, allowing that we have assumed an approxi- mately correct value for this loss. All uncertainty could be dispelled, however, were the radiation and conduction losses measured directly. This could be accomplished by finding experimentally the temperature of the outside shells of the stoves, and calculating the external losses of heat by the laws of radiation and conduction; but in order to do this satisfac- torily, it would be necessary to divide the shells of the stoves into zones, and determine the temperature and calculate the losses from each zone separately a rather long operation, but one worth doing. DRYING AIR BLAST. The advantage of dried blast has been already discussed at length in these calculations. The advantage is due primarily to the higher temperature obtained when the moisture has been removed. The means adopted commercially for drying the air have 326 METALLURGICAL CALCULATIONS. been those of refrigerating the uncompressed air, before its entrance into the blowing cylinders. This has the great ad- vantage of furnishing the cylinders with cold air, and, there- fore, of greatly increasing their blowing capacity, since the weight or quantity of air blown is proportional to the absolute temperature of the air entering the cylinders. Illustration. Outside air being at 30 C., what increase in the amount of air furnished by blowing engines will result if the temperature of the air is artificially reduced to 5 C. ? How much slower can the engines be driven, in the second case, in order to furnish the same weight of air as before? The two temperatures are respectively 273 + 30 and 2735 absolute, that is, 303 and 268. The engines, if run at uniform speed, would furnish 303-7-268 =1.13 times as much air in the second case, or 13 per cent. more. If the engines were slowed down to 268-^-303 = 0.884 of their former speed they would furnish the same amount of air; that is, they could be run 11.6 per cent, slower. In fact, they could be run more than 11.6 per cent, slower in the second case, and yet supply the same quantity of air, because at the slower running the delivery efficiency is somewhat higher. The disadvantage of cooling the uncompressed blast is that it must be cooled much more, to eliminate from it a given per- centage of its moisture, than if it were first compressed, and to obtain nearly dry air the moisture must be practically frozen out. Illustration: Air at 30 C. and 85 per cent, humidity is to be cooled until 95 per cent, of its moisture is eliminated, with- out compression; to what temperature must it be cooled? From the tables of aqueous tension, we find that the maxi- mum tension which aqueous vapor can exert at 30 C. is 31.5 31 5 millimeters, which means practically that ^ ' of a cubic oU pr meter of moisture accompanies ' of a cubic meter of air proper. If the humidity is 85 per cent., then this same quantity of air is accompanied by 31 5 = 0.0352 cubic meters PRODUCING, HEATING AND DRYING BLAST. 327 of moisture, or, per cubic meter of air measured dry 0. 0352 -*- - = 0.0368 cubic meters. If 95 per cent, of this is' removed by cooling, then the mois- ture left, per cubic meter of air measured dry, is 0.0368X0.05 = 0.00184 cubic meters. And the relative tensions of air and moisture, to attain this dryness, must have been reduced to 1.0000:0.00184. Since the sum of these tensions is always 760 mm. in the un- compressed air, the actual tensions of air and moisture will be 758.6:1.4, that is, the temperature must be reduced until the moisture present can exert only 1.4 mm. pressure. This, on examining the tables of tension of aqueous vapor is found to be less than C., in fact -15 C. Problem 62. Air at 30 C., carrying 85 per cent, of its possible amount of moisture, is cooled to C., and the moisture condensed to liquid water at that temperature. Barometer 760 mm. Required : (1) The percentage of the moisture condensed. (2) The amount of heat to be abstracted from each cubic meter of original moist air (refrigerating effect). (3) The percentage of moisture which would be condensed if the temperature were reduced to 5 C. (4) The total heat to be abstracted, per cubic meter of orig- inal air, in the latter case. Solution : (1) From the preceding illustration we can take the figures that in the moist air taken, each cubic meter of air proper is accompanied by 0.0368 cubic meter of moisture. In the cooled 328 METALLURGICAL CALCULATIONS. air at 0, the relative volumes of air proper and residual moisture will be as their relative tensions, viz.: 755.4:4.6 or 1:0.0061. The moisture accompanying the given quantity of air is, there- fore, reduced from 0.0368 to 0.0061, showing a condensation of 0.0307, equal to (2) The air at 30 contains, as before figured out, air proper and moisture in the proportions 1:0.0368. or, in 1 cubic meter of moist air, in the proportions 0.9645:0.0355. We have, therefore, to calculate the heat to be extracted from 0.9645 cubic meter of air proper, falling 30 to 0, and from 0.0355 cubic meter of water vapor, falling 30 to and 83.4 per cent, of the latter condensing to liquid at 0. The figures are, remembering that the volumes heretofore handled are at 30: Air proper: 070 . 9645 X X 0.3038 X 30 = 7 . 920 Calories Moisture, if all uncondensed: oyq 0.0355X^X0.3445X30 = 0.332 oUo Heat of condensation: 0.0355X0.834XX0.81X606.5 = 13.105 " oUo _ Total = 21. 357 " (2) (3) If the temperature were reduced to 5 C., the tension of the residual moisture would be 3.4 mm. of mercury, and PRODUCING, HEATING AND DRYING BLAST. 329 that of the air proper 756.6 mm., their relative volumes would be in the same ratio, viz.: 756.6:3.4 or 1:0.0045. showing that out of 0.0368 cubic meter of moisture originally accompanying 1 cubic meter of air proper, 0.0323 had been condensed, or 0393 0.878 = 87.8 per cent. (3) (4) The condensation is seen to be 87.883.4, or 4.4 per cent, more, if the air is cooled to 5 C. A considerably larger amount of refrigeration, however, is required in this case, because the moisture would all be frozen. The 1 cubic meter of moist air at 30, containing, as before calculated, 0.9645 cubic meter of air proper and 0.0355 cubic meter of moisture, would have 87.8 per cent, of the latter condensed to ice, or 0.03117 cubic meter, and, therefore, 0.00433 cubic meter re- maining uncondensed. Air proper, 30 to -5 C.: 070 . 9645 X 57^ X 0.3037 X 35 = 9 . 237 Calories oUo Moisture, if all uncondensed, 30 to 5 C.: 070 0.0355X5^X0.3438X35 = 0.385 " oUo Condensation of 0.03177 m 3 to liq. at - 5 C.: 070 0.03177X5^X0.81X605 = 14.025 " oUo Freezing of same, at 5 C. : 27Q 0.03177X^X0.81X80 OUO (4) The conclusion is, that an increased condensation of 4.4 per cent, has been obtained by an increase in the refrigerating requirement of 25.50221.357 = 4.145 Calories, or nearly 330 METALLURGICAL CALCULATIONS. 20 per cent. Another way of stating the comparison, is that when not freezing the moisture condensed, about 4 per cent, of the moisture was condensed from each cubic meter of air for one Calorie of refrigeration, whereas, the removal of ad- ditional moisture by cooling below zero requires nearly one Calorie refrigeration for each additional per cent, of moisture eliminated. A practical conclusion is, that the expense of refrigeration might easily be justified down to C., and yet be unjustified by the results when cooling below 0. Mr. James Gayley has, I believe, patented the scheme of refrigerating the air in two stages; first, nearly to zero, re- moving the moisture thus condensed as liquid, and then cooling the nearly dry air further, eliminating more moisture as ice. In this manner, the amount of refrigeration required is reduced by the latent heat of solidification of the larger part of the moisture, and cooling below zero becomes profitable. The saving in the above example would be 80 Calories per kilogram on all the moisture condensed at 0, viz.: 0.0216X80 = 1.728 Calories, cutting down the refrigeration required from 25.502 to 23.774 Calories, that is, enabling 4.4 per cent, additional drying to be produced for 2.4 Calories additional refrigeration. This scheme of Mr. Gayley is founded on sound scientific as well as prac- tical considerations. The idea of cooling the compressed blast by moderately cool water, recently proposed, is also a very practical idea and founded on sound scientific principles. At a given tem- perature moisture cannot exert more than a maximum vapor tension. It follows that if we start with air saturated with moisture, at a given temperature, and compress it to double its initial tension, keeping its temperature constant, about half of its moisture must condense out. If, at the same time, it is artificially cooled, then more than half of its moisture will condense as liquid. If we start with air not saturated with moisture, the com- pression, temperature being constant, will increase the tension of the moisture present until the air becomes saturated, after which increased pressure will cause condensation. PRODUCING, HEATING AND DRYING BLAST. 331 Illustration: Air at 30 C., 85 per cent, saturated with mois- ture, is compressed. At what effective pressure does it become saturated with moisture, temperature remaining constant? The moisture present is exerting 85 per cent, of its maximum vapor tension at this temperature. Therefore, the tension must be increased in the ratio of 85 to 100, to make the air saturated, viz.: in the ratio 1 to 1.177. The effective pressure necessary to be applied is, therefore, 1.177 1.000 = 0.177 atmospheres (2.6 pounds per square inch). Illustration: If air at 30 C., 85 per cent, saturated with moisture, is compressed to one atmosphere effective pressure and its temperature kept 'constant, what proportion of its moisture will condense? Before compression, the tension of the moisture being 31. 5 X 0.85 = 26.8 mm., the relative volumes of air proper and mois- ture are as 733.2 : 26.8 or, as 1 : 0.0367 After compression, the tension on the mixture being two atmospheres (1520 mm.), and the tension of the uncondensed moisture being it maximum, or 31.5 mm., the relative volumes of air and uncondensed moisture will be as 1488.5 : 31.5, or, as 1 : 0.0212. The proportion of the original moisture remaining uncondensed is therefore, - - 578 - 57 - 8 per cent - and the amount condensed out = 42.2 per cent. Problem 63. Air at 30 C., and 85 per cent, saturated with moisture is compressed to one atmosphere effective pressure (760 mm. of mercury), and simultaneously cooled by river water to 10 C. Barometer 730 mm. Required : (1) The percentage of the original moisture, condensed. 332 METALLURGICAL CALCULATIONS. (2) The weight of moisture remaining in the air, expressed in grams per cubic meter of dry air at standard conditions (i.e., per 1.293 kilograms of air proper). Solution : (1) Tension of uncompressed moist air = 730 . mm. Tension of moisture present 31.5X0.85 = 26.8 " Tension of air proper, uncompressed = 703.2 " Relative volumes of air proper and moisture = 1 : 0.0381 Tension of compressed moist air 730 + 760 = 1490.0 " Tension of uncondensed moisture (maximum tension at 10) . = 9.1 " Tension of air proper, when compressed = 1480.9 " Relative volumes of air proper and uncondensed moisture = 1 : 0.0061 Proportion of moisture condensed: 0.0381-0.0061 . n 0.0381 = ' 84 = a4 Per Cent ' (1) (2) The relative volumes of air proper and uncondensed moisture are, as found above, 1 : 0.0061, And the relative specific gravities of air proper and moisture are as the standard weights of 1 cubic meter of each, viz.: as 1.293 : 0.81. It follows, therefore, that 1.293 kilos, of air proper (1 cubic meter at standard conditions) must be accompanied by, 0.0061X0.81 = 0.0049 kg. of moisture, or * = 4.9 grams. (2) The original moist air contained, similarly considered, 0.0381X0.81 = 0.0309kg. 30.9 grams. CHAPTER VII. THE BESSEMER PROCESS. The outlines of this famous process are known to every educated person; the mechanical and most of the chemical details are familiar to most technologists; the exact way to run the converter is the source of income to hundreds of superin- tendents of works, and yet the quantitative side of the chemical and physical operations involved is mastered by very few. To state the case briefly, melted pig iron is put into the con- verter, numerous air jets are blown through, the impurities of the iron carbon, silicon, manganese and, in a special case, phosphorus oxidize relatively faster than the iron, and the final product is usually nearly pure iron. This is recarburized to steel by spiegel-eisen. During the blow very little free oxygen escapes from the converter, and the gases produced are principally nitrogen, carbon monoxide and some carbon di- oxide, while some hydrogen may come from the decomposition of the moisture of the air. The silicon, manganese, phosphorus and iron form silica, manganous oxide (MnO) principally, fer- rous oxide (FeO) principally, phosphorus pentoxide, P 2 O 5 , which go into the slag, while a little Fe 2 3 , Mn 3 O 4 and SiO 2 escape as fume. The applications of calculations to this process are numerous and important. They include such subjects as the amount of air theoretically required per ton of iron, the dimensions and power of the blowing engines, the weight of slag produced, the balance sheet of materials, the balance sheet of heat evolved and distributed, the radiation losses, the discussion of the efficiency of the various impurities as heating agents in the process. AIR REQUIRED. Basing our calculations on the analysis of the pig iron used, and assuming it to be blown to pure iron, we must next as- sume the probable loss of iron itself in the blow. This varies 33? 334 METALLURGICAL CALCULATIONS. considerably, from 1 to 10 per cent, on the pig iron used in ordinary practice, but as much as 20 to 25 per cent, in some carelessly run " Baby " Besserners in steel-casting foundries. The silicon all oxidizes to SiO 2 ; iron mostly to FeO, and a small part, say not over one-tenth, to Fe 2 O 3 ; manganese mostly to MnO, a small part, up to one-fourth, may form Mn 2 O 3 ; phosphorus forms only P 2 O 5 ; carbon burns mostly to CO, but from one-fifth, at times nearly one-half, burns to CO 2 . When all the calculated oxygen has been found, the*blast to contain it can be calculated, if it is assumed that no free oxygen es- capes from the converter; at times, however, up to one-third of all the oxygen blown in may thus escape, but this is very exceptional, ordinarily less than one-fifth thus escapes, and often none at all. Problem 64. Pig iron containing 3.10 per cent, carbon, 0.98 silicon, 0.40 manganese, 0.101 phosphorus and 0.06 sulphur is blown in an acid-lined converter, to metal practically free from carbon, silicon and manganese, but no sulphur or phosphorus is elimi- nated. To get the minimum and the maximum amounts of air which could be needed, make the following assumptions: Case 1. Case 2. Per cent, of iron lost by oxidation. . . ._. ... ... 1 15 Proportion of iron oxidized to Fe 2 O 8 . . ...none one-tenih Proportion of Mn oxidized to Mn 2 3 . ...none one-fifth Proportion of C oxidized to CO 2 . /. ..one-fifth one-half Proportion of O 2 escaping in the gases '.none one-third Requirement: (1) Find the weight of dry air needed per metric ton of pig iron blown, in each case, and its volume at C. Express the results also in pounds and cubic feet per ton of 2,000 pounds. Solution : Case 1. Oxygen needed per 1,000 kg. of pig iron: 00 C to CO 2 6.2 kg.X || - 16.53 kg. CtoCO.. ....24.8 " X = 33.07 " THE BESSEMER PROCESS. 335 00 Si to SiO 2 ....................... 9.8 " X^i = 11.20 kg. Zo MntoMnO ...................... 4.0 " X^| = 1.16 " oo FetoFeO ....................... 10.0 " X = 2.86 " 64.82 " N 2 accompanying = 216.07 " Air needed = 280.89 " Volume at C. ....... i = 217.2 m 3 Volume needed per 1000 oz. Av = 217.2 ft 3 Volume needed per 2000 Ibs. Av = 6,950 ft 3 Case 2. Oxygen needed" per 1000 kg. of pig iron: qo C to CO 2 15.5 kg.X j^ = 41.33 kg. CtoCO.... 15.5 " X^ = 20.67 " iZ Si to SiO 2 .. . 9.8 " x||= 11.20 " Zo 1 fi MntoMnO 3.2 " X^ = 0.93 " 55 Mn to Mn 2 Q 3 . . 0.8 " X ^= 0.34 " FetoFeO.. .135.0 M X~ 38.57 " OD FetoFe 2 3 15.0 " X^= 6.43 " O 2 unused (one-half sum of above) = 59.73 " 179.20 ' N 2 accompanying = 597.33 " Air used = 776.53 ' Volume at C = 600 m 3 Volume used per 1000 oz. Av = 600 ft 3 Volume used per 2000 Ibs. Av = 19,200 ft 3 336 METALLURGICAL CALCULATIONS. For temperatures of the air above C., a corresponding increase in the volume used would be found. Since this is net air received by the converter an allowance for loss of 10 to 25 per cent, (in exceptional cases 50 per cent.) would be needed to get the piston displacement of the blowing engines. The above figures are the maximum and minimum for this quality of pig iron only, blown in an acid-lined converter; other quali- ties of pig iron might require a little more or less, and if blown in a basic-lined converter considerably more, to oxidize the phosphorus. The detailed calculations can be made in each specific case. AIR RECEIVED. The converse of the preceding proposition is to take an actual case, in a Bessemer converter, and to calculate how much air is being received. This will serve as a check on the blowing engines, since the volume received, divided by the piston displacement, gives the volume efficiency of the blowing plant. To make the calculation we need to know the weight and analysis of the pig iron and the analysis of the blown metal, in order to find the weights of impurities oxidized, also the average composition of the escaping gases, to find the propor- tion of carbon burning to CO 2 and of unused oxygen; also the composition of the slag, to get therefrom the weight of iron oxidized and the weight of slag, if practicable, but this can sometimes be calculated; also the weight and composition of the fume, if it is considerable. The temperature of the air entering the blowing cylinders its hygrometric condition and the barometric pressure should also be noted. Problem 65. At the South Chicago works of the Illinois Steel Co. (see paper by Howe, in Trans. American Institute of Mining Engi- neers, XIX. [1890-91], p. 1127), the charge weighed 22,500 pounds, and contained 2.98 per cent, carbon, 0.94 silicon, 0.43 manganese, 0.10 phosphorus, and 0.06 sulphur. After blowing 9 minutes 10 seconds the bath contained 0.04 per cent, carbon, 0.02 silicon, 0.01 manganese, 0.11 phosphorus and 0.06 sulphur. The slag formed contained 63.56 per cent, silica, 3.01 alumina, 21.39 FeO, 2.63 Fe 2 3 , 8.88 MnO, 0.90 CaO and 0.36 MgO, of THE BESSEMER PROCESS. 337 which the APO 3 , Cal and MgO and part of the SiO 2 come from the lining. The gases, analyzed dry, show an average com- position during the blow of CO 2 5.20 per cent. CO 19.91 H 2 1.39 N 2 ...73.50 and were free from fume. The piston displacement during the blow was 190,406 cubic feet, air in engine room 36 C., humidity 50 per cent., barometer 756 millimeters. Requirements: (1) The weight of oxygen acting on the bath during the blow, and the theoretical volume of air at standard conditions to which this would correspond, per 2,000 pounds of metal blown. (2) The volume of moist air at the conditions of the engine room, received by the converter during the blow, and the vol- ume efficiency of the blowing machine and connections. (3) The weight of slag produced and the loss in weight of the lining by corrosion during the blow. Solution: (1) The percentages of impurities left in the bath are so small that we can take them as equivalent to the same percentages reckoned on the original weight of the bath If they had been larger it would be necessary to assume an ap- proximate loss of iron during the blow, find the final weight of the bath and reckon, the percentages on this revised weight. Making this assumption, we know at once the weights of carbon, silicon and manganese oxidized, but we do not know the weight of iron lost. That follows, however, from a con- sideration of the slag, for the manganese and iron in the slag are derived only from the metallic bath, and the analysis of the slag practically gives us the relation between the weights of manganese and iron in it; since we know the weight of the former oxidized, the weight of iron lost can be calculated. Thus the slag contains : MnO 8.88 per cent = 6.88 per cent. Mn FeO 21.39 " = 16.64 " Fe as FeO Fe 2 3 2.63 " = 1.84 " Fe as Fe 2 3 338 METALLURGICAL CALCULATIONS. The loss of manganese being 0.42 per cent. = 94.5 pounds, the loss of iron is : Ifi fi4 94.5 X-^ '- = 228.6 pounds Fe as FeO O.oo 94.5 X = 25-3 pounds Fe as Fe 2 O 3 b.oo The remaining item still undetermined is the weight of carbon oxidizing to CO 2 and to CO. The gas analysis shows 5.20 per cent. CO 2 to 19.91 per cent. CO, and ^since equal volumes of each of these gases contain equal weights of carbon, it follows 5 20 that ' of the total carbon is present in the gas as CO 2 , Zo.iL and the rest as CO. Since the total carbon oxidized is 22,500 X 0.0294 = 661.5 pounds, we have 661.5 X = 137.0 pounds C burning to CO 2 = 524.5 " " CO Weight of oxygen absorbed by the bath : C to CO 2 ........... 137.0 pounds X 8/3 = 365.3 pounds CtoCO ............ 524.5 " X 4/3 =699.3 SitoSiO 2 .......... 207.0 " X32/28 =236.6 MntoMnO ......... 94.5 " X 16/55 = 27.5 " FetoFeO .......... 228.6 " X 16/56 = 65.3 " FetoFe 2 3 ....... 25.3 " X48/112 = 10.8 1404.8 " N 2 corresponding = 4682.7 Dry air corresponding = 6087.5 Volume at C = 75,483 ft 8 Weight of O 2 per 2000 pounds = 124.9 lbs.(l) Volume of air per 2000 pounds = 6,710 ft 3 (l) (This result is for comparison with data of Problem 64). THE BESSEMER PROCESS. 339 (2) The nitrogen in the gases can be obtained from its vol- ume relation to the carbon, and from this we can calculate the real volume of blast used. Weight of carbon in 1 cubic foot of gases: (0.0520 + 0.1991) X 0.54 = 0.1356 oz. Av. Volume of gases produced at standard conditions: 661.5X16 _ 7gQ , 7 f ,3 0.1356 78 ' 57 J Volume of nitrogen at standard conditions: 78,057X0.7350 = 57,372 ft 3 Weight = 57,372X1.26 = 72,289 oz. Av. = 4,518 Ibs. The question now is, how much nitrogen is contained in each cubic foot of air in the engine room. Knowing that, we are prepared to calculate the volume of this actually received by the converter:, Barometric pressure = 756 m.m. Tension of moisture (44x0.5) = 22 " Tension of air present = 734 Tension of nitrogen present (734X0.792) = 580 Weight of nitrogen in 1 cubic foot : 97*3 = 0.8495 OZ .Av Volume of air actually received : 4,518X16 0.8495 Volume efficiency of machinery: 85,095 85,095 ft.3 (2) 190,406 = 0.447 - 44.7 p. c. (2) (3) The slag contains 8.88 per cent, of MnO, equal to 6.88 per cent, of Mn, as already calculated. But 94.5 pounds of 340 METALLURGICAL CALCULATIONS. manganese is oxidized, therefore, the weight of slag produced is: 94.5^-0.0688 = 1374 pounds. (3) Weight of Si0 2 in slag (1374 X 0.6356) = 873 pounds SiO 2 from the Si of bath (207.0 + 236.6) =444 SiO 2 corroded from the lining = 429 CaO, A1 2 O 3 and MgO (1374X0.0427) = 59 Loss in weight of lining = 488 " (3) BLAST PRESSURE. It is necessary to use sufficient blast pressure to overcome the static pressure of the metallic bath, plus that of the slag formed, also the back pressure in the converter, to give the necessary velocity to the air in the tuyeres and to overcome friction in the same. When the tuyeres are near to the sur- face of the bath, pressures of 1 or 2 pounds will run the small converter, but the orcjmary converter with bottom tuyeres re- quires from 15 to 30 pounds pressure per square inch (1.054 to 2.108 kg. per square c.m.). We will consider the latter, the more frequent and the more complex case to discuss. The metal lies 12 to 24 inches (30 to 60 c.m.) deep in the converter. Since its specific gravity melted is about 6.88 (Rob- erts and Wrightson), the ferro-static pressure which it exerts is practically 0.25 pounds per square inch for each inch depth of metal, or 0.00688 kilos, per square centimeter for each centi- meter depth. The slag lying on the metal has a specific gravity melted of approximately half that of the metal. Its amount may vary from 5 to 10 per cent, of the weight of the metal treated, in an acid-lined converter, up to from 15 to 35 per cent, in a basic- lined vessel. Taking into account its lower specific gravity, its depth in the converter may be, therefore 10 to 20 per cent, the depth of metal in an acid-lined vessel, and 30 to 70 per cent, in a basic-lined converter; but the static pressure exerted would be only in direct pioportion to the relative weights; i. e., 5 to 10 or 15 to 35 per cent, of that exerted by the metal. The static pressure of the slag may, therefore, be reckoned as 0.125 pounds per square inch for each inch in depth, or 0.00344 THE BESSEMER PROCESS. 341 kilos, per square centimeter for each centimeter depth, and the depth of slag as lying between the following extremes : Acid Lined. Basic Lined. f ., finches.. 12 to 24 12 to 24 Depth of metal | Centimeters.. 30 to 60 30 to 60 ^ , , . /Inches 1.2 to 4.8 3.6 to 16.8 \ Centimeters. . 3.0 to 12.0 9.0 to 42.0 The probable depth of slag can be calculated in any par- ticular case, when the composition of the metal to be blown is known, its approximate depth in the vessel, and the approxi- mate composition of the slag to be formed. The back pressure of gases in the converter itself, that is, their static pressure, will vary with the shape of the converter and the size of the free opening for their escape into the air. A measurement at the Pennsylvania Steel Go's, works gave 0.275 pounds per square inch, but it is not stated just how the measurement was made. If we know approximately the volume of gas which must escape from the converter and from its temperature and the time and the size of the outlet calculate its velocity, the static pressure giving it this velocity can be calculated as , V 2 in which, if V is in feet per second, 2g = 64.3 and the resultant pressure is in feet of the hot gas; if V is in meters per second, 2g = 19.6, and h is in meters of the hot gas. Knowing the approximate specific gravity of the hot gas (weight of 1 cubic foot in pounds or of 1 cubic meter in kilograms) the static pressure is obtainable in pounds per square foot or kilograms per square meter. Illustration: The gases escaping from a converter are 78,057 cubic feet (standard conditions), and weigh 0.0801 pounds per cubic foot (standard conditions). They escape from the con- verter at an average temperature of 1,500 C., and the opening is 24 inches in diameter. What is the gaseous back pressure in the converter? Time of blow 9 minutes 10 seconds. 342 METALLURGICAL CALCULATIONS. Volume of gas at 1,500: 1500 + 273 78,057 X 273 Volume per second: 506,940^-550 Area of outlet : 2X2X0.7854 Velocity (assuming 0.9 coefficient) 921.7-^0.9 + 3.1416 Head of hot gas giving velocity: h = 326X326 64.3 = 506,940 ft 3 = 921.7 " = 3.1416 ft 2 = 326 ft. per second = 1,653 ft. Pressure of this column per square foot : = 20.4 pounds 070 1.653X^X0.0801 1773 Per square inch 0.14 This solution omits one consideration; the velocity of the gases in the body of the converter is neglected. This is some- what counterbalanced by the great friction of the gases against the sides of the converter, so that the one item tends to neu- tralize the other. If the interior were 8 feet in diameter, the velocity of the gases therein would average only some 20 feet per second, showing the above corrections to be practically negligible, since the pressure thus represented would be only 0.4 per cent, of the total obtained above. The pressure necessary to force the blast through the tuyeres is calculable on principles similar to the above; the differences are that the blast, at temperatures varying from 100 C. in the blast box to possibly 200 at its entrance into the metal, is divided up into fifty or 150 streams of approximately 1 centi- meter (0.4 inch) in diameter, the length of tuyeres being some 50 centimeters (20 inches). The formula similar to that used THE BESSEMER PROCESS. 343 for chimney draft, or rather, friction al resistance in a chimney, applies to this case. V 2 KL in which h is the head in terms of the air passing, V is its velocity, 2g the gravitation constant, L the length of the tuyere, D its diameter, and K the coefficient of friction, which latter is for relatively smooth flues 0.05 (Grashof), and may be so assumed here. Problem 66. In the converter mentioned in Problem 65, where 22,500 pounds of metal was blown in 9 minutes 10 seconds, using as therein calculated 85,095 cubic feet of air, at 36 C., and producing 1,374 pounds of slag, assume the inside diameter of the converter as 7 feet, and that the bottom contains fourteen tuyere blocks, each containing eleven openings of 0.5 inch diameter each; blocks 24 inches long. Assume back pressure in converter 0.14 pounds per square inch, total blast pressure in equalizing reservoir 27 pounds per square inch. Temper- ature of air in the tuyeres 150 C. Required: (1) The pressure needed to overcome the head of metal and slag. (2) The pressure absorbed in friction in the tuyeres. (3) The pressure represented by the velocity of the blast in the tuyeres. (4) The loss of pressure from the reservoir to the blast-box. (5) The distribution of the total pressure. (6) The length of the blow if the blast pressure were reduced to 20 pounds. (7) The length of the blow if the pressure were maintained at 27 pounds, but twenty-one tuyere blocks (each with eleven J-inch holes) were used. Solution: (1) At the start there is 22,500 pounds of melted metal, the volume of which will be 22,500 22,500 52.3 cubic feet 6.88X62.5 430 344 METALLURGICAL CALCULATIONS. The depth of metal, the inside diameter being 7 feet, is CO O CO O bZ ' 6 = 1.356 feet 7X7X0.7854 38.5 = 16.4 inches Static pressure = 16. 4 X 0.25 = 4.1 Ibs. per sq. in. The slag, formed during the first half of the blow, weighs 1,374 pounds, and has a volume of 1374 1374 = 6.4 cubic feet 3.44X62.5 215 The depth of slag, at its maximum, will be 6.4-38.5 = 0.167 feet = 2.0 inches Static pressure 2.0X0.125 = 0.25 Ibs. per sq. in. The static pressure during the blow will, therefore, be 4.1 pounds to start with, increasing during the first half of the blow to 4.35 pounds, and staying practically constant at that, and, therefore, will average ,4.1 + 4.35 , 4.35 2X2 + ^~ (2) Each of the 14X11 = 154 tuyeres receives 85,095 -r- 154-:-550 = 1.005 cubic feet of air per second, measured at 36 C. At 150 C. this volume is 1.375 cubic feet And the velocity in the tuyere: Q - K 0.5X0.5X0.7854 ^_ 1.375-5 -- - = 1009 feet per second The head absorbed in friction in the 24-inch tuyeres will be , 1009X1009^0.05X2 " -- X Changing this pressure of air at 150 C. to pounds per square inch we have: THE BESSEMER PROCESS. 345 Weight of 1 cubic foot of air at = 0.0808 pounds Weight of 1 cubic foot of air at 150 = 0.0522 " Weight of air column = 37,893X0.0522 = 1978 " Pressure in pounds per square inch = 13.7 (2) (3) The pressure absorbed as velocity has already been expressed in getting the friction in the tuyeres. The velocity head is simply: V 2 1009X1009 1 h = 2F 64.3 = 1 which becomes in pressure 15,835X0.0522 = 826.6 pounds per square foot = 6.73 pounds per square inch (3) (4) The remaining part of the 27 pounds pressure used is lost between the blast reservoir and the entrance to the tuyeres. It is 27.00- (13.70 + 5.73 + 4.29 + 0.14). = 3.14 pounds. (4) (5) Distribution of blast pressure: Fall between reservoir and tuyeres = 3.14 pounds = 11.6% Absorbed in friction in the tuyeres =13.70 " = 50.7% Absorbed in velocity in the tuyeres = 5.73 " = 21.2% Static head of liquid bath =4.29 " 15.9% Velocity head of issuing gases = 0.14 " = 0.6% 27.00 =100.0% (6) All the items of absorption of pressure are proportional to the square of the velocity of the gases, excepting the static pressure of the bath. It remains constant at 4.29 pounds. If the total pressure were reduced to 20 pounds, there would be only 20 4.29 = 15.71 pounds pressure to give velocity and overcome friction, instead of 27 4.29 = 22.71 pounds. The relative quantities of air blown through in a given time in the two instances would be practically proportional to the square roots of the two effective pressures, i.e.'. \/22Tl : VIsTTl = 1 : 0,832 and the times of the blows inversely as the latter: 550 sec. 4- 0.832 = 673 seconds = 11 min. 13 sec. (6) 346 METALLURGICAL CALCULATIONS. (7) If the tuyere area were increased 50 per cent., then the velocity of the air in the tuyeres would be decreased one-third, assuming the amount of air passing to be unchanged. This would decrease the pressure absorbed in friction, and in giving velocity in the tuyeres to (0.67) 2 = 0.444 of its former amount. The 19.43 pounds previously absorbed in these two items would then become 19.43X0.444 = 8.61 pounds, and the total pressure needed to run the converter just as fast as before would be 27- (19.43- 8.61) = 16.18 pounds per square inch. If, however, the pressure were maintained at 27 pounds, giving still 27 4.29 = 22.71 pounds to overcome f national resist- ances and to give velocity, then the velocity and consequent amount of air blown through by this 22.71 pounds pressure would increase in proportion to the square root of these two available pressures; i.e., be as Vl6.18-4.29 : \/27- 4.29 = 1 : 1.38 The duration of the blow would be just that much shorter ; i.e. : 550 sec. -s-1.38 = 398 seconds = 6min.38sec. (7) FLUX AND SLAG. No flux is used in the acid-lined converter, and the silica, iron oxides and manganese oxide formed in the converter unite to a silicate slag which corrodes the lining and thus takes up more silica. The slag being analyzed, its weight is ob- tained by considering the percentage of manganese which it contains, because the weight of manganese oxidized is known definitely from the analysis of the bath; it is usually all oxi- dized. Calculation of the weight of slag cannot be based upon the silica, because an unknown amount comes from the lining ; nor upon the iron, because the weight of iron left in the con- verter is not definitely known. Having the weight of the slag, analysis tells us the total weight of silica in it, as also the amount of iron. The silica in the slag minus that formed from silicon in the pig iron, gives silica corroded from the lining. In the basic Bessemer converter, phosphorus is nearly en- tirely eliminated from the metal, so that, assuming none to be volatilized, the amount going into the slag is known, and using the slag analysis the weight of slag can be calculated. In THE BESSEMER PROCESS. 347 this process the lining is mainly dolomite, containing CaO and MgO, in proportion easily determined by analysis. The weight of slag being known, the amount of corrosion of the lining can be determined from the percentage of magnesia therein, which may be assumed as practically coming entirely from the lining ; it cannot be told from the CaO in the slag, because nearly pure CaO is added during the blow, and some of it, a variable amount, gets blown out of the converter. For the same reason it is not possible to base a good calculation of the weight of slag on the lime alone which is added, be- cause of the indefinite proportion of it which is blown out. The weight of slag may also be gotten from the silica or man- ganese oxide in it, assuming these to come almost entirely from the oxidation of silicon or manganese. Lime must be added as flux, in the basic converter, to pro- tect the lining and to make the slag so basic that the percentage of silica in it is below 15 per cent., phosphoric acid below 20 per cent., and lime over 50 per cent. These considerations must be balanced in each particular case. Illustration: Pig iron blown in a basic-lined converter con- tained 1.22 per cent, silicon, 2.18 phosphorus, 1.03 manganese and 3.21 carbon. It is blown until all of these and 2.00 per cent, of iron are oxidized, and burnt lime is added to form slag during the blow. Composition of the burnt lime: MgO, 1.00 per cent.; SiO 2 , 2.00 per cent.; CaO, 97 per cent. How much lime should be added per 10 metric tons of pig iron charged? The slag-forming ingredients from the oxidation of the bath, and the addition of X kilos, of lime, are SiO 2 10,000X0. 0122 X 60/28 = 261.4 kg. P 2 O 5 . 10,000X0.0218X142/62 = 499.3 " MnO 10,000X0. 0103 X 71/55=133.0 " FeO 10,000X0. 0200 X 72/56=257.1 " fCaO XX 0.9700 = 0.97 X \ MgO XX 0.0100 = 0.01 X [SiO 2 XX 0.0200 = 0.02 X Weight for slag = X + 1150.8 kg. Corrosion of the lining will undoubtedly increase this weight, so some allowance should be made, say to increase it 5 per 348 METALLURGICAL CALCULATIONS. cent., probably an outside figure. Of this 5 per cent., half can be considered lime and half magnesia. The total weight of slag will then be 1.05 X-f 1208.3, and of the ingredients prin- cipally in question: SiO 2 = 261. kg. + 0.02 X MgO = 28.7 " +0.035 X CaO = 28.7 " + 0.995 X To make our slag 50 per cent. CaO will require the addition of enough to make 28. 7 + 0. 995 X = 0.50 (1.05 X + 1208.3) X = 1224 kg. To make a slag with at most 15 per cent, of SiO 2 requires 261.4 + 0.02 X = 0.15 (1.05 X + 1208.3) X = 583 kg. To make a slag with at most 20 per cent, of P 2 O 5 requires 499.3 = 0.20 (1.05 X + 1208.3) X = 1227 kg. The larger of these three amounts would be used, with 10 per cent, added to cover lime dust blown out, making 1350 kg. added, and the calculated composition of the slag: CaO 1250 kg. = 50.1 per cent. MgO 723 " = 2.9 SiO 2 286 " = 11.4 P 2 O 5 499 " = 20.0 FeO 257 " = 10.3 MnO 133 " = 5.3 Total 2497 RECARBURIZATION. When the bath has been blown to nearly pure iron, melted spiegeleisen is run in, to add the necessary carbon and man- ganese. Knowing the approximate composition and weight of the bath, and the composition of the melted Spiegel, a simple arithmetical calculation would give the amount of the latter to be added, assuming no loss of carbon or manganese in the THE BESSEMER PROCESS. 349 operation. But experience shows that there is some loss, and that the carbon and manganese in the finished metal are always lower than the calculated amount. An interesting field is open here for calculating the loss of manganese and carbon and the amount of oxygen which must have been in the metal to cause these losses. A tabulation of many such calculations gives the metallurgist the necessary data for assuming the average amounts of carbon and manganese lost during recarburization, under different conditions of working, such as letting the metal stand before pouring or pouring at once, turning on the blast 5 or 10 seconds to mix up the bath, etc. Problem 67. At the end of the blowing the converter of Problem 65 con- tained 21,283 pounds of metal of the composition 0.04 per cent, carbon, 0.02 silicon, 0.01 manganese, 0.11 phosphorus, 0.06 sulphur, an unknown amount of oxygen (probably < 0.3 per cent.) and the rest iron. There is added to it 2,500 pounds of spiegeleisen, containing 4.64 per cent, carbon, 0.035 silicon, 14.90 manganese, and 0.139 phosphorus. The finished metal contained 0.45 per cent, of carbon, 0.038 silicon, 1.15 manganese, 0.109 phosphorus and 0.059 sulphur. Assume no iron oxidized. Required: (1) A balance sheet of materials before and after recarburizing. (2) The proportions of carbon and manganese going into the finished metal. Solution: (1) Blown Gases Metal Spiegel. Steel. or Slag. C 8.5 116.0 106.5 18.0 Si 4.3 0.9 9.0 -3.8 Mn 2.1 372.5 271.2 103.4 P 23.4 3.5 25.8 1.1 S 12.8 .... 14.0 - 1.2 Fe 21,232 2007 23,239 The differences in the sulphur, phosphorus and silicon are within the limits of error of the data, but there is no doubt as to the loss of carbon and manganese. (2) The proportions of the two elements in question going into the finished steel are: 350 MET A LL URGICAL CALCULA TIONS. Carbon 106.5^-124.5 = 0.85 = 85 per cent. Manganese 271.2-^-374.6 = 0.72 = 72 The calculated percentages in the finished steel should have been, and actually were: Carbon 0.53- 0.45, loss = 0.08 per cent. Manganese 1.58- 1.15, loss = 0.43 Concerning oxygen removed, if we assume the loss of carbon and manganese to be due to their combining with oxygen dis- solved in the bath, to form CO and MnO, the percentage of oxygen thus absorbed is: By carbon 0.08x16/12 = 0.11 per cent. By manganese 0.43X16/55 = 0.12 0.23 [The next chapter will consider the thermo-chemistry of the Bessemer process.] CHAPTER VIH. THERMOCHEMISTRY OF THE BESSEMER PROCESS. The feature of the Bessemer operation which strikes the observer as most wonderful, is that cold air is blown in great quantity through melted pig iron, and yet the iron is hotter at the end than at the beginning. If the observer will reflect a moment, however, he can see that if nothing but fuel, on fire, was in the converter, it would certainly be made much hotter by the air blast ; in similar manner, the oxidation or combustion of part of the ingredients of the pig iron furnishes all the heat required for the process. Ten tons of pig iron contains, for example, some 350 kilograms, of carbon which is all burnt out in the bussemerizing, furnishing heat equal to the combustion of some 400 kilograms of coke a not insignificant quantity, since it is burned and its heating power utilized in a very few minutes. ELEMENTS CONSUMED. The usual ingredients of pig iron are: Iron 90.0 to 95.0 per cent Carbon 2.5 " 4.5 Manganese 0.5 " 4.0 Silicon 0.5 " 4.0 Phosphorus 0.01 " 4.0 Sulphur 0.01 " 0.5 Some of the unusual constituents are nickel, chromium, titan- ium, aluminium, vanadium, tungsten, copper and zinc; all of these are rare, and there is seldom present as much as 0.5 per cent, of any one except in unusual cases. In the Bessemer operation, carried out with the usual silica lining, iron, carbon, manganese and silicon are freely oxidized, but phosphorus and sulphur remain practically unchanged. In the basic Bessemer, lined with burnt dolomite and tar, phos- 351 352 METALLURGICAL CALCULATIONS. phorus is also freely oxidized at the end of the operation, but sulphur is only slightly diminished the more, the more man- ganese is in the slag. After all the oxidizable impurities are eliminated, iron itself oxidizes in much larger quantity, oc- casioning great loss if the blast is permitted to continue. Iron oxidizes during the blow mostly to FeO, which enters the slag as ferrous silicate, and partly to Fe 2 3 . The brown fume which escapes in large amount if the blow is continued too long contains iron as Fe 2 3 . Fe to FeO 1173 Calories per kg. bf Fe Feto Fe 2 O 3 1746 FetoFe 3 4 1612 The amount of iron oxidized can be gotten from the weight and percentage composition of the slag; also from the com- parison of the weights of materials used and weight of ingots produced, knowing the weight of other impurities oxidized. Carbon oxidizes mostly to CO gas, and partly, especially in the early part of the blow, to CO 2 gas. The proportion of each of these formed can only be known by analyzing the gases produced at various stages of the blow. The proportionate volumes of CO and CO 2 express the proportionate amounts of carbon forming each respective gas. A shallow bath allows more CO 2 to pass, on essentially the same principle that a deep layer of fuel on a grate favors the production of CO. The heat evolved by oxidation of carbon is : C to CO 2430 Calories per kg. of C CtoCO 2 8100 Manganese oxidizes quickly and mostly to MnO. If the metal is a little overblown, Mn 2 O 3 in small amount is found in the slag, while Mn 2 O 3 is also present in the fumes. The heat evolved in these oxidations is : Mn to MnO 1653 Calories per kg. of Mn MntoMn 2 O 3 2480 (?) " The last figure is estimated; it has not yet been determined experimentally. Silicon oxidizes rapidly and early in the blow to SiO 2 , form- ing silicate slag with the metallic oxides formed. Its heat THE BESSEMER PROCESS. 