ENGINEERING LIBRARY THE THEORY OF HEAT ENGINES ELEMENTARY APPLIED MECHANICS BY ARTHUR MORLEY, M.Sc., M.I.MECH.E. AND WILLIAM INCHLEY, B.Sc., A.M.I.MECH.E. With 285 Diagrams, numerous Examples and Answers. Crown 8vo, 35. net. LABORATORY INSTRUCTION SHEETS IN ELEMENTARY APPLIED MECHANICS BY ARTHUR MORLEY, M.Sc., M.I.MECH.E. AND WILLIAM INCHLEY, B.Sc., A.M.I.MECH.E. 8vo, is. %d. net. i LONGMANS, GREEN & CO. LONDON, NEW YORK, BOMBAY, AND CALCUTTA. THE THEORY OF HEAT ENGINES BY WILLIAM INCHLEY B.SC., A.M.I.MECH.E. LECTURER IN ENGINEERING IN UNIVERSITY COLLEGE, NOTTINGHAM WITH 246 DIAGRAMS AND NUMEROUS EXAMPLES LONGMANS, GREEN, AND CO. 39, PATERNOSTER ROW, LONDON NEW YORK, BOMBAY, AND CALCUTTA 1913 All rights reserved ENGINEERING LIBRARV PREFACE THIS book has been written mainly for engineering students, and covers the ground required for University and similar examinations in the theory of heat engines. It will also be found suitable for students reading for the examinations of the Institution of Civil Engineers, the Institution of Mechanical Engineers, the City and Guilds of London Institute, and the Board of Education, and should prove useful to the engineer who desires a thorough knowledge of the theory of the subject. The Author feels that no apology is necessary for adding yet another book on this subject because, although many excellent books exist which deal solely with one or two special branches of the subject, there are very few which deal with the subject as a whole. An attempt has here been made to give in a complete and concise form the thermodynamical and mechanical principles of the subject ; to that end all purely descriptive matter has been designedly omitted. Many numerical examples are fully worked out in the text, and the student is urged to read all of these, and to work out for himself the examples at the ends of the chapters in order to obtain a thorough know- ledge of the subject. Those marked (L.U.) are taken from various London University papers for the B.Sc. (Engineering) examination. Many important researches in the subject have been noticed, references to which are freely given. In particular, the Proceedings of the Institution of Mechanical Engineers, the Institution of Civil Engineers, and the Reports of the Gaseous Explosions Committee of the British Association have, with permission, been freely drawn upon for this purpose. The Author's thanks are due to many friends for hints and suggestions, particularly to Professor A. Morley, M.Sc., and Professor W. Robinson, M.E., and also to Mir. R. H. King, B.Sc., Mr. J. S. Robinson, B.Sc., and Mr. T. P. G. Stone, who have so kindly checked many of the numerical examples. It is too much to hope that, with so many numerical examples, this edition will be free from errors ; any intimation of these or suggestions for future consideration will be cordially appreciated. W. INCHLEY. UNIVERSITY COLLEGE, NOTTINGHAM, September, 1913. 9693S8 CONTENTS CHAPTER I THERMODYNAMICS AND PROPERTIES OF GASES PAGE First law of thermodynamics Laws of permanent gases Work done during expansion Adiabatic expansion and compression Relations between /, v, and T Rate of heat reception Entropy 1-22 CHAPTER II HOT-AIR ENGINES Classification Graphic representation of work done during change in volume Cycles of operation Carnot's cycle Garnet's principle Conditions for maximum efficiency Second law of thermodynamics Stirling's engine Ericsson's engine Joule's engine 2 3~39 CHAPTER III PROPERTIES OF STEAM Generation of steam under constant pressure Relations between /, v, and t Total heat External work done Internal energy Superheated steam Throttling or wire-drawing Measurement of dryness of steam Theory of throttling calorimeter Entropy of steam Calculation of dryness fraction after expansion Mollier diagram Total heat-pressure diagram 40-68 CHAPTER IV THEORY OF THE STEAM ENGINE Work done during adiabatic expansion of steam Perfect steam engine working on Carnot's cycle Non-expansive engine Rankine cycle T diagram for Rankine cycle Effect of using superheated steam Engine in which the steam is kept dry and saturated during expansion The regenerative steam engine Effect of clearance on mean effective pressure Binary vapour engine . . 69-102 CHAPTER V THEORY OF THE STEAM ENGINE (continued) Actual indicator diagram Wire-drawing, clearance and cushioning Initial con- densation and re-evaporation Temperature range of cylinder walls Indicated weight of steam Saturation curve Missing quantity Methods of drawing T< diagram from pv diagram Valve leakage Most economical ratio of expansion Steam jacket Effect of superheating Diagram factors Steam consumption, the Willans Law lOS^S? viii CONTENTS CHAPTER VI COMPOUND EXPANSION Advantages Compound expansion with and without receiver Ratio of cylinder volumes Effect of varying cut-off in high-pressure cylinder Effect of throttling at admission to high-pressure cylinder Effect of varying cut-off in low-pressure cylinder Initial loads Cylinder dimensions Combination of indicator diagrams 138-152 CHAPTER VII MECHANICAL REFRIGERATION Types of mechanical refrigerating machines Coefficient of performance The cold air machine Reversed Joule engine or Bell-Coleman machine Warming machine Vapour compression machines Choice of a refrigerating agent . 153-170 CHAPTER VIII FLOW OF STEAM THROUGH ORIFICES AND NOZZLES Adiabatic flow through an orifice Weight of steam discharged Flow of super- heated steam Flow through nozzles Design of nozzles Use of Mollier diagram Effect of friction Theory of injectors Types of injectors . . 171-185 CHAPTER IX THEORY OF THE STEAM TURBINE Function of a turbine Impulse and reaction Single-stage turbines Multi-stage turbines Turbines with one or more stages, each compounded for velocity- Multi-stage reaction turbines The Parsons turbine Losses in steam turbines Effect of pressure, superheat and vacuum on efficiency Exhaust steam turbines Governing 186-214 CHAPTER X THEORY OF AIR COMPRESSORS AND MOTORS Transmission of power by compressed air Methods of reducing losses Simple, two-stage, and three-stage air compressors Effect of clearance Simple, two- stage, and three-stage air motors Efficiency of compressors and motors . 215-232 CHAPTER XI COMBUSTION Combustion of hydrogen, carbon and sulphur Minimum amount of air required for the complete combustion of I pound of solid or liquid fuel, or I cubic foot of gaseous fuel Calculation of quantity of air supplied from the analysis of the flue or exhaust gases Calculation of mean specific heat Heat carried away by products of combustion and excess air Calorific value of solid, liquid, and gaseous fuels Boiler draught Theory of producer gas 233-256 CONTENTS ix CHAPTER XII HEAT TRANSMISSION PAGE Transmission through flat plates Efficiency of heating surface Transmission through walls of a thick tube Effect of high gas speeds Estimation of the temperature of the gases leaving a boiler The most efficient rate of combustion Heat transmission through condenser tubes 257-270 CHAPTER XIII THEORY OF THE GAS ENGINE General considerations Constant volume and constant pressure, four-stroke and two-stroke cycles Atkinson cycle Otto cycle After burning Effect of strength of mixture Scavenging Ignition Governing Study of the indicator diagram Rate of heat reception and rejection from/z> diagrams T diagram for ideal Otto cycle Methods of drawing T diagram from/z/ diagram Losses in gas engines Standard cycle for internal combustion engines Heat trans- mission through cylinder walls 271-311 CHAPTER XIV THEORY OF THE INTERNAL COMBUSTION ENGINE ASSUMING THE SPECIFIC HEAT A LINEAR FUNCTION OF THE TEMPERATURE Explosion at constant volume Internal energy of gases at high temperatures Measurement of internal energy and specific heat at high temperatures Rate of heat reception with variable specific heat Adiabatic expansion with variable specific heat Reduction of efficiency when the working fluid is replaced by one of greater specific heat Calculation of ideal efficiency 312-330 CHAPTER XV THEORY OF THE OIL ENGINE The Diesel engine Hornsby engine Other types of heavy oil engines Petrol engines 33*-3& CHAPTER XVI TESTING OF INTERNAL COMBUSTION ENGINES Commercial tests Scientific tests Indicated and brake horse-power Fuel con- sumption Method of drawing up heat account and balance 339~353 CHAPTER XVII STEAM ENGINE AND BOILER TRIALS Commercial tests Scientific tests Indicated and brake horse-power Steam con- sumption Condition of steam at engine stop valve Method of drawing up heat account and balance Steam boiler trials Fuel consumption and calorific value Rate of water evaporation Temperature and analysis of flue gases Efficiency and heat account of a boiler 354-3^9 x CONTENTS CHAPTER XVIII VALVE DIAGRAMS AND VALVE GEARS Slide valves Piston displacement curve Rectangular valve diagram Reuleaux, Zeuner, Oval and Bilgram valve diagrams Choice of a valve diagram Analytical solution Earliest' cut-off possible with simple eccentric valve gear Meyer expansion valve gear Link motions Stephenson, Gooch, and Allan link motions Hackworth's radial valve gear Marshall's valve gear Joy's valve gear 370-409 CHAPTER XIX TWISTING MOMENT DIAGRAMS Twisting moment for any crank angle Inertia of reciprocating parts Method of drawing twisting moment diagram Inertia of the connecting-rod Comparison between exact and approximate effects Kinetic energy of the connecting-rod Graphical method for finding twisting moment exerted by the connecting-rod Function of the fly-wheel Cyclic variation of speed 410-433 CHAPTER XX BALANCING Centrifugal force Dynamical load on a shaft Method of balancing any number of rotating weights in one plane Method of balancing any number of rotating weights in more than one plane Primary balancing Balancing of locomotives Secondary balancing 434-451 CHAPTER XXI GOVERNORS Function of the governor Watt and Porter governors Modified Proell governor Stability and sensitiveness Hunting Spring-loaded governors, Hartnell, Hartung, Proell Curves of controlling force 452-472 ANSWERS TO EXAMPLES . . . . .'.'... 473~479 STEAM TABLES 480-481 TABLE OF GLAISHER'S FACTOR . .* 482 MATHEMATICAL TABLES 483-487 INDEX 489 INTRODUCTION UNITS SOME of the information given in this introduction will be known to all readers, while the whole of it may be already known to others ; it is placed here for convenient reference, and on beginning the study of the Theory of Heat Engines, the student will do well to be thoroughly con- versant with the various units used in both the British and the Metric systems. Units of Work. The British engineer's unit of work is the foot-pound, being the amount of work done when a force of one pound weight acts through a distance of one foot in its own direction. The unit of work in the metric system is the work done when a force of one kilogramme acts through a distance of one metre ; it is called the kilogramme-metre. Units of Heat. The British thermal unit (B.Th.U.) is the quantity of heat required to raise the temperature of one pound of water i F. It is numerically equal to about 778 foot-pounds. Another thermal unit which finds increasing favour with British engineers is the Centigrade heat unit (C.H.U.), being the quantity of heat required to raise the temperature of one pound of water i C. ; it is equal to 778 X i'8 or 1400 foot-pounds. Since the specific heat of water is not quite constant and equal to unity, it follows that the quantity of heat required to raise the temperature of one pound of water i F. or i C. is not the same at high as at low temperatures ; but the difference is so small that in practical calculations it may safely be neglected. Calorie. The gramme-calorie is the quantity of heat required to raise the temperature of one gramme of water i C. The kilo-calorie is the quantity of heat required to raise the temperature of one kilogramme of water i C. 252 gramme-calories are equivalent to one B.Th.U., and one kilo-calorie is equivalent to 3*96 B.Th.U. Power. The British engineer's unit of power is the horse-power (usually written H.P.), being the power expended when working at the rate of 550 foot-pounds per second, or 33,000 foot-pounds per minute. The French horse-power (force de cheval) is the power expended when working at the rate of 75 kilogramme-metres per second. The electrical engineer's unit of power is the watt, being the rate of working when one ampere flows under a pressure of one volt^ i.e. when work is being done at the rate of one joule per second. One British horse-power is equal to 746 watts. xii INTRODUCTION Kilowatt. The watt is an inconveniently small unit for practical purposes, hence the electrical engineer usually estimates power in kilowatts, one kilowatt being equal to 1000 watts. Board of Trade Unit. The Board of Trade unit of electric supply is one kilowatt-hour, being the quantity of work done in one hour when working at the rate of one kilowatt, or one thousand joules per second. Specific Heat. The usual definition of specific heat may be expressed as the ratio Quantity of heat required to raise the temp, of unit mass of a substance i Quantity of heat required to raise the temp, of unit mass of water i As defined above, the specific heat of a substance is a pure number, and it is immaterial in what units the quantity of heat is expressed. The specific heat of a substance is also frequently expressed as the number of B.Th.U., or foot-pounds, required to raise the temperature of one pound of the substance i F. Specific Heat of Gases. The specific heat of a gas is not a constant quantity (see Chap. XIV.) ; it varies with the temperature. The results of most experiments on the energy of gases have been expressed in the form of tables of formulae giving the specific heat (referred to unit mass of the gas as above) in terms of the temperature. It would appear preferable for most purposes to exhibit them in terms of the internal energy per unit of volume. This is the form most convenient for purposes of thermo- dynamic calculations, and it has the further advantage that it expresses the actual quantity measured ; and in most