Sold by Book Department MINING AND SCIENTIFIC PRESS 42O MARKET STREET SAN FRANCISCO Catalogue of Technical Books on request 2_ MINING AND MINE VENTILATION A PRACTICAL HANDBOOK ON THE PHYSICS AKD CHEMISTRY OF AND MINE VENTILATION FOR VOCATIONAL SCHOOLS, AND FOR THOSE QUALIFYING FOR MINE FOREMAN AND MINE INSPECTOR CERTIFICATES BY JOSEPH J. WALSH Mine Inspector, Wilkes-Barre, Pa. ILLUSTRATED NEW YORK D. VAN NOSTKAND COMPANY 25 PARK PLACE 1915 77/30 / Copyright, 1915 BY JOSEPH J. WALSH THE SCIENTIFIC PRESS ROBERT DRUMMOND AND COMPANY BROOKLYN. N. Y. PREFACE IN adding to the number of text-books on Mine Venti- lation the author aims to provide new material and to dwell more fully on the fundamental theories and laws of ventilation, and to furnish, if possible, to the student a more suggestive method of study in a more graphic form. While ventilation experts practically agree upon the essential theorems in ventilation, it is believed, however, that the subject may yet hold a new attractiveness and be more readily mastered if a few important principles, which are now generally misunderstood by the student, are mag- nified. The method of determining the size of fan, etc., to ven- tilate a mine under given conditions, together with certain facts pertaining to the water gauge, and the chapter on Mine Fires are entirely new features. The author wishes to express his gratitude to the Robinson Ventilating Co. of Pittsburgh, Pa.; The American Blower Company, and their manager, Thomas W. Fitch, Jr., Detroit, Mich.; The Jeffrey Manufacturing Co., Columbus, Ohio; The Colliery Engineer, Scranton, Pa.; M. B. King, Expert Assistant in Industrial Education, Harrisburg, Pa.; and The Taylor Instrument Co., Roches- ter, N. Y., for many illustrations, tables and other^infor- mation used in this book. J. J. W. WILKES-BARRE, PA., June 1, 1915. iii 345970 CONTENTS PAGE CHAPTER I MATTER Matter and its Properties. Hooke's Law CHAPTER II MOTION, VELOCITY AND FORCE Motion. Newton's Laws of Motion. Velocity. Force. Parallel- ogram of Forces CHAPTER III GRAVITATION Newton's Law of Gravitation. Weight. Effect of a Constant Force. Formulas for Falling Bodies 12 CHAPTER IV LIQUIDS AND LIQUID PRESSURE Specific Gravity, or Relative Density. How to Find the Specific Gravity of Bodies Lighter than Water, of Liquids. Table of Densities CHAPTER V HEAT Thermometer. Conversion of Thermometer Readings. Table of Melting-points. Table of Temperatures 27 v vi CONTENTS CHAPTER VI GASES PAGE The Atmosphere. Composition of the Atmosphere. Atoms and Molecules. Elements. Table of Elements. Density. Specific Gravity. Table of Gases 32 CHAPTER VII GASES Chemical Compounds. Mechanical Mixtures. Chemical Sym- bols. Atomic weight. Molecular Weight. Chemical Equations. Hygrometer and Its Use. Table of Water Vapor Contained in Saturated Air. Absolute Humidity. Relative Humidity. Dew Point. How to Find Relative Humidity and Table. Diffusion of Case? 39 CHAPTER VIII BAROMETER The Aneroid and Mercurial Barometers. Atmospheric Pres- sure. Use of Barometer in Mines. Use of Barometer in Determining Altitudes. Table of Altitudes. Barometer Indications. Effect of Temperature and Pressure on Vol- ume of Gases. Charles' Law. Boyle's Law. Absolute Zero. Absolute Temperature. Calculation of the Weight of a Gas at Different Temperatures and Pressures 52 CHAPTER IX GASES Acetylene Gas. Safety Lamps. Occlusion of Gases. Properties. Physical and Chemical Properties of Air. Carbon Monoxide. Carbon Dioxide, how Produced. Effect of Black Damp on Atmospheres Containing Fire Damp. Marsh Gas. Detection of Fire Damp. Ethane. Ethylene. Sulphurated Hydrogen. Table of Chemical Analysis of Mine Air 68 CONTENTS vii CHAPTER X SPECIFIC HEAT PAGE Heat Capacity. Table of Specific Heat. Measurement of Specific Heat 82 CHAPTER XI MINE VENTILATION Ventilation. Pressure Defined. Water Gauge. Calculations. Laws of Friction. Table of Velocity Pressure 87 CHAPTER XII MINE VENTILATION Natural Ventilation. Water and Steam Jet System of Ventila- tion. Furnace Ventilation. Ventilation by Means of Fan. Robinson Fan. Table of Dimensions and Volumes. The Sirocco Fan. Table of Quantities. Jeffrey Fan. Theoreti- cal Water Gauge. Cost per Horse-power. Installation of 3 Fan. Motive Column. Splitting of Air Currents. Regula- tors. Resistance 104 CHAPTER XIII FORMULAS Formulas and Their Application. Coefficient of Friction. Transposition of Formulas 141 CHAPTER XIV MINE FIRES Suggestions to Prevent Mine Fires. Suggestions for Guidance after a Fire or Explosion. Sealing a Mine Fire. Effect Produced by Sealing a Mine Fire. Useful Tables and Formulas 152 SUMMARY. . .175 CHEMISTRY OF MINING AND MINE VENTILATION CHAPTER I PROPERTIES OF MATTER 1. Matter. The term " matter " is one which has a very wide meaning. To say that " matter is that which occupies space/' adds little if anything to our common understanding of the term. Matter includes all things which exist of which we can become aware by our sense of sight, touch, taste, smell and hearing. There are numer- ous different kinds of matter and they are usually indi- cated by the term SUBSTANCE. Thus air, coal, iron, wood, water, etc., are different kinds of matter, also different substances. Matter may be classified under three distinct heads: namely, solids, liquids, and gases. 2. Properties of Matter. Matter is possessed of certain peculiar qualities which serve to define i't. These prop- erties are either GENERAL or SPECIFIC. General Properties are those found in all matter, such as EXTENSION, DIVISIBILITY, IMPENETRABILITY, POROSITY, INERTIA, INDESTRUCTIBILITY. Specific properties are those found in certain kinds of matter only, such as DUCTILITY, HARDNESS, MALLEABILITY. 3. Extension. All bodies have extension in three direc- 2 MINING kft'D 'MTNE VENTILATION tions, and occupy space, commonly called length, breadth, and thickness. The absence of any one of these three dimensions is sufficient to prove that the thing under consideration is not matter. Hence, lines and surfaces are not bodies in the physical sense. 4. Impenetrability. This property means that no two bodies, however small, can occupy the same space at the same time. When a stone is dropped into a tumbler full of water some of the water overflows. If the volume of the stone is one cubic inch, exactly one cubic inch of water is displaced. A nail driven into a block of wood pushes the substance of the wood together; the wood now occupies only part of the space it originally occupied. 5. Porosity. All matter is pprous, that is, the par- ticles of matter of which a body is composed do not fill the entire volume occupied by it. The molecules of a body are spherical therefore there is space between them. Hence, a blotter will absorb ink, lime will absorb carbon dioxide, without change of volume. Glass, iron and other hard substances are known to be porous. 6. Compressibility. The compressibility of a substance is evidence of its porosity. Gases are very compressible, solids to a much less degree, and liquids are almost incom- pressible. If the pressure upon a gas is doubled, tem- perature remaining the same, its volume is diminished one-half. While changing the pressure upon water in the same manner its volume diminishes only ^-Wir- 7. Indestructibility. Matter can be made to assume different forms as the result of PHYSICAL and CHEMICAL CHANGES. Sometimes the change is only temporary, as in the freezing of water or in the melting of iron. Such changes are called PHYSICAL CHANGES. In this case the substance does not lose its identity, but may be restored by merely mechanical means to its original state when the PROPERTIES OF MATTER 3 original temperature is resumed. But often the change is permanent, as in the burning of coal, the rusting of metals; in this case the original cannot be restored by mechanical means. Changes in which a substance thus loses its identity are called CHEMICAL CHANGES. Matter may be changed by crushing, burning, cooking and mixing with other substances; but MATTER ITSELF CANNOT BE DESTROYED. The number of atoms in the uni- verse is exactly the same now as it was hundreds of years ago. 8. Divisibility. Divisibility is that property of matter which indicates that a body can be divided into smaller parts without changing the matter of which it is composed. 9. Inertia. Inertia is the tendency possessed by a body to remain at rest or in motion. A body cannot put itself in motion or bring itself to rest. To do either, it must be acted upon by some force outside of itself. Use is made of inertia, as in driving on the head of a hammer by striking the end of the handle. The violent jar to a water pipe on suddenly closing the faucet is due to the inertia of the stream. 10. Elasticity. This property exhibited by matter indi- cates that if a body be distorted within its elastic limit it will resume its original form when the distorting force is removed. Apply pressure to a rubber ball, stretch a rubber band, bend a piece of steel. In each case the original form is changed, but the body readily recovers from the strain on the removal of the stress, and will resume its original form if not distorted beyond its elastic limit. All bodies, whether solids, liquids, or gases, when reduced in volume by addition of pressure, regain or partly regain their volume when this added pressure is removed. A piece of wood may be compressed to half its volume and 4 MINING AND MINE VENTILATION when released it expands, but does not nearly return to its original volume. It has been compressed beyond its ELASTIC LIMIT. Liquids and gases have no ELASTIC LIMIT. No amount of compression can permanently change their volume; they always return to their original volume when the dis- torting pressure is removed. HOOKE'S LAW. Whenever the forces that produce dis- tortions in any body are within the elastic limit, the distor- tions produced are directly proportional , to the forces that produce them. That is, if a one-pound weight be suspended from a spring balance, and the stretch of the spring measured, after which a four-pound weight be suspended in the same manner, it will be found that the four-pound weight will stretch the spring four times that of the one- pound weight. 11. Cohesion. When we try to break a piece of wood, we are conscious of a force tending to hold the parts to- gether. Hence, COHESION gives to solids their stabilty of form. All bodies are made up of small particles called molecules, and COHESION is the natural attraction that these particles have for each other. It is measured by the force required to pull them apart. The COHESION is not as strong in liquids as in solids. In fact, it is not sufficient to maintain the form, yet the molecules in a drop of water hanging from the roof of a mine have sufficient attraction for each other to support the weight of the drop, unless it becomes so large that the weight is greater than the COHESION. In gases the molecules are so far apart that there is very little COHESION between them. On account of this gases cannot be moved by a pull, they must be moved by a push or pressure. Air cannot be pulled through the airways of a mine; it is moved by reason of pressure. PROPERTIES OF MATTER 5 Likewise water, must be moved by pressure; its force of COHESION is not sufficient to allow it to be pulled. QUESTIONS 1. What different forms can water be made to assume by changing its temperature? 2. Why can the head of a hammer be driven on the handle better by striking the end of the handle against a stone than by striking the head against the stone? 3. How would you find the volume of a piece of coal by displacement? 4. Under how many heads is " matter " classified? What are they? 5. Can matter be destroyed? 6. What is Hooke's law of elasticity? Give an example of its application. 7. If the pressure upon a volume of gas is doubled, what change takes place in the volume of the gas, and to what extent? 8. If the pressure upon a volume of water is doubled, to what extent is the volume reduced? 9. What is meant by the term " inertia "? 10. If a hoisting rope on a shaft is stretched beyond its elastic limit will the rope recover from the strain after the stress is removed? 11. What do you understand by the term " COHESION "? 12. Why cannot air be pulled through a mine? 13. (a) If a stone, one-half cubic foot in volume, be dropped in a vessel filled with water, how much water is displaced? (6) If water weighs 62.5 Ibs. per cubic foot, what weight of water does the stone lift? (c) How much less will this stone weigh under water? CHAPTER II MOTION, VELOCITY AND FORCE 12. Motion. Motion is a change of place, and is the opposite of rest. Or, motion is the change in the relative position of a body with respect to some point or place. When a body moves in a path which constantly changes in direction, it is said to move in a curve. Strictly speaking all bodies moving in space are constantly changing in di- rection. A ball dropped from a balloon moves toward the center of the earth, but as the earth itself is moving around the sun, the path of the ball must be in a curving direction. For this reason a stone dropped into a deep shaft will strike the side before it reaches the bottom; however, for all practical purposes the slight curvature referred to may be neglected. 13. Newton's First Law of Motion. Every body con- tinues in its state of rest, or uniform motion in a straight line, except in so far as it may be compelled by force to change that state. Newton states in this law that a state of uniform motion is just as natural as a state of rest. This is at first difficult to realize, because rest seems the natural state and motion the enforced one. But difficulty is at once dispelled as soon as one begins to inquire into the causes that hinder the movement of a body artificially set in motion. A rifle ball soon stops because resistance of the air continually lessens its speed, and finally gravity draws it down upon the earth. A baseball rolling upon a level field soon stops because, 6 MOTION, VELOCITY AND FORCE 7 in moving forward, it must repeatedly rise against the attraction of gravity in order to pass over minor obstacles such as pebbles and mounds. There is also much surface friction. If it were possible to fire a rifle ball into space very remote from the attraction of the solar system, it would travel for ages, because no attraction or atmosphere would resist its progress. 14. Newton's Second Law of Motion. The second law reads: " Change of motion is proportional to force applied and takes place in the direction of the straight line in which the force acts. Thus if a cannon ball is shot horizontally along a level surface and another ball allowed to drop vertically from the mouth of the cannon, they will both strike the surface at the same instant. This shows that the force which gives the cannon ball its horizontal move- ment has no effect on the vertical force, which compels both balls to fall to the surface." 16. Newton's Third Law of Motion. To every action there is always an equal and opposite reaction. To illustrate, in Newton's own words: " If you press a stone with your finger the finger is also pressed by the stone. And if a horse draws a stone tied to a rope the horse will be equally drawn back toward the stone; for the stretched rope, in one and the same endeavor to relax or unstretch itself, draws the horse as much toward the stone as it draws the stone toward the horse." There must be, and always is, a pair of forces equal and opposite. Horse and stone advance as a unit because the mus- cular power of the horse exerted upon the ground exceeds the resistance of the stone. In springing from a boat we must exercise caution, because the force necessary to shove the body out of the boat reacts and tends to push the boat from the shore. 8 MINING AND MINE VENTILATION 16. Velocity. Velocity is the rate of motion. When a body moves over equal spaces in successive equal times its motion is uniform; if it travels unequal spaces in suc- cessive equal times its motion is variable. For example, an engine controlled by a governor runs, practically speak- ing, with a uniform velocity. On the other hand the motion of a stone falling down a shaft is variable, for its speed is increasing each second as it descends. 17. Force. Force may be denned as that which tends to produce motion or to hinder the motion of a body. When two forces act on a body along the same line and in the same direction the resultant force is simply their sum, and it acts in the same direction as the forces. The same is true when there are more than two forces acting in the same direction. When two forces act on a body along the same line, but in opposite directions, their resultant equals their dif- ference, and it acts in the direction of the greater force.. For example, if an engine pulls on a train of cars with a force of 3000 Ibs. and another pushes at the back with a force of 4000 Ibs., the resultant force applied to move the train is 7000 Ibs. But if one pulls with a force of 1000 Ibs. in one direction and the other with a force of 800 Ibs. in the other direction, the resultant force tending to move the train forward is only 200 Ibs. Therefore, the resultant of two forces acting in the same straight line, but in op- posite directions, is the difference of the given forces and acts in the direction of the greater. 18. Parallelogram of Forces. Forces may be repre- sented by lines drawn to the same scale. EXAMPLE. Suppose the force E (Fig. 1), to be 3 Ibs. and acting along the line AC toward C and at right angles to the force F, which is 2 Ibs. and acting toward B. Represent force E by line AC drawn to scale, say one MOTION, VELOCITY AND FORCE 9 inch equals 2 Ibs., in like manner draw AB representing force F. Complete the parallelogram by drawing the dotted lines CD and BD (parallel to AB and AC respectively). The magnitude and direction of the resultant of the two forces E and F will be equal to G and in the direction of line AD. When the angle is a right angle, as in the present case, the diagonal AD is the hypotenuse of the right- angled triangle ACD; the force of G, being equal to the hypotenuse, is found as follows: = 3.606. If the angle is not a right angle, the resultant can be found by measurement or by the principles of trigonom- etry. EXAMPLE. Suppose a sheave wheel and hoisting drum, as shown in Fig. 2, the rope passing from the drum around the sheave to the cage and making an angle of 30 with a vertical line. If the weight of the cage is 10 tons, (a) Wha, force will then be on the shaft of the sheave wheel? (6) In what direction will the resultant force act? Solution. As the weight of the load is 10 tons the tension at any point along the rope is 10 tons, consequently there is a force of 10 tons acting from the sheave to the drum and also a force of 10 tons acting from the sheave to the load. Produce the lines AB and BC to D, thus we have the point of application and direction of the forces. Using a scale of 1 inch = 5 tons, lay off from D a distance equal to 2 inches or 10 tons along lines DA and DC, then com- 10 MINING AND MINE VENTILATION plete the parallelogram by drawing line FE parallel to line DC and line GE parallel to line DA. Next draw line DE, which will be the direction in which the resultant force acts, and the length of DE, using the same scale as given above, will equal 19.5; therefore if the parts FIG. 3. are not in motion the weight on the shaft of the sheave wheel is 19.5 tons. Velocity can also be represented graphically. For example, if a man rows a boat across a stream with a uni- form velocity of 4 miles per hour, and the stream flows with a uniform velocity of 3 miles per hour, the direction taken by the boat can be determined by the velocities. If the boat starts from A, Fig. 3, the path of the boat MOTION, VELOCITY AND FORCE 11 may be found by laying off AB to represent the velocity of 4 miles per hour, and BC to represent the velocity of the stream, 3 miles per hour. Then AC will be the direction the boat will take. If the width of the stream is known line AC can be readily found. QUESTIONS 1. Find the resultant of 30 Ibs. north and 40 Ibs. east. Represent forces and resultant graphically. 2. What is motion? 3. Why will a stone not fall down a deep shaft without striking the side of the shaft? 4. What is Newton's first law of motion? 5. If a man rows a boat across a river at the rate of 2 miles per hour and the river is flowing at the rate of 3 miles per hour, show graphically the direction of the boat. 6. If a cannon ball is shot horizontally along a level surface and another ball is allowed to drop from the mouth of the cannon at the same instant, neglecting the resistance offered by the air, which will strike the ground first? 7. A rope runs from a hoisting drum at an angle of 45 and passes over a sheave wheel which is directly over the center of the shaft; on the shaft end of the rope there is a cage weighing 8 tons; what is the force on the sheave wheel? 8. What is velocity? 9. If two forces act on a body along the same line and in the same direction, each force is equal to 100 pounds, what is the resultant force? 10. If two forces act on a body along the same line, but in opposite direction, one force is equal to 50 Ibs., and the other 100 Ibs., what is the resultant force and in what direction does it act? CHAPTER III GRAVITATION 19. The power called gravitation is the name given to the attractive force between different bodies. It is this power that prevents the earth, moon, and other heavenly bodies from swerving outside their paths in space. The force of the great law of gravitation is so evenly and con- stantly applied that, hundreds of years in advance, the places of planets in space and the exact hour, minute and second when eclipses will happen can be foretold. Regardless of this what gravitation is is not yet known. It is only known that it acts instantaneously over distances whether great or small, and no known substance interposed between two bodies has power to interrupt their gravitational tendency toward each other. While the term GRAVITATION is applied to universal attraction existing between particles of matter, the more restricted term GRAVITY is applied to the attraction that exists between the earth and bodies upon or near its surface. 20. Newton's Law of Gravitation. The law may be stated as follows: First, that every particle of matter in the universe attracts every other particle directly as its mass or quantity of matter. Second, that the amount of this attrac- tion increases in proportion as the square of the distance between the bodies decreases. 21. Weight. The weight of a body is the measure of the attraction that exists between the earth and that body. Bodies weigh most at the surface of the earth. Below the 12 GRAVITATION 13 surface the weight decreases as the distance to the center decreases. Above the surface the weight decreases as the square of the distance increases. According to the above rule a body that weighs 100 Ibs. at the surface of the earth will weigh nothing at the center, the body being attracted equally in all directions. EXAMPLE. If the radius of the earth is 4000 miles and a body on the surface weighs 200 Ibs., what will it weigh 1000 miles below the surface? Solution. At 3000 miles from the center it will weigh 150 Ibs. 4000-1000 = 3000 and 4000 : 3000 : : 200 : 150. EXAMPLE. If the same body was carried 1000 miles above the surface or 5000 miles from the center of the earth, it will weigh 128 Ibs. 5000 2 : 4000 2 ::200 : 128. Therefore it can be seen that the weight of a body at any place is the attraction between it and the earth at that place. If two bodies have the same weight at a given place they must also have the same mass. THE DIRECTION OF THE EARTH'S ATTRACTION. When a plumb line is suspended from a certain point, the line is said to be vertical. Vertical lines suspended at different points on the earth's surface, if continued, would meet approximately at the earth's center, hence they are never strictly parallel, though practically so, provided they are near to each other. 