353 of oxidation has been usually taken as 7830 Calories per kilo- gram, but recent investigations have thrown doubt on this figure, Berthelot having found as low as 6414 Calories, and Mr. H. N. Potter, 7595 for crystalline silicon, equal to 7770 for amorphous silicon. Under these circumstances the best course is probably to use the middle value ad interim, and consider Si to SiO 2 7000 Calories per kg. of Si We hope that the exact figure will soon be determined. Phosphorus oxidizes to P 2 O 5 , and only towards the end of the blow. It is practically completely eliminated, going into the slag as calcium phosphate : P to P 2 5 5892 Calories per kg. of P Sulphur is not eliminated at all in the acid-lined converter. In the basic converter it is reduced in amount in the last few minutes, while phosphorus is disappearing, and partly escapes, mostly as SO 2 in the gases. The presence of a very basic slag is necessary, but the sulphur, while possibly going into the slag, does not remain there, but passes into the gases. The heat of oxidation of the unusual elements sometimes present are, as far as known, Ni to NiO 1051 Calories per kg. of Ni Ti to TiO 2 4542 " " Ti Al to A1 2 O 3 7272 " " Al ZntoZnO 1305 " " Zn Vto V 2 O 5 4324 . " V ^WtoWO 3 .... 1047 " " W CrtoCrW 2344 " " Cr Some of the above values are a little uncertain, and none of them include the heat of combination of the oxide formed with the slag. HEAT BALANCE SHEET. Taking o C. (32 F.) as a base line, we may express the total heat contents of the pig iron, steel, gases, slag, blast, etc., 354 METALLURGICAL CALCULATIONS. from this temperature. This method is simpler than to take the bath at any one high temperature and to reckon from there. The items of heat available during the blow are: Heat in the body of converter at starting. Heat in the melted pig. iron used. Heat in the spiegleisen or ferro-manganese added. Heat in hot lime added (sometimes in basic process). Heat in the blast, if warm on entering. Heat developed by oxidation of the bath. Heat developed by formation of the slag. The items of heat distribution are: Heat in the body of the converter at finishing. Heat in the steel poured. Heat in the slag at finishing. Heat in the gases escaping. Heat in the fume. Heat in the slag or metal blown out. Heat absorbed in decomposing moisture of the blast. Heat to separate the constituents of the bath. Heat conducted away by supports, blast pipe, etc. Heat conducted to the air in contact with converter. Heat radiated during the blow. These two columns should balance each other if all the items of each are correctly determined. Heat in Converter Body at Starting. If the converter were quite cold when the pig iron was run in and the blow started, this item would be zero. But such a case would result disastrously, since the heat absorbed by the converter during the blow would be more than any ordinary heat could afford to lose. It is, therefore, customary, when starting for the first time, to build a fire in the converter and turn on a little air blast, so as to bring the inside up to bright redness, say 900 to 1000 C. The outside shell would, under these conditions, be at about 200, and the mean temperature of the converter lining, say 400. If the converter were in regular operation, one charge being introduced as soon as the other was finished, then the heat in the body of the converter at start- ing could be practically regarded as equal to the heat in the THE BESSEMER PROCESS. 355 same at finishing, assuming the heats are running regularly An estimate of the heat in the converter body at starting is therefore only necessary when the converter is first started up, or when it is allowed to stand some time between blows. Illustration: Assume a converter weighing, without charge, 25 tons, of which 5 tons is iron work and 20 tons silica lining. The mean specific heat of iron (for low temperatures) being 0.1 1012 + 0.000025t + 0.0000000547t 2 , and for silica 0.1833 + 0.000077t, calculate the heat contained in the body of the con- verter. (1) When the temperature of the outside shell is 200 and that of the lining averages 400 (converter warmed up for starting). (2) When the temperature of the outside shell is 300 and that of the lining averages 725 (converter empty at end of a blow) . Solution: (1) Heat in 5000 kg. of iron work: 0.11012 + 0.000025 (200) +0.0000000547 (200) 2 = 0.11731 0.11731X200X5000 = 117,310 Calories. Heat in 20,000 kg. of silica lining: 0.1833 + 0.000077 (400) = 0.2141 0.2141X400X20,000 = 1,712,800 Calories. Total heat in converter body = 1,830,110 It may be remarked that this would require the consumption of at least 250 kilograms of coke, with a calorific power of 8,000 Calories, to warm the converter to this extent. (2) Heat in 5,000 kg. of iron work: 0.12354X300X5,000 = 185,310 Calories. Heat in 20,000 kg. of silica lining: 0.2391 X 725X20,000 =3,466,950 Total = 3,652,260 This condition is assumed as representing the .converter when just emptied and immediately refilled. In this case, in regular working, the heat in the converter body is practically the same at the beginning and at the end of a blow, and the heat losses through it are only those due to conduction to the air and ground and radiation. 356 METALLURGICAL CALCULATIONS. Heat in Melted Pig Iron Used. This quantity depends on the temperature at which the pig iron is run into the converter. If the iron is high in silicon, which would tend to produce a hot blow, it may be run in somewhat cool; but if low in silicon it should be run in much hotter. The minimum quantity of heat contained in a kilogram of melted pig iron may be put at 250 Calories; the maximum, for very hot pig iron, 350 Calories; about 300 Calories would be a usual average figure. This may be easily determined ex- perimentally in any given case by granulating a sample in water in a rough calorimeter. Heat in Metallic Additions. Melted spiegeleisen is usually not very hot when run into the converter. It may contain 250 to 300 Calories per kilogram; say an average of 275. Ferro-manganese is sometimes added red-hot, at 800 to 900 C. At this temperature it would con- tain 120 to 135 Calories per kilogram, assuming a mean specific heat of 0.15. Heat in Preheated Lime. Taking the mean specific heat of CaO as 0.1715 + 0.00007t, it would contain 154 Calories per kilogram at 700, and 211 Calories at 900. The heat content can be calculated for any known temperature at which the lime is used. Heat in Warm Blast. No Bessemer converters are run by hot blast, but the air pressure used is so great (20 to 35 pounds per square inch) that the blast is warmed by compression in the cylinders. If air at 1 atmosphere tension (ordinary air) be compressed to 2 or to 3 atmospheres tension, giving effective blast pressures of 1 to 2 atmospheres, the air is heated 60 or 103, respectively, above its initial temperature. While some of this heat may be lost in the cylinder and conduits, yet the air, unless artificially cooled, passes to the tuyeres at 25 to 50 C. above the outside temperature, and thus imparts some heat to the converter. Heat Developed by Oxidation. We have already discussed the thermochemical data required for this calculation. To use the data we must find the weights THE BESSEMER PROCESS. 357 of each ingredient oxidized (not forgetting the iron itself) and the nature of the oxide it forms. This is deduced from the analysis of slag, gases and steel produced, as compared with those of the pig iron and additions used. Heat of Formation of Slag. The slag is a mechanical mixture or mutual solution of sili- cates of iron and manganese, with a little alumina (up to 5 per cent.) and lime and magnesia from nothing up to 5 per cent. In the basic process a large amount of calcium phos- phate is present, representing over half of the entire slag, while magnesia is present in considerable amount, and alumina is almost absent. Having, in the preceding column, calculated the heat of oxidation of all the metals oxidized in the converter, it re- mains to calculate the heat of combination of these to form slag. The silicate of alumina contributes nothing, since it occurred combined in the lining or lime added. The same can be said of the ,lime and magnesia in the acid process slags. The FeO and MnO are then to be considered, and the only thermochemical data we have are: (MnO, SiO 2 ) = 5,400 Calories. (FeO, SiO 2 ) = 8,900 These data are for 71 or 72 parts of MnO or FeO respectively, uniting with 60 parts of SiO 2 . Since in acid slags there is always proportionately less of the bases, we should utilize the above heats of formation by expressing them per unit weight of MnO or FeO going into combination. These figures are: Per kg. of MnO = 5,400-^-71 = 76 Calories. Per kg. of FeO = 8,900^72 = 124 From these figures the heat of formation of the slag can be calculated: Any Fe 2 O 3 present in the slag would be calculated as its equivalent weight of FeO (by multiplying by 1444-160). Illustration: A Bessemer slag contained by analysis: SiO 2 47.25 per cent. A1 2 O 3 3.45 FeO 15.43 MnO 31.89 CaO, MgO 1.84 358 METALLURGICAL CALCULATIONS- What is its heat of formation per kilogram? Solution: The APO 3 , CaO and MgO are to be neglected, for reasons already given. The heat of combination, per kilo- gram of slag, is therefore, FeO uniting with SiO 2 = 0.1543X124 = 19.1 Cal. MnO uniting with SiO 2 = 0.3189 X 76 = 23.2 " Total = 42.3 " In basic Bessemer slags the conditions are much more com- plicated. The content of P 2 O 5 is not under 14 per cent., runs as high as 25 per cent., and averages 19 per cent.; the CaO averages 45 per cent., limits 35 to 55; the silica is usually below 12 per cent., and averages 6 to 8 per cent.; magnesia is present from 1 to 7 per cent., average about 4 per cent. In such slags we should first of all assume the P 2 O 5 to be combined with CaO as 3CaO. P 2 O 5 , containing 168 of CaO to 142 of P 2 O 5 , (1,183 to 1) and having a heat of formation from CaO and P 2 5 of 159,400 Calories per molecule, or 1123 Calories per unit weight of P 2 O 5 . Next the iron and manganese present may be calculated to FeO and MnO respectively, and treated as to their combination with SiO 2 , the same as in an acid slag. Alumina may be considered in such basic slags as an acid, and equivalent to 120/102 of its weight of silica. These allowances will leave considerable lime and either an excess or deficiency of silica; if an excess, we can assume it combined with lime and magnesia, with a heat evolution equal to 476 Calories per kilogram of silica; if a deficiency, we let the calculations stand without further modification. Illustration: Slag made at a Rhenish works contained: SiO 2 7.73 per cent. P 2 O 5 21.90 APO 3 3.72 Fe 2 O 3 1.00 FeO 4.73 MnO 2.05 CaO 50.76 MgO 4.00 CaS 1.71 What is its heat of formation per unit of slag? THE BESSEMER PROCESS. 359 Solution: The 21.90 parts of P 2 5 would be combined with 21.90X168/142 = 25.91 parts of CaO. This leaves 50.76 25.91 = 24.85 of CaO as either free, dissolved CaO or partly combined, in the slag. The 2.05 MnO would correspond to 2.05X60/71 = 1.73 SiO 2 ; the 4.73 FeO to 4.73X60/72 = 3.94 SiO 2 ; the 1.00 Fe 2 O 3 to 1.00X60/80 = 0.75 SiO 2 ; a total SiO 2 requirement for these three bases of 6.42 per cent. The SiO 2 present is 7.73 per cent., adding to which the SiO 2 equiva- lent to the A1 2 O 3 present (3.72X120/102 = 4.38), we have 12.11 per cent, of summated silica. The ratio of summated FeO to summated SiO 2 is considerably below the ratio 72 to 60, we can, therefore, consider the summated FeO as all com- bined with silica. The summated FeO is : FeO = 4.73 per cent. FeO equivalent of MnO = 2.08 FeO equivalent of Fe 2 3 = 0.90 " Total = 7.71 " And the SiO 2 combining with this as FeO. SiO 2 is 7.71X60/72 = 6.42 per cent. The excess of summated silha free to combine with lime is 12.11 6.42 = 5.69 per cent. As there is 24.85 of CaO and 4.00 of MgO for it to combine with, the heat of this combina- tion must be calculated on the SiO 2 going into this combination. We then have the formation heat of the slag as: P 2 O 5 to 3CaO . P 2 5 21.90 X 1123 = 24,594 Cal. MnO to MnO . SiO 2 2.05X 76= 156 " FeO to FeO .SiO 2 5.63 X 124= 698 " SiO 2 to 3CaO . SiO 2 5.69 X 476 = 2,708 " 28,156 ' This equals 281.6 Calories per unit weight of slag, forming a very important item in the heat balance sheet, particularly when the slag is large in amount. Heat in Converter Body at Finishing. This item reaches its maximum at the end of the blow, and would be equal to that calculated for the beginning of the blow, 360 METALLURGICAL CALCULATIONS. with the exception that some of the lining, silica or dolomite has been corroded and passed into the slag, carrying with it its sensible heat. Heat in Finished Steel. This should be determined experimentally in each particular case. If not so determined an average value may be assumed, based on the following considerations: The finishing tempera- ture averages 1650 C. ; at this heat average Bessemer steel will contain a total of 350 Calories of heat per kilogram. If the temperature is determined by a pyrometer, a correction of 1/5 Calorie can be made for every degree hotter or colder than 1650. Heat in Slag. Some experimental data are badly needed, concerning the heat in slags of different composition at different temperatures. At present it is necessary to make guesses, wherever the heat in the slag produced is not directly determined. At a finishing temperature of 1650 it is likely that the slag contains 550 Calories per kilogram; with a variation of 1/4 Calorie for each degree hotter or colder than 1650. Heat in Escaping Gases. The amount of these gases can only be determined satis- factorily from their analysis and the known weights of carbon oxidized. Direct estimation from the piston displacement is of much less exactness, because the slip and leakage at these high pressures may reach 25 to 50 per cent, or even more. The temperature of the gases is only slightly less than that of the bath, that is, some 1350 at starting and 1650 at finishing. Outside in the air the Bessemer flame may be much hotter than this, but that is due to further combustion of CO to CO 2 out- side the converter, and should be disregarded. Where extra air is blown upon the surface of the bath, as in " baby " converters, it is quite possible that the gases in the converter and the es- caping gases may be considerably hotter than the bath itself These variations must be taken into consideration. The most satisfactory condition is to insert a pyrometer tube into the opening of the converter and measure the temperature directly. The heat carried out by the gases can then be calculated ac- THE BESSEMER PROCESS. 361 curately, using the proper mean specific heats to these high tem- peratures already used in these calculations. Heat in Escaping Fume. This is mostly oxides of iron and manganese, with some- times silica. It is relatively small in amount. Its quantity being known, consider it at the same temperature as the gases, with a mean specific heat of 0.40 if free from silica, and 0.35 if siliceous. Heat in Slag or Metal Blown Out. These can be counted as equal to the heat in an equal quantity of slag or metal at the finishing temperature. Heat Absorbed in Decomposing Moisture. Knowing the amount of moisture blown in with the blast, a proper allowance is 29,040 -f- 9 = 3,227 Calories for every kilo- gram of moisture thus blown in. It is probable that this moisture is all decomposed, its hydrogen appearing in the gases. In the absence of data as to the hygrometric condition of the blast, the amount of moisture entering may be inferred and calculated from the amount of hydrogen in the gases. Heat to Separate Constituents of Bath. We here meet the question, how much heat of combination exists between the bath and the various ingredients which are removed carbon, silicon, manganese, phosphorus, sulphur. Le Chatelier believes manganese to exist in the bath as Mn 3 C, requiring 80 Calories per kilogram of manganese to decompose it. Silicon and carbon have, as far as has at present been determined, no sensible heat of combination with iron. Sulphur requires 750 Calories to separate each kilogram from iron. Phosphorus requires, according to Ponthiere, 1397 Calories to separate each kilogram from iron, but the reliability of this datum is doubtful. Until more reliable tests are made it is perhaps better to omit this item than to use it, although it must be of great importance if as large as Ponthiere states it to be. Heat Conducted Away by Supports. This is a very difficult quantity to determine, being con- ditioned by the size of the supports, their cooling surface and 362 METALLURGICAL CALCULATIONS. the kind of connection they have with other objects. The heat which would pass to the blast pipe is practically returned to the converter by the incoming blast. The heat passing into the supports is perhaps best found by taking their temperature at different places, and calculating the heat loss from their sur- face by radiation and conduction to the air. This amounts practically to considering them as part of the outer cooling surface of the converter; the calculation of these surface losses is given in the two following paragraphs: Heat Conducted to the Air. This is a function of the extent of outside surface, its tem- perature, the temperature of the air, and the velocity of the air current. Measurement will give the extent of surface in contact with the air and the average velocity of the air cur- rent; the surface being rough iron, the coefficient of transfer conductivity may be taken as k = 0.000028 (2 + V~v). where v is the air velocity in c.m. per second, and k is the heat con- ducted per second in gram-calories, from each square c.m. of surface per 1 difference of temperature. The temperature of the outer surface should be carefully measured, so that a reliable average is obtained, and the air velocity likewise aver- aged, since it has considerable influence on the heat lost to the air. Illustration: A converter has an outside surface of 50 square meters, at an average temperature during the blow of 200 C. the average air current being 1 meter per second, and out- side air 30 C. What is the heat loss by conduction to the air in kilogram Calories per minute? Solution : Coefficient of transfer conductivity : 0.000028 (2+ VlOO) = 0.000336 . Heat loss per 1 difference, per second, in gram- calories : 50X10,000X0.000336 = 168 calories. Heat 'loss per 170 difference, per minute, in kg.-calories: 168 X 170 X 60 -T- 1000 = 1714 calories. THE BESSEMER PROCESS. 3t53 Heat Radiated During the Blow. This is a function of the temperature of the outside shell, the mean temperature of the surroundings of the converter and the nature of the metallic surface. As the surface is oxidized iron, it would lose about 0.0141 gram-calories from each square centimeter per second, if at a temperature of 100 and the surroundings at 0; or practically 1 gram-calorie per second from each square meter, for every 100,000,000 of numerical dif- ference between the fourth powers of the absolute tempera- ture of the radiating surface and its surroundings. (See Metal- lurgical Calculations, Part 1., p. 185). Illustration : Assuming the surroundings of the converter at 30, in the preceding illustration, what amount of heat is radiated per minute in large Calories? Solution: The absolute temperatures in question are 273 + 30 = 303, and 273 + 200 = 473. The difference of their fourth powers is : 473 4 303 4 = 41,626,500,000, which, divided by 100,000,000, gives 416.265 gram-calories lost per second per each square meter of radiating surface. The radiation loss for the whole surface per minute is, therefore: 416.265 X 50 X 60 4- 1000 = 1249 kg.-Calories. While the assumption made as to the temperature of the out- side shell is doubtless only approximate, yet if the tempera- ture of the same is carefully determined the radiation loss can be accurately calculated. If the outside of the converter were polished, this radiation loss might be reduced nearly nine tenths. Problem 68. From the data and results of calculation of Problem 65 (see pages 310, 317 and 323,) we see that 22,500 pounds of pig iron and 2,500 pounds of spiegeleisen produced 24,665 pounds of steel, there being eliminated during the blow and recar- burization : Carbon 679.5 pounds (140.7 to CO 2 ) Silicon 203.2 Manganese 197.9 Iron 253.9 " (25.3 to Fe 2 O s ) 364 METALLURGICAL CALCULATIONS. The gases contain : CO 2 5.20 per cent. CO 19.91 H 2 1.39 N 2 73.50 The slag contains: SiO 2 ...63.56 A1 2 O 3 3.01 FeO 21.39 Fe 2 O 3 2.63 MnO 8.88 CeO 0.90 MgO 0.36 Make average assumptions for requisite data not given. Required: A balance sheet of heat evolved and distributed. Solution: The items of this balance sheet have already been discussed in detail. We will apply them to this specific case: Heat in Body of Converter at Starting: Assuming that this is a blow in regular running, the heat may be taken at any reasonably approximate quantity, because the same quantity with only a slight deduction will be allowed as contained in the same on finishing. We will, therefore, take a figure already calculated, 8,034,970-pound Calories, as the heat in the con- verter body at starting. Heat in Melted Pig Iron: We will take it at 300 Calories per pound, or a total of 300X22,500 = 6,750,000 Calories. Heat in Spiegeleisen: 2,500 X 300 = 750,000 Calories. Heat in Blast: This may safely be considered as warmed by compression and entering the converter at 60 C. The amount of blast received altogether is calculated thus: Carbon oxidized = 679.5 pounds. Volume of CO and CO 2 formed: 679.5X16^0.54 = 20,133 cu. ft. Volume of gases ^TT = 80 ' 179 " U.^Oll Volume of nitrogen 80,179 X 0.735 = 58,932 " Volume of air proper in blast = 74,408 Volume of hydrogen in gases . 80,179X0.0139= 1,115 Volume of moisture in gases = 1,115 " THE BESSEMER PROCESS. 365 Assuming the blast at 60 C., the heat in it is: Air 74,408X0.3046 = 22,665 oz. Cal. per 1 H 2 O 1,115X0.3790 = 423 " 23,088 - 23,088X60 = 1,385,280 oz. Cal. = 86,580 Ib. Cal. Heat of Oxidation: C to CO 2 140.7X8100 = 1,139,670 Calories C to CO 538.8X2430 = 1,309,280 Si to SiO 2 203.2X7000 = 1,422,400 Mn to MnO 197.9X 1653 = 327,130 Fe to FeO 228.6 X 1173 = 268,150 Fe to Fe 2 O 3 25.3X1746 = 44,170 4,510,800 Heat of Formation of Slag: The 197.9 pounds of manganese oxidized forms 255.5 of MnO. Hence the weight of the slag is 255.5 -T- 0.0888 = 2877 pounds. The slag, therefore, con- tains also 2877X0.6356 = 1829 pounds of SiO 2 , of which 203.2 X 60/28 = 435 pounds came from the silicon oxidized, and 1794 pounds from the lining. The lining will also have lost the A1 2 3 , CaO and MgO in the slag, equal to 2877 X 0.0427 = 123 pounds. The total iron going into the slag, 253.9 pounds, is equivalent to 326.4 pounds of FeO. The heat of formation of the slag will therefore be: FeO 326.4X124 = 40,474 Calories. MnO 255.5X 76 = 19,418 Total 59,892 Heat in Converter at Finishing: This will be the same as at starting, less the heat in 1794 + 123 = 1917 pounds of lining, which was corroded and entered the slag. Assuming this to have been on the inner surface, at an average temperature of 1500, the heat in it, using the mean specific heat of silica, will have been 1917X0.2988X1500 = 851,200 Calories. 366 METALLURGICAL CALCULATIONS. And the heat in the converter body at the finish: 8,034,970851,200 = 7,183,770 pound Calories Heat in Finished Steel: Taking its temperature as 1650, with 350 Calories per unit, we have 24,665X350 = 8,632,750 Calories. Heat in the Slag: 2877X550 = 1,582,350 Heat in Escaping Gases: These have already been calcu- lated as consisting of Nitrogen 58,932 cubic feet. Hydrogen 1,115 " " Carbon monoxide 15,964 " Carbon dioxide 4,169 " The first three have the same heat capacity per cubic foot, so assuming their temperature 1550: N 2 , H 2 , CO 76,011X534.5 = 40,627,900 oz. Cal. CO 2 4,169X947.1 = 3,948,250 44,576,150 = 2,786,000 Ib. Cal. Absorbed in Decomposing Moisture: The 1,115 cubic feet of hydrogen in the gases represent so much steam or water vapor decomposed. Since 1 cubic foot = 0.09 ounces, the heat absorbed is 1,115X0.09X29,040 = 2,914,160 oz. Cal. = 182,130 Ib. Cal. Heat Conducted to the Air: Assuming the conditions worked out in the illustration under this heading, this item would be approximately, for 9 min. 10 sees., in Ib. Cal. 1,714X2.204X9.167 = 34,630 Ib. Cal. Heat Lost by Radiation: Making similar assumption, we have 1,249X2.204X9.167 = 25,240 Ib. Cal. THE BESSEMER PROCESS. 367 Recapitulation. Lb. Cal. Heat in converter body at starting 8,034,970 melted pig-iron 6,750,500 spiegeleisen 750,000 blast 86,580 Heat of oxidation 4,510,800 " formation of slag 59,890 Total on hand and developed 20,192,740 Heat in converter body at finish 7,183,770 finished steel 8,632,750 slag 1,582,350 escaping gases 2,786,000 Heat absorbed in decomposing moisture 182,130 Heat conducted to the air 34,630 Heat lost by radiation 25,240 Total accounted for 20,426,870 Another way of expressing this balance is to itemize the avenues of heat evolution and utilization, as follows: Lb. Cal. Received from converter body 851,200 Received by oxidation 4,510,800 Received by formation of slag 59,890 Total 5,421,890 Lb. Cal. Used, excess of heat in gases over blast 2,699,420 Used, excess of heat in steel and slag over pig iron and spiegel 2,714,600 Decomposition of moisture 182,130 Radiation and conduction . 59,870 Total 5,656,020 CHAPTER IX. THE TEMPERATURE INCREMENT IN THE BESSEMER CONVERTER. In the preceding chapter, we have studied the generation of heat in the Bessemer converter, arid its distribution. We saw in that analysis that in a typical operation, nearly one-half of the heat generated during the blow is carried out by the hot gases, about half is represented in the increased temperature of the contents of the converter, while only about 5 per cent, is lost by radiation, etc. In the present paper we wish to analyze still further this question of increased temperature, which is so vitally necessary for the proper working of the process, and to calculate the relative efficiency of the various substances oxidized in causing this rise of temperature. While at no time in the Bessemer operation is only one sub- stance being oxidized, yet we can get the best basis for our computations by assuming a charge in operation and only one substance oxidized at a time. Whatever substance is in ques- tion, let us assume one kilogram burnt in a given short period of time, generating the heat of its combustion. With the bath at a given temperature, the air, at say 100 C., comes in contact with it, bearing the necessary oxygen. If nothing was oxidized, the oxygen and nitrogen would simply be heated to the temperature of the bath, and" pass on and out, while the bath would be meanwhile losing heat also by radiation. It is evident then, that unless at least as much heat as the sum of these two items is generated, the bath will cool, and that only the excess of heat above this requirement is available for increasing the temperature ot the bath and resulting gases. The proper procedure for us will therefore be to calculate, in each case, the chilling effect of the air entering, subtract this from the heat generated, and the residue is net heat avail- able for raising the temperature of the contents of the con- 368 THE BESSEMER CONVERTER. 369 verier and the gases, and supplying radiation losses. The latter are proportional to the time, and therefore to the amount of air used, assuming blast constant. SILICON. This is burnt out in the first part of the process, during which the temperature begins low and ends high. We will there- fore assume two temperatures, and calculate the thermal increment at each. We will take 1250 and 1600. The net heat is absorbed by the bath, slag and nitrogen. Oxygen necessary to burn one kilogram of silicon : IX 32^28= 1.143 kg. Nitrogen accompanying this oxygen = 3.810 Weight of air needed = 4.953 " Volume of air = 4.953-^-1.293 = 3.831 m 3 Specific heat, 100 to 1250, per m 3 = 0.3395 Specific heat, 100 to 1600, per m 3 = 0.3489 Chilling effect of air at 100, bath at 1250 = 3.831X0.3395X1150 = 1496 Cal. Shilling effect of air at 100, bath at 1600 = 3.831 X 0.3489 X 1500 = 2043 Cal. Heat generated per kilogram of silicon = 7000 Cal. This is, however, for cold oxygen and cold silicon burning to cold solid silica. Under the conditions prevailing we have melted silicon at the temperature of the bath, oxidized by hot oxygen to hot silica, giving a slightly different heat of com- bination, calculated as follows: Heat in melted silicon at 1250 = 480 Cal. " " oxygen required, at 1250 = 334 " " " silica, at 1250 = 750 " Heat of oxidation at 1250 = 7,000 + 480 + 334750 = 7,064 " The difference from 7,000 is so small as to be within the possible error of the 7,000 itself, and we can therefore make the calculations with all the accuracy they allow, taking the ordinary heats of oxidation from the tables. One factor of heat generation has, however, not been men- tioned, viz.: the heat of combination of silica with oxides of 370 METALLURGICAL CALCULATIONS. iron and manganese to form slag. This is 148 Calories per kg. of silica when forming iron silicate, and 90 when forming man- ganese silicate, which quantities would become 317 and 193 Calories respectively when calculated per kg. of silicon oxi- dizing. The question arises, however, whether it is fair to credit all this to the silica, because this generation of heat by slag formation is really a mutual affair, chargeable to the credit of both silica and the other oxides; we should, therefore, not charge it all up to the credit of silica formation, and we do not know what part to charge to the credit of silica if we do not charge it all. In this dilemma it may be well to remem- ber that silicon is probably oxidized before the iron and man- ganese, and that the heat of formation of the slag is therefore more properly considered as being generated afterwards, and therefore may be practically credited entirely to the oxidation of iron and manganese. Resume for oxidation of 1 kilo, of silicon: Cal. Heat generated = 7,000 Chilling effect of the blast, 100 to 1250 = 1,496 Chilling effect of the blast, 100 to 1600 = 2,043 Available heat, bath at 1250 = 5,504 Available heat, bath at 1600 = 4,957 If we assume a radiation loss proportional to the length of the blow, i. e., proportional to the air blown in, we can find out from average blows that this amounts to about 50 Calories per cubic meter of blast used. The radiation loss during com- bustion of 1 kilogram of silicon would therefore be Cal. 3.831X50 = 192 Leaving net available heat at 1250 = 5,312 At 1600 = 4,765 The above quantity of heat is expended in raising the tem- perature of 99 kg. of bath, 2.143 kg. of silica, and 3.810 kg. of nitrogen gas, from their initial temperature. At the tempera- tures of 1250 and 1600, respectively, the heat capacity of these products of the operation will be, per 1 C. rise: THE BESSEMER CONVERTER. 371 Specific Heat Heat Capacity Products. at 1250 at 1600 at 1250 at 1600 Bath, 99 kg 25 0.25 24.8 24.8 SiO 2 , 2.14 kg 0.37 0.43 0.8 0.9 N 2 , 3.02m 3 .. 0.37 0.39 1.1 1.2 Totals. 26.7 26.9 Theoretical rise of temperature : 5,312 -v- 26.7 = 199 (bath at 1250) 4,765^26.9 = 177 (bath at 1600) Average rise, per average 1% of -silicon =188 C. MANGANESE. As high as 4 per cent, of manganese may be oxidized during the blow, and therefore this heat of combustion is sometimes important. We will calculate the net heat available for rais- ing temperature and the rise of temperature per 1 per cent, of manganese oxdized, i. e., for 1 kilo, of manganese per 100 kilos, of bath. Oxygen necessary 1 X 16/55 = 0.291 kg. Nitrogen accompanying this = 0.970 Weight of air used = 1.261 " Volume of air used = 0.975 m 3 Chilling effect of air at 100 on bath at 1250 = 0.975X0.3395X1150 = 381 Cal. Heat generated per kg. of manganese = 1653 " Heat of formation of MnO.SiO 2 1.291kg. MnOX 76 = 98 " Total heat developed = 1751 " Heat available = 1751381 = 1370 " Radiation loss = 0.975X50 = 49 " Net available heat = 1321 " Heat capacity of 99 kg. of bath per 1 = 24.8 " Heat capacity of 2.4 kg. of slag = 0.7 " Heat capacity of 0.8 m 3 nitrogen = 0.3 " Heat capacity of products per 1 = 25.8 " 372 METALLURGICAL CALCULATIONS. Theoretical rise of temperature : 1321^25.8 = 51 C. This is, in round numbers, one-fourth the efficiency of silicon. IRON. While it is not desired to oxidize iron, and while it is rela- tively less oxidizable than silicon or manganese, yet some is always oxidized because of the great excess of iron present in the converter. The amount of iron thus lost is variable, and some of it, towards the end of the blow, may be oxidized to Fe 2 3 instead of FeO, the larger part, however, oxidizes to FeO. We will make the calculations for both oxides, per kilo- gram of iron. Formation of FeO. Oxygen necessary 1X16/56 = 0.286 kg. Nitrogen accompanying this = 0.953 " Weight of air used = 1.239 " Volume of air used = 0.958 m 3 Chilling effect of air at 100 on bath at 1250 = 0.958X0.3395X1150 = 374 Cal. Chilling effect of air at 100 on bath at 1600 = 0.958 X. 3489X1500 = 501 " Heat generated per kg. of iron =1,173 " Heat of formation of FeO.SiO 2 = 1.286kg. FeO X 124 = 159 " Total heat developed = 1,332 " Net heat available at 1250 = 1332 374 = 958 " Net heat available at 1600 = 1332 501 = 821 " Radiation losses = 0.958X50 = 48 " Net available heat at 1250 = 910 " Net available heat at 1600 = 773 " Heat capacity of 99 kg. of bath per 1 = 24.8 " Heat capacity of 2.4 kg. of slag per 1 = 0.7 " Heat capacity of 0.8 m 3 of nitrogen = 0.3 " Heat capacity of products per 1 = 25.8 * Theoretical rise of temperature: 910^25.8 = 36 C. 773 H- 25.8 = 30 C. This is only about one-sixth as efficient as silicon. THE BESSEMER CONVERTER. 373 Formation of Fe 2 0*. Weight of air used = 1.859 kg. Volume of air used = 1.438 m 3 Chilling effect of air at 100 on bath at 1600 = 1.438X0.3489X1500 = 753 Cal. Total heat developed by oxidation = 1746 " Heat of formation of slag = 159 " Total heat developed = 1905 " Heat available = 1905753 = 1152 " Radiation losses = 1.438X50 = 72 " Net heat available = 1080 " Heat capacity of products = 26 " Theoretical rise of temperature : 1080^26 = 42 C. TITANIUM. While titanium is an unusual constituent of pig iron, yet it is conceivable that titaniferous pig iron might be made and blown to steel. If so, the following calculation, based on a quite re- cently determined value for the heat of oxidation of titanium, will show that the titanium is a valuable heat producing sub- stance, being, in fact, weight for weight three fourths as efficient as silicon. Oxygen needed 1x32/48 = 0.667 kg. Nitrogen accompanying this = 2.222 " Air used = 2.888 " Volume of air needed = 2.250 m* Chilling effect of air at 100 on bath at 1250 = 2.250X0.3395X1150 = 878 Cal. Heat generated per kg. of Ti = 4542 " Heat of formation of slag unknown Net heat available = 4542878 = 3664 " Deducting 144 for radiation losses = 3520 " Heat capacity of products per 1 C. = 26.5 " Theoretical rise of temperature : 3520-5-26.5 = 133 C. 374 METALLURGICAL CALCULATIONS. ALUMINIUM. This metal is also rarely found in pig iron, yet when present it would be a powerful heat producer, as the following calcu- lations show: Oxygen needed 1x48/54 = 0.889 kg. Nitrogen = 2.963 " Air =3.852 " Volume of air = 2.964 m 3 Chilling effect of air at 100 on bath at 1250 = 2.964X0.3395X1150 = 1157 Cal. Heat generated per kg. of Al = 7272 " Heat of formation of slag uncertain Heat available = 72721157 = 6115 " Deducting for radiation losses = 5967 " Calorific capacity of products per 1 C. = 26.6 " Theoretical rise of temperature : 5967 -r 26.6 = 224 C. If a blow was running cold, and ferro-silicon was not on hand to add in order to increase its temperature, ferro-alum- inium, or aluminium itself, would be a good substitute in the emergency. NICKEL. It is hardly probable that nickeliferous pig iron would be blown to steel, because of the waste of valuable nickel in the slag; yet if 1 per cent, of nickel were thus oxidized, calculations similar to the preceding would show a net rise in temperature of the contents of the bath of about 33 C. CHROMIUM. Quite recently some chromiferous pig iron has been blown in the Bessemer converter, and those in charge were hampered by the lack of data as to how chromium would behave during the blow and its heat value to the converter, Technical litera- ture will probably soon contain an account of the practice which has been developed at the Sparrow's Point works of the Maryland Steel Company. No thermo-chemist has, as yet, determined the heat of formation of chromium slag. From THE BESSEMER CONVERTER. 375 what we know of the chemical reactions of chromium it is likely that this heat is considerable. Neglecting the heat of formation of slag, we have the following approximation to its heating efficiency, assuming it to be oxidized when the bath is near to its maximum temperature: Oxygen needed 1x48/104 = 0.462 kg. Nitrogen entering = 1.540 " Air used = 2.002 " Volume of air = 1.548 tn 3 Chilling effect of air at 100 on bath at 1600 = 1.548X0.3489X1500 = 810 Cal. Heat of oxidation = 2,344 " Net heat = 2344810 = 1,534 " Deducting for radiation losses = 1,457 " Calorific capacity of products per 1 = 26.1 " Theoretical rise of temperature : 1457-26.1 =56C. CARBON. This element commences to be oxidized in large amount only towards the middle of the blow, when the temperature of the bath is high, because of the previous oxidation of silicon. It will be about right, therefore, to estimate the bath at an average temperature of 1600 during the elimination of carbon. The product is mostly CO, but partly CO 2 . We will, therefore, calculate for each of these possible products separately. The net heat available, after allowing for average radiation losses, is used to increase the temperature of the products, i. e., of the bath, the nitrogen and the CO or CO 2 . Oxidation to CO 2 . Oxygen required =1x32/12 = 2.667 kg. Nitrogen accompanying = 8.889 " Air used = 11.556 " Volume of air = 8.937 m 3 Chilling effect of air at 100 on bath at 1250 = 8,937X0.3395X1150 = 3,489 Cal. Heat of oxidation = 8,100 " 376 METALLURGICAL CALCULATIONS. Heat available = 8,1003,489 = 4,611 Cal. Radiation losses = 8.937X50 = 447 " Net heat available = 4,164 " Heat capacity of 99 kg. bath = 99X0.25= 24.80 " Heat capacity of 7.05 m 3 N 2 = 7.05X0.37 = 2.61 " Heat capacity of 1.90m 3 CO 2 = 1.90X0.88= 1.70 " Heat capacity of products, per 1 C. = 29.11 " Theoretical rise of temperature : 4,164-*- 29.11 =143C. Since carbon burns to CO 2 principally at the beginning of the blow, while the bath is cold, we see that carbon thus con- sumed is about three-quarters as efficient as an equal weight of silicon in raising the temperature of the bath. Oxidation to CO. Oxygen needed 1 X 16/12 = 1.333 kg. Nitrogen accompanying = 4.444 " Air used = 5.777 " Volume of air = 4.469 m 3 Chilling effect of air at 100 on bath at 1600 = 4.469X0.3489X1500 = 2,339 Cal. Heat of oxidation = 2,430 " Heat available = 2,4302,339 91 " Radiation losses = 4.469X50 = 233 " Net heat available = 91233 =142 " Heat capacity of 99 kg. of bath = 99X0.25 = 24.8 " Heat capacity of 3.5 m 3 of N 2 \ ,. . ft o Q _ 9 * Heat capacity of 1.9 m 3 of CO / M*fl Heat capacity of products = 26.9 " Theoretical rise of temperature : -142 -r- 26.9 = 5C. The result is, therefore, that when carbon burns, as it mostly does, to CO, and the temperature of the bath is high, there is practically no further rise of temperature, for the heat of oxi- dation is barely sufficient to counteract the chilling effect of the air and to supply radiation and conduction losses. In the above calculations, no allowance was made for heat THE BESSEMER CONVERTER. 