cases the lower limit of tempera- ture is near that of the room. The rate of change with temperature of the energy so determined is sometimes called the " true " or " instantaneous " specific heat, and sometimes the " thermal capacity " of the gas. The " Gaseous Explosions " Committee of the British Association l suggest that this quantity should be called the " volumetric heat," which, if adopted, should include in its significance that the measurement to which it relates should be made at constant volume and refer to unit volume of the gas. The term " specific heat " could then be restricted to its usual meaning, which refers to unit mass of the substance. They further recommend that the volumetric heat be referred to the gramme-molecule under standard conditions, which is nearly the same for all gases, namely 22*25 litres, and that the zero of temperature from which energy is reckoned be 1 00 C. in order that steam may be included on the same basis as other gases. The volumetric heat of a gas may also be expressed in foot-pounds per cubic foot by multiplying calories per gramme-molecule by 3*96, since One calorie per gramme-molecule = 3-96 foot-pounds per cubic foot. 1 Seethe First Report of this Committee, Section G, Dublin, 1908. THE THEORY OF HEAT ENGINES CHAPTER I THERMODYNAMICS AND PROPERTIES OF GASES i. The First Law of Thermodynamics: Keatand energy are mutually convertible, and Joule's - equivalent is the rate of exchange. The value of .this equivalent -(usually denoted by the letter J) is 778 foot-pounds are equivalent to one' bniisli .thermal unit, or 1400 foot-pounds are equal to one Centigrade heat unit. 1 The function of a heat engine is to convert heat energy into mechanical energy p , which operation is much more difficult to perform than to convert mechanical energy into heat energy. If a heat engine converted all the heat energy supplied to it into mechanical energy, it would convert H units of heat into JH units ,of mechanical energy, where J represents Joule's equivalent. Hence if W represents the number of units of work done we may write W = JH No engine, however, can convert all the heat it receives into work, as will be seen later, the maximum efficiency (Art. 51) ever reached being certainly not much greater than 45 per cent., the actual efficiency of the commercial engine being considerably less (see p. 346). The Working Fluid. In all heat engines the working fluid is either a gas or a vapour ; when a liquid is converted into the gaseous state it becomes a vapour, and, as such, possesses properties similar to those of a gas. The higher the temperature to which a vapour is heated the more closely does it approximate to a gas ; in fact, all gases are merely vapours at temperatures far above the boiling-point of the corresponding liquid. All vapours at ordinary temperatures can be condensed or liquefied by the application of pressure at constant temperature, but experiment shows that if a gas is above a certain temperature, it cannot be liquefied by pressure alone. This temperature, for any particular gas, is called the critical temperature of that gas. Hence, unless a gas is first cooled below its critical temperature it cannot be liquefied by pressure alone. The critical temperatures of oxygen, hydrogen, nitrogen, air, etc., are so very low, however, that at the working temperatures used in heat engines these gases may be considered as permanent gases. 2. The First and Second Laws of Permanent Gases. The first law (Boyle's Law) states that in a perfect gas the absolute pressure is 1 See Robinson's " Gas and Petroleum Engines," or any text-book on physics. i B 2 THE THEORY OF HEAT ENGINES [CHAP. i. inversely proportional to the volume when the temperature remains constant. If p denotes the pressure and v the volume, we may write or pv = a. constant. The second law (Law of Charles') says that under constant pressure, equal volumes of different gases expand equally for the same increment in temperature ; also, if a gas be heated under constant volume, equal increments of its pressure correspond to equal increments of temperature. For example 492 cubic feet of gas at 32 F. become 491 cubic feet at 31 F., 460 cubic feet at o F., and finally, if it follows this law, the gas will have no volume at 460 F. As a matter of fact, any actual gas would change its physical state before reaching so low a temperature. 3. Absolute Temperature. The absolute temperature of a sub- stance is its temperature reckoned from absolute zero. If ? fx=.*the ternp/ritiif pri the ordinary thermometer scale and T =*tne absolute 'temperature, then i ** vf * ' ^ ^ ^-f*' 460 'on the Fahrenheit scale, and "^ '' : " * "' T =7*4- 273 on the Centigrade scale. The absolute zero of temperature as obtained by the above reasoning (460 F. or 273 C.) corresponds very nearly to the absolute zero obtained from the purely thermodynamic considerations discussed in Art. 4, and the above values will be taken in making calculations which involve the absolute temperature. 4. Connection between the Pressure, Volume, and Tempe- rature of a Gas. Boyle's Law states that / cc - when the temperature remains constant ; Charles' Law states that / cc T when the volume remains constant. Combining the two laws, we have for a given weight of gas pvt* T or /z; = RT ...... (i) where p = absolute pressure in pounds per square foot. v = volume in cubic feet. T = absolute temperature (Fahrenheit or Centigrade). R = a constant depending on whether T is expressed on the Fahren- heit or on the Centigrade scale. For dry air the numerical value of the constant R is 53*18 when T is measured on the Fahrenheit scale ; it may be obtained as follows : Consider one pound of air at normal temperature and pressure (N.T.P.) : under these conditions the weight of one cubic foot of air is known to be 0*0807 pound, hence the volume of one pound is Q . g 07 or 12*391 cubic feet. Standard atmospheric pressure is 147 X 14*4 or 2116 pounds per square foot, and the normal temperature is 32 F. or 492 absolute, hence 2ii6 X 12-391 492 = 53 'i 8 foot-pounds per pound of air. ART. 5] THERMODYNAMICS AND PROPERTIES OF GASES 3 5. The Third Law of Permanent Gases : The specific heat at constant pressure is constant for any gas. Let C p = specific heat at constant pressure, and C v the specific heat at con- stant volume. Now at constant volume the gas does no external work when heated, hence, all the heat supplied is utilised in increasing its stock of internal energy ; when heated at constant pressure the gas expands and does external work equal to the pressure multiplied by the change in volume. Suppose i pound of gas to be heated at constant pressiire p from absolute temperature T x to absolute temperature T 2 , and let v be the volume of the gas at temperature T I} and v 2 tne volume at temperature T 2 , then- Heat taken in by the gas = C P (T 2 T x ) and Work done by the gas = p(v 2 z^) == R(T 2 T t ) from (i) Art. 4 Also Increase in internal energy = heat taken in work done (i) 6. The Fourth Law of Permanent Gases: When a per- fect gas expands without doing external work, and without taking in or giving out heat (and therefore without changing its stock of internal energy), its temperature does not change. The actual gases met with in practice are not perfect, and for all real gases this law is not perfectly true; for instance, Dr. Joule and Lord Kelvin found that air in expanding freely through a porous plug without doing work became cooled J C. for each atmo- sphere fall in pressure. From this law we see that whatever may be the change in the pressure and volume of a perfect gas when it expands under the above conditions, its store of internal energy remains unaltered, and hence the internal energy of a perfect gas depends only on its temperature. Suppose one pound of a perfect gas to be heated at constant volume from absolute temperature Tj to absolute temperature T 2 . Then Heat taken in = work done -j- increase in internal energy C0(T 2 TI) = o -J- increase in internal energy i.e. increase in internal energy = Ct?(T 2 Tj) ...... (i) Now in Art. 5 we saw that the increase in internal energy was equal to (C^ R)(T 2 Tj), hence we may say that the expression C V (T 2 Tj) represents the increase in internal energy, no matter how the pressure and volume may change during the process. 7. Relation between the Specific Heats C P and C v > From Art. 5, the increase in internal energy of one pound of perfect gas when heated at constant pressure from T x to T 2 is From Art. 6, the increase in internal energy of one pound of perfect gas when heated at constant volume is 4 THE THEORY OF HEAT ENGINES [CHAP. i. Equating these quantities we have (Cp - R) (T 2 - T,) = C,(T 2 - Tj) Cp R = C which may be written Q, C B = R ........ (i) or the difference between the specific heats is constant, and for dry air is equal to 53 'i 8 foot-pounds per pound of air. Equation (r) may also be written The ratio 7^ is very important, and is usually denoted by y, hence C w R v- I= c v or C *= For dry air Regnault found C p 0-2375 B.Th.U. per pound, which is equal to 0-2375 X 778 or 184-8 foot-pounds per pound. The same authority found Cv 0*1691 B.Th.U. per pound, which is equivalent to 0-1691 X 778 or 131*6 foot-pounds per pound, hence C p 0-2375 184-8 y or 7^= 2 or ^=1-404 Y C v 0*1691 I31'6 the value of y is not a constant quantity, however, because the specific heat at constant volume (C v ) varies with the temperature (see Chap. XIV.). 8. Work done by a Gas expanding according to Boyle's Law. Isothermal or Hyperbolic Expansion. Suppose the expansion takes place from the initial state /j, v, and T l5 to the final state / 2 > v i> anc ^ T 2 . The law of the expansion curve is pv = a constant = , say ; for a small change in volume 8v during which the average pressure is/ (Fig. i) the work clone 8W=/X 80 Let W denote the work done during the expansion, then w=y^'. Since pv = k = RT or W=/ 1 z; 1 log e ^ or RT lo ge -^ ' ' ' ' (l) which may be written, W =A^i lo ge r ....... (2) final volume where r is the ratio of expansion, U initial vo i ume ART. 9] THERMODYNAMICS AND PROPERTIES OF GASES 5 9. Work done by a Gas expanding according to the Law pv n = a Constant. As in Art. 8, let the expansion take place from the state /!, vi, and T I} to the state / 2 > ^2- an( ^ ^2> tnen Volume. FIG. i. The law of the expansion curve is pv n = a constant = /, say, k 7 r i = k\- .0t- Li n i n I - 7jl~n _ and since A^j =A? ; 2> equation (i) becomes i I (0 Also for a perfect gas/z> = RT, and (2) may be written TI To) N ^tnr (3) 6 THE THEORY OF HEAT ENGINES [CHAP. i. This expression for work done may be put in several forms, as follows : Since W = ^-^'l~ n - z'i~ n j we may write i ;/ w= Jto n i Again, p =/z (5) Pl Substituting (5) in (4), we have If all pressures are measured in pounds per square foot, and all volumes in cubic feet, the work done as calculated from either (2), (4), or (6), will be in foot-pounds. If all pressures are measured in kilograms per square metre, and all volumes in cubic metres, the work done as calculated from the above formulae will be in kilogram-metres. For most purposes, particularly when under examination, the student is advised to remember and use the simple form^-^ - _2_2. n i 10. Adiabatic Expansion and Compression. When a gas expands without gaining or losing heat, and does an amount of work equal to the difference between its initial and final internal energy, the expansion is said to be adiabatic ; conversely, if a gas is compressed without either heat being supplied to it or taken away from it, its change of internal energy being equal to the work done on it, the compression is said to be adiabatic. The value of " ;/" in the general/^" = a constant may be found as follows : When heat is supplied to a gas we have the fundamental relation discovered by Dr. Joule heat supplied = work done + increase in internal energy Let SH denote a small quantity of heat supplied, and ST and 8v the resulting small increments of temperature and volume, then If the expansion takes place without gain or loss of heat as in adiabatic operations, SH = o, hence = o ST p ART. 10] THERMODYNAMICS AND PROPERTIES OF GASES 7 hence in the limit : dT p Now for a perfect gas /^ = RT =(C P C r )T (Art. 7). Differentiating we have ( Try-i Substituting the value of -- from (i) in (2) gives -/ - or Integrating we have log/ = y log z; + a constant or log p + y log z/ = constant or pv y = constant. Hence the law of an adiabatic expansion or compression curve is /~ pv* = constant, where y=~. {^v Alternative Method. This result may also be found as follows : From (3) Art. 9, n i Now if the gas changes in temperature from T l to T 2 its internal energy is diminished by the amount C^Tj T 2 ) from Art. 6, -n "D or ^^ (Tj - T 2 ) since C v = ^^ (Art. 7 , equation (2)) Hence equating the loss of internal energy to the work done, the condition for adiabatic expansion is secured when _Ji_ / TI - T 2 ) = -^-(T! - T 2 ) y i v 1 n i v l that is when n=.y. Hence the expansion or compression will be adiabatic when v* = constant. 8 THE THEORY OF HEAT ENGINES [CHAP. i. ii. Relations between Pressure, Volume and Temperature during the Expansion or Compression of a Gas according to the Law pv n = Constant. NOW ) ^ = constant and since for a perfect gas pv = RT or -^ = R (a constant) 2 (2 \ ' Substituting (3) in (2) gives Substituting (5) in (2) gives Hence we may write the expression for work done, equation 6, Art. 9 (7) ;/ 12. Collection of the Formulae proved above, For isothermal or hyperbolic expansion, W =/!?! log r ......... (2) For adiabatic expansion, W = ART. n] THERMODYNAMICS AND PROPERTIES OF GASES 9 For adiabatic expansion, A ^2 1 ., (9) x/!/ If the expansion is not adiabatic but follows the general law/z; w = a constant, the equations (3) to (9) inclusive will hold if n is used instead of y. p v "^constant where n is /ess than /. p v= constant where n * /.. p v n ^constant where n is greater than I. Volume v. FIG. 2. 