22. Laws of Falling Bodies. A body that is moving under the influence of gravity alone is a FREELY FALLING BODY. This condition can be obtained only in a vacuum, as the air constantly offers a resistance to the passage of any body through it. 14 MINING AND MINE VENTILATION If an iron ball and a piece of paper are dropped from the same height, the ball will strike the ground first. This is not because the ball is heavier, but because the resistance of the air has a greater retarding effect upon the paper. If the same ball and paper were placed in a glass tube from which all the air has been extracted, and allowed to fall as before, they would both fall with the same veloc- ity and reach the bottom at the same instant. Some bodies do not fall, but ascend. For example, a balloon in the air or a cork under water. This is not because the earth does not attract them, but because an equal bulk of the air or water immediately above the body contains a greater mass of matter and is therefore more strongly attracted by the earth than the body itself. The balloon and the cork are, by reason of the greater mass of air and water above them, consequently exchanging places, the greater mass sinking and forcing the smaller mass upward. 23. Effect of a Constant Force. Whenever a body is falling freely under the influence of gravity only, regard- less of its size, it will fall in the first second 16.08 ft., and its velocity at the end of the first second, will be 32.16 ft. per second. This latter number is always denoted by g, and is the constant accelerating force exerted on all freely falling bodies. It should be understood that g varies at different points on the earth's surface, it being a little greater at the poles than at the equator. Careful experimenting has determined that at New York the acceleration of a freely falling body is, as stated, 32.16 ft. The distance through which a freely falling body will move in a given time is equal to 16.08 multiplied by the square of the time in seconds. EXAMPLE. The distance a body will fall in 2 seconds equals 16,08 times 2 2 or 16.08 times 4 or 64.32 ft. In GRAVITATION 15 3 seconds it will fall 16.08 times 3 2 or 16.08 times 9, or 144.72 ft. 24. Formulas for Falling Bodies. The relations ex- pressed by these formulas are usually known as the laws of falling bodies. They apply strictly to those bodies which fall without being hindered by the air or anything else. Let t number of seconds a body falls; v = velocity at the end of the time; h = distance that a body falls ; g = force of gravity, or accelerating force due to the attraction of the earth (0 = 32.16). EXAMPLE. If it requires 10 seconds for a stone to fall down a shaft, what is its velocity at the end of the 10th second, assuming that the air offers no resistance? Solution. v = gt or 32.16X10 = 321.6 ft. per second. EXAMPLE. If the shaft mentioned in the above problem is 1608 ft. deep, how long will it take a stone to fall from top to bottom? Solution. 1 = or = \ \ flf J2h / = \ = \ \ g \ 1608X2 or 10 seconds. oo ^ g \ 32.16 EXAMPLE. If a falling body has a velocity of 321.6 ft. per second, how long had it been falling at that instant? v 321.6 solution. t = . t = 00 g = 10 seconds. g 32.16 EXAMPLE. A stone dropped down a shaft has a velocity of 321.6 feet when it strikes the bottom, how deep is the shaft? 16 MINING AND MINE VENTILATION Solution. v 2 321.6X321.6 103426.56 2g 2X32.16 =1608 ft., depth of shaft. EXAMPLE. A body falls down a shaft which is 1608 feet deep, what will be its velocity at the end of the fall? Solution. v = \/2gh. v=v / 2X32.16X 1608 = 321.6 ft., velocity per sec. EXAMPLE. How far will a body fall in 10 seconds? Solution. h = %gt 2 or h = 16.08 Xt 2 . 16.08X100 = 1608 ft. EXAMPLE. If a ball is thrown vertically upward with an initial velocity of 321.6 ft. per second, (a) how long a time will elapse before it reaches the earth again? v 2 Solution. h = -=-, h = = 1608 feet. To find the time required to reach a height of 1608 ft. v 321.6 = - = = ^-^ = 10 seconds. g 62.1D As it will take the same length of time for the ball to fall to the earth the total time consumed in going both directions will be 10X2 = 20 seconds. Ans. GRAVITATION 17 QUESTIONS 1. If a stone thrown upward returns to the ground in 4 seconds, how high does it ascend? 2. A cannon ball is fired horizontally from the top of a cliff 200 ft. high. In how many seconds will it strike the plain at the foot of the cliff? 3. Define (a) gravitation; (6) gravity. 4. What is Newton's law of gravitation? 5. If an iron ball weighs 100 Ibs. on the surface of the earth, what will it weigh 1000 miles below the surface? 6. What will a 100-lb. ball weigh 1000 miles above the surface of the earth? What will it weigh at the center of the earth? 7. If two plumb lines suspended at different points on the earth's surface were projected through the earth where would they meet? 8. Will an iron ball weighing 2 Ibs. fall with a greater velocity than a smaller ball weighing 1 lb.? 9. How far will a freely falling body fall in the first second? What will be its velocity at the end of the first second? 10. A rifle ball is shot vertically upward with a velocity of 1500 ft. per second. In what time will it reach the ground, neglecting the friction of the air? 11. How far must a ball fall in order to acquire a velocity of 321.6 ft. per second? 12. A stone dropped down a shaft strikes the bottom in 4 seconds. What is the depth of the shaft? 13. A stone falls down a shaft 400 ft. deep. In what time will it strike the bottom? 14. Explain what is meant by accelerating force, and at what velocity will it cause a freely falling body to move at the end of the first second of its fall? 18 MINING AND MINE VENTILATION 15. Why does a balloon ascend in the air? 16. Why does not the attraction of the earth cause fire damp to lodge in low places in a coal mine? 17. When we speak of the weight of a body to* what do we refer? 18. Why is it a body has no weight at the center of the earth? 19. A stone 1 cubic foot in volume and weighing 180 Ibs. is under water. If a man lifts the stone while under water what weight does he lift? 20. What is the velocity of a freely falling body at the end of 12 seconds? 21. If a stone thrown vertically upward reaches its maximum height in 2 seconds in how many seconds will it fall to the starting point? CHAPTER IV LIQUIDS AND LIQUID PRESSURE 25. Liquids offer great resistance to forces tending to diminish their volume. Water is reduced only 0.00005 of its volume by a pressure of one atmosphere. A gas is reduced to one-half its volume by the same pressure. The case of sea-water is of special interest on account of the influence of its compressibility upon the ocean level. Tait, in his extended investigation of this property in connection with the deep-sea exploration, computed the loss of volume due to the compression of each layer of ocean water by the superincumbent mass, and found the level of the sea to be more than 600 ft. below that which would exist in the case of a strictly incompressible fluid. 26. Pressure on the Side of a Vessel. When a liquid is contained in a vessel, the sides being vertical, the pres- sure at any point of a side depends upon its distance from the surface of the liquid. The total pressure on the sides of the vessel is the sum of all these pressures, which vary from zero at the surface to a maximum at the bottom. RULE. The pressure of a liquid upon any submerged surface is equal to the weight of a column of the liquid having the area of the surface for its baseband the depth of the center of gravity of the given surface, below the surface of the liquid, for its height. This rule applies to all submerged surfaces whether vertical, horizontal or inclined, plane or curved. If the surface is the horizontal base of the vessel the height of the column will be the total depth of the liquid. 19 20 MINING AND MINE VENTILATION EXAMPLE. A vessel is filled with water. Its base is 2 ft. by 2 ft. and 5 ft. high. What is the total pressure on the base? Solution. 5 X (2 X 2) X 62.5 = 1250 Ibs. A cubic foot of water weighs about 62.5 Ibs. or 1000 ozs. NOTE. In plane surfaces the center of gravity is the center of area. The center of gravity of a triangle is a point two-thirds of the distance from any angle to the middle point of the opposite side. The pressure per square inch due to any head of water may be found by multi- plying the head or vertical height of the water by .434. This number is obtained as follows: 62. 5 -j- 144 = .434. * EXAMPLE. What is the pressure per square inch on the bottom of a column standing full of water, the ver- tical height being 500 ft.? Solution. .434X500 = 217 Ibs. 27. Specific Gravity, or Relative Density. The density of a body depends both upon its mass and its volume. If we were to select some one substance as a standard and compare the density of every other substance to that standard, we should obtain a set of results called the rela- tive densities of substances. The most suitable standard is water, therefore it is used for the purposes of determining the density or specific gravity of solids and liquids. The density of water being 1, the weight of any solid or liquid can be readily found if the specific gravity of the solid or liquid be known. EXAMPLE. If the specific gravity of anthracite coal is 1.4, what is its weight per cubic foot? Solution. As the relative densities of water and coal are as 1 : 1.4, meaning that the coal is 1.4 times the weight of water, therefore, as water weighs 62.5 Ibs. per cu.ft., a cubic foot of the coal will weigh 62.5X1.4 = 87.5 Ibs. 28. How to Find the Specific Gravity. As the density LIQUIDS AND LIQUID PRESSURE 21 of water varies with its temperature as well as with its purity, the temperature 4 C. or 39 F. is taken for the standard density because the density of water is greatest at that temperature. In order to get the most accurate results, distilled water at the temperature given must be used. DEMONSTRATION. Weigh a piece of coal in the air and note its weight; weigh again, letting the coal hang in a vessel of water, and the scale will be found to read less. The operation may be expressed by the following formula: ~ _ _ Weight of the body in air _ Difference of the weight in water and air' or In this example W is the weight of the body in air, W" its apparent weight in water. EXAMPLE. A piece of coal weighs 48 ozs. in the air and weighs 9 ozs. in water, what is its specific gravity? Solution. 4 o Q = 1.23 sp.gr. 29. How to Find the Specific Gravity of Bodies Lighter than Water. If the body be lighter than an equal body of water and will not sink, it must be fastened to a heavy body in order to submerge it. The specific gravity can then be found as follows: Weigh the body in air (W), then weigh a heavy sinker in water and call its apparent weight S. Tie the sinker to the body and weigh them both in water. Call the apparent weight W". Compute the specific gravity from the formula W 22 MINING AND MINE VENTILATION LAW OF FLOATING BODIES. A floating body displaces a volume of liquid that has the same weight as the floating body. EXAMPLE. A piece of wood weighs 4 ozs. in air (W), a sinker registers 5 ozs. in water (S), and the two when tied together and submerged register 3 ozs. (W"). It is noticed that the wood not only displaces its own weight of water, but buoys up 2 ozs. of the weight of the sinker; therefore the wood displaces 4+2 ozs. of water, hence its specific gravity is 4 4 = |=.67. 4+(5-3) 6 30. How to Find the Specific Gravity of Liquids. The specific gravity is accurately obtained by means of the specific gravity bottle. Any bottle with a small neck having a fixed mark around it can be used. First, weigh the bottle when empty (a); then fill with water to the fixed mark and weigh (6). The difference will be the weight of the water (b a). Fill the bottle with the liquid of which the specific gravity is required and weigh (c) ; the difference (c a) gives the weight of the same volume of the liquid; then the specific gravity will be Weight of liquid _c a Weight of equal volume of water 6 a' EXAMPLE. Grammes Bottle+water 65 Bottle 15 Weight of water 50 Bottle + calcium chloride solution 75 Bottle 15 Weight of solution 60 LIQUIDS AND LIQUID PRESSURE 23 Therefore the specific gravity or relative density of the calcium chloride solution = f$ = 1.2 (taking water as 1) . For practical purposes this method would be slow and tedious, and in such cases the hydrometer is employed. This instrument (Fig. 4) consists of a bulb attached to a long stem and is weighted at the bottom with mercury or small lead shot so that it will float upright in liquid. The stem is graduated, usually with a paper scale inside the glass. The reading on the stem corre- sponding to the level of the liquid in which the hydrometer is inserted can be easily read. The densities and specific gravities in table A are averages of results found by different observers. FIG. 4. QUESTIONS 1. Why do liquids buoy up objects immersed in them? 2. State the law of floating bodies. 3. A certain bottle when filled with water weighs 156 gms.; when filled with an oil it weighs 148 gms. If the empty bottle weighs 73 gms. find the specific gravity of the oil. 4. A boat displaces 580 cu.ft. of water; find the weight of the boat. 5. A tank 5 ft. deep and 10 ft. square is filled with water. What is the pressure on the bottom of the tank? What on one side? 6. How high must the reservoir of a city's water sys- tem be above any point to produce a pressure of 50 Ibs. per square inch at that point? 24 MINING AND MINE VENTILATION TABLE A Density Sp.gr. Density in Lbs. per Cu.ft. Coal, anthracite (varies) Charcoal (oak) 1.5 0.57 93.75 35 6 Ice 917 57 3 Sandstone 2 35 146 8 Aluminum. 2 57 160 6 Glass 2 60 162 5 Quartz 2 65 165 6 IMarble 2 65 165 6 Granite 2 75 171 8 Iron (gray cast) . 7 08 442 5 Zinc (cast) 7 10 443 7 Tin (cast) 7 29 455 6 Iron (wrought) 7.85 490 6 Brass (yellow) 8.44 527.5 Brass (red) 8 60 537 5 Nickel 8.60 537.5 Copper (cast) 8 88 555 Silver (cast) Lead (cast) 10.45 11 34 653.1 708 7 Mercury 13.6 850.0 Gold 19 3 1206 2 Platinum. .... 21.45 1340 . 6 Water (pure 39 F.) 1.00 62.5 7. What is the vertical depth of a column of water which counterbalances a column of mercury 30 ins. deep when the liquids are placed in the U-tube? 8. Why does a hydrometer float vertically in a liquid? 9. A boy can lift 75 Ibs. How many cubic inches of coal, the sp.gr. of which is 1.4, can he lift? 10. A block of wood is 1 ft. square and 2 ft. long. Its sp.gr. is .65. How much pressure would be required to keep it under water? 11. How do you find the specific gravity of a liquid? 12. How do you find the specific gravity of a solid which is lighter than water? LIQUIDS AND LIQUID PRESSURE 25 13. Which offers the greater resistance to compression, liquids or gases? 14. If a cubic foot of anthracite coal weighs 90 Ibs. what is its specific gravity? 15. A cubic foot of sandstone (sp.gr. 2.35) is suspended in water by a rope. What is the tension on the rope? What will it be when it is lifted from the water?* 16. A shaft mine 500 feet deep is allowed to fill with water. A certain section of the mine was squeezing prior to the water entering. To what extent will the water aid in stopping the squeeze? 17. If a piece of anthracite coal weighs 50 ozs. in the air and its apparent weight in water is 15 ozs., what is the specific gravity of the coal, and what is its weight per cubic foot? 18. If a body lighter than water weighs 15 ozs. in the air and a sinker weighs 25 ozs. in water and the body and the sinker fastened together weigh 20 ozs. in water, what is the specific gravity of the body? 19. If the weight of a certain liquid is 10 ozs. and the weight of an equal volume of water is 12 ozs., what is the specific gravity of the liquid? 20. An engineer reporting on a certain tract of coal land discovered that 180 acres contained coal, the seam being flat and 7 feet thick throughout the entire property. How many tons of coal are on this property if the specific gravity of the coal is 1.4? 21. A block of wood 1 ft. square and 2 ft. long is pushed down into water until its upper side is 6 ins. below the surface. What is the upward pressure upon the bottom of the block? What is the downward pressure of the water on the top of the block? How much pressure is required to keep the block in place if its specific gravity is .65? How much pressure would be required to keep it 26 MINING AND MINE VENTILATION at a depth of 2 ft.? Ans. 187.5 Ibs., 62.5 Ibs., 43.75 Ibs., 43.75 Ibs. 22. A cake of ice 6 ft. square and 2 ft. thick is floating on a lake. How much will it settle in the water if a man weighing 180 Ibs. stands upon it? Ans. .96 inch. CHAPTER V HEAT THE commonly used unit to measure the quantity of heat generated by the burning of coal or other substance is called the British Thermal Unit (B.T.U.). It is equiv- alent to the amount of heat required to raise the temper- ature of 1 Ib. of water 1 degree of the Fahrenheit scale, or 1 B.T.U. is equivalent to 778 foot-pounds. When heat is added to a body, whether solid, liquid or gaseous, the vibration of the molecules composing the body increases. This increased molecular motion will require an increased space between the molecules, and the body grows larger in volume that is, it expands and cooling a body will diminish its molecular motion and reduce its volume. The vibratory movement will cease only when a body is deprived of all its heat. Changes in temperature are detected and measured by the thermometer. To determine the actual amount of change in temperature in any case and to make it possible to compare the records of one thermometer with those of another, the thermometers must be similarly con- structed. To do this we must have one or more easily determined temperatures, called the FIXED POINTS. (1) Careful experimenting has shown that the temperature at which pure ice melts is practically constant, and (2) that the temperature of steam as it comes from boiling water is likewise constant when the pressure upon the water is constant. Then to establish the fixed points the bulb and part of the stem of the thermometer are filled with mercury and are placed in a vessel containing finely 27 28 MINING AND MINE VENTILATION broken ice and allowed to remain until there is no further change in the final position of the top of the mercury. The top of the mercury is then marked; this point is called the freezing-point of water or the melting-point of ice. The thermometer is then put into a steam generator and left until the mercury ceases to expand; this point is then marked and is called the boiling-point. On the Fahren- heit thermometer the freezing point of water is marked 32 degrees and the boiling-point 212 degrees. On this scale the difference between the two fixed temperatures is divided into 180 degrees. The centigrade scale differs from the Fahrenheit in making the freezing-point and the boiling-point 100, the space between being divided into 100 equal parts. This thermometer is the one in general use among scientific men. Water boils when its vapor escapes with sufficient pressure to overcome the pressure of the atmosphere upon its surface. Hence the boiling-point depends upon the pressure of the atmosphere or the vapor within a vessel such as a steam boiler. The boiling-point is lower as the pressure is decreased and higher as the pressure is increased. Warm water will boil under the receiver of an air pump or on top of a high mountain, the decreased pressure allow- ing the free movement of the molecules. At a point in South America, 9350 ft. above sea level, water boils at such a low temperature that it is not hot enough to cook potatoes. 31. Exception to the General Rule of Expansion. Generally speaking water expands and contracts in the manner common to all liquids, but between the tempera- tures (32 and 39 IT.) it presents a remarkable and most important exception. If water at the freezing-point is warmed its volume steadily decreases until 39 F. is reached, but when it is further heated water expands as other liquids do, up to its boiling-point. HEAT 29 Conversion of thermometer readings from one scale to another: C. to F., multiply by 9, divide by 5, add 32. F. to C., subtract 32, multiply by 5, divide by 9. EXAMPLE. Convert 350 C. into the corresponding Fahrenheit reading. Solution.-F. = +32 * 662. o EXAMPLE. Convert 662 F. into the corresponding centigrade reading. . The temperature of a melting solid remains unchanged from the time melting begins until the body is entirely melted. TABLE B TABLE OF AVERAGE MELTING-POINTS Ice . . C or 32 00 F. Sulphur .. . 115.1 C or 239.18 F. Lead. 326 c or 618 8 F Silver . . 950 C or 1742 F. Copper. . . . .1100 c or 2012 F. Iron 1500 c or 2732 F. Platinum Cast iron (gray) Steel. . . . . 1900 . . . 1275 . .1375 c. c. c. or or or 3452 F. 2327 F. 2507 F. TABLE C APPROXIMATE TEMPERATURES Just glowing in the dark, about. 525 C. or 977 F. Dark red .................... 700 C. or 1292 F. Cherry red ................... 910 C. or 1670 F. Bright cherry red ............. 1000 C. or 1832 F. Orange ...................... 1160 C. or 2120 F. White ....................... 1300 C. or 2372 F. Dazzling bluish white ......... 1500 C. or 2732 F. Bunsen flame ................ 1500 C. or 2732 F. Electric arc . . . 3500 C. or 6332 F. 30 MINING AND MINE VENTILATION 32. A freezing mixture can be made by mixing 1 part of salt with 3 parts of snow or cracked ice. The ice in contact with the salt is melted, the heat necessary for the melting being withdrawn from the objects near by. The salt is dissolved and the temperature falls to the freezing- point of the salt solution, which is lower than that of water. In this manner substances are frozen, for example ice cream. QUESTIONS 1. What effect (a) does expansion always have upon the density of a body? (6) Contraction? (c) Name an important exception to the general rule that expansion accompanies a rise in temperature. 2. What are the fixed points (a) on a Fahrenheit ther- mometer? (6) On a centigrade thermometer? (c) How are they marked? 3. Why does ice float in water? 4. Is boiling water over a gas flame receiving any heat? 5. If the bulb of a thermometer be plunged into hot water the mercury at first falls; why? 6. How is it possible to heat water above the ordinary boiling-point? 7. Convert (a) C. into the corresponding Fahrenheit reading; (b) 212 F. into the corresponding centigrade reading. 8. From the time a piece of cast iron starts to melt until it is all melted does the temperature change? 9. Do water pipes burst when they freeze or when they are thawed? 10. Explain why water boils at a lower temperature under reduced pressure. 11. A piece of ice is floating for a time in warm water. Does the water lose heat? Does the ice receive heat? HEAT 31 Does the temperature of the water change? Does the temperature of the ice change? 12. What is the temperature of the Bunsen flame? 13. When a body expands due to a rise in temperature, do the molecules increase in size? 14. Why will water boil at a lower temperature on a high mountain than at sea level? 15. When will the vibratory movement of the mole- cules of which a body is composed cease? 16. At what temperature is water at its greatest den- sity? 17. Explain why the specific gravity of ice is less than water. 18. Convert (a) 32 F. into the corresponding centi- grade reading; (6) 60 C. into the corresponding Fahren- heit reading. 