377 required to separate carbon from its combination with iron, or for the variation in the heat of combination of carbon with oxygen from the combination heats at ordinary temperatures. The former is not known, or perhaps is very nearly zero. The heat of oxidation of liquid carbon at 1250 to CO 2 or at 1600 to CO is calculated as follows: Oxidation 1 kg. C to CO 2 at = 8,100 Cal Heat to raise 1 kg. C to 1250 = 505 Cal. Heat to liquefy 1 kg. at 1250 = 129 " Heat to raise 2.67 kg. O 2 to 1250 = 779 " Heat to raise reacting substance to 1250 = 1,413 Heat in 3.67 kg. CO 2 at 1250 = 1,493 " Heat of reaction at 1250 (8,100+1,413-1,493) = 8,020 " For production of CO, at 1600, the correction is larger, as is seen from the following : Oxidation 1 kg. C to CO at = 2,430 Cal. Heat to raise 1 kg. to 1600 = 680 Cal. Heat to liquefy 1 kg. C at 1600 = 156 " Heat to raise 1.33 kg. O 2 to 1600 = 554 " Heat to raise reacting substances to 1600 = 1,390 " Heat in 2.33 kg. of CO at 1600 = 1,104 " Heat of reaction at 1600 (2,430 + 1,3901,104) = 2,716 " The use of this corrected value makes the oxidation of C to CO give a small net heat development, with consequent slight rise of temperature, instead of the slight cooling effect before calculated. The conditions are so nearly even, however, that a slight increase of temperature or slowing up of the blow would wipe out the heat excess. PHOSPHORUS. This is the last important element to be considered, and is always eliminated after the carbon, at the maximum bath tem- perature, which we will assume, for calculation, at 1600. The heat generated, at ordinary temperatures, is 5892 Calories per kilogram of solid phosphorus. Per kilogram of liquid phos- phorus it would be only 5 Calories more, or 5897 Calories. For the reaction at 1600 we would have a different value, probably 378 METALLURGICAL CALCULATIONS. some 500 Calories more, but the necessary data concerning the specific heats of P and P 2 O 5 are not known, and we must omit this calculation. The heat of combination of iron and phos- phorus is also a doubtful quantity. Ponthiere places it as high as 1,397 Calories per kilogram of phosphorus, but this ap- pears altogether improbable since another experimenter could obtain no heat of combination at all. As concluded in another place, I advise for the present omitting this questionable quantity. The phosphorus pent-oxide forms 3CaO . P 2 O 5 with the lime added, but since there is always more lime present than cor- responds to these proportions (3CaO : P 2 5 :: 168 : 142), the calculation of heat of formation of the slag must be based on the amount of P 2 O 5 formed (1123 Calories per kilogram of P 2 5 ). This amounts to a considerable item. On the other hand, the lime needed for slag is put in, usually preheated, for the sole purpose of combining with the P 2 O 5 . It seems, there- fore, only right to charge the phosphorus with the heat re- quired to raise this lime to the temperature of the bath. The lime added averages three times the weight of P 2 O 5 formed, and is preheated usually to about 600. Assuming these con- ditions, the following calculations can be made per kilogram of phosphorus oxidized : Oxygen required = 1.29 kg. Nitrogen accompanying = 4.30 Air used = 5.59 " Volume of air = 4.32m 3 Heat of formation of slag, 2.29 kg. P 2 O 5 X1123 = 2572 Cal. Heat of oxidation of phosphorus = 5897 Total heat developed = 8469 " Chilling effect of air at 100 on bath at 1600 = 4.32X0.3489X1500 = 2261 Chilling effect of lime (600 to 1600) (2.29 X 3) X 0.328X1000 = 2253 Chilling effect of blast and lime = 4514 Heat available = 84694514 = 3955 Radiation losses = 4.32X50 = 216 Net heat available = 3739 " THE BESSEMER CONVERTER. 379 Heat capacity 99 kg. of bath = 99 X 0.25 = 24.8 kg. Heat capacity 3.4m 3 of N 2 = 3.4X0.39 = 1.3 " Heat capacity 6.9 kg. of slag = 6.9X0.3 = 2.1 " Heat capacity of products, per 1 = 28.2 " Theoretical rise of temperature : 3739^-28.2 = 133 C. If the lime were added cold, its cooling effect would be 883 Calories greater, the net heat available would be 883 Calories less, and the calculated rise of temperature 31 less, or 102 C. Using the preheated lime we can regard phosphorus as being practically two-thirds as efficient, weight for weight, as silicon; with cold lime, about one-half as efficient. RESUME. Heat effect of oxidizing 1 kilogram of element. or f u J ^ a || a & 1 1 aj tfc^ ^^ ^ +4 J 1 'o $ 11 * fc, (O Qq Silicon 7 000 7 000 1 688 Manganese ... , . . . . 1 653 98 1 751 430 Iron (to FeO) 1,173 159 1,332 422 Iron (to Fe 2 O 3 ) . . 1,746 159 1 905 825 Titanium . .4,542 4,542 1 022 Aluminium . . . 7 272 7 272 1 305 Nickel 1 051 159 1 210 378 Chromium 2 344 2 344 887 Carbon (to CO 2 ) . . . 8.100 8 100 3 936 Carbon (to CO) 2 430 2 430 2 572 Phosphorus 5 897 2 572 8 469 ( 2 477 2,253* * Chilling effect of lime added, preheated to 600 a 'lt.* It? I ! il sSf If ^ O -Si 1> 5,312 188 1,321 51 910 33 1,080 42 3,520 133 5,967 224 832 33 1,457 56 4,164 143 -142 5 3,739 133 It must be observed that the above table is for comparison only, it cannot be used for an actual case, such as when 1 per cent, of silicon, 3 of iron, 4 of carbon and 2 of phosphorus are 380 METALLURGICAL CALCULATIONS. oxidized. In such a case, the rise in temperature would be only very roughly: - From silicon 1 X 188 = 188 From iron 3 X 33 = 99 From carbon - say = From phosphorus 2X 133 = 266 Total = 553 C. It is to be recommended that in each specific case the calcu- lation be made for the specific conditions obtaining, such as temperature of the metal at starting, temperature of the blast, time of the blow (as far as this affects radiation and conduc- tion losses), proportion of carbon burned to CO 2 , free oxygen in the gases, moisture in the blast, temperature and quantity of lime added, corrosion of lining. When all these items and conditions are taken into account there will be room for only small discrepancy between the calculated and the observed rise of temperature. The chief items needing experimental re- search at present are: The specific heat of the melted bath, the specific heat of the slag, the heat of combination of various elements comprising the bath, the heat of formation of the slag, and the heat of oxidation of some of the rarer elements. Such establishments as the Carnegie Institution could not do the cause of metallurgy better service than to subsidize metallur- gical laboratories for the determination of such data. CHAPTER X. THE OPEN-HEARTH FURNACE. By the above title we mean to designate not only the re- generative gas furnaces for making steel but also those for reheating purposes; in other words, regenerative or recupera- tive reverberatory furnaces. Prominent among these is the Siemens-Martin furnace, with complete gas and air preheating regenerative chambers. In all these furnaces the charge is heated, or kept hot, partly by direct contact with the gaseous products of combustion and partly by radiation from the flame and the sides and roof of the furnace. It was the Siemens brothers who first insisted on the relatively great importance of the radiation principle, in distinction to the direct impinge- ment of the flame on the material, pointing out that a luminous flame radiates from all parts of its volume, while a hot, solid body radiates only from its surface, and direct impingement interferes with the development of perfect combustion and communicates heat relatively slowly at best. GAS PRODUCERS. Gas producers are the usual adjunct for open-hearth fur- naces, excepting where natural gas or blast furnace gas is available. They may be placed far from the furnace, when they deliver cool gas to the regenerators, or close to the fur- nace, delivering comparatively hot gas to the regenerators, or even be made part of the furnace itself, delivering their hot gas immediately to the ports of the furnace. The latter is un- doubtedly the most economical arrangement where practicable. The following generalizations concerning the relations of the gas producer to the open-hearth furnace may be made: Pro- ducers furnish 4,300 to 4,600 cubic meters of gas per metric ton of coal used (150,000 to 160,000 cubic feet per short ton) ; the gas produced runs 3 to 8 per cent. CO 2 , 5 to 20 per cent. H 2 . 20 to 30 per cent. CO, and 50 to 60 per cent. N 2 ; its calorific 381 382 METALLURGICAL CALCULATIONS. power is 750 to 1000 Calories per cubic meter (47 to 63-pound Calories, or 85 to 115 B. T. U. per cubic foot); its calorific power represents 60 to 90 per cent, of the calorific power of the fuel used; in steel-making processes, the keeping of the furnace up to proper heat requires the gasifying of 25 to 35 kilograms (50 to 80 pounds) of coal- per hour, in the producers, for each ton of metal capacity of the furnace; good producers gasify 60 to 65 kilograms of coal per hour per each square meter of gas-producing area (10 to 15 pounds per hour per each square foot); a furnace therefore requires some 0.4 to 0.6 square meter (4 to 6.5 square feet) of gas-producing air in the producers for each ton of metal capacity of the furnace. FLUES TO FURNACE. In conducting the gases to the furnace, the flues or conduits should be of ample size. If too small the gas must pass through them with high velocity, requiring considerable draft to give them this velocity, which the chimney, or blower may or may not be capable of furnishing. Producers are almost always worked by a steam blower, furnishing mixed air and steam and a plenum of pressure in the upper part of the producer, which suffices to send the gas through the conduits under a slight pressure, and thus avoids any sucking in of air through crevices in the conduits. With too sm'all conduits the result- ing friction and high velocity required may give the blower more work than it can do, and thus entail demands for draft upon the furnace stack. A reasonable rule is to give the flues such cross-sectional area that the hot gas which must pass through them shall have a velocity between 2 and 3 meters per second. REGENERATORS. The dimensions of the regenerators are of the first import- ance to the working of the furnace. They should have suf- ficient length in the direction the gas currents are passing, so that the gases may be properly cooled or heated ; they should have sufficient cross-sectional area of free space, so that the velocity of the gases through them is not too great; they must have sufficient thermal capacity, so that they can absorb the requisite quantity of heat. Length. From 4 to 6 meters (13 to 20 feet) is a suitable THE OPEN-HEARTH FURNACE. 383 length in the direction of the gas currents. This permits the hot products to become properly cooled before going to the chimney, and the gas or air to be properly heated before en- tering the furnace. The shorter length may be used when the regenerator is of large cross-sectional area, with slow velocity of gas currents through the free spaces; the longer when the regenerator is rather restricted in cross-section and the gas currents have somewhat high velocity. Cross-Section. The free-space sectional area should be such that the gases where hottest should not have a calculated velocity of over 3 meters (10 feet) per second, and if calculated for 2 meters (6.5 feet) will give much better results as regards transmission of heat to the checker work. In this manner, knowing how much gas must go through the regenerator and what its maximum temperature will probably be, the cross- section area of free passage space can be calculated. The rela- tion of this to the cross-section of the entire stove must next be considered, and this is entirely a question of how the checker work is built up. If the bricks are stacked close together the free space may be reduced to as much as one-half the total; as ordinarily stacked it may be 60 to 80 per cent, of the total; if perforated bricks are used, as in blast furnace firebrick- stoves, the area of free space averages one-half the total ; with ordinary bricks the average is 70 per cent. This is a question which is very variously worked out in different furnaces, and to which not as much scientific thought has been given as should be. The thickness of the bricks influences greatly the rela- tive amount of free space and filled space, and the rate at which the generator heats up or cools off. In a regenerator of given length and cross-section closer packing of the bricks gives more heat absorbing surface, increases the velocity of the gases and diminishes the cross-sectional area of each passage and of the sum of all the passages; some of these factors in- crease the efficiency of the regenerator, others tend to decrease it, and there are, therefore, several independent variables to be considered in finding the best arrangement for highest effi- ciency. A numerical solution is indeed a possibility, but is too involved for an elementary presentation of the subject. Relative Sizes. The relative sizes of gas and air regenera- tors is a question of importance which admits of easy solu- 384 METALLURGICAL CALCULATIONS. tion by calculation. So far we have treated the pair of re- generators together, and discussed the sum of their cross-sec- tions as deduced from the volume of products passing through at an assumed maximum temperature and allowable velocity. The regenerators at one end of a furnace are, however, usually divided into a pair or set, one for heating gas and the other for heating air. This is not usual where natural gas is used because of the deposition of soot in the regenerator by the latter when it is heated, but nine out of ten open-hearth fur- naces preheat their gas as well as the air. The heating capacity of the regenerators should be divided in proportion to the calorific capacities of the gas and air simultaneously heated. The problem is therefore to find the heat capacity per degree of the gas and air used, or, more exactly, the total heat capacity of each of these between the temperature at which they enter the regenerators and that at which it is desired that they should enter the furnace. Problem 69. An open-hearth furnace uses producer gas containing, by volume, at it reaches the regenerators: CO 26.97 per cent. CO 2 4.37 CH 4 0.33 H 2 13.00 NH 3 0.21 H 2 S 0.10 N 2 54.01 Air . 1.03 Each cubic meter, measured at 20 C. and 720 m. m. baro- metric pressure, is accompanied by 73.22 grams of moisture, as determined by drawing through a drying tube and weighing the moisture. The air used is at 20 C., 720 m.m. barometer, and three-quarters saturated with moisture. A maximum of 10 per cent, more air is used than is theoretically necessary to com- pletely burn the gas (assuming NH 3 to burn to H 2 O and NO 2 , and allowing for the air already present in the gas). The gas and air may be both assumed as coming to the regenerators at 20 C., and to be heated in the regenerators to 1200 C. THE OPEN-HEARTH FURNACE. 385 Required. (1) The relative volumes of gas and air passing through the gas and air regenerators. (2) The total amounts of heat necessary to be furnished to each, per cubic meter of gas used. (3) The relative sizes of the two regenerators. Solution. (1) Taking 1 cubic meter of the dry gas, as rep- resented by the analysis, the combustion of its combustible in- gredients is represented by the equations: 2CO +0 2 = 2CO 2 CH 4 + 2O 2 = CO 2 + 2H 2 O 2H 2 +O 2 =2H 2 O 4NH 3 + 7O 2 =6H 2 O + 4NO 2 2H 2 S +3O 2 = 2H 2 O + 2SO 2 And since molecules represent volumes, each unit volume of CO, CH 4 , H 2 , NH 3 and H 2 S is seen to require respectively 0.5, 2, 0.5, 1.75, or 1.5 volumes of oxygen. The 1 cubic meter of dry gas therefore requires oxygen as follows: CO 0.2697X0.5 = 0.1349m 3 CH 4 0.0437X2.0 = 0.0874 " H 2 0.1300X0.5 = 0.0650 " NH 3 0.0021X1.75 = 0.0037 " H 2 S 0.0010X1.5 = 0.0015 " Total = 0.2925 ' Air needed = 0.2925-^0.208 = 1.4062m 3 Add 10 per cent, excess = 1.5468 " Air present in gas = 0.0103 " Air to be supplied = 1.5365 " Each cubic meter of dry gas, at any given conditions of tem- perature and pressure, would require 1.5365 cubic meters of dry air at the same conditions of temperature and pressure. This, however, is not exactly the relation required, for the reason that the gas is accompanied by considerable moisture, which in reality adds to its volume, while the air is also moist, add- ing to its volume. Two corrections must therefore be ap- plied; first, to calculate the volume of the moisture accom- panying 1 cubic meter of (assumed) dried gas; the second, to calculate the volume of moisture accompanying 1.5365 cubic 386 METALLURGICAL CALCULATIONS. meters of (assumed) dry air. Assuming our gas and moist air both at 20 and 720m.m. pressure, the volume of moisture accompanying 1 cubic meter of (assumed) dry gas is the vol- ume of 73.22 grams of moisture at these conditions, which is 73.22-hlOOO-HO.Sl X 2y !_t 2 X = 0.1024 m The air being 0.75 saturated with moisture at 20, the tension of this moisture will be 17.4X0.75 = 13 m.m. The air proper is therefore under 720 13 = 707 m.m. tension instead of 720 m.m. and the volume of the moist air containing 1.5365 cubic meters of (assumed) dry gas is therefore: 1.5365X^5 1.5648 cubic meters. The relative volumes of actual (moist) gas and actual (moist) air used are therefore : 1.1024:1.5684 = 1.0000:1.419 (1) (2) To calculate the heat necessary to raise gas and air from 20 to 1200 per cubic meter of gas used, the best preliminary is to calculate the composition of the gas ; including its mois- ture. Since 1 cubic meter of (assumed) dry gas is accom- panied by 0.1024 cubic meter of water vapor, the sum being 1.1024 cubic meters, we can calculate the real percentage com- position to be: CO ............................. 24,47 per cent. CO 2 ............................. 3.96 CH 4 ....................... ..... 0.30 H 2 .............................. 11.7 NH 3 ..................... ....... 0.19 H 2 S ............................ 0.09 N 2 .............................. 48.99 Air ............................. 0.93 H 2 ............................ 9.29 Of the above quantities the (assumed) dry gas in 1 cubic meter of (actual) moist gas is 0.9071 c.m. The air used for its com- bustion will therefore be: THE OPEN -HEARTH FURNACES 387 0.9071X1.5365 = 1.3938 tn 3 dry air. 0.9071X1.5648 = 1.4080 m 3 moist air. 1.40801.3938 = 0.0142 m 3 of moisture. The heat required by the 1 cubic meter of (actual) moist gas is found as follows: Volume XMean Specific Heat 20 1200 CO 0.2447 H 2 0.1179 ' X 0.3359 = 0.2895 Calories. N 2 0.4899 Air 0.0093 CO* 0.0396 X 0.6384 = 0.0253 " H 2 O 0.0929 X 0.5230 = 0.0486 " CH 4 0.0030 X 0.6484 = 0.0019 " NH 3 0.0019 X 0.5752 = 0.0011 " H 2 S 0.0009 X 0.5230 = 0.0005 " Mean cal. capacity per 1 = 0.3669 " Total calorific capacity 20 1200 0.3669X1180 = 432.9 Calories. The calorific capacity of the moist air simultaneously heated through the same range will be. Air 1.3938X0.3359 = 0.4682 Calories. H 2 O 0.0142X0.5230 = 0.0074 " Sum = 0.4756 Total calorific capacity 20 1200 0.4756X1180 = 561.2 Calories. The air regenerator should therefore have 561.2-5-432.9 = 1.30 times the heating power or cross-section of the gas re- generator; i. e., 30 per cent. more. Or the combined capacity of the pair of regenerators should be divided so as to give 57 per cent, to the air regenerator and 43 per cent, to the gas regenerator. In ordinary practice it is usual to allow about 60 388 METALLURGICAL CALCULATIONS. and 40 per cent, respectively; it is better to calculate ahead for the specific case in hand, if the composition of the gas to be used is known. VALVES AND PORTS. No very exact rule can be given as to the size of the gas and air valves, or those leading the products to the chimney. If made too large they are cumbersome to operate and apt to warp; if made too small they give undue obstruction to the flow of gas. A general rule is to calculate the free opening, such as to give the gases passing through a velocity between 3 and 5 meters (10 and 16 feet) per second, allowing, of course, for the average temperature of the gas or air products of com- bustion passing through them. It may be remarked that while water-seal valves are very convenient, the water is evap- orated where in contact with gas or air, and diminishes the heating efficiency of the furnace, the use of a non- volatile oil or a fine sand would appear perferable to water. The ports are a very important part of the furnace, and may be designed in many different styles for various ways in which a furnace is to be worked. Their cross-section, however, can be calculated when we know the volume of gas or air leaving the regenerators and their temperature, or the volume of the products of combustion entering the regenerators and their temperature. They should be so designed that the velocity of the gases through them is not over 20 meters per second, while 10 meters per second is a better velocity to use. A long fur- nace can admit of higher velocities at the ports than a short one; but in any case the higher the velocity the farther com- plete combustion will occur from the ports, and if the velocity is too high for the length of the furnace combustion may even be continued in the opposite regenerators and less than the maximum occur in the furnace. This is a condition to be scrupulously avoided if possible. Problem 70. Producer gas of the following composition: CO 24.47 per cent. NH 3 0.19 per cent. CO 2 3.96 " N 2 48.99 CH 4 0.30 " Air 0.93 H 2 11.79 " H 2 9.29 H 2 S.. . 0.09 THE OPEN-HEARTH FURNACE. 389 is burned with 1.408 times its volume of moist air (see Prob- lem 69). The furnace treats 50 metric tons of steel in 12 hours, using 17.5 tons of coal in the producers, from which 15 tons of carbon pass into the gas. The gas and air pass out of the regenerators at 1200, and the products of combustion (as- sumed complete) pass into the opposite regenerators at 1400. Assume a maximum velocity of the hot gas and air as 10 meters per second, as they pass through the ports. Required. (1) The volume of gas and air at 20 C. and 720 m.m. barometer used by the furnace per second. (2) The areas of the gas and air ports. (3) The velocity of the products entering the opposite ports. Solution. (1) The carbon in 1 cubic meter of the gas at standard conditions is CO 0.2447 CO 2 0.0396 CH 4 0.0030 0.2873X0.54 = 0.1551kg. Gas used in 12 hours (standard conditions) : 15,000^-0.1551 = 96,710m 3 Per second = 2.24 m 3 Gas used at 20 C. and 720 m.m. : 2.24 X X y = 2 - 53 m3 P er second - (D Air used (standard conditions): 2.24X1,408 = 3.15m 3 Air used at 20 C. and 720 m.m. : 2.53 X 1,408 = 3.56 m 3 per second. (1) (2) The volume of gas used per minute, as it issues from the ports at 1200, is ' and of air 3.15X " X " = 17.9 " and assuming a maximum velocity for each of 10 meters per second, the areas of the ports must be: 390 METALLURGICAL CALCULATIONS. Gas ports 1 . 28 m 2 Air ports 1.79 " Sum 3.07 (2) (3) There is usually contraction when gases burn, the pro- ducts having less volume than the gas and air used. In- specting the equations of combustion of CO, CH 4 , H 2 , H 2 S and NH 3 given in Problem 69, we can construct the following table of relative volumes concerned and the ensuing contraction: CO CH 4 H 2 IPS NH 3 Volumeused 1.01.0 1.0 1.0 1.0 Oxygen used 0.52.0 0.5 1.5 1.75 Gases combining 1.53.0 1.5 2.5 2. 75 Volume of products ...1.03.0 1.0 2.0 2.5 Contraction 0.5 0.0 0.5 0.5 0.25 Using 1 cubic meter of producer gas the contraction resulting from its combustion with an excess of air is CO 0.2447X0.5 = 0.12235m 3 CH 4 = 0.00000 " H 2 0.1179X0.5 = 0.05895 " H 2 S 0.0009X0.5 = 0.00045 " NH 3 0.0019X0.25= 0.00050 " Total contraction = 0.1822 Since the volume of gas, plus air used, is 2.408 m 3 , the volume of the products, at standard conditions, is 2.408-0.182 = 2.226 m 3 per cubic meter of gas used under standard conditions. The volume of products per minute, at standard conditions, is, therefore, 2.226X2.24 = 4.986 m 3 . And at 1400 and 720 m.m. pressure: THE OPEN-HEARTH FURNACE. 391 Since the sum of the area of gas and air ports is 2.91 m 2 , the velocity of the products in these ports will be 32.7-:-3.07 = 10.7 m. per second. (3) In both the calculations of the size of the ports and the velocity of the products we have assumed the tension of the gas, air or products in the ports to be the prevailing atmos- pheric tension. This may or may not be exactly true, because the air or gas may be under a slightly less tension, being drawn into the furnace by the stack draft. If the pressure inside the furnace, with doors closed, is greater or less than atmos- pheric pressure, the tension of the gases in the entrance ports will be correspondingly greater or less than the atmospheric pressure, while the tension of the products will probably always be less than atmospheric pressure, because of the stack draft. Under ordinary conditions these corrections are too small to need to be taken into consideration. LABORATORY OF FURNACE. The laboratory consists of the open space enclosed between the hearth, sides, ends and roof. Its dimensions vary with the intended capacity of the furnace and the ideas of the designer. If a hearth is to contain, say, 50 tons of melted steel, which weighs some 7 tons per cubic meter (425 pounds per cubic foot), there will be contained in the furnace at one time 7 cubic meters, or 260 cubic feet of steel. The deeper this lies the more slowly it will be heated or oxidized by the flame, and there- fore there is a limiting depth of, say, 50 centimeters, or 20 inches, which it is not advisable to exceed, while a more shallow bath will result in faster working. Assuming a depth of 40 centimeters (16 inches), the volume divided by the depth will give the area of the bath: 7^-0.4 = 17.5 square meters 260-^-1.25 = 208 square feet. We can then either choose a convenient width, consistent with a practicable roof span, and derive the length, or choose a length and derive the width, or choose a certain ratio of length to width, and derive both. If the width is 3 meters the 392 METALLURGICAL CALCULATIONS. length must be 5.8; if the length is assumed 5 meters, the width is 3.5; if the ratio of length to width is 2 to 1, the length figures out 5.92 meters and the width 2.96. These dimensions are those of the bath of metal, and each should be increased by at least 1 meter to get the area of the hearth inside the walls, thus allowing 0.5 meter clear space all around the metal. If a furnace is short it should be wide and the roof high, in order to give cross-sectional area and thus diminish the velocity of the gases over the hearth. The gases attain their maximum temperature in the laboratory, theoretically, some 1700 to 1900, and their velocity depends solely on the vertical cross-sectional area of the laboratory or body of the furnace. In Problem 70, for instance, about 5 cubic meters of products of combustion (measured at standard conditions) pass through the furnace per second. At 1800 this volume would be 38 cubic meters, and if the laboratory were 4.5 meters wide by 1.5 meters high above the level of bath, there would be 7.25 square meters of cross-section, and the velocity of the gases would be 38 -J- 7.25 = 5.2 meters per second. This would allow barely 1 second for the hot gases to pass over the bath, which would result in a low rate of heating and probable in- complete combustion, for the gas can only burn as it gets mixed with air, and it is hardly likely that 100 per cent, of it would get mixed with air and consumed in 1 second. Such could only be attained by sub-division of the gas and air and very intimate mixture at the ports. Much of the economy undoubt- edly attained by raising the roof of open-hearth furnaces is due to the slowing-up of the gas currents in the laboratory, though it is usually ascribed to avoidance of contact of flame and bath, increased heating by radiation, etc. In the writer's opinion the raising of the roof from 1 to 2 meters, let us say, thus doub- ling the vertical cross-sectional area, cutting in half the velocity of the gases through the furnace, and doubling the period in which they are able to combine, and to radiate or impart heat to the furnace walls and charge is the principal reason for the increased economy observed. An equally important improvement is lengthening the dis- tance between ports. There is a limit to the width of the furnace, set by the practicable arch for the roof; there is also a limit to the height of roof, set by the increasing distance of THE OPEN-HEARTH FURNACE. 393 the gases from the hearth ; when both these factors have reached their maximum, further efficiency of utilization of the heat of combustion can only be secured, as far as the body of the furnace is concerned, by lengthening the hearth. There is no mechanical limit, and in every case the distance between the ports and the velocity of the gases should be such that complete combustion takes place in the furnace laboratory before the products pass into the regenerators. Problem 71. H. H. Campbell gives in the Transactions of the American Institute of Mining Engineers, 1890, analyses made at the Pennsylvania Steel Co.'s works, as follows: Gas Burned, Products, Entering Furnace. Leaving Furnace. CO 2 5.5 per cent. 3.1 per cent. O 2 2.3 " 0.7 " CO 8.2 " 7.1 CH 4 7.3 " 0.0 " H 2 .... 39.8 " 11.6 " N 2 36.9 " 77.5 " Required. (1) The proportion of the calorific power of the fuel developed while passing through the body of the furnace. (2) The proportion of the air necessary for complete com- bustion which was used. Solution. (1) One cubic meter of the gas contains the fol- lowing weight of carbon: (0.055 + 0.082 + 0.073) X 0.54 = 0.1134 kg. One cubic meter of products contains: (0.071 + 0.031) X 0.54 = 0.0551 kg. Therefore, volume of products per 1 cubic meter of gas: 0.1134 + 0.0551 = 2.06m 3 . Calorific power of 1 cubic meter of gas: CO 0.082X3062 = 251 Calories CH 4 0.073X8623 = 629 H 2 0.398X2613 = 1040 1920 394 METALLURGICAL CALCULATIONS. Calorific power of 2.06 cubic meters of products: CO 0.071X3062= 217 Calories. H 2 0.116X2613 = 303 520 520X2.06 = 1071 Heat developed in the furnace: 1920 - 1071 = 849 Calories. Proportion of the possible heat development: 849^-1920 = 0.442 = 44.2 per cent. (1) (2) The 1 cubic meter of gas needed, to burn its combustible constituents, the following amount of oxygen: CO 0.082X0.5 = 0.041m 3 CH 4 0.073X2.0 = 0.146 " H 2 0.98 X0.5 = 0.199 " Sum = 0.386 " Oxygen present in gas = 0.023 ' Oxygen needed from air = 0.363 " Air = 0.363^0.208 = 0.745 " The 2.06 m 3 of products of incomplete combustion require for their combustion: CO 0.071X2.06X0.5 = 0.0731 m 3 oxygen. H 2 0.116X2.06X0.5 = 0.1195 " Sum = 0.1926 ' Oxygen present in the products = 0.0070 " Oxygen needed from the air = 0.1856 " Air needed to complete combus- tion = 0.892 ' Total air needed for complete combustion = 1.745 Air supplied in the furnace = 0.853 ' Percentage supplied: 0.853-^1.745 - 0.489 = 48.9 per cent. (2) THE OPEN-HEARTH FURNACE. 395 It is almost needless to remark that with less than half the air necessary for complete combustion supplied, a high calorific intensity of flame and a high utilization of the calorific power of the fuel are impossible. More air should have been used and the furnace made longer, so as to secure perfect combus- tion in the furnace, and not have over half the possible de- velopment left to take place in the regenerators or stack. CHIMNEY FLUES AND CHIMNEY. The gases pass into the chimney flues at from 150 to 450. If their volume at an assumed average temperature of, say, 300 is calculated, they can be given an assumed velocity of 2 to 3 meters (5 to 10 feet) per second, and thus a suitable cross- sectional area of the chimney flues obtained. The stack will work best with a velocity of 5 meters per second, and thus its cross-sectional area may be calculated. A height of 25 to 30 meters (75 to 100 feet) is sufficient for most furnaces. MISCELLANEOUS. Some other data useful in figuring up the dimensions and running conditions of modern open-hearth steel furnaces are the following, taken mostly from an article by H. D. Hess, in the Proceedings Engineering Club of Philadelphia, January, 1904: Average coal consumption, in pounds, per hour per ton of metal capacity of the furnace, 55 to 80. Cubical feet of space in one pair of regenerators, per ton of metal capacity of the furnace, 30 to 75. Cubic feet of space in one pair of regenerators per pound of coal consumed per hour, 0.5 to 1.0. A correllation and combination of data of this sort, with details as to the actual working of the furnaces, would point the way towards a general solution, which would furnish the best condition for every possible case, with strict consideration for all the variables involved. CHAPTER XL THERMAL EFFICIENCY OF OPEN-HEARTH FURNACES. The ordinary open-hearth steel furnace receives cold pig iron, cold scrap, warm ferro-manganese, cold limestone and cold ore, it receives cold air and moderately warm producer gas, and it furnishes melted steel and slag at the tapping heat. The larger part of the usefully applied heat is that contained in the melted steel, for it must be melted in order to be cast, and when once taken away from the furnace the latter is done with it. The total heat available for the purposes of the furnace and which should be charged against it consists of the following items: (1) Heat in warm or hot charges. (2) Heat in warm or hot gas as it reaches the furnace. (3) Heat in warm or hot air as it reaches the furnace. (4) Heat which could be generated by complete combustion of the gas. (5) Heat of oxidation of those constituents of the charge which are oxidized in the furnace. (6) Heat of formation of the slag. The items of distribution of this total will be as follows: (1) Heat in the melted steel at tapping. (2) Heat absorbed in reducing iron from iron ore. (3) Heat absorbed in decomposing limestone added for flux. (4) Heat absorbed in evaporating any moisture in the charges. These first three items constitute the usefully applied heat, and their sum measures the net thermal efficiency of the furnace. (5) Heat absorbed in reducing ferric oxide to ferrous oxide. (6) Heat in the slag. (7) Heat lost by imperfect combustion. 396 EFFICIENCY OF OPEN -HEARTH FURNACES. 397 (8) Heat in the chimney gases as they leave the furnace. (9) Heat absorbed by cooling water. (10) Heat lost by conduction to the ground. (11) Heat lost by conduction to the air. (12) Heat lost by radiation. (1) HEAT IN WARM CHARGES. If the pig iron is charged melted instead of cold an immense amount of thermal work is spared the furnace, and it should be charged with all the heat (reckoning from C. as a base line) which is in the melted pig iron as it runs into the fur- nace. This will average 275 Calories per unit of pig iron, but should be actually determined calorimetrically in each specific instance wherever possible. The net thermal efficiency of the furnace will figure out higher with cold charges than with melted pig iron, because, with a possible flame tempera- ture of 1,900 C. in the furnace, heat is absorbed much more rapidly by cold charges than by hot ones, and a larger per- centage of the available heat will be thus usefully applied. Scrap is almost always charged cold, but if any of it is hot its weight and temperature should be known and the amount of heat thus brought in charged against the furnace. Or a small piece may be dropped into a calorimeter and its heat content per unit of weight measured directly, and thus the heat in all the hot scrap used may be estimated. Ferro-manganese is often added cold, but usually is pre- heated to cherry redness (about 900) in another small fur- nace, in order that it may dissolve more quickly in the bath. Knowing its weight, temperature and specific heat, the heat which it brings into the furnace can be calculated ; a better plan is to drop a piece into a calorimeter and measure the actual heat in a sample of it. Limestone and ore are almost invariably put into the fur- nace cold. If used warm the heat in them can be determined by the methods just described. (2) HEAT IN THE GAS USED. By this is meant, not the heat in the gas after it is heated by the regenerators, but its sensible heat as it reaches the furnace. This applies only to furnaces where the producers 398 METALLURGICAL CALCULATIONS. or gas supply are independent of the furnace. Where the pro- ducers are an integral part of the furnace it is impracticable to consider them separately from the furnace, and the efficiency of the whole plant, including the producers, must be consid- ered together. But where the gas supply from whatever source comes to the furnace from outside, and reaches the furnace warm, its sensible heat is to be charged against the furnace as part of the heat which the furnace must account for. If the gas comes from producers its amount is satisfactorily found from the known weight of carbon gasified per hour, or per furnace charge, and the weight of carbon contained in unit volume of gas, as calculated from its analysis. If gas comes from a common main which supplies several furnaces, or is simply natural gas, its amount can only be roughly estimated by measuring the area of the gas supply pipe or flue and meas- uring the velocity of flow by a pressure gauge or Pitot tube or anemometer. None of these methods just mentioned are satisfactorily accurate, and there is great need of simple methods for determining accurately the flow of gases in flues or pipes,. If the velocity of warm gas is determined suitable correction for its temperature must be made to reduce it to volume at standard conditions. (3) HEAT IN THE AIR USED. If the air coming to the furnace is warm its sensible heat must be charged against the furnace. If the air is warmed, however, before it goes into the regenerators by waste heat from the furnace itself, then its sensible heat should not be charged against the furnace, because that would amount to charging the furnace twice with this quantity of heat. Such preheating it in reality only a part of the regenerative princi- ple, even though it may not be done in regenerators, but, for instance, by circulating air around a slag-pot or through the hollow walls of the furnace. If the air used is moist its mois- ture should not be omitted in the calculation. The amount of air used is best determined by a comparison of the analyses of gas and chimney products, and a calculation based on the carbon contents of each and the known volume of gas used. EFFICIENCY OF OPEN-HEARTH FURNACES. 399 (4) HEAT OF COMBUSTION. Under this head comes the principal item of heat available for the furnace. In reckoning it we should calculate the total heat which could be generated by the perfect combustion of the gas used, to CO 2 , N 2 and H 2 O vapor. If there is in reality imperfect combustion, as is shown by analysis of the chimney gases, that is a defect of operation of the furnace which should be written down against it. Problem 71 showed an actual case in which there was very incomplete combustion in the body of the furnace, but where combustion was afterwards completed in the regenerators. In such a case the same principle applies; the furnace must be charged with the total calorific power of the fuel used, and incomplete combustion can be charged against the furnace as a whole only on the basis of uncon- sumed ingredients in the chimney gases the products finally rejected by the furnace. If there is poor combustion in the body of the furnace and combustion is only completed in the re- generators, the furnace will not give as high net thermal effi- ciency as if combustion were complete above the hearth. (5) OXIDATION OF THE BATH. The oxidation of carbon, iron, silicon, manganese and some- times phosphorus and sulphur, add a not inconsiderable amount to the heat resources of the furnace. Carbon should be burnt to CO 2 , iron is usually oxidized to FeO, manganese to MnO, silicon to SiO 2 , phosphorus to P 2 O 5 , and sulphur to SO 2 . All of these oxidations generate heat, and, moreover, heat which should be very efficiently utilized, being produced in direct contact with the metallic bath ; it should all be charged against the furnace as part of its available heat. (6) FORMATION OF SLAG. The metallic oxides, produced unite with each other, and with the lime and silica of ore used and lining of the hearth to produce the slag, the heat of formation of which can be calcu- lated and counted in as available heat. (1) HEAT IN MELTED STEEL. This is a large item in the work done by the furnace ; in fact, usually the largest single item. It should be determined 400 METALLURGICAL CALCULATIONS. calorimetrically when possible; if this is not done its tem- perature should be known and its composition, in order to compare it with the calorimetric experiments of others, and thus derive a probable value for its heat contents. Not many experimental values in this line have so far been published, and a very much needed investigation is one upon the total heat in melted steels of different compositions at different temperatures. Values from 275 to 350 Calories per kilogram have been observed. (2) HEAT OF REDUCTION OF IRON FROM ORE. This is an item which appears whenever ore is used to facili- tate oxidation of the bath. The weight and composition of the charges and the products will easily show how much iron has been reduced. The ore used is almost always hema- tite, less frequently magnetite; hydrated iron oxides are not used for obvious reasons. The heat absorbed is 1,746 Calories per kilogram of iron reduced from Fe 2 O 3 and 1,612 Calories per kilogram from Fe 3 O 4 . (3) DECOMPOSITION OF LIMESTONE FLUX. If limestone is charged raw, as is usually done in order to avoid the dusting caused by using burnt lime, then the furnace is called upon to burn this limestone in place of the lime kiln. The heat absorbed may be taken as either 451 Calories per kilogram of CaCO 3 decomposed. 1,026 Calories per kilogram of CO 2 driven off. 806 Calories per kilogram of CaO produced. (4) EVAPORATION OF MOISTURE IN THE CHARGES. If the ore, flux, scrap or ore are put into the furnace wet their moisture must be evaporated. The correct figure for this evaporation is the latent heat at ordinary temperatures, viz.: 606.5 Calories per kilogram. This allows for the heat re- quired to convert into cold vapor of water, and puts the H 2 O thereafter upon the same basis as all the other gas going out of the furnace. The chimney gases carry out sensible heat, and the H 2 O in them can be calculated as carrying out a cer- tain amount of heat as gas, reckoning from zero, and thus the correct chimney loss obtained. It is incorrect either to EFFICIENCY OF OPEN-HEARTH FURNACES. 401 charge the latent heat of vaporization as chimney loss or to charge the sensible heat of the water vapor in the chimney gases to heat absorbed in evaporating water in the furnace. It is also incorrect to do as is frequently done, viz.: to calcu- late the heat required to evaporate the moisture to water vapor at 100 637 Calories and say that this is the heat to evaporate the moisture. With almost no moisture in the gases, the moisture of the charges would commence to evaporate at once, while they were yet cold, and the moisture is no more evaporated at 100 or to vapor at 100 than it is to 200 or 500. The only safe course is to confine the evaporation heat to that necessary to convert the moisture into cold vapor, and let its sensible heat as it escapes as vapor at any other tem- perature be reckoned in with the sensible heat of the chimney gases. (5) REDUCTION OF ORE INTO THE SLAG. While considerable of the iron in the ore used is reduced to the metallic state, yet often the larger part is reduced merely to the state of FeO, and as such goes into the slag. The amount so reduced can be determined by subtracting the iron reduced from ore from the total iron in the ore used; the differences gives the iron from the ore going into the slag as FeO. The weight of FeO corresponding is then easily calculated. The heat absorbed in this partial reduction is: 446 Calories per kilogram of FeO reduced from Fe 2 8 . 