13. Effect of " n " on the Slope of the Expansion or Com- ssior = k, pression Curve. The slope of the curve is given by jK and since and yn+l From this expression we see that as " n " increases 4- increases also, the effect on the slope of the curve being as shown in Fig. 2. io THE THEORY OF HEAT ENGINES [CHAP. i. 14. Rate of Heat Reception or Rejection assuming Constant Specific Heats. Let " dp " be an indefinitely small change of pressure accompanying an indefinitely small change in volume " dv" Then Hh is the rate of change of pressure with respect to volume. Also if ** dB. " be the small quantity of heat given to the gas during the above small changes of pressure and volume, and dT the small change in temperature, will represent the rate at which the expanding gas receives heat per unit change in volume. Now for a perfect gas we have for any kind of expansion = R Differentiating we get But heat supplied = work done + increase in internal energy, 7TT , T I /"* JT uti = pdv -+ Lxv"- A Substituting in (2) the value G, = _ f , and the value of -^ from (i). we have If now the expansion takes place according to the general law pv n = we have dp kn = -- -T or kv~ n X nv-i But kv~ n p dp np Hence = -- J dv v Substituting this value of - in (3) we have ^ == ) 7^1 V ^ d?H y ?2 / v or W**'yZri - ; * ' <4) ART. 14] THERMODYNAMICS AND PROPERTIES OF GASES n Thus we see that for an adiabatic expansion or compression in which 7TT ft = y, ~j- = o. This result is obvious, since heat is neither received nor dv rejected during any adiabatic operation. Alternative Proof. This result may also be obtained in the following manner : Let H = the amount of heat supplied or rejected during the operation, and let the change of state be from/!, z^, T 1? to/ 2 , ?' 2 , T 2 . Then H = work done -f- change in internal energy, _P\ V \ A 7> 2_ _R //!#! _ AVv ni \ R R ) n i y i _ A y 2 y * i x y-i or H ^ y __ i X work done ......... (5) Hence i.e. when " n " is less than " y," the gas is receiving heat during the expansion ; but if the curve is steeper than the adiabatic, i.e. if " n " is greater than " y," the gas is losing heat during the expansion. If the expansion is isothermal (n = i) the heat received is equal to the thermal equivalent of the work done from (5). Similarly, if the gas is compressed isothermally, an amount of heat equal to the work done on the gas must be taken away from the gas. Also, if the gas is being com- pressed and " n " is less than " y," the rate of heat rejection will be positive, i.e. heat will be taken from the gas ; if, however, " n " is greater than " y " the rate of heat rejection will be negative, and the gas will be receiving heat during the compression. 12 THE THEORY OF HEAT ENGINES [CHAP. i. EXAMPLE i. The temperature of i pound of air is observed to fall from 600 F. to 300 F. while it expands adiabatically, doing 39,445 foot- pounds of work. Find Cv and Q>. Since the expansion is adiabatic, no heat is supplied to or taken away from the air during the process. Hence the work done will be equal to the loss of internal energy during the expansion, namely C v X (fall in temperature). Expressing both quantities in heat units we have loss of internal energy work done, r Now pr = 1*4 (Art. 7) \-SV hence C p = o'i6g X 1*4 = 0-237 B.Th.U. per pound. EXAMPLE 2. One cubic foot of gas at a pressure of 300 pounds per square inch absolute expands to 60 pounds per square inch absolute, the 1*2 law of expansion being pv = constant. Find the volume at the end of the expansion, and the work done during expansion. 1-2 1-2 1-2 ft-. 1-2 1*2 log Z> 2 log 5 from which z/ 2 = 3-823 cubic feet. P-\VI P<2^z 144(300 X i 60 X 3*823) Work done = - = "" r2 T = 50,820 foot-pounds. The work done might also be found without calculating v% by using equation (5), Art. 12, i.e. n i 300 X 144 X if A ~^2 I 1 Hi i == 50,820 foot-pounds as before. EXAMPLE 3. One pound of dry air (volume 12-39 cubic feet) at atmospheric pressure requires compressing to a pressure of 200 pounds per square inch absolute : which will be the more economical, to compress isothermally or adiabatically ? ART. 14] THERMODYNAMICS AND PROPERTIES OF GASES 13 ( i ) Isothermal compression. 14-7 X 12 39 ~ = ' J cublc feet - Work done on the gas = Vn) = r 44 X 147 X 12-39 log e = 144 X 147 X 12-39 log e 13-6 = 2116 X 12-39 X 2-607 = 68,320 foot-pounds. (2) Adiabatic compression. Using (5), Art. 12, g= I'4 I 0-4 = 72,630 foot-pounds. The negative sign simply means that work is done on the gas. Hence, by compressing isothermally we save 72,630 68,230 = 4,400 foot-pounds. EXAMPLE 4. Ten cubic feet of air at 90 Ibs. per square inch abs. and at 65 F. are expanded to four times the original volume, the law of expan- sion being /z/ 12 = const. Given <,= 130-3 ft.-lbs. per lb., and C p = 183*4 ft.-lbs. per lb. : find (1) The temperature of air at the end of expansion. (2) The work done in ft.-lbs. (3) The amount of heat which must have been given by, or been rejected to, an external source during the cycle. T 2 = 0-707^ = 0-707(65 + 460) = 0-707 X 525 = 371*2 absolute F. = 371*2 460 = 88*8 F. = temperature of air at the end of expansion. = 90 X (o'7o7) 5 = 15-84 Ibs. persq. in. .-. Work done W = A"i-/^ n i 90 X 10 15*84 X 40 = 144 * - 0-25 = 576 X (900 633*6)=576x 266-4=153,446 ft.-lbs. (3) From (5), Art. 14. Heat given H = W X = 57,542 ft.-lbs. = 73-9 B.Th.U. 14 THE THEORY OF HEAT ENGINES [CHAP. i. EXAMPLE 5. Air is drawn into a cylinder and compressed adiabati- cally to a pressure of 75 Ibs. above its original pressure (15 Ibs. per sq. in. abs.), and is then expelled at this pressure into a receiver; its original temperature was 60 F. In the receiver the compressed air cools down to its original temperature, and, in order to maintain a uniform pressure in the receiver, an equal weight of compressed air is constantly drawn off and expanded isothermally in a working cylinder down to 15 Ibs. pressure. Calculate (a) the work spent per Ib. of air in the compressor ; (b) the work done per Ib. of air in expanding; (V) the temperature of the air as it enters the receiver. (a) Now /i^i =/2 z '2 P er cubic foot at 60 F. / 2 9 .*. i*4logz' 2 1*4 log z/!^ log 15 log 90 /. 1-4 log z> 2 = 1*1761 i'9542 = 0-7781 0-7781 .-. log v 2 = -- j~ = 0-5559 = 1-4441 /. # 2 = 0-2781 cubic ft. Work done per cycle =f> 2 v 2 p&i 4*7 - = 3*5(/2^2 A^i 1 4 I = 3-5 X 144(90 X 0-2781 15) = 3'5 X 144(25-02915) = 3-5 X 144 X 10-029 = 5065 ft-lbs. Taking the volume of i Ib. of air at 60 F. = 13 cubic ft., we have Work done per Ib. of air == 5065 X 13 = 65,845 ft.-lbs. (b) Work done per cubic foot in expanding =/? log t r>> log, 55 = 15 X 144 X 2-303 X 0-7782 = 3870 ft.-lbs. .*. Work done per Ib. = 3870 X 13 = 50,310 ft.-lbs. '" And ^ = 60 + 460= 520 absolute .-. T 2 = 52ox 67 log T 2 = log 520 + | log 6 = 2-7160 + 1 X 07782 = 2*7160 -}- 0*2223 .-. log T 2 = 2*9383 .*. T 2 = 867*6 abs. = 867*6 460 = 407-6 F. EXAMPLE 6. 12*39 cubic feet of air at atmospheric pressure are com- pressed to a pressure of 200 pounds per square inch absolute. Find the quantity of heat which must be added to or taken away from the air during the operation (a) when the compression is isothermal ; (b) when the com- pression is according to the law pv = constant ART. is] THERMODYNAMICS AND PROPERTIES OF GASES 15 () From Example 3 the work done on the gas during compression is equal to 68,320 foot-pounds. Hence the heat taken away from the gas during compression is equal to 68,320 foot-pounds, or ' Wo* done = = i47 X 144 X i2-3 9 C /2oo ^~~\ 1-2 i ( \i47/ ) = 131,086(1 1-545} = 71,440 foot-pounds. Hence the work done on the gas during the compression is 71,440 foot-pounds. From (5), Art. 14, the heat taken away from the gas during the compression is y n H = _ X work done 1*4 1-2 = 1-4-! -< 71,440 35>7 20 foot-pounds or H 1400 15. Entropy ; Definition. The entropy (denoted by ) of a sub- stance is that thermal property of it which remains constant when the substance neither gains nor loses heat from external sources, as in adi- abatic operations, and which is increased or reduced when heat is given to or taken from the gas. Consequently adiabatic operations are often called isentropiC) i.e. operations at constant entropy. When a pound of any substance takes in or gives out a quantity of heat H at the absolute tem- perature T, its gain or loss of entropy is measured by the quantity T~ so that the quantity of heat H = T X change in entropy, i.e. H = TX<. If the temperature is not constant we may define the change of entropy FT by the general expression ^-^r , each element 8H of heat supplied or rejected being divided by the absolute temperature the substance had at 16 THE THEORY OF HEAT ENGINES [CHAP, i the time. For an adiabatic, SH = o, since the change of heat is nothing, i.e. the substance expands or is compressed without gain or loss of heat (Art. 10). Consider the indicator diagram shown in Fig. 3. The area of this diagram represents the work done to some scale. The temperature entropy diagram is shown alongside. During the isothermal expansion DA on the p.v. diagram, the change of entropy is represented by da on the T0 diagram; during the adiabatic or isentropic expansion AB, the entropy is constant, and is represented by ab\ similarly, during the isothermal compression BC the change of entropy is &, whilst the adiabatic or isentro- pic compression CD is represented by cd. The diagram " dabc" is called the " Temperature-Entropy " diagram and corresponds to the indicator or " Pressure-Volume " diagram DABC. isothermal 1 Entropy FIG. 3. Again , when a quantity of heat H passes from one body at absolute temperature TI to another body at a lower temperature T 2 , the warm body TT TJ loses entropy ^r and the colder body gains entropy -. Now, since T 2 1] 12 is less Tj the gain of entropy of the system as a whole is _ t-, T 2 T!~ T!T 2 If one pound of any substance gains heat by the small amount SH the temperature remaining constant at T during this small change, its gain of entropy as we have already seen will be or in the limit ART. 1 6] THERMODYNAMICS AND PROPERTIES OF GASES 17 If now the pound of substance is raised in temperature from T 2 to Tj the total gain of entropy will be given by where C = specific heat of the substance ; P'lC^T T! JT 2 -1- ==C1Q ^T 2 ...... (0 It should be noted that entropy units are the same whether H and T are both measured on the Fahrenheit or on the Centigrade scales. 16. General Expression for the Change of Entropy of a Perfect Gas when passing from the State p b v l5 T^ to the State p 2 , V 2 , T 2 . Since the energy equation for a perfect gas is given by H = C, (T! - T 2 ) +p (v 2 - Vl ) (Art. 10) we have for a small change 8H = C.8T +/80 ....... (i) Dividing both sides of the equation by T we have SH ST /, Tp~ - V^WTrT T rnOV or in the limit Now ^ = Substituting (3) in (2) we have 2 . . (2) Integrating we have f^ 2 ^/H f^z dT f vz dv r 1 I _i T? i rrV V_/7J / , ,, 1 XV I J TJ r 7 T : i y .,! z/ or gain of entropy ^ 2 ^i = ^w log 2 + R log .... (4 ) Li e 7^ Now R = C^ C v , hence Cplog, ..... (6) Also since Substituting (7) in (6) we have , . p< v% . 2 (pi = Cy log r~ C^> log ! . (8^ c i8 THE THEORY OF HEAT ENGINES [CHAP. i. Another expression can be found for the change of entropy as follows : Substituting C v = C p R in (4) we have 2 - k. = (Q. - R) iog. + R log. (9) Hence in calculating; the change of entropy of a perfect g as when passin g from the state /j, v lt 1\, to the state / 2 > ^2> ^2> we may use either (4), (8), or (9), i.e. If British units are used, R, C p and C v should all be expressed either in B.Th.U. or in C.H.U., v% and v l in cubic feet, / 2 an d Pi m pounds per square foot ; the change of entropy will then be measured in " units of entropy." No name has been universally accepted for the unit of entropy, but Professor Perry has proposed to measure entropy in Ranks. In the case of an isothermal change of state from /j, z'j, to / 2 > z; 2 Tj = T 2 , and equation (4) becomes \ v ) loge . . (lo) If the change takes place at constant volume, v z = v and (4) becomes T 2 92 > Art " ' z t>l VI 2' and R = C (y i) from (2), Art. 7 hence EXAMPLE. Find the change in entropy when one pound of air at 32 F. and atmospheric pressure changes in volume to 2 cubic feet with a temperature of 539 F. given C p = 0-2375 and C v = 0-1691. We may use either of the equations (4), (8), or (9), Art. 16. Here /i = 147 X 144 = 2116 pounds per square foot T! = 32 -j- 461 = 493 absolute T 2 = 539 H- 461 = 1000 absolute #2 = 2 cubic feet. Using equation (4), Art. 16, we get. R = C p C v = 0-2375 0-1691 = 0*0684 or 53-18 foot-pounds , , 1000 . -_ = 0-1691 log e - + 0-0684 = 0*1196 0-1247 = 0*0051 rank. 20 THE THEORY OF HEAT ENGINES [CHAP. i. Using equation (8), Art. 16, we get = RT 2 = _2 53_1: * iQoo __ 26,590 pounds per square foot = 0*1691 log = 0*4280 0*4331 = Also by equation (9) Art. (16) f 0-2375 log e 0*0051 rank. , 1000 = 0-2375 log - = 0*1680 0*1731 = 0*0051 rank. The gain of entropy (i) being negative means that during the change of state the air loses entropy by the amount 0-0051 rank. 17. Work done by an Expanding Gas from Consideration of the Temperature Entropy Diagram. Let 1\ be the initial temperature of the gas and T 2 the final temperature after ex- pansion, and let AB (Fig. 4) re- present the temperature-entropy curve for the expansion. Further, suppose SH is a small quantity of heat supplied at any tempera- ture T, and giving rise to a small change of entropy 8(f>. Then && =?5 JEittropy ft. FIG. 4. The total area under the curve AB is = ohL or Consider the elementary strip of the diagram whose width is S and height T, then the area of this strip is TS<, and from the above this represents the small quantity of SH supplied. the total heat supplied during the expansion. Similarly if the gas is compressed from temperature T 2 to temperature T]_ it will reject an amount of heat equal to the above. Suppose now the gas undergoes a complete cycle of changes so that its temperature, pressure, and volume are the same at the end as at the beginning of the cycle. The temperature-entropy diagram will then form a closed figure as in the case given in Fig. 3. ART. 17] THERMODYNAMICS AND PROPERTIES OF GASES 21 Let Hj be the amount of heat supplied to the gas, and H 2 the amount rejected by the gas during the cycle, then, in order to get the area of the closed figure we must integrate over the whole cycle, and we obtain JT^ == Hj H 2 Now H! H 2 is equal to the heat converted into work, hence we see that the area of the temperature-entropy diagram represents to some scale the work done in a complete cycle. EXAMPLES I 1. Find the volume of 3 pounds of air when at a pressure of 70 pounds per square inch absolute and at a temperature of 75 F. [Take C p = 183*4 foot-pounds and C,, = 130*2 foot-pounds.] 