19. Seventy-six degrees is called summer temperature on the Fahrenheit thermometer. What will be its reading on the centigrade thermometer? CHAPTER VI GASES 33. The Atmosphere. The earth is surrounded by a great mass of gas commonly known as the ATMOSPHERE or ATR. The estimated height to which the atmosphere extends has not been definitely fixed, but observation on meteors show that it really extends to a height of at least 100 miles, and indeed at that height it is sufficiently dense to cause the rapid combustion of a meteor passing through it. This great volume of gas rests upon the earth. The weight of the whole mass is such that it presses on every square inch of the earth's surface at sea level with a weight equal to 14.7 Ibs. At higher elevations the pres- sure is not so great. The pressure of the entire mass of the whole atmosphere may be approximately found by multiplying 14.7 by the number of square inches on the whole surface of the earth. In round numbers we might say that it is five thousand million of millions of tons. 34. Composition of the Atmosphere. Pure dry air is chiefly a mixture of oxygen, nitrogen and carbon dioxide, containing nearly four volumes or parts of nitrogen to one part of oxygen. Figures that are still more exact, and which are frequently used by the chemist when cal- culating the amount of oxygen in a given volume of air, are as follows: Per Cent. Carbon dioxide (CO 2 ) 0.03 Oxygen (O 2 ) 20.93 Nitrogen (N 2 ) 79.04 32 GASES 33 These percentages are those commonly used and refer to parts by volume that is, 100 cubic feet of air contain 0.03 cu.ft. of carbon dioxide, 20.93 cu.ft. of oxygen and 79.04 cu.ft. of nitrogen. By weight the percentages of oxygen and nitrogen are different, for in 100 Ibs. of dry air there are approximately 23 Ibs. of oxygen and 77 Ibs. of nitrogen. Ordinary air is not perfectly dry, but con- tains some water vapor. Besides oxygen, nitrogen and carbon dioxide, air con- tains five so-called rare gases which contribute about 1 per cent of the total volume. These gases are about the same as nitrogen and are considered as nitrogen in most calculations. All of the gases found in pure air are without color, smell or taste. Pure dry air contains oxygen and nitro- gen in the same proportions by volume all over the globe, at either sea level or high altitudes. 35. Atoms and Molecules. We often speak of atoms as if an atom of matter could exist. We do so simply because such an expression helps to describe and interpret chemical action. Atoms do not as a rule exist in the un- combined state. As soon as atoms are freed from combina- tion they at once unite with some other atom or atoms. When atoms unite the combination is called a MOLECULE. Hence a molecule is formed by the chemical union of two or more atoms. The atoms forming a molecule may be like or unlike. If the atoms in a molecule are atoms of the same element or kind, then the molecule is a mole- cule of an element; but if the atoms of different elements are combined, then the molecule is the molecule of a com- pound. All matter consists of molecules and the mole- cules are made up of atoms. We may define an ATOM as the smallest conceivable division of an element, and a MOLECULE as the smallest part of a compound, or of an 34 MINING AND MINE VENTILATION element which can exist in a free state and manifest the properties of the compound. Thus the smallest particle of marsh gas that can exist is a molecule of marsh gas, but a molecule of marsh gas contains smaller particles still, viz., atoms of carbon and hydrogen. 36. Elements. An elementary body consists of a simple substance which cannot be analyzed or reduced to parts that have properties other than those peculiar to itself. An element is a substance composed wholly of like atoms; oxygen, nitrogen, hydrogen, gold, silver, iron, etc., are all elements neither of which can be divided chemically into two or more substances; other substances can be added to them, but we cannot get simpler substances from them. TABLE D TABLE OF THE MOST IMPORTANT ELEMENTS Name. Symbol. Approximate Atomic Weight. Oxygen o 16 Nitrogen N 14 Hydrogen H 1 Carbon c 12 Sulphur s 32 Iron Fe 56 Lead. Pb 207 Gold Au 197 Copper. Cu 63 5 Chlorine Cl 35.4 Calcium. Ca 40 Aluminium Al 27 Mercury He 200 Nickel Ni 58 Rhodium Rh 103 Silver Ag 108 Sodium Na 23 Tin. . Sn 119 Tungsten W 184 Zinc . Zn 65 GASES 35 37. Density. Density is compactness of mass and has reference to the amount of matter in a given volume. When the density of a gas is spoken of it is understood to be compared with hydrogen gas as a standard taken as 1. Thus the density of air is 14.4 and of oxygen 16.0. That is, air and oxygen are respectively 14.4 and 16 times as heavy as hydrogen. 38. Specific Gravity. When the specific gravity of a gas is mentioned it is understood that the comparison is made with air as a standard. Thus the specific gravity of carbon dioxide is 1.527 and marsh gas 0.555, one being approximately 1J times as heavy, and the other half as heavy as air, the specific gravity of air being 1. Specific gravity is tbe measure of the density of a body. The density or specific gravity of all gases is affected by the temperature and pressure; if the temperature be increased the density is reduced and if the temperature be decreased the density is increased. The pressure also affects the volume and therefore the weight, if the gas be free to expand or contract. Hence the comparison of all densities and specific gravities is understood to have been made at the same standard temperature and pressure, namely, 60 F. and 30" barometer. The units of measure are as follows: For solids and liquids, as has been stated, 62.5 Ibs., the weight of 1 cu.ft. of water. For gases, .0766, the weight of 1 cu.ft. of air (temperature 60 F., barometer, 30"). EXAMPLE. If the specific gravity of carbon dioxide is 1.527, what is the weight of a cubic foot of the gas? Solution. 1.527 X. 0766 = .1169. EXAMPLE. Find the weight of 5 cu.ft. of marsh gas at a temperature of 60 F. and a pressure due to 30 ins. of barometer, the gas having a specific gravity of 0.559. Solution. 0.559 X .0766 X 5 = .2141. 36 MINING AND MINE VENTILATION "todpOo w It* " 00 "^ 1>- (N t~- Q} OO t>- t^- . ft EH' W ooooo o ooo o o o o o 0* CO lO CO 00 IP lO IQ C 1 ! iS- 00 ^^ t^ ^^ Os *O O^ ^O Oi '"^ C^ O^ O^ T-H T-H rH r-H O +J is 2*8 i-i (M TJ< CO OCO>OOO *-! COCOOO^HCO COOOOlN 1 * COGCOdCO %% COCOCOCOCO ^<*Tf^-^ cOt-Oi I-H CO COCOCOTtOi COCOt>l>I> >.- (IN COCO-*>OO 000005O'-H t-OOOOOSO I It OCOCGCOCOO OOOOXGOC 0505OS050S 050SOI0505 GASES 47 < i M -*f >O CO 00 Oi O *- i CO'OiO ICICCOCDCD CO COCCCDCCCO CO O OC50C5O OOOOO OO'-H-H'-H 00 OOOOOOOOOi -O5O>O5O50i O5O5O5OiO 48 MINING AND MINE VENTILATION any healthy person it is because the humidity is too low and water should be evaporated to bring the moisture up to the right degree. In other words water instead of coal should be used to make rooms comfortable when the tem- perature has reached 68 degrees. " Humidity causes the temperature, as shown by the thermometer, to vary as much as 35 degrees from the tem- perature as felt by your body. If it were not for the moisture in the air it would be too cold to live in. The reason for this is that if the air is dry the heat goes through it without warming it. If the air is moist it stops the ra- diated heat and warms it so that humidity acts as a check and prevents the heat from passing through the air. The dry air allows too much radiation from the body and too rapid evaporation makes us feel cold." The cooling effect produced by a wind does not neces- sarily arise from the wind being cooler, for it may, as shown by the thermometer, be actually warmer, but arises from the rapid evaporation it causes from the surface of the skin. Without moisture in the air there would be no life. The lack of humidity causes discomfort, ill health, catarrhs, colds and other diseases of the mucous mem- brane. It is supposed that colds are taken (in winter) by the sudden change in temperature in stepping out of doors, but as a matter of fact the change in humidity is much more harmful. In buildings heated by steam and hot water, with an average temperature of 70 degrees, the relative humidity averages about 30 per cent; in stepping from this atmosphere to an outside humidity of about 70 per cent the violent change is productive of harm, particu- larly to the delicate mucous membrane of the air pas- sages. The pneumonia period is the season of artificial heat in living rooms. The relative humidity should never be lower than 60 to 65 per cent. GASES 49 By the use of the following table the relative humidity of the air can be determined from the hygrometer readings. How TO FIND RELATIVE HUMIDITY BY THE TABLE Look in column on left for the nearest degree to the dry-bulb reading, then go horizontally along until the column is reached, on the top of which is the difference between the dry and wet-bulb thermometers, in which column the relative humidity will be found. EXAMPLE. Dry-bulb reading is 62; the wet-bulb 53; the difference is 9. Find 62 in column on left, run the eye horizontally along the column on top of page until 9 is reached, when the relative humidity will be found to be 54 per cent. 47. Diffusion of Gases. Fill two jars with gas, one with carbon dioxide and the other with hydrogen, and place them mouth to mouth, the jar containing the heavier gas (carbon dioxide) beneath the jar contain- ing hydrogen; while in this position it would appear that the lighter gas in the upper jar would rest on the heavier gas in the lower jar; it will be found, however, that this is not the case, as in a short time the gases will intermix and the composition of the gas in each jar will be the same. Further, these gases will never separate again into a heavy and light layer as they were before mixing. All gases when in proximity to each other mix or spread one into the other. The greater the difference between the densities of the gases the quicker they mix. This property of gases mixing will account for the fact that carbon dioxide is not always found on the floor of a mine, but is sometimes well diffused in the atmosphere. If two liquids which do not act chemically upon each other be mixed and allowed to stand, it will be found after a short time that the heavier liquid has settled to the bottom. 50 MINING AND MINE VENTILATION LAW of DIFFUSION OF GASES. The rate of diffusion of gases varies inversely as the square roots of their densi- ties. EXAMPLE. The density of hydrogen being 1, that of carbon dioxide being 22, their relative rates of diffusion will be inversely as Vl : \/22 which is as 1 : 4.69. That is, hydrogen will diffuse 4.69 times as quickly into carbon dioxide as carbon dioxide will diffuse into hydrogen. EXAMPLE. If the density of marsh gas is 8, and of air 14.4, what is the rate of diffusion? Ans. 1 : 1.34. Thus we see from the foregoing examples that 4.69 volumes of hydrogen will diffuse in the same time as 1 volume of carbon dioxide, and 1.34 volumes of marsh gas in the same time as 1 volume of air. QUESTIONS 1. What is a chemical compound? 2. What is a mechanical mixture? 3. Define atomic weight. 4. What is a chemical symbol and why are they used? 5. What is the molecular weight of carbon dioxide and how is it found? 6. The formula for carbon dioxide is C(>2; in a volume of this gas what part of its weight is carbon? 7. Can matter be destroyed? 8. When we speak of the relative humidity of the air being high what is meant? 9. If the relative humidity of the air entering a bitu- minous mine is low in what condition would you expect to find the coal dust? 10. What effect has humidity on the explosive limits of marsh gas and air? 11. Describe the hygrometer. What is it used for? 12. What is meant by diffusion of gases? GASES 51 13. Which will diffuse into air more quickly, marsh gas or carbon monoxide? 14. When liquids which do not act chemically are mixed what takes place? 15. Will the diffusion of two gases be quicker if the difference between their densities is great or small? 16. What is the difference between relative humidity and absolute humidity? 17. Will evaporation be fast or slow when the rela- tive humidity is high? 18. When are dust explosions more liable to occur in bituminous mines, summer or winter? Why? 19. When evaporation is rapid does the wet-bulb ther- mometer of the hygrometer rise or fall? 20. What is the specific gravity of saturated aqueous vapor? 21. There are two intake shafts at a mine, the rela- tive humidity of the air in one is 90 per cent and in the other 40 per cent. The depths and other conditions being similar through which will the most air flow? Why? 22. What should be the relative humidity of the air in living rooms? 23. If the dry-bulb of a hygrometer reads 60 degrees and the wet-bulb 55 degrees, what is the relative humidity? 24. A steam jet is placed in an upcast shaft, the tem- perature is increased 8 degrees and the relative humidity is increased 40 per cent. Does the increased relative humidity assist in increasing the quantity of air. Explain. 25. Will a wet road in a mine dry sooner if the velocity of the air is high or if it is low? Why? CHAPTER VIII BAROMETER 48. There are two kinds of barometers in use, the mercurial barometer and the aneroid barometer. The aneroid, owing to its portable form and great sensitive- ness in responding to changes in pressure of the atmos- phere, is to-day in more general use than any other form of barometer. It will denote a change much quicker than the mercurial barometer. In measuring altitudes, owing to its portability, sen- sitiveness and ease with which approximate results may be obtained, it is highly valuable to the engineer and sur- veyor. The illustration (Fig. 6) shows the general construction of the movement with its elastic metallic box called the vacuum chamber, A. The chamber is constructed with two circular discs of thin corrugated German silver firmly soldered together at the edges, forming a closed box as shown in Fig. 7. The air is exhausted from this box, which causes the top and bottom discs to close together as shown in Fig. 8. The pressure of the air upon the outside surface of an ordinary size chamber is equal to a force of about 60 Ibs. The vacuum chamber A is firmly fixed to the circular metal base B by a post upon its center projecting through the base plate. An iron bridge C spans the chamber, resting upon the base plate by means of the two pointed screws, c'c. 52 BAROMETER 53 (These screws are used to regulate the tension upon the chamber A.) To the bridge C is fixed the mainspring D, which is forced down by mechanical means sufficient to insert the knife-edge piece, e. As this knife edge is fastened (by means of a central pillar) to the top disc of chamber A, the mainspring D FIG. 6. .(W FIG. 7. FIG. 8. when released lifts the upper part of the chamber, drawing the two discs asunder so that the box again has the appear- ance as shown in Fig. 7. As this forms a perfect balance (the power of the main- spring opposing the atmospheric pressure upon the vacuum chamber) any variation in air pressure will now be shown by a movement up or down of the elastic chamber. A decrease in pressure will allow the mainspring to over- come the power of the vacuum, the action then being 54 MINING AND MINE VENTILATION upwards, and an increase of air pressure will produce the contrary result. This vertical action of the vacuum cham- ber is multiplied and converted to a horizontal movement of the indicating hand by a series of mechanical movements. As there is sometimes a settlement of some of the metal parts and springs which alters the position of the indicating hand, it is advisable, whenever an opportunity offers, to compare the readings of an aneroid with a stand- ard mercurial barometer. If they do not agree the aneroid may be adjusted by turning the small adjusting screw until the indicating hand pn the dial coincides with the height of the mercury column. FIG. 9. FIG; 10. In the best made instruments the main lever G is made of a composite bar of two metals, steel and brass, the quan- BAROMETER 55 tity of each metal being altered until it is correctly com- pensated for any change in temperature. This averts the necessity of making allowances for changes in tem- perature, as is necessary in reading a mercurial barometer. The divisions upon the scale of an aneroid barometer represents inches and fractions of an inch of atmospheric pressure, the scale being determined by comparison with a mercury column as explained. The words "storm," "fair," "rain," etc., upon the dial are simply relatives, as it does not follow that the weather conditions indicated by these words will neces- sarily exist when the indicating hand of the barometer points to them. The meaning, for example, when the hand points to the word "fair," is that the atmospheric pressure at that time is favorable to fair weather. The mercurial barometer consists of a glass tube about 34 ins. long. The tube is closed at one end and filled with mercury; it is then inverted with its lower end constantly below the surface of the mercury in a vessel fixed at the bottom. Figs. 9 and 10 show the mercurial and aneroid barometers ready for use. 49. Atmospheric Pressure. The average pressure of the atmosphere at sea level is 14.7 Ibs. per square inch. This is called the pressure of 1 ATMOSPHERE. If the area of a cross-section of the barometer tube is 1 sq.in. there should be 30 cu.ins. of mercury in a column 30 ins. high. As a cubic inch of mercury weighs .49 Ib. the whole column would weigh 30 X. 49 = 14.7 Ibs. EXAMPLE. If the barometer reads 28 inches, what is the atmospheric pressure? Solution. 28 X. 49 = 13.72 Ibs. per sq.in. If a liquid less dense than mercury is used the column will be correspondingly longer. Hence, if water be used instead of mercury the column at sea level would be 56 MINING AND MINE VENTILATION 408 inches, since mercury is 13.6 times as heavy as water. EXAMPLE. If the mercury column is 29 inches, what would be the height of a water column under the same pressure? 13.6X29 = 394.4 ins. Ans. The pressure of the atmosphere on the top of a high mountain is less than at sea level and it is greater at the bottom of a shaft than at the top. If the barometer reads 29 inches at the top of a shaft and at the bottom it reads 30 inches, the shaft is about 900 feet deep, as the difference of 900 feet in altitude means a rise or fall of approximately 1 inch of barometer. 50. Use of a Barometer in a Mine. The barometer is an instrument of great importance in mines where explo- sive gases are generated, as an increase or decrease in the atmospheric pressure is instantly indicated by the aneroid (not so quickly by the mercurial). Hence, in case the barometer drops 1 inch this will mean a reduction of .49 Ib. on every square inch of surface in the mine, thereby allowing the occluded gases in the coal to escape more freely. At some of the anthracite mines of Pennsylvania the barometer is located in a convenient place on the surface and the readings are recorded three times each day. In case the barometer indicates a decrease in pressure, the person in charge of the mine is notified of the impending danger. 51. Use of the Aneroid in Determining Altitudes. When a compensated barometer is used it is not necessary to make allowance for temperature. Before taking an altitude reading the of the altitude scale should always be opposite 31 inches on the barometer dial. BAROMETER 57 For example, suppose the aneroid indicated a pressure of 29 inches, and if we ascend a hill and the hand (by reason of a decreasing pressure) moves to 25 inches, the method of determining the difference in altitude is as follows: The value of 29 inches with the of the outer rim at 31 is about 1800 feet, while the value of 25 inches under the same conditions is 5850 feet. 5850-1800 = 4050, the difference in altitude. Now suppose the barometer indicates a pressure of 29 inches, but instead of having the feet at 31 inches, we move the milled ring so that the feet is standing opposite 29 inches. If the observer then ascends a mountain until the hand moves to 25 inches the altitude registered will be only 3750 feet, or 300 feet in error. The graduations on the altitude scale of an aneroid gradually diminish in size. The first inch of pressure, from 31 inches to 30 inches, represents an ascent of about 900 feet, while an inch of pressure, from 26 inches to 25 inches, represents about 1050 feet. Difference in altitude cannot be accurately determined by means of the barometer in the mine workings, because there is always a difference in pressure between the intake and return airways. If in case of an exhaust fan ventilating a mine, the barometer is carried through the intake and back through the return to the fan, it will be found that the barometer gradually falls as the distance to the fan decreases. Stepping from an intake airway through a door to the return airway will cause a fall in the barometer equal to the difference in pressure between the two airways. Under such conditions the barometer might show a difference of 58 MINING AND MINE VENTILATION 100 feet in elevation, while in reality the elevations of both roads are the same. TABLE H PROFESSOR AIREY'S TABLE OF ALTITUDES Barom- eter in Inches. Height in Ft. Barom eter in Ins. Height in Ft. Barom- eter in Ins. Height in Ft. Barom- eter in Ins. Height in Ft. 31.00 28.28 2500 25.80 5000 23.54 7500 30.94 50 28.23 2550 25.75 5050 23.50 7550 30.88 100 28.18 2600 25.71 5100 23.45 7600 30.83 150 28.12 2650 25.66 5150 23.41 7650 30.77 200 28.07 2700 25.61 5200 23.37 7700 30.71 250 28.02 2750 25 . 56 5250 23.32 7750 30.66 300 27.97 2800 25.52 5300 23.28 7800 30.60 350 27.92 2850 25.47 5350 23.24 7850 30.54 400 27.87 2900 25.42 5400 23.20 7900 30.49 450 27.82 2950 25.38 5450 23.15 7950 30.43 500 27.76 3000 25.33 5500 23 . 1 1 8000 30.38 550 27.71 3050 25.28 5550 23.07 8050 30.32 600 27.66 3100 25.24 5600 23.03 8100 30.26 650 27.61 3150 25.19 5650 22.98 8150 30.21 700 27.56 3200 25.15 5700 22.94 8200 30.15 750 27.51 3250 25.10 5750 22.90 8250 30.10 800 27.46 3300 25.05 5800 22.86 8300 30.04 850 27.41 3350 25.01 5850 22.82 8350 29.99 900 27.36 3400 24.96 5900 22.77 8400 29.93 950 27.31 3450 24.92 5950 22.73 8450 29.88 1000 27.26 3500 24.87 6000 22.69 8500 29.82 1050 27.21 3550 24.82 6050 22.65 8550 29.77 1100 27.16 3600 24.78 6100 22.61 8600 29.71 1150 27.11 3650 24.73 6150 22.57 8650 29.66 1200 27.06 3700 24.69 6200 22.52 8700 29.61 1250 27.01 3750 24.64 6250 22.48 8750 29.55 1300 26.96 3800 24.60 6300 22.44 8800 29.50 1350 26.91 3850 24.55 6350 22.40 8850 29.44 1400 26.86 3900 24.51 6400 22.36 8900 29.39 1450 26.81 3950 24.46 6450 22.32 8950 29.34 1500 26.76 4000 24.42 6500 22.28 9000 29.28 1550 26.72 4050 24.37 6550 22.24 9050 29.23 1600 26.67 4100 24.33 6600 22.20 9100 29.17 1650 26.62 4150 24.28 6650 22.16 9150 29.12 1700 25.57 4200 24.24 6700 22.11 9200 29.07 1750 26.52 4250 24.20 6750 22.07 9250 29.01 1800 26.47 4300 24.15 6800 22.03 9300 28.96 1850 26.42 4350 24.11 6850 21.99 9350 28.91 1900 26.37 4400 24.06 6900 21.95 9400 28.86 1950 26.33 4450 24.02 6950 21.91 9450 28.80 2000 26.28 4500 23.97 7000 21.87 9500 28.75 2050 26.23 4550 23.93 7050 21.83 9550 28.70 2100 26.18 4600 23.89 7100 21.79 9600 28.64 2150 26.13 4650 23.84 7150 21.75 9650 28.59 2200 26.09 4700 23.80 7200 21.71 9700 28.54 2250 26.04 4750 23.76 7250 21.67 9750 28.49 2300 25.99 4800 23.71 7300 21.63 9800 28.43 2350 25.94 4850 23.67 7350 21.59 9850 28.38 2400 25.89 4900 23.62 7400 21.55 9900 28.33 2450 25.85 4950 23.58 7450 21.51 9950 BAROMETER 59 BAROMETRIC INDICATIONS. The barometer not only indicates an increased or decreased pressure on the occluded gases in the pores of the coal, or determines the height of mountains or depth of shafts, but also forecasts the weather. A rapid fall or a rapid rise of the barometer indicates that a strong wind is about to blow and that this wind will bring with it a change in the weather. What the nature of the change will be will depend upon the direction from which the wind blows. If an observer stands facing the wind the locality of low barometric pressure will be at his right and that of high barometric pressure at his left. With low pressure in the west and high pressure in the east, the winds will be from the south; but with low pressure in the east and high pressure in the west, the wind will be from the north. A slow but steady rise indicates fair weather. A slow but steady fall indicates unsettled or wet weather. A rapid rise indicates clear weather with high winds. A very slow fall from a high point indicates wet and unpleasant weather without much wind. A sudden fall indicates a sudden shower or high winds or both. When a barometer falls considerably without any precise change of weather it may be certain that a storm is raging at a distance. A stationary barometer indicates a continuance of existing conditions, but a slight tap on the barometer face will likely move the hand a trifle, indicating whether the tendency is to rise or fall. The principal maximum barometer pressure occurs before noon and the principal minimum after noon. EXAMPLE. If the barometer reading is 30 inches, (a) what is the pressure per square inch on the face and 60 MINING AND MINE VENTILATION sides of a chamber in a mine? (6) If the reading is 28 inches, what is the pressure? Am. (a) 14.7 Ibs. (6) 13.72 Ibs. By the above example it is readily seen that as the pressure per square inch is less with a barometer of 28 inches than with a barometer of 30 inches, a larger volume of gas will escape from fissures and pores of the coal, thus rendering the mine atmosphere more dangerous than if the barometer stood at 30 inches. 