341 Calories per kilogram of FeO reduced from Fe 3 O 4 . (6) HEAT IN SLAG. This is usually a small item in open-hearth practice, but may amount to a very considerable one in the method of running with large ore charges, as in the Monell process. The varia- tions of composition of the slag, and especially in the tempera- ture at which it is run off, are so large that the heat in the slag should always be determined calorimetrically for each specific case. If assumptions have to be made, 450 to 550 Calories per kilogram of slag would be assumed. The weight of slag is seldom taken, although it could in most cases be done if desired. If the weight is not known it may be calcu- lated from the known weight of either iron, manganese or 402 METALLURGICAL CALCULATIONS. phosphorous going into it, as seen from the balance sheet and the percentages of these elements in the slag as shown by ana- lysis. (7) Loss BY IMPERFECT COMBUSTION. This is based upon the unconsumed ingredients of the chim- ney gases, as shown in an analysis. From this the calorific power of the unburnt gases in 1 cubic meter can be calculated. If then we know the volume of chimney gases per unit of charge, we get the heat loss by imperfect combustion per unit of charge. The volume of chimney gas is found by means of the carbon in it, which must all come from the carbon in the gas used, plus the carbon oxidized out of the charges, plus the carbon of CO 2 , driven off raw limestone used as flux. Or, putting it in another way, the total carbon going into the furnace in any form, less the carbon in finished steel, must give the carbon in the chimney gases. This divided by the weight of carbon in unit volume of chimney gas gives 'the volume of the latter, per whatever unit of charge is used as the basis of calculations. This volume times the calorific power of unit volume of chimney gas, gives the total heat lost by imperfect combustion. (8) SENSIBLE HEAT OF CHIMNEY GASES. The temperature of these gases should be taken as they enter the chimney flue. Their amount is determined as ex- plained under the previous heading. The water vapor con- tained should not be overlooked, being reckoned simply as vapor or gas in exactly the same category as the other gases. The analysis of the chimney gases being usually given on dried gas, a separate determination of the moisture carried per unit volume of such dried gas is necessary. If this is not done an approximation can be made by considering all the hydrogen in the gas burned to form water vapor, and add in the moisture of the air used and the moisture in the charge. (9) HEAT LOST IN COOLING WATER. This is a very variable amount, and must be determined for each furnace by measuring the amount of water used per unit of time and its temperature before reaching and after leaving EFFICIENCY OF OPEN-HEARTH FURNACES. 403 the furnace. Doors are frequently water-cooled, also ports, where the heat is fiercest, and sometimes a ring around the hearth at the slag line. (10) Loss BY CONDUCTION TO THE GROUND. This is a quantity extremely difficult to measure or to esti- mate. If a closed vessel filled with water were put into the foundations the rate at which its temperature rose might give some idea of the rate at which heat passed in that direction per unit of surface contact. At present, lacking all reliable data, we must put this item in the " by difference "class. (11) Loss BY CONDUCTION TO THE AIR. This is an amount which can be observed and calculated with some approach to satisfaction. The sina qua non for this par- pose is a FeYy radiation pyrometer, by which the temperature of the outside of the furnace at different parts can be accurately determined. Then the velocity of the air blowing against the furnace, if it is in a current of air, is observed with a wind gauge, and its temperature before reaching the furnace. With these data and by the methods of calculation before explained in these calculations (Part I, Chap. VIII) the heat lost to the air may be calculated. (12) RADIATION Loss. Having determined the temperature of the outer surface of the furnace and measured its extent, as above explained, the radiation loss can also be calculated, knowing the mean tem- perature of the surroundings, by the principles of radiation, having due regard to the nature of the radiating surface. Tables of specific radiation capacity of different substances (fire-brick, stone, iron) will be found at the reference just given above. Problem 72. Jiiptner and Toldt (Generatoren und Martinofen, p. 73) ob- served the following data with regard to an open-hearth steel furnace charge: Weight of cold charges, at 26 C 3,745 kg. Weight of hot charges, at 700 C 1,700 " Total weight of charge 5,445 " 404 METALLURGICAL CALCULATIONS. Average composition of charge C = 1.07 per cent. Si = 0.50 " Mn = 1.33 Coal gasified in producers 1,980 kg. Carbon gasified from coal 47.13 per cent. Average composition of dried gas used CO 2 3.81 " O 2 0.98 CO 23.82 CH 4 0.42 " H 2 8.75 " N 2 62.22 " Moisture accompanying each m 3 of gas = 82 grams. Temperature of gas reaching furnace. 165 C. Temperature of air used 26 C Moisture accompanying each m 3 of air 12 grams. Barometer 717 m.m. Steel produced 5,191 kg. Cpmposition of steel C = 0.12 per cent. Si = 0.04 Mn = 0.19 " Temperature of steel when tapped 1410 C. Heat in 1 kg. steel, by calorimeter (to C.) . . 277 Cal. Composition of slag SiO 2 45.65 per cent. FeO 33.60 " MnO 18.21 CaO 2.54 " Weight of slag. . . 425 kg. Temperature of slag when issuing 1,410 C. Heat in 1 kg. slag, by calorimeter (to 0) 560 Cal. Composition of the stack gases (dry) CO 2 11.12 per cent. O 2 6.78 " N 2 82.10 " Temperature of the gases in chimney flue 500 C. Requirements: (1) A balance sheet of materials entering and leaving the furnace. (2) A balance sheet of the heat available and its distribution. (3) What excess of air above what is necessary for complete combustion is used, and what per cent, of all the available heat of the furnace is thereby lost? EFFICIENCY OF OPEN-HEARTH FURNACES. 405 (4) What is the thermal efficiency of the furnace? Solution: (1) BALANCE SHEET OF MATERIALS. Slag. 25 62 115 11 132 Charges. . Steel. Metal, 6,445 kg. c, 58 " 6 Si, 27 " 2 Mn, 72 " 10 Fe, 5,288 " 5,173 Limestone, 20 " CaO, 11 " .... c, 2.5 " .... o, 6.5 " .... Hearth, SiO 2 132 " .... Gas, 7,884 " c, 933 " .... 0, 2,003 " .... H, 118 " .... N, 4,830 " .... Air, 16,026 " 0, 3,812 " .... N, 12,195 " .... H, 19 " Totals, 29,507 " 5,191 80 425 Gases. 52 2.5 6.5 933 2,003 118 4,830 23,891 NOTES ON THE BALANCE SHEET. The distribution of carbon, silicon, manganese and iron is governed by the known amounts of these elements present in the steel, the rest of the carbon going into the gases (as CO 2 ), and the manganese, silicon and iron passing into the slag (as MnO, SiO 2 and FeO, respectively.) The amount of limestone used was not stated, but was de- duced from the fact that the slag was said to weight 425 kilos, and to contain 2.54 per cent, of CaO, which makes the CaO 11 kilos, and assuming pure limestone, this would bring in 9 kilos, of CO 2 , which appears on the balance sheet as 2.5 kilos, of carbon and 6.5 kilos, of oxygen, contributed to the gases. 406 METALLURGICAL CALCULATIONS. The weight of silica contributed by the hearth is deduced from the fact that the slag must contain the silicon, manganese and iron oxidized from the charge, as SiO 2 , MnO and FeO,the CaO of the flux and the SiO 2 contributed by the hearth, and its total weight is 425 kilos. The ingredients of the slag must, therefore, be Kg. SiO 2 25X60/28 = 53.5 MnO 62X71/55= 80.0 FeO 115X72/56 = 147.9 CaO = 11.0 . Sum 293 From hearth (difference) 132 Total slag 425 The gas used we find by starting with the fact that 1,980 kilos, of coal is used, from which 47.13 per cent, of carbon enters the gases. This makes carbon in the gases 933 kilos. Each cubic meter of dry gas, as analyzed, contains 0.2805 cubic meter of CO 2 , CO and CH 4 added together, and there- fore, 0.2805X0.54 = 0.1515 kilos, of carbon. The volume of dry producer gas used is therefore, at standard conditions, 933-^0.1515 = 6,160 cubic meters, containing by its analysis: CO 2 6,160X0.0381 = 235 m 3 = 465 kg. O 2 " X 0.0098 = 60m 3 = 86" CO " X 0.2382 = 1,467 m 3 = 1,840 " CH 4 " X 0.0042 = 26m 3 = 19" H 2 " X 0.0875 = 539 m 3 = 49" N 2 " X 0.6222 = 3,833 m 3 = 4,830 " 6,160 m 3 = 7,297 The moisture is 82 grams per each cubic meter of gas, meas- ured at 26 and 717 m.m. pressure; but the 6,160 cubic meters of gas at standard conditions would be 7,175 cubic meters at those conditions of temperature and pressure, and therefore be accompanied by 7,175X82 = 588,350 grams = 588 kg. of H 2 O vapor. EFFICIENCY OF OPEN -HEARTH FURNACES. 407 We can now enter the gas on the balance sheet either as so much CO 2 , CO, H 2 O, etc., or else resolve it into its essential constituents C, H, O and N, which course we have followed on the balance sheet. The carbon in the gas, is 933 kg. by assumption, 'the oxygen is 8/11 the CO 2 , all the O 2 , 4/7 the CO and 8/9 the H 2 O, a total amounting to 2,003 kilos; the hydro- gen is 4/16 the CH 4 , all the H 2 and 1/9 the H 2 O, a total of 118 kilos. The air supplied is best found from the volume of the chim- ney gases. The total carbon entering these is 52 + 2.5 + 933 = 987.5 kilos., as seen from the balance sheet. Each cubic meter of dry chimney gas contains 0.1112 m 3 of CO 2 , carrying 0.1112X0.54 = 0.0600 kilos, of carbon. The volume of dry chimney gas at standard conditions is therefore 987.5-^0.0600 = 16,458 m 3 . This contains 16,458X0.8210 = 13,512 m 3 of N 2 , and since 3,833 cubic meters came in with the gas 9,679 m 3 must have come in with the air, corresponding to 12,220 m 3 of dry air at standard conditions. This would consist of 12,195 kilos, of N 2 and 3,660 kilos, of O 2 . To find the moisture present the volume of this air at 26 and 717 m.m. pressure would be 14,230 m 3 , and would be therefore accompanied by 14,230X12 = 170,760 grams = 171 kg. of H 2 O. This consists of 19 kilos, of hydrogen and 152 kilos, of oxygen, the latter increasing the total oxygen in the air used to 3 660 + 152 = 3,812 kilos. (2) HEAT BALANCE SHEET. Heat Available. Per- Cal. centages. Heat in the warm charges 189,210 = 2.45 Sensible heat of air used 99,480 = 1 . 29 Sensible heat of gas used 360,550 = 4.68 Heat of combustion of gas .6,202,300 = 80.44 Heat of oxidation of the bath 833,600 = 10. 81 Heat of formation of slag 24,200 = 0.31 Total. . 7,709,340 = 100.00 408 METALLURGICAL CALCULATIONS. Heat Distribution. In melted steel at tapping 1,437,900 = 18.65 Decomposition of limestone 9,200 = 0. 12 Sensible heat of slag 238,000 = 3.09 Sensible heat of chimney gases 3,118,450 = 40.45 All other losses, not classified 2,905,790 = 37.69 Total 7,709,340 = 100.00 Notes on the Heat Balance Sheet. The warmed charges weighed 1,700 kilos, at 700, and the cold charges 3,745 kilos at 26. Taking C. as the base line the sensible heat in these is Cat. 1,700X0.15X700 = 178,500 3,745X0.11 X 26 = 10,710 Sum = 189,210 The air used contains, at standard conditions, 12,220 m 3 of air and 171 -f- 0.81 = 211 m 3 of water vapor. These carry at 26, heat as follows: Col. 12,220X0.3037X26 = 98,490 211X0.3439X26 = 990 Sum = 99,480 The gas used, coming in at 165 C., carries in heat as follows: Col. O 2 , CO, H 2 , N 2 5,899 m 3 X 0.3075 = 1,814 CO 2 235 m 3 X 0.4063 = 95 CH 4 26 m 3 X 0.4163 = 11 H 2 O 726 m 3 X 0.3648 = 265 Average calorific capacity per 1 = 2,185 Heat content 2, 185 X 165 = 360,525 The heat of combustion is that of the combustible ingredi- ents of the gas used to CO 2 and H 2 O vapor: EFFICIENCY OF OPEN-HEARTH FURNACES. 409 Col. CO 1,467 m 3 X 3,062 = 4,492,000 CH 4 26 m 3 X 8,598 = 223,500 H 2 539 m 3 X 2,613 - 1 486,800 Sum = 6,202,300 The heat of oxidation of the bath is from the various sub- stances oxidized: Cal. C to CO 2 52X8,100 = 421,200 Si to SiO 2 25X7,000 - 175,000 Mn to MnO 62x1,653 = 102,500 Fe to FeO 115X1,173 = 134,900 Sum = 833,600 The heat of formation of the slag is the heat of combination of 80 kilos, of MnO, 148 kilos, of FeO and 11 kilos, of CaO, with 186 kilos, of SiO 2 . This will be, since the bases are largely in excess of the silica: 186X130 = 24,200 Cal. The figure 130 is an average for the heat of combination of 1 kilo, of SiO 2 with about 2 parts of FeO to 1 part MnO. The heat in the steel at tapping is simply its weight multi- plied by its heat contents per kilo. (5,191x277). The heat in the slag is similarly obtained (425x560). The decomposition of limestone represents 9 kilos, of CO 3 liberated, and the heat absorbed is 9x1,026. The sensible heat in the chimney gases is obtained by first noting that their volume (measured dry), as already obtained, is 16,458 cubic meters. The CO 2 , 11.12 per cent., becomes, therefore, 1,843 m 3 ; the O 2 , 1,116 m 3 ; N 2 , 13,512 m 3 , while the H 2 O accompanying this will be 9 times the weight of hydrogen passing into the gases, which is 9X (118+17) = 1,215 kilos. = 1,500 m 3 . A simpler way to get the volume of the water vapor is to observe that it is always equal to the volume of hydrogen going into it, and, therefore, in this case would be (1 18+ 17) -T- 0.09 = 1,500 m s . The heat carried out by these gases would therefore be: 410 METALLURGICAL CALCULATIONS. N 2 + 2 = 14,628 m 3 X 0.3165 = 4629.8 Cal. per V CO 2 = 1, 843 m 3 X 0.4800 = 984.4 " H 2 = 1,500 m 3 X 0.4150 = 622.5 " Calorific capacity = 6236.9 " Total capacity = 3,118,450 " per 500 The heat balance sheet as a whole discloses the fact that in this furnace the fuel only supplies some 80.5 per cent, of the total heat available, and that about 10.8 per cent, is furnished by the oxidation of the bath itself. On the other hand, the melted steel accounts for 18.6 per cent., while chemical reactions absorb almost none, giving a net thermal efficiency of slightly under 19 per cent. The other important items are 40.5 per cent, of the total heat lost up the chimney, and 38 per cent, lost by radiation and conduction. Such data as these point the way to avenues of possible saving or avoidance of waste of heat. (3) The excess of air is obtained directly from the chimney gases. The 1,116 m 3 of oxygen, unused, represents 1,116-s- 20.8 = 5,365 m 3 of air in excess, which leaves 16,4585,365 = 11,093 m 3 of air which came in and was used. The per- centage of air used in excess of that which was necessary was: 5,365-4-11,093 = 0.4845 = 48.5 per cent. (3) No properly run open-hearth regenerative gas furnace should ever have such a large excess of air, for it cuts down the tem- perature of the flame and leads to high chimney losses. The air used brought in 171 kilos of water vapor, and there- fore the excess air brought in 48 5 171X 7I^-H = 56 kilos - of water 14o. O the volume of which, at standard conditions, would be 56^-0.81 = 61 cubic meters. The excess air, with its accompanying water, going into the chimney at 500, carried out heat as follows: Cal. 5,365X0.3165X500 = 849,000 61X0.4150X500 = 12,650 861,650 EFFICIENCY OF OPEN-HEARTH FURNACES. 411 Representing 861,650-5-7,709,340 = 0.112 = 11.2 per cent. (3) (4) The thermal efficiency has been already added up as 18.65 + 0.12 = 18.77 per cent. (4) Problem 73. In the open-hearth furnace of the preceding problem, assume that the calculations therein made showed that, per heat of steel produced, 5,191 kilograms, there entered and left the furnace the following volumes of gases, measured at standard conditions: Producer Gas. Air. Chimney Gases. CO 2 235m 3 1,830m 3 O 2 60m 3 3,833m 3 1,116m 3 CO 1,467m 3 CH 4 26m 3 H 2 539m 3 N 2 3,834m 3 9,679m 3 13,512m 3 H 2 O 726m 3 211m 3 1,500m 3 The temperature of air used was 26, of producer gas 165, of chimney gases 400. The excess of air used was 48.5 per cent. The gas and air entered the laboratory of the furnace preheated to 1,100, and the products of combustion entered the regenerators at 1,450. Items of heat balance sheet: Available. Calories In warm charges 189,210 Sensible heat of air used at 26 90,480 Sensible heat of gas used at 165 360,550 Heat of combustion of the gas 6,202,300 Heat of oxidation of the bath 833,600 Heat of formation of slag 24,200 7,700,340 Distribution. In melted steel at tapping at 1,410 1,437,900 In slag at tapping at 1,410 238,000 Decomposition of limestone 9,200 Sensible heat in chimney gases at 400 3,065,350 All other losses, not classified . . . . 2,949,890 7,700,340 412 METALLURGICAL CALCULATIONS. Required : (1) The thermal efficiency of the regenerators. (2) The thermal efficiency of the laboratory of the furnace. (3) The temperature of the flame. . (4) The change in (3) if only the theoretical amount of air for combustion were used. Solution : (1) The products of combustion, entering the regenerators at 1,450, carry into them the following amounts of heat: Calories. CO 2 1,830X0.689 = 1,261 O 2 + N 2 14,628X0.342 = 5,003 H 2 O 1,500X0.557 = 836 Calorific capacity per 1 = 7,100 Calorific capacity per 1,450 = 10,295,000 Calories. The same gases entering the chimney flue at 400 carry with them the following amounts : Calories. CO 2 1,830X0.458 = 838.1 O 2 + N 2 14,628X0.314 = 4,593.2 H 2 1,500X0.400 = 600.0 Calorific capacity per 1 = 6,031.3 Calorific capacity per 400 = 2,412,500 Calories. The gas used for combustion, entering the regenerators at 165 and leaving it at 1,100, carried into the regenerators 360,550 Calories, as already given in the balance sheet, and carried out at 1,100 the following: Calories. CO 2 235X0.612 = 143.8 CH 4 26X0.620 = 16.1 O 2 , N 2 , H 2 , CO 5,899X0.333 = 1,964.4 H 2 O 726X0.505 = 366.6 Calorific capacity per 1 = 2,490.9 Calorific capacity per 1,100 = 2,780,000 Calories. Heat abstracted from regenerators: 2,780,000360,550 = 2,419,450 Calories. EFFICIENCY OF OPEN-HEAR! H FURNACES. 413 The air used, entering the regenerators at 26, carries in as sensible heat 90,480 Calories, as already given in the balance sheet, and issuing from them at 1,100 carries out the fol- lowing : Calories. 2 + N 2 13,512X0.333 = 4,499.5 H 2 O 211X0.405= 85.5 Calorific capacity per 1 = 4,585.0 Calorific capacity per 1,100 = 5,043,500 Calories. Heat abstracted from regenerators : 5,043,50090,500 = 4,953,000 Calories. The thermal efficiency of the regenerators may now be cal- culated from three standpoints. There is no doubt that the gas and air take from the regenerators, and return to the body of the furnace 2,419,450 + 4,953,000 = 7,372,450 Calories. This is, therefore, the usefully returned heat, and the ratio of this to the heat received by the regenerators measures their efficiency qua regenerators. The three figures obtained for this efficiency depend on what is to be considered as the heat chargeable against the regenerators. Are they to be charged with all the heat in the hot products at 1,450 (10,295,000 Calories), or only with the heat left in the regenerators by these products leaving at 400 (10,295,0002,412,500 = 7,882,500 Calories), or per- haps only with the heat carried in less a certain assumed amount representing the minimum temperature to which it is desirable to cool the products before they enter the chimney? If we charge the regenerators with all the heat brought in by the products their thermal efficiency figures out: = 0.72 =72 per cent. (!) If we charge them with the heat left in the regenerators, by the products, their efficiency is: 7,372,450 75^600 = 0-94 =94 per cent. (1) 414 METALLURGICAL CALCULATIONS. leaving 7 per cent, of the heat chargeable against them lost by radiation from their walls and conduction to the ground. If we think that the first calculation gives too low an effi- ciency, because the gases must leave the regenerators hot, in order to be used for chimney draft, and therefore some or all of the heat in the chimney gases should not be charged against the regenerators, on the other hand, the second calculation may represent too high an efficiency, because the gases may be dis- carded to the chimney at a higher temperature than is neces- sary to provide the requisite chimney draft, and this excess of chimney temperature and consequent heat loss is a defect of the regenerators which they should be charged with. If we assume that a chimney gives very nearly its maximum drawing capacity with the gases entering it at 300, it would be per- fectly proper to charge the regenerators with all the heat which could be given out by the products in cooling from 1,450 to 300, heat which they should have entirely absorbed. In the case in hand, the products at 300 would contain (by calcu- lations similar to those already made) 1,777,400 Calories, leaving 10,295,0001,777,400 = 8,517,600 Calories chargeable against the regenerators, as the heat which they should have absorbed or intercepted. Measured by this standard, their efficiency is: The losses from the regenerators in this view would be 7 per cent, (about) by radiation and 7 per cent, by unnecessary chim- ney loss. Comparing these three methods of considering efficiency, the third appears to the writer as the fairest, and the one which gives the metallurgist the most reliable criterion of the real work which his regenerators are doing for him, and the best basis of comparison when considering the work of different regenerators or of the same regenerators under different con- ditions. (2) The laboratory of the furnace, the space enclosed be- tween the hearth, roof, side walls and ports, receives from the entering preheated gas and air their sensible heat, and the heat generated by combustion. In the case in point these items total, as already calculated: EFFICIENCY OF OPEN-HEARTH FURNACES. 415 Calories. Sensible heat in preheated gas ......... 2,722,300 Sensible heat in preheated air .......... 5,043,500 Heat generated by the combustion ..... 6,202,300 Total ........................... 13,968,100 Hot products at 1,450 take out ....... 10,295,000 Heat left in the laboratory ............ 3,673,100 These figures show that the laboratory of the furnace ap- propriates to its own purposes 3,673,100 Calories out of the 13,968,100 Calories poured into it. This part of the furnace, therefore, qua laboratory, has an efficiency in this respect of (2) This is the datum which would be useful to the metallurgist in comparing the efficiencies of differently shaped laboratories, such as those with differently shaped hearths, differently shaped roof, differently arranged ports, etc. This conception of effi- ciency is that taken by Damour, and repeated by Queneau in his book on " Industrial Furnaces." We must be careful here not to compare the heat left in the laboratory with the heat of combustion alone. This would give 3,673,100 - ' 59 - 59 But this is not the heat absorption of the laboratory alone, but is a function of the furnace as a whole, and depends largely on the efficiency of the regeneration accomplished by the regen- erators. If we wish to obtain an idea of the perfection with which the laboratory of the furnace appropriates the heat pass- ing into and through it, we must simply compare what it ap- propriates with what was sent into it. The percentage of this appropriation measures the efficiency of the laboratory for abstracting heat for the purpose of heating itself a datum highly useful to know if the furnace is used simply for keeping a given working space up to a given temperature for an in- 416 METALLURGICAL CALCULATIONS. definite time. If the heating of the furnace charge to the furnace heat is only a minor part of the useful work of the furnace, and keeping it at that temperature is the chief function of the furnace, then the relation of the heat thus appropriated and utilized to the total heat available to the laboratory, is a measure of the perfection of construction of the laboratory. This is the view of Damour and Queneau, and seems in some re- spects plausible. However, if we are examining this question thus in detail, it appears to the writer that the view just explained and the conclusions derived thereform may really be the very opposite of the truth of the matter, and lead to entirely erroneous con- clusions. If two exactly similar open-hearth furnaces are constructed, with the sole difference that the body of one is thicker walled than the body of the other, it is perfectly true that maintaining the same temperature in each will require more gas in the thin- walled furnace, but that it will abstract and radiate a considerably larger proportion of the heat passing through it than the thick- walled furnace. Yet, according to Damour and Queneau's principle, the thin-walled furnace would be considered as being the more efficient laboratory of the two. The truth is, we must either compute the heat left in the -laboratories per unit of time per cubic meter of stock space in order to compare different furnaces, or else to get absolute efficiency compute the ratio of the heat passing into the laboratory with that absorbed usefully by the charge, viz.: in this case: 1,447,100 A 1A/1 1A A 0.104 = 10.4 per cent. 13,968,100 (2) (3) The temperature of the flame is that to which the pro- ducts of combustion can be raised by the maximum heat which they contain. When perfectly consumed they contain the heat pre-existing as sensible heat in the hot gas and air, plus the heat of combustion. We have already calculated this total as 13,968,100 Calories. The mean heat capacity of the products of combustion, per 1 between and t, is: EFFICIENCY OF OPEN-HEARTH FURNACES. 417 CO 2 1,830 m 3 (0.37 +0.0002210 = 677 + 0.4026t O 2 + N 2 14,628 m s (0.303 + 0.000027t) = 4,428 + 0.3950t H 2 O 1,500 m 3 (0.34 +0.00015t) = 510 + 0.2250t Sum = 5,615 + 1.0226t Therefore, 13,968,100 ~ 5,615 +1.0226t Whence, t = 1,749 (3) (4) If the excess air were not used, the 1,116 cubic meters of oxygen in the products, together with its corresponding quan- tity of nitrogen, 4,249 cubic meters, would be absent from the products of combustion, as would also some of the 211 cubic meters of water vapor. Since the oxygen in the air was 3,833 m 3 and the unused excess 1,116, the proportion of the 211 m 3 of water vapor belonging to the excess air was and the final products are CO 2 1,830 m 3 , N 2 9,263 m 3 and H 2 1,439 m 3 . The heat available will be the same as before from com- bustion, the same from preheated gas, but less from preheated air, because of the absence of the excess air. Since the air formerly used brought in 5,043,500, the proportion of this carried by the excess air is 5,043,500 X L -= 1,468,500, leaving heat brought in by air in the second case the difference, viz.: 3,575,000, and the total heat in the products 13,968,100 1,468,500 = 12,499,600 Calories. The mean heat capacity of the diminished quantity of pro- ducts, per 1, is, by the same methods as previously used, 3,973 + 0.8685t, and, therefore, 12,499,600 3,973 + 0.8685t 418 METALLURGICAL CALCULATIONS. a gain of nearly 400 in maximum temperature, by cutting off the 40 per cent, excess of air which was being used. Problem 74. The new style Siemens furnace for small plants has the gas producers built in as part of the furnace, between the two re- generators. The furnace has only one regenerative chamber at each end for preheating the air used for burning the gas, while the latter comes to the ports hot directly from the pro- ducers. This plant is compact, economical of fuel, allows high temperatures to be reached, and occupies little floor space. As compared with the old style separate furnace and producer plant, it occupies about half the floor space, costs about 60 per cent, as much to build and uses about 60 per cent, as much coal. They are largely used abroad for melting steel for cast- ings in small foundries and for melting pig iron for castings to be subsequently annealed to malleable castings. Such a furnace melted a charge of 3,000 kilograms of cast iron in 4 hours, using 750 kilograms of coal. The cast iron was charged cold, at 0, and run out melted at 1,450, con- taining 300 Calories of sensible heat per kilogram. The coal and gases were of the following compositions: Coal C 75, H 5, O 12, H 2 O 2; ash 6. Producer Gas CO 20, CO 2 5, CH 4 2, H 2 16, N 2 57. Chimney Gas CO 2 19, O 2 1.8, N 2 79.2. The air coming to the furnace is at and dry ; a steam blower is used to run the producer, using 1 kilo, of steam per 6 kilos, of air blown in, and the raising of this steam requires 0.1 kilo of coal used under boilers. The ashes produced weigh 75 kilos, per heat run. Temperature of the hot producer gas entering the furnace laboratory 600, of preheated air 1,000, of pro- ducts leaving laboratory 1,400, of products entering chimney flue 350. Required : (1) A heat balance sheet of the furnace as a whole. (2) The net thermal efficiency of the furnace. (3) The net thermal efficiency of the laboratory of the fur- nace. (4) The net thermal efficiency of the regenerators. EFFICIENCY OF OPEN-HEARTH FURNACES. 419 (5) The theoretical flame temperature in the furnace. Solution : (1) Heat balance sheet per charge of 3,000 kilos. Heat Available. Calories. Calorific power of coal used in producers 5,248,500 = = 2.3 ohms. Using this resistivity, the voltage drop across the electrodes, at starting, will be V c = 2.3X4X0.0250 = 0.23 volts. (1) (0.0250 amperes goes across each square centimeter and the plates are 4 centimeters apart.) (2) With H 2 SO 4 only 5 per cent, and copper contents, say, 20 per cent, because of solution of copper by free acid, we would have ~ = 0.21 + 0.05 = 0.26 Using this resistivity, the voltage drop is V* = 3.6X4X0.0250 = 0.36 volts. (2) In actual practice an increase of 0.01 volt above this might be expected, because of the formation of a film of slime on the surface of the anode. (3) The solution at starting has a specific gravity of 1.20 and a specific heat of 0.875. (Combination of data from Landolt-Bornstein-Meyerhoffer's Tabellen.) The water value as heat absorber of a column of liquid 1 cm. 2 and 4 cm. long, between the electrodes, is therefore 1.20X4X0.875 = 4.2 calories per 1 C. Heat equivalent of the electric energy expended 0.23X0.0250X0.2389 = 0.00138 calories per 1". 562 METALLURGICAL CALCULATIONS. Rate of rise of temperature of solution . 00138 -5- 4 . 2 = 0.00033 per second = 0.0198 " minute = 1.2 " hour (3) = 28.8 " day (4) With half the thickness of anode and cathode to be sup- plied with heat, the copper having a specific gravity of 8.9 and specific heat of 0.093, the water value of the copper concerned per square centimeter area of electrolyte section is (-^- + -^-)x 8.9X0.093 = 2.276 calories per 1 \ 2i 2i / Rate of rise of temperature of solution plus electrodes: 0.00138 ^-(2. 276 + 4. 2) = 0.00021 per second = 0.013 " minute = 0.78 " hour = 18.7 " day (4) Energy Lost in Contacts. This is an extremely variable and yet important factor in the economics of copper refining. No one item in copper refining will at the present time better repay close attention and study of methods of improvement. A proper contact should cause a very small voltage drop to begin with, and should be capable of being kept efficient that is the chief requirement. Con- tacts are often made on the hit or miss style, and arranged in such position as to receive spattering or drippings from the electrolyte; such may cause immense losses of electrical energy as they rapidly pass from bad to worse. B. Magnus (Electrochemical and Metallurgical Industry, De- cember, 1903), and L. Addicks (idem., January, 1904) have given us valuable information on this point. The weight of the plates, when resting on the contacts, improves the contact; in such cases the anode contacts are best when the tank is first set up and the cathode contacts worst. In such cases a screw clamp, to put several hundred pounds of mechanically applied pressure on the joint, should be adopted, so as to make the contacts " best " all the time. THE METALLURGY OF COPPER. 563 Figures given by Magnus as a carefully determined average of results in a western refinery are: Current 4,000 amperes Current density (per sq. ft.). . 11 Total voltage per tank 0.230 volts Energy used per tank 0.920 kw. Per Drop bus-bar to anode rod. 0.0270 volts Cent. anode rod to anode _ Q QOQQ _ 147 hanger 0.0060 " anode hook to anode. 0.0008 " anode to cathode. . . . 0.1782 " = 0.1782 = 77.5 " cathode to cathode rod 0.0045 " J " cathode rod to bus- L = 0.0180 = 7.8 bar.. 0.0135 " 0.2300 100.0 It thus appears that in this tank the loss in contacts was some 20 per cent of the total voltage employed (assuming the anode rods to be of such design that their resistance was neg- ligible). Magnus fitted up a similar tank with mercury cup contacts, and found a drop from bus-bar to anode rod of 0.0050 volt instead of 0.0270, and cathode rod to bus-bar 0.0050 instead of 0.0135 volt. If it were possible in the case of above tank to make these improvements alone, the saving on voltage per tank would be 0.0305 volt, or 13.3 per cent of the voltage used, which means a saving of 13.3 per cent of the power required to run the tank. Problem 120. In an electrolytic refinery system of 200 tanks the resistance of the contacts was, on an average, 0.0368 volt per tank, and 0.2567 volt across two electrodes; 3,800 amperes was passed through the series, and the voltage at the terminals of the dynamo was 67 volts. Power costs $157 per kilowatt-year, Efficiency of deposition of copper 82 per cent. Required : (1) The cost of power per ton (2,000 pounds) of copper refined. (2) The cost per ton of copper produced, of the power lost 564 METALLURGICAL CALCULATIONS. by (a) , resistance of the conductor bars (b) , by the contacts and (c) , consumed in the electrolyte itself. (3) If interest on capital is 6 per cent, what expenditure would be justified on means for (a), reducing the conductor losses (b), reducing the contact losses (c), reducing the elec- trolyte losses. (4) If the ampere efficiency of deposition could be increased 5 per cent by reducing the current density 20 per cent, would it pay? Solution : (1) Copper deposited in 1 tank per day 3800X0.82 = 3118 oz. = 195 Ibs. In 200 tanks 200X195 = 39,000 " Power used 3800X67+1000 = 255 kw. Cost of power per day 255 X $157 +-365 = $109.70 Cost of power per 2,000 Ibs. copper $109.70 + 19.5= $5.63 (1) (2) The voltage lost in the conductors is 67 - 200(0.0368 + 0.2567) = 67 - 58.7 = 8.3 volts and the power thus lost 3800X8.3+1000 = 31.54 kw. Cost per year 31.54X $157 = $4,952.00 Cost per day $4952+ 365 = $13.57 Cost per ton of copper produced = $0.70 (2a) Power lost in the contacts 200 X . 0368 X 3800 + 1000 = 27.97 kw. Cost, per year = $4,391.00 Cost, per day = $12.03 Cost, per ton = $0.62 (26) Power consumed in the electrolyte 200 X . 2567 X 3800 + 1000 = 195 kw. Cost, per year = $30,615.00 Cost, per day = $83.88 Cost, per ton = $4.30 (36) THE METALLURGY OF COPPER. 565 (3) The cost of the power lost per year, divided by 0.06, will give the capital investment representing that yearly outlay. This would be Cost of Power Capitalized. Conductor resistances $4,952 $82,530 Contact resistances 4,391 73,180 Electrolyte resistances 30,615 510,250 In the specific cases in point each 1 per cent of the power consumption which could be saved by improvements of a per- manent character, not subject to depreciation, would* justify capital expenditures in the plant up to the following amounts: Conductor .resistances -. $825 Contact resistances 732 Electrolyte resistances 5,100 If the improvements were liable to depreciation in use, such as larger tanks, contact clamps, etc., the depreciation must be added to the inte'rest to find the justifiable capital expenditure. Assuming depreciation 10 and 20 per cent, respectively, this plus interest amounts to 16 and 26 per cent, and the justified expenditures must be less than the following amounts for 1 per cent saving of power in each direction: 10% 20% Depreciation. Depreciation. Conductor resistances $309 .50 $190 . 50 Contact resistances 274 . 50 169 . 00 Electrolyte resistances 1,913 . 50 736 . 00 There is here a wide field for experiment and improvement. In the case in point, making all allowances, one would be justi- fied in expending in permanent plant for each tank at least $1.00 for every 1 per cent reduction which could be made in the power loss in the conductors, $0.85 for every 1 per cent gained in reducing contact resistances and $3.68 for each 1 per cent reduction in the resistance of the electrolyte. To be more specific, assume the main conductors to be proportioned to carry 2,000 amperes per square inch of cross-section. Their resistance in the case in point is 8. 3 -5- 3, 800 = 0.0022 ohm, cross-section 1.9 square inches, resistance per running inch 0.00000067 -f- 1.9 = 0.00000035 ohm, and total length 0.0022-v- 0.00000035 = 6,286 inches = 524 feet. 566 METALLURGICAL CALCULATIONS. The weight, at 62.5 X 8.9 -v- 1,728 = 0.322 Ibs. per cubic inch is 6,286X1.9X0.322 = 3,846 Ibs. which at $0.20 per Ib. costs 3,846X0.20 = $769.20 Since there is practically no depreciation (except change in market value) in these copper conductors, if we increased the cross-section of the conductors 100, 200, 300, and 400 per cent, respectively, we would decrease their resistance 50, 67, 75 and 80 per cent, respectively, and arrive at the following figures:' Per Ton of Copper Produced. Per Cent \ Interest Per Cent Decreased on Increased Resistance Increased Increased Decreased Size of of Cost of Cost of Cost of Conductors. Conductors. Plant. Plant. Power. Gain. 100 50 $769 $0.0065 $0.35 $0.35 200 67 1,538 0.013 0.47 0.46 300 75 2,307 0.019 0.52 0.50 400 80 3,077 0.026 0.56 0.53 900 90 6,923 0.058 0.63 0.57 1,900 95 14,613 0.123 0.67 0.54 We thus see that for this plant, with power costing $157 a kilowatt-year and copper bar $0.20 per pound, a great reduc- tion in the power costs up to $0.57 per ton of copper produced can be made by making the main conductors ten times as large ; that is, by making them carry only 200 amperes per square inch of section instead of 2,000. At a locality like Great Falls, Montana, where the water power does not cost over one-tenth the above price, let us say $15.70 per kilowatt-year, the decreased cost of power, in the above table would be one-tenth the above figures there would be only $0.07 per ton of copper to save if all were saved, and it can easily be seen that the gains would be for 100 per cent increased size $0.029 "200 " " " 0.034 "300 " " " 0.033 400 " " " 0.030 THE METALLURGY OF COPPER. 567 We reach the maximum saving in this case by increasing the size 200 per cent, giving a current density of 660 amperes per square inch of conductor section. It can easily be seen that the cost of copper bars, rate of in- terest, and cost of power are all three factors in determining the economic size to be given the conductors. In general they are made much smaller than they should be because of lack of capital to invest in them. As soon as a refinery is run- ning and paying dividends part of its profits should be ap- plied to permanent investment in heavier copper conductors in order to reduce the running costs. (4) If the ampere efficiency were increased 5 per cent and the current density 20 per cent, all other data being constant except those affected by these changes, the output of copper would be 87 per cent of the theoretical output of 3,040 amperes instead of 82 per cent of the theoretical output of 3,800 am- peres. The output of copper per day would therefore be, in place of 39,000 pounds: 3,040X0.87X200-;- 16 = 33,060 Ibs. per day. The power cost will be reduced 20 per cent by the lower amperage used, that is, be $5.63X0.80 = $4.50 per ton a sav- ing of $1.13. Whether it will pay to do this depends of what the other fixed charges of the plant are. Decreasing the out- put 15 per cent increases the fixed charges per ton of copper produced 12 per cent. It is then a question whether 12 per cent of the fixed charges in the first case is greater or less than $1.13. If the fixed charges amounted to, say, a usual figure of $5.00 per ton, decreasing the output 15 per cent would in- crease this charge per ton of copper produced 12 per cent, or $0.60; but if there is a power saving of $1.13 there is a mar- gin of $0.53 per ton in favor of making the change. If the power cost were only $0.56 per ton of copper it would just barely pay to make the change. Everything depends upon the relative cost of power and of the other fixed charges. Energy Lost in Conductors. We have already touched on this subject, but it deserves still greater attention. The resistivity of copper is 0.00000167 ohm per centimeter cube, or 0.00000067 ohm per inch cube. 568 METALLURGICAL CALCULATIONS. Given a conductor of certain length its resistance in ohms is 0.00000167 multiplied by its length in centimeters and divided by its cross sectional area in square centimeters. The similar calculation can be made, for the inch units. The drop in po- tential in the bus-bars is their resistance multiplied by the amperes passing, and the power consumption is the voltage drop into the amperes passing or the resistance in ohms into the square of the amperes passing. The watts thus expended may be expressed in kilowatts, or transposed into horse-power by dividing by 746. The heat generated in the conductors is, per second, in gram calories, the number of watts expended in them multiplied by 0.2389. The rate of rise of temperature of the conductors, leaving out radiation and conduction to the air, will be, per second, the gram calories generated in the con- ductors divided by the heat capacity of the conductors per de- gree, which in the case of copper is volume in cubic centi- meters into 0.837. This rise in temperature will continue until by radiation and conduction to the air the conductors lose as much heat per second as the current generates in them. This will vary with the shape of the cross section of the conductors, because the loss of heat is proportional to the surface exposed, and the area of exposed surface for a given cross section is greater the thinner the conductor and is a minimum for a round conductor. The amount of heat thus lost can be cal- culated by the principles explained in Metallurgical Calcula- tions, Part I., pp. 172-186. If it is desired to get along with the minimum heating of the conductors, they should be as flat and thin as possible; if heating can be ignored they may be of the cheap round shape. It should not be forgotten also that the resistivity of copper increases as it gets hotter in direct proportion to its absolute temperature (C + 273), so that the heating effect really aggravates itself from this increased resistance. Problem 121. A copper conductor is desired to carry 4,000 amperes, 300 meters. Round bars sell at 18 cents per pound. Interest rate 10 per cent. Cost of electric power as delivered $50 per kilo- watt year. Average temperature of surrounding air 20 C. ; air still. Conductivity of the copper 600,000 reciprocal ohms per centimeter cube. Specific gravity of copper 8.9, specific heat 0.094. THE METALLURGY OF COPPER. 