2. If I pound of air at 32 F. has its volume doubled at constant atmospheric pressure, what is its final temperature ? How much external work is done during the expansion and how much heat must be supplied during the expansion? [Cp = 0*2375.] 3. A cylinder contains 0*5 cubic foot of gas at 15 pounds per square inch absolute. Find the work expended in compressing it to a pressure of 90 pounds per square inch absolute, the law of compression being pv r constant. 4. The area of an engine piston is 100 square inches. If the length of cylinder occupied by gas is 18 inches when the pressure is 120 pounds per square inch absolute, find the work done by the gas in driving the piston through a distance of 2 feet. Take the law of expansion as/z/ 1 ' 5 = constant. 5. Find the work done by the gas in Question 4, if the gas is kept at constant temperature during the expansion. 6. In a gas engine cylinder 5 cubic feet of gas and air at 14*7 pounds per square inch absolute are compressed into a clearance space of I cubic foot. If the compression is adiaba'ic, (a) what is the pressure at the end of the compression stroke? and (I)) how many foot-pounds of work must be expended in the compression of the charge ? [7= i -4-1 7. If 0*1 pound of gas occupying 0*5 cubic foot is expanded in a cylinder at constant pressure of 150 pounds per square inch absolute until its volume is I cubic foot, and is then expanded adiabatically to 5 cubic feet, find the temperature of the gas, (a) at the end of the constant pressure stage, (b] at the end of the adiabatic expansion, and calculate the heat expended and the work done during each portion of the process. Take Cp 198 foot-pounds and C v = 144 foot-pounds. [L. U.] 8. The temperature of the mixture of gas and air in a gas engine at the end of the admission stroke is 90 F. and the pressure 15 pounds per square inch absolute. The clearance volume is 4*6 cubic feet, and the total volume of clearance plus piston dis- placement is 12 cubic feet. Assuming adiabatic compression^ 1 ** = constant, determine the temperature at the end of the compression stroke. If the pressure after ignition is 240 pounds per square inch, find the temperature in the cylinder. [L. U.] 9. In a certain oil engine the piston displacement is 0*395 cubic foot and the volume of the clearance space 0*210 cubic foot, and the pressure of the charge at the instant compression begins is 13 pounds per square inch absolute. Find the compression pressure and the temperature reached at the end of the compression stroke if the temperature of the charge at the instant compression began was 264 F. Assume the law of compression to be/z/ 1 ' 39 = constant. [L. U.] 10. If 20 cubic feet of dry air are compressed adiabatically from 15 pounds per square inch absolute and 60 F. to 225 pounds per square inch absolute, find the temperature after the compression and the work expended. [Take 7 = 1*4.] 11. If 13 cubic feet of air at 60 F. and 200 pounds per square inch absolute are expanded to a pressure of 25 pounds per square inch absolute, calculate the final volume and the work done during the expansion (a) if the expansion is isothermal, (b] if the expansion is adiabatic. 12. The law of the expansion curve of a gas engine indicator diagram is found to be pv 1 '* 1 = constant. Assuming =? = 1-37 find the rate of heat reception ^-. If the 22 THE THEORY OF HEAT ENGINES [CHAP. i. law of the compression curve is ^z/ 1 ' 24 = constant, what is the rate of heat reception during compression ? If the piston speed is 600 feet per minute when the pressure on the expansion curve is 150 pounds per square inch absolute, what is the rate of heat reception per second at this instant ? 13. Air at 15 pounds per square inch absolute is drawn into a cylinder and com- pressed adiabatically to 135 pounds per square inch absolute, and is then expelled at this pressure into a receiver ; its original temperature was 50 F. In the receiver the compressed air cools to 50 F., and in order to maintain a uniform pressure in the receiver an equal weight of compressed air is drawn off constantly and expanded isother- mally down to 15 pounds per square inch absolute. Calculate (a) the work spent per cubic foot of air in the compressor ; (b) the work done per cubic foot of air in expanding ; (c) the temperature ot the air as it enters the receiver. 14. If i pound of air occupying 3 cubic feet at 15,950 pounds per square foot, and absolute temperature 900 F., expands at constant temperature to a volume of 12 cubic feet, find its pressure after expansion, the heat taken in, and the gain in entropy. 15. If 42*46 cubic feet of air at pressure 676 pounds per square foot and absolute temperature 539 F., be compressed isothennally to volume IO'62 cubic feet, what is its pressure after compression, the work done on it, the heat taken from it, and the loss of entropy ? 16. Ten cubic feet of air at 65 F. and 90 pounds per square inch absolute are expanded to 4 times the original volume, the law of the expansion curve being /z/ 1 ' 25 = constant. Given C p = 130-2 foot-pounds, find the change of entropy. [Take 7 = I '4-] CHAPTER II HOT-AIR ENGINES 18. Classification and General Remarks. Hot-air engines may be divided into two classes : (i) external and (2) internal combustion engines. In the first class the working substance is atmospheric air which receives heat from an external furnace by conduction through the containing vessel or heater, in the same way that steam in a steam boiler receives heat through the plates or tubes of the boiler. In the second class the working substance is the products of combustion (a mixture of gases) of fuel, maybe solid, liquid, or gaseous, which takes place within the engine itself. In its widest sense, then, the second class includes both gas and oil engines, which are however dealt with separately in Chapters XIII. and XV. Compared with a steam engine using saturated steam, an air engine has the advantage that the temperature and pressure of the working substance are independent of one another. In any heat engine which uses a saturated vapour as the working substance, the upper temperature limit must be, for mechanical reasons, comparatively low, in consequence of the exceedingly high pressures which accompany very high temperatures. In an air engine, however, it is possible to use an upper temperature limit greatly in excess of that permissible in a steam engine, and it will easily be seen that if the lower temperature limit is not raised, an increase of thermodynamic efficiency results, since the efficiency of a heat engine depends upon the working range of temperature (Art. 21). Although by using superheated steam in a steam engine we may raise the upper temperature limit, yet the steam continues to take in the greater part of its heat at the comparatively low temperature of evaporation in the boiler, and in consequence the thermodynamic efficiency is correspondingly reduced, since for maximum efficiency a heat engine must take in all its heat at the highest temperature and reject all at the lowest temperature (Art. 24). In the case of the external combustion air engine, there must be a considerable drop in temperature between the temperature of combustion in the furnace and the temperature of the air in the containing vessel. This drop in temperature is, of course, essential in order to have rapid conduction of heat through the walls of the heater, but it results in a lower efficiency. Internal combustion engines, therefore, have the advantage that the temperature of combustion in the engine itself is the upper limit in the thermodynamic cycle. The drawback to the use of the external combustion engine is the large heating surface of metal which is required to supply the air with 23 THE THEORY OF HEAT ENGINES [CHAP. n. Volume K. FIG. 5. heat, and this large surface, always kept at a very high temperature, soon burns out. 19. Graphic Representation of the Work done during the Change in Volume of a Fluid. Let Fig 5 represent the pressure- volume diagram, in which the ordinates represent the pressure of the working fluid, and the abscissae the volume. The distance OE represents the initial volume of the working fluid, and EA the initial pressure. The fluid expands along any curve such as ABC to a volume OF and pressure FC, the area under this curve representing to some scale the work done by the fluid during the expansion. Let compression then take place along the curve CDA, during which work is done on the fluid repre- sented to the same scale by the area under this curve. The shaded area, therefore, represents the difference of these quantities, which is equal to the net amount of work done by the fluid during the process. Such a diagram is called the pressure-volume or indicator diagram. In general terms Heat taken in = work done -f- heat rejected Area EABCF = shaded area ABCD + area EADCF 20. Cycles of Operation. The working substance receives heat from the heat source, expands in a cylinder doing useful work, and then rejects the heat which is not used to a cold body or condenser. The working substance then receives heat again, and the same cycle of opera- tions is performed, so that after each cycle the substance returns to the same state of pressure, volume, and temperature as it started from ; in other words, it passes through a closed cycle. There are thus three essential organs of a heat engine: (i) a hot body; (2) a cold body; (3) the working fluid. 21. Carnot's Cycle. Carnot, in 1824, showed that the amount of heat which could be converted into mechanical work by an ideal perfect heat engine using a perfect gas as the working fluid, depended solely upon the working range of temperature. Since no gas is perfect, and in practice a perfect engine cannot be constructed, the Carnot cycle affords a ready means of comparing the actual performance of any engine, with the best theoretical performance possible under the same working range of tem- perature. Fig. 6 shows at (a) the pressure-volume or indicator diagram of an engine working on this cycle. Starting at point a the gas expands isother- mally at constant temperature T 1? heat being supplied to keep the tem- perature constant, from volume v a to volume v b . The supply of heat is then shut off and the gas expands adiabatically to volume v e the temperature ART. 2l] HOT-AIR ENGINES falling to T 2 . On the return stroke of the piston the gas is compressed isothermally from volume v c to volume v& at constant temperature T 2 , heat being taken from it to keep the temperature constant, the cycle being completed by compressing adiabatically from volume 'v& and temperature T 2 to volume v a and temperature T x . A closed cycle is thus obtained, the pressure, volume, and temperatures of the gas being the same at the end as at the beginning of the cycle. \e Volume v. EnZropu 0. fb) FIG. 6. To find the Proper Place to stop the Isothermal Compression. The point d must be so chosen that an adiabatic drawn through it will pass through the starting point a. By Art. 1 1 we have, by equation (4) for the cooling during the adiabatic expansion stage from b to c. Also for the heating during the adiabatic compression stage from d to a we have Hence and therefore ^- /z Vb) \v _ ~ This shows that the point d must be so chosen ti\tii the ratio of isothermal compression is the same as the ratio of isothermal expansion. 26 THE THEORY OF HEAT ENGINES [CHAP. n. Efficiency of the Cycle. Let r denote the ratio of isothermal expansion or compression, then we have during the stage ab Heat taken in = RTj log e r = work done by the gas be No heat taken in or rejected cd Heat rejected = RT 2 Iog 6 r = work done on the gas da No heat taken in or rejected Now efficiency ^ heat converted into mechanical work heat supplied _ RTj log, r-RTj lo ge r RTi lo ge r Derivation of this Result from the Temperature-Entropy Diagram. The temperature-entropy diagram is shown in Fig. 6, at (b], since the area of the temperature-entropy diagram represents the work done (Art. 17). The work done by the gas during the stage ab is represented by area fabe = 0T X . The work done on the gas during the stage cd is represented by area fdce = the effect of those parts in causing work to be performed is equal. The first law of thermodynamics (Art. i) might lead one to think that an engine can convert all the heat it receives into useful work, but in con- FIG. 6A. sequence of the second law only a fraction of the heat supplied can be so converted (Art. 21). The ratio heat converted into useful work heat taken in by the engine is always less than unity, and is called the efficiency of the engine con- sidered as a heat engine. 26. Rankine's Statement of the Second Law represented Graphically. In Fig. 6A let AjBjCiDi ; A 2 B 2 C 2 D 2 ; A 3 B 3 C 3 D 3 , etc., represent a series of isothermal curves for temperatures, T 1} T 2 , T 3 , etc , such that 1\ T 2 = T 2 T 3 = T 3 T 4 = ST, and so on, there being equal intervals of temperature ST between successive isothermals, and let a series of adiabatic curves A 1 A 2 A 3 A 4 ; B 1 B 2 B 3 B 4 ; C 1 C 2 C 3 C 4 , etc., be drawn to cut these isothermals in such a manner that the areas A 1 ^ 1 B 2 A 2 ; A 2 B 2 B 3 A 3 ; BiQCgBa, etc., are all equal. Consider the portion of the diagram A 1 B 1 B 2 A 2 . This represents an engine indicator diagram working on the Carnot Cycle (Art. 21). ART . 