52. Effect of Temperature and Pressure on Volume of Gases. CHARLES' LAW. It has been found by experi- ment that under constant pressure all gases expand or contract equally for equal changes of temperature. More explicitly, a gas expands or contracts 1/491 of its volume at 32 F., for every degree through which it is heated or cooled. This means that 491 cubic feet of gas at 32 F. becomes 492 cubic feet at 33 F., 490 at 31 F. BOYLE'S LAW. It has been found by experiment that under constant temperature the volume of a gas is inversely proportional to the pressure. This means that doubling the pressure halves the volume, and vice versa. If a gas is under a certain pressure and the pressure is dimin- ished to J, f, |, etc., of its original pressure, the gas will increase in volume 2, 4, 8, etc., times. Its tension increas- ing the volume will decrease at the same rate. EXAMPLE. If 6 cu.ft. of air be under a pressure of 10 Ibs. (a) what will be the volume when the pressure is 20 Ibs.? (6) when the pressure is 5 Ibs., temperature remain- ing the same? (a) vol. = ^5= 3 cu.ft. 6X10 (6) vol. = - = 12 cu.ft. BAROMETER 61 It should be remembered that, when the temperature remains the same, the volume of a given quantity of gas varies inversely as the pressure. By means of the following formula the increased or decreased volume of a gas, due to an increase or decrease in the temperature, can be found, pressure remaining con- stant. Let v volume of gas before heating; v' = volume of gas after heating; t = temperature corresponding to volume v ; t' = temperature corresponding to volume v f . Thus, 459+O v = v (459+0 EXAMPLE. If 10 cubic feet of air at a temperature of 40 is heated under constant pressure until the temper- ature reaches 150, what is the new volume? , 459+150 (609) ,. "-" 459+40 =1 X (499) = 12.20 cu.ft. 53. Absolute Zero. The atoms and molecules of all bodies are in a constant state of vibration. An increase of heat will increase this vibratory movement and a decrease of heat will have the opposite effect. This vibratory movement will continue, however, until a temperature 459 F. below zero is reached, at which point the movement will cease. 459 F. is called absolute zero. No one has ever succeeded in depriving a body of all its heat or cooling it to absolute zero, though some exper- iments have come within 10 of it; nevertheless, it has a meaning and is used in many formulas. EXAMPLE. If 10,000 cu.ft. of air enters a mine at a 62 MINING AND MINE VENTILATION temperature of 30 F., what volume will leave the mine if the temperature in the return airway is 70 F. Ans. 10,818 cu.ft. nearly. ABSOLUTE TEMPERATURE. If the absolute temperature of a gas is known the ordinary temperature may be found by subtracting 459 from the absolute temperature. EXAMPLE. If the absolute temperature of a quantity of air is 500 F., the ordinary temperature is 500-459 = 41 F. The pressure, volume, temperature or weight of air can be found by the following formulas, in which P = pressure per square inch ; V = volume of air in cubic feet; T = absolute temperature ; W = weight of the air; .37052TF7 7 T = P PV .37052TF PV ~.37052r EXAMPLE. If 20 cubic feet of air weighs 2 pounds and the temperature is 60 F., what is the pressure or tension in pounds per square inch? .37052TFT' .37052X2X519 Solution. P = = = - ort = 19.23 per V A\J sq.in. nearly. Ans. EXAMPLE. The temperature of a certain quantity of a ir is 60 F. Its weight is 2 Ibs. and the pressure per square inch is 19.23 Ibs. What is the volume? BAROMETER 63 .37052F7 7 .37052X2X519 on ., Solution. V = -- p -- = - ~To~23~~ ~~ = C Ans. EXAMPLE. If 20 cubic feet of air have a (pressure) tension of 19.23 Ibs. per square inch and weighs 2 Ibs., what is the temperature? rp_ Solution. 19.23X20 .37052TF-. 37052X2 EXAMPLE. If 20 cubic feet of air have a tension of 19.23 Ibs. and a temperature of 60 F., what is the weight? PV 19.23X20 2 lbs ' An8 ' If a certain quantity of gas be heated through any number of degrees and the volume remains the same the tension or pressure will increase. For example, if a volume of gas is confined in a cylinder and if the gas is heated, its tendency to expand is prevented by the sides of the cyl- inder, consequently, the tension or pressure of the gas is increased. It will be found that for every increase of temperature of 1 F. there will be an increase of ^ir of the original tension at 32 F. If the gas is free to expand adding heat will increase the volume and the tension will remain constant. EXAMPLE. If a quantity of gas is heated from 30 F. to 70 F., the volume remaining constant (that is the volume enclosed so it cannot expand), what is the result- ing tension if the original tension was 14.7 Ibs. per square inch? 64 MINING AND MINE VENTILATION p = original tension; t = original temperature; ' = any temperature; p' = corresponding tension. .(459+Q P P (459 +t) ' _ (459+70) .(529) P = EXAMPLE. If a quantity of gas is heated under con- stant volume from 30 F. to 100 F., what is the resultant tension, the original tension being equal to one atmos- phere? The term " a pressure of one atmosphere " is sometimes used as a unit of pressure; it means 14.7 Ibs. Thus, two atmospheres mean a pressure of 29.4 Ibs. per sq.in. 54. Calculation of the Weight of a Gas at Different Temperatures and Pressures. The weight per cubic foot of any gas at different temperatures and pressures can be found by the following formula: Let W = weight in pounds; V = volume in cubic feet; B = barometric pressure; S = specific gravity; T = absolute temperature. EXAMPLE. If 250 persons are employed in a mine and each person is allowed 200 cubic feet of air per minute, what is the weight in tons of the air passing through the mine in 10 hours, the temperature being 60 F. and the barometer 30 inches? x Solution. 250X200X60X10 = 30,000,000 cu.ft. BAROMETER 65 weight per cu , t . EXAMPLE. What is the weight of 100 cubic feet of carbon dioxide gas at a pressure of 30 inches and a tem- perature of 30 F.? NOTE. The constant, 1.3253, is the weight in pounds of one cubic foot of air at 1 absolute temperature (F.) and 1 inch barometer. w Solution. W QUESTIONS 1. What do we mean when we say a barometer is com- pensated? 2. What is the meaning of the words " stormy," " fair," " rain," etc., upon the dial of an aneroid barometer? 3. Which will denote a change in pressure more quickly, the aneroid or mercurial barometer? 4. What is the weight of a cubic inch of mercury? 5. Why is mercury used in a barometer instead of some other liquid? 6. What is the average pressure of the atmosphere per square inch at sea level? 7. If 2 cubic feet of air are under a pressure of 50 Ibs. per square inch, (a) what will be the pressure when the volume is increased to 5 cubic feet? (6) to 3 cubic feet? 66 MINING AND MINE VENTILATION 8. If 20 cubic feet of air have a tension of 6 Ibs. per square inch, (a) what is the volume when the tension is 5 Ibs.? (6) 10 Ibs.? (c) 15 Ibs.? 9. The weight of 1 cubic foot of air at a temperature of 60 F. and under a pressure of 1 atmosphere (14.7 Ibs. per square inch) is .0766 lb., what would be the weight per cubic foot if the volume be compressed until the tension is 6 atmospheres, temperature remaining the same? 10. If in the last example the air had expanded until the tension was 10 Ibs. per square inch, what would have been its weight per cubic foot? 11. If 10 cubic feet of air at a temperature of 60 F. and a pressure of 1 atmosphere are compressed to 4 cu.ft. (temperature remaining the same) what is the weight of a cubic foot of the compressed air? 12. When 5 cubic feet of air at a temperature of 40 F. are heated under constant pressure up to 150 F., what is the new volume? 13. What is the weight of 100 cubic feet of air at a temperature of 60, barometer 30 inches? 14. What is the weight of 100 cubic feet of marsh gas (conditions same as question 13)? 15. What is the weight of (a) 100 cubic feet of carbon dioxide, (6) 100 cubic feet of carbon monoxide (temper- ature and pressure same as in question 13)? 16. Which is the more dense, (a) air or marsh gas, (b) air or carbon dioxide? 17. There are two shafts connected under ground and so located that the barometer reading at the top of the first shaft is 29 ins. and at the top of the second shaft 30 ins.; the temperature in the first shaft is 60 and in the second 100, in what direction will the air move (natural ventilation) ? 18. If water were used in the construction of a barom- BAROMETER 67 eter, what would be the height of a water barometer when the mercurial barometer stands at 30 ins.? at 25 ins.? 19. Explain the principle involved in using the barom- eter to measure elevations. 20. Why do we not feel the pressure of the atmosphere? CHAPTER IX GASES 55. Acetylene Gas. Acetylene burns in the air with a smoky, luminous flame, but when air is mixed with the gas as it issues from a small opening such as the jet of a miner's lamp, the mixture burns with a brilliant white flame which does not smoke, in view of which the lamp is now used quite extensively in the mines. The flame is much smaller than an ordinary gas flame of the same lighting power. Acetylene is generated by putting calcium carbide into a flask and allowing water to drop slowly upon the carbide. A pound of calcium carbide yields about 5 cubic feet of acetylene gas. The formula for acetylene is C 2 H 2 . Acetylene is slightly poisonous, though very much less so than carbon monoxide. Investigations made by the Bureau of Mines with acetylene generated from carbide such as is used in a miner's lamp, indicate that there is little if any chance of men being poisoned because of the use of acetylene in mines. Acetylene, of course, is suffo- cating, as are carbon dioxide, nitrogen and hydrogen. Calcium carbide is made by heating a mixture of lime and coke in an electric furnace. It is a hard, brittle solid; its specific gravity is 2.2. Owing to its action with water, it should be packed in air-tight cans. Fig. 11 shows the style lamp used in the mines for lighting purposes. The acetylene lamp will burn in air that contains only 68 GASES 69 10 to 11 per cent of oxygen, a proportion which is much too low to support the flame of an ordinary oil lamp. For this reason objection has been made to the use of acety- lene lamps in mines, because they may not warn the miners that the atmosphere is so low in oxygen as to cause them immediate harm. If a man exerts himself in such atmos- phere his labored breathing warns him that the air is not FIG. 11. fit to breathe. The best authorities agree that a man will live without serious inconvenience in an atmosphere where the oxygen is reduced to 10 per cent. Deficiency of oxygen becomes a real danger when it is as low as 7 or 8 per cent. The acetylene flame is extinguished before the danger point is reached and the suggestion that it does not give adequate warning by extinction in an atmosphere low in oxygen has been disproved not only scientifically but practically. 70 MINING AND MINE VENTILATION PERCENTAGE TO WHICH OXYGEN MUST BE REDUCED TO EXTINGUISH VARIOUS FLAMES Coirbustible. Percentage of Oxygen. Candle 16 to 17 Benzine 16 to 17 Hydrogen 7 to 8 Acetylene 10 to 11 Petroleum 16 to 17 56. Safety Lamps. It frequently happens that an explosive mixture of gases accumulates in coal mines. An ordinary lamp brought in contact with this mixture would cause an explosion. To prevent this and still make it possible to use a light, Sir Hum- phry Davy devised a form of lamp (Fig. 12) in which the flame is en- tirely surrounded with wire gauze. Whenever the lamp is brought into an inflammable mixture of gases some of the mixed gas will enter the lamp and burn there. But the heat is ab- sorbed by the gauze to such an extent that the gas outside the lamp does not receive heat enough to ignite until the gauze becomes so heated that it cannot take any more heat from the burning gas; the flame will then pass through the gauze and ligct the gas in the surrounding atmosphere. The standard adopted as a limit of safety was iron wire gauze with 784 meshes per square inch, the wires being about ^V inch in thickness. In a dangerous atmosphere the entire space within the gauze becomes occupied with flame; under such condition the lamp should be removed FIG. 12. GASES 71 carefully from the gaseous mixture, making no quick move- ments while doing so. Modifications of the Davy lamp have come into use, chiefly with a view to surrounding the flame with glass so as to increase the effective radiation of light; but in each case ingress and egress of air are effected through one or more thicknesses of wire gauze. The lamps most commonly used are the Davy, Clanny and Wolf. The features most desired in a safety lamp are (1) safety in strong currents; (2) maximum illumi- nating power; (3) security of lock; (4) so constructed that it can be relighted without opening the lamp; (5) simplicity of construction. 67. Occlusion of Gases. A gas is occluded when it is absorbed and pent up in the pores of any substance. Hydrogen is absorbed freely by several metals, especi- ally platinum and palladium. Gases exist in varying quantities in coal seams; those most commonly occluded in the coal are marsh gas or methane, carbon dioxide, nitrogen, oxygen and ethane. The pressure of the occluded gases is sometimes as high as 12 to 15 atmospheres. In newly-exposed coal faces the gas can be heard and felt exuding from the pores. Many cases are recorded where the flow of gas from coal seams was so strong that a formerly reasonably safe atmosphere became in a short time explosive. The writer has many times heard it said that in some mines gas exudes with such force from the coal seams that it prevents the movement of the air. This is due, not to the force with which the gas is emitted, but to the volume of gas given off. To remove a large body of firedamp it would require a pressure much greater than the average mine fan now in operation can produce. 72 MINING AND MINE VENTILATION 58. Properties. All gases conform and behave uni- formly with changes of pressure and with changes of tem- perature. Thus if the pressure on a certain volume of marsh gas be doubled the volume will be reduced one-half, and if the pressure on a volume of carbon dioxide be doubled the volume will also be reduced one-half, temperature remaining the same. Also if the temperature of several gases be increased one degree the amount of increase in volume will be the same in all, pressure remaining the same. There is an equal number of molecules in equal volumes of all gases at the same temperature and pressure. Therefore, since one molecule of oxygen weighs 16 times more than one molecule of hydrogen, 100 molecules of oxygen will weigh 16 times more than 100 molecules of hydrogen. 59. Physical Properties of Air. Air when pure is col- orless, tasteless, odorless and transparent. It can be liquefied by pressure at a very low temperature. It is 14.4 times as heavy as hydrogen. 60. Chemical Properties of Air. The formula for air is ON4. Its oxygen supports combustion, the energy of which is checked by the diluting nitrogen. When air con- taining carbon dioxide is passed through lime-water the carbon dioxide renders the clean liquid milky in appear- ance. 61. Carbon Monoxide. This gas is sometimes called white damp (CO). When burned, a blue flame, such as is produced by the burning of anthracite coal, can be seen. Carbon monoxide is produced when carbon is burned in a limited supply of air. 62. Properties. Carbon monoxide is a very pois- onous gas; it' is doubly dangerous because its lack of odor prevents its detection. The gas is a little lighter than air, its density being 14. GASES 73 It does not support combustion, but is combustible. It burns with a pale blue flame and yields carbon dioxide (CO2) as the sole product of its combustion. One per cent of it in the air is fatal to life. The best antidote is the free inhalation of pure oxygen. When a lighted lamp is placed in an atmosphere con- taining this gas the flame brightens and lengthens into a more or less slim taper with a bluish tip. 63. How Produced. Carbon monoxide is a product of the incomplete combustion of carbonaceous fuel when the supply of air is limited. Mine fires and explosions of powder and fire damp are the principal sources of this gas in mines. 64. Explosive Properties. Carbon monoxide mixed with air is explosive, but explosions of mixtures of carbon monoxide and air in mines are very rare. However, if an atmosphere contains 15.5 per cent of carbon monoxide, it will explode, but such a large percentage of carbon monoxide is seldom found in the gases from a mine fire. A mixture of carbon monoxide and air containing too little carbon monoxide to be explosive may become explo- sive by the addition of enough marsh gas, even if the pro- portion of marsh gas in the mixture be below the explo- sive limit of marsh gas and air. 65. Carbon Dioxide. CO2, commonly called " black damp," is a colorless gas and is about one and one-half times heavier than air, its density being 22. On account of its weight it can be displaced and poured from one vessel to another. This gas diffuses very slowly on account of its high density, therefore it often accumulates in low places in mines. The gas is soluble in water volume for volume at ordinary temperatures and pressures. It is not as dangerous as carbon monoxide. It can be detected by 74 MINING AND MINE VENTILATION an ordinary lamp; the light becomes dim and appears to pull away from the wick. If air containing carbon dioxide is passed through lime-water the liquid becomes milky. The exhausts from gasoline engines used in mines are sometimes so arranged that they exhaust through lime-water. The carbon dioxide unites with the lime in the water and is thereby prevented from being discharged into the atmosphere. If a known volume of dry air is forced through a known weight of lime-water the increase in weight of the water will be the weight of carbon dioxide in the volume of air used. 66. How Produced. Carbon dioxide is formed when carbon or any substance containing carbon is burned in a plentiful supply of air; thus mine fires, explosions of gas, burning of lamps and explosions of powder are the prin- cipal producers of carbon dioxide by combustion in mines. It is also produced when vegetable and animal matter decays and by the breathing of men and animals. As the gas is very soluble in water it is largely carried into a mine in this manner and when the water evaporates the gas escapes. The gas is incombustible and will not sup- port combustion. Animals die when put into this gas. The supply of oxygen is cut off in a manner similar to drowning. A small quantity of the gas in the air pro- duces headache, and if the quantity be increased suf- ficiently death results by suffocation. 67. Effect of Black Damp on Atmospheres Contain- ing Fire Damp. It has been found by experiment that atmospheres containing only 13 per cent of oxygen may be explosive when enough methane is also present. Con- sequently the atmosphere in one part of the mine may contain black damp enough to put out an oil flame and be non-explosive, but farther on in the mine where more GASES 75 methane is present an electric spark or a flame may cause an explosion. 68. Marsh Gas. Marsh gas (CH 4 ) is a colorless, odorless, tasteless gas; it is slightly soluble in water. It is one of the lightest known substances, its density being 8. It is produced by the decay of vegetable matter con- fined under water in the absence of air. It is found to a greater or less extent in all coal seams, and when mixed with air in the following proportions forms fire damp. NOTE. Marsh gas is also known as methane or light carbureted hydrogen; either term may be used when re- ferring to the gas. Lowest explosive limit: Volume of marsh gas = 1 Volume of air =5.5 6.5 Percentage of gas in mixture ^ XI 00 = 15.38 per cent. o . o Greatest explosive force: Volume of marsh gas = 1 Volume of air =9.5 10.5 Percentage of gas in mixture T7 = X 100 = 9 . 52 per cent. 1U . o Highest explosive limit : Volume of marsh gas = 1 Volume of air = 13 14 Percentage of gas in mixture ^X 100 = 7. 14 per cent. 76 MINING AND MINE VENTILATION On account of its low density marsh gas diffuses very rapidly with air, forming fire damp. This mixture, owing to its lightness, ascends and lodges along the roof of the mine. Some idea of the enormous quantity of marsh gas that may be carried from a mine by the ventilating current is shown by the following statement: In the main return airway of a certain mine is passing 150,000 cu.ft. of air per minute; this air contains 1 per cent of marsh gas, hence the total amount of gas expelled from the mine in 24 hours is 150,000 X 60 X 24 X. 01 =2, 160,000 cu.ft. An explosive mixture of marsh gas (or methane) and air ignites if heated to a temperature of about 1300 F. If the flame be cooled below this temperature it goes out. Sulphureted hydrogen when mixed with the quantity of air necessary for complete combustion will ignite at a temperature of about 600 degrees Fahrenheit, while ethane, ethylene and carbon monoxide will, under the same con- ditions, ignite at about 1300 degrees Fahrenheit. The relative humidity of the mine air will affect the explosive limits of marsh gas and air, thus a percentage of marsh gas that would just make under-saturated air explosive would be totally inexplosive in air saturated with watery vapor. It has also been found by experi- ment that a mixture of marsh gas and air that is outside the explosive limits is rendered explosive by an increase of pressure. A heavy blast in a mine might create sufficient pressure to render an inexplosive mixture explosive. 