569 Aluminium conductors to replace same have conductivity of 375,000 reciprocal ohms, specific gravity 2.60, specific heat 0.230; cost of round bars 36 cents per pound. Required : (1) The cross-sectional area of the conductors of copper which will give the minimum cost of power and interest on the investment. (2) The same for aluminium. (3) The running temperature of the conductors of copper above the room temperature. (4) The same for aluminium. Solution : (1) Let S = cross section of conductor, in sq. cm. Resistance of copper conductor 0. 00000167 X (300 X 100) -S = 0.0501 oh ^ o Power expended in the conductor: 0.0501 801,600 ^ X (4,000) 2 = ^ watts. o o Cost of power thus expended, per year 801,600 40,080 A -r- 1,000X50 = ^ dollars. S Weight of copper in conductor: (300 X 100) X S X 8 . 9 -*- 1 ,000 = 267 X S kilograms. Cost of conductor, if round 267. SX 0.18X2.205 = 106 XS dollars. Interest on cost of conductor, at 10% = 10.6XS dollars. Sum of cost of power and interest on investment: 4 ' 080 +( 10.6XS) S Solving for S a minimum, we find S = 61.5 sq. cm. (1) Total costs, substituting S above $652 (for power) +$652 (interest) = $1,304 per year. 570 METALLURGICAL CALCULATIONS. (2) Resistance of aluminium conductor 0. 00000267 X (300X100)^-8 = ^ 801 ohm. o Power expended in the conductor 0.0801 ^x, nnn v, 1,281,600 g X (4,000) 2 = g - watts. Cost of power thus expended, per year 1,281,600 64,080 , - ~ - -^1,000X50 == ^ dollars. o o Weight of aluminium in conductor (300 X 100) X 8x2.64-1,000 = 78x8. kilograms. Cost of conductor, if round 78. 8. X 0.36X2.205 = 62x8. dollars. Interest on cost of conductor, at 10% = 6.2x8. dollars. Sum of cost of power and interest on investment: Solving for S = a minimum: S = 101.2 sq. cm. (2) Total costs, substituting S above $633 (for power) +$627 (interest) = $1,260 per year, (3) Power expended in copper conductor = 13,035 watts. bl.o Heat generated in conductor, per second 13,035X0.2389 = 3,114 gm. cal. Area of conductor surface (300 X 100) X 3. 14 (x/61.5 4- 0.7854) = = 30,000X27.8 (cm. circumference) = 834,000 sq. cm. THE METALLURGY OF COPPER. 571 Loss of heat per second by contact with still air, per 1 difference 0.000056X834,000 = 46 . 7 calories. Rise of temperature, excluding radiation loss 3,114*46.7 = 67. The radiation loss must, however, be considered. From copper it is 0.00068 gram calories per second per sq. centi- meter, if the surroundings are at and the hot body at 100, this is approximately 0.0000068 gram calories per each degree in this range. We have the combined conduction and radia- tion losses per 1 per second . 000056 + . 000007 = . 000063 calories and the rise in temperature of the conductor 3,1 14 -KG. 000063X834,000) = 59. The conductors would therefore run at a temperature of 20 + 59 = 79 C. (3) = 174 F. A correction of the second order would be to take into con- sideration the fact that at 79 the copper would have less con- ductivity than at 20, its resistivity at 79 being 0. 00000167 X = 0.0000020 ohm. This would make the required section for minimum expense 68 sq. cm. instead of 61.5, and would, by increasing the sur- face, make a different rise in temperature. The power ex- pended would be the equivalent of 3,373 gram calories per second, and the revised temperature of the conductor would be (3,373 -r-52. 54) +20 = 64 + 20 = 84 C. (4) Power expended in the aluminium conductor =12,665 watts. Heat generated in conductor, per second 12,665 X . 2389 = 3,026 calories. 572 METALLURGICAL CALCULATIONS. Area of conductor in square centimeters = 1,074,000 sq. cm. Conduction and radiation losses, per 1 (0.000056 + 0. 000010) X 1,074,000 =70.88 grm. cal. Rise in temperature of conductor 3,026-^70.88 = 43. Working temperature of conductor 43 + 20 = 63 C. The aluminium conductor thus is seen to work cooler than the copper, and to admit of a lower minimum of working costs. v Electric Smelting of Copper Ores. The use of electrothermal processes for smelting copper ores has been proposed in recent years, and in a few cases experi- mentally tested. The whole question depends on the recog- nition and appreciation of a few fundamental facts. By far the greater bulk of all copper ores are sulphide ores, and in many cases the sulphur and iron present are sufficient heat producers to allow of the smelting of the ore simply by the heat of their oxidation, if the operation is skillfully conducted. Such is the basic principle of pyritic smelting, and whenever it can be applied it is very economical to do so, and electric processes have no chance of an application. Other ores of copper contain so small an amount of the metal that the result of the smelting down of the whole to a fluid mass, by any method of fusion, would not repay the cost of the operation. Some such ores can be treated by aqueous and other methods not involving fusion, with a margin of profit to pay for the operation. The only field for electrothermic processes appears to be in the smelting down of ores carrying too little sulphur to be " pyritically " smelted, and which require, as usually treated, the use of carbonaceous fuel to assist the fusion. Wherever such fuel is expensive, and electric power may be obtained at a low price, an electrothermic process might be theoretically possible and profitable. Such conditions may easily occur in the vicinity of copper mines. Situated frequently among the mountains, remote from THE METALLURGY OF COPPER. 573 railways and cheap fuel, they frequently are near large water powers which would furnish cheap electric power. Some may be even so situated that concentration of a very lean sulphide ore to matte by use of fuel, or by mechanical concentration, would be unprofitable, and yet a concentration or simple melt- ing to matte by electrically-generated heat be profitable. There are, so far as the writer knows, no electrothermal processes yet in commercial operation on copper ores, yet they have been tested experimentally with promising results. Vat- tier, for instance, conducted experiments at the Li vet works of the Compagnie Electrothermique Keller et Leleux, on April 23, 1903, in the presence of several distinguished metallurgists, such as Mr. Stead, of Middlesborough ; Mr. Allen, of Sheffield, and Mr. Saladin, of the Creusot works. The ores were Chilian ores sent by the Chilian Government to test the possibility of electric smelting, taken from the " Vulcan " mine, owned by G. Denoso, and low-grade ore from Santiago. At the mines, coke costs $20 per ton, but the slopes of the Andes afford a fine opportunity for developing large water powers cheaply; it is estimated that the total cost of electric power delivered should not exceed $6 per kilowatt-year. A detailed account of these tests can be found in the appendix to the Report of the Can- adian Commission on Electric Smelting, 1904. Problem 122. Vattier mixed his rich and poor Chilean ores to a charge consisting of Per Cent. Copper 5. 10 Sulphur. ...... 4. 13 Iron : 28.50 Manganese .7 . 64 Silica 23.70 Alumina 4 . 00 Carbonic acid 4.31 Lime 7.30 Magnesia 0.33 Phosphorus 0.05 This was charged directly into a shaft furnace 1.8 meters long, 0.9 meter wide and 0.9 meter high. The melted material 574 METALLURGICAL CALCULATIONS. ran out into a forehearth 1.2 meters long, 0.6 meter wide and 0.6 meter high. Two carbon electrodes, 30 centimeters square by 170 centimeters long, were used in the smelting chamber, and two electrodes, 25 centimeters square by 100 centimeters long, were put in the forehearth to reheat it for tapping. The matte obtained analyzed: Per Cent. Silica 0.80 Alumina . 50 Iron 24.30 Manganese 1 . 40 Sulphur 22.96 Copper 47.90 The slag contained : Silica ..27.20 Alumina 5 . 20 Lime 9.90 Magnesia 0.39 Iron 32.50 Manganese 8.23 Sulphur 0.57 Phosphorus 0.06 Copper 0.10 The current used was 4,750 amperes at 119 volts; power factor = 0.9; 8,000 kilograms of ore mixture were smelted in 8 hours; consumption of electrodes 50 kilograms. Required : (1) A balance sheet of materials entering and leaving the furnace per hour. (2) A balance sheet of the heat development and distribution in the furnace. (3) The saving in cost of treatment per ton of ore, assuming electrodes to cost 4 cents per kilogram, coke $20 per metric ton, and that when coke is used one-third of its calorific power leaves the furnace in the escaping hot gases. Assume all other costs similar, except that air blast costs $0.10 per ton of ore smelted additional. Electric power $6 per kilowatt-year. (4) If the resulting slag were smelted electrically for ferro THE METALLURGY OF COPPER. 575 silicon, at an expenditure of 0.75 electric horse power-year per ton of ferro-silicon obtained, and other smelting costs were $10 per ton, how much ferro-silicon would be obtained per ton of copper ore treated, and what would be its cost? Solution : (1) Balance sheet per 1,000 kg. of ore. Charges Matte Slag Gases Ore: 1,000 kg. Silica 237 kg. 1 236 Alumina 40 " 0.5 39.5 Iron 285 " 25 260 Manganese 76 " 1.5 74.5 Sulphur 41 " 24 5 , 12 Copper 51 " 50 1 .... Carbonic acid 43 " 43 Lime 73 " 73 Magnesia 3 " 3 Oxygen '. 137 " 2 74 61 Electrode : Carbon.. 6 " 6 104 766 122 Composition of Slag. Calculated. Analyzed. Silica 29.5 per cent. 27.2 per cent. Alumina 5.0 " 5.2 Lime 9.2 " 9.9 Iron 32.5 " 32.5 Manganese 9.3 " 8.2 Sulphur 0.6 " 0.6 Heat Development. Electric energy per hour: 4,750 X 1 19 X . 9 = 509 kilowatt-hours. 509X860 = 437,750 Cal. per hour. Oxidation of carbon: 6X4,300 = 25,800 " " a 463,550 a 576 METALLURGICAL CALCULATIONS. Heat Distribution. Heat in matte : 104X270 = 28,080 Calories. Heat in slag: 796X400 = 318,400 Heat in gases = 30,000 " Loss by radiation and conduction = 87,070 " 463,550 The radiation and conduction loss is 19 per cent of the total heat generated in the furnace a satisfactory showing. The useful effect of the furnace is nearly 75 per cent, reckoning as usefully applied heat that contained as sensible heat in the melted matte and slag. Nearly 90 per cent of this usefully applied heat is contained in the slag. (3) If coke were used instead of electric heat, enough coke would have to be used to furnish 433,550 Calories plus the heat in the gases. Since the latter is assumed to be one-third of its calorific power, the coke must have a calorific power of 650,325 Calories, which at 7,000 Calories per kilogram would require 93 kg. of coke per 1,000 of ore smelted. The costs of coke and blowing engine would therefore be Coke, 93X0.02 = $1.96 Engine, 0.10 Sum $2.06 The cost of electrodes and electric energy, which would replace coke and blowing engine, would be: Electrodes, 6 kg. X $0.04 =$0.24 Power 509kw.X A= 0.35 Sum, $0.59 Saving, $1.47 (4) The resulting slag contains, according to the balance sheet, 260 kg. of iron, 74.5 kg. of manganese, and 236 kg. of silica equal to 126 kg. of silicon. It should produce, if com- THE METALLURGY OF COPPER. 577 pletely reduced, 460 kg. of alloy (omitting the carbon it might contain) of the approximate composition: Fe 260 kg. = 56 per cent. Mn 74 kg. = 13 Si 126 kg. = 28 a 460 kg. Cost of power $6.00 Xj = $4.50 per ton. Other costs 10.00 " Total costs per ton $14 . 50 This cost is very low, because of the very low figure assumed for power cost. However, the whole calculation points to the' possibility of great saving in some localities in smelting down copper ores by electrically applied heat, and the possi- bility of cheaply producing, at the same spot, valuable ferro- silicon as a by-product. CHAPTER II. THE METALLURGY OF LEAD. The extracting of lead from its ores and the refining of the crude metal to commercial lead, constitutes the metallurgy of lead. The chief ore is galena, lead sulphide, PbS; but the carbonate, cerussite, PbCO 2 , and the sulphate, Anglesite, PbSO 4 , occur at some places in important quantities; the silicate, phos- phate, molybdate, tungstate, chloride and native lead are rare minerals. The reduction of the oxidized lead ores, such as often occur at the outcrops of sulphide veins, is very simple; carbon, the cheapest and most universal reducing agent, reduces them satisfactorily at a red heat, with some loss of lead by volatiliza- tion. The sulphide of lead, however, is not reducible by carbon; it requires other treatment. If we tabulate the most common sulphides according to their thermochemical heats of formation, expressed per unit weight of sulphur held in combination (32 kilograms), we find the common metals arranged as follows: Calories. Potassium (K 2 , S) 103,500 Calcium (Ca, S) 94,300 Sodium (Na 2 , S) 89,300 Manganese (Mn, S) 45,600 Zinc (Zn, S) 43,000 Cadmium (Cd, S) 34,400 Iron (Fe, S) 24,000 Cobalt (Co, S) 21,900 Copper (Cu 2 , S) 20,300 Lead (Pb, S) 20,200 Nickel (Ni, S) 19,500 Mercury (Hg, S) 10,600 Hydrogen (H 2 , S) 4,800 Silver (Ag 2 , S) 3,000 578 THE METALLURGY OF LEAD. . 579 A glance at this table shows us that some metals unite with sulphur more energetically than lead does, and others less energetically, and points to the theoretical possibility of de- composing lead sulphide by the agency of copper, cobalt, iron, cadmium, zinc, etc. Of these agents, iron is, of course, the cheapest and most available, and can be used to reduce lead sulphide with formation of iron sulphide. PbS + Fe = FeS + Pb - 20,200 + 24,000 Showing an excess heat development per 207 parts of lead set free, of 24,000-20,200 = 3,800 Calories. If lead sulphide and iron are brought together at a red heat, at which heat the sulphide is molten, they react energetically with evolution of heat enough theoretically to raise the temperature of the 207 kg. of lead and 88 kg. of iron sulphide some 280 C. The reaction is fairly complete if sufficient reducing agent is present and time is allowed, so that it can be used in assaying to deter- mine lead by fire assay, but in commercial practice more or less undecomposed lead sulphide always remains in the iron sul- phide, forming a sort of double sulphide or iron-lead matte. It is interesting to note, en passant, that lead vigorously reduces silver sulphide, the liberated silver alloying with the excess of lead. This reaction is accompanied by a large evolu- tion of heat, as the table shows, is the basis of the ordinary assaying methods for silver ores, and is used commercially to extract silver from its ores wherever lead 'is plentiful. The affinity of lead for oxygen is a no less interesting sub- ject, since it concerns not only the reduction of oxide ores but also the roasting of sulphide ores to oxide, and very largely controls the refining of impurities from lead by methods in- volving oxidation. The heat of combination of some of the more common elements with oxygen, expressed per unit weight of 16 kilograms of oxygen held in combination, is as follows: Calories. Magnesium (Mg, O) 143,400 Calcium (Ca, O) 131,500 Aluminium J (AP, O 3 ) . 130,870 Sodium (Na 2 , O) 100,900 Silicon i (Si, O 2 ) 98,000 580 ;" METALLURGICAL CALCULATIONS. Calories. Manganese (Mn, O) 90,900 Zinc (Zn, O) 84,800 Tin }(Sn, O 2 ) 70,650 Iron (Fe, O) 65,700 Iron J(Fe 2 , O 3 ) 65,200 Nickel (Ni, O) 61,500 Hydrogen (H 2 , O) 58,060 Antimony J(Sb 2 , O 3 ) 55,630 Arsenic J(As 2 , O 3 ) 52,130 Lead (Pb, O) 50,800 Carbon |(C, O 2 ) 48,600 Bismuth J(Bi 2 , O 3 ) 46,400 Copper (Cu 2 , O) 43,800. Sulphur }(S, O 2 ) 34,630 Sulphur i(S, O 3 ) 30,630 Carbon (C, O) 29,160 Mercury (Hg, O) 21,500 Silver (Ag 2 , O) 7,000 Reduction of Lead Oxide. An inspection of above table shows that lead oxide is a weak oxide, weaker than the oxides of most of the common metals. It is reduced to metallic lead by many reagents with evolution of heat. It is also reduced by some weaker reagents with absorption of heat, provided that the oxide formed is gaseous and the necessary heat energy is supplied from outside. For instance, PbO + C = Pb + CO -50,800 +29,160 involves an absorption of 21,640 Calories, or several times as much heat as is necessary to raise the reacting substances, PbO and C, to the reacting temperature, a red heat. The reaction is endothermic, absorbing heat, and therefore only progresses in measure as the necessary calories are supplied from outside. The fact that all the substances concerned are liquid or solid except the product CO, gives a predisposing cause which facilitates the reaction, that is, the reaction can only go one way, since the CO gas escapes from the sphere of action as soon as formed. THE METALLURGY OF LEAD. 581 Oxidation of Lead Sulphide. When PbS is roasted, that is, heated to redness with free access of air, both the sulphur and the lead tend to oxidize, generating a large heat of oxidation, against which there is absorbed only the comparatively feeble heat of decomposition of PbS. The equation may be discussed as 2PbS + 3O 2 = 2PbO + 2SO 2 - (20,200) + 2 (50,800) +2 (69,260) The net heat evolved in the equation is 240,120 40,400 = 199,720 Calories, which is 33,290 Calories per unit weight (16 kg.) of oxygen used, or 418 Calories for each kilogram of lead sulphide oxidized. This great heat of oxidation, evolved in roasting, is quite sufficient, in fact more than sufficient, to provide the heat required for self-roasting without the use of any other fuel, the chief difficulty is really to keep the charge from getting too hot and melting everything down to a liquid before the roasting is anywhere near complete. In the ordinary hand-worked reverberatory roaster, the oxidation is so slow that the fire on the grate really controls the temperature on the hearth, and the temperature of the roasting ore can be regu- lated accordingly. Where the roasting is done quickly, by an air blast, as in " Pot Roasting," the temperature is kept down somewhat by previously roasting off part of the sulphur, or by liberally wetting the charge, or by using limestone in with the ore, to absorb by its decomposition into CaO and CO 2 a large part of the excess heat while the melting down of the charge is prevented by the mechanical interference presented by inter- mixed, inert and infusible material, such as silica, lime, etc., which simply prevents the really melted globules of sulphide from running together and thus melting down to a liquid mass. An interesting variation is the roasting of lead sulphide to sulphate; some of this always forms, due to the high formation heat and difficult decomposability of the sulphate. PbS + 2O 2 =PbSO< - (20,200) + (215,700) The net heat evolved is 215,700-20,200 = 195,500 Calories, or 48,875 Calories per unit weight (16 kg.) of oxygen used, or 818 Calories for every kilogram of lead sulphide so oxidized. 582 METALLURGICAL CALCULATIONS. This is a high heat of oxidation, and explains the great ten- dency to form sulphate observed during roasting. If the con- ditions could be found whereby only this reaction occurred, the sulphate roasting could be easily made automatic without out- side fuel. There is needed, at the present time, a careful lab- oratory investigation of the conditions, mechanical, physical, chemical and thermal, for the roast exclusively to sulphate just as a bit of badly needed metallurgical information. Double Reactions. A large part in the metallurgy of lead is played by double reactions, such as the following: PbS + PbSO 4 = 2Pb + 2SO 2 PbS + 2PbO = 3Pb + SO 2 PbS + 3PbSO 4 = 4PbO + 4SO 2 On roasting lead sulphide for a short time, either lead or lead sulphate or mixtures of these are formed, according to the temperature and excess of air provided. With large excess of air and low temperature, and especially in presence of infusible materials which act as catalyzers (i.e., which promote the union of SO 2 with O, and consequent formation of SO 3 and PbSO 4 ), sulphate may be formed almost exclusively. If the temperature is then raised, the tendency of PbSO 4 to react upon the undecomposed PbS rapidly gets stronger, until at an orange heat this takes place rapidly and fairly completely, forming the one single gaseous product, SO 2 . PbS + PbSO 4 = 2Pb + 2SO 2 - (20,200) - (215,700) +2(69,260) Net deficit, 97,380 Calories. PbS + 2PbO = 3Pb + SO 2 - (20,200) - 2(50,800) + (69,260) Net deficit, 52,540 Calories. In both these cases the well-known phenomena of an endo- thermic reaction are manifest the high temperature and strong firing necessary, and the formation of a single gaseous product from non-gaseous materials, assisting the reaction. The reaction : PbS + 3PbSO 4 = 4PbO + 4S0 2 - 20,200 - 3(215,700) +4(50,800) +4(69,260) Net deficit, 187,060 Calories, THE METALLURGY OF LEAD. 583 is supposed to take place in " pot roasting " where excess of air is blown through the finely divided material, but it is too endo- thermic a reaction to take place in a pot-roasting operation to more than a very subsidiary extent. Oxidation Refining. Impure lead is refined or " softened " by oxidation at a red heat. We must expect a great deal of lead to be oxidized in this operation, simply because of its pre- ponderating mass, but the impurities present will oxidize rela- tively faster or slower than lead in proportion as their affinity for oxygen is relatively greater or less. The skimmings or slags obtained during softening are always principally com- posed of lead oxide, PbO, but they come off containing, in order, zinc, tin, antimony, arsenic, bismuth and small quantities of silver. During cupellation down to silver, which is con- tinued oxidation until all the lead is oxidized, bismuth oxide is concentrated in the last parts of lead oxide formed, which may also carry silver in small amount; before this happens, how- ever, the lithage, formed is almost chemically pure. The converse of these oxidation reactions also holds, viz. : differential reduction. Taking a softening skimming rich in antimony, for instance, it is possible by mixing it with a small amount of reducing agent, such as carbon, to reduce out of it all the silver and considerable of the lead which it contains without reducing much antimony. This leaves the remaining unreduced material desilverized and poorer in lead, or richer in antimony, and on subsequent reduction of this by excess of reducing agent a rich antimony-lead alloy is obtained. The easy reducibility of lead oxide is complementary to the slow oxidation of lead ; both facts are clear from the heat of formation of the various metallic oxides, and both are extensively utilized in the metallurgy of lead. THE VOLATILITY OF LEAD. The melting point of lead is 326, its mean specific heat in the solid state 0.02925 + 0.000019t, heat in melted lead at its melting point 11.6 Calories, latent heat of fusion 4.0 Calories, heat in just melted lead 15.6 Calories, specific heat in the liquid state 0.042 and approximately constant boiling point at normal atmospheric pressure about 1,800 C., latent heat of vaporization, calculated by Trouton's rule (23 T) 47,680 584 METALLURGICAL CALCULATIONS. Calories per molecular weight = 230 Calories per kilogram, assuming the vapor monatomic, specific gravity of vapor 103.5, referred to hydrogen gas at the same temperature and pres- sure, or, theoretically, 9.315 kg. per cubic meter at C. and 760 mm. pressure, as a standard datum. The question of the volatility of lead at other temperatures than 1,800 C. is highly important in the smelting of lead ores, yet is practically an unknown quantity. The following is an at- tempt to calculate these data, so important in practical metal- lurgy. The vapor tension curve of mercury is known for very low and up to comparatively high pressures. A rule has been observed between mercury and water vapor, in that the absolute temperatures at which these two substances have the same vapor tensions are found to stand in the ratio 1.7 to 1 through a large range of temperatures. Since lead vapor is in all prob- ability monatomic, like mercury vapor, we will deduce the vapor tension curve of lead from that of mercury, using the constant ratio derived from the two temperatures at which their respective vapors have atmospheric tension, viz.: TPb 1,800 + 273 2,073 THg "~ 357 + 273 630 The following table gives the most reliable data for the vapor tension curve of mercury, and the corresponding data calculated for lead, assuming the constant ratio 3.3 between the absolute temperatures at which they have the same vapor tension : Tension of Vapor Mercury Lead mm.ofHg. C ' C 0.0002 625 0.004 33 735 0.045 67 844 0.28 100 954 1.47 133 1,064 5.73 167 1,173 18.25 200 1,283 50. 233 1,393 106. 267 1,502 THE METALLURGY OF LEAD. 585 Tension of Vapor Mercury Lead mm. of Hg. C C 242. 300 1,612 484. Atmospheres. 333 1,722 760. =1.0 357 1,800 849. =1.1 367 1,841 1588. =2.1 400 1,951 4.3 450 2,116 8.0 500 2,280 13.8 550 2,445 22.3 600 2,609 34. 650 2,774 50. 700 2,938 72. 750 3,103 102. 800 3,267 137.5 850 3,436 162.' 880 3,525 From the above table a vapor pressure curve of mercury and lead can be constructed. An examination of the data shows that lead is certainly volatile to a minute extent at a low red heat, and that a current of inert gas passing across the surface of melted lead at that temperature certainly carries away vapor of lead; at the melting point of silver the tension is only about one-quarter of a millimeter, or one three thousandth of an atmosphere, yet this means that each cubic meter of inert gas carries off one three-thousandth of its volume, or one-thirtieth of 1 per cent of its volume, of lead vapor. At 1,300, the tem- perature of a commercial zinc retort, or of a lead smelting furnace, the tension is about one-fortieth of an atmosphere, which would mean that any other gas or vapor could carry 2.5 per cent of its volume of lead vapor with it. It must be remembered, moreover, that such gas saturated with lead vapor if suddenly cooled does not deposit its excess of lead vapor as liquid lead, but that a suspension similar to hoar- frost almost inevitably results, the so-called " lead fume," which is simply molecularly divided liquid or solid lead carried in suspension by a current of gas. The lead vapor is thus al- most entirely carried out of the furnace without condensation and deposition. Looking at the higher pressures, we can understand why an 586 METALLURGICAL CALCULATIONS. explosive reaction results when PbO is reduced by finely di- vided aluminium, the " alumino thermic " reaction. The im- mense heat liberated in the reaction, 3PbO + 2Al = 3Pb + Al 2 3 , 220,000 Calories, raises the products to an electric furnace tem- perature, to some 3,000 C., at which temperature the lead vapor has a maximum tension, according to our table, of 60 atmospheres. No wonder, then, that when Tissier tried this test for the first time, in 1857, using a piece of sheet aluminium weighing less than 3 grams (0.1 ounce), " le creuset a e*te brise en mille pieces et les portes du fourneau projetees au loin " the crucible was broken into a thousand pieces and the doors of the furnace blown to a distance. ROASTING OF LEAD ORES. The principal operations in the metallurgy of lead are the roasting of the ore, its reduction to metal and the refining of the crude metal. The chief ore of lead being galena, PbS, the operation of roasting it in air converts it partly into PbO and partly into PbSO 4 . Since PbS is easily fusible, and is volatile per se at a yellow heat, it is necessary to roast carefully and slowly, avoiding high temperatures, which first make the ore sticky or pasty and afterwards fuse it, thus practically stopping the roasting reaction. The only manner in which rapid roasting can be done is by having the ore mixed with so much infusible inert matter, like lime, that the globules of melted sulphide cannot run together, but are kept isolated and continue to oxidize on their surfaces, the mass being meanwhile kept " open " or porous by the inert, infusible material, to allow the rapid passage of air through the mass. This is the principle of " pot roasting " the most radical improvement of recent years in the metallurgy of lead. Problem 123. A Savelsberg " pot roaster " treats 5 tons of ore mixture, consisting of 100 parts lead ore, 10 parts quartzose silver ore, 10 parts spathic iron ore, 19 parts limestone. The lead ore is galena ore, containing 78 per cent of lead and 15 per cent of THE METALLURGY OF LEAD. 587 sulphur; the silver ore may be called silica sand; the spathic iron ore FeCO 3 ; the limestone CaCO 3 . The mixture is mois- tened with 5 per cent of its weight of water. Air blast is kept constant at 7 cubic meters per minute, blower displacement at 15 C., and the operation lasts 18 hours, leaving 2 per cent of sulphur in the product, which may be assumed as undecom- posed PbS. Gases produced contain approximately 10 per cent of SO 2 and 5 per cent free oxygen; 50 kg. of charcoal is used to start the operation, and is assumed to be pure carbon. Required : (1) The weight. of product and its percentage composition. (2) The complete analysis of the gases escaping. (3) The efficiency of delivery of the blower. (4) The proportion of the heat generated by the oxidation of the ore which is absorbed in the decomposition of the lime- stone. (5) The proportion of the heat generated in the pot used in evaporating the moisture of the charge. (6) The proportion of the heat generated in the pot carried away by the gases at an average temperature of 300 C. (7) Make a heat balance sheet of the whole operation. Solution : (1) The components of the 5-ton (=5,000 kilos) charge are: Galena ........... 3597 kg. = Pb 2806 kg. S : 540 kg. Silver ore ........ 360 " = SiO 2 : 360 " Iron ore .......... 360 " = FeO : 224 " CO 2 : 136 " Limestone ........ 683 " = CaO: 382 " CO 2 : 301 " Approximate composition of product: ooq PbO ............... ....... .2806X = 3023 kg. SiO 2 ....................... = 360 " FeO ....................... = 224 " CaO ....................... = 382 " 3989 " But, since product contains 2 per cent of sulphur as unde- composed PbS, the above weight is only 99 per cent of the 588 METALLURGICAL CALCULATIONS. weight of the roasted ore, because the lead combined with this sulphur has been calculated above to PbO, and the O in PbO is only half the weight of the S in PbS. The above weight is short, therefore, by an amount equal to one-half the weight of the sulphur present, and since the latter is 2 per cent of the roasted ore the above weight is 1 per cent short. The cor- rected weight of the roasted ore is, therefore, . 3,989 -J- 0.99 = 4,029kg., and its sulphur content 81 kg., corresponding to 521 kg. of lead, or 602 kg. of PbS. This leaves present, as PbO, 2,806- 521 = 2,285 kg. of lead, equal to 2,285X223/207 = 2,461 kg. of PbO. Revised composition of product: 61 per cent PbS 602 " =15 SiO 2 360 " = 9 FeO 224 " = 5 CaO.. . 382 " = 9 4029 " (1) (2) The charge gives to the gases H 2 O 5000 X . 05 = 250 kg. CO 2 (50 X 3. 67) + 136 + 301 = 620 " S 540 - 81 = 459 " and the blast gives to the charge O 2461 - 2285 = 176 kg. The volume of SO 2 produced from 459 kg. of sulphur, which makes 918 kg. of SO 2 , will be, at standard conditions, (0.09XTH = 319 cubic meters. The free oxygen in the gases is half the volume of the SO 2 , and, therefore, will be exactly one-quarter of its weight. The volumes of H 2 O and CO 2 present in the gases, at standard conditions, are THE METALLURGY OF LEAD. . 589 H 2 O 250^- /O.OQXy) = 309 cubic meters. CO 2 620 -r- (o.OQXy) = 313 " The nitrogen present is that in the air used, or 10/3 the weight of oxygen in the air. The latter is the weight of oxygen in the PbO of the roasted charge, plus the oxygen necessary to burn the 50 kg. of charcoal, plus the oxygen oxidizing sulphur to SO 2 and the free oxygen in the gases. O in PbO = 176 kg. 00 O to burn C 50X ff = 133 " 00 O to burn S. 459X^ = 459 " oZ Free O 2 in the gases = 230 " Total 998 ' N 2 corresponding = 3327 " Composition of escaping gases: Per Cent. Nitrogen 3327 kg. = 2609 cu. meters = 70.3 Oxygen 230 " = 159 " " = 4.3 SO 2 918 " = 319 " " =8.6 CO 2 620 " = 313 " " = 8.5 H 2 O vapor 250 " = 309 " " = 8.4 3709 100. (2) (3) Oxygen delivered by blast 998 kg. Nitrogen in the blast 3327 " Air delivered 4325 " Volume at = 4325-1-1.293 = 3345 cu. meters. ooo Volume at 15 = 3345 X ~ = 3529 " Z i o 590 METALLURGICAL CALCULATIONS. Displacement of blower at 15 7X60X18 = 7560 " oc on Efficiency of blower = = 0.467 = 46.7 per cent. (3) oou (4) The heat generated in the oxidation of the ore consists of the heat of formation of the PbO formed, plus that of the SO 2 , and less that of the PbS decomposed. This is not the complete heat balance of the whole operation, but simply that concerned with the oxidation of the galena. Heat of oxidation of Pb to PbO 2285 kg. Pb.X245 = + 559,825 Cal. Heat of oxidation of S to SO 2 459kg. SX 2164 =+ 993,275 " Heat of formation of PbS decomposed 2285kg. PbX 98 =- 223,930 " Roasting heat = +1,329,170 " The decomposition of the limestone, producing 301 kg. of CO 2 , absorbs 301X1,026 = 308,825 Cal. The proportion of the roasting heat thus absorbed is 0.232 = 23.2 per cent. (4) (5) The 5-ton charge is moistened with 5 per cent of its weight of water, or 250 kgs., in order to keep down the tem- perature. This absorbs, simply in becoming vapor, 250X606.5 = 151,625 Cal., Which, on the heat generated in the roasting operation, is 151,625 1,329,170 - 0.114 - 11.4 per cent. (5) (6) If the gases pass away at an average temperature of 300, and the air comes in at an average temperature of 15, the THE METALLURGY OF LEAD. 591 gases cool off the pot to the extent of the difference between these two heat capacities: Heat in the air entering: 3345 m 3 X [0.303 + 0.000027(15)] X 15 = 15,225 Cal. Heat in gases escaping: ^n^i- X [0.303 + 0.000027(300)] - 861.1 Cal. per 1. u loy m ) SO 2 319m 3 X[0.36 + 0.0003(300)] = 143.6 CO 2 313m 3 X [0.37 + 0.00022(300)] = 136.5 H 2 O 309m 3 X [0.34 + 0.00015(300)] = 119.0 1260.2 Total = 1260.2X300 = 378,060 Cal. Proportion of the total roasting heat thus carried out: 362,835 1,329,070 0.280 = 28.0 per cent. (6) (7) Heat Available. Calories. Sensible heat of air blast, at 15 15,225 Combustion of charcoal 405,000 Net heat generated by the roasting 1,329,170 Total 1,749,395 Heat Distribution. Sensible heat in the hot gases 378,060 = 22 per cent. Decomposition of carbonates : Calcium carbonate 308,825 = 18 " Iron carbonate. . 76,980 =4 " Evaporation of moisture 151,625 =9 " Sensible heat of product, at 400. . . . 400,000 = 23 " Loss by radiation and conduction. . . 433,905 = 25 " 1,749,395 REDUCTION OF ROASTED ORE. If the roasted ore is reduced in reverberatory furnaces, the chances are that the undecomposed sulphide it contains will react with the oxide or sulphate, forming metal and SO 2 , and 592 METALLURGICAL CALCULATIONS. thus eliminating almost all of the sulphur left in the roasted ore; this results in the formation of no matte, or at most of very little rich matte. If the roasted ore is reduced in shaft furnaces, the carbonaceous fuel and strong reducing atmos- phere of the same, tend to reduce oxides to metal and sul- phates to sulphides before they reach the temperature at which they can react on sulphide, thus cutting out very largely the double reaction, and throwing into matte the larger part of the sulphur left in the roasted ore. This is highly advan- tageous if the ore has copper in it, even if only in small amount, as the matte will concentrate the copper in it, is easily sep- arated from the metal and the slag, and forms a valuable by- product. Since most lead ores contain some copper, enough to make it pay to save it, the roasting of these is never pushed to completion, and the reduction in low shaft furnaces, run at moderate temperature, is pursued with the object of pro- ducing a copper-iron matte. Any arsenic or antimony left in the roasted ore is likely also to form spiess, a compound of arsenic and antimony with iron, copper, nickel, cobalt, lead and silver, which is heavier than matte, but lighter than lead, and separates out in the cooling pots between these. It is a highly undesirable product, because of its complexity and the difficulty of satisfactorily and cheaply separating its constitu- ents; it is always to be advised to remove arsenic and anti- mony as completely as possible in the roasting operation, even at the expense of removing too much sulphur, and then to supply the sulphur needed in the smelting operation by mixing with the dead-roasted ore some raw sulphide ore free from arsenic and antimony, if such can be obtained. In calculating the charge for such a smelting operation, the roasted ore must be charged with such amounts of iron ore and limestone as will, together with the gangue of the ore and the ash of the fuel used, make an easily fusible slag. To produce such a slag is of fundamental importance, since it must melt and become thinly liquid at the moderate temperature neces- sarily prevailing in a lead furnace. Experience has shown that a typical lead slag will contain about 30 per cent SiO 2 , 40 per cent FeO and 20 per cent CaO, with 10 per cent allowed for other ingredients, such as APO 3 , etc., or in more general terms, that SiO 2 , FeO and CaO are best present in the propor- THE METALLURGY OF LEAD. 593 tions 30 : 40 : 20. The other limitations are that a certain amount of hand-picked slag is always returned to the furnace for re-smelting and purification from matte, and that the fuel, coke, cannot melt more than a certain amount, say seven times its weight of inert material. With these conditions in mind, we will work out a typical smelting problem, the data for which are taken from Hofman's Metallurgy of Lead. Problem 124. A mixture of lead carbonate ore and galena (which repre- sents pretty closely a partially roasted sulphide ore) is smelted with the addition of iron ore and limestone. The charges for the furnace consist of 1,000 pounds of burden (material to be smelted) and 150 pounds of coke (containing 10 per cent ash), and the 1,000 pounds consist of 100 pounds return slags and 900 pounds of lead ore, iron ore and limestone. The percentage composition of these materials is : Lead Ore. Iron Ore. Limestone. Ash of Coke. SiO 2 32.6 4.3 2.7 40.3 FeO 14.8 72.4 4.5 26.5 MnO 4.3 1.7 CaO 2.2 3.1 37.3 6.9 MgO 5.3 11.9 2.4 APO 3 :.. 2.5 .... .... 20.4 BaO 1.5 ZnO 2.4 The iron present in these mate- S 4.4 rials is really present mostly as As 0.5 Fe 2 O 3 and only partly as FeO, but Pb 20 . 7 the analysis is expressed usually as Cu 2.9 FeO in order to facilitate the slag Ag 0. 17 calculations. The analyses, as given, are not expected to add up to 100. In making the calculations, assume that the slag to be pro- duced contains SiO 2 : FeO: CaO in the ratio 30 : 40 :20; that the FeO here expressed includes the MnO, calculated to its FeO equivalent, and the CaO includes the MgO, BaO, and ZnO calculated to their CaO equivalent. Assume also that all the ZnO of the charge goes into the slag, all the sulphur into matte, all the arsenic into speiss, of the formula Fe 5 As, all the silver and lead into the lead bullion. 594 METALLURGICAL CALCULATIONS. Requirements : (1) The proportions of lead ore, iron ore and limestone to be used in the 900 pounds of these to a charge. (2) The weight and composition of the slag formed. (3) The weight and composition of matte formed. (4) The weight and composition of speiss formed. (5) The weight and composition of lead bullion formed. (6) A balance sheet of the essential materials entering and leaving the furnace. Solution : (1) There are really only two unknown weights to be de- termined, for the sum of the three weights required is to be 900. Similarly, the ratio 30 : 40 : 20 really gives two condi- tions to be fulfilled, since there are practically two ratios to be worked to. The simplest method of solution is, undoubt- edly," the algebraic one, letting x and y represent two of the ores, 900 (x + y) the other, and then working out the weights of SiO 2 , FeO and CaO in the slag, expressed in terms of x and y. The ratio 30 : 40 : 20 then gives us two independent equations, containing the two unknowns, and the problem is solved. Let x = weight of lead ore. y = weight of iron ore. . 900 x y = weight of limestone. 15 = weight of ash of coke. 100 = weight of return slags. Calculation of FeO going into Slag. Sulphur in 100 parts ore 4.4 Ibs. Copper in 100 parts ore 2.9 " Sulphur united with Cu to form Cu 2 S 0.7 " Sulphur left over to form FeS 3.7 " Iron needed to form FeS = 3.7 x|| = 6.5 " o5a 72 FeO corresponding to this Fe = 6.5X^ = 8.4 " oo Arsenic in 100 parts of ore 0.5 " Iron required to form Fe 5 As = 0.5X-^~ = 1.9 " i O THE METALLURGY OF LEAD. 595 72 FeO corresponding to this Fe = 1.9X- = 2.4 Ibs. oo Total FeO disappearing in matte and speiss =10.8 " FeO left over to go into slag = 14.8- 10.8 = 4.0 " Therefore x parts of lead ore contributes to the slag: SiO 2 0.326 x FeO 0.040 x MnO 0.043 x = 0.044 x FeO. CaO 0.022 x MgO 0.053 x = 0.074 x CaO. APO 3 0.025 x BaO 0.015 x = 0.006 x CaO. ZnO 0.024 x = 0.017 x CaO. ' The FeO equivalent of MnO is 72/71 the MnO; the CaO equivalent of MgO is 56/40 the MgO, of BaO is 56/141 the BaO, of ZnO is, 56/81 the ZnO. These ratios are the chem-' ically equivalent values of these oxides, as taken from their molecular weights. Adding all these together, our lead ore contributes to the slag the equivalent of SiO 2 0.326 x, FeO 0.084 x, CaO 0.119 x. The y parts of iron ore similarly contributes to the slag: SiO 2 0.043 y, FeO 0.741 y, CaO 0.031 y. The 900 x y parts of limestone contributes likewise : SiO 2 0.027 (900- x-y) FeO 0.045 (900 -x-y) CaO 0.373(900 -x-y) MgO 0.1 19 (900- x-y) = 0.167(900 - x - y) of CaO. Therefore, total contribution of the limestone to the slag: SiO 2 0.027(900 - x - y), FeO 0.045(900 -x-y), CaO 0.540 (900- x-y). Contribution of ash of fuel to the slag, in similar manner: SiO 2 6.1 FeO 4.0 CaO 1.4 596 METALLURGICAL CALCULATIONS. Adding all these together, we have in the slag: SiO 2 ......................... 30.4 + 0.299 x + 0.016 y FeO ......................... 44.5 + 0.039 x + 0.696 y CaO ......................... 