2 6] HOT-AIR ENGINES 29 Let Q ^ quantity of heat supplied during the isothermal expansion Then work done = area of A^BjjAa = Q X ^^^ (see Art. 21). T -T and heat rejected = Q Q X L ~ 1 ' 4 = constant, find the mean pressure for the cycle. A sketch of the indicator diagram is shown in Fig. 14, the work done per cycle being represented by the shaded area. Let the volume at a be i cubic foot, then the volume at c and d is 15*3 cubic feet, and / a =/ d x \~Y) = *4 X (i5'3) 1<4 = 6 37'8 pounds per square inch. Since the i cubic feet, and Since the ratio of expansion is 7-5, the volume at b is -7"^== 2-04 ' 4 /' i V 4 = 637*8 X ( ) = 38 pounds per square inch Now work done during the cycle = shaded area = area/^ + area gbce area/htd? = 144 X 637-8 X 1-04 + r * 4 ^ t (637*8 X 2-0438 X i5'3) - p^T(637-3 X i-i4X 15-3) = 95>47 2 + 259,200 152,496 = 202,176 foot-pounds. THE THEORY OF HEAT ENGINES [CHAP. n. work done in foot-pounds Mean pressure = \ -, : ^-. 7 stroke volume in cubic feet 202,176 = 14,138 pounds per square foot = 98*1 pounds per square inch. f 3 EXAMPLE 5. If in Example 4 the temperature at the end of the suction stroke is 60 F., estimate the efficiency. [^ = 0-2375, C v = 0-1691.] f-i = 520 X (i5'3) ' 4 = 1548 absolute And for a perfect gas Also T " T 6 Vb . /. 1^= laX since /o=/6 = 1548 X ~= 3158 absolute. Tfc = T c ; 0-4 = 3,158 X I -f- ) = 1410 absolute. ART. 29] HOT-AIR ENGINES 39 Heat supplied per pound of air = 0-2375(3158- 1548) = 382-37 B.Th.U. Heat rejected per pound of air = C,(T C - T d ) = 0-1691(1,410 520) = 150*5 B.Th.U. _ heat supplied heat rejected Efficiency = - *- - ^5 = - heat supplied _ 382-37 i5 '5 382-37 = 0*606 or 60' 6 per cent. The engine working on this cycle has been developed, not as an air engine, but as the Diesel oil engine (see Art. 194). EXAMPLES II 1. In a Stirling air engine working between temperatures of 700 F. and 80 F., the ratio of isothermal expansion is 2. Calculate the ideal efficiency when (a) the engine is fitted with a perfect regenerator ; (l>) when the efficiency of the regenerator is O'9. Take C p = 0-2375 and C v = 0-1691. 2. Compare the efficiencies of: (a) A Stirling engine with perfect regenerator in which the maximum pressure is 140 Ibs. per square inch absolute and minimum pressure 15 Ibs. per square inch absolute, and limits of temperature 75 E. and 7 F. ; and (b) a perfectly reversible steam engine working between the same limits of pressure. 3. If a perfect air-engine works on the Carnot cycle between limits of temperature 780 F. and 50 F., estimate its efficiency at the ratio of adiabatic expansion. 4. In a Joule air engine the maximum temperature is 600 F. and the initial pressure 1 80 Ibs. per square inch. If the ratio of expansion be 3, calculate its efficiency. 5. What will be the efficiency of the engine in question (i) if no regenerator is fitted? 6. A Stirling engine, with perfect regenerator, works between pressures of 135 pounds per square inch absolute, and 15 pounds per square inch absolute, and temperatures 550 F. and 50 F. respectively. Calculate the mean effective pressure on the piston. 7. An air engine works on an ideal cycle in which heat is received at constant pressure, and rejected at constant volume. The pressure at the end of the suction stroke is 14 Ibs. per square inch absolute, and temperature 40 C. The ratio of compression is 13 and the ratio of expansion 6-5. If the expansion and compression curves are adiabatic pv\"i = constant, find the mean pressure for the cycle and its efficiency. 8. An ideal air engine works on the following cycle : air is taken in at atmospheric pressure 147 pounds per square inch and at temperature 55 F., and is then compressed adiabatically so that at the end of the stroke the pressure is 550 pounds per square inch absolute. Heat is then taken in at constant pressure and then the air expands adiabatically, the ratio of expansion being 5. The air is exhausted at the end of the expansion stroke, the heat being rejected at constant volume. Estimate the efficiency. (Take C p = 0-2375 and Ct, = 0-1691.) CHAPTER III PROPERTIES OF STEAM 30. Generation of Steam under Constant Pressure. This is the process which takes place in a steam boiler when its engine is working, the steam being withdrawn at the same rate as it is generated. Suppose we have i pound of water at a temperature of / F., and occupying a volume of V w cubic feet, to which heat is applied, then the following operations will result : (1) The temperature of the water will rise to some temperature f F., at which steam will commence to form. The value of t will depend upon the pressure (which remains constant throughout) to which the water is subjected. The quantity of heat so supplied is called the sensible heat. If the specific heat of water be assumed constant and equal to unity the sensible heat added will be /-/ B.Th.U. The specific heat, however, is not quite constant, 1 but the variation is so small that for practical purposes it may be neglected, at any rate the specific heat of water may be assumed equal to unity for approximate calculations. (2) If the application of heat be continued, steam will be formed at a constant temperature f F. until all the water has been evaporated. During this stage when the steam is formed in contact with the water the steam is said to be saturated. When all the original pound of water has been turned into steam, the heat supplied during evaporation is called the latent heat of the steam at the particular temperature f, the steam being called dry saturated steam. (3) If more heat be added to the dry saturated steam its temperature will rise to, say, t F. The quantity of heat so added is sensible heat, and will be approximately equal to 'i - t) B.Th.U. where C P is the mean specific heat of the steam at constant pressure. Steam may therefore exist either as saturated steam, which may be either dry or wet, or as superheated steam. The temperature t corre- sponding to the particular pressure at which the steam is formed is known as the temperature of saturation. Superheated steam which is well above the temperature of saturation approaches in properties to a perfect gas, and was called by Rankine steam gas. 1 See " Steam Tables," by Marks and Davis, Longmans, Green & Co. 40 ART. 31] PROPERTIES OF STEAM 31. Relation between Pressure and Temperature of Satu- rated Steam. There is no simple law connecting pressure and tempera- ture. The classical experiments on this subject were carried out by Regnault in the year 1847, as a result of which he gave the law , ......... (0 where 147 p = absolute pressure in pounds per square inch. and t = temperature in degrees Fahrenheit. Rankine expressed the relation by the formula 39^,945 log / = 6-1007 --!g--* v y"* (2) where p = absolute pressure in pounds per square inch, and T = absolute temperature on the Fahrenheit scale. In 1908, Thiesen expressed the relation by the formula (/-J-459'6) log ^ =5'409(/ 212) 3*71 X io~ 10 {(689 / t ) 4 477 4 } (3) 1470 where p = absolute pressure in pounds per square inch, and t = temperature in ordinary Fahrenheit degrees. 32. Relation between Pressure and Volume of Dry Satu- rated Steam. The volume of i pound of dry saturated steam at any assigned pressure is a quantity known as the specific volume, and is extremely difficult to measure by direct experi- ment. It may, however, be calculated as follows : Consider a perfect heat engine using steam as the working fluid , between limits of temperature T! j and T 2 , and working on the Carnot \ cycle. The indicator diagram is \ shown in Fig. 15. The heat taken in ^ per pound of steam is its latent heat L, and since the efficiency is 1 ~" % (Art. 21) the area of the diagram, or the work done per pound will be L. B.Th.U., or JL T Volume. FIG. 15. - foot-pounds Suppose now, that the temperature range is very small, say from T to T ST, the fall in temperature being a small amount 8T, and the corre- sponding fall in pressure being a small amount /. The indicator diagram will now be represented by the narrow strip dbdc in which length of diagram ab = V s V w , and height of diagram bd = S/ 42 THE THEORY OF HEAT ENGINES [CHAP. in. Hence the area abdc on the work done will be 8XV,-V tf ) ......... (i) But by (i) the work done is also equal to TL^ Ji, T equating (i) and (2) we get 8J(V S -V W ) = ]L 8 ^ V.-V. + tt" ....... (3) This is approximately correct when ST, and therefore S/, are small finite quantities. If now 8T be diminished indefinitely the limiting value of (3) becomes in which V s = volume of i pound of dry saturated steam in cubic feet at absolute temperature T. V w = volume of i pound of water in cubic feet. x/T* The value of -rr is obtained from the temperature-pressure curve, being the tangent to the curve at the particular temperature and pressure under consideration. The student will find it instructive to plot the temperature- pressure curve from the steam tables given on page 480, and from that curve work out the volume of i pound of dry saturated steam at a few different pressures, comparing his calculated volumes with the volumes tabulated there. Equation (4) may also be used to find the change in the freezing-point of water due to change in pressure. The following example will illustrate this It is known that i pound of water at 32 F. changes in volume from 0*016 cubic foot to 0*0174 cubic foot on solidifying, and gives out its latent heat 144 B.Th.U., and from (4) dT .,, T where V = volume of i pound of ice in cubic feet at 32 F., or 493 absolute and w = water dT (0*0174 0*016)40-? , .'. = 2 _Zzz!> = 0*0000065 dp 778 X 144 Hence at atmospheric pressure when dp = 2116 pounds per square foot dT = 2116 X 0*0000065 = 0*0135 F. i.e. the freezing-point of water changes 0*0135 F. for every atmosphere change in pressure, so that at a pressure of 1 1 atmospheres the freezing- point would be 32*135 F. ART. 34] PROPERTIES OF STEAM 43 33. Total Heat of Evaporation of Saturated Steam. The total heat H of i pound of saturated steam is always reckoned from 32 F. For stages (i) and (2), Art. 30, we have total heat H = sensible heat + latent heat H = /4 + L (i) For steam at a temperature f F. this becomes approximately H = (/_ 32 ) + L The values of H found by Regnault may be approximately expressed by the equation H=io8 2 +o'3/B.Th.U (2) A more accurate value for the total heat of evaporation of dry saturated steam is H = 1150-3 + 0-3745 (* 212)- 0-00055 (/ 212)2 B.Th.U.i (3) If the temperature is measured in C., the value of H according to Regnault may be approximately written H = 607 + o-3/ C.H.U (4) The above equations only refer to dry saturated steam. If the steam is wet, the quantity of heat supplied during the formation of the steam will not be equal to the latent heat L because all the water will not have been evaporated. Suppose i pound of the wet steam contains x pound of steam, the remainder (i x) being water mechanically suspended in it; then x is called the dry ness fraction of the steam, and H' = ^ + *L (5) 34. External Work done during Evaporation at Constant Pressure. Let P be the absolute pressure, in pounds per square foot, at which the steam is generated, the volume will change during evaporation from that of i pound of water (V w ) to that of i pound of dry saturated steam (V s ), and the external work done will be Pressure (pounds per square foot) X change in volume (cubic feet) i.e. E = P (V s V w ) foot-pounds, or p v(V s V w ) heat units. If the external work is required in B.Th.U. the value of J is 778 ; if J be taken as 1400 the result will be in C.H.U. 35. Internal Energy of Steam. When steam is generated at constant pressure, the heat of evaporation as defined in Art. 33 is not the total heat the steam possesses. During evaporation external work has to p be done of amount j (V s V*) heat units, and of the latent heat L, sup- plied, this quantity is not found in the steam ; the difference is known as the internal latent heat, and is usually denoted by p. The internal latent heat may therefore be written Latent heat external work done during evaporation p=L -j(V s -V^ (i) 1 Marks and Davis, " Steam Tables." 