69. Detection of Fire Damp. Fire damp is detected by means of the safety-lamp. The lamp should be raised cautiously in a vertical position to the place where gas is suspected. Some prefer testing with the ordinary work- ing flame, while others prefer a much smaller light. If gas be present it will flame inside the gauze. If it is desired GASES 77 to test for a small percentage of gas in the atmosphere the wick must be pulled down until a very small light appears above the burner. In this case the presence of gas is manifested by a non-luminous cap above the flame. When a Davy lamp burning sperm or lard oil is em- ployed the height of the cap produced in any percentage of gas will vary slightly, depending on the original flame used. The results obtained will be more uniform if the wick is drawn down until a very small light remains. The percentage of gas in the atmosphere can then be calculated as follows : P = percentage of gas in the air; h = height of gas cap in inches; Thus, if the lamp indicated \ inch cap the percentage of gas present is = 2.6 per cent of gas. 1.25 ins. .8 " .6 " .5 " 78 MINING AND MINE VENTILATION NOTE. By experiments conducted by Mr. Beard he discovered that in an unbonneted Davy lamp the height of the flame cap was 1/36 of the cube of the percentage of gas producing the cap. 70. Ethane. Ethane (C2H6) is a member of the marsh- gas series. It is a colorless, odorless and tasteless gas with properties very similar to those of marsh gas; it is rarely found in mines. It is produced by dry decomposition of vegetable matter, and is explosive when mixed with air. 71. Ethylene. Ethylene (C2H 4 ), or olefiant gas, is formed by the destructive distillation of wood and coal. It is a colorless gas and has a pleasant odor. It burns with a bright yellow flame and is one of the illuminating constituents of coal gas. 72. Sulphurated Hydrogen. Sulphureted hydrogen (H2S) is seldom found in mines in large quantities. It is a colorless gas and has an odor of rotten eggs. It is poisonous. When breathed in small quantities produces headache and larger quantities renders one unconscious. It explodes violently when mixed with air to about seven times its volume. The gas is soluble in water one volume of water dissolving about three volumes of the gas at ordi- nary temperature and pressure. A familiar example of the action of this gas is seen in its effect upon silver, which becomes covered by a bluish- black deposit after being exposed for a short time to air containing the gas. The gas occurs in the waters of sulphur springs; it is often found in the air in sewers and is produced by the decay of organic matter containing sulphur. GASES 79 TABLE I SAMPLES OF MINE AIR EXAMINED AND RESULTS OF THE EXAMINATION State. County. Kind of Coal. Volume of Air Current, Cu.ft. per Min. CO 2 in Sample, per Cent. CH 4 in Sample, perCent. Pennsylvania Luzerne Anthracite 18,100 0.07 1.34 1 < ' ' < i 18,100 .07 1.34 " ' 25,760 .07 1.16 " " " 178,560 .0-3 .76 < < " " 178,560 .09 .78 " 11 tl 90,000 .06 1.01 " " " 90,000 .05 1.04 " 11 " 140,344 .08 .35 " " Anthracite 140,344 .08 .37 " 11 " (a) .04 .02 " " 11 (a) .05 .04 " " < < (a) .05 2.34 " 11 " (a) .07 2.37 " H " 44,200 .04 1.57 " 11 " 44,200 .02 1.60 " Lackawanna " 28,764 .33 .52 < t 11 1 1 28,764 .30 .50 " " " 21,000 .28 .76 " " 11 21,000 .27 .75 " Luzerne " 17,136 .13 1.27 " " 11 17,136 .10 1.27 " " '< 60,060 .16 2.29 " 14 '* 23,760 .14 2.20 ' ' M ' ' 23,760 .13 2.19 lt 11 " 13,600 .17 3.06 ' ' < < < < 13,600 .16 3.05 1 ' Williamson Bituminous 36,190 .24 .14 1 ( H " 41,580 .35 .21 ' f ' ' ' ' 54,225 .05 .09 " " " 16,240 .37 .21 1 1 ' ' ' ' 13,650 .44 .19 t( " < < 23,870 .05 .00 t < Jackson < < 30,800 .05 .02 (a) Still air. 80 MINING AND MINE VENTILATION TABLE I Continued Volume of COa in CH 4 in State. County. Kind of Coal. Air Current, Cu.ft. per Min. Sample perCent. Sample, per Cent Pennsylvania Jackson Bituminous 55,300 .11 .04 < < < 20,844 .31 .00 t t Franklin 90,396 .05 .03 1 1 64,288 .10 .24 ' l 6,000 .10 .35 Colorado Fremont (a) .19 .34 1 1 (a) .29 .41 i ( 27,090 .05 .21 West Virginia 14,000 .05 .88 (a) Still air. QUESTIONS 1. What is calcium carbide? 2. How is it made and for what is it used? 3. How should it be stored? 4. What is acetylene gas? 5. Describe the acetylene flame. 6. What precautions should be observed in using acety- lene gas as an illuminant in the mine? 7. What is the formula for acetylene gas? 8. What is the specific gravity of acetylene gas? 9. Does the flame from an acetylene lamp give off smoke? 10. What are the dangers to be met with in the use of acetylene lamps in the mine? 11. What is meant by the term " occluded gases "? 12. What are the physical properties of air? 13. What is the chemical formula for air? 14. How is carbon monoxide produced? 15. How may carbon monoxide be changed to carbon dioxide? GASES 81 16. How is carbon dioxide produced and what is its density? 17. Where is carbon dioxide usually found? Why? 18. How is marsh gas produced? (a) What is its den- sity? (6) What is its chemical formula? 19. What is fire damp? 20. What per cent of gas is in the mine atmosphere when it is (a) at its lowest explosive limit? (6) Highest explosive limit? (c) When the mixture is such that the explosive force is greatest? 21. What effect has the relative humidity of the atmos- phere on the explosive limits of fire damp? 22. How is fire damp detected? 23. If a cap of one inch appeared on your safety lamp, what is the per cent of gas in the atmosphere? 24. What is the formula and density of ethane? How is it produced? 25. What is the density of sulphur eted hydrogen and how is it produced? 26. If CO gas is passing over a fire how may it be reduced to CO 2 ? 27. An ordinary wick-fed flame goes out when the proportion of oxygen in mine air is reduced to about 17 per cent. Will an acetylene flame burn in this percentage of oxygen? 28. What is changed when a gas is compressed? the size of the molecules or the distance between them? 29. How much must the volume of air in a pneumatic drilling hammer be compressed to drive it at a pressure of 45 Ibs. per square inch? 30. How deep must water be in a vessel so that its press- ure upon the bottom may be the same as that of the atmos- phere? CHAPTER X SPECIFIC HEAT 73. By specific heat is meant the quantity of heat necessary to raise the temperature of a substance one degree compared with the amount of heat necessary to raise the temperature of an equal weight of water one degree. Place 1 Ib. of shot in one test tube, and 1 Ib. of iron filings in another similar tube. Raise both to the same temperature by placing them in a vessel of hot water. Into each tube pour equal weights of water that has been cooled to 32 F. by means of ice. Take the temperature of the water in each and it will be found that the filings have given the water the greater amount of heat, as is shown by the higher temperature of the water in the tube con- taining the filings. This experiment shows that iron has a greater amount of heat than lead at the same temperature. 74. Heat Capacity. If equal weights of water, iron and mercury are so placed that each will receive as many heat units per minute as the other, at the end of a given time a thermometer will show that the mercury has been warmed through about 30 times and the iron about 9 times as many degrees as the water. This shows that a given weight of mercury requires about ^V and an equal weight of iron i as much heat to* warm it one degree as an equal weight of water requires; therefore all substances do not have the same heat capacity. 82 SPECIFIC HEAT 83 Water being the standard the heat capacity of all sub- stances are compared with its heat capacity, from which we get a set of ratios known as specific heats. The following table gives the average specific heat of the most common substances, water being the standard: TABLE OF SPECIFIC HEAT Air (at constant pressure) . 237 Water. 1.000 Alcohol 0.620 Copper 0.094 Iron 0.113 Lead 0.031 Mercury 0.033 Silver 0.056 Ice 0.502 Steam 0.480 Aluminum . 214 Tin 0.055 Zinc 0.094 Hydrogen (at constant pressure) ...... 3.406 75. Measurement of Specific Heat. A convenient method of measuring the specific heat of a body is the METHOD OF MIXTURE. W'hen two bodies that are at differ- ent temperatures are put together the temperature of one will fall and that of the other will rise until they reach the same temperature. It will also be noticed that the heat absorbed by the cool body in heating is exactly the amount given out by the hot body in cooling. This prin- ciple may be stated in its simplest form as follows: HEAT GAINED = HEAT LOST. The quantity of heat absorbed by the cool body in heating = mass X change in temperature X specific heat. 84 MINING AND MINE VENTILATION The quantity of heat given out by the hot body in cooling = mass X change in temperature X specific heat. Thus, t = temperature change; s = specific heat; EXAMPLE. Two pounds of fine shot at 90 were poured into 1 Ib. of water at 15, and the resulting temperature was 20. What is the specific heat of the shot? Since the specific heat of water = 1 = 2X70Xs; therefore, s = r-7 = .036 , specific heat of the shot. If the same quantity of heat is im parted to equal weights of water and fine shot the temperature of the shot will be about 28 times higher than that of the water: Specific heat of water _L__oc_ Specific heat of shot .036 With the exception of the gas hydrogen, water has the highest heat capacity that is, the largest specific heat of all substances. On this account water is well suited for conveying heat in the warming of buildings. For a similar reason the presence of a large quantity of water prevents a rapid change in the temperature of the air in contact with it, hence large bodies of water moderate the climate in their vicinity. SPECIFIC HEAT 85 QUESTIONS 1. When two liquids having different temperatures are mixed, what is the relation between the quantity of heat lost by the warmer and the quantity of heat gained by the cooler liquid? 2. What is the meaning of specific heat? 3. If 12 Ibs. of water at 16 F. and 72 Ibs. of metal at 100 F. when mixed give a final temperature of 30 F., find the specific heat of the metal. 4. A piece of nickel at 100 F. was dropped into an equal weight of water at 32 F. and the resulting temper- ature was 10. Find the specific heat of the nickel. 5. If an equal weight of water and iron at the same temperature be so placed that each receive the same amount of heat per minute, after five minutes which will be the higher in temperature? 6. What substance is used as the standard for comput- ing specific heat? 7. The specific heat of iron is higher than lead. How would you prove this statement? 8. If equal masses of water, iron and lead are so placed that each receive the same number of heat units per min- ute, (a) which will show the highest temperature? (6) the lowest? 9. Why is water well suited for conveying heat in the warming of buildings? 10. Would alcohol be a better heat conveyor than water? Why? 11. A piece of silver at 194 F. weighing 200 ozs. is put into a volume of water at a temperature of 50 F. If the resulting temperature is 64.76 F., what is the weight of the water? 86 MINING AND MINE VENTILATION Sp.ht. X mass X temperature change = sp.ht.X mass X temp, change Therefore, (.056X200) X (194-64.76) -f- (1 X64.76-50) = 98 ozs., wt. of water. CHAPTER XI MINE VENTILATION 76. Ventilation. The movement of air through a mine is caused by a difference in pressure between the intake and return airways. The velocity at which the air moves and the quantity of air passing through the airways of a mine will depend on this difference in pressure together with the resistance offered to the movement of the air by the rubbing surface of the airways. The most important point to be considered in con- nection with mine ventilation, after the proper quantity of air has been decided upon, is the determination of the mine resistance or the pressure that will be necessary to overcome the resistance offered by the mine. A mistake is frequently made by installing a fan designed to deliver a large quantity of air at a water gauge insuffi- cient to overcome the resistance of the mine. A fan having certain dimensions, producing a 2-in. water gauge, may at one mine cause to be circulated 100,000 cu.ft of air, while at another mine the same fan would produce only 50,000 cu.ft. of air with the same water gauge, namely, 2 ins. The general impression among mining students is that the water gauge generated at a fan is due to the mine resistance; this is not true; the fan produces the water gauge and the mine resistance consumes it or that part of it which is necessary to overcome the friction offered by the mine. 87 88 MINING AND MINE VENTILATION It does not matter whether a fan is ventilating a mine or running in the open atmosphere, the water gauge will be the same in both cases if the revolutions remain the same. A fan designed to produce 50,000 cu.ft. of air with a 1-inch water gauge might be placed at a mine, the airways in which may be of ample size and yet the quan- tity of air might be less than one-half the volume expected. This is due to the fact that the water gauge produced by the fan is not sufficient to overcome the mine resistance and produce sufficient velocity. Therefore it will readily be seen that in order to cause air to move through this mine at a greater velocity the water gauge or pressure must be increased; this can only be done by the fan or other means employed for the purpose of producing venti- lation. 77. Pressure Defined. Air in motion in a mine is under the influence of three distinct pressures, namely, the VELOCITY, STATIC and DYNAMIC or TOTAL PRESSURES. The VELOCITY PRESSURE is that pressure which is required to create the velocity of flow. The STATIC PRESSURE, sometimes termed the FRIC- TIONAL PRESSURE, is that pressure required to overcome the resistance offered to the flow. The TOTAL PRESSURE, also termed the DYNAMIC or IMPACT PRESSURE, is the sum of the static and velocity pressures. QUESTION. The water gauge produced by a fan is ^ in., the airway is 5 ft. by 5 ft. What must be the rub- bing surface in this mine to prevent the air from moving faster than 1 ft. per minute? Solution. -& or MINE VENTILATION . 89 NOTE. See Chapter XIII, for formulas and values for k, which is the coefficient of friction. It is plain that in order to increase the velocity in this mine it will be necessary to erect a fan capable of producing a greater water gauge, because in this case nearly the entire water gauge generated by the fan is consumed in overcoming the mine resistance and only a small part of it is left to produce a velocity. Hence the static pressure required to overcome the resistance of the mine in ques- tion is equal to about J-in. water gauge, and any addi- tional pressure that might be added will be wholly consumed in producing velocity and overcoming the additional friction caused by the increased velocity. The static water gauge or pressure can be found by the use of two water gauges, one of which should be piped into the airway with the end of the pipe opening at right angles to the direction of the air current, and the other gauge close to the same place with the end of the pipe opening pointing against the current so that the air can rush into the end of the pipe. While the gauges are in this position it will be noticed that one of the gauges will show a higher reading than the other. The difference in the readings is the velocity pressure, or the pressure producing the velocity. WATER GAUGE. The water gauge, Fig. 13, consists of a glass tube bent in the form of a U, both ends of which are open. When it is desired to measure the difference of pressure between two airways, one of the ends is inserted in a small hole bored in a door or brattice between the intake and return airways. When in this position the two ends of the gauge are subjected to two different press- ures, the atmospheric pressure on the intake side if the fan is exhausting, and a pressure less than the atmosphere on the return side. This difference in pressure causes the 90 MINING AND MINE VENTILATION water to drop in the tube on one side of the gauge and to rise a corresponding distance on the other side. The difference of the level of the water in the two tubes can be read by means of the scale attached, as shown in the figure. FIG. 14. FIG. 13. ANEMOMETER. The anemometer, Fig. 14, is an instru- ment used for measuring the velocity of air currents in mines and the ventilators of public buildings. The instru- ment consists of a delicately constructed fan wheel which revolves in a circular frame. Placed in an air passage the instrument registers automatically the rate at which MINE VENTILATION 91 the air is traveling through it. The revolutions of the wheel are recorded by means of several pointers or hands on the face of the instrument. The large hand makes one revolution for each hundred revolutions of the wheel. One revolution of the large hand per minute is equivalent to a velocity of 100 ft. per minute. ' Anemometers indicate satisfactorily velocities up to 10,000 ft. per minute, and each instrument is supplied with a chart of correction for different velocities. 78. Calculations. The relation between fan speed, pressure, volume of air delivered and power required have been fully verified by tests and will be found convenient for reference by those interested in mine ventilation. 1. The volume of air delivered by a fan varies directly as the number of revolutions, resistance remaining the same; that is, if a fan running 80 R.P.M. delivers 100,000 cu.ft., how much air will be delivered if the revolutions are increased to 160? Solution. 80 : 160:: 100,000 : X X = 200,000 cu.ft. 2. The water gauge or pressure produced by a fan varies directly as the square of the speed. If a fan run- ning at 80 R.P.M. produces 1 in. water gauge, what water gauge will be produced if the revolutions are increased to 160? Solution. 80 2 : 160 2 :: 1 in. : X. X = 4 in. w.g. 3. The water gauge or pressure required to force air through a mine varies directly as the square of the volumes. If with 1 in. water gauge 100,000 cu.ft of air are passing through a mine per minute, what water gauge will be required to pass 200,000 cu.ft. per minute? Solution. 100,000 2 : 200,000 2 :: 1 : X. X = 4 in. w.g. 92 MINING AND MINE VENTILATION 4. The power required to drive a fan varies directly as the cube of the revolutions. If it requires 25 H.P. to run a fan 80 R.P.M., what power will be required to run the fan 160 R.P.M.? Solution. 80 3 : 160 3 : : 25 : X. X = 200 H.P. 5. The power required to ventilate a mine varies as the cube of the volume of air passing. If it requires 25 H.P. to circulate 100,000 cu.ft. of air through a mine, what H.P. will be required to circulate 200,000 cu.ft.? Solution. 100,000 3 : 200,000 3 ::25 : X. Z = 200 H.P. 6. To find the size of motor or engine required to drive a fan under average mine conditions, multiply the number of cubic feet of air by the water gauge and divide this product by 4500. If a fan is delivering 100,000 cu.ft. of air at 2 in. water gauge, what size motor or engine will be required to drive it? , .. 100,000X2 . Solution. r =44.4 H.P. NOTE. The above formula bases the equipment at 71 per cent mechanical efficiency, or, which is the same thing : 100,000X2X5.2 33,000 X. 71 = 44.4 H.P. nearly. 7. The horse-power of an engine is found by means of the following formula: MINE VENTILATION 93 PXLXAXN 33,000 = H.P. P = mean effective pressure; L = length of stroke in feet; A = area of piston in square inches ; N = number of strokes per minute. EXAMPLE. If a 16 in. by 18 in. engine is running 150 R.P.M. and the mean effective pressure is 40 Ibs., what is the H.P.? NOTE. The number of revolutions at which an engine runs per minute multiplied by 2 will equal the number of strokes. 40X1^X201X300 irkft ~ TJ o Solution. 330QO -= 109.6 H.P. 8. The electric H.P. consumed by a direct-current motor is found by means of the following formula: VXA V = volts; A = amperes; 746 = number of watts in one H.P. EXAMPLE. If a direct-current motor is using 100 amperes, 250 volts, what is the H.P. input? Solution. 10 * 25 = 33.5 H.P. 74o 9. The mechanical efficiency of a ventilating equip- ment is the ratio of the actual H.P. consumed to the actual H.P. applied. 94 MINING AND MINE VENTILATION EXAMPLE. If the actual H.P. of an engine is 44.4 and the effective H.P. is 31.5, what is the mechanical efficiency? 31 5 Solution. ~rr~ = 71%. nearly. 10. The theoretical water gauge of a fan is computed by means of the following formula: 32.16X5.2' V = peripheral speed of fan in feet per second ; .078 = weight of a cubic foot of air; 32.16 = g. acceleration due to gravity; 5. 2 = pressure per square foot for 1 in. water gauge. EXAMPLE. If a fan is running 80J ft. peripheral speed per second, what is the theoretical water gauge? 80FX.078 Solution. 32 1 5 2 = 3+ms. water gauge. 11. The manometric efficiency of a fan is the ratio of the theoretical water gauge to, the actual water gauge developed by the fan. EXAMPLE. If a fan running at a peripheral speed of 80J ft. per second produces an actual water gauge of 2 ins. and the theoretical water gauge is 3 ins., as found in Example 10, what is the manometric efficiency of the fan? 2" Solution. 577 = 66f per cent manometric efficiency. o MINE VENTILATION 95 12. The volumetric capacity of a fan is the ratio of the actual volume produced to the cubical contents of the fan multiplied by the number of revolutions. EXAMPLE. The cubical contents of a fan 10 ft. in diameter and 5 ft. wide is 392.7 cu.ft and running at 100 R.P.M. = 39,270 cu.ft. If the actual volume delivered by the fan is 78,540, what is its volumetric capacity? Solution. * _ =200 per cent. >u ) I U EXAMPLE. The quantity of air delivered by a fan is 150,000 cubic feet per minute at a water gauge of 3i inches. If the efficiency of the plant is 65 per cent and the average steam pressure on the piston is 45 Ibs. per square inch, what size engine will be required to do the work ? Solution. The foot pounds of work done on the air per minute are 150,000X3.25X5.2 = 2,535,000. The efficiency of the engine being 65 per cent, the foot pounds developed by the engine must be 2,535,000 X 100 = uo The foot pounds developed by the engine are, piston speed in feet per minute X pressure per square inch on piston X area of piston. So that, taking the piston speed to average 400 feet per minute, the area of the piston is 150,000 X 3.25 X 5.2 X 100 _ 16 65 X 400 X 45 X. 7854 79. The ventilating pressure may be expressed in inches of water gauge or in pounds per square foot. For 96 MINING AND MINE VENTILATION instance, a water gauge of 2 ins. is equal to 2X5.2 or 10.4 Ibs. per square foot. Should it be necessary to express the pressure per square foot in inches of water gauge, simply divide the pressure per square foot by 5.2. The number 5.2 is found by dividing 62.5, the weight of a cubic foot of water, by 12. EXAMPLE. If the water gauge is 2J ins., (a) what is the pressure per square foot? (6) If the area of the airway is 30 sq.ft., what is the total pressure? Solution. (a) 2.5X5.2 = 13 Ibs. per sq.ft. (6) SOX 13 = 390 Ibs. 80. First Law of Friction. When the velocity remains constant the total pressure required to overcome friction varies directly as the extent of the rubbing surface. This law means that if the rubbing surface be doubled the pressure must also be doubled in order to pass the air at the same velocity. EXAMPLE. If an airway 10 ft. by 10 ft. and 1000 ft. long is increased in length to 2000 ft., how much addi- tional pressure must be added to pass the same quantity of air? Solution. As the rubbing surface is doubled the press- ure will therefore have to be doubled in order to pass the same quantity. EXAMPLE. Find the rubbing surface of an airway, the sides of which are 10 ft. by 6 ft. and 2000 ft. long. Solution. 10+10+6+6 = 32 ft. distance around the air- way. 32 X 2000 = 64,000 sq.ft. Ans. EXAMPLE. Suppose in the above example the sides of the airway were 15 ft. by 4 ft., the length being the same, what would be the rubbing surface? MINE VENTILATION 97 Solution. 15+15+4+4 = 38 ft. distance around the air- way. 38X2000 = 76,000 sq.ft. Ans. EXAMPLE. If 20,000 cu.ft. of air passes per minute through an airway 1000 ft. long, what must be the increase in pressure to pass the same quantity through the same airway if the length is increased to 1500 ft.? Solution. Since the rubbing surface is increased 1.5 times, it follows that, according to the first law of friction, the pressure must also be increased 1.5 times. The form of the airway in a mine has considerable effect on the amount of rubbing surface, as will be shown by the following example: EXAMPLE. Suppose there are three airways, the length of each 1000 ft.; one airway being 8 ft. by 8 ft., another 4 ft. by 16 ft., the third being circular, the diameter of which is 9.026 ft., what is the rubbing surface and area of each? Solution. 