487.6 -0.421 x -0.509 y According to assumption, these ingredients as thus summated should bear the relations 30 : 40 : 20. We therefore have OQ 30.4 + 0.299 x + 0.016 y = (487.6 - 0.421 x- 0.509 y) 44.5 + 0.039 x + 0.696 y = ^ (487.6 - 0.421 x- 0.509 y) Z(j Whence x = lead ore = 523 Ibs. y = iron ore = 274 " 900 - x - y = limestone = 103 " (1) (2) With the weights of materials used, as calculated, we can calculate the weights of the ingredients of the slag as: Lead Ore. Iron Ore. Limestone. Coke Ash. Total. SiO 2 .......... 170.5 11.8 2.8 6.1 191.2 FeO .......... 20.9 198.4 4.6 4.0 227.9 MnO ......... 22.5 4.7 ........ 27.2 CaO .......... 11,5 8.5 38.4 1.0 59.4 MgO .......... 27.7 ____ 12.3 0.4 40.4 A1 2 O 3 ......... 13.1 ........ 3.1 16.2 BaO .......... 7.8 ............ 7.8 ZnO.. 12.6 12.6 582.7 Percentage Composition and Check on Ratio. SiO 2 32.8percent FeO. 39.0 MnO 4.7 " = 4.8 per cent FeO. CaO 10.2 MgO 6.9 " = 9.7 " CaO. APO 3 2.8 BaO 1.3 u = 0.5 " CaO. ZnO 2.2 " = 1.5 CaO. Summated SiO 2 : FeO : CaO = 32.8 : 43.8 : 21.9 = 30 : 40 : 20 THE METALLURGY OF LEAD. 597 (3) Sulphur in 523 Ibs. of lead ore = 23.0 Ibs. Copper in 523 Ibs. of lead ore : = 15.2 " Sulphur to form Cu 2 S with this Cu = 3.8 " Sulphur to form FeS = 23;0 - 3.8 = 19.2 " Fe to form FeS = 19.2 X ~| = 33.6 " FeS in matte =19.2 + 33.6 =52.8 " Cu 2 S in matte = 15.2 + 3.8 = 19.0 " Total weight of matte = 71.8 " (3) Composition of matte Cu 2 S = 26.5 per cent = Cu =21 per cent. FeS = 73.5 " Fe = 47 S =22 " (3) (4) Arsenic in 523 Ibs. of lead ore = 2.6 " 9&0 Fe to form Fe 5 As =2.6X -~ = 9.7 7o Weight of speiss formed = 12.3 " (4) Composition : Fe 79 per cent As 21 " (4) (5) Lead in 523 Ibs. of lead ore = 108.3 Ibs. Silver in 523 Ibs. of lead ore = 0.9 " 109.2 " (5) Composition: Pb 99.2 per cent. Ag 0.8 " (5) (6) Balance sheet of materials entering and leaving per 1,000 Ibs. of burden: Charges. Bullion. Matte. Speiss. Slag. Gases. Lead Ore 523 SiO 2 170.5 170.5 FeO 77.4 ... Fe 33.6 Fe 9.7 Fe O20.9 O 13.2 MnO 22.5 ... 22.5 CaO 11.5 11.5 598 METALLURGICAL CALCULATIONS. Charges. MgO APO 3 Bullion. 27.7 ... 13.1 ... Matte BaO 7.8 ... .... ZnO 12.6 ... .... S 23.0 ... 23.0 As 2.6 ... .... Pb 108.3 108.3 .... Cu 15.2 ... 15.2 Ag 0.9 0.9 Iron Ore 274 SiO 2 11.8 ... FeO 198.4 ... , . MnO 4.7 ... .... CaO 8.5 ... .... Limestone 103 SiO 2 2.8 ... . . . FeO 4.6 ... .... CaO MgO 38.4 ... 12.3 o Ash of Fuel 15 SiO 2 6.1 ... ^ FeO 4.0 ... .... CaO 1.0 ... MgO A1 2 O 3 0.4 ... 3.1 ... .... Return Slag 100 ... Speiss. Slag. Gases. 27.7 13.1 7.8 12.6 2.6 11.8 198.4 4.7 8.5 2.8 4.6 38.4 12.3 6.1 4.0 1.0 0.4 3.1 100.0 109.2 71.8 12.3 682.7 13.2 THE ELECTROMETALLURGY OF LEAD. As far as the present, the use of electrical methods in the metallurgy of lead has been confined to the Salom process of cathodic reduction of galena and the Betts process of refining crude lead bullion. Electrothermic methods of smelting, and even of roasting, are among future possibilities, as are also the leaching by suitable solvents and electrical precipitation of lead from the solutions, but they have not been commercially practiced. The Salom process puts the powdered galena on a " hard- THE METALLURGY OF LEAD. 599 lead " plate, in a solution of dilute sulphuric acid, and using a " hard-lead " anode, passes a strong current through, reducing PbS in situ to Pb with formation of H 2 S (with some hydro- gen) at the cathode and oxygen at the anode. It was run . commercially on a fair scale at Niagara Falls. For further details reference may be made to Transactions American Elec- trochemical Society, Vol. I, p. 87; II, p. 65; IV, p. 101. The Betts process consists in using impure lead as anode in a solution of lead fluo-silicate, PbF 2 . SiF 4 , strongly acid with HF. The refining plant and its operation are quite sim- ilar to a copper refining plant. The cathodes are sheet steel, greased, and the dense sheet lead deposited is' stripped off from time to time. Problem 125. In a Salom apparatus powdered galena, density of powder 3.5, is placed in a layer 0.5 mm. thick on a revolving lead table having an effective treatment area of 2.25 square meters. Current density 330 amps, per square meter of cathode surface. Time of treatment of the layer 90 minutes. Electrolyte dilute H 2 SO 4 . Resistance of cell 0.001 ohm. Heats of formation: (Pb, S) 20,300; (H 2 , S) 4,800 (as gas); (H 2 , O) 69,000. Assume reduction to Pb complete. Required : (1) The efficiency of application of the amperes passing to the reduction of the galena, and the percentage of ampere efficiency lost by the evolution of hydrogen. (2) The average composition of the gases coming from the cell and their volume per hour, at normal pressure and 20 C., assuming the electrolyte saturated with H 2 S, H 2 and O 2 at starting. (3) The working voltage absorbed, if the efficiency of re- duction of galena were 100 per cent. (4) The working voltage if the cell were kept running after all the galena was reduced. (5) The average working voltage of the cell in actual opera- tion and the distribution of this. (6) The proportion of the power required by the cell which could be generated by burning the gases produced in a gas engine (if such were possible) at a thermo-mechanical efficiency of 100 per cent. 600 METALLURGICAL CALCULATIONS, (1) Current used 2.25X330 = 742 amperes. Lead theoretically reducible in 90 minutes 0.00001035X103.5X60X90X742.5 = 4295 grams. Galena in the apparatus : 2.25 X 10,000 X 0.05 X 3.5 = 3938 " Lead under treatment: 907 3938X^ = 3410 Ampere efficiency of the treatment: = 0.794 = 79.4 per cent. Per cent of amperes evolving hydrogen = 20.6 (1) (2) If all the current evolved IPS, the gases evolved would be: 2H 2 S O 2 2 parts 1 part. If all the current evolved H 2 , the gases would be: 2H 2 O 2 2 parts 1 part. If 79.4 per cent of the current evolves H 2 S and 20.6 per cent hydrogen, the average gases evolved will be: H 2 S 1.588 parts = 62.93 per cent. H 2 0.412 " = 13.73 O 2 1.000 " = 33.33 " (2) The volume evolved per hour is found as follows: Oxygen produced by 1 ampere hour 0.00001035X8X60X60 = 0.29808 grams. Produced by 742.5 amperes, in one hour 0.29808X742.5 = 221.3 Volume at 760 mm. and C. 221.3-1.44 = 153.7 litres. Volume of gases produced per hour 153.7X3 = 461. = 0.461 cubic meter Volume at 20 0.461 X 27 :Lt 2 = 0.495 (2) Zi o THE METALLURGY OF LEAD. 601 (3) Voltage for ohmic resistance 0.001X742 = 0.742 volts. Chemical work done if only H 2 S is formed, per formula PbS + H 2 O = H 2 S + O + Pb - 20,300 - 69,000 +4,800 = - 84,500 Cai. Per chemical equivalent concerned = 42,250 " Voltage absorbed in chemical work 42,250-^23,040 = 1.83 volts Total voltage drop in the cell 1.83 + 0.74 =2.59 " (3) (4) Chemical work done if only H 2 is liberated = 69,000 Cal. Voltage absorbed in chemical work ^2^23,040= 1.49 volts. Total voltage drop in the cell = 2.23 (4) (5) Voltage absorbed in chemical work if 79.4 per cent of the current evolves H 2 S and 20.6 per cent H 2 : (1.83X0.794) + (1.49X0.206) = 1.76 volts. Total voltage drop, in average running = 2.50 " (5) Logically, the 1.76 volts absorbed in chemical decomposition in average running is properly calculated thus, from the energy evolved: (42,250X0.794) + (34,500x0.206) 23,040 (6) Heat of combustion of 1 cubic meter of the average gas : H 2 S 0.5293 m 3 X5,513 = 2918 Cal. H 2 0. 1373 m 3 X 2,614 = 359 " Sum = 3277 Calorific power of gas produced per hour: 3277X0.461 = 1511 Calories. Watt hours producible from this at 100 per cent thermo-mechan- ical efficiency : 1511^0.860 = 1757 watts. 602 METALLURGICAL CALCULATIONS. Current used by the cell, average running: 742.5X2.50 = 1856 watts. Proportion theoretically regainable: = 0.95 = 95 per cent. (6) The reason why more power is theoretically regainable than is used in doing chemical work in the cell, is that the gas engine burns the sulphur of the PbS ultimately to SO 2 , and this is the source of great energy, while only a relatively small amount of energy was necessary to convert the PbS into H 2 S ready for combustion. Problem 126. Lead bullion refined by the Betts process is 96.73 per cent lead, and the refined lead is practically pure. The anodes are 1.5 inches thick, and weigh, with lugs for suspension, 275 pounds. They are left in the tanks an average of nine days, running with 135 amps, per plate, whose active surface is 1,700 square inches. Space between anode and cathode 1J inches, specific resistance of solution 10 ohms. Each tank has twenty-two anodes and twenty-three cathodes, and produces an average of 545 pounds of lead per day. Power costs $50 per electric horsepower-year, as delivered to the tanks. Drop per tank, due to resistance of contacts, 0.15 volt; specific gravity of pure lead, 11.35. Required : (1) The weight of lead theoretically deposited by the cur- rent per day. (2) The ampere efficiency of the deposition. (3) The voltage drop per tank. (4) The amount of anode scrap to be remelted in percent of the total anode weight. (5) The average thickness of the cathode deposit per day. (6) The power costs per ton of impure lead refined. Solution: (1) Lead theoretically deposited by 1 ampere per day . 00001035 X 103 . 5 X 60 X 60 X 24 = 92.55 grams. THE METALLURGY OF LEAD. 603 Per anode per day, dissolved 92.55X135^1000 = 12.5 kg. = 27.5 Ibs. Per tank, per day, deposited 27.5X22=605 " (1) (2) Efficiency of deposition ^J = 0.90 = 90 per cent. (2) DUO (3) Amperes used per plate = 135 Current density per square inch = 135^1700= 0.0794 amperes Current density per square centimeter = 0.0794-6.25 = 0.0127 Distance of plates apart = 1.17 inches = 2.93 cm. Resistance of 1 cm. cube of electrolyte =- 10 ohms. Voltage drop in electrolyte 10 X 2 . 93 X . 0127 = 0.37 volt. Voltage drop in connections = 0.15 Total voltage drop per tank = 0.52 " (3) (4) Dissolved from anode in 9 days 545 -v- 22X9 = 223 Ibs. Anode corroded in 9 days, assuming slime to fall off it 223 -=-0.9673 = 231 Ibs. Anode scrap = 275 - 231 = 44 44 Percentage of anode scrap 7^ = 0.16 = 16 per cent. (5) Deposit of lead one side of a cathode, per day 545^-44 = 12.4 Ibs. = 5.636 kg. Area of 1 side of a cathode = 850 sq. inches. = 5312 sq. cm. Deposit, per day, per sq. cm. 5636^5312 = 1.06 grams. 604 METALLURGICAL CALCULATIONS. Volume of this deposit 1 . 06 -v- 1 1 . 35 = 0.093 cu. cm. Therefore thickness deposited per day = 0.93 mm. (5) (6) Bullion refined, per tank, per day 545 -=-0.9673 = 563 Ibs. Power required per tank 2970X0.52 = 1544 watts. 2.06 e.h.p. Cost of power per year = 50X2.06 = 103. dollars. per day = 103^-365 = 0.28 9nnn per ton of lead treated 0.28X^ = 0.99 * (6) CHAPTER III. THE METALLURGY OF SILVER AND GOLD. The processes most used for the extraction of silver and gold from their ores are the smelting of the ores with copper ores to matte and eventually blister copper, followed by electro- lytic refining of the latter, or smelting with lead ores, followed by desilverization of the resulting lead bullion, cupellation of the rich lead, or electrolytic refining of the same. In all of these cases the silver or gold is a comparatively minute constitutent of the ore, intermediate products and final metal, so that the metallurgy of silver or gold, after these methods, up to the production of impure silver or gold, or silver or gold bullion, is practically the metallurgy of copper or lead both of which subjects we have considered at length. When we obtain products in which the silver or gold is the chief constituent, we approach a condition in which calculations upon the precious metal contained may be based, but here again there are not the same conditions of economy so promi- nent in the metallurgy of lead or copper. It does not matter greatly, for instance, whether we use 50 cents' worth of one fuel or a dollar's worth of another in melting 1,000 ounces of silver worth over $500; the question of which fuel is the cleanest or most convenient is of greater importance than the cost of the fuel. The object of this introduction is to show that, until it comes to handling nearly pure silver or gold, close calculations as to amounts of reagents, running of furnaces, etc., are not prac- ticable for silver or gold, per se. ELECTROLYTIC REFINING OF SILVER BULLION. The parting of gold and silver by acids is rapidly giving place to the cleaner and less expensive electrolytic separation. The bullion may be of quite variable composition. The follow- ing analyses of two samples give an idea of this variation. 605 606 METALLURGICAL CALCULATIONS. Cupelled Silver. Impure Silver. Ag 98.69 74.08 Pb 1.09 3.71 Cu 0.12 20.23 Fe 0.09 1.01 Au 0.0023 0.05 The electrolytic refining proceeds easiest with the nearly pure silver, because the impurities going into solution accumu- late much more slowly, and the solution therefore requires purifying less frequently. On the other hand, the solution of a large amount of copper, iron or lead adds electromotive force to the circuit, since silver only is deposited, and thus decreases the electrical work to be done. Problem 127. Two varieties of silver bullion are refined electrolytically carrying No. 1. No. 2. Ag 98.69 74.08 Pb 1.09 3.71 Cu 0.12 20.23 Fe 0.09 1.01 Au 0.01 0.05 The slimes from each carry in percentages: No. 1. No. 2. Ag 60 55 Pb 10 5 Cu 5 15 Fe '.... 3 5 Au 22 20 In each case all the gold goes into the slimes. A current density of 200 amperes per square meter of electrode surface is used, and the plates are 2 centimeters thick. Specific gravity of No. 1, 10.15; of No. 2, 9.79. There is anode scrap equal to 12 per cent of the weight of the plates. The working space between the electrodes is 4 cm. 'total current 220 amperes per tank, specific resistance of electrolyte 20 ohms. Deposited silver, 19.85 kg. per tank per day. THE METALLURGY OF SILVER AND GOLD. 607 Required : (1) The ampere efficiency of the deposition of silver. (2) The weight of anode corroded in a tank per day. (3) The deficit in weight of silver in the electrolyte in each tank per day. (4) The electromotive force contributed to the circuit by the solution of impurities in the case of each bullion. (5) The total drop of potential across each tank, adding 10 per cent for loss in contacts. (6) The horse power-hours required per kilogram of silver deposited in each case. Solution : (1) Silver deposited theoretically by one ampere, per day 0.00001035X108X60X60X24 = 96.58 grms. 220 amperes deposit 96.58X220 =21,248 " = 21.248 kg. 19 85 Ampere efficiency = 0.934 = 93.4 per cent. (1) This means that 6.6 per cent of all the silver which the cur- rent is capable of depositing is prevented from depositing by the nitric acid present, forming AgNO 3 . The electrolyte is silver nitrate with copper nitrate, and contains always about 1 per cent of free nitric acid, which thus acts chemically upon the deposited silver (or acts to prevent its deposition to this extent, whichever way we wish to look at it). (2) The weight dissolved, per 100 grams of anode corroded, is, assuming all the gold, to appear in the slimes: No. 1. Anode. Slimes. Dissolved. Ag ............. ......... 98.69 0.03 98.66 Pb ...................... 1.09 ____ 1.09 Cu ............ . ........ . 0.12 ____ 0.12 Fe ...................... 0.09 ..... 0.09 Au ........ .............. 0.01 0.01 No. 2. Anode. Slimes. Dissolved. Ag ................. ..... 74.08 0.14 77.94 Pb ...................... 3.71 0.01 3.70 608 METALLURGICAL CALCULATIONS. Anode. Slimes. Dissolved. Cu 20.23 0.04 20.19 Fe.. 1.01 0.01 1.00 Au 0.05 0.05 .... The amount of current necessary to dissolve these weights will be the sum of the current which would be necessary to dissolve each constituent separately. These will be found in ampere-hours by dividing each by its electrochemical chemical equivalent X 3,600. No. 1. Ampere-Hours. Ag 98.66^-0.001118X60X60 = 24.51 Pb 1. 09 -5- 0.00107 X60X60 = 0.28 Cu 0.12 -0.00033 X60X60 = 0.12 Fe.. . 0.09-hO. 00029 X60X60 = 0.09 25.00 No. 2. Ag .............. 77.94-4.025 Pb .............. 3.70^3.856 Cu .............. 20.19-hl.185 Fe .............. 1.00^-1.044 38.32 Since there is disposable 220X24 5,280 ampere-hours in a tank per day, there will be corroded the following weights of anodes per day in tanks containing Nos. 1 and 2, respectively; No. 1 ..... 100 x- = 21,120 Grams. No. 2 100 X = 13,779 (3) There are dissolved in a tank, per day, the following weights of silver: No. 1 anodes: 21,120X0.9866 = 20,837 Grams No. 2 anodes: 13,779X0.7794 = 10,739 " Deposit in each tank = 19,850 " Surplus in No. 1 tank = 987 " Deficit in No. 2 tank = 9,111 " (3) THE METALLURGY OF SILVER AND GOLD. 609 (4) The heats of formation of the acid and salts concerned are: (Ag, N, O 3 , Ag) = 23,000 cal. = 213 cal. per gram Ag. (Pb, N 2 , O, Ag) = 98,200 " = 474 " * Pb. (Cu, N 2 , O 8 , Ag) = 81,300 " = 1,278 " " Cu. (Fe, N 2 , O 6 , Ag) = 43,900 " = 784 " " Fe. (H, N, O 3 , Ag) = 48,800 " = 48,800 " " H The heat balance per 100 grams of anodes No. 1 is: Solution of Ag, 98.66 X 213 = 21,015 cal. Pb, 1.09 X 474 = 517 " Cu, 0.12X 1,278 = 153 " Fe, 0.09 X 784 = 71 " Heat evolved 21,756 cal. Deposition of Ag, 93.99 X 213=20,020 " Liberation of H 2 0.06X48,800 = 2,928 " Heat absorbed = 22,948 Heat deficit 1,192 " Since this is per 25 ampere-hours [see (2)] and 1 ampere hour at 1 volt = 860 calories, the voltage which this deficit of energy will absorb in the case of anodes No. 1 is: 1,192^-25^-860 = 0.06 volt. (4) The similar calculation for anodes No. 2 gives per 100 grams corroded : Solution of Ag, 77.94 X 213 = 16,601 cal. Pb, 3.70 X 474 = 1,754 " ." Cu, 20.19X1,278 =25,803 " Fe, 1.00 X 784 = 784 " Heat evolved 44,942 cal. Deposition of Ag, 98.11 X 213 = 20,897 " Liberation of H 2 0.064x8,800= 2,928 " Heat absorbed 23,826 Surplus heat 21,117 " Voltage generated = 0.98 volt. (4) 610 METALLURGICAL CALCULATIONS. (5) Resistance of each element of electrolyte of 1 square cm. area is 20X4 = 80 ohms. The current passing through this is 220 -f- 10,000 = 0.022 amperes. The drop of voltage, due to the resistance of the electrolyte, is, therefore, 0.022X80 = 1.76 volts. Adding decomposition voltage we have In case of No. 1 anodes, 1.76 + 0.06 =- 1.82 volts. Add for resistance of contacts 0.18 " Working voltage 2.00 " (5) In case of No. 2 anodes, 1.76 - 0.98 = 0.78 " Add for resistance of contacts 0.18 " Working voltage 0.96 " (5) (6) Horsepower required per tank: 220X2.00 Anodes No. 1 : ^7; - = 0.59 hp. 750 Anodes No. 2: Power per kilogram of silver deposited: Anodes No. 1, 0.59-7-19.85 = 0.020 hp-days = 0.48 hp-hours. (6) Anodes No. 2, 0.28^19.85 = 0.014 hp-days. 0.34 hp-hours. 6) Problem 128. The Wohlwill process for refining gold bullion operates with AuCl 3 solution, strongly acid with HC1. Design a plant for refining bullion having the analysis: Au .............. 60.3 Fe ............... 2.2 Ag ..... :..'.'. ..... 7.0 Ni ............. ..2.0 Cu .............. 6.5 Pb ............... 7.0 Zn .............. 15.0 Assume the following data to start with: Current density, 1,000 amperes per square meter. Use ten tanks in series. Tanks available, 500 mm. X 500 mm. X300 mm. deep, inside, THE METALLURGY OF SILVER AND GOLD. 611 Electrodes about 4 cm. apart. Starting sheets 1 mm. thick. Anodes 20 mm. thick. Specific gravity of anodes 17.5. Gold present in solution 50 grams per liter. Specific resistance of electrolyte 5 ohms. Specific gravity of the electrolyte 1.15. Loss of electromotive force at contacts one-half the loss due to the resistance of the solution. Anode products, AuCl 3 , AgCl, CuCl, ZnCP, Fed 2 , NiCl 2 , PbCl 2 . Required: (1) The size and dimensions of electrodes and number in a tank. . (2) Current used and total voltage of the system. (3) Gold deposited per day. (4) Anodes corroded per day, assuming corrosion uniform. (5) Deficit of gold in the tanks per day. (6) If anodes are left in until 9/10 dissolved, how long will they last? (7) If cathodes are removed every 24 hours, what is the average time of treatment? (8) At 6 per cent per annum, what is the interest charge per kilogram of gold under treatment in the plant, gold being worth 72.9 cents per gram. Solution : (1) Allowing 1 cm. clearance at the sides, the plates must not be over 48 cm. in width across the tank. The electrolyte must not be nearer than 2 cm. to the top of the tank, and the plates ought to be 8 centimeters above the bottom, to allow space for slimes to accumulate. The immersed depth of plates is, therefore, 40 cm., and the superficial area 960 sq. cm. on a side. With spaces 4 cm., anodes 2 cm. thick and cathodes 0.1 cm., using one more cathode than anode, we have the spaces equal in number to twice the anodes, and, therefore, (2 X anodes) -f (anodes + 1) 0.1 + (2 anodes + 4) = 50, whence the number of anodes figure out 4.9. (1) 612 METALLURGICAL CALCULATIONS. Choosing the nearest whole number, 5, there will be 6 cath- odes, and the working spaces will be 10, and their width 50 -5CO- 6(0.1) ^ 94cm (1) (2) Ignoring the edges of the plates, the working surface per tank is 960X2X5 = 9,600 sq. cm., and the current 9,600 -r- 10,000X1,000 = 960 amperes. (2) The voltage to overcome ohmic resistance is - *- 97 volts - Add 50 per cent loss at contacts = . 98 " Sum = 2.95 The voltage corresponding to the chemical action occurring can be calculated from the energy of formation of the AuCl 3 decomposed minus the energy of formation of the salts formed. More simply, although not so logically, it may be derived from the voltages of decomposition of these compounds, allowing for the proportions of each formed and decomposed. These are: (Au, Cl 3 ) = 27,200-^-3-^23,040 = 0.393 volt. (Ag, Cl) = 29,000^1^23,040 =1.257 '" (Cu, Cl) = 35,400^-1-^-23,040 =1.536 " (Zn, Cl 2 ) = 113,000-^-2^-23,040 =2.448 " (Fe, Cl 2 ) = 100,100-^2-^23,040 =2.170 * (Ni, Cl 2 ) = 93,900 -i- 2 ~- 23,040 =2.037 a (Pb, Cl 2 ) = 77,900 -r- 2 -f- 23,040 =1.690 These are the voltages which would be generated at the anode in case each one of these salts was the only salt being formed. But, since we know the exact composition of our anode dissolved, we can calculate the corresponding voltage generated at the anode: Au, 0.603X0.393 = 0.237 volt. Ag, 0.070X1.257 = 0.088 " Cu, 0.065X1.536 = 0.101 " Zn, 0.150X2.448 = 0.367 * Fe, 0.022X2.170 = 0.048 THE METALLURGY OF SILVER AND GOLD. 613 Ni, 0.020X2.037 = 0.041 volt. Pb, 0.070X1.690 = 0.118 " Sum 1.000 Absorbed at the cathode 0.393 " Total voltage generated 0.607 " Absorbed in electrolyte and connections 2.95 " Working voltage per tank 2.34 " Total for the system 23.4 " (2) (3) Au deposited by 960 amperes in ten tanks per day: 1Q7 0. 00001035 X^ X 60 X 60X24X960X10 = 563,729 gr. = 563.729 kg. (4) One gram of anode requires for its corrosion: Au, 0.6034-197, 4- 34-0.00001035 Ag, 0.603 -=-108 4-0.00001035 Cu, 0.065-:- 63.6 4-0.00001035 Zn, 0.150^- 65 4- 2 4- 0.0000 1035 Fe, 0.022 4- 56 4-24-0.00001035 Ni, 0.020^- 59 4-2-0.00001035 Pb, 0.070 4-207 -f-24- 0.00001035 = 0.01703 4-0.00001035 = 1,645 amp. seconds. Current available per day: 960 X 60 X 60 X 24 X 10 = 829,440,000 amp. seconds. Anodes corroded per day: 829,440,0004-1,645 504,220 grams. 604.220 kilograms (4) (5) Gold deposited 563.729 Gold dissolved = 504.220X0.603 = 304.045 Deficit of gold in the plant per day = 259.684 u (5) (6) Weight of anodes: 960X4X17.5X5X10 = 3,360,000 grams. 3,360 kilograms. One-tenth scrap 336 " Weight corroded = 3,024 614 METALLURGICAL CALCULATIONS. Days to corrode them: o 094 |= 5.93 = practically 6 days. (6) (7) Average time of treatment, removing one-sixth each day: (1 + 2 + 3 + 4 + 5 + 6) -f-6 = 3.5 days. (7) (8) Value of gold in plant is, in regular running, value of cathodes for all the time plus value of anodes for 3.5-f-6 = 58 per cent of the time. Value of cathodes : 960X0.1X19.2 (sp. gr. gold) X6X 10X0.729 = $80,578 Value of gold in anodes : 3,360,000X0.603X0.729 = 1,477,012 Average value of anode gold under treatment: 1,477,012 X (3.5 -5-6) = 861,590 Sum of cathode gold continually in stock and average value of anode gold under treatment: $80,578 + $861,590 = $942,168 Interest per annum at 6 per cent = 56,530 Interest charge per day = 155 Interest charge per kg. of anode refined: $155 -^ 504 . 22 = $0.31 per kg. Interest charge per kg. of gold under treatment: $155 -^ 304 . 045 = $0.51 per kg. (8) THE VOLATILIZATION OF SILVER AND GOLD. It is known that both of these metals can be vaporized; it can easily be done in the electric arc. If they are placed in a vacuum, in quartz vessels, they show signs of metal vapor- izing at 680 and 1,070, respectively, although what vapor tension this corresponds to we do not exactly know, but we can surmise it to be a small fraction of a millimeter. If heated, still higher in a vacuum, they show the phenomenon of ebul- lition, or boil, at 1,360 and 1,800, respectively, although here again we do not know what pressure this corresponds to, THE METALLURGY OF SILVER AND GOLD 615 except it is the pressure of the melted metals above the spot where the vapor commences to form beneath the surface, which might be 5 to 10 millimeters of melted metal, say 10 millimeters of mercury, if the bath of metal were shallow. From these data, obtained by Schuller, Krafft and Bergfeld, it is estimated that the boiling points of the two metals under atmospheric pressure are 2,040 and 2,530, respectively, on the assumption that the temperature interval between first signs of vaporization in a vacuum and boiling in a vacuum is equal to the interval between the latter temperature and the ordinary boiling point. (This is said to be true of mercury.) O. P. Watts has estimated the boiling point of silver and gold as 1,850 and 2,800, respectively, meaning probably at atmospheric pressure; but his estimate is based on such as- sumptions that they cannot lay claim to as much accuracy as the data and estimates above cited. If we go back to the very probable rule, that at equal frac- tions of the normal boiling points, expressed in absolute tem- peratures, metals have the same vapor tensions, we can com- pare silver and gold with mercury, whose vapor tension is known through a wide range, and get quite probable values for the vapor tensions of these metals through a large range of temperature. The following table is from nearly zero ten- sion to about the temperature at which ebullition is noticeable in a vacuum: Tension of Vapor. Mercury. Lead. Silver. Gold. mm. of Hg. C C C C 0.0002 625 729 942 0.0005 10 658 766 987 0.0013 20 691 802 1,031 0.0029 30 724 839 1,075 0.0063 40 757 876 1,120 0.013 50 790 913 1,165 0.026 60 822 949 1,209 0.050 70 855 986 1,254 0.093 80 888 1,023 1,298 0.165 90 921 1,059 1,343 0.285 100 954 1,096 1,387 616 METALLURGICAL CALCULATIONS. Tension of Vapor. Mercury. Lead Silver. Gold. mm. of Hg. C ' C C C 0.478 110 987 1,133 1,432 0.779 120 1,020 1,169 1,476 1.24 130 1,053 1,206 1,520 1.93 140 1,086 1,243 1,565 2.93 150 1,119 1,280 1,611 4.38 160 1,151 1,316 1,654 6.41 170 1,184 1,353 1,699 9.23 180 1,217 1,390 1,743 Inspecting the above table, we see that apparently about 0.0002 mm. of mercury tension is sufficient to make a metal show signs of vaporization. The corresponding temperatures at which silver and gold show 0.0002 mm. tension are, from the above table, 729 and 942, respectively, whereas it is said to have been observed for these metals at 680 and 1,070. The divergence is not wide, considering the lack of exact data involved. Mercury is said to show ebullition in a vacuum at 180, lead at 1,250, silver at 1,360 and gold at 1,800. From the above table we see these metals having the same vapor tension at 180, 1,217, 1,390, and 1,743, respectively. The agreement is encouraging. The table in continuation is for temperatures from those causing ebullition in a vacuum to those required to boil the metals at atmospheric pressure: Tension of Vapor. Mercury. 1 ^>ead. Silver. Gold. mm. of Hg. C c C C 9.23 180 ! 1,217 1,390 1,743 14.84 190 1,250 1,427 1,788 19.90 200 1,283 1,463 1,832 26.35 210 ,316 1,500 1,877 34.70 220 ,349 1,537 1,921 45.35 230 ,382 1,574 1,965 58.82 240 ,415 1,610 2,010 75.75 250 ,448 1,647 2,055 96.73 260 ,480 1,684 2,099 THE METALLURGY OF SILVER AND GOLD. 617 Tension of Vapor. Mercurv. Lead. Silver. Gold. mm. of Hg. C ' C C C 123. 270 1,513 1,720 2,144 155. 280 1,546 1,757 2,188 195. 290 1,579 1,794 2,233 242. 300 1,612 1,830 2,277 300. 310 1,645 1,867 2,322 369. 320 1,678 1,904 2,366 451. 330 1,711 1,941 2,410 548. 340 1,744 1,977 2,455 663. 350 1,777 2,014 2,500 760. 357 1,800 2,040 2,530 For tensions of metallic vapors over 1 atmosphere, such as may easily occur in high temperature work, particularly in electric furnaces, the following data may be useful: Tension of Vapor. Mercury. Lead. Silver. Gold. Atmospheres. C C C C 1.0 357 1,800 2,040 2,530 2.1 400 1,951 2,197 2,722 4.25 450 2,116 2,380 2,945 8. 500 2,280 2,564 3,167 13.8 550 2,445 2,747 3,390 22.3 600 2,609 2,931 3,612 34.0 650 2,774 3,114 3,835 50. 700 2,938 3,298 4,057 72. 750 3,103 3,481 4,280 102. 800 3,267 3,665 4,502 137.5 850' 3,436 3,848 4,725 162. 880 3,525 3,958 4,858 The last table may not be of much immediate practical value, but it is interesting scientific information. The pre- ceding tables are, however, technically and commercially important. They point particularly to the wastefulness of melting silver or gold in open furnaces where furnace gases pass over the metal. Such gases, at temperatures above 700 for silver and 950 for gold, i.e., even passing over unmelted metal, can cause volatilization and loss of weight, because they 618 METALLURGICAL CALCULATIONS. evaporate the metal on exactly the same principle that a current of dry air evaporates ice or water. They also explain why silver volatilizes with lead in the cupellation operation. At 1,000 C. lead has a vapor tension of about 0.62 mm. of mer- cury, and silver about 0.07 mm., and therefore, a considerable loss of silver is sure to occur with the lead vapors passing off. Gold at the same temperature has only 0.0007 mm. tension, and, therefore, proportionately less of it is lost, say only one-fiftieth as much as the silver loss, reckoning from the proportionate tensions and the relative densities of the two vapors. In calculating such metallic vapor losses it is important to remember that the metals are monatomic when in the state of vapor, and- that the molecular weights of their vapors are simply their atomic weights. The hypothetical weight of a cubic meter of such metallic vapor at standard conditions, C. and 760 mm. pressure, is therefore, 0.09kg.x at miC 9 Weight . and from this theoretical datum the weight of any volume at any temperature and any tension can be calculated. Illustration: Calculate the weight of silver vapor contained in 1 cubic meter of furnace gases, if saturated with silver vapor and at 1,100 C. Solution: Since the tension of silver at this temperature is 0.28 mm., "the question resolves itself into this: What is the weight of 1 cubic meter of silver vapor at 1,100 and 0.28 mm. pressure. One cubic meter at standard conditions would be 0.09 X (108 -5-2) = 4.86 kilograms; therefore, at these given conditions, it contains: = 3.5 grams. It is true that the silver vapor, being heavy, mixes slowly with the furnace gases, but, nevertheless, the gases rushing over and coming in contact with the silver may become nearly saturated with this vapor, and carry considerable away. In THE METALLURGY OF SILVER AND GOLD. 619 the Bessemerizing of copper matte to blister copper, by blowing air directly through it, as much as 30 per cent of all the silver present may be thus vaporized from the bath as soon as the silver has taken the metallic form. Every smelter and refiner of the precious metals should be familiar with these facts, and draw conclusions from them useful to the conduct of his business. The specific heat of these metallic vapors is 0.225 per cubic meter (calculated to standard conditions), and the latent heat of vaporization is approximately twenty-three times the normal boiling point expressed in absolute temperature for one atomic weight of the metal as explained in Part I of these calculations. For silver and gold we would have the latent heats: Ag, 23 (2,040 + 273) = 53,200 Cal. per 108 kg. = 493 " I kg. Au, 23 (2,530 + 273) = 64,470 " 197 kg. = 327 " 1 kg. The above latent heats of vaporization are for boiling at normal atmospheric pressure ; for vaporization at other pressures and corresponding temperatures, correction must be made for the difference in specific heats of the metallic vapor and liquid metal. CHAPTER IV. THE METALLURGY OF ZINC. (INCLUDING CADMIUM AND MERCURY.) At the present time, the metallurgy of zinc is briefly com- prehended in the following statements: The chief ore is zinc sulphide, ZnS, infusible at ordinary furnace heats, non-volatile, easily roasted to ZnO ; ' the roasting is done principally in me- chanically stirred furnaces, the ore being in small pieces, because it is non-porous, compact, and roasts slowly; the roasted ore, principally ZnO, is mixed with an excess of carbon as a re- ducing agent, and heated in closed fire-clay retorts having condensers attached; zinc vapors begin to come off at 1033 C., and come off rapidly at the working temperature of the charge, say 1200 to 1300; the zinc vapor and carbon monoxide pass into the condensers, and as they cool deposit the zinc, some in the form of fine dust (like hoar frost) , most of it as liquid drops; the cadmium in the ore and some lead, if present, distil over with the zinc, constituting its chief impurities. Arsenic is sometimes present in the condensed product. Iron does not distil over, but some is absorbed from ladles and moulds in which the liquid zinc may be handled and cast. The chief operations with which calculations may be con- cerned are the roasting, reduction by carbon, condensation of the vapors, possibility of blast-furnace extraction, of electric furnace reduction, electrolytic extraction, electrolytic refining. Roasting of Sphalerite. Before roasting, the crude ore is crushed and concentrated. Pure ZnS contains 67 per cent zinc and 33 per cent sulphur; its specific gravity is 3.9 and it has fine cleavage, so that it crushes easily, producing much fines or slimes. In the Joplin district, in Missouri, the largest zinc field in America, the ore as mined averages some 4.3 per cent of zinc, equal to 6.4 per cent of pure blende, ZnS, and is concentrated by jigging to 620 THE METALLURGY OF ZINC. 621 heads carrying about 60 per cent of zinc (90 per cent of ZnS), containing some 70 per cent of the zinc in the ore, and tails carrying about 1.35 per cent of zinc (2 per cent of ZnS), repre- senting 30 per cent of the zinc in the ore. The concentrates average 5 per cent of the weight of the ore; concentration ratio 20 to 1. Cost of such crushing and concentration 20 to 40 cents per ton of ore milled. (Ingalls.) The average com- position of these concentrates is: Zinc 60 per cent = 90 per cent ZnS. Iron 2 " = 3 per cent FeS. Silica 7 Sulphur 31 100 Zinc sulphide begins to be oxidized by air at a dull red heat, say, 600 C., and if the supply of air is kept up the oxidation is rapid, generating a large amount of heat and consequent high temperature. 2ZnS + 3O 2 = 2ZnO + 2SO 2 The question as to how high a temperature would theoret- ically result if sphalerite were, thus burned is an interesting one. Ingalls quotes Hollaway as giving 1992 C. We will investigate this point as being of interest and value in connection with practical roasting. Problem 129. Pure ZnS is oxidized by air. The heats of formation con- cerned are: (Zn, S) = 43,000 (Zn, O) = 84,000 (S, O 2 ) = 69,260 The specific heats of the materials concerned are: Sm to O ZnS = 0.120 + 0.00003t Zn O = 0.1212 + 0.0000315t SO 2 (1 m 3 ) = 0.36 +0.0003t O 2 or N 2 ( " ) = 0.303 + 0.000027t Assuming that the blende is first heated to 600, to start the roasting, and then the air supply put on, the exposed roasting 622 METALLURGICAL CALCULATIONS. surface being sufficient for rapid oxidation, and the tempera- ture too high to form SO 3 : Required: (1) The theoretical temperature at the roasting surface at starting, if all the oxygen passing is utilized. (2) The same, when the operation has continued to its max- imum temperature. (3) The same, if three-fourths the oxygen passing is utilized. (4) The same, if half the oxygen passing is utilized. (5) The same, if the resulting gases contain only 5 per cent of SO 2 gas. Solution : (1) The reaction gives, thermally: Decomposition of ZnS 43,000 Cal. Formation of ZnO +84,800 " " SO 2 +69,260 " Net heat evolution 111,060 Cal. This would be concerned in the oxidation of 97 kg. of ZnS to 81 kg. of ZnO and 64 kg. of SO 2 (= 22.22 m 3 ), and requiring 3X16 = 48 kg. of O 2 = 208 kg. of air (containing 160 kg. = 127 cubic meters of N 2 ). Heat in 97 kg. of ZnS at 600 = 8,032 Cal. Total heat in the products = 119,092 " Mean heat capacity of the products per 1, from to t: 81kg. ZnO = 9.8172 + 0.002552t 22.22m 3 SO 2 = 8.0000 + 0.006667t 127 m 3 N 2 = 38.4810 + 0.003429t. Sum = 56.2982 + 0.012648t Therefore, theoretical surface temperature, at starting, 119,092 _ 1KftRor m = 56.2982 + 0.012648t = (2) When the surface of the oxidizing material has attained its maximum temperature, it is practically at t, and the pro- ducts of combustion will contain 111,060 Calories plus the heat in 97 kg. of ZnS at t. We then have THE METALLURGY OF ZINC. 623 111,060 + 11.24t + 0.00291t 2 56.2982 + 0.012648t 1780. (2) It must be emphasized that this is the theoretical maximum under the most favorable conditions as to oxidation and exact air supply, conditions never realized in practice. (3) The excess of oxygen would be 48 x J = 16 kg. = 11.11 m 3 . This would be the most favorable possible proportion for mak- ing sulphuric acid from the gases, since it would just serve to oxidize the SO 2 to SO 3 in the acid chambers, the complete re- actions being: Roasting -ZnS + 2O 2 = ZnO + SO 2 + O Acid chambers- SO 2 + O = SO 3 . The ll.llm 3 of oxygen corresponds to 53.4m 3 of excess air, whose mean heat capacity is 16.1903 + 0.001443t. Adding this to the mean heat capacity of the products we have: 111,060 + 11.24t + 0.00291t 2 72.4885 + 0.014091t (4) This is more nearly the practical conditions, and apply- ing the principles above explained, we have: 111,060 + 11. 24t + 0.00291t 2 ^ 104.8691 + 0.016977t (5) If the gases contain 5 per cent of SO 2 gas, their total volume per formula weight of SO 2 produced, in kilograms, is: 22.22-0.05 = 444.4m 3 and the volume of excess air they contain is: 444.4 - 22.2 - 127 = 295.2 m 3 the heat capacity of this excess air, per average 1 is: 89.4456 + 0.0079701 and 111,060 + 11. 24t + 0.00291t 2 145.7438 + 0.024947t = 731 (5) It follows from this analysis and these results, that effective auto-roasting of zinc sulphide is practicable provided that the ore is finely divided, so as to expose large surface, oxidize quickly, and utilize the oxygen of the air efficiently. Without these conditions, auto-roasting is impracticable. 624 METALLURGICAL CALCULATIONS. It ought to roast satisfactorily by pot roasting, if the proper conditions as to size of ore, speed of blast, thickness of pot walls, etc., can be found. Problem 130. Average blende concentrates, containing 90 per cent ZnS, 3 per cent FeS, 7 per cent SiO 2 , are roasted by the use of 30 per cent of coal, in a furnace with a hearth 135 feet effective length, the unroasted sphalerite remaining being 2.6 per cent of the roasted material. The iron is roasted to Fe 2 O 3 . The coal carries 75 per cent of carbon, 5 per cent of hydrogen, 8 per cent of oxygen, 1 per cent of sulphur and 11 per cent of ash. The chimney gases average 2 per cent of SO 2 and escape from the furnace at 300 C. Charge is drawn from the furnace at 1000 C. Furnace roasts 40,000 pounds per day. Width of furnace, 12 feet; height, 8 feet. Required : (1) The composition of the roasted ore. (2) The heat generated by the roasting of 2,000 pounds of the ore. (3) The heat generated by the combustion of the fuel. (4) The heat in the hot charge as drawn. (5) The heat in the chimney gases. (6) The heat lost by radiation and conduction. , (7) The heat loss calculated to pound-calories per square foot of outer surface per minute. Solution : (1) The 2.6 per cent of ZnS in the roasted ore is per cent of it, and not of the raw ore. The simplest method of getting the composition of the roasted ore is to represent by x the quantity of ZnS remain unoxidized per 100 of raw ore. We then have: Weight of ZnS oxidized, per 100 raw ore = 90 x " " ZnO formed = (90 - x) X S - 75.2 - 0.835 x y / Fe 2 O 3 " == 3 X =2.7 THE METALLURGY OF ZINC. 625 Weight of SiO 2 remaining =7.0 " roasted ore =84.9- 0.835 x " " " " also = x- 0.026 Therefore 84.9- 0.835 x = x- 0.026 Whence x = 2.2 Percentage of the ZnS oxidized nrv 09 9Q = 0.976 = 97.6 per cent. Composition and weight of the roasted ore: ZnS = 2.2 = 2.6 ZnO = 73.4 = 86.1 Fe 2 O 3 = 2.7 = 3.1 SiO 2 = 7.0 = 8.2 Weight 85.3 Zinc contents = 60.0 = 70.3 per cent. (1) (2) Zinc sulphide oxidized to ZnO 2000 X (0 . 90 - . 022) = 1,756 Ibs. Heat of oxidation of 97 Ibs. of ZnS to ZnO and SO 2 (from Prob. 129) = 111,060 Ib.-Cal. Heat of oxidation of 1 Ib. 1,145 " " " " " 1,756 Ibs. = 2,010,620 " (fc, (3) Heat of combustion of 1 Ib. of fuel to CO 2 and H 2 O vapor: C to CO 2 0.75X8100 6075 Ib'.-Cal. Available Hydrogen 0.05 - 0.01 = 0.04 H to H 2 O condensed 0.04X34,500 = 1380 " " Calorific power to H 2 O condensed 7455 Water formed 0.05X9 = 0.45 Latent heat of condensation 0.45X606.5 . = 273 Calorific power to H 2 O vapor = 7182 626 METALLURGICAL CALCULATIONS Coal used per 2000 Ibs. of ore 600 Ibs. Calorific power = 600X7182 = 4,309,200 Ib.-cal. (3) Total heat generated in the furnace 2,010,620 + 4,309,200 = 6,319,820 " " Proportion of total generated by the roasting 2,010,620-^6,319,820 = 0.325 = 32.5 per cent. (4) Charge, as drawn, per 2000 Ibs. of raw ore: ZnS 44 Ib. ZnO 1468 " Fe 2 3 54 " SiO 2 140 " 1706 Heat in this, at 1,000 C. ZnS 44X0.150 = 6.6 ZnO 1468X0.153 = 224.6 Fe 2 O 3 54X0.344 = 18.0 SiO 2 140X0.260 = 36.4 285.6X1000 = 285,600 Ib.-Cal. (4) (5) Sulphur going into the gases: S from ZnS =1756X32/97 = 579.4 Ib. S " FeS = 60X32/88 = 21.8 " S " coal = 600 X 0.01 = 6.0 " Weight of S = 607.2 ' " " O for SO 2 = 607.2 " " " SO 2 1214.4 " Volume of SO 2 (1214.4 X 16) -^ 2.88 = 6,747 cubic feet. Volume of chimney gas 6,747-^-0.02 = 337,350 " Volume of CO 2 in it (600X0.75X16)^1.98 = 3,636 " a Volume of H 2 O in it (600X0.45X16)^-0.81 = ' 5,333 " a Volume of N 2 and excess air =321,634 " a THE METALLURGY OF ZINC. 627 Heat in these gases at 300 C. CO 2 3,636X0.436 = 1,585 oz.-cal. per 1 H 2 O 5,333X0.385 = 2,057 " " " SO 2 6,747X0.450 = 3,036 " " " N 2 andO 2 321,634X0.311 =100,028 " " " 106,706 " " = 6,669 Ib.