44 THE THEORY OF HEAT ENGINES [CHAP. HI. The internal or intrinsic energy of steam is the name given to the sum of the internal latent heat, and the sensible heat of the water i.e. internal or intrinsic energy (I) = p -{- 7z . . . (2) or I = H E . . (2A) The total heat of evaporation may therefore be written H = h + L orH = ^ + p+E ...... (3) The latent heat of steam varies with the temperature at which it is produced, and for practical calculations, when steam tables are not available, it may be estimated from the approximate formula L= 1114 o'7/B.Th.U ....... (4) where / is the temperature of saturation in F. or L=i437 o'7T ....... (4 A) where T is the absolute temperature of the steam, i.e. t -f- 460. The total heat of dry saturated steam may therefore be approximately expressed as H= + L = ('32) + 1114 o-7/ = 1082 + o'3/ B.Th.U. per pound . . . . (5) a result already given in Art. 33. It will be observed from equation (4) that the latent heat decreases as the temperature (and therefore the pressure) increases, but from (5) the total heat always increases as the temperature (and therefore the pressure) increases. If the temperature is measured in C. the approximate formula for latent heat becomes L=6o7 o'7/C.H.U ....... (6) where t is the temperature of saturation in C. The total heat of dry saturated steam may therefore be approximately expressed as = /+ 607 o'7/ = 607 -j- o'3/ C.H.U. per pound ..... (7) 36. Specific Volume and Internal Energy of Wet Steam. The specific volume, or volume of i pound, of wet steam will be less than that of dry steam at the same pressure because of the difference between the densities of the steam and the water it contains. Let V W s be the volume of i pound of wet steam in cubic feet, then, using the same notation as before V m = (volume of dry steam + volume of water) in i pound of the mixture = *V, + (i-*)V, . . . , . ......... (i) or neglecting the volume of the water ( i x) V W) y W8 = xV 8 approximately ...... (2) ART. 37] PROPERTIES OF STEAM 45 The external work done during the formation of i pound of wet steam at constant pressure will be E' = pressure per square foot X change in volume (cubic feet) E' = P{*V S + (i *)V W V w } foot-pounds xV w ) foot-pounds .......... (3) or E' = y(V. V w ) heat units . , -, . . ,- . . . . . . (4) Now the total heat of evaporation is by (i) Art. 33 H' = h + xL therefore the internal, or intrinsic energy, is I' = H - E' P# = h + xL -- (V s V w ) heat units ... (5) 37. Superheated Steam. The specific heat of steam at constant pressure (C p ) was first determined by Regnault, who only worked with steam at atmospheric pressure. The figure he obtained was 0*4805. Since then, an enormous amount of research has been carried out, having for its object the determination of the specific heat at various pressures. The results obtained may be briefly summed up as follows : (1) The mean value of 0*4805 obtained originally by Regnault is only true for steam at atmospheric pressure, that pressure being the only one he used in his experiments. (2) The specific heat is a function of both the pressure and the temperature. (3) The value of Cp increases with the pressure but decreases as the temperature rises. In the neighbourhood of atmospheric pressure recent researches confirm Regnault's value 0*48; at 100 pounds per square inch absolute and at the temperature of saturation corresponding to this pressure, the value appears to be about 0*57, falling to 0*48 at a temperature of about 700 F. At the temperature of saturation, corresponding to a pressure of 200 pounds per square inch absolute, C p =o"j falling to 0*49 at a temperature of 700 F. Complete curves showing the variation of C p with pressure and temperature will be found in Marks and Davis' " Steam Tables." Total Heat of Superheated Steam. The total heat of i pound of superheated steam will be the heat of evaporation per pound of dry saturated steam, plus the additional quantity of heat supplied during superheating. This latter quantity will be a function of the temperature, and may be expressed as #VT where t = the temperature of saturation and /! = the temperature of the superheated steam The total heat may therefore be written as follows : H = h + L + CpJT heat units ... . . . (i) 4 6 THE THEORY OF HEAT ENGINES [CHAP. in. For purely academic problems we may assume C p to be constant, the expression for total heat under this assumption being H = h + L + Cpfa /) heat units .... (2) If the temperatures are measured on the Centigrade scale, H will be in C.H.U. ; if on the Fahrenheit scale, H will be in B.Th.U. 38. Throttling or Wire-drawing of Steam. If wet steam be allowed to expand freely without doing work, its final and initial velocities being equal, and without receiving or rejecting heat, it will become drier ; if the steam be dry to commence with it will become superheated after the expansion. Free unresisted expansion of this nature may be called expansion at constant heat (since there is no interchange of heat), and the steam is said to be wire-drawn or throttled. Let /! = initial pressure of the steam before throttling. . / 2 fi na l pressure of the steam after throttling. H! = total heat of the saturated steam per pound at pressure^. H 2 = total heat of saturated steam of the same quality per pound at pressure / 2 - Then H x is greater than H 2 , and since no heat is supplied or taken away during the unresisted expansion, it follows that for each pound of steam (Hj H 2 ) heat units are available for dryng the steam if it were originally wet, or for superheating the steam if it were originally dry ; and further, if (H x H 2 ) heat units are more than necessary to completely dry the steam, the excess will be available for superheating. If the steam be wet after throttling we have, per pound, Heat before throttling = heat after throttling where the suffixes i and 2 refer to the condition of the steam before and after throttling respectively. EXAMPLE i. Calculate the total heat of evaporation, and the internal energy of i pound of saturated steam at a pressure of 100 pounds per square inch absolute, (a) when the steam is dry, and (b) when the dryness fraction of the steam is 0*8. Given temperature of saturation 328 F., and the specific volume of dry saturated steam at this pressure and tempera- ture 4-229 cubic feet. Latent heat at 3 2 8 F. = 1114 07 X 328 = 1114 229-6 = 884-4 B.Th.U. (a) Total heat H == h + L = (328-32)4-884-4 = 1 180-4 B.Th.U. From Art. 34 p External work done E = (V V w ). ART. 38] PROPERTIES OF STEAM 47 Here P = 144 X 100 pounds per square foot, and the volume of i pound of water (V w ) is o'oi6 cubic foot, hence 144 X ioo E =-^^8 (4*229 - 0-016) 144 X ioo = 773 X 4 ' 213 = 77' 9 B.Th.U. .-, Internal energy I = H E (Art. 35 (2 A)) = 1180-4 77-9 = 1002-5 B.Th.U. (b) H = /fc + *L = (328 -32) + 0-8 X 884-4 = 2964-707-5 = 1003-5 B.Th.U. From (4) Art. 36 External work done E = -y (V, V w ) 144 X ioo X 0-8 if- ~ X4 ' 213 = 62-3 B.Th.U. Internal energy I = H E = 1003-5 62-3 = 941-2 B.Th.U. EXAMPLE 2. In a steam boiler 9-5 pounds of steam are generated per pound of coal burned. The boiler pressure is 155*3 pounds per square inch gauge and the temperature of the feed water is 90 F. If the dryness fraction of the steam is 0*98, and the calorific value of the coal is 14,500 B.Th.U. per pound, calculate the efficiency of the boiler. Here the absolute steam pressure = 155-3 -f- 147 = 170 pounds per square inch. From " Steam Tables " (p. 481) we find that at this pressure, h = 340*7 and L = 8547 hence, total heat of evaporation per pound H 1 = /i-{- xL = 340-7 +0-98 X 8547 = 340-7 + 837-6 = 1188-3 B.Th.U. But the feed temperature is 90 F., therefore each pound of water con- tains 90 32 or 58 B.Th.U. when fed into the boiler, hence Heat supplied by the boiler per pound of steam = 1188 3 58 = 1130-3 B.Th.U. Efficiency of boiler = II3 i 4 3 5 ^ 9>5 = 0-7405 or 74*05 per cent. EXAMPLE 3. Measurements from an indicator diagram taken on a steam engine show that at a certain instant during expansion the volume 48 THE THEORY OF HEAT ENGINES [CHAP. in. of the steam is 1*15 cubic feet and the pressure 50 pounds per square inch absolute. If the weight of steam in the cylinder is 0*15 pound what is the quality of the steam at that instant ? From " Steam Tables " we see that at a pressure of 50 pounds the volume of i pound of dry saturated steam is 8*5 1 cubic feet. If, therefore, the steam in the cylinder were dry its volume would be 0*15 X 8-51 = 1-27 cubic foot. Hence, neglecting the volume of the water in the steam Dryness fraction = - o'88 1*27 EXAMPLE 4. A vessel of 3 cubic feet capacity is full of steam at a pressure of 20 pounds per square inch absolute and of dry ness fraction o'8. It is coupled up to a steam pipe in which steam of dryness 0-9 is at a pressure of 190 pounds per square inch. Steam is admitted into the vessel from the pipe until the pressure in the vessel is 190 pounds absolute. Find the weight of steam admitted and the final dryness fraction of the mixture. Given Pressure. A L Specific volume. 20 I 9 196*1 350*4 960-0 846-9 20-08 2'40 We must first find the weight of steam in the vessel at a pressure of 20 pounds per square inch, as follows : Neglecting the volume of the water in the steam, we have Volume of i Ib. of dryness fraction 0-8 = 20*08 X o'8 = 16*064 cubic feet Actual volume ojf steam = 3 cubic feet .*. weight of steam = , , = 0*1868 pound Let W = weight of steam admitted in pounds and x = final dryness fraction of the mixture. Then the total heat of o* 1868 pound of steam at 20 pounds pressure, and W pounds at 190 pounds pressure before admission into the vessel is 0-1868(196-1 + 0-8 X 960) + W(35o-4 + 0-9 X 846-9) B.Th.U. = 0-1868(196-1 + 768) + W( 3 5o-4 + 762-2) B.Th.U. After mixing the total heat contents in the vessel will be (\V -|- 0-1868X350-4 + x X 846-9) B.Th.U. Assuming no loss of heat to take place during the mixing and neg- lecting the thermal capacity of the vessel, these two quantities will be equal, ... (W+o-i868)(35o-4+^X846'9)=o-i868X964'i+WXiii2-6 . (i) In equation (i) there are two unknowns, x and W, we must therefore find another equation connecting them. If the final contents of the vessel were dry saturated the volume would be (W + 0-1868)2-4 cubic feet. ART. 39] PROPERTIES OF STEAM 49 The volume actually occupied is 3 cubic feet. Hence neglecting the volume of the water present, 3 /_ x 0-1868) (2) in (i) gives = o-i868X 9 6 4 -i+WXni2-6 35 o- 4 W + 65-45 + = 180-09 + HI2-6W 35o'4\V + 65-45 + 1058-6 = 180-09+ iii2-6W 762'2\V = 943-96 W= 1-238 pounds. Substituting this value of W in (2) gives 2-4(1-238 + 0-1868) = 0-877. EXAMPLE 5. Boiler steam of dryness 0*97 and at 340 pounds per square inch absolute (/= 429 F.) is wiredrawn to 200 pounds per square inch absolute (/ = 382 F.). Find the dryness fraction on the engine side of the reducing valve. Solving this problem without the use of " Steam Tables " we have Latent heat at 429 F. = 1114 0-7 X 429 = 814 B.Th.U. Latent heat at 382 F. = 1114 0-7 X 382 = 847 B.Th.U. Let #2 == dryness fraction after wiredrawing, then (429 - 32) + (0-97 X 814) = (382 - 32) + * 2 X 847 397 + 789-6 = 350+847*2 847*2 = 836-6 * 2 z= 0-987 39. Measurement of the Dryness of Steam. There are several methods in use for determining the dryness fraction of steam, some of which are more accurate than others. The greatest uncertainty experienced, in all cases, is the extreme difficulty of obtaining a representative sample of the steam to be tested (see Art. 209). The simplest method is to blow a certain weight of steam into a known weight of water, and measure the rise in temperature produced. Then by equating the heat lost by the steam to the heat gained by the water the dryness fraction can be calculated. Let w = weight of steam condensed in pounds. x = dryness fraction of the steam. W = weight of water in pounds. /! = initial temperature of the water in F. / 2 = final temperature of the water in F. W' = weight of vessel containing the water in pounds. s = specific heat of the material of which the vessel is made. Then, heat lost by steam = w(xL + tt 2 ) B.Th.U. (where f F. is the temperature of the steam under test) and heat gained by vessel and water = (W + jW')(/ 2 /j) B.Th.U. E 50 THE THEORY OF HEAT ENGINES [CHAP. in. Equating these two quantities we have w(*L + /-/ 8 ) = (W.+ *W')(4-/i) . . . (i) an equation from which x may be calculated. The above theory assumes that the specific heat of water is constant and equal to unity, which of course is not quite true. For all practical purposes, however, the error involved is negligible, particularly in com- parison with the uncertainty of the sample of steam taken. If instead of /, /i and / in (i) we write h, h and /; 2 , the sensible heats at these tem- peratures, we have h - /y = (W + sW)(A z - h) ... (2) an equation which is quite accurate. Another method frequently employed is similar to the above, but the steam and water are not allowed to come into contact. The steam under test flows through a pipe (or series of tubes) around which water is cir- culating. The inlet and outlet temperatures of the water are measured, together with the weight of water flowing through the instrument in any convenient time and the weight of steam condensed in the same time. By equating the heat lost by the steam and the heat gained by the water the dryness fraction of the steam may be calculated. Let w =.weight of steam condensed in pounds per minute. /' = temperature of condensed steam in F. W = weight of cooling water in pounds per minute. t = temperature of the steam in F. /! = inlet temperature of cooling water in F. / 2 = outlet temperature of cooling water in F. Then, the heat lost by the steam = w(xL -}- h h') B.Th.U. and heat gained by the water = W(/& 2 h) B.Th.U. .-. w(xL + h H] = W(^ 2 /2 a ) ..... (i) or, neglecting the variation in the specific heat of water, -t') = W(t z -t 1 ) ...... (2) In Art. 38 it has been shown that steam becomes drier after throttling, and in certain cases, superheated. This principle has been utilised in the throttling calorimeter, first designed by Professor Peabody, which enables the dryness of steam to be conveniently and accurately measured. 40. Theory of the Throttling Calorimeter. This calorimeter is an instrument for the experimental determination of the dryness of steam. Fig. 16 shows the arrangement. The pipe B is screwed into the main steam pipe A, and steam admitted through the valve C expands through a small hole into the calorimeter D. The thermometer E gives the temperature of the steam after the expansion, while the manometer F gives the pressure (above atmospheric) of the steam after expansion. From Art. 38 it will be seen that if the steam in the main pipe is nearly dry, after expansion it will be dried and then super heated, the amount of superheat being obtained from the reading of the thermometer E, and the temperature of saturation corresponding to the pressure after expansion. ART. 40] PROPERTIES OF STEAM Let / 3 = reading of thermometer E. /2 = temperature of saturation corresponding to pressure / 2 after expansion (obtained from steam tables). h = sensible heat at pressure/! before expansion. Lj = latent heat of steam at pressure p. Then if x = dryness fraction required From (i) x may be calculated. In practice the values of h^ L 1} H 2 , and tables. will be taken from steam Main Sream Pipe Exhausr steam FIG. 1 6. Throttling calorimeter. Limitation of the Instrument. The action of the calorimeter depends upon the fact that the steam after expansion is superheated. The calori- meter would be useless if the steam were so wet that after expansion it remains wet. The theoretical limit is obviously reached when after ex- pansion the steam is just dry, and t B = / 2 ; this usually occurs when the dryness fraction is about 0-94 to 0*95. The calorimeter is very reliable for steam having a dryness fraction of about 0*98, and since any good modern steam boiler will give steam of this dryness, it is used almost universally for measuring the dryness of steam leaving a boiler. If the steam contains more than 5 per cent, of moisture (dryness fraction 0-95) its dryness may be measured as indicated in Art. 39, or by means of a separating calorimeter combined with a throttling calorimeter. EXAMPLE i. In a test of a condensing plant the following data were obtained : (a) Steam condensed per hour 1552 pounds. (b) Temperature of exhaust vapour entering condenser 1047 F. (c) Circulating water per minute 476 pounds. (d) Temperature of circulating water as it enters the condenser 60 F. (e) Temperature of circulating water as it leaves the condenser 90 F. (f) Temperature of air pump discharge 95*5 F. 52 THE THEORY OF HEAT ENGINES [CHAP. in. Calculate the dryness fraction of the exhaust steam as it enters the condenser. L 104 . 