1. (8+8+8+8) X 1000 = 32,000 sq.ft. rubbing surface, Area = 64 sq.ft. 2. (4+4+16+16) X1000 = 40,000 sq.ft. rubbing surface, Area = 64 sq.ft. 3. 9.026X3.1416X1000 = 28,356 sq.ft. rubbing surface, Area = 64 sq.ft. 81. Second Law of Friction. When the velocity and rubbing surfaces remain the same, the pressure required to force air through the airways of a mine increase and decrease inversely as the sectional area of the airways increase or de- crease. 98 MINING AND MINE VENTILATION This law means that if the velocity and rubbing surface remain the same, the pressure per square foot that will be necessary to maintain this velocity will increase as the sectional area decreases, and as the sectional area increases the pressure will decrease. Hence if the sectional area be reduced to J, J, etc., of its original area, the pressure per square foot must be increased 2, 4, etc., times in order to maintain the same velocity, and if the sectional area be increased 2, 4, etc., times the pressure per square foot necessary to main- tain the same velocity will be reduced to J, J, etc., of the original pressure. The rubbing surface remaining the same. EXAMPLE. If it requires a pressure of 10.4 Ibs. to main- tain a velocity of 1000 ft. per minute in an airway 8 ft. by 8 ft., what pressure per square foot will be required to maintain the same velocity in an airway 4 ft. by 4 ft. rubbing surface remaining the same? Solution. 8' X 8' = 64 sq.ft. area. 4 X4 =16 sq.ft. area. 16 : 64:: 10.4 : X, or X = 41.6 Ibs. per sq.ft. EXAMPLE. If it requires a pressure of 5 Ibs. per square foot to pass air through an 8 ft. by 10 ft. airway with a certain velocity, what pressure per square foot will be required to pass air through a 6 ft. by 8 ft. airway with the same velocity? The rubbing surface remaining the same. EXAMPLE. If it requires a pressure of 2 Ibs. to force air through a 10 ft. by 10 ft. airway, what pressure per square foot will be required to pass air through an airway 5 ft. by 5 ft. at the same velocity? The rubbing surface remaining the same. MINE VENTILATION 99 82. Third Law of Friction. The pressure required to overcome friction varies as the square of the velocities or quan- tities when the rubbing surface and the area of the airway remain the same. This law means that if the sectional area and rubbing surface remain the same the pressure per square foot will vary as the square of the velocity or quantity. EXAMPLE. If it requires a pressure of 5 Ibs. to pro- duce a velocity of 400 ft. per minute in a certain airway, what pressure will be required to produce a velocity of 500 ft. in the same airway? Solution. 400 2 : 500 2 ::5 : X, or X = 7.S Ibs. EXAMPLE. If 10 Ibs. pressure produce a velocity 350 ft. per minute, what pressure will be required to produce a velocity of 700 ft. per minute in the same airway? 40 Ibs. Ans. TABLE J TABLE OF PRESSURE PER SQUARE FOOT DUE TO DIFFERENT VELOCITIES OF THE AIR Feet per Minute. Pressure in Lbs. per sq.ft. 100 006 150 014 200 025 300 057 400 102 500 159 600 230 700 312 800 408 900 517 1000 ; 638 1500 1.437 2000 2.555 2500.. 3.991 100 MINING AND MINE VENTILATION Feet per Minute. Pressure in Lbs. per sq.ft. 3000 5.750 3500 7.825 4000 10.220 4500 12.937 5000 15.970 5500 19.298 6000 23.000 6500... 26.976 7000 31.302 7500 35.937 8000 40.886 8500 46.155 9000 51.750 9500 57.744 10000 63.883 FORCE OF AIR. To ascertain the force in pounds per square foot of an air current, multiply the square of the velocity of the air in feet per second by .0023. QUESTIONS 1. What causes air to move through a mine? 2. A fan is running at 50 revolutions per minute, the water gauge being 1 in., and is producing 80,000 cu.ft. of air at a certain mine; a similar fan is in operation at another mine running at 50 revolutions and has a water gauge of 1 in. and is producing only 50,000 cu.ft. of air. What is the cause of the difference in quantity? 3. If the water gauge reading is 2J ins., what is the pressure per square foot? 4. The area of an airway is 60 sq.ft., the water gauge reading is 2 ins. What is the total pressure? 5. If the water gauge reading at a mine is 1J ins. and 5.2 Ibs. pressure per sq.ft. are consumed in overcoming the mine resistance, what is the velocity? 6. Define static and velocity pressure. 7. What is the first law of friction? MINE VEOTllftON 101 8. An airway is 7 ft. high, 12J ft. wide and 5400 ft. long. What is the rubbing surface? 9. Does the mine resistance or the fan produce the water gauge? 10. A fan running at 50 revolutions produces a water gauge of 1 in. while ventilating a large mine. If the mine is cut off and the fan allowed to run in the open atmos- phere at the same speed, what then will be the water gauge? 11. If a fan running 50 revolutions per minute pro- duces 50,000 cu.ft. of air per minute, what quantity will this fan produce when running 100 revolutions per minute? 12. If a fan running at 40 revolutions per minute pro- duces a 2-in. water gauge, what water gauge will be pro- duced when the fan is running 80 revolutions per minute? 13. If a 1-in. water gauge causes 50,000 cu.ft. of air to flow through a mine, what water gauge will be necessary to pass 100,000 cu.ft. of air through the same mine? 14. If it requires 20 H.P. to run a fan at 40 revolutions per minute, what horse power will be required to run the fan at 80 revolutions per minute? 15. If it requires 10 H.P. to produce 50,000 cu.ft. of air per minute, what H.P. will be required to produce 100,000 cu.ft. of air? 16. A fan is delivering 50,000 cu.ft. of air per minute, the water gauge is 1 in. What power motor or engine is required to do this work? 17. A direct-current motor is consuming 100 amperes at a voltage of 500. What is the H.P.? 18. If a 10-ft. fan is running at 120 revolutions per minute, what is the theoretical water gauge? 19. If the actual water gauge produced by the fan in Question 18 is 1.3 ins., what is the manometric efficiency? 20. If a fan 12 ft. in diameter and 5 ft. wide, running at 100 revolutions per minute, is delivering 90,000 cu.ft. 102 MINING AND MINE VENTILATION of air per minute, what is the volumetric capacity of the fan? 21. If you were about to order a fan to ventilate a mine, what points should be considered? 22. If a fan while ventilating a mine produces a 2-in. water gauge, what will be the water gauge if the mine is cut off and the fan run at the same speed in the open atmosphere? 23. How is the static pressure of a mine found? 24. If (in Question 22) the mine is cut off by means of a stopping so arranged and constructed that the fan can get no air, if the revolutions remain the same, what will be the water gauge? 25. If it requires a pressure of 10 Ibs. to produce a velocity of 500 ft. per minute in a certain mine, what pressure will be required to produce a velocity of 800 ft. per minute? 26. If a 2-in. water gauge produces a velocity of 300 ft. per minute, what velocity will a 4-in. water gauge pro- duce? 27. If it requires 40 H.P. to run a fan 80 revolutions per minute, how fast will the fan run if 50 H.P. is applied? 28. If a fan running at 80 revolutions per minute delivers 100,000 cu.ft. of air per minute, at how many revolutions per minute will it have to run to produce 200,000 cu.ft of air per minute? 29. If a 14 in. by 16 in. engine is running 100 R.P.M. and the mean effective pressure is 40 Ibs., what is the H.P.? 30. If the effective horse-power of a ventilating equip- ment is 20.5 and the actual horse-power is 32, what is the mechanical efficiency? 31. If the effective horse-power of a ventilating equip- ment is 40 and the pressure producing ventilation is 10 Ib. per square foot, what is the quantity? MINE VENTILATION 103 32. The quantity of air delivered by a fan is 150,000 cu.ft. per minute and the water gauge is 2 ins. What is the effective horse-power? 33. If it requires 27 H.P. to produce 50,000 cu.ft. of air per minute, what quantity will 64 H.P. produce? 34. If 20 H.P. will produce 40,000 cu.ft. of air per minute, what horse-power will be required to produce 80,000 cu.ft. per minute? 35. A fan is 12 ft. in diameter and 5 ft. wide; it is run- ning 100 revolutions per minute; the actual volume of air delivered by this fan is 80,000 cu.ft. per minute; what is its volumetric capacity? 36. A fan 10 ft. in diameter is running 100 revolutions per minute. What is the theoretical water gauge? 37. If the theoretical water- gauge of "a fan is 3 ins. and the actual water gauge is 2| ins., what is the mano- metric efficiency of the fan? 38. There are two airways, one of which is 6 ft. by 6 ft. and the other 4 ft. by 9 ft., each being the same length, namely, 2500 ft. Through which airway will the larger quantity of air pass under the same pressure? 39. The water-gauge reading at a fan is 2 ins. If the static pressure at this mine is 8 Ibs. per square foot, what is the velocity pressure? 40. The pressure produced by a fan is .4 of an inch water gauge, the mine airway is 4 ft. by 4 ft. What should be the length of this airway in order to prevent the air moving faster than 1 ft. per minute? CHAPTER XII MINE VENTILATION 83. Mine Ventilation. Every precaution should ba taken to keep large airways, and a mine should not be permitted to get into a condition requiring more than a 3-in. water gauge pressure to ventilate it. However, a great number of old mines in operation to-day require a much larger gauge, the cause being due to long airways, small sectional areas and unequal splitting of the air cur- rent. The fans "erected at. those mines when the airways were short and little resistance offered to the movement of the air are now unfit for the work they are expected to perform. It is the custom with some mine operators when order- ing a fan to designate a certain size fan, even making out detailed specifications for it and state that the fan must perform a certain work in the number of cubic feet of air per minute and the water gauge against which the fan must work, when as a matter of fact the mine conditions require an equipment entirely different. The want of knowledge of such matters has led many mine operators into getting a fan that does not fit the mine conditions. The matter of circulating and conducting air through the workings of a mine is a very easy matter if in the first place a fan be erected that will work economically against the mine conditions. VENTILATING CURRENTS, How PRODUCED. Ventilating currents are produced by natural heat, by water falling, or a water-jet, by a steam jet, by a furnace and by a fan. 104 MINE VENTILATION 105 84. Natural Ventilation. Natural ventilation is pro- duced in a mine when there is a difference in elevation between the intake and outlet airways and a difference in temperature between the two columns: For the purpose of illustration, let Fig. 15 represent two shafts the tops of which are at different elevations. During cold weather the shaft AB will be the downcast because the imaginary column of outside air from A to WINTER FIG. 15. E is heavier than the air column from C t to level of shaft AB, the difference in weight being due to the difference in temperature. In the summer time the outside air being warmer than that of the mine the shaft CD will be the down- cast, and the shaft AB the upcast. This system of ventilation works fairly well during the seasons of extreme heat and cold, but during the spring and fall when the temperatures inside and out are about equal, natural ventilation is inef- fective. 106 MINING AND MINE VENTILATION 85. Water and Steam Jet System of Ventilation. This system of ventilation, while not very successful, is sometimes applied in cases of emergency. The jet is so arranged that water is sprayed and allowed to fall down the intake shaft. In connection with this system a steam jet is sometimes used. The steam jet is arranged so as to blow in the upcast. During the year 1852 a committee of the House of Commons at England reported: " That any system of ventilation depending on complicated machinery is unde- sirable, since under any disarrangement or fracture of its parts the ventilation is stopped or becomes inefficient. " That the two systems which alone can be considered as rival powers are the furnace and the steam jet. " Your committee is unanimously of opinion that the steam jet is the most powerful and at the same time least expensive method for the ventilation of mines.' 5 86. Furnace Ventilation. Furnaces are placed at the bottom of the upcast and are usually constructed of brick with air chambers on either side to prevent heating the surrounding strata. The heated air passing . over the furnace and entering the upcast is, by reason of its rarefied state, lighter than the cool air in the downcast shaft and is consequently forced upward. The quantity of air produced by a furnace depends principally on the amount of heat generated together with the depth of the furnace shaft. 87. Ventilation by Means of Fan. If a fan while working on a mine is exhausting air therefrom, the fan is then, due to centrifugal force, creating a partial vacuum at its center or axis; the extent of this vacuum depends on the peripheral or rim speed of the fan. The peripheral speed at which a fan should run depends altogether on its construction. While some fans may stand a rim speed MINE VENTILATION 107 of 16,000 ft. per minute, others will not stand more than 5000 ft. per minute. When the inlet of the fan is connected to the mine the only air that can get to the fan must pass through the mine, and hence the ventilating current is maintained as long as the fan runs. When the fan is running the pressure A Robinson Reversible Fan. of the air is always less at the inlet of the fan than outside, and the difference between this pressure and the pressure of the atmosphere is the pressure producing ventilation, or the extent to which a vacuum is approached by the fan. Many differently constructed fans are being used for the purpose of ventilating mines, the most prominent of which are those manufactured by the Robinson Ventilat- 108 MINING AND MINE VENTILATION ing Company, American Blower Company (Sirocco), Jeffrey Manufacturing Company and others possessing similar features. THE ROBINSON FAN. The Robinson fan, one of the late developments in fans for mine ventilation, Fig. 16, shows the runner of this fan. The blades are curved to pass the air through the fan with the least friction or loss in power. By reason of the blade arrangement the air FIG. 16. is readily changed from its horizontal direction as it enter? the wheel; hence, it is claimed, great economy is obtained by this fan. It is strongly constructed and is capable of standing any desired speed. The following table, prepared by the Robinson Ven- tilating Company, shows the approximate quantity of air delivered by their fan when running at different speeds and having different dimensions: MINE VENTILATION 109 TABLE K Diameter, Ins. Width, Ins. R.P.M. Volume. 18 17 2500 21,000 24 20 1900 35,000 30 23 1500 45,000 36 26 1200 60,000 42 29 1000 75,000 48 32 800 90,000 54 35 700 100,000 60 40 650 120,000 66 45 575 140,000 72 50 500 170,000 84 55 420 210,000 96 60 350 260,000 120 65 275 350,000 144 70 225 420,000 168 75 200 450,000 192 80 180 500,000 216 85 170 575,000 240 90 160 650,000 264 92 150 700,000 288 95 140 800,000 300 100 130 900,000 312 105 120 1,200,000 The following illustration, Fig. 17, shows a Robinson disc fan operated by electric motor and chain drive. It is used where the development does not justify the installa- tion of a centrifugal fan and is highly efficient when placed in an airway to boost along feeble currents. It is easily installed and can be moved from place to place as the condition of the mine may require. In order to reverse the air current it is only necessary to change the direction of rotation. THE SIROCCO FAN (Fig. 18) The special advantages presented by this fan are: (1) large inlet area; (2) uniform action over the whole periphery, due to the large number 110 MINING AND MINE VENTILATION FIG. 17. FIG. 18. MINE VENTILATION 111 of blades; (3) absence of whirlpool motion of the entering air before reaching the fan blades, thereby avoiding the expenditure of power on unnecessary work; (4) the blades are so constructed and arranged that the power consuming eddies are minimized. Instead of the work being done by 12 or 16 blades, as in the majority of old fans, the Sirocco has 128 blades in the double-inlet type of fan, thereby securing uniformity of action around the entire circumference. Fig. 19 shows a single-inlet reversible fan and fan drift installed at a drift mouth, and Fig. 20 shows a double- inlet reversible fan on a shaft mine. It frequently happens that a fan installed at a mine cannot create sufficient pressure to cause the proper volume of air to circulate through the remote parts of the mine. To remedy this difficulty a booster fan (Fig. 21) is some- times installed at a convenient point in the airway. In all such installations, however, the pressure produced by the booster must be above the pressure of the air current at the point of installation. If the booster is unable to produce a greater pressure than is already in the air cur- rent, then it will be unable to increase the volume. Below is given a table of approximate capacities of various sizes of Sirocco fans against varying resistances. As stated, no definite rule can be given by which the quan- tity of air a fan will cause to flow through a mine can be calculated, unless the exact mine conditions are known. A fan listed in the tables, given herewith, as being capable of producing 150,000 cu.ft. of air per minute at a 2-in. water gauge might be placed at a mine in which the air- ways are such that the pressure is only sufficient to over- come friction and cause little or no velocity. If economy and efficient ventilation are desired it is absolutely necessary that the manufacturer know the 112 MINING AND MINE VENTILATION MINE VENTILATION 113 114 MINING AND MINE VENTILATION MINE VENTILATION 115 116 MINING AND MINE VENTILATION TABLE L Volume Cu.ft. Per Min. Static Pressure in w.g. Approx. H.P. Re- quired. Size and Speed of Fan Wheel. Single Inlet. Double Inlet. Diam. Width. R.P.M. Diam. Width. R.P.M. ins. ft. in. ft. in. ft. in. ft. in. 40,000 1 8 7 0X3 6 124 5 0X5 172 1 11 6 6X3 3 154 4 6X4 6 224 U 16 6 0X3 205 4 0X4 310 2 21 6 0X2 6 235 4 0X3 6 358 60,000 1 16 8 0X4 124 5 6X5 6 185 if 24 8 0X3 156 5 6X4 6 224 2 32 7 6X2 10 190 5 0X4 6 282 3 48 7 0X2 6 250 5 0X3 8 346 80,000 1 21 9 0X4 6 102 6 6X6 6 154 2 42 8 6X3 2 170 6 0X5 235 3 63 8 0X2 10 220 6 0X4 288 4 84 8 0X2 8 250 5 6X3 8 366 100,000 2 53 10 0X3 6 143 7 0X5 204 3 80 9 6X3 2 182 6 6X4 4 270 4 107 9 0X2 10 222 6 0X4 4 333 5 133 9 0X2 8 248 6 0X3 8 376 125,000 2 66 11 0X4 130 8 0X5 8 176 3 100 10 6X3 6 166 7 6X5 232 4 133 10 0X3 2 200 7 0X4 8 285 5 167 10 0X3 222 7 0X4 322 150,000 2 80 12 0X4 4 120 8 6X6 4 167 3 120 11 6X3 10 151 8 0X5 4 220 4 160 11 0X3 6 182 7 6X5 268 5 200 11 0X3 4 202 7 6X4 8 298 175,000 2 93 13 0X4 8 110 9 6X6 8 149 3 140 12 6X4 2 138 9 0X5 8 193 4 187 12 0X3 8 168 8 6X5 4 236 5 234 11 6X3 6 195 8 6X5 263 MINE VENTILATION TABLE L Continued 117 Size and Speed of Fan Wheel. Volume Cu.ft. Per Min. Static Pressure in w.g. Approx. H.P Re- quired. Single Inlet. Double Inlet. Diam. Width. R.P.M. Diam. Width. R.P.M. ins. ft. in. ft. in. ft. in. ft. in 200,000 3 160 13 0X4 4 136 9 6X6 4 182 4 214 12 6X4 2 160 9 0X5 8 222 5 266 12 0X3 10 186 9 0X5 4 248 6 320 11 6X3 8 214 8 6X5 288 250,000 3 200 14 6X5 120 10 6X7 166 4 266 14 0X4 6 144 10 0X6 4 200 5 334 13 6X4 2 166 10 0X6 222 6 400 13 0X4 188 10 0X5 6 247 300,000 3 240 16 0X5 6 108 11 6X7 8 151 4 320 15 6X5 129 11 0X7 182 5 400 15 0X4 8 148 11 0X6 8 202 6 480 14 6X4 6 168 10 6X6 4 232 350,000 4 374 16 6X5 4 122 12 0X7 4 168 5 468 16 0X5 140 11 6X7 195 6 560 15 6X4 10 157 11 0X6 8 224 7 655 15 0X4 8 175 10 6X6 4 254 400,000 5 532 17 0X5 4 132 12 0X7 8 186 6 640 16 6X5 2 148 11 6X7 4 214 7 746 16 0X5 165 11 6X7 230 8 854 15 6X4 10 182 11 6X7 244 mine conditions or the pressure that will be consumed in circulating the quantity of air desired. Fig. 22 shows a Sirocco Ventura Disc Mine-fan. The Ventura Disc-Mine Fan is the latest achievement in the development of this class of apparatus. It is especially adapted to the ventilation of drift mines and for develop- 118 MINING AND MINE VENTILATION FIG. 22. MINE VENTILATION 119 merit work on new operations. It should be carefully noted, however, that speed considerations limit the application of this type of fan to mines where a high-water gauge is not required. JEFFREY FAN. Fig. 23 shows the extreme sizes and construction of the Jeffrey fan wheel. The high effici- ency developed by this fan is primarily due to the relative FIG. 24. position and curvature of the blades, which are so arranged that the air is discharged in a forward direction, and each blade is backed up by an auxiliary blade which prevents eddy currents and the slippage of air. The conical scoops, by their special form and position prevent the gushing of air from the inlet when working against a high water gauge. Fig. 24 shows plan and side view of a double inlet exhaust reversible fan with explosion cover, on a shaft mine. 120 MINING AND MINE VENTILATION TABLE M PERIPHERAL SPEEDS OF FAN REQUIRED FOR THEORET- ICAL WATER GAUGE Ins. Water Gauge. Velocity, Ft. per Min. Ins. Water Gauge. Velocity, Ft. per Min. i 4 1390 5 6216 1 1966 5J 6369 1 2407 &i 6520 1 2780 51 6666 If 3108 6 6809 if 3405 61 7087 If 3677 7 7355 2 3931 71 7613 2* 4170 8 7863 2* 4395 8| 8105 21 4610 9 8340 3 4815 9 8568 3i 5012 10 8791 3^ 5201 10i 9008 3! 5383 11 9220 4 5560 111 9427 4i 5731 12 9630 4 5897 12* 9828 4| 6059 13 10023 The following table gives a comprehensive idea of the results obtained from various sizes of Jeffrey mine fans. It is understood that the proportions of the fans may be changed to meet other conditions, that is, a fan may be built wider to handle economically a larger volume at the same gauge, or on the other hand the fan may be built narrower to handle a smaller volume at the same gauge while the speed of the fan remains constant. MINE VENTILATION 121 TABLE N DOUBLE INLET Water Gauge, Ins. Volume. R.P.M. H.P. 2-ft. Fan. 1 A 6,000 260 | i 8,000 367 1 f 10,000 448 1.8 i 12,000 520 3 11 15,000 633 5.5 2 17,000 734 8 4-ft. Fan. i 25,000 183 3 I 31,000 224 6 i 36,000 259 9 H 44,000 317 17 2 50,000 366 25 3 62,000 450 46 6-ft. Fan. i 40,000 122 5 1 56,000 172 14 H 70,000 211 24 2 80,000 244 36 3 100,000 294 67 8-ft. Fan. 1 75,000 124 17 U 91,000 152 30 2 106,000 176 47 3 129,000 215 86 4 150,000 248 133 10-ft. Fan. 1 100,000 100 22 1| 123,000 123 42 2 141,000 141 62 3 173,000 173 115 4 200,000 200 178 12-ft. Fan. 1 125,000 84 28 2 177,000 119 79 3 218,000 146 145 4 250,000 168 222 5 280,000 188 311 122 MINING AND MINE VENTILATION TABLE N Continued DOUBLE INLET Water Gauge, Ins. Volume. R.P.M. H.P. 14-ft. Fan. 1 150,000 70 33 2 214,000 100 95 3 261,000 122 174 4 300,000 140 267 5 338,000 158 376 16-ft. Fan. 1 175,000 63 39 2 245,000 88 109 3 200,000 108 200 4 350,000 126 311 5 386,000 139 430 18-ft. Fan. 1 200,000 56 45 2 283,000 79 125 3 344,000 96 230 4 400,000 112 355 5 443,000 124 492 6 485,000 136 647 20-ft. Fan. 1 225,000 50 50 2 315,000 70 140 3 380,000 85 250 4 450,000 100 400 5 495,000 110 550 6 540,000 120 720 The most important factors to be considered in the construction of mine airways are the main intake and return shafts. All the air entering the mine must pass through those openings, and if their sectional areas are small, the velocity will necessarily be high and a large part of the total pressure generated by the ventilating equipment will be consumed in the shafts. In the mine MINE VENTILATION 123 workings it is different. In case the consumption of pres- sure is high, the mine can be divided into districts and ventilated by separate and independent air currents; by this means the velocity is reduced in that part of the mine in which the air current is divided. By reason of this reduction in velocity the pressure consumed is also reduced. Part of the pressure thus saved is converted into velocity A Jeffrey Double Inlet Exhaust Reversible Fan. pressure and the remainder is used up in overcoming the friction, caused by reason of the increased volume obtained by splitting. In the case of shafts the pressure consumed remains constant while the velocity remains constant, and nothing can be done with the ventilating arrangement in the mines that will reduce the pressure consumed in the shafts, and at the same time maintain the same quantity of air. It is therefore recommended that the main airways be of sufficient area to allow the passage of the desired volume of air at a velocity not exceeding 1000 ft. per minute. 124 MINING AND MINE VENTILATION Any reduction in velocity brought about by an increase in the area of the airways will reduce the pressure and horse-power necessary to maintain the same quantity. The following examples will better illustrate the extrav- agant use of power by reason of high velocities: EXAMPLE 1. The return air shaft of a mine is 10 ft. by 10 ft. in section and 1000 ft. deep; the quantity of air desired is 400,000 cu.ft. per minute. What pressure and horse-power will be required to do the work? ksv 2 .00000001 X 40,000 X4000 2 P = ~^~'' 100 - = 64 Ibs. per sq.ft. 400,000X64 33,000 - =776 H ' R In the following example we will change the sectional area of the shaft, using the same quantity and depth: EXAMPLE 2. The return air shaft of a mine is 12 ft. by 15 ft. in section, and 1000 ft. deep; the quantity of air desired is 400,000 cu.ft. per minute. What pressure and horse-power will be required to do the work? ksv 2 .00000001 X 54,000 X 2222 2 ., P = = ~ ~~ - = 14.8 Ibs. per sq.ft. 400,000X14.8 33,000 H ' r '' , or a saving under the second condition, of 776 179.3 = 596.7 H.P. At a cost of $62.08 per horse-power per year a saving of 596.7 H.P. will amount to $37,043. The consumption of power in the ventilation of mines is an important item in the burden account, full of possi- bilities for saving. At a certain mine in *d . .3 -i MINE VENTILATION 125 J9d jndui -tT] - CO t>- I s * CO TI !> lO O^ *-O CO C^ ^^ |> CO i l i I CO d rH IO IO s 00 00 t^ CO T-H 1-1 T-H CO Tj< ui no ui jiy jo A O^ CO^ -^ 00^ O; cxT TjT co" cT rj^ T-H rtl CO Tj< rH CO -^ T-H 10 i> co oo co o 10 o o 10 10 r> TT ^> = <1 per cent mechanical efficiency. 1UU ri.r. To find the manometric efficiency of the fan: _, 3.75 w.g. = 80 per cent manometric efficiency. 130 MINING AND MINE VENTILATION SUMMARY Diameter of fan 10 feet Width of fan 6 feet Revolutions of fan per minute 171 Actual water gauge 3 inches Theoretical water gauge 3.75 ins. Quantity of air delivered per minute . . 150,000 cu.ft. Volumetric capacity 200 per cent Horse-power output 71 Horse-power input 100 Mechanical efficiency 71 per cent Manometric efficiency 80 per cent If it now be required to proportion a fan for a 5-in. actual water gauge instead of a 3-in., and 150,000 cu.ft. of air, using the same diameter of fan, namely, 10 ft., it is evident that the fan will have to run faster to generate the larger gauge; then if we figure on the same volume ratio it will be necessary to build the fan narrower. 89. Motive Column. The motive column is a column of air in the downcast shaft the weight of which is the difference between the weight of the downcast and upcast columns; therefore if the length of the motive column be subtracted from the downcast column, the remaining portion of the downcast column will be equal in weight to the upcast column. EXAMPLE. At a certain mine there are two shafts 500 ft. deep. The temperature in the downcast shaft is 50 F., and the temperature of the upcast air is 150 F.; what is the motive column? Solution. M = MINE VENTILATION 131 t = temperature of air in the upcast; t' = temperature of air in the downcast; D = depth of shaft in feet; M = motive column. (150-50) . x 500 = 82.1 feet motive column. (459 + 150) If in the above example the barometer pressure is equal to 30 ins. and it is desired to find the pressure per square foot, the weight of a cubic foot of air in the down- cast shaft must first be found. Thus, 1.3253 XB ~~ (459+0 ' in which B = the barometric pressure in inches; t = temperature of the air in the shaft; W = weight per cubic foot. Applying formula: w 1.3253X30 n7ftl W = (459+50) = ' 07811b - Now if the height of the motive column is 82.1 feet and the weight of a cubic foot of air in the downcast shaft is .0781 lb., the pressure per square foot is: 82. 1 X .0781 = 6.41 Ibs. Ans. EXAMPLE. The temperature of the air in a downcast shaft is 60 F. and in the upcast shaft 170 F.; the shafts are 900 ft. deep. If the barometer reading is 30 ins., what is the pressure per square foot? EXAMPLE. There are two shafts 600 ft. deep. The 132 MINING AND MINE VENTILATION temperature in the upcast shaft is 180 F., and the tem- perature of the downcast air is 40 F.; what is the motive column? EXAMPLE. What is the weight of a cubic foot of air at a temperature of 60 F. when the barometer reading is 30 ins.? 90. Splitting the Air Current. By splitting air means the dividing of the main intake current into two or more separate currents, the purpose of which is to ventilate the different independent districts of a mine with air that is not vitiated by the smoke or gases from another dis- trict. The advantage derived from splitting the air is as follows : (1) A larger volume of air with the same power. The extent of the increase in volume will depend on how nearly equal the splits are. (2) Purer air circulated through the working faces. (3) An explosion or fire in one district is not likely to affect the other districts. (4) A fall of roof affects only the section in which it occurs. (5) The velocity of the air is kept within a reasonable limit in a greater portion of the mine. Fig. 25 shows an air bridge or overcast. Such struc- tures are necessary when it is desired to pass one current of air over or under another current. The sides and floors of air bridges are usually constructed of concrete. By the use of overcasts the main air current can be divided and conducted across one another for the purpose of ven- tilating the different districts of a mine. Sometimes an undercast bridge is employed for the same purpose as an overcast, but there is the liability of water flooding them and blocking the air. MINE VENTILATION 133 The following examples will show the effect of splitting a continuous air current into several splits: EXAMPLE. An airway is 6 ft. by 10 ft. and 16,000 ft. long. What power will be required to circulate 24,000 cu.ft. of air through this airway? Solution. _ ks OOQ . av rroo f\f\f\ (30) p (32) p=. (33) p = . (34) FORMULAS FOR FINDING THE CUBIC FEET OF AIR PER MINUTE (35) q = av. (36) ^ (37) #33,000 w P P (39) 3 = FORMULAS FOR FINDING THE RUBBING SURFACE IN SQUARE FEET (40) s = ol. (41) s = . () -. ,.., H33,000 u (44) S = -' (45) S = ' FORMULAS 145 FORMULAS FOR FINDING THE UNITS OF POWER PER MINUTE (46) ^ = #33,000. (47) u = pav. (48) u = Pv. (49) u = pq. (50) u = ksv*. FORMULAS FOR FINDING THE VELOCITY IN FEET PER MINUTE (5D = *. (52) ,-Jj /ro\ I * /r A\ /lOO, 000 ( 53 ) v = \T n - (54) v = pa , KK . #33,000 /K _ N (55) w = ^ -. (56) w (57) . = |. (58) t; = W w 03'^ QQ > K 00 ^03-^ m > d 8 Ml bo gl in ^i3o 11 g| |ll il 3 a a s -S |S(S .S D.C or^ 3s 5 ^ < 5J 5 * Jj 21 8f 11.95 114 22.80 16 228 45.6 10 2 71 9.85 95 18.90 15 190 37.9 9^ 2i 71 8.00 78 15.60 13 156 31.2 8| 2 61 6.30 62 12.40 12 124 24.8 8 H 4.85 48 9.60 10 96 19.2 n If 5 4.15 42 8.40 8 84 16.8 61 li 41 3.55 36 7.20 71 72 14.4 5! If 41 3.00 31 6.20 7 62 12.4 51 11 4 2.45 25 5.00 6 50 10.0 5 H 3* 2.00 21 4.20 6 42 8.4 4^ l 3 1.58 17 3.40 51 34 6.8 4 1 21 1.20 13 2.60 4| 26 5.2 3^ t 21 .89 9.7 1.94 4 19.4 3.88 3 I 2 .62 6.8 1.36 3| 13.6 2.72 21 ft If .50 5.5 1.10 2f 11.0 2.20 U ^ H .39 4.4 .88 21 8.8 1.76 ii 11 .30 3.4 .68 2 6.8 1.36 H 1 H .22 2.5 .50 1-2 5.0 1.00 1 * i .15 1.7 .34 1 3.4 .68 1 4 3 4 .10 1.2 .24 * 2.4 .48 ^ MINE FIRES 163 TABLE P TABLE OF MANILLA ROPE Diam., Inches. Circ., Inches. Weight per Ft., Pounds. Breaking Load, Pounds. Diam., Inches. Circ., Inches. Weight per Ft., Pounds. Breaking Load, Pounds. .239 1 .019 560 1.91 6 1.19 25,536 .318 1 .033 784 2.07 61 1.39 29,120 .477 11 .074 1,568 2.23 7 1.62 32,704 .636 2 .132 2,733 2.39 u 1.86 36,288 .795 2* .206 4,278 2.55 8 2.11 39,872 .955 3 .297 6,115 2.86 9 2.67 47,040 1.11 3| .404 8,534 3.18 10 3.30 54,208 1.27 4 .528 11,558 3.50 11 3.99 61,376 1.43 4i .668 14,784 3.82 12 4.75 68,544 1.59 5 .825 18,368 4.14 13 5.58 75,712 1.75 51 .998 21,952 4.45 14 6.47 82,880 NOTE. The strength of manilla rope is very variable. The strength of pieces from same coil may vary 25 per cent. A few months of exposed work weakens ropes 20 to 50 per cent. QUESTIONS 1. What is the rubbing surface of an airway 8 ft. 6 ins. by 6 ft. 9 ins. and 3000 ft. long? Ans. 91,500 sq.ft. 2. A circular airway is 15 ft. in diameter and 1200 ft. long; this airway is equally divided into two compart- ments by a partition (the thickness of partition may be neglected). What is the rubbing surface in this airway? Ans. 92,548.8 sq.ft. 3. If 50,000 cu.ft. of air is passing per minute in an airway at a velocity of 8 ft. per second, assuming the air- way to be square, find its area and perimeter. Ans. (A) 104.16 sq.ft. (0) 40.8 ft. 164 MINING AND MINE VENTILATION 4. An airway is 10 ft. wide and 6 ft. high and 5000 ft. long. What pressure will be required to pass 60,000 cu.ft. of air per minute through this airway? Ans. 53J Ibs. per sq.ft. 5. Explain the constant 5.2 used in connection with water gauge and pressure calculations. 6. An airway is 8 ft. by 14 ft. and 5790 ft. long, the velocity is 1 ft. per minute. What amount of pressure in inches of water gauge will be necessary to overcome the friction in this airway? Ans. .0000087 in. 7. A mine airway is 7 ft. by 10 ft. and 6720 ft. long; if the water gauge is 2 ins., what is the velocity? Ans. 399 -f- ft. per min. 8. An airway is 8 ft. wide at the top and 10 ft. wide at the bottom, and 7 ft. high. How much air will pass through this airway if the anemometer makes 165 revo- lutions per minute? Ans. 10,395 cu.ft. per min. 9. An airway is of triangular shape, the sides are 6, 8 and 10 ft. If the velocity of the current is 425 ft. per minute, what is the quantity? RULE. From one-half the perimeter of the airway subtract each side separately; multiply together the three remainders thus found and half of the perimeter and extract the square root of the product. The result is the area required. A = area, a, 6, and c the three sides and o the sum of the three sides. or ^8) (12^10) =24 sq.ft. 24X425 = 10,200 cu.ft. per min. MINE FIRES 165 10. Explain the term coefficient of friction as used in mine ventilation. 11. The quantity of air passing through a mine is 60,000 cu.ft. per minute and the water gauge is 2 ins. What is the horse-power producing ventilation? Ans. 18.9 H.P. 12. An airway is 12 ft. by 8 ft. and 4000 ft. long. If 28,800 cu.ft. of air passes through this airway per minute, what is the pressure, rubbing surface and horse-power? Ans. 3 Ibs. per sq.ft. 160,000 sq.ft. 2.61 H.P. 13. There are two splits in a mine as follows: Split A is 4 ft. by 12 ft. and 6000 ft. long; Split B is 6 ft. by 8 ft. and 10,000 ft. long. If the quantity of air entering the mine is 10,000 cu.ft. per minute, what amount will pass through each split? Ans. (A) 5470 cu.ft. (B) 4530 cu.ft. 14. The velocity of the air passing through a mine is 200 ft. per minute when the water gauge is .25 in. What is the water gauge when the velocity is 400 ft. per minute? Ans. 1 in. 15. How much must the pressure be increased in order to double the quantity of air? Ans. 4 times. 16. If it requires 3 H.P. to produce 20,000 cu.ft. of air per minute, what horse-power will be required to pro- duce 40,000 cu.ft. per minute? Ans. 24 H.P. 17. In order to double the quantity of air passing through a mine, how much will the horse-power have to be increased? Ans. 8 times. 18. An airway is 6 ft. by 7 ft. through which 28,000 cu.ft. of air is passing per minute. With the same power and length how much air will pass through an airway 5 ft. by 5 ft.? Ans. 18,190 cu.ft. per min. 166 MINING AND MINE VENTILATION 19. 100,000 cu.ft. of air passes through an airway 6 ft. by 5 ft. and 10,000 ft. long. Three splits were then made as follows: Split A, 6 ft. by 6 ft., 2000 ft. long; Split , 6 ft. by 5 ft., 4000 ft. long, and Split C, 6 ft. by 4 ft., 6000 ft. long. What quantity of air will pass in each split while the pressure remains the same? Ans. (A) 52,488 cu.ft. (B) 29,447 cu.ft. (CO 18,065 cu.ft. 20. If an anemometer registers 30,000 ft. velocity per hour in an airway 8 ft. by 5 ft., what quantity of air is passing per minute? Ans. 20,000 cu.ft. per min. 21. The weight of a cubic foot of air is .0766 Ib. and the water gauge is 1.5 ins. What is the height of the motive column? Ans. 101.8+ ft. 22. If 155,650 cu.ft. of air enters a mine at a temper- ature of 32 F., what is the volume leaving the mine, the temperature being 65 F.? Ans. 166,090 cu.ft. 23. If 32,000 cu.ft. of air is entering a mine at the inlet, and 34,000 cu.ft. leaving it at the outlet, what is the cause of the increase in quantity? 24. If the theoretical water gauge at a fan is 4 ins. and the water gauge actually produced is 2 ins., what is the manometric efficiency of the fan? Ans. 50 per cent manometric efficiency. 25. If while running at 50 revolutions per minute a fan delivers 60,000 cu.ft. of air, how much air will be deliv- ered if the revolutions are increased to 100 per minute? Ans. 120,000 cu.ft. 26. A fan running 40 revolutions per minute pro- duced 1 in. water gauge. What will be the water gauge when the revolutions are 60 per minute? Ans. 2.25 in. w.g. 27. If with 1 in. water gauge 50,000 cu.ft. of air are MINE FIRES 167 passing through a mine per minute, what water gauge will be required to pass 100,000 cu.ft. per minute? Ans. 4 in. iy.gr. 28. If it requires 15 H.P. to run a fan 40 revolutions per minute, while ventilating a mine, what power will be required to run it 80 R.P.M.? Ans. 120 H.P. 29. If it requires 15 H.P. to circulate 40,000 cu.ft. of air per minute through a mine, what horse-power will be required to circulate 80,000 cu.ft.? Ans. 120 H.P. 30. If a fan is producing 150,000 cu.ft. of air per minute under average mine conditions with a 2-in. water gauge, what size motor or engine will be required to drive it? Ans. 66f H.P. 31. Name the five causes which go to make up the total resistance met with by the air while flowing through a mine. 32. What effect has the sudden contraction and expan- sion of an airway on the quantity of air passing through a mine? 33. An engine is making 300 strokes per minute, the area of the piston is 201 ins. What is the H.P. if the length of the stroke is 1 ft. and the mean effective pres- sure is 40 Ibs.? Ans. 109.6 H.P. 34. If a direct-current motor is using 150 amperes and 240 volts, what is the horse-power? Ans. 48.25 H.P. 35. If the actual horse-power of a ventilating equip- ment is 150 and the effective horse-power is 110, what is the mechanical efficiency? Ans. 73J per cent. 36. If the rim speed of a fan is 90 ft. per second, what is the theoretical water gauge? Ans. 3.7 w.g. 37. If the theoretical water gauge of a fan is 3 ins. and the actual water gauge developed by the fan is 2 ins., what is its manometric efficiency? Ans. 66f per cent. 38. If the horse-power of a ventilating equipment is 168 MINING AND MINE VENTILATION 30, and from which only 50 per cent useful effect is obtained, what is the quantity of air produced if the water gauge reading is 2.3? Ans. 41,388 cu.ft. per min. 39. If the input horse-power of a fan engine is 40 and the output horse-power is 28, what is the mechanical effi- ciency? Ans. 70 per cent. 40. If while running 100 R.P.M. a fan produces 40,000 cu.ft. of air, how much air should the same fan produce when running 125 R.P.M.? Ans. 50,000 cu.ft. 41. A fan is running 50 R.P.M. and is producing 100,000 cu.ft. of air per minute at a water gauge of 1 in. and horse- power of 50. What will be the quantity of air, revolution of the fan and the horse-power of the engine if the water gauge is increased to 4 ins.? Ans. 200,000 cu.ft. per min. 100 R.P.M. 400 H.P. 42. A fan is 10 ft. in diameter and 6 ft. wide, and is running 100 R.P.M. If the actual volume of air delivered by the fan is 80,000 cu.ft., what is its volumetric capacity? Ans. 169.7 per cent. 43. If the total pressure required to maintain a velocity of 400 ft. per minute in a certain airway is 100 Ibs., what is the rubbing surface? Ans. 31,250 sq.ft. 44. The length of an airway is 2000 ft., the perimeter 32 ft. What is the rubbing surface? Ans. 64,000 sq.ft. 45. A gangway is 9 ft. by 9 ft. If the water gauge shows f in. and the velocity of the air is 280 ft. per minute, what is the rubbing surface? Ans. 201,466 sq.ft. 46. If a power of 120,000 foot-pounds per minute is required to maintain a velocity of 500 ft. per minute in a certain airway, what is the rubbing surface? Ans. 48,000 sq.ft. MINE FIRES 169 47. 80,000 cu.ft. of air is passing through a certain airway, the velocity is 300 ft. per minute and the pressure equal to 1 in. water gauge. What is the rubbing surface? Ans. 770,370 sq.ft. 48. An airway is of such length that it requires a total pressure of 140 Ibs. to maintain a velocity of 500 ft. per minute. What is the rubbing surface? Ans. 28,000 sq.ft. 49. The area of an airway is 36.5 ft. and a pressure of 3 Ibs. per square foot is required to maintain a velocity of 600 ft. per minute. What is the rubbing surface? Ans. 15,208 sq.ft. 50. If it requires 30 H.P. to maintain a velocity of 400 ft. per minute in a certain airway, what is the rubbing surface? Ans. 773,437 sq.ft. 51. A circular airway has a radius of 5 ft., its length is 1000 ft. What is the rubbing surface? Ans. 31,416 sq.ft. 52. If the total pressure required to produce a velocity of 400 ft. per minute is 80 Ibs., what will be the units of power? Ans. 32,000 units. 53. An airway is 6 ft. by 6 ft., the water gauge is 1 in., the velocity is 300 ft. per minute. Find the units of power per minute? Ans. 56,160 units. 54. If it requires 8.5 Ibs. pressure per square foot to pass 100,000 cu.ft. of air per minute through an airway, what are the foot-pounds of work per minute? Ans. 850,000 foot pounds. 55. If 40 H.P. is consumed to ventilate a certain mine, what is its equivalent in units of work per minute? Ans. 1,320,000 units. 56. Find the foot-pounds of work necessary to venti- late a certain mine if the area of the airway is 36 ft., the 170 MINING AND MINE VENTILATION velocity 400 ft. per minute and the pressure 10 Ibs. per square foot. Am. 144,000 foot pounds. 57. Find the units of work necessary to maintain a velocity of 400 ft. per minute in an airway, if the rubbing surface is 200,000 sq.ft. Ans. 256,000 units. 58. An airway is 6 ft. by 6 ft. and 2000 ft. long, the quantity of air passing is 150,000. Required, the units of work. Ans. 69,444,444 units. 59. The pressure producing ventilation is equal to 1.5 in. water gauge, the area of the airway is 25 ft. and the velocity of the air is 400 ft. What are the units of work? Ans. 78,000 units. 60. If the area of an airway is 60.4 ft., the velocity is 325 ft. per minute, what is the quantity? Ans. 19,630 cu.ft. 61. In a certain airway the rubbing surface is 172,000 sq.ft. and a velocity of 500 ft. per minute is maintained by a water gauge of 2 ins. What is the quantity? Ans. 41,346 cu.ft. 62. In order to produce a certain quantity of air, the units of work required equal 125,000 and the pressure 11 Ibs. What is the quantity? Ans.' 11,363+ cu.ft. 63. An airway is 10 ft. by 10 ft. and 2000 ft. long, through which a certain quantity of air is passing under a pressure of 12 Ibs. per square foot. What is the quan- tity? Ans. 86,600 cu.ft. 64. The rubbing surface of an airway is 180,000 sq. ft. and a velocity of 400 ft. is maintained by a pressure equal .to 1 in. water gauge. What is the quantity? Ans. 44,307+ cu.ft. 65. The units of work and pressure necessary to pro- duce a certain quantity of air equals, units 210,000, pres- sure 8 Ibs. per square foot. What is the quantity? Ans. 26,250 cu.ft. MINE FIRES 171 66. If the water-gauge reading at a fan is 2.3 ins., what is the pressure per square foot? Ans. 11.96 Ibs. per sq.ft. 67. An airway has a rubbing surface of 80,000 sq.ft., the velocity is 400 ft. per minute, the quantity of air pass- ing is 40,000 cu.ft. What is the pressure? Ans. 2.56 Ibs. per sq.ft. 68. If the horse-power producing ventilation is 40, the quantity of air passing through an airway is 80,000 cu.ft., what is the pressure? Ans. 16.5 Ibs. per sq.ft. 69. The area of an airway is 100 sq.ft., the velocity of the air is 400 ft. If it requires 20 H.P. to produce this velocity, what is the pressure? Ans. 16.5 Ibs. per sq.ft. 70. The area of an airway is 80 sq.ft., the rubbing surface is 70,000 sq.ft. If the quantity of air passing is 100,000 cu.ft., what is the pressure? Ans. 27.3 Ibs. per sq.ft. 71. If the total pressure producing ventilation in an airway 10 ft. by 10 ft. is 800 Ibs., what is the water gauge? Ans. 1.53 in. w.g. 72. If it is required to pass 20,000 cu.ft. of air per minute, (a) what is the pressure per square foot if 6 H.P. is required? (6) What is the water gauge reading? Ans. (a) 9.9 Ibs. per sq.ft. (6) 1.9 in. w.g. 73. The quantity of air passing per minute in a mine is 112,000 cu.ft., the effective power of the fan is 40 H.P. What is the pressure? Ans. 11.78 Ibs. per sq.ft. 74. If it requires 10 H.P. to maintain a velocity of 200 ft. per minute in an airway 6 ft. by 8 ft., what is the pres- sure? Ans. 34.37 Ibs. per sq.ft. 75. If it requires 30 H.P. to produce a velocity equal to 400 ft. per minute, what is the total pressure? Ans. 2475 Ibs. 172 MINING AND MINE VENTILATION 76. If the airway (in Question 75) is square and its diagonal is 40 ft., what is the pressure per square foot? Ans. 3.09 Ibs. per sq.ft. 77. If the units of work required to produce a velocity of 300 ft. per minute equals 660,000, what is the total pressure? Ans. 2200 Ibs. 78. The quantity of air passing through an airway is 36,000 cu.ft., the rubbing surface is 78,000 sq.ft. and the area of the airway is 36 sq.ft. What is the total pressure? Ans. 1560 Ibs. 79. If the units of work necessary to force 100,000 cu.ft. of air through an airway equals 1,320,000, what is the pressure per square foot? Ans. 13.2 Ibs. per sq.ft. 80. What is the horse power in Question 79? Ans. 40 H.P. 81. An airway is 6 ft. by 10 ft. and 2400 ft. long. If the air is moving at a velocity of 500 ft. per minute, what is the pressure per square foot and the total ventilating pressure? Ans. 6.4 Ibs. per sq.ft. 384 Ibs. total pressure. 82. The quantity of air passing through an airway is 100,000 cu.ft. per minute, the area of the airway is 125 sq.ft. and the rubbing surface is 175,000 sq.ft. What is the pressure per square foot and the total pressure pro- ducing ventilation? Ans. 17.92 Ibs. per sq.ft. 2240 Ibs. total pressure. 83. An airway 10 ft. by 10 ft. is 1 mile long; the quan- tity of air passing is 150,000 cu.ft. per minute, the total pressure is 500 Ibs. What is the coefficient of friction? Ans. .00000000105. 84. If in the above example the water gauge was 2J ins., what would be the coefficient of friction? I Ans. .00000000273. MINE FIRES 173 85. If (in Question 83) the horse-power producing ventilation is 40, what is the coefficient of friction? Ans. .0000000018. 86. If it requires 99,000 units of work to produce a velocity of 300 ft. in an airway 6 ft. by 8 ft. and 10,000 ft. long, what is the coefficient of friction? Ans. .000000013. 87. A square airway is 2000 ft. long, the water gauge is 1 in., the total pressure is 520 Ibs., the quantity of air passing is 10,000 cu.ft. What is the coefficient of friction? Ans. .00000065. SUMMARY SOME of the most important statements made in this book are summarized as follows: The average height of the barometer in the United States at sea level is 29.92 ins. A cubic foot of dry air at 32 F. at sea level weighs 0.080728 Ib. The lowest United States barometer reading was taken at Galveston, Texas, during the year of flood, when the barometer reached 28.48 ins., or nearly f Ib. per square inch below normal. A sudden rise in the barometer is nearly as threatening as a sudden fall, because it shows that the level is unsteady. An accurate aneroid will show the altitude of a table, if lifted from the floor to the top of the table. The barometer falls lower for high winds than for heavy rain. The temperature of 111 below zero was taken at St. Louis, Mo., at an altitude of 48,700 feet. The temperature of the sun is estimated to be 14,072 F. The highest known average monthly temperature ever observed is that of 102 F. for July at Death Valley, Cali- fornia. The lowest is 60 F. for January at Siberia. Absolute zero is 459.4; above this temperature every- thing scientifically contains heat. If every particle of moisture in the air were precipitated, it would cover the entire globe to a depth of nearly 4 ins. Sounds travel far when the humidity is high. 175 176 SUMMARY Aqueous vapor is a gas much lighter than air; its atomic weight is 9. When this vapor mixes with air the mixture becomes lighter than dry air. When air is flowing through a mine there must be a difference in density between the intake and return air. The water gauge produced by a fan is usually about 70 per cent of its theoretical water gauge. That bituminous coal dust may be rendered inert by the proper application of moisture has been shown both by laboratory tests and by the absence of explosions at mines in which moisture is present in the proper propor- tion to the quantity of dust produced. Methane, or any other gas, when once thoroughly mixed with air, will not separate from the mixture. Black damp is not carbon dioxide alone, but a mixture of carbon dioxide and nitrogen. Lights grow dim or go out in an atmosphere containing carbon dioxide because of the low percentage of oxygen in the atmosphere and not because of the presence of carbon dioxide. The effect on a person breathing carbon dioxide is not always due to the carbon dioxide present, but is some- times due to the lack of oxygen. Air may be what is termed chemically pure and yet cause distress if its temperature and relative humidity are high. An atmosphere must hot be assumed to be non-explosive because it does not contain enough oxygen to support the combustion of an oil-fed flame. The presence of a fatal percentage of carbon monoxide is not indicated by the lamp flame. All mine air contains water vapor, the proportion de- pending chiefly upon the temperature of the air and amount of water present along the airways. INDEX Absolute temperature, 62 Absolute zero, 61 Acceleration, 14 Acetylene lamp, 69 Air bridge, 132, 133 Air, composition of, 32, 33, 72 effect of expansion, 60 height of, 32 humidity of, 42, 43 moisture in, 76 pressure of, 32, 55 properties of, 72 weight of, 36 Altitude, table of, 58 Analysis, anthracite coal, 161 bituminous coal, 161 Anemometer, 90, 91 Aneroid barometer, 52, 53, 54, 55 Area of circle, 158 of ellipse, 158 of sector, 158 of trapezoid, 158 of triangle, 158 Atmosphere, composition of, 32, 33 Atmospheric pressure, 32, 55, 64, 71 Atom, 33, 61 Atomic weight, 36, 37, 40 Avoirdupois weight, 159 Barometer, 52, 53, 54, 55 indications, 59 use of, 56, 57 Bituminous coal analysis, 161 Black damp, 74 Boiling, 28 Boyle's law, 60 British thermal unit, 27 Calcium carbide, 68 Calculations, 91 Carbon monoxide, 72 explosive properties, 73 how produced, 73 properties of, 72, 73 dioxide, 73, 74 how produced, 74 Carbureted hydrogen, 36, 75 Centigrade scale, 28, 29 Charles' law, 60 Chemical compound, 39 equations, 41 properties of air, 72 symbols, 39, 40 Coal, analysis of, 161 carbon in, 161 Coefficient of friction, 141 Cohesion, 4 Common names of gases, 36 Compressibility, 2 Constant force, 14 Cost per horse-power, 124, 125 Davy safety lamp, 70, 77 Density, 21, 35, 45 Dew point, 45 177 178 INDEX Diffusion of gases, 49 Divisibility, 3 Efficiency of fan, 93, 94, 95 Elastic limit, 3 Elasticity, 3 Elements, 34 table of, 34 English measure, length, 159 surface, 159 volume, 159 Entering a mine after an ex- plosion, 153, 154, 155 Ethane, 78 Ethylene, 78 Evaporation, 42 Examination questions, 146, 163 Exchange of heat, 83 Expansion by heat, 28, 60, 61 Explosion, 153, 154, 155 Extension, 1 Fahrenheit scale, 28, 29 Falling bodies, 15, 16 Fans, 106, 107, 108, 109 calculations of, 91 installation of, 126, 127, 128, 129 manometric efficiency, 93, 94, 129 mechanical efficiency, 93, 94, 129 volumetric capacity, 95, 128, 129 Fire in mines, 152 damp, 75 detection of, 76 percentage of gas in, 75 Force, 8 effect of, 14 parallelogram, 8, 9, 10 Formulas, 36, 141, 142, 143, 144, 145 transposition of, 146 Freezing, 28 mixture, 30 -point, 28 Friction, laws of, 96, 97, 99 Furnace ventilation, 106 Gas caps, 77 Gases, 32, 39 acetylene, 68, 69 calculation of weight, 37, 64, 65 density of, 35 diffusion of, 49 effect of temperature, 63 general properties of, 72 moving of, 4 occlusion of, 71 specific gravity of, 36 table of, 36 Gravitation, 12 laws of, 12 Gravity, 12 Heat, 27 capacity, 82 measurement of, 27, 83, 84 specific, 82, 83 Height of flame cap, 77 Hooke's law, 4 Horse-power, 124, 125, 160 cost of, 125 Humidity, absolute, 44, 45 of the air, 42, 45, 48 relative, 45, 76 tables, 46, 47 Hydrogen, 40 Hydrometer, 23 Hygrometer, 43 INDEX 179 Ignition point, 76 Impenetrability, 2 Indestructibility, 2 Inertia, 3 Installation of fan, 126 Iron, specific gravity, 24 Jeffrey fan, 118, 123 drift, 119 table of capacities, 121, 122 Lamps, safety, 70, 71 acetylene, 69 Laws of diffusion of gases, 49 of falling bodies, 13 of floating bodies, 22 of friction, 96, 97, 99 of gravitation, 12 Length, units of, 159 Liquids and liquid pressure, 19 expansion of, 19, 28 measure, 160 Manometric efficiency, 94, 129 Marsh gas, 36, 75, 76, 77 amount in air to explode, 75 how produced, 75 temperature of ignition, 76 Matter defined, 1 properties of, 1 Maximum density of water, 28 Mechanical mixture, 39 Melting-point, table of, 2Q Methane, 36, 75 Mine air, samples of, 79, 80 Mine fires, 152 sealing, 155, 156, 157 suggestions to avoid, 152, 153 suggestions to extinguish, 153, 154, 155 Moisture, 42, 43 Molecular weight, 41 Molecules, 33, 61, 72 Motion, 6 laws of, 6, 7 Motive column, 130, 131 Natural ventilation, 105 Newton's laws of gravitation, 12 of motion, 6, 7 Nitrogen, 36 Occlusion of gases, 71 Oxygen, 36 action on flame, 68, 69, 70 Parallelogram of forces, 8, 9, 10 Percentage composition of gases, 41 Physical properties of air, 72 Porosity, 2 Pressure of atmosphere, 55 defined, 88, 89 of gases, 62, 63, 64 of liquids, 9, 20 table of, 99, 100 Properties of matter, 1, 72 Regulators, 134, 135 Relative density, 20 humidity, 76 Resistance, 126, 127, 136, 137 effect of in mines, 137, 138 Robinson fan, 107, 108, 109 table of volumes, 109 Rules and formulas, 158, 159, 160, 161 Safety lamps, 70, 71 Saturated water vapor, 45 Sirocco fan, 109, 110, 111, 112, 113, 114, 115, 118 180 INDEX Specific gravity, 35,. 36 bottle, 22 of liquids, 22, 23 of solids, 20, 21 table of, 24 Specific heat, 82, 83 measurement of, 83 table of, 83 Splitting of air currents, 132 advantages of, 132 Static pressure, 88 Steam jet, 106 Sulphureted hydrogen, 76, 78 Table of specific gravity, 24-36 of air analysis, 79, 80 of altitudes, 58 of cost per horse-power, 125 of elements, 34, 36 of gases, 36 of hoisting ropes, 162 of humidity, 46, 47 of manilla ropes, 163 of melting-point, 29 of specific heat, 83 of strength of ropes, 162, 163 of temperatures, 29 of theoretical water gauge, 120 of velocity pressure, 99, 100 of weight, 24, 36 of weight of water vapor, 44 Temperature, absolute, 62 effect on volume, 60 estimation of, 29 Tension, 63 Thermometer, 27, 28 Troy weight, 159 Vacuum, 106, 155 Vapor, 44, 45 Velocity, 8, 123 of air current, 99, 123 measurement of, 90, 91 pressure, 99 Ventilation, 87, 88, 89, 104 by fan, 106, 107 by furnace, 106 by steam jet, 106 by water jet, 106 how produced, 104, 105, 106, 107 Water, 28 boiling-point, 28 expansion of, 19, 28 gauge, 89 gauge calculations, 91 gauge, theoretical, 94, 120, 127 jet, 106 maximum density, 21 specific gravity, 21 vapor, table of, 44 weight of, 20, 161 Weather indications, 55, 59 Weight of gases, 36 White damp, 36, 72, 73 Zero, absolute, 61 D. VAN NOSTRAND COMPANY 25 PARK PLACE NEW YORK SHORT-TITLE CATALOG OP OF SCIENTIFIC AND ENGINEERING BOOKS This list includes the technical publications of the following English publishers : SCOTT, GREENWOOD & CO. JAMES MUNRO & CO.. Ltd. CONSTABLE & COMPANY, Ltd. TECHNICAL PUBLISHING CO. ELECTRICIAN PRINTING & PUBLISHING CO., for whom D. Van Nostrum! Company are American agents. Descriptive Circulars sent on request e AUGUST, 1915 SHORT-TITLE CATALOG OP THE Publications and Importations OP D. VAN NOSTRAND COMPANY 25 PARK PLACE Prices marked with an asterisk (*) are NET All bindings are in cloth unless otherwise noted Abbott, A. V. The Electrical Transmission of Energy 8vo, *$5 oo A Treatise on Fuel. (Science Series No. 9.) i6mo, o 50 Testing Machines. (Science Series No. 74.) i6mo, o 50 Adam, P. Practical Bookbinding. Trans, by T. E. Maw.iamo, *2 50 Adams, H. Theory and Practice in Designing 8vo, *2 50 Adams, H. C. Sewage of Seacoast Towns 8vo, *2 oo Adams, J. W. Sewers and Drains for Populous Districts... .8vo, 250 Adler, A. A. Theory of Engineering Drawing 8vo, *2 oo Principles of Parallel Projecting-Line Drawing 8vo, *i oo Aikman, C. M. Manures and the Principles of Manuring. . . 8vo, 2 50 Aitken, W. Manual of the Telephone 8vo, *8 oo d'Albe, E. E. F. Contemporary Chemistry i2mo, *i 25 Alexander, J. H. Elementary Electrical Engineering I2mo, 2 oo Allan, W. Strength of Beams under Transverse Loads. (Science Series No. 19.) i6mo, o 50 Allan, W. Theory of Arches. (Science Series No. n.). . i6mo, Allen, H. Modern Power Gas Producer Practice and Applica- tions i2mo, *2 50 Gas and Oil Engines 8vo, *4 50 Anderson, J. W. Prospector's Handbook i2mo, i 50 D. VAN NOSTRAND COMPANY'S SHORT-TITLE CATALOG 3 Andes, L. Vegetable Fats and Oils 8vo, *4 oo Animal Fats and Oils. Trans, by C. Salter 8vo, *4 oo Drying Oils, Boiled Oil, and Solid and Liquid Driers . . .8vo, *5 oo Iron Corrosion, Anti-fouling and Anti-corrosive Paints. Trans, by C. Salter 8vo, *4 oo Oil Colors and Printers' Ink. Trans, by A. Morris and H. Robson 8vo, *2 50 Treatment of Paper for Special Purposes. Trans, by C. Salter i2mo, *2 50 Andrews, E. S. Reinforced Concrete Construction i2mo, *i 25 Theory and Design of Structures 8vo, 3 50 Further Problems in the Theory and Design of Struc- tures 8vo, 2 50 Andrews, E. S., and Heywood, H. B. The Calculus for Engineers i2mo, i 25 Annual Reports on the Progress of Chemistry. Eleven Vol- umes now ready. Vol. I, 1904, to Vol. XI, 1914, 8vo, each *2 oo Argand, M. Imaginary Quantities. Translated from the French by A. S. Hardy. (Science Series No. 52.) i6mo, o 50 Armstrong, R., and Idell, F. E. Chimneys for Furnaces and Steam Boilers. (Science Series No. i.) i6mo, o 50 Arnold, E. Armature Windings of Direct Current Dynamos. Trans, by F. B. DeGress 8vo, *2 oo Asch, W., and Aseh, D. The Silicates in Chemistry and Commerce 8vo, *6 oo Ashe, S. W., and Keiley, J. D. Electric Railways. Theoreti- cally and Practically Treated. Vol. I. Rolling Stock i2mo, *2 50 Ashe, S. W. Electric Railways. Vol. II. Engineering Pre- liminaries and Direct Current Sub-Stations i2mo, *2 50 Electricity: Experimentally and Practically Applied. i2mo, *2 oo Ashley, R. H. Chemical Calculations i2mo, *i oo Atkinson, A. A. Electrical and Magnetic Calculations. .8 vo, *i 50 Atkinson, J. J. Friction of Air in Mines. (Science Series No. 14.) i6mo, o 50 Atkinson, J. J., and Williams, E. H., Jr. Gases Met with in Coal Mines. (Science Series No. 13.)-. i6mo, o 50 4 D. VAN NOSTRAND COMPANY'S SHORT-TITLE CATALOG Atkinson, P. The Elements of Electric Lighting i2mo, i 50 The Elements of Dynamic Electricity and Magnetism. i2mo, 2 oo Atkinson, P. Power Transmitted by Electricity i2mo, 2 oo Auchincloss, W. S. Link and Valve Motions Simplified 8 vo, * i 50 Austin, E. Single Phase Electric Railways 4to, *s oo Ayrton, H. The Electric Arc 8vo, *5 oo Bacon, F. W. Treatise on the Richards Steam-Engine Indica- tor 1 2mo, i oo Bailes, G. M. Modern Mining Practice. Five Volumes. 8 vo, each, 3 50 Bailey, R. D. The Brewers' Analyst 8vo, *5 oo Baker, A. L. Quaternions i2mo, *i 25 Thick-Lens Optics i2mo, *i 50 Baker, Benj. Pressure of Earthwork. (Science Series No. 56.) i6mo, Baker, I. Levelling. (Science Series No. 91.) i6mo, o 50 Baker, M. N. Potable Water. (Science Series No. 61) . i6mo, o 50 Sewerage and Sewage Purification. (Science Series No. 18.) i6mo, o 50 Baker, T. T. Telegraphic Transmission of Photographs. i2mo, *i 25 Bale, G. R. Modern Iron Foundry Practice. Two Volumes. i2tno. Vol. I. Foundry Equipment, Material Used *2 50 Vol. II. Machine Moulding and Moulding Machines *i 50 Ball, J. W. Concrete Structures in Railways 8vo, *2 50 Ball, R. S. Popular Guide to the Heavens 8vo, *4 50 Natural Sources of Power. (Westminster Series) 8vo, *2 oo Ball, W. V. Law Affecting Engineers 8vo, *3 50 Bankson, Lloyd. Slide Valve Diagrams. 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Trans, by J. P. Millington 8vo, *7 50 Blyth, A. W. Foods: Their Composition and Analysis. ..Svo, 7 50 Poisons: Their Effects and Detection 8vo, 7 50 Bockmann, F. Celluloid i2mo, *2 50 Bodmer, G. R. Hydraulic Motors and Turbines i2mo, 5 oo Boileau, J. T. Traverse Tables Svo, 5 oo Bonney, G. E. The Electro-plater's Handbook i2mo, i 20 Booth, N. Guide to Ring-Spinning Frame i2ino, *i 25 Booth, W. H. Water Softening and Treatment Svo, *2 50 Superheaters and Superheating and their Control. ..8vo, *i 50 Bottcher, A. Cranes: Their Construction, Mechanical Equip- ment and Working. Trans, by A. Tolhausen. .. .4to, *io oo Bottler, M. Modern Bleaching Agents. Trans, by C. Salter. 1 2 mo, *2 50 Bottone, S. R. Magnetos for Automobilists i2mo, *i oo Boulton, S. B. Preservation of Timber. (Science Series No. 82.) i6mo, 050 Bourcart, E. Insecticides, Fungicides and Weedkillers . . . Svo, *4 50 Bourgougnon, A. Physical Problems. (Science Series No. 113.) i6mo, o 50 Bourry, E. Treatise on Ceramic Industries. Trans, by A. B. $earle Svo, *s oo Bowie, A. J., Jr. A Practical Treatise on Hydraulic Mining. Svo, 5 oo Bowles, 0. Tables of Common Rocks. (, Science Series.) .i6mo, 050 D. VAN NOSTRAND COMPANY'S SHORT-TITLE CATALOG 7 Bowser, E. A. Elementary Treatise on Analytic Geometry. i2mo, i 75 Elementary Treatise on the Differential and Integral Calculus i2mo, 2 25 Bowser, E. A. Elementary Treatise on Analytic Mechanics, i2mo, 3 oo Elementary Treatise on Hydro-mechanics i2mo, 2 50 A Treatise on Roofs and Bridges *2 25 Boycott, G. W. M. Compressed Air Work and Diving. .8vo, *4 oo Bragg, E. M. Marine Engine Design i2mo, *2 oo Design of Marine Engines and Auxiliaries ,(In Press.) Brainard, F. R. The Sextant. (Science Series No. ioi.).i6mo, Brassey's Naval Annual for 1911 8vo, *6 oo Brew, W. Three-Phase Transmission 8vo, *2 oo Briggs, R., and Wolff, A. R. Steam-Heating. (Science Series No. 67.) i6mo, o 50 Bright, C. The Life Story of Sir Charles Tilson Bright. .8vo, *4 50 Brislee, T. J. Introduction to the Study of Fuel. (Outlines of Industrial Chemistry.) 8vo, *3 oo Broadfoot, S. K. Motors Secondary Batteries. (Installation Manuals Series.) i2mo, *o 75 Broughton, H. H. Electric Cranes and Hoists *g oo Brown, G. Healthy Foundations. (Science Series No. 80.). i6mo, o 50 Brown, H. Irrigation 8vo, *5 oo Brown, Wm. N. The Art of Enamelling on Metal i2mo, *i oo Handbook on Japanning and Enamelling i2mo, *i 50 House Decorating and Painting i2mo, *i 50 History of Decorative Art i2mo, *i 25 Dipping, Burnishing, Lacquering and Bronzing Brass Ware i2mo, *i oo Workshop Wrinkles 8vo, *i oo Browne, R. E. Water Meters. (Science Series No. 8i.).i6mo, o 50 Bruce, E. M. Pure Food Tests i2mo, *i 25 Bruhns, Dr. New Manual of Logarithms 8vo, cloth, 2 oo Half morocco, 2 50 Brunner, R. Manufacture of Lubricants, Shoe Polishes and Leather Dressings. Trans, by C. Salter 8vo, *s oo 8 Buel, R. H. Safety Valves. (Science Series No. 21.) . . . i6mo, o 50 Burley, G. W. Lathes, Their Construction and Operation, 121110, I 25 Burstall, F. W. Energy Diagram for Gas. With text. ,.8vo, *i 50 Diagram sold separately ''i oo Burt, W. A. Key to the Solar Compass i6mo, leather, 2 50 Buskett, E. W. Fire Assaying i2mo, *i 25 Butler, H. J. Motor Bodies and Chasis 8vo, *2 50 Byers, H. G., and Knight, H. G. Notes on Qualitative Analysis 8vo, *i 50 Cain, W. Brief Course in the* Calculus 12 mo, *i 75 Elastic Arches. (Science Series No. 48.) i6mo, o 50 Maximum Stresses. (Science Series No. 38.) i6mo, o 50 Practical Dsigning Retaining of Walls. (Science Series No. 3.) i6mo, 050 Theory of Steel-concrete Arches and of Vaulted Struc- tures. (Science Series.) i6mo, o 50 Theory of Voussoir Arches. (Science Series No. 12.) i6mo, o 50 Symbolic Algebra. (Science Series No. 73.) i6mo, o 50 Carpenter, F. D. Geographical Surveying. (Science Series No. 37.) i6mo, Carpenter, R. C., and Diederichs, H. Internal-Combustion Engines 8vo, *s oo Carter, E. T. Motive Power and Gearing for Electrical Ma- chinery 8vo, 3 50 Carter, H. A. Ramie (Rhea), China Grass .i2mo, *2 oo Carter, H. R. Modern Flax, Hemp, and Jute Spinning. . 8vo, *3 oo Bleaching, Dyeing and Finishing of Fabrics Jvo, *i oo Cary, E. R. Solution of Railroad Problems With the Use of the Slide Rule i6mo, *i oo Cathcart, W. L. Machine Design. Part I. Fastenings . . . 8vo, *3 oo Cathcart, W. L., and Chaffee, J. I. Elements of Graphic Statics and General Graphic Methods 8vo, *3 oo Short Course in Graphic Statics i2mo, *i 50 D. VAN NOSTRAND COMPANY^ SHORT-TITLE CATALOG 9 Caven, R. M., and Lander, G. D. Systematic Inorganic Chem- istry 1 2mo, *2 oo Chalkley, A. P. Diesel Engines 8vo, *3 oo Chambers' Mathematical Tables 8vo, i 75 Chambers, G. F. Astronomy i2mo, *i 50 Charpentier, P. Timber 8vo, *6 oo Chatley, H. Principles and Designs of Aeroplanes. (Science Series.) i6mo, o 50 How to Use Water Power i2mo, *i oo Child, C. D. Electric Arcs 8vo, *2 oo Child, C. T. The How and Why of Electricity i2mo, i oo Christian, M. Disinfection and Disinfectants 121110, *2 oo Christie, W. W. Boiler-waters, Scale, Corrosion, Foaming, 8vo, *3 oo Chimney Design and Theory 8vo, *3 oo Furnace Draft. (Science Series.) i6mo, o 50 Water, Its Purification and Use in the Industries. .8vo, Church's Laboratory Guide. Rewritten by Edward Kinch . 8vo, *2 50 Clapperton, G. Practical Papermaking 8vo, 250 Clark, A. G. Motor Car Engineering. Vol. I. Construction 8vo, *3 oo Vol. II. Design (In Press.) Clark, C. H. Marine Gas Engines i2mo, *i 50 Clark, J. M. New System of Laying Out Railway Turnouts, 121110, i oo Clarke, J. W., and Scott, W. Plumbing Practice. Vol. I. Lead Working and Plumbers' Materials .. 8vo, *4 oo Vol. II. Sanitary Plumbing and Fittings (In Press.) Vol. III. Practical Lead Working on Roofs (In Press.) Clerk, D., and Idell, F. E. Theory of the Gas Engine. (Science Series No. 62.) i6mo, o 50 Clevenger, S. R. Treatise on the Method of Government Surveying i6mo, mor., 2 50 Clouth, F. Rubber, Gutta-Percha, and Balata 8vo, *5 oo Cochran, J. Treatise on Cement Specifications 8vo, *i oo Concrete and Reinforced Concrete Specifications 8vo, *2 50 10 D. VAN NOSTRAND COMPANY'S SHORT-TITLE CATALOG Coffin, J. H. C. Navigation and Nautical Astronomy. .i2mo, *s 50 Colburn, Z., and Thurston, R. H. Steam Boiler Explosions. (Science Series No. 2.) i6mo, o 50 Cole, R. S. Treatise on Photographic Optics i2mo, i 50 Coles-Finch, W. Water, Its Origin and Use 8vo, *5 oo Collins, J. E. Useful Alloys and Memoranda for Goldsmiths, Jewelers i6mo, o 50 Collis, A. G. High and Low Tension Switch-Gear Design. 8vo, *3 50 Switchgear. (Installation Manuals Series.) i2mo, o 50 Coombs, H. A. Gear Teeth. (Science Series No. 120). . .i6mo, o 50 Cooper, W. R. Primary Batteries 8vo, *4 oo Copperthwaite, W. C. Tunnel Shields 4to, ~g oo Corey, H. T. Water Supply Engineering 8vo (In Press.) Corfield, W. H. Dwelling Houses. (Science Series No. 50.) i6mo, o 50 Water and Water-Supply. (Science Series No. 17.). . i6mo, j 50 Cornwall, H. B. Manual of Blow-pipe Analysis 8vo, *2 50 Cowell, W. B. Pure Air, Ozone, and Water i2mo, *2 oo Craig, J. W., and Woodward, W. P. Questions and Answers about Electrical Apparatus i2mo, leather, i 50 Craig, T. Motion of a Solid in a Fuel. (Science Series No. 49.) i6mo, o 50 Wave and Vortex Motion. (Science Series No. 43.) . i6mo, o 50 Cramp, W. Continuous Current Machine Design 8vo, *2 50 Greedy, F. Single-Phase Commutator Motors 8vo, *2 oo Crocker, F. B. Electric Lighting. Two Volumes. 8vo. Vol. I. The Generating Plant 3 oo Vol. II. Distributing Systems and Lamps Crocker, F B., and Arendt, M. Electric Motors 8vo, *2 50 and Wheeler, S. S. The Management of Electrical Ma- chinery i2mo, *i oo Cross, C. F., Bevan, E. J., and Sindall, R. W. Wood Pulp and Its Applications. (Westminster Series.) 8vo, *2 oo Crosskey, L. R. Elementary Prospective 8 vo, i oo Crosskey, L. R., and Thaw, J. Advanced Perspective 8vo, i 50 Culley,J. L. Theory of Arches. (Science Series No. 87.). i6mo, 050 Dadourian, H. M. Analytical Mechanics 8vo. *s oo Danby, A. Natural Rock Asphalts and Bitumens 8vo, *2 50 D. VAN NOSTRAND COMPANY'S SHORT-TITLE CATALOG 11 Davenport, C. The Book. (Westminster Series.) 8vo, *2 oo Davey, N. The Gas Turbine 8vo, *4 oo Da vies, F. H. Electric Power and Traction 8vo, *2 oo Foundations and Machinery Fixing. (Installation Manuals Series.) i6mo, i oo Dawson, P. Electric Traction on Railways 8vo, *g oo Deerr, N. Cane Sugar 8vo, 7 oo Deite, C. Manual of Soapmaking. Trans, by S. T. King. -4to, *5 oo De la Coux, H. The Industrial Uses of Water. Trans, by A. Morris 8 vo, *4 50 Del Mar, W. A. Electric Power Conductors 8 vo, *2 oo Denny, G. A. Deep-Level Mines of the Rand 410, *io oo Diamond Drilling for Gold *5 oo De Roos, J. D. C. Linkages. (Science Series No. 47.). . . i6mo, o 50 Derr, W. L. Block Signal Operation Oblong i2mo, *i 50 Maintenance of Way Engineering (In Preparation.) Desaint, A. Three Hundred Shades and How to Mix Them. 8vo, 8 oo De Varona, A. Sewer Gases. (Science Series No. 55.)... i6mo, o 50 Devey, R. G. Mill and Factory Wiring. (Installation Manuals Series.) i2mo, *i oo Dibdin, W. J. Purification of Sewage and Water 8vo, 6 50 Dichman, C. Basic Open-Hearth Steel Process 8vo, *3 50 Dieterich, K. Analysis of Resins, Balsams, and Gum Resins 8vo, *3 oo Dinger, Lieut. H. C. Care and Operation of Naval Machinery i2mo. *2 oo Dixon, D. B. Machinist's and Steam Engineer's Practical Cal- culator i6mo, mor., i 25 Doble, W. A. Power Plant Construction on the Pacific Coast. (In Press.) Dommett, W. E. Motor Car Mechanism i2mo, *i 25 Dorr, B. F. The Surveyor's Guide and Pocket Table-book. i6mo, mor., 2 oo Draper, C. H. Elementary Text-book of Light, Heat and Sound i2mo, i oo Draper, C. H. Heat and the Principles of Thermo-dynamics, New and Revised Edition i2ino, 2 oo 12 D. VAN N08TRAND COMPANY S SHORT-TITLE CATALOG Dron, R. W. Mining Formulas i2mo, i oo Dubbel, H. High Power Gas Engines 8vo, *s oo Duckwall, E. W. Canning and Preserving of Food Products .8 vo, *s oo Dumesny, P., and Noyer, J. Wood Products, Distillates, and Extracts 8 vo, *4 50 Duncan, W. G., and Penman, D. The Electrical Equipment of Collieries 8 vo, *4 oo Dunstan, A. E., and Thole, F. T. B. Textbook of Practical Chemistry i2mo, *i 40 Duthie, A. L. Decorative Glass Processes. (Westminster Series) 8vo, *2 oo Dwight, H. B. Transmission Line Formulas 8vo, *2 oo Dyson, S. S. Practical Testing of Raw Materials 8 vo, *5 oo and Clarkson, S. S. Chemical Works 8vo, *7 50 Eccles, W. H. Wireless Telegraphy and Telephony. ... (In Press.) Eck, J. Light, Radiation and Illumination. Trans, by Paul Hogner 8vo, *2 50 Eddy, H. T. Maximum Stresses under Concentrated Loads, 8vo, i 50 Edelman, P. 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