-Cal. = 2,000,700 " " 300 (5) (6) Summary of heat distribution : Heat available, per 2,000 Ib. ore = 6,319,820 Ib.-Cal. Heat in hot ore drawn, 235,600 ' chimney gases, 2,000,700 Loss by radiation and conduction, 4,083,620 6,319,820 ' (6) (7) Approximate periphery of furnace : 12 + 12 + 8 + 8 = 40 feet. Approximate outer surface, including base: 40X135 = 5,400 sq. feet. Ore roasted per hour: 40,000-24 = l,6671b. Heat radiated and conducted away per hour: 1)667 = 3 > 360 > 000 Ib.-Cal. Per square foot outside area, per hour: 3,360,000-^5,400 = 622 Ib.-Cal. per minute = 10.6 " " (7) Reduction of Zinc Oxide. Since metallic zinc boils under atmospheric pressure at 930 C., and carbon does not begin to reduce zinc oxide until 1033 is reached, the zinc reduced is necessarily obtained in the state of vapor. To make the reaction proceed fast, a temperature of the charge inside the retorts of 1100 to 1300 C. at the end 628 METALLURGICAL CALCULATIONS. is necessary. A large amount of heat is absorbed in the reac- tion, which must be supplied as a current or flow of heat through the walls of the retort in order to keep the reaction and re- duction going. Since the walls of the retort are fire-clay, aver- aging 4 centimeters (1.6 inch) thick, there must be a consider- able difference of temperature (temperature head) between the inside and outside of the retort in order to keep up the flow of heat inward. THERMOCHEMICAL CONSIDERATIONS. The heat of formation of zinc oxide, at ordinary tempera- tures, is 84,800 Calories for a molecular weight, 81 kilos, of oxide, as determined by thermochemical experiment. This means that cold zinc uniting with cold oxygen, and the hot product cooled down to the same starting temperature, results in the evolution of heat stated. Cold carbon uniting with cold oxygen to cold carbonous oxide, CO, gives similarly 29,160 Calories per molecular weight, 28 kilograms, of gas formed. These facts are expressed thermochemically as (Zn, O) = 84,800 (C, O) -. 29,160 The equation of reduction becomes ZnO + C '= Zn + CO - 84,800 +29,160 = - 55,640 This means that if we start with cold zinc oxide and cold carbon, the heat absorbed is 55,640 Calories ending up with cold products, zinc and CO at ordinary temperatures. What is actually done in practice, however, is to first heat the ZnO and C to a high temperature, at least to 1033, and then to supply the heat of the reaction at that temperature, pro- ducing zinc vapor and CO gas, both at 1033. The process of reduction, therefore, resolves itself plainly into two steps: (1) heating the charge up to the reacting temperature, (2) supply- ing the latent heat of the chemical change at that temperature. Heating up the Charge. The charge usually contains more carbon than is theoretically necessary for reduction of the zinc oxide, because some is con- sumed by air in the retort, some in reducing iron oxides and THE METALLURGY OF ZINC. 629 some is left behind unused; it is cheaper to lose carbon than unreduced zinc oxide. However, making the calculation on the theoretical amount of carbon only (anyone can modify it for any excess of carbon used in any specific case) we need to know the heat necessary to raise these reacting substances to the temperature of reaction. The heat in 1 kilogram of zinc oxide at various temperatures was determined in the writer's laboratory as 0.1212t + 0.0000315t 2 . The heat in 1 kilogram of carbon at temperatures above 1000 is represented by 0.5t 120. We have the heat necessary to raise our 81 kilograms of ZnO and 12 kilograms of carbon to 1033 as ZnO: 0.1212 (1033)4-0.0000315 (1033) 2 X81 = 159X81 = 12,879 Cal. C : [0.5 (1033) - 120] X 12 = 396X12= 4,752 " Sum = 17,631 This quantity represents the heat per 81 of oxide and 12 of carbon, or 93 of charge containing 65 of zinc. Per 1000 kilo- grams of oxide, containing 800 kilos of zinc, it would be 17,631X^5 = 217,670 Calories. ol It should be noted that this quantity is actually proportional to the weight of the charge, ore plus reducing agent, and inde- pendent of the amount of zinc in it. A ton of poor ore will practically require as much heat to bring it up to the reaction temperature as a ton of rich ore, so that these costs are propor- tional to the weight of charge treated and not to the weight of zinc it contains. If the charge is heated electrically, the amount of electric power needed for heating up can be calculated, assuming an average loss of 10 to 30 per cent of the total heat by radiation and conduction from the furnace. Illustration : A retort contains 300 kilograms of charge mixture and is heated electrically to 1033, the reaction temperature at an effi- ciency of 75 per cent, by an electric current of 250 horse-power. How long will the heating-up period last. 630 METALLURGICAL CALCULATIONS. Solution : Heat needed in. the charge, 217,670X0.3 = 65,300 Calories. Heat to be supplied = 65,300^-0.75 = 87,070 Calories. Power applied supplies 635X250 = 158,750 Calories per hour. Time required 87,070-r-158,750 = 0.55 hour = 33 minutes. If the charge is heated by furnace gases outside the retort we must take into consideration that the fire-clay is a poor con- ductor, that the rate of transmission of heat through the walls falls off as the -charge becomes hot, and that the outside sur- face of the retort must be kept well above 1033 in order to get the charge to that temperature in a reasonable time. The conductance of fire-clay for high temperature is 0.0031; that is, 0.0031 gram-calories pass through each square centimeter per second, if one centimeter thick, per 1 C. difference of tem- perature. When the charge is cold, the inner surface of the retort may be reduced in temperature to a low red heat, say, 500, and towards the end of the heating-up period its tempera- ture becomes at least 1033, while the outer surface is kept continually at 1200, let us say, by the furnace gases. During the heating-up period there is a difference of temperature producing heat flow of 700, for a short time, down to, say, 200, which we may average up as 300 heat difference. The heat conductivity of the material of the retort and the thick- ness of its walls have everything to do with the rate at which the heat can get through and the charge be heated up to the reaction temperature. Problem 131. An oval zinc retort is 130 centimeters long, 30 centimeters diameter outside, one way, and 15 centimeters the other, and its walls are 3 centimeters thick. It is charged with 30 kilo- grams of ore and 12 kilograms of reduction material. The temperatures of the charge in the retort and of the gases out- side the retort were as follows: In Retort Outside Difference At starting. 0C. 1067C. 1067C. In 0.5 hour 350 1067 717 " 1.0 " 600 1067 467 " 1.5 hours 781 1067 286 " 2.0 " 814 1100 286 THE METALLURGY OF ZINC. 631 In Retort Outside Difference 2.5 " 869 1100 231 " 3:0 " 924 1110 187 " 3.5 " 957 1155 198 ." 4.0 '" 935 1166 231 " 4,5 " 935 1138 203 " 5.0 " 946 1144 198 " 5.5 " 946 1155 209 " 6.0 979 1166 187 " 6.5 ' 1001 1177 176 " 7.0 " 1034 1177 143 Average, 319 Take 159 Calories per kg. as the heat required to bring the ore to 1034 and 396 for the reduction material. Required : The average heat conductivity in C. G. S. units of the mate- rial of the retorts, in the lange given, assuming the inner sur- face of the retort to be at the same temperature as the charge. Solution: Heat passing through the retort walls in 7 hours: 159X30 = 4770 Calories 396X12 = 4752 9522 Per second = 0.378 " = 378 gram-cal. Surface of retort: Periphery V30X15X3.14 = 66 cm. Area of sides, 66X130 = 8580 sq. cm. Heat passing through each square cm. per second 378 + 8580 = 0.044 cal. Heat passing per 1 difference 0.044-319 = 0.00014 cal. Since thickness is actually 3 centimeters, conductance in C. G. S. units is: 0.00014X3 = 0.00042. 632 METALLURGICAL CALCULATIONS. Correction : This coefficient of conductance is far too low. The reason is that the inner temperature, the temperature of the charge, is always lower than the temperature of the inside surface of the retort, and the difference of temperature between the outer' and inner walls of the retort must have been far less than the aver- age, 319. If we take the coefficient of conductance as deter- mined by experiment for firebrick, viz., 0.0031, then the average difference of temperature between the inner and the outer walls of the retort would be: 0.00042 9X 0.0030 ~ C ' This is more likely, than that the conductance of the retort material should be So extraordinarily low. In fact, we are led to the conclusion that the poor heat conductivity of the charge itself is the chief obstacle to its rapid heating, and that pre- heating of the charge would be very advisable if it could be done by some of the waste heat of the gases leaving the furnace. Distillation of the Charge. The driving off of the zinc is altogether a different operation from the heating up of the charge to the stated reduction tem- perature. It is an endothermic chemical operation, comparable to the boiling of water at a constant temperature, the latent heat of the chemical reaction is exactly comparable to the latent heat of vaporization. The temperature must be kept up to the temperature of reduction in order for the reaction to take place at all, and then heat-calories must be supplied at that temperature to keep the reaction going, and the reduction proceeds pari passu with the quantity of heat supplied at that constant temperature. The question now is: what is the latent heat of this reaction at the reaction temperature. We must for this purpose know the heat of formation of the substances involved, ZnO and CO, at 1033. The method of calculating these is too long to insert here, but may be learned from the writer's book on Metallurgical Calculations, Part I, p. 51. We have the heats of formation from the elements as they exist at 1033: THE METALLURGY OF ZINC. 633 (Zn, O) 1033 = 112,580 (C, O) 1033 = 30,091 and the reaction at 1033 ZnO + C = CO + Zn - 112,580 + 30,091 = -82,489 This is seen to be nearly 50 per cent greater than the heat of the reaction calculated to ordinary temperatures, from the ordinary heats of formation as taken from thermochemical tables. The metallurgist should understand this difference, for it is one of the utmost importance in thermochemical calcula- tions, if we want our calculations to represent actual conditions and to check up with practice. This amount of heat is proportional to the zinc oxide re- duced or to the zinc distilled from the charge, but not to the weight of ore charge itself. This requirement will, therefore, be greater, per retort full of material, the richer the charge is in zinc. It is a constant requirement for a given output of zinc, and not per given weight of ore treated. The heat required for the reduction, therefore, as distin- guished from the heating-up period, is oo 400 1* = 1,018 Calories per kg. of ZnO reduced ol 82 '^ 89 = 1,269 " " " " Zn distilled off o5 = 1,269,000 " " ton " " Furnishing this reduction heat, at the high temperature re- quired, is the larger part of the heat needed in the whole pro- cess. We see that it is some 4.7 times the amount of heat needed to raise the materials to the reduction temperature. Problem 132. The retort charge of Problem 131 was kept at the reduction temperature for 14 hours, the temperature outside the retort being gradually raised to 1300 and the average difference in temperature between the gases and the charge being 147, and there being distilled from the charge in that time 18 kilograms of zinc. 634 METALLURGICAL CALCULATIONS. Required: The average heat conductance in C. G. S. units of the mate- rial of the retort from the above data and assumptions. Solution : The heat furnished in the 14 hours was as follows: 1,269X18 = 22,842 Calories. Heat furnished per hour: 22,842-14 = 1,632 Heat furnished per second: = 0.453 = 453 calories Heat passing each sq. cm. of retort surface per second: 453 -t- 8,580 = 0.053 calories. Per 1 difference: 0.053^-147 = 0.00036 " Conductance, in C. G. S. units: 0.00036X3 = 0.00108 Remarks: This conductance calculates out 2.5 times as great as from the data on the heating-up period, the reason of the higher value being that the charge is at nearly uniform tem- perature during this reduction period, and therefore the dif- ference between the temperature taken in the' middle of the charge and the true temperature of the inner walls of the retort is less than before, and the error from this source less. If we make the same assumption as in the correction to the previous problem, i.e., take the conductance of the retort material as 0.0030, the difference in temperature of the outer and inner walls of the retort during this period would calculate out 147X0.00108 0.0030 while the center of the charge was then 147 53 = 94 cooler, on an average, than the inner wall of the retort from which it was deriving its heat. These figures appear reasonable, and the writer would con- THE METALLURGY OF ZINC. 635 elude, from the data so far available, that the conductance 0.003 probably represents a good approximation to the correct value for zinc retort material, but that, in order to use it, we should have more experimental data as to the average difference between the temperature of the inner walls of the retort and the temperature of the charge at various points in the retort. Problem 133. A zinc ore containing 50 per cent of zinc is mixed with 40 per cent of its weight of small anthracite coal, and retorted in a Belgian furnace. The recovery of zinc was 82 per cent of the zinc content of the ore. The consumption of anthracite to heat the furnace was 2.25 tons per 10 of ore. The anthracite contained 90 per cent of carbon and 10 per cent of ash, and had a calorific power of 7500. Required : (1) The total consumption of fuel per 1000 of zinc obtained. (2) The efficiency of transfer of heat from the furnace gases to the charge. Solution : (1) Zinc charged, per 1000 of zinc obtained 1 000 -H 0.82 = 1220 Ore charged, per 1000 of zinc obtained 1220-^0.50 = 2440 Coal charged with ore 2440X0.40 = 976 Coal burned on grate 2440X2.25 = 5490 Total coal used, per 1000 of zinc obtained 976 + 5490 =6466 (1) (2) Calorific power of coal burned 5490X7500 = 41,170,000 Heat required to raise ore to reduction point 2440 X 159 = 387,960 636 METALLURGICAL CALCULATIONS. Heat required to raise fuel to reduction point Ash 98 X 159 = 15,580 Carbon 878 X 396 = 347,690 Total to raise charge to reduction point 751,230 Heat absorbed in distilling away 1000 of zinc 1,269,000 Total heat utilized = 2,020,230 Thermal efficiency of utilization of the fuel = 0.049 = 4.9 per cent. (2) Problem 134. Natural gas from lola, Kan., has the following composition: CH 4 93 per cent. H 2 2 CO 1 C 2 H 4 1 N 2 3 " It is used in a zinc retort furnace, at an efficiency of transfer of heat to the charge of 4.9 per cent, as calculated in Prob. 133, working a charge which absorbed 2,000,000 Calories per 1000 kilograms of zinc distilled off. Required: The volume of natural gas required to be used to displace the anthracite fuel used per 1000 kilograms of zinc produced. The cubic feet of gas per 1000 Ibs. of zinc produced. Solution : Calorific power of the gas CH 4 0.93X 8623 = 8009 H 2 0.02X29030 = 581 C 2 H 4 0.01X14365 = 144 CO 0.01 X 3062 = 31 8765 THE METALLURGY OF ZINC. 637 This result may be called Calories per cubic meter of gas, or ounce-calories (1 C.) per cubic foot according to whether it is desired to work in metric units or the English system. Cubic meters of gas required, per 1000 kilograms of zinc produced: 2,000,000 0.049 Per 1000 Ibs. of zinc produced: 2,000.000 0.049 X 16 ^8765 = 75,200 cubic feet. Electric Smelting of Zinc Ores. In an interesting communication to the American Electro- chemical Society (Vol. XII, p. 117), Gustave Gin calculates the electric power necessary for the smelting of several varieties of zinc ore, assuming that the zinc vapor and other gases pass out of the furnace at the usual high temperature 1200 C. Mr. Gin makes his calculations of heat required on the basis of molecular weight of zinc compound reduced; that is, for 81 parts of ZnO and 65 parts of Zn. Calculating to 1200, he finds the heat in the products Zn and CO to be In 65 parts Zn 26,720 Cal. u 28 " CO.. . 8,280 " Sum 35,000 Cal. whereas we calculate for the same quantities In 65 parts Zn 37,695 Cal. "28 " CO ....: 8,945 " 46,640 Cal. The difference between these numbers is principally in the latent heat of vaporization of zinc,. which Gin assumes as 15,370 Calories, but which by a method of evaluation used by the writer figures out 27,670 Calories, and almost exactly the same value has been obtained by a different method by W. McA. Johnson. Assume then, that we start with cold materials and end with 638 METALLURGICAL CALCULATIONS. the products of the reaction leaving the retort at 1200, the sum total of usefully applied heat is the heat of the reaction calculated at ordinary temperature plus the sensible heat of the products at 1200, or Heat absorbed by reaction, ordinary temperature. . . .84,800 Cal. Heat in necessary products, at 1200 46,640 Total 131,440 To this must be added the sensible heat in the residue left in the retort, to get the total heat which has been applied to the charge. If the charge were pure ZnO, with the theoretical amount of carbon, the residue would be nil, but in practice there is always a residue of gangue with unused carbon. Mr. Gin, having calculated the heat requirement on the above basis, then makes his calculations of power required per ton of ore smelted in the following ingenious way: The weight of each component of 1000 kg. of ore is divided by its mole- cular weight, and thus the number of molecular weights of material in one ton of ore determined, which, so to speak, gives a kind of chemical formula for the ore. An example will make this clear. Composition of a calcined calamine zinc ore : ZnO 40.50 per cent. ZnSiO 3 38.07 Fe 2 O 3 9.60 APSiO 5 8.10 CaO 2.80 If we divide the weight of each compound present by its molecular weight we get the relative number of molecules present in the ore, and the formula for the ore: KG. IN 1000 KG. ZnO 405 -4- 81 = 5.0 molecular weights. ZnSiO 3 380.7 -f-141 = 2.7 Fe 2 3 96. -h 140 = 0.6 APSiO 5 81 -7-162 = 0.5 tt CaO 28 -h 56 = 0.1 " THE METALLURGY OF ZINC. 639 The ore may therefore be represented by the formula 5ZnO + 2.7ZnSi0 3 + 0.6Fe 2 O 3 + 0.5APSiO 5 + 0. ICaO. Assuming that the Fe 2 3 becomes FeSiO 3 , and that there is to be added enough CaO to form CaSiO 3 with the rest of the SiO 2 of the zinc silicate, we will need 1.5 CaO to do it, and since there is 0.1 CaO present, 1.4 CaO must be added, which represents 1.4X56 = 79 kg. of CaO. To form CO with the oxygen combined with the zinc and iron, reducing the latter to FeO, will require: for 5 ZnO 5.0 C. 2.7ZnSiO 3 2.7 " " 0.6 Fe 2 O 3 0.6 " Sum 8.3 C. = 100 kg. If we use twice the theoretical amount of carbon needed for reduction we have the following balance sheets; weight in kilo- grams being enclosed: ORE ADDED CHARGE 5 ZnO (405) 5 ZnO (405) 2.7 ZnSiO 3 (381) 2.7 ZnSiO 3 (381) 0.6Fe 2 O 3 (96) .... 0.6 Fe 2 O 3 (96) O.SAPSiO 5 (81) O.SAPSiO 5 (81) 0.1 CaO (28) 1.4 CaO (79) 1.5 CaO (79) 16.6 C (200) 16.6 C (200) The distribution of this charge will be as follows: CHARGE. GASES. RESIDUE. 5 ZnO (405) 5 Zn (325) 2.7ZnSi0 3 (381) 2.7 Zn (175) 0.6Fe 2 O 3 (96) 1.2 FeSiO 3 (158) 0.5 APSiO 5 (81) 0.5 APSiO 5 (81) 1.5 CaO (79) 1.5 CaSiO 3 (174) 16.6 C (200) 8.3 CO (232) 8.3 C (100) The essential reactions involved in the reduction are expressed by the reaction formula: 5ZnO + 2.7ZnSiO 3 + 0.6Fe 2 O 3 + 1.5CaO + 8.3C .2FeSiO 3 +1.5CaSiO 3 + 8. 640 METALLURGICAL CALCULATIONS The heat requirements per ton of ore treated may therefore be figured out as: Calories. Decomposition of 7.7ZnO = 84,800X7.7 = 652,960 2.7ZnSiO 3 into ZnO and SiO 2 = 10,000X2.7 = 27,000 Decomposition of 0.6Fe 2 O 3 into FeO and O = 64,200X0.6 - 38,520 Heat absorbed in decompositions = 718,480 Formation of 1.2 FeSiO 3 from FeO and SiO 2 =8,900X1.2 = 10,680 Formation of 1.5 CaSiO 3 from CaO and SiO 2 = 17,850X1.5 = 26,770 Formation of 8.3 CO = 29,160X8.3 =222,030 Heat evolved in formation heats 259,480 Net heat of chemical reactions absorbed 459,000 Sensible heat in 7.7 Zn = 46,640X7.7 = 359,130 8.3 CO = 8,945X8.3 = 74,240 " " 8.3 C = 5,760X8.3 = 47,810 11.2 FeSi'O 3 0.5 APSiO 5 1.5 CaSiO 3 = 413 kilograms = 460X413 = 200,000 Sensible heat in products and residue = 681,180 AoM heat absorbed in chemical reactions = 459,000 Total heat requirement of the charge = 1,140,180 Important principle : The heat absorbed in an electric furnace by chemical reactions is electric energy utilized at an efficiency of 100 per cent. It is only part of the sensible heat of the materials being treated which is lost by radiation and con- duction. Heat losses by radiation and conduction in electric furnace processes should be expressed upon the total energy of the current diminished by the heat absorbed in chemical reac- tions, and not upon the total energy of the current. This principle has been expressed most clearly by Mr. F. T. Snyder, THE METALLURGY OF ZINC. 641 of Chicago, the pioneer electric furnace zinc metallurgist of America. Expressed thus, electric furnaces give higher efficiencies the greater the absorption of heat in chemical reactions taking place within them. In mere physical processes, such as melting, electric furnaces on a large scale give 75 per cent net efficiency, with 25 per cent loss by radiation and conduction. In the above case, figured out for 1000 kg. of zinc ore, 40 per cent of the total heat requirement is absorbed as chemical heat at 100 per cent efficiency, and 60 per cent is needed as sensible heat. This 60 per cent then represents the net sensible heating effect, and the loss of heat by radiation and conduction must be calculated as one-third of this quantity, and not one-third of the total net heat requirement. We therefore have : Calories. Net heat requirements for chemical reactions ......... 459,000 , " sensible heat ............... 681,180 Loss by radiation and conduction ................... 227,060 Gross heat requirement of the furnace ............... 1,367,240 Kilowatt hours of current required Mr. Gin's figures have been modified by the writer, as ex- plained above, and anyone consulting his paper on this subject may make similar modifications to the other cases cited in that paper. Mr. F. T. Snyder, of the Canada Zinc Company, at Van- couver, B. C., has operated the first practical electric zinc smelting furnace in America. Treating mixed lead and zinc ores, Mr. Snyder uses the furnace and process protected in his United States patents of July 2, 1907. (See Electrochemi- cal and Metallurgical Industry, V. 323, 489.) The lead is obtained liquid, and the zinc also condensed to the liquid state before leaving the furnace proper, the heat of condensation being partly absorbed by water cooling the condensers and partly by the descending charges, which carry it back into the focus of the furnace. Under these circumstances, Mr. Snyder reports that he has attained unexpectedly low results as to 642 METALLURGICAL CALCULATIONS. power requirement. It will be recalled, that in discussing the question of the electric smelting of zinc ores in general, and Mr. Gin's process in particular, we assumed the zinc vapor and CO gas to escape from the furnace at a minimum of 1033 C. If, as in Mr. Snyder's furnace, they escape at about 500, the zinc liquid, the heat in these hot products is reduced very con- siderably, particularly that in the zinc. Mr. Snyder states in discussing Mr. Gin's paper (loc. cit.) that he has smelted pure zinc oxide at an expenditure of 1050 kilowatt hours per 1000 kilograms of oxide. Let us see how this coincides with the theoretical figures: Calories Heat value of 1050 kw-hours .......... . . = 1050X860 = 903,000 Heat of the chemical reactions, assumed to take place at ordinary temperatures ............. 1000X687 = 687,000 Sensible heat in products, at 500 Zinc: 800 X 80 = 64,000 CO: 342X152 = 52,000 = 116,000 Leaving, by difference for radiation, conduction and cooling water = 100,000 Mr. Snyder does not give details as to the exact working of his furnace, but his claim to reduce a ton of zinc ore with 1000 kw-hours is seen to be a possibility, if he can remove the zinc as liquid zinc from the furnace, not lose too much heat in cool- ing water, and get most of the heat of condensation of the zinc vapor usefully returned by the descending charges into the working focus of the furnace. Snyder's furnace, working as claimed, would show a useful efficiency of on the current used, with 12 per cent losses. The real heat lo'sses, however, should be expressed upon the energy used less that absorbed in chemical reactions, or as 100,000 loss on 216,000 used for sensible heat, = 0.46 = 46 per cent. THE METALLURGY OF ZINC. 643 This shows that the furnace loses by radiation, conduction and cooling water 46 per cent of the energy developed as sen- sible heat in the furnace, but the latter item is only 24 per cent of the total energy applied to the furnace. In short, about three-quarters of all the energy consumed by the furnace is utilized in the heat of the chemical reactions produced; of the other one quarter, half of it is represented by the sensible heat of the products leaving the furnace and half is lost by radiation, conduction and cooling water. Zinc Vapor. The condensation of zinc from the state of vapor follows the same laws as regulate the condensing of all gases. If it is by itself in a cooler space, it condenses by contact with the walls as finely divided zinc dust (blue powder) if the walls are cold, as liquid zinc if the walls are hot, and it keeps on condensing until the tension of the remaining vapor is equal to the maxi- mum tension of zinc vapor at the temperature of the condensing vessel. This condition having been reached, no more will con- dense until the condenser is cooled. At any temperature, there- fore, an amount of zinc remains uncondensed corresponding to the maximum vapor tension of zinc at that temperature. If other indifferent gases are present, the zinc vapor sustains only part of the atmospheric pressure, or pressure on the whole mixture, so that condensation cannot begin until a lower tem- perature is reached; that is, a temperature at which the maxi- mum vapor tension of the zinc equals its partial vapor tension in the gas mixture. At this point condensation begins and con- tinues as the temperature falls, the gas mixture always being saturated with zinc vapor. The phenomenon is precisely simi- lar to the production of rain by the cooling of air containing moisture. As to what the vapor tension of zinc is, the data are scanty. The boiling, point under atmospheric pressure is 930 C., where its vapor tension is 760 mm. of mercury column. According to Barus, the maximum tension increases 6.67 mm. for each 1 C. increase of temperature. This is a difficult determination, but corresponds satisfactorily with the calculated increase from an- alogy to water and mercury. Water, boiling at 100 C. (373 absolute), varies 1 in boiling 644 METALLURGICAL CALCULATIONS. point for 27.20 mm. variation in vapor tension ; mercury, boiling at 357 C. (630 absolute) varies 1 in boiling point for 12.66 mm. variation in vapor tension. The ratio of the two varia- tions is seen to be somewhere in the inverse ratio of the two boiling points. 630 27.20 373 12.66 If we compare mercury with zinc, with the two boiling points 357 and 930 (630 and 1203 absolute), and the observed va- riations in vapor tension at the normal boiling point of 12.66 mm. and 6.67 mm. per 1 rise in boiling temperature, the two ratios are 1203 12.66 _ 630 ~ "6^7" This coincidence justifies us completely in assuming that the vapor tension curve of zinc may be calculated directly from that of mercury, by assigning for any given vapor tension an absolute temperature to the zinc vapor of 1.91 times that of mercury vapor. For cadmium we may use the similar ratio of the normal boiling points in absolute degrees: 273 + 780 1053 273 + 357 630 and thus calculate the cadmium curve from the mercury curve. From incipient vaporization to ebullition in a vacuum. Tension of vapor Mercury Cadmium Zinc mm. of Hg. C. C. C. 0.0002 183 248 0.0005 10 200 267 . 0.0013 20 216 286 0.0029 30 233 305 0.0063 40 250 324 0.013 50 267 344 0.026 60 283 363 0.050 70 300 382 THE METALLURGY OF ZINC. 645 Tension of vapor Mercury Cadmium Zinc mm. of Hg. C. C. C. 0.093 80 317 401 0.165 90 333 420 0.285 100 350 439 0.478 110 367 458 0.779 120 383 477 1.24 130 400 496 1.93 140 417 516 2.93 150 433 535 4.38 160 450 554 6.41 170 467 573 9.23 180 483 592 Before proceeding further, we would remark that cadmium melts at 320 and zinc at 419; that both can, therefore, show vaporizing phenomena nearly 150 below their melting points. At 9 mm. tension, 180 C., mercury begins to show the phenom- ena of ebullition, and we would, therefore, expect cadmium to " simmer " at 483 and zinc at 592. Neither of these obser- vations has as yet been experimentally investigated, so far as the author knows. From ebullition in vacuum to normal boiling point. Tension of vapor Mercury Cadmium Zinc mm. of Hg. C. C. C. 9.23 180 483 592 14.84 190 500 611 19.90 200 517 630 26.35 210 533 649 34.70 220 550 668 45.35 230 567 687 58.82 240 584 706 75.75 250 600 726 96.73 260 617 745 123. 270 634 764 155. 280 650 783 195. 290 667 802 242. 300 684 821 300. 310 700 840 646 METALLURGICAL CALCULATIONS. Tension of vapor Mercury Cadmium Zinc mm. of Hg. C. C. C. 369. 320 717 859 451. 330 734 878 548. 340 750 897 ' 663. 350 767 915 760. 357 780 930 The above table contains the more important data for the actual condensation of zinc, cadmium and mercury vapors in practice. Thus, if a mixture of zinc or cadmium vapor with an equal volume of indifferent gas goes into a condenser, the partial pressure of the metallic vapor being in this case only half an atmosphere, or 380 mm., no metal will commence to condense until the temperature of the gases is reduced to 862 for zinc and 720 for cadmium. If mercury vapor from a roaster is mixed with 19 times its volume of other gas, so that it only forms 5 per cent of the mixture, its partial tension will be only 5 per cent of 760 mm. = 38 mm., and no mercury will commence to condense until the temperature of the gas mixture is reduced to 224. These temperatures are exactly analogous to the phenomenon of the dew point of moist air, the tempera- ture at which rain forms. From normal boiling point to high pressures. Tension of vapor Mercury Cadmium Zinc atmospheres. C. C. C. 1.0 357 780 930 2.1 400 851 1012 4.25 450 934 1107 8. 500 1018 1203 13.8 550 1101 1298 22.3 600 1185 1394 34.0 650 1268 1489 50. 700 1352 1585 72. 750 1435 1680 102. 800 1519 1776 137.5 850 1602 1871 162. 880 1652 1928 THE METALLURGY OF ZINC. 647 The latter table may have several special applications. If zinc, for instance, were placed in a closed electric furnace filled with indifferent gas and capable of supporting 100 atmospheres pressure, zinc could be kept in the liquid state up to some 1700 C., and its alloying with other metals greatly facilitated. In the making of brass, for instance, much zinc is lost by vola- tilization at the melting point of the copper used, yet a pressure on the furnace of four atmospheres would entirely prevent loss of zinc at the melting point of copper. Induction electric furnaces could easily be run inside a pressure vessel at this temperature, and the pressure be relieved gradually as the temperature of the melted alloy was reduced. To keep zinc from boiling, or to keep it in the liquid state, at 1300, not a high temperature, would require a pressure of nearly 14 at- mospheres to prevent it from boiling ad libitum. Problem 135. Roasted zinc sulphide ore, consisting practically of pure ZnO, is reduced by carbon, and the reduction gases passed into a condenser. Assume the reaction to be ZnO + C = Zn + CO. Required : (1) The temperature at which zinc commences to condense from the above gas mixture. (2) The proportion of the zinc condensed out for each 1 reduction of temperature below this " dew point." (3) The proportion of the zinc escaping uncondensed, if the gas escapes from the condenser at 600 C. (4) How would these data be affected, if the reduction was taking place at Denver, 1500 meters above sea level, barometer 560 mm? Solution : (1) Since zinc vapor has been shown to have a specific grav- ity of 32.5 referred to hydrogen gas at the same temperature and pressure, and the molecular weight of hydrogen gas, for- mula H 2 , is 2, the molecular weight of zinc vapor must be 2X32.5 = 65. This coincides with the atomic weight of zinc, and therefore zinc vapor is monatomic and the symbol of its molecule is Zn. The gas mixture in the above equation con- 648 METALLURGICAL CALCULATIONS. tains, therefore, one molecule each of its constituents, and, therefore, each gas supports half the atmospheric pressure. This is ordinarily expressed by saying that the gas is half one constituent and half the other, meaning that if the two gases were separated and each measured under the prevailing normal pressure, the volume of each would be half the volume of the gas mixture. The statement that each supports half the nor- mal pressure is the more scientific and logical, for in a gas mixture each gas certainly possesses the volume of the mixture, but under only a fraction of the pressure on the mixture, said fraction being identical with the fraction of the volume of the whole which it would constitute, if measured separately under the pressure supported by the mixture. Consulting the table, we find zinc vapor to have a pressure of 380 mm. at 862. This would, therefore, be the dew point, at which the zinc would commence to condense. (1) (2) At this condensing temperature, a difference of 19 in the temperature corresponds to 82 mm. difference in the maxi- mum vapor tension, and, therefore, in this gas mixture at this temperature, a reduction of 1 in its temperature will reduce the tension of the zinc vapor 82-7-19 = 4.3, or practically 4 mm. 4 This will result in the condensation of -^^ th. of the zinc, be- ooU cause its tension was 380 and was reduced to 376, and, there- 376 fore, -r^r- th. of it remained uncondensed. The proportion con- OoU densed for the reduction of 1 in temperature is, therefore, a trifle over 1 per cent. (2) (3) Leaving the condenser at 600, the tension of the zinc vapor escaping would be, from the table, 11.6 mm. The ten- sion of the CO gas would, therefore, be 760- 11.6 = 748.4 mm. For every cubic meter of mixed gases, at 862, containing a cubic meter of zinc vapor at that temperature and 380 mm. pressure, there will escape a fraction of a cubic meter of mixed gas at 600, containing zinc vapor at 11.6 mm. pressure. The fractional volume can be calculated from the tension on the CO. 1 m 3 CO at 862 and 380 mm. pressure, reduced to 600 and 748.4 mm. pressure, becomes THE METALLURGY OF ZINC, 649 600 + 273 380 _ X 862 + 273 X 74O ~ This is, then, the actual volume of gas mixture escaping, for each actual 1 m 3 of gas mixture in the condenser at the " dew point," 862. The uncondensed zinc vapor is, therefore, 0.381 m 3 at 600 and 11.6 mm. tension. This would be, at 862 and 380 mm. tension 862 + 273 11.6 _ ' 3i LX 600 + 273 X W There, therefore, escaped uncondensed 0.015 = 1.5 per cent of the zinc. (3) A quicker solution, not so easy to understand, however, is 11.6 748.4 = -0.015 = 1.5 per cent uncondensed. (4) If the operation takes place at such elevation above sea- level that the barometer stands normally at 560 mm., then the partial pressure of the zinc vapor in above case would be 280 mm. instead of 380 mm. and the " dew point " or temperature at which condensation commences would be 835 instead of 862, as under previous conditions. At this temperature, a difference of 1 in the temperature corresponds to a difference of 3 mm. in the tension of the zinc vapor, producing therefore a condensation of the zinc present, per 1 fall of temperature, of 3-^-280 = 0.011 = 1.1 per cent, instead of 1.2 per cent. If the temperature of the escaping gases is 600, with the saturated zinc vapor at 11.8 mm. tension and the carbonous oxide, CO, at 56011.8 = 548.2 mm., the proportion of the original quantity of zinc escaping uncondensed is 11. 8-5- 548.2 = 0.022 = 2.2 per cent, instead of 1.5 per cent. Condensation of Zinc and Mercury Vapors. The temperature at which a metallic vapor, like zinc or mer- cury, commences to condense, depends upon the vapor tension curve of the metal and the amount of other uncondensable gas 650 METALLURGICAL CALCULATIONS. with Which it is mixed. These intermixed gases reduce the pressure upon the metallic vapor, and lower the temperature at which the vapor becomes saturated vapor. The phenomenon is exactly analogous to the " dew point " of air and the precipita- tion of rain. Problem 136. A roasted zinc ore contains 15 per cent. Fe 2 O 3 and 70 per cent. Zn O. It is reduced by excess of carbon, in a retort. Required: (1) The average composition, by volume, of the gas mixture entering the condenser, assuming no CO 2 in it. (2) The " dew point " of the mixture at which it begins to deposit zinc. (3) The percentage of zinc escaping condensation, if the gases leave the condenser at 600 C. Solution: (1) Per kilogram of zinc there is used, assuming complete reduction: 1-7- (0.70X65/81) = 1.78kg. Oxygen in 1.78 kg. of ore: As Zn O = 1X16/65 = 0.246 kg. As Fe 2 O 3 = 1.78X0.15X48/160 = 0.080 " Sum = 0.326 " Co formed = 0.326X28/16 = 0.57 " Volume of CO = 0.57-^-1.26 = 0.45m 3 Volume of Zn = 1.00-7-2.93 = 0.34 m 3 Sum = 0.79 m 3 Percentage composition of gases : CO = 57 per cent. Zn = 43 * (1) (2) If the barometric pressure is assumed normal, then the zinc vapor supports 760X0.43 = 327mm and the temperature at which the mixture becomes saturated with zinc vapor is found from the table on page 620 to be 847. This is 83 below the normal boiling point of zinc. THE METALLURGY OF ZINC. 651 (3) From 847 down, the vapor in the retort will be always saturated, its tension being found from the tables. The propor- tion of original zinc condensed, or remaining uncondensed, cannot be inferred simply from the vapor tension at any tem- perature. For instance, the vapor tension of the zinc at 600 C. is 12 mm of mercury column. The CO gas leaving the con- denser at this temperature will, therefore, be at a tension of 76012 = 748 mm, and will be accompanied by 12/748 of its volume of zinc vapor; as it came into the condenser it was ac- companied by 327/433 of its volume of zinc vapor. The ratio of these two quantities or proportions represents the real frac- tion of the zinc escaping uncondensed; that is, the proportion escaping condensation, in terms of the total zinc concerned, is 12/748-^327/423 = 0.016 -f- 0.773 = 0.008 = 0.8 per cent. The same result can be calculated in several ways. One is based on the datum that at any given temperature zinc vapor is 65-^28 = 2.3 times as heavy as CO gas. Problem 137. Numerous attempts have been made to reduce Zn O continu- ously in a blast furnace, and condense the zinc from the gases. In such a case, the gases will consist of zinc vapor, carbon monoxide and nitrogen, and the deficit of reduction heat must be obtained by the burning of carbon to CO at the region of the tuyeres. A low charge column must be used in order that the gases pass out hot enough to carry the zinc. Assume the gases to escape at 900 C., and the furnace to lose by radiation and conduction, etc., an amount of heat equal to the sensible heat in the blast used. Required: (1) The amount of fixed carbon to be charged into the fur- nace, per 2000 Ib. of zinc. (2) The composition of the gases leaving the furnace. (3) The temperature at which these gases will begin to de- posit zinc or become saturated with zinc vapor. (4) The proportion of the zinc they carry which will escape from the condensers, as vapor, at 600 C. (5) The pressure necessary to apply to the furnace to keep 652 METALLURGICAL CALCULATIONS. the zinc in the melted state at the minimum temperature of re- duction, 1033 C. Solution: (1) If the reaction is assumed as practically ZnO + C = Zn + CO there is needed for reduction, per 2000 Ib. of zinc 2000 X (12/65) = 369 Ib. There is in this reaction a deficit of heat, which is per 65 parts of zinc concerned simply (Zn, O)-(C, O) = 84,800-29,160 = 55,640. This deficit can only be made up by burning more C to CO right at the tuyeres. If two more atoms of carbon were burned we would have a little more than enough heat. We will, there- fore, assume the reaction at the tuyere region as the using of one atom of carbon for reduction and of double as much for supplying the heat deficit, and since one volume of oxygen is accompanied by 3.81 volumes of nitrogen, the whole reaction will be Zn O + 3 C + 2 + 3.81 N 2 = Zn + 3 CO + 3.81 N 2 . The amount of fixed carbon necessary is approximately three times that necessary for reduction only and amounts, per 2000 Ib. of zinc, to 36X~22=11081b. (1) (2) The furnace gases, therefore, contain, by volume, Zinc vapor = 1 volume = 12.8 per cent. Carbon monoxide =3 " = 38.4 " Nitrogen =3.81 " = 48.8 " (2) (3) The gases will be saturated with zinc vapor when the partial pressure of the zinc, which is 760X0.128 = 97mm is equal to the maximum tension of zinc at the temperature attained. This is at 745 C., as seen from inspecting the vapor tension tables, or 185 below the normal boiling point of zinc. (3) THE METALLURGY OF ZINC. 653 (4) The relative volume of zinc vapor to uncondensable gas in the original gases is In the gases issuing from the condenser at 600 it would be w - The proportion of the zinc escaping condensation under these conditions would, therefore, be !L16 = . 109 = 10 . 9 per cent. (4) By cooling the gases to the melting point of zinc, it would be possible to leave uncondensed only = 0.0014 = 0.14 per cent. (5) If it was a question only of keeping pure zinc from vaporiz- ing at 1033, in the absence of other gases, we could at once consult the vapor tension table, and find 2.2 atmospheres as the actual tension of zinc vapor at 1033, corresponding to 1.2 atmospheres effective pressure upon the furnace. But in the gas mixture coming from the furnace the zinc is practically one-eighth, by volume, of the gases, which means that it sus- tains one-eighth of ' the pressure upon the whole. It would, therefore, take a tension of 2.2X8 = 17.6 atmospheres to keep all this zinc in the liquid state, corresponding to an effective pressure of 16.6 atmospheres. (5) This is an entirely impracticable working condition and shows the practical impossibility of the schemes upon which much money have been spent for reducing zinc ore to liquid zinc in a blast furnace run under high pressure. Problem 138. In the metallurgy of mercury, the average ore contains 2 per cent, of Hg S and is roasted by the use of 10 per cent, of its 654 METALLURGICAL CALCULATIONS. weight of wood having an average composition of water 20 per cent., carbon 32, hydrogen 5.3, oxygen 42.7. Assume that just enough air enters to completely burn the sulphur and car- bon, that the ore and air used are dry. Required: (1) The percentage composition by volume of the roaster gases. (2) The temperature at which the mercury will commence to condense. (3) The proportion of the mercury present escaping as vapor if the gases pass out of the condensers at 100 C., at 50 C., at 15 C. Solution: (1) Per 1000 kg. of ore, containing 20 kg. of Hg S, there must be burned 20X32/232 = 2.8 kg of sulphur. 100X0.32 = 32 " " carbon. requiring of oxygen : 2.8X1 = 2.8kg. 32 X 32/12 = 85.3 " Sum = 88.1 Nitrogen accompanying = 293 . 7 Air = 381.8 ' Composition of the gases: Hg = 17.2 kg = 17.2 ^ 9.00 = 1.91 m 3 = 0.5 per cent. SO 2 = 5.6 " = 5.6-7-2.88= 1.91 " = 0.5 H 2 O= 68.0 " = 68.0 -=-0.81= 83.95 " = 22.1 N 2 = 293.7 " = 293.7 -v- 1.26 = 233.10 " = 61.3 CO 2 = 117.3 " = 117.3 ^ 1.98 = 59.24 " = 15.6 380.11 " = 100.00 (1) (2) In reality, each of these gases possesses the volume of the gas mixture, and is at the above percentage of the normal barometric pressure upon the mixture. The tensions of the different constituents of the mixture are, therefore, THE METALLURGY OF ZINC. 655 Hg 760X0.005 = 3.8mm SO 2 760 X 0.005 = 3.8 " H 2 O 760 X 0.221 = 168.0 " N 2 760 X 0.613 = 465.9 " CO 2 760X0.156 = 118.5 " 760. " The mercury will, therefore, commence to condense when the temperature of the gas mixture is that of mercury vapor at a maximum tension of 3.80 mm. From the vapor tension table of mercury we find this to be 156 C., or 201 below the nor- mal boiling point of mercury. Above this temperature the gas mixture is not saturated with Hg vapor and no condensation can occur. (2) (3) In the gas mixture, the mercury vapor is equivalent to 3.8/760 of the volume of the whole, or 3.8/756.2 of the volume of the other gases. This latter proportion is 0.00503. At 100 C., the uncondensed mercury vapor can have a tension of 0.285 mm, leaving 759.715 to the other gases, and giving the ratio of mercury vapor to other gases 0.000375. Since the other gases have remained constant in amount, the proportion of mercury remaining condensed is 0.000375 -f- 0.00503 = 0.075 = 7.5 per cent. If the temperature of the mixture is reduced to 50 C., the mercury vapor remaining can exert a tension of only 0.013 mm and the water vapor only 92 mm. The other gases present will therefore sustain 760- (92 + 0.013) = 668 mm. In the origi- nal mixture the Hg and H 2 O vapors are, respectively, equiva- lent to Hg 3.8 ^ 587.2 = 0.0065 H 2 O 168.0 -^ 587.2 = 0.2860 of the volume of N 2 , CO 2 and SO 2 together. In the gas escaping at 50 the Hg and H 2 O will be equiva- lent to Hg 0.013^-668 = 0.00002 H 2 O92. -668 = 0.1377 of the volume of the same gases. The proportions of the Hg and H 2 O escaping uncondensed at 50 will, therefore, be 656 METALLURGICAL CALCULATIONS. Hg 0.00002 -J- 0.0065 = 0.0031 = 0.31 per cent. H 2 O 0.1377 -f- 0.2860 = 0.4815 = 48.15 per cent. By exactly similar reasoning, using the maximum tensions of Hg and H 2 O at 15 (0.0009 mm and 13 mm, respectively), the proportions of each to the N 2 + CO 2 + SO 2 are Hg 0.0009 -f- 747 = 0.0000012 H 2 O 13. -f- 747 = 0.0174 >;'* " ' . 1" and the proportions escaping condensation would be ; Hg 0.0000012 -f- 0.0065 = 0.00018 = 0.018 per cent. H 2 O 0.0174 -f- 0.2860 = 0.06064 = 6.064 per cent. (4) The differences between these percentages and 100 .will be the proportions of these materials condensed to the liquid state. METALLIC MIST OR FUME. When zinc or mercury are in the state of vapor their atoms are each separate from other atoms. Chemically speaking, their vapor molecules are mon-atomic. On condensing to the liquid state we do not know whether the atoms come together into compound molecules or whether they simply come closer together. In either case, the molecules of the liquid metal, as the temperature is reduced low enough to form them, exist at first as isolated molecules, constituting the liquid metal in its finest possible state of sub-division, so small that it is for the time being practically . still a gas. If, however, we give these liquid molecules the possibility of uniting to liquid masses, we get the latter. This is a matter principally of reducing the sur- face tension which holds the liquid molecules each to itself as a sphere. Contact with a rubbing surface, with dust particles, filtration through cloth, or even electrification (perhaps a mag- netic field) reduce this surface tension and cause agglomera- tion into liquid masses which precipitate. The amount of liquid or solid particles thus held in suspen- sion and escaping with the current of uncondensable gas is en- tirely independent of the quantity escaping as true vapor, which has been calculated above, except in so far as they are both functions of the amount of said uncondensable gas. The veloc- ity of the escaping current is a large factor in the amount of mist, because the carrying power of gas for mist particles varies THE METALLURGY OF ZINC. 657 probably with the cube of its velocity, so that halving the velocity of the issuing gas will diminish greatly this source of loss. Large settling chambers in which the mist can de- posit, because of stagnation of the gas current, and large rub- bing surfaces, are effective means for catching or depositing the mist. Filtration through bags is still more effective. In the case of zinc, the solidifying point is 420, while con- densation from the state of vapor begins at, say, 860. In this interval only can the condensed particles, as a mist, collect together to liquid zinc. If the temperature passes quickly through this range, the particles have little chance to agglomer- ate into liquid masses, and the proportion which chills into solidified mist particles is increased. These solidified mist particles, analogous to " hoar frost " in nature, form the well known " blue powder " of the zinc works. It settles in large settling chambers, and is in extremely fine particles, mostly under 0.01 mm in size. It is quite easy to collect all the zinc in this form if the vapor is chilled suddenly and the resulting gas settled properly or filtered. The loss of mercury or zinc as mist or hoar frost is there- fore, to be considered entirely apart from the loss as true vapor; it may be smaller than the latter and it may be larger ; its amount varies particularly with the nature of the settling apparatus and the velocity of the gases as they escape. The theory as to its amount under any given set of conditions would be very difficult to elaborate, and the data for doing so with exactness are practically unknown. The best that can be done at present is to study the loss as mist or hoar frost in actual practice, experimentally, and tabulate the actual results as a guide for future metallurgical use. CHAPTER V. METALLURGY OF ALUMINIUM. The two essential principles here involved are " differential reduction " as used in the electric furnace purification of alumina, and " electrolytic furnace operation," as illustrated in the decomposition of the alumina by electrolysis in the manner usually practiced. Only the latter problem will be covered here. ELECTROLYTIC FURNACE REDUCTION OF ALUMINA. The Hall process is beautifully simple and technically admir- able. Al 2 O 3 is found to dissolve in melted alkaline-aluminium double fluorides; it is as pretty a case of solution, so far as ap- pearances go, as dissolving a spoonful of powdered sugar in a glass of distilled water. The melting point of the fused fluorides is reduced by the solution of alumina, just as that of water is reduced by dissolving salt. In passing the electric current the constituents of the dissolved alumina appear at the electrodes, oxygen at the anode and aluminium at the cathode. The best practical arrangement is to use a carbon-lined pot, with molten aluminium in the bottom as the cathode, upon it a few inches depth of the bath, and dipping into this the carbon anodes. The writer has given most of the technical details of this operation in his treatise on "Aluminium," and Prof. Haber has published extensive laboratory studies of the process in the Zeitsckrift fur Elektrochemie. With a solvent salt consisting of melted sodium fluoride and aluminium fluoride, such as called for in one of the Hall patents, with alumina dissolved therein, and using carbon anodes, the electrolytic elements of the process are simplicity itself. The bath contains sodium, aluminium, fluorine and oxygen, and the anode is carbon. Under these conditions those elements or com- pounds will form at the electrodes which cannot further react upon the bath material; in other words, those materials most stable in contact with the bath material or electrodes. A mo- 658 THE METALLURGY OF ALUMINIUM. 659 ment's reflection explains what happens, and what must happen. At the cathode, sodium cannot be liberated because metallic sodium reacts chemically on this bath, separating out alumin- ium; therefore,- the electrolytic reducing tendency at the sur- face of the cathode can only expend itself in separating out aluminium. At the anode, fluorine cannot be liberated because fluorine acts strongly upon alumina even when cold, converting it into fluoride and expelling its oxygen; therefore, the electro- lytic perducing tendency at the surface of the anode will tend to set free oxygen. But oxygen cannot be set free at a carbon surface at a cherry-red heat because of its inevitable tendency to unite with the carbon to form CO. The electrolytic agency at the surface of the carbon anode must, therefore, cause the formation of CO. The whole electrolytic operation results in the removal of A1 2 O 3 from the bath and the formation of alumin- ium and carbon monoxide. The thermochemical relations, as far as known, agree abso- lutely with the, above explained experimental results. The materials in presence of each other are sodium fluoride, alumin- ium fluoride, aluminium oxide and carbon. The heats of formation of these, per molecule and per chemical equivalent concerned, are as follows: (Na, F) = 109,720 '= 109,720 per chemical equivalent. (Al, F 3 ) = 275,220 = 91,740 " (Al 2 , O 3 ) = 392,600 = 65,430 " (C, O) = 29,160 = 14,580 " The heat of formation of carbon tetra-fluoride is unknown, but is probably small, since it is so difficult to form. From the last column, which largely governs the work done by the current, we see that the current does far the least work when it decomposes alumina; in fact, it does still less, by 14,580 calories, because of the assistance rendered by carbon uniting with the oxygen. Now, although the electrical current does not consistently adhere to the doctrine of " least work," yet it does in this case because forced to do so by the chemical relations of sodium, aluminium, fluorine, oxygen and carbon at the tem- perature of the bath, as explained in the preceding analysis. It is probable that in electrolysis the chemical relations of the possible products control what the current does rather than the 660 METALLURGICAL CALCULATIONS. thermochemical relations alone, but in most cases the two condi- tions or controlling circumstances coincide in their influence and lead to identical results. This is the case in the process in question ; it is absolutely normal and is explainable by either method of reasoning. If the operation is run with a small vessel and correspond- ingly small current, the heat necessarily evolved by the passage of the current is small compared to the conduction and radia- tion losses and must be supplemented by outside heating to keep the contents at proper temperature. If the size of the operation is increased in all its items and dimensions, the neces- sarily generated internal " resistance " heat will suffice to keep the bath at the requisite temperature, when the enlargement is done on a certain scale. If enlarged past this point, too much internal heat is unavoidably generated and means must be used to artificially cool the pot. Problem 139. An electrolytic vessel is composed of a block of carbon 25 cm cube, with a cavity 10 cm square by 10 cm deep inside. The cavity has a round carbon, 5 cm diameter, dipping into it. The vessel weighs 30 kilograms; the fused bath in the cavity 2 kg; the carbon immersed in it 0.1 kg. The specific heat of the car- bon is 0.5, of the bath 0.3, at the running temperature. An ex- periment showed that the bath material cooled off, at the work- ing temperature, at the rate of 10 C. per minute, the walls of the vessel at an average rate of 2 C. per minute, when left to cool by themselves. Required: (1) The number of watts which must be converted into heat in the vessel in order to maintain it at the working temperature. (2) Assuming 75 per cent, of the theoretical ampere efficiency to be obtained, what amperes passed through the pot will keep it at working temperature if the working voltage is kept at 10 volts? Solution: (1) The 30 kg of vessel material losing heat at the rate of 2 per minute, with specific heat of 0.5, gives a heat loss per minute of 30X0.5X2 = 30 Calories. THE METALLURGY OF ALUMINIUM. * 661 Similarly, the immersed carbon in the cavity and the bath material itself lose 0.1X0.5X10 = 0.5 Calories. 2.0X0.3X10 = 6.0 The total heat loss is, therefore, 36.5 Calories per minute. To maintain the temperature constant the current must fur- nish this heat, and since 1 watt is 0.239 gram calories per sec- ond, the watt energy thus converted into heat must be 36.5X1000 -,- 0.239X60 == 2545watts (D (2) If all the amperes passing through separated out metal the voltage absorbed in decomposition in the bath would be from the thermochemical heats of formation of chemical equiva- lent quantities of AP O 3 and CO: 65,430-14,580 = 2.2 volts. 23,040 If 75 per cent, of the amperes are efficient, the voltage thus ab- sorbed is 2.2X0.75 = 1.65 volts. The voltage disappearing in overcoming ohmic resistance will then be 10-1.65 = 8.35 volts and the current which when passed will keep the pot at working temperature will be 2545 g-^5 = 305 amperes. (2) The above-used principles are applicable to any kind of elec- trolytic-furnace operation. INDEX. ADDICKS, L., on energy loss at electrical contacts 562 Air, composition of 4 Air, received by converter 336 Air, required by converter 333 Alloys, heats of formation 37 Alloys, thermophysics of 109 Alumina, action in blast furnace slag 256 Alumina, behavior in the blast furnace 240 Alumina, reduction of 658 Alumina, thermophysics of 123 Aluminates, heats of formation of 35 Aluminium alloys, thermophysics of 113 Aluminium compounds, heats of formation of 18, 37 Aluminium compounds, thermophysics of 139 Aluminium conductors, economical size of 570 Aluminium conductors, heating of, in use ' 572 Aluminium, electrical conductivity of 569 Aluminium, heat of oxidation of 353, 374 Aluminium, metallurgy of 658 Aluminium, reduction of lead oxide by 586 Aluminium, thermophysics of 80 Amalgam furnace, efficiency of , 107 Amalgams, heats of formation of 37 Ammonia gas, mean specific heat of 142 Anode, insoluble, use of 551 Anode, soluble, use of iron as 548 Anode, use of matte as 539 Anthracite coal, combustion of 220, 221 Antimonides, heats of formation of 23 Antimonides, thermophysics of , 136 Antimony alloys, thermophysics of 101 Antimony, behavior in the blast furnace 240 Antimony, compounds, heats of formation of 23 Antimony, thermophysics of 95 Aqueous vapor, maximum tension of ' Ill Argo, Col., reverberatory smelting at 501 Arsenic, behavior in the blast furnace 240 Arsenic compounds, heats of formation of 33 Arsenic compounds, thermophysics of 136, 138 Arsenic, thermophysics of 88 Arsenides, heats of formation of 23 Ashcroft's process for copper ores 537 Atacamite, formula of 537 Atomic weights 1 BALANCE sheet of blast furnace 235 Barium compounds, heats of formation of 18-40 Barium compounds, thermophysics of 113-144 Barium, thermophysics of 97 Bauxite, efficiency of furnace, in melting 127 Beeswax, thermophysics of 142 Belgian furnace, distillation of zinc ore in a 635 Bergfeld, on the boiling of silver and gold 615 663 664 INDEX Beryllium compounds, heats of formation of 18-40 Beryllium compounds, thermophysics of 113-144 Beryllium, thermophysics of 76 Bessemerizing copper matte 525 Bessemer process, problem concerning 11 Bessemer process, the 333 Betts' refining process for lead 598, 599, 602 bi-carbonates, heats of formation of 29 Bismuth alloys, thermophysics of . . . . 110 Bismuth compounds, heats of formation of 18-40 Bismuth compounds, thermophysics of 113-144 Bismuth, thermophysics of ; 102 Bi-sulfates, heats of formation of 33 Bituminous coal, combustion of 219, 221, 222, 223, 226 Blast, amount used in blast furnace 244 Blast, dry, Gayley process of producing , 330 Blast, dry, the rationale of 309 Blast engines, efficiency of 493, 590 Blast furnace, balance sheet of 235 Blast furnace, carbon needed in the 277 Blast furnace gas, in gas engines 226 Blast furnace gases, surplus power from 272 Blast furnace, heat balance sheet of . 281 Blast furnace, reduction of zinc ores 651 Blast furnace, temperature in 223 Blast furnace, utilization of fuel in the 268 Blast, heating of, apparatus for 318 Blast, hot, the function of 308 Blast pressure required in Bessemer process 340 Blast, warm, use in pyritic smelting 487 Blast, work done in producing 313 Blister roasting of copper matte 523 Blowing engines, efficiency of 299 Boiling of metals in a vacuum 614 Boiling of zinc 641 Borates, heats of formation of 34 Boron compounds, heats of formation of 18-40 Boron compounds, thermophysics of 138 Boron, thermophysics of 77 Brass, thermophysics of Bringing forward of copper matte . 514 Bromides, thermophysics of 133 Bromine, thermophysics of 89 Bronze, thermophysics of Ill Bullion, gold, electrolytic refining of 606 Bullion, silver, electrolytic refining of 605 Burgess process of refining iron Butte, Mont., copper concentrates from 479 Butte, Mont., roasting furnace at 482 CADMIUM alloys, thermophysics of 113 Cadmium compounds, heats of formation of 18-40 Cadmium compounds, thermophysics of 131-144 Cadmium, thermophysics of 93 Cadmium, vapor tension of 644 Caesium compounds, heats of formation of 18-40 Caesium, thermophysics of 96 Calcium compounds, heats of formation of 18-40 Calcium compounds, thermophysics of 131-144 Calcium oxide, heat absorbed in reducing 293 INDEX 665 Calcium oxide, thermpphysics of . . 124 Calcium, thermophysics of 83 Calorific power of fuels, calculation of 503 Calorimeter, test for specific heat by 225 Calorimeter, test for temperature by 224 Campbell, H. H., furnace data by 393 Canada, electrical processes in 435 Canada Zinc Co., operations of the 641 Carbides, heats of formation of 25 Carbon, amount needed in the blast furnace 277 Carbon, heat of oxidation of 352, 375 Carbon, thermophysics of 77 Carbonates, heat absorbed in decomposing : . : 291 Carbonates, heats of formation of 28 Carbonates, thermophysics of 137 Carbonic oxide, thermophysics -of 122 Carbonous oxide, thermophysics of 121 Cathodes, use of copper matte as 545 Cathodes, use of lead sulfide as 598 Cement copper, composition of 548 Cerium compounds, heats of formation of 18, 40 Cerium, thermophysics of 97 Chalcopyrite, composition of ' 474 Charge, blast furnace, calculation of . . . . : 250 Chimney draft . 185 Chimney draft, problems concerning 192, 194, 224, 230, 231 Chimney for open-hearth furnace 395 Chlorates, thermophysics of . 139 Chlorides, heats of formation of 27 Chlorides, thermophysics of 131 Chlorine, thermophysics of 82 Chromates, thermophysics of 138 Chromium, heat of oxidation of 353, 374 Chromium, thermophysics of 84 Coal, combustion of powdered Cobalt, behavior in the blast furnace 240 Cobalt compounds, heats of formation of 18-40 Cobalt compounds, thermophysics of 131-144 Cobalt, thermophysics of 87 Coehn, improved Hoepf ner apparatus 558 Coke-oven gases, combustion of 219-228 Coke, use of, in pyritic smelting 488 Columbium, thermophysics of 90 Compression of blast, work done in 313 Concentration of zinc ores, cost of 621 Condensation of metallic vapors 649, 654 Conduction of heat, principles of 200 Conductivity for heat, tables of 203, 211 Conductors, electrical, economic size of 564 Conductors, electrical, heating of, in use 568 Constants, thermochemical 38-40 Contacts, electric, energy loss at . 562 Copper alloys, thermophysics of Ill Copper, behavior in the blast furnace 240 Copper, calculations on metallurgy of 469, 619 Copper chloride, electrolysis of 537, 557 Copper compounds, heats of formation of 18-40 Copper compounds, thermpphysics of 131-144 Copper, electrical conductivity of ........: 567 Copper, electrolytic refinery of 558 666 INDEX Copper matte, composition of 469, 471 Copper, principles of the metallurgy of 472 Copper ores, electric smelting of 572 Copper solutions, physical properties of 611 Copper solutions, resistivity of : . . . 559 Copper, thermophysics of 87 Crushing of zinc ores, cost of 621 Cupola, thermal efficiency of : 115 Cyanates, heats of formation of 36 Cyanides, heats of formation of 35, 36 Cyanides, thermophysics of 136 DAMOUR, on furnace efficiency , 416, 423 Decomposition of moisture of blast 293 Dehydrating charges, heat absorbed in 290 Dell wick-Fleischer gas producer 179 Distillation of zinc from ore 632 Dried blast, rationale of action of 309 Dried blast, temperatures attained with 312 Drying air blast commercially 325 Drying charges, heat absorbed in 290 Drying of wet peat 227 Ducktown pyrrhotite, smelting of 489 Dulong and Petit's laws of specific heats 70 Dulong's laws of heat of combustion of coals 48 EFFICIENCY of electric furnaces 640 Efficiency of electric heating 441 Efficiency of furnaces, discussion of 104 Efficiency of open-hearth furnaces 396 Eissler's hydrometallurgy of copper 469 Eldred process of combustion 55 Electric furnaces, efficiency of 640 Electric melting of steel 440 Electric reduction of iron ore 429 Electric refining of iron 446 Electric smelting of copper ores 572 Electric smelting of zinc ores 637 Electrical contacts, energy loss at 562 Electrolytic furnace, definition of 534 Electrolytic furnace for reducing aluminium . . . . 658 Electrolytic refining of copper : . . . 558 Electrolytic refining of gold 610 Electrolytic refining of iron 446 Electrolytic refining of lead 598 Electrolytic refining of silver 605 Electrolytic refining, principles of 536 Electrometallurgy of copper 534 Electrometallurgy of iron and steel 429 Electrometallurgy of lead '. 598 Electrothermal processes, definition of 535 Ely, Vermont, copper ore from 475 Ethylene, mean specific heat of 142 Evans-Klepetko furnace, calculations on 482 FEED-WATER heater, utility of 225 Ferro-cyanides, heats of formation of 36 Fire-clay, heat conductivity of 630 Flues, size of 382 Fluorides, heats of formation of 26 INDEX 667 Fluorides, thermophysics of 134 Flux, amount needed in the blast furnace 251 Flux, behavior in the blast furnace 243 Flux, comparison of values of 263 Flux, required in Bessemer converter 346 Freeland, W. H., on smelting pyrrhotite.' 489, 515 Fuels, behavior of, in the blast furnace 236 Fuels, calorific power of 46 Fuels, comparison of values of 261 Fuels, utilization of, in the blast furnace 268 Fume, metallic 656 Furnaces, efficiency of 104 Furnaces, electric, efficiency of 640 Furnaces, electrolytic, definition of 534 Furnaces, electrolytic, for reducing alumina 658 Fusion of latent heats of, discussion 71 Fusion of latent heats of, tables 118-144 Fusion of matte 500 Fusion of slag 500 GALENA, roasting of 581 Gallium, thermophysics of 88 Gas, artificial 145 Gas, engine operation 226 Gas, mixed, producers for 158 Gas, natural, combustion of 10, 47, 53, 219 Gas, natural, from lola, Kansas 636 Gas producers 381 Gases, composition of, from smelting furnace 497 Gases, pressure corrections for 7 Gases, relative volumes of Gases, temperature corrections for 5 Gases, thermophysics of 141 Gases, waste, heat in 287 Gayley process of drying blast 330 Genoa, Marchesi's process at 539 Gin, G., in electrometallurgy of zinc 637 Glass, thermophysics of 141 Gold, behavior in the blast furnace 240 Gold bullion, refining of 610 Gold compounds, heats of formation of 1840 Gold compounds, thermophysics of 131-144 Gold, metallurgy of 605 Gold, thermophysics of 100 Gold, vapor tension of 616 Gordon, F. W., slag calculations by 261 Goutal's method for calorific power of coal 503 Great Falls, Montana, copper refining at 566 Gruner's ideal working of a blast furnace 274 HABER, on reducing alumina 658 Hall, C. M., on reducing alumina 658 Halske and Siemens' copper process 555 Heat balance sheet of Bessemer converter 353 Heat balance sheet of blast furnace 281 Heat balance sheet of electrical furnace Heat consumed, in reductions 292 Heat of formation of chemical compounds 18-40 Heat of formation of slag. 500 Heat of formation of some nitrates 510 668 INDEX Heat of formation of some oxides 579 Heat of formation of some sulfides 578 Heat properties of matte 500 Heat properties of slag 500 Heat units 14 Heats of formation ' !.-... 18 Herreshoff furnace, calculations on 489, 515 Hess, H. D., on open-hearth furnaces 395 Hixon's notes on matte smelting , . . 526 Hoepfner's copper chloride process 557, 558 Hofman's Metallurgy of Lead 593 Holla way, J., on bessemerizing matte . 525 Holla way, J., on oxidation of sphalerite 621 Hot blast, sensible heat in 285 Howe, H. M., notes on a bessemer charge 336 Hydrates, heats of formation of 19 Hydrides, heats of formation of 24 Hydrocarbons, combustion of 45, 217 Hydrocarbons, heats of formation of 24, 217 Hydrogen compounds, heats of formation of 18-40 Hydrogen compounds, thermophysics of 131-144 Hydrogen, thermophysics of 75 Hyposulfites, thermophysics of 136 ICE, specific heat of 119 Ideal working of a blast furnace, Gruner's 274 Ingalls, on cost of treating zinc ores 621 Ingalls, on oxidation of sphalerite 621 Insoluble anodes, use of 551 Iodides, thermophysics of 133 Iodine, thermophysics of 95 lola, Kansas, natural gas from 636 Iridium, thermophysics of 99 Iron alloys, thermophysics of 113 Iron compounds, heats of formation of 18-40 Iron compounds, thermophysics of 125, 128-130 Iron, electric refining of 446 Iron, heat of oxidation of 352, 366 Iron ores, electrical reduction of 429 Iron oxide, discussion of reduction of 69 Iron oxide, problems concerning reduction of 128, 129 Iron oxides, heat absorbed in reducing '. 292 Iron persulfate process for copper ores 555 Iron sulf ate solutions, resistivity of 559 Iron sulfide, heat absorbed in reducing 293 Iron, thermophysics of 85 Iron, use as soluble anode 548 Isabella, Tenn., pyritic smelting at 489 Isabella, Tenn., smelting of matte at. 515 JANNETTAZ, "Les Cenvertisseurs pour le Cuivre" 526 Joplin, zinc ore from 620 Jiiptner, data on furnace charges 403 KOCH, on warm blast in smelting \ ^7 Krafft, on the boiling of silver and gold . 614 LABORATORY of open-hearth furnaces 391 Laboratory of furnaces, efficiency of .' . . 414 La Lustre smelter, warm blast at 487 INDEX 669 Landis, W. S., laboratory determinations by 500 Latent heats of fusion, discussion of 71 Latent heats of fusion, tables of 75-103, 118 Latent heats of vaporization, discussion of Latent heats of vaporization, tables of 75-103, 118-144 Lead alloys, thermophysics of 109 Lead compounds, heats of formation of 18-40 Lead compounds, thermophysics of 131-144 Lead, electrometallurgy of 598 Lead, metallurgy of , 578 Lead, physical constants of ' 583 Lead, refining of 583 Lead, thermophysics of 101 Lead, use as insoluble anode 551 Lead, vapor tension of Lime, amount needed in Bessemer converter 347 Lime, heat in preheated 356 Lithium compounds, heats of formation of 18-40 Lithium compounds, thermophysics of 131-144 Lithium, thermophysics 76 MAGNESIUM compounds, heats of formation of 18-40 Magnesium compounds, thermophysics of 131-144 Magnesium, thermophysics of 80 Magnus, B., on energy losses at electrical contacts 562 Manganese, behavior in the blast furnace 240 Manganese compounds, heats of formation of 18-40 Manganese compounds, thermophysics of 131-144 Manganese, heat of oxidation of 371 Manganese, oxides, heat absorbed in reducing Manganese, thermophysics of 85 Manhes, P., on bessemerizing matte 525 Marchesi, on electrolytic treatment of matte 539, 541 Marsh gas, mean specific heat of 142 Martin process Matte, bessemerizing of 525 Matte, blister roasting of 523 Matte, bringing forward of 514 Matte, copper, composition of 469 Matte, electric furnace production of 574 Matte, electrolytic treatment of 539 Matte, grade obtained in smelting 474 Matte, thermophysical properties of 500 Maximum tension of aqueous vapor Mayer, F., Das Bessemern von Kupf ersteinen " 526 Mercury alloys, thermophysics of 113 Mercury compounds, heats of formation of 18-40 Mercury compounds, thermophysics 131-144 Mercury, metallurgy of 653 Mercury, thermophysics of 100 Mercury, vapor tension of 584 Metallic mist 656 Methane, mean specific heat of Mine water, extraction of copper from Mist, metallic 656 Moisture, heat absorbed in decomposing 293 Molybdenum compounds, heats of formation of 18-40 Molybdenum compounds, thermophysics of 131-144 Molybdenum, thermophysics of ; 91 Mond gas 169 670 INDEX Mond gas, superheating of 175 Monell process, problem concerning 424 Morgan producer, problem concerning 163 Mount Lyell, gases of furnace at 497 Mount Lyell, pyritic smelting at 488 NATURAL gas, combustion of 10, 47, 53, 219 Natural gas, from lola, Kansas 636 Niagara Falls, Salem process as used at 599 Nickel alloys, thermophysics of 113 Nickel, behavior in the blast furnace : 240 Nickel compounds, heats of formation of 18-40 Nickel compounds, thermophysics of 131-144 Nickel, heat of oxidation of 353, 374 Nickel, thermophysics of 86 Nickeliferous pyrrhotite, reduction of 435 Nitrates, heats of formation 30, 609 Nitrates, thermophysics of 137 Nitrides, heats of formation of 23 Nitrogen, thermophysics of 78 Oil, fuel, combustion of 219 Oil furnace, efficiency of 106 Open-hearth furnaces 381 Ore, behavior of, in the blast furnace 239 Ore, comparison of values of 265 Ore, electrical reduction of 429 Ores, copper, direct treatment by electrolysis 536 Osmium, thermophysics of 98 Oxides, heat absorbed in reducing 292 Oxides, heats of formation of ' 18, 579 Oxides, thermophysics of 118 Oxygen, thermophysics of 79 PALLADIUM compounds, heats of formation of 18-40 Palladium, thermophysics of 92 Paraffin, thermophysics of 142 Parrot Works, Butte, Bessemer process at 525 Peat, kiln drying of 227 Peters, "Modern Copper Smelting" 469, 526 Peters, "Principles of Copper Smelting" 489, 484, 566 Phosphates, heats of formation of 33 Phosphates, thermophysics of 138 Phosphides, heats of formation of Phosphorus, behavior in the blast furnace 240 Phosphorus, heat of oxidation of 353, 377 Phosphorus, oxide, heat absorbed in reducing 293 Phosphorus, thermophysics of 81 Pig-iron, heat of formation of 285 Pig-iron, heat in melted 289 Platinum alloys, thermophysics of 113 Platinum compounds, heats of formation of. 18-40 Platinum compounds, thermophysics of 131-144 Platinum, thermophysics of 99 Ports of an open-hearth furnace 388 Potassium compounds, heats of formation of 18-40 Potassium compounds, thermophysics of . 131-144 Potassium, thermophysics of 83 Pot roasting, applicability to sphalerite : 621 Pot roasting, principle of 586 INDEX 671 Potter, H. N., on heat ( Powdered coal, combust Power obtainable from : Power required to comp Power, surplus, from bl; Pressure of blast, metric Pressure of blast, requii Page Problem 1 8 Problem 2 10 Problem 3 11 Problem 4 47 Problem 5 54 Problem 6 106 Problem 7 107 Problem 8 108 Problem 9 114 Problem 10 114 )f formation of s ion of ilica 353 ,223 502 313 272 317 340 Page 460 461 461 462 462 464 465 475 477 479 482 489 497 501 510 515 530 537 541 546 549 550 552 556 558 560 563 568 573 586 593 599 602 606 610 621 624 630 633 635 636 647 650 651 653 660 229 381 483 221 furnace gases. . . iress blast ast furnace gase; >ds of measuring ed in the Bessei Problem 51 . s ner proces Page 245 s Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 52. Problem 53 . Problem 54 . Problem 55 259 .... 267 . . . . 268 270 Problem 56 . Problem 57. Problem 58. Problem 59. Problem 60. Problem 61 . Problem 62. Problem 63 . Problem 64 . Problem 65. Problem 66 . Problem 67. Problem 68 . Problem 69 . . . . -272 .... 294 .... 305 .... 315 319 . 322 .... 327 .... 331 .... 334 336 . ... 343 . ... 349 . ... 363 384 Problem 103. Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17. ... Problem 18 Problem 19. ... Problem 20. ... Problem 21 Problem 22 Problem 23 Problem 24 . . 114 . . 115 .. 127 .. 128 . . 129 . . 149 . . 154 . . 163 . . 169 . . 179 . . 182 . . 192 . . 194 208 Problem 104 Problem 105 Problem 106 Problem 107 Problem 108 Problem 109 Problem 110 Problem 111 Problem 112 Problem 70 . Problem 71 . Problem 72 . Problem 73. Problem 74. Problem 75. Problem 76 . Problem 77 . Problem 78. Problem 79 . Problem 80 . Problem 81 . Problem 82 . Problem 83 . Problem 84 . Problem 85 . Problem 86 . Problem 87 . Problem 88 . Problem 89 . Problem 90 . Problem 91 . ... 388 . 393 . ... 403 . ... 411 . ... 418 424 . ... 430 . ... 435 . ... 442 443 . ... 446 . ... 452 . ... 452 . ... 453 . ... 454 . ... 454 . ... 455 . ... 456 . ... 456 . ... 456 . ... 457 457 Problem 113 Problem 114 Problem 115 Problem 116 Problem 117 Problem 118 Problem 119 Problem 120 Problem 121 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29. ... Problem 30 Problem 31 Problem 32 . . 215 . . 219 . . 219 . . 220 . . 221 . . 221 . . 222 . . 223 Problem 122 Problem 123 Problem 124 Problem 125 Problem 126 Problem 33 Problem 34 Problem 35 Problem 36. ... Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42. ... Problem 43 Problem 44. ... . . 223 . . 223 . . 224 . . 224 . . 225 . . 225 . . 226 . . 226 . . 227 . . 227 . . 228 . . 229 Problem 127 Problem 128 Problem 129 Problem 130 Problem 131 .. . Problem 132 Problem 133 Problem 134 Problem 92 . Problem 93 . tion of 458 . ... 460 Problem 135 Problem 136 Problem 137 Problem 138 Problem 139 Problem 45 229 Problem 46 . . 230 Problem 47. ... PRODUCER gas, Producers gas . . 231 combus . .219,227,228, 246 Pvritic smelting 672 INDEX Pyritic smelting, reaction at the focus 498 Pyritic smelting, slag produced by , 487 Pyritic smelting, temperature attained in 485 Pyrrhotite, electrical reduction of 435 Pyrrhotite, smelting of 489 QUENEAU, on furnace efficiency 416, 423 RADIATION, heat 213 Radiation, from a Bessemer converter 533 Radiation, from a blast furnace 289 Radiation, from an electrolytic furnace 660 Radiation, from a roasting furnace 479, 482, 627 Radiation, from a smelting furnace 506 Radiation, heat, tables of ; 213 Radiation, heating by 392 Radium, thermophysics of 102 Reactions, double, in copper metallurgy . . 474 Reactions, double, in lead metallurgy 582 Recarburizatipn, in the Bessemer converter 348 Reduction of iron ores electrically 429 Reduction of roasted lead ore 591 Reduction of zinc oxide 627 Refining, electrolytic, of copper 558 Refining, electrolytic, of gold 610 Refining, electrolytic, of iron 446 Refining, electrolytic, of lead 598 Refining, electrolytic, of silver 605 Refining, electrolytic, principles of 536 Refining, of impure lead 583 Regenerators, dimensions of 382 Regenerators, efficiency of 412 Resistivity of some solutions 559 Reverberatory smelting 501 Rhodium, thermophysics of 92 Richard, T. A., on pyrite smelting Roasting, Bessemer, of lead ores 586 Roasting, heat generated in 477, 479 Roasting of lead sulfide 581, 586 Roasting of sphalerite 620 Roasting, removal of sulfur in 473 Roasting, smelting of copper matte 523 Rubber, thermophysics of . 142 Rubidium, thermophysics of 90 Ruthenium, thermophysics of : . . . 91 SALOM, P. G., process for copper matte ', 545 Salom, P. G., process for reducing galena 598 Saulte Ste. Marie, electrical reduction at , 435 Savelsberg pot-roasting operation 586 Schnabel "Handbook of Metallurgy" 469, 526 Schuller, on the boiling of silver and gold . 615 Selenides, heats of formation of Selenium, thermophysics of 89 Siemens and Halske's copper process 555 Siemens-Martin process 381 Siemens new-style furnace : Silica, behavior of, in the blast furnace 240 Silicates, heats of formation of 31 INDEX 673 Silicates, thermophysics of 139 Silicides, heats of formation of 26 Silicon compounds, heats of formation of 18-40 Silicon compounds, thermophysics of 139 Silicon, heat of oxidation of 353, 369 Silicon, thermophysics of 81 Silver alloys, thermophysics of : 113 Silver, analyses of impure 606 Silver, behavior in the blast furnace 240 Silver, compounds, heats of formation of 18-40 Silver compounds, thermophysics of 131-144 Silver, metallurgy of : 605 Silver, thermophysics of : 93 Silver, vapor tensions of 616 Slag, amount produced in the blast furnace 251 Slag, blast furnace, composition of 252 Slag, blast furnace, heat in ''. 289 Slag formed in the Bessemer converter 346 Slag, fusion of, in the copper furnace 500 Slag, heat of formation of 286, 357 Slag, summation of ingredients of 254 Slag, thermophysics of 144 Smelting electric of copper ores 572 Smelting of lead ores .:......... 591 Smelting of zinc ores , 637 Smelting, ordinary, of copper ores . 499 Smelting, pyritic, rate of Snyder, F. T., on efficiency of electric furnaces 640 Snyder, P. T., on electric zinc smelting 641 Sodium compounds, heats of formation of 18-40 Sodium compounds, thermophysics of 131-144 Sodium, thermophysics of 79 Solutions, extraction of copper from 548 Specific heats of chemical compounds - 131-144 Specific heats of products of combustion 51 Specific heats of the elements 75-103 Speiss, formation of 592 Spence furnace, calculations on a 479 Sphalerite, roasting of 620 Steel, electrical production of 440 Steel furnace, efficiency of Steel, thermophysics of 113 Sticht, analysis of furnace gases by 497 Sticht, on pyritic smelting 484 Stolberg, Marchesi's process at ^ 539 Strontium compounds, heats of formation of 18-40 Strontium compounds, thermophysics of. 131-144 'Strontium, thermophysics of '. . . 90 Sulfates, heats of formation of Sulfates, thermophysics of 136 Sulfides, heats of formation of 21, 578 Sulfides, thermophysics of 134 Sulfur, available in ordinary smelting 474 Sulfur, available in pyritic smelting . 486 Sulfur, behavior in blast furnace 237 Sulfur, condition of, in roasted ore 475, 479, 507 Sulfur, heat absorbed in separation of 293 Sulfur, removal by partial roasting 473 Sulfur, thermophysics of Sulfuric acid solutions, resistivity of 559 43 674 INDEX TANTALUM, thermophysics of 97 Tellurides, heats of formation of 22 Tellurium compounds, heats of formation of 18-40 Tellurium, thermophysics of 96 Temperature, theoretical, of combustion 50 Temperature, thermochemistry of high 61 Temperatures in the Bessemer converter 368 Temperatures in the blast furnace 308 Temperatures, using dry blast 312 Tension, maximum, of aqueous vapor. 121 Thallium compounds, heats of formation of 18-40 Thallium compounds, thermophysics of 131-144 Thallium, thermophysics of 101 Thermochemical constants of bases and acids 38 Thermochemistry, applications of . . . 13 Thermochemistry, calculations in 41 Thermochemistry of the Bessemer process 351 Thermophysics of alloys 109-1 13 Thermophysics of chemical compounds 118-144 Thermophysics of the elements 75-103 Thermit process, calculations 43 Thermit process, temperatures in Thorium, thermophysics of 103 Tin alloys, thermophysics of 109 Tin compounds, heats of formation of 18-40 Tin compounds, thermophysics of. 131-144 Tin, thermophysics of 94 Tissier, reduction of lead oxide by aluminium 585 Titanates, heats of formation of 34 Titanium, heat of oxidation of 353, 373 Titanium, thermophysics of Toldt, data on furnace charge 403 Tungstates, heats of formation of Tungsten compounds, thermophysics of . . '. 131-144 Tungsten, heat of oxidation of 353 Tungsten, thermophysics of 97 ULKE, modern electrolytic copper refining 469 Units, heat, definition of 14 Uranium, thermophysics of 103 VACUUM, volatility of gold and silver in a 614 Valves for open-hearth furnaces Vanadium, heat of oxidation of 353 Vanadium, thermophysics of 84 Vancouver, B. C., electric zinc smelting at 641 Van Liew, on bessemerizing matte 530 Vapor, aqueous, maximum tension of 121 Vaporization, latent heats of, discussion 73 Vaporization, latent heats of, tables 75-103, 118-144 Vapor, metallic, calculation of weight of 618 Vapor, tension of cadmium 644 Vapor, tension of gold 615 Vapor, tension of lead 583 Vapor, tension of mercury Vapor, tension of silver 615 Vapor, tension of zinc 643 Vattier, on electric smelting of copper ores 573 Volatility of lead 583 Volatility of silver and gold * 614 INDEX 675 Voltages of decomposition 547 Vulcanite, thermophysics of 142 WATER, cooling, heat in 290 Water gas 177 Water gas, combustion of 219 Water, heat of formation of 15 Water, thermophysics of 119 Water vapor, maximum tension of 121 Watts, O. P., on boilingpoints of silver and gold 615 Weightman, A. T., on treatment of copper matte 545 Welch process of copper concentration 514 Wohlwill process of refining gold bullion 610 Work done by blowing engines 313 ZINC alloys, thermophysics of 110 Zinc, behavior in the blast furnace 240 Zinc compounds, heats of formation of 18-40 Zinc compounds, thermophysics of 131-144 Zinc, distillation of 632 Zinc, distilling furnace, efficiency of 108 Zinc, electric reduction of 637 Zinc, heat of oxidation of . 353 Zinc, metallurgy of 620 Zinc oxide, blast-furnace reduction of 651 Zinc oxide, electric reduction of 626 Zinc oxide, reduction of 63, 627 Zinc oxide, specific heat of 621 Zinc, reduction in a blast furnace 651 Zinc sulfide, roasting of 620 Zinc sulfide, specific heat of 621 Zinc sulfide, temperature of roasting 621 Zinc, thermophysics of 87 Zinc, vapor of 643 Zirconium, thermophysics of 90 F RETURN CHEMISTRY TlBRARY ^- 100 Hildebrand Hall 642-3753 LOAN PERIOD 1 1 MONTH 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS Renewable by telephone DUE AS STAMPED BELOW ~~ FORM NO. 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