7 = 1114 07 X 1047 = 1114 72-3 = 10407 B.Th.U. Heat lost by the steam per min.=l|p{(io4795-5)4-(^Xio407)}B.Th.U. Heat gained by the circulating water per minute = 476 (90 60) B.Th.U. Assuming no loss of heat, the heat lost by the steam will be equal to the heat gained by the water, hence ^{(1047 95*5) + (* X 10407)} = 476(90 60) 1552(10407^ + 9-2) = 476 X 30 X 60 104073: = 522-2 9-2 4? #== -=0-529 or 5 2-9 per cent. 10407 EXAMPLE 2. The following data were obtained from a test with a combined throttling and separating calorimeter : Water collected in separating calorimeter 4*5 pounds, steam condensed after leaving throttling calorimeter 45*5 pounds. Steam pressure in main steam pipe 150-3 pounds per square inch gauge, barometer 30 inches, temperature of steam in throttling calorimeter 290 F., reading of manometer 4 inches, estimate the dryness of the steam in the main steam pipe. Separating Calorimeter. The moisture extracted from 50 pounds of steam is 4-5 pounds; the remaining 45-5 pounds of steam which has thereby been partially dried is then passed through the throttling calorimeter. Throttling Calorimeter. Absolute pressure of the steam admitted = gauge pressure + atmospheric pressure. = ISO'S + 147. = 165 pounds per square inch. Absolute pressure in calorimeter = 4-1-30 = 34 inches of mercury. = 34 X 0-49. ==167 pounds per square inch. From steam tables we find at 165 pounds absolute, h = 338-2, L = 856-8 at 16-7 pounds absolute, H = 1152-7, /= 218-5 F. Hence assuming the specific heat of steam to be 0-48, /&!-!- *Li = H 2 + 0-48 (/ 3 -/ 2 ) 338-2 +x X 856-8= 11527 +0-48(290 218-5) = 1152-7 + 34-3 = 1187*0 856-8.*= 1187 338-2 848-8 *- 856-8 = 0*990 or, in each pound of steam passing through the throttling calorimeter there is o'oi pound of water, hence in 45-5 pounds there will be 45*5 X 0*0 1 = 0-455 pound of water. .-. total water in the 50 pounds of steam = 4-5 -f- 0-455 = 4'955 pounds. and dryness fraction = 2-2S5 0-900 or 90%. ART. 42] PROPERTIES OF STEAM 53 41. Entropy of Steam. It has been shown in Art. 15, that when i pound of any substance is heated at constant pressure from temperature T T! to temperature T 2 the gain of entropy is C p Iog 6 ^ . If the substance be water, and we assume the specific heat C P to be constant and equal to unity, this expression becomes T Gain of entropy = log e =% For convenience in practice it is usual to measure the entropy of water and steam from some arbitrary temperature, the temperature universally used being 32F. or oC. The entropy of water at the above absolute temperature T will therefore be the change of entropy when i pound of water is heated from 492 (i.e. 32 + 460) to T, and is written If now the pound of water at temperature T be converted into dry saturated steam at the same temperature, the change of entropy during evaporation will be I ......... w The entropy of i pound of dry saturated steam at absolute temperature T will therefore be T L or s = log e -- f- - (where T and L are in Centigrade units) (4) 42. Entropy of Superheated Steam. The increase of entropy during the superheating of steam at constant pressure is frequently calcu- lated on the assumption that the specific heat is constant. On this assump- tion the gain of entropy during the superheating of dry saturated steam from temperature T to temperature T' will be %** f (Art. 1 5) The entropy of i pound of superheated steam at absolute temperature T' will therefore be The entropy given in steam tables is calculated by taking account of the variability of the specific heat of both water and steam, but for aca- demic purposes, or in cases where steam tables are not available, equation (3), Art. 41, may be used for dry saturated steam, and (i) above for super- heated steam. EXAMPLE. Calculate from first principles the entropy of i pound of water and i pound of steam at the following temperatures : 110 F., 2i2F., 280 F., 320 F., and 366 F. 54 THE THEORY OF HEAT ENGINES [CHAP. in. At 110 F., the absolute temperature 15110X461 = 571. >. ^w = iog f ifi= 2-3026 iog 10 m = 2*3026 X 0*0638 = 0*1469 Similarly we find at 212 F. <, = 2*3026 Iog 10 ||| = 2*3026 X o- -352 = 0*3113 280 F. < w = 2-3026 Iog 10 |ff = 2-3026 X 0-1788 = 0*4117 320 F. cf) w = 2*3026 Iog 10 JfJ = 2*3026 X 0*1999 = 0*4606 366 F. s = 0*1469 -{- 1*8161 = 2*0230 2i2F. & = 0-31 13 + 1-4353 = 1*7466 28oF. ^ = 0-4117 + 1-2338 = 1-6455 320 F. s = 0*4606 + 1*1382 = 1*5988 .#, = 0-5163 + 1-0384 = 1-5447 43. Temperature-Entropy Diagram for Steam. Draw the log-curve ob (Fig. 17) for i pound of water, according to the equation T w = log e . Let the water be turned into steam at a temperature T x , then the gain of entropy during evaporation is T^T being represented by ae. Similarly at any other temperature T 2 , the gain of entropy during evapora- tion is Ffr or be. In this manner the saturation curve cd (Fig. 17) may L 2 be plotted. The two curves ob and cd must meet in some point where the latent heat is zero, which point will be at the critical temperature. If the steam be now superheated to temperature T 3 , the change of T entropy during superheating will be 0*48 log e ^r represented by ce (Fig, 18) * 2 if we assume the specific heat to be constant and equal to 0*48. If, there- T fore, from the point c in Fig. 17 we plot the curve 0*48 log e ^r we get the 1 2 curve ce Fig. 18, which shows the change of entropy during superheating. The curve ce is called a constant pressure line because it represents the ART. 43] PROPERTIES OF STEAM _ T 2 55 493 FIG. 17. Curve of entropy for dry saturated steam. <* FIG. 18. Curve of entropy for superheated steam. 56 THE THEORY OF HEAT ENGINES [CHAP. in. entropy of superheated steam at the pressure corresponding to the tem- perature of saturation T 2 460. 44. Entropy of Wet Steam. When i pound of wet steam is pro- duced from water, the quantity of heat supplied is not equal to the latent heat of i pound of steam, because all of the steam is not evaporated. Let T be the temperature of the steam and x its dryness fraction, then the quantity of heat supplied during this partial evaporation will be xL and the change of entropy will be where L denotes the latent heat of i pound of dry saturated steam at temperature T. The total entropy ac ~. = x ab If now a number of isothermal lines on the diagram are divided in this constant ratio, and the points so obtained joined by a smooth curve, a line of constant dry- ness d, Fig. 19, is ob- tained. A number of these constant quality lines are shown drawn on the temperature en- tropy diagram in Fig. 21. 45. Constant Vo- lume Lines. Consider the isothermal line ab, If only - of a FIG. 19. Line of constant quality. Fig. 19 pound of water has been evaporated into dry steam, the change of entropy will be , and this will be represented by the point c such that ART. 46] PROPERTIES OF STEAM 57 The volume of the steam will be - of the volume at point b^ and the volume of the mixture ( i ~) pound of water and - pound of steam, will be for all practical purposes equal to - of the volume at b since the volume of ( i - ) pound of water is very small compared with the volume of i pound of steam, par- ticularly at low pressures. In the case of high-pres- sure steam, where the specific volume is low, the proportional error made in neglecting the volume of water will be greater than when the specific volume is higher, hence for absolute accuracy the point c must be so chosen as to include the volume of the water. In order to draw any constant volume line we may therefore pro- ^ ceed as follows : On the saturated steam line find (by means of steam tables) the points where the volume of the steam is i, 2, 3, 4, 5, etc., cubic feet. Draw iso- thermals through these points. Fig. 20 shows a constant volume line drawn for i cubic foot ; the length ac is made \ of #2, dc \ of e z i so ill I M i linirrm lin i in I 1 i u ART. 46] PROPERTIES OF STEAM 59 bh sented by the point h, such that the initial dryness fraction =^i and at the end of expansion, the dryness fraction would be _< X2 ~0 Let the dryness fraction before expansion be denoted by x^ then T x L Entropy of the steam at T l = log e -f- -Tjr- 1 . . (i) and entropy of the steam at T 2 = log e ^ -j- - - (2) FIG. 22. Isentropic or adiabatic expansion of steam. Now the entropy remains constant throughout the expansion or ^ . (3) EXAMPLE. By means of an entropy chart determine (a) The dryness fraction at the end of expansion when one pound of dry saturated steam at 366 F. expands adiabatically to 225 F. (b) The dryness fraction at the end of expansion when one pound of wet steam with dryness fraction 075 and temperature 380 F. expands adiabatically to 240 F. 6o (a) THE THEORY OF HEAT ENGINES [CHAP, in, 366 F. = 366 + 461 = 827 abs. = Tj 225 F. = 225 + 461 = 686 abs. = T 2 366 = 1114 0*7 X 366= 1114 256-2 = 857-8 ^25 =HI4 07 X 225 = 1114 I57*5 = 956-5 Let x = dryness fraction after expansion then^ = - (see Fig. 23). __AF-f-FE AD 1*0372 686 i'3943 = 0-878 T i = 380 + 461 = 841 and T 2 = 240 + 461 = 701 L 380 = 1114 0-7 x 3 8o = 1 114 266 = 848 ^240 = z I T 4 0'7 X 240 =1114168 = 946 82. 686 841 701 \ FIG. 23. FIG. 24. pp In Fig. 24 = initial dryness fraction = 0-75 ^075 and final dryness fraction =^=^ AD AD 841 -- AD 946 701 o'i82i -f- 07562 I "3495" ART. 47] PROPERTIES OF STEAM 61 The student should check the above results by means of the tempera- ture-entropy diagram. 47. Gain of Entropy due to Throttling or Wire-drawing. There are two cases to be considered. In the first case, the steam after throttling may be either wet or it may be just dry saturated ; in the second case the steam may be initially dry it will then be superheated after throttling (Art. 38). First Case, in which the Steam is not Superheated after Throttling. Let the steam have an initial dryness fraction x^ and let throttling take place be- tween absolute temperatures ' T! and T 2 . The tempera- ture - entropy diagram is shown in Fig. 25. The condition of the steam be- fore throttling is represented by the point c such that T be k/ ~bl The expansion is not is- entropic because no work is done, so that the condi- tion of the steam after throttling will be repre- sented by the point d, such that x _fd 2 fm where x 2 denotes the dry- 493 or32F are FIG. 25. Throttling of steam. ness fraction after throttling down to temperature T 2 . The length ed will therefore represent the gain in entropy, which may be calculated as follows : Entropy at T,) =/ > reckoned from 1 2 ) Entropy at T 2 j _ _ #2^2 reckoned from T 2 ) J T 2 .-. gain of entropy = ed =fd fe (i) But since the heat contents remain the same, the total heat after throttling will be the same as the total heat before throttling, i.e. T 2 492 492 -\ -T 2 + 62 THE THEORY OF HEAT ENGINES [CHAP. HI. Substituting this value for x 2 in (i) we get ~LT~ ' ) T! ' TI Alternative Method. The above result may be obtained by a more direct method as follows Reckoning all heat quantities from the absolute zero of temperature we have Total heat per pound before throttling = area odbch Total heat per pound after throttling = area oafdk .*. area oabch = area oafdk and since the area oaf eh is common, it follows that ce = area edkh Now the area fbce represents the work done per pound of steam on the Rankine cycle (see Art. 57), namely (Ti - T 2 ) /. area edkh = - T 2 log* - 1 B.Th.U. (Art. 57) 2 (T : - T 2 )(i + ^) - T 2 loge ^ II / 2 , or ed = - T 493 or32F -t Shaded areas are equal FIG. 26. Steam superheated after throttling. Case when the Steam is Dry Saturated before Throttling. The condition d of the steam at pressure p and temperature Tj before throttling is represented by the point c on the saturated steam line Fig. 26. After throttling down to a pressure P 2 the condition of the steam is represented by the point d on this constant pressure line in the superheated region of the temperature-entropy dia- gram, the isothermal / being drawn for saturated steam at the temperature T 2 corre- sponding to the final pres- sure p z . The length gm represents the gain in entropy and may be found as follows : ART. 48] PROPERTIES OF STEAM 63 Entropy at pressure A I = f v=i f n 4. fc reckoned from T 2 5 Entropy at pressure / 2 ] _ , _ , , reckoned from T 2 $ /" ^+ 0-48 log, ^ A 2 A 2 where T' 2 is the final temperature of the superheated steam after throttling. .-. gain of entropy = gm =fm fg T T' T L = =? + 0-48 log* Tp* - log e ^ - ^ (2) 1 2 *2 -^ A! Now the total heat after throttling is equal to the total heat before throttling, i.e. H 2 + 0-48 (T' 2 -T 2 ) = T 1 -492+Li - (3) where H 2 denotes the total heat per pound of dry saturated steam at pressure / 2 , being obtained from steam tables, or in their absence from H 2 = T 2 492 + L 2 From equation (3) the final temperature of the steam can be calculated and the substitution of its value in (2) gives the gain of entropy. It should be noticed that reckoning the total heat from the absolute zero of temperature, Total heat per pound before throttling = area oabch Total heat per pound after throttling = area oafedk a nd since the area oafgh is common it follows that area/fog* = area gedkh. EXAMPLE. Find the gain of entropy when dry steam at a pressure of 210 pounds per square inch absolute is wire-drawn to a pressure of 30 pounds per square inch absolute. First find the temperature after throttling, using equation (3) H 2 + 0-48 (T' 2 - T 2 ) = T! - 492 + L! Substituting the values of H 2 , T 2 , T l and L! from steam tables we get- 1163-9 + 0*48 T' 2 0-48 X 710-3 = 846 492 +839-6 from which T' 2 = 7 70 F. absolute. From (2) we find the gain of entropy to be 945' 1 , oi 770 846 839-6 ^ * _L '48 log e loge ATTX^*/^ I " D I* *%* O TT/^v*^ = I '335 + 0-0388 0-1749 0-9924 = O'2O2 48. Heat-Entropy Diagram for Steam. Mollier Diagram. In this diagram the rectangular co-ordinates are entropy and total heat. A 6 4 THE THEORY OF HEAT ENGINES [CHAP. in. diagram of this description, using British units, is supplied with Marks and Davis' "Steam Tables." 1 As indicated in Fig. 27 this diagram is only drawn for steam in the neighbourhood of the saturation curve, the range of pressure used being large enough to include all cases likely to be met with in practice. Vertical lines on the diagram are lines of constant O 3 entropy and horizontal lines are lines of constant heat. The nearly straight lines running diagonally upwards from left to right are lines of constant pressure; below the saturation curve, the nearly straight lines running downwards from left to right are lines of constant quality, whilst 1 A heat-entropy diagram is also published by Oliver and Boyd, Edinburgh, price 3v Saturated steam, however, is a vapour, not a perfect gas, and although the law of expansion may be represented by the empirical formula pv n = constant, it would be meaningless to say that the index " n " for steam is the ratio of the two specific heats, because as the expansion goes on the steam becomes wetter (Art. 46), and it is impossible to say what the ratio of the specific heats may be ; in any case it would be a variable quantity. The question arises then, what value of " n " is to be taken ? Dr. Zeuner gives the value where x = the initial dryness fraction of the steam. If the steam be initially dry the law of the expansion curve becomes ^1-135 = constant ....... (2) Rankine gave an approximate value of " " as ^ which is too small if the steam is initially dry, and corresponds to the case when the steam has an initial dryness fraction of about 0*75. Let the steam expand from the state p^v^ to the state p 2 v 2 ; the work done will be *!***=* to **9) ....(,) In using this equation, the final volume v z will have to be calculated by first finding the dryness fraction after expansion (see Art. 46) and then multiplying the dryness fraction by the volume the steam would have if it was dry and saturated at the pressure^- Another expression for the work done follows directly from the defini- tion of an adiabatic expansion given in Art. 10, namely, Work done = change of internal energy. Using the notation of Art. 35 this becomes W = I x I 2 heat units ....... (2) = (Hi Ei) (H 2 E 2 ) heat units .... (3) in which the pressures are measured in pounds per square inch, and the volumes in cubic feet. 69 THE THEORY OF HEAT ENGINES [CHAP. iv. EXAMPLE T. Calculate the work done when i pound of steam expands adiabatically from 150 pounds per square inch absolute to 16 pounds per square inch absolute, (a) when the steam is initially dry and saturated, (b) when the initial dryness fraction is 0*8. From steam tables, p. 480, we find the following A t. A. L. V. ISO 358-5 330-2 863-2 3-OI2 16 216-3 184-4 967-6 24-79 From (3) W = H! - H a - (a) Here H x = 330-2 + 863-2 = 1193-4 B.Th.U. From a temperature-entropy diagram, dryness fraction after expansion (Art. 46) = 0-874 H 2 = 184-4 + 9 6 7' 6 X 0-874 = 184-4 + 845-7 = 1030-1 B.Th.U. z/ 1 = 3-012 cubic feet and ^ 2 = 24-79 X 0-874 = 21-67 cubic feet .-. W= 1193-4 1030-1 ^1(150 X 3' 012 l6 X 21-67) W = 163-3 i9'S = i43'8 B.Th.U. (b) By the method of Art. 46 we find x 2 = 0-725. Here a isothermal aiT, H x = 330*2 + o'8 X 863-2 = 330-2 -j- 690-6 = 1020-8 H 2 = 184-4 + 9 6 7' 6 X 0-725 == 184-4 -j- 701-5 = 885-9 z/j = 0-8 X 3*012 = 2-41 cubic feet ^2 = '7 2 5 X 24-79 = i?'97 cubic feet W == 1020-8 885-9 771(150 X 2-41 16 X i7'97) = 134-9 13-7 = 121-2 B.Th.U. 51. Perfect Steam Engine working on the Carnot Cycle. Fig. 29 shows the pressure- volume diagram for the cycle. Stage "ab." Heat is sup- plied to water at a and converts it into steam at constant pressure ; the steam expands isothermally at constant pressure and at constant temperature T r Stage " be." When the water is all evaporated to dry saturated steam, the source of heat is re- moved and the steam allowed to continue the expansion adiabati- cally down to temperature T 2 . Stage "cd." The steam is now compressed isothermally at temperature T 2 , being condensed by rejecting heat to the condenser. Stage "da." The cycle is Isothermal at T 2 Volume, V. FIG. 29. pv diagram for steam engine working in Carnot cycle. completed by stopping the condensation of the steam at some point d and ART. 51] THEORY OF THE STEAM ENGINE 71 then compressing adiabatically. If the point d has been chosen correctly the cycle will be completed by restoring the working fluid to its initial condition as water at temperature Tj_. Since the cycle is closed and therefore reversible, all the heat being taken in at the higher temperature T x , and all rejected at the lower temperature T 2 , the efficiency will be Tl ~ T2 (as in Art. 21) Let L! be the latent heat of the steam at temperature T!, then the amount of heat supplied during the cycle will be L x and the work done is T, (I) Instead of assuming the working fluid to consist wholly of water at a and steam at b, ab (Fig. 29) might be taken to represent the partial evaporation of an original mixture of water and steam. In this case if #& denotes the dryness fraction of the mixture at b and x a the dryness fraction at #, the amount of heat supplied during the cycle will be (x b x a )Li and work done would be rp ^^ rp #& ^L X ^ - If the above cycle could be realised in practice we should have a thermodynamically perfect steam engine using saturated steam and like any other perfect heat engine, its efficiency would depend solely ^ upon the working range of temperature T x T 2 . The ideal efficiencies of an engine working on this cycle have been worked out for several cases, assuming condensation to take place at the lower temperature limit of 520 absolute, i.e. 60 F. and tabulated below. Absolute pressure, Ibi. per square inch. Temperature, Ti Fffiripnrv n/mciency 728T7 753"57 0-284 0-308 '80 772-87 0-326 100 120 788-63 802-06 0-339 0-350 140 813-83 0-359 160 824-34 0-368 180 833-89 o-375 200 842-64 0-381 220 850-70 0-387 2 4 858-40 0*393 260 865-60 0-398 280 872*10 0-402 300 878'5o\ 0-407 52. Limits of Temperature Permissible. The lower limit T 2 is the temperature of the available water supply, and cannot of course be 72 THE THEORY OF- HEAT ENGINES [CHAP. iv. less than 492 (32 F.). To the higher temperature T 1 a limit is set in practice by the pressure, which becomes very high for comparatively low temperatures and causes mechanical difficulties with regard to the strength of the working parts of the engine and to the lubrication of the same. As seen above, even with the high pressure of 300 Ibs. per square inch (which is about the limit in practice), the temperature is only 878 absolute or say 420 F., and taking 60 F. as the lowest temperature, the ideal efficiency would be about 40 per cent. It is impossible to attain this result in practice because no steam engine works on this cycle, and many of the causes of imperfection cannot be removed. From a thermodynamic point of view the worst thing to be considered is the irreversible drop in temperature between the furnace and the boiler. The combustion of the fuel gives a very high temperature, but a great portion of the heat which could be converted into work is immediately lost, as it were, by the fall in temperature which takes place before the conversion of heat into work occurs. 53. How closely the Process is carried out in Practice. The first stage " ab" (Fig. 29) may be carried out as in theory. Second Stage, " be." There is no reason why the expansion should not be approximately adiabatic, given an isolated cylinder and high piston speed. Third Stage, " cd " may be performed as in theory by exhausting into the condenser, in which case the final pressure and temperature in the cylinder after expansion must be the same as the pressure and temperature in the condenser. Fourth Stage, "da." The cycle cannot be completed by the adiabatic compression owing to the existence of the separate condenser. The best that can be done is to return the condensed water through an economiser or feed-water heater to the boiler by means of a feed pump or injector. Actually the exhaust is stopped before the end of the exhaust stroke, and the steam compressed or cushioned for mechanical reasons, and to reduce the loss of power due to clearance. This does not materially effect the thermodynamic efficiency. Again, the expansion is never carried right down to the condenser or back pressure, as below a certain pressure, the work done in the engine cylinder is less than that lost by engine friction and condensation. The indicator diagram of an engine working on this cycle (assuming no clearance) is shown in Fig. 30. " ab" represents the admission of the high-pressure boiler steam, "" is the point of cut off, "be" is the expansion line (adiabatic), and 'W the exhaust. Let/! = initial pressure in pounds per square foot at " a " or " b " A = final or exhaust pressure at " c" or " d" v = initial volume (at " b ") before expansion r = ratio of expansion. Then the work done per pound during admission = A 7 'i ( J ) Then the work done per pound during expansion ) _ Pi^\pz rv i \ to volume rv ~~ _ Work done on the steam during exhaust =p 2 rv l . . (3) by the feed pump = (A A) v > (4) Heat supplied = (ft // 2 ). (5) ART. 53] THEORY OF THE STEAM ENGINE Hence the nett work done per pound of steam 73 (6) and the efficiency is nett amount of work done heat supplied n i (A A) v ~ (7) This is also considered by a different method in Art. 55. The above reasoning is for complete expansion, i.e. the steam expands until its temperature is the same as the temperature of the condenser. If Volume, FIG. 30. the expansion is incomplete, as it is invariably in practice, equation (2) still holds if / 2 is taken as the pressure after expansion, and in equation (4) the pressure in the condenser must be substituted instead o case calling /& the back pressure the efficiency becomes m n i A)V, (8) Incomplete expansion is illustrated in Fig. 30, the expansion being stopped at some point " e" It results in a loss of work done on the piston represented by the toe of the diagram shown shaded. The theoretical efficiency as calculated above always falls short of the Carnot efficiency because of the absence of the fourth stage, viz. the adiabatic compression to the original temperature Tj. Without the 74 THE THEORY OF HEAT ENGINES [CKAP. iv. compression some of the heat is taken in at temperatures varying from T 2 to y _ 'p T, therefore as shown in Art. 24 the efficiency must fall short of ^ - AI The efficiency of the engine working in this 'way without compression is, however, much greater than is actually obtained in practice even when the same limits of temperature are employed, as will be seen in Art. 57. 54. Efficiency of an Engine working" without Expansion. This was approximately done a _ A in the early Newcomen engine. Consider i pound of water at temperature T 2 . The indicator diagram is shown in Fig. 31. The water at " d" is heated to T l5 being converted into steam and expanding along " ab" at pressure/!, it is then condensed along "be" _ causing instantaneous fall in Volume V. pressure until at "d" there FIG. si.-jte, diagram for non-expansion engine. is the ori g inal water at tem - perature T 2 and pressure / 2 . Heat taken in during stage " ab "= L A - ...... (i) Heat rejected " cd" =%! /i 2 ..... (2) Work done during cycle = (^ / 2 )( v * v >) (3) Heat supplied = Hj 7i z . . . . (4) Therefore efficiency = (/1 ~f_f )( V * ~ V . FIG. 32. T$ diagram for non-expansion engine. entropy diagram for the particular case under consideration is drawn directly on it. The area of this diagram, Fig. 32, will represent to scale the work done during the cycle. 55- Steam Engine working without Compression but with Complete Expansion. Rankine Cycle. This has already been considered from a pressure-volume point of view in Art. 53. This cycle (known as the Rankine-Clausius cycle) is of the utmost importance in the discussion of steam engines, because it represents the ideal performance of an engine working with no sudden drop in pressure at release, and with cylinders and pistons which are perfect non-conductors of heat. As HP nr indicated in Art. 53, the efficiency is less than - - because heat is supplied during the fourth stage by a non-reversible process, otherwise the cycle is reversible. The engine takes in the greater part of its heat at the upper temperature limit T l5 but some is taken in between T 2 and Tj. So far as the actions occurring in the engine cylinder are concerned the cycle may be con- sidered as reversible, if the feed water be considered as taking up its heat 76 THE THEORY OF HEAT ENGINES [CHAP iv. in an infinite number of instalments at temperatures ranging from T 2 to Tj. In each little instalment the expression T T 2 ~~T represents the efficiency of the transformation into work of the small quantity of heat, say SQ, which is taken in by the working substance at any particular temperature T. The amount of heat converted into work is therefore for each instalment SQ(T-T 2 ) T and the total quantity of heat converted into work will be ^ ........ ( During this stage of the cycle the whole heat taken in is sensible from T 2 to T I} and equals h^ h 2 . The whole heat taken in per pound of working substance is the sensible heat plus the latent heat at tempera- ture Tj. Let Lj = latent heat at temperature Tj. Then the total work done per pound of working substance, expressed in heat units is J. this gives w = ^ - 1 2 - T 2 log ^ + ^ (Tl ~ T2) . . . (i) Assuming the specific heat of water as being constant and equal to unity, in which case ^Q = dT, writing h h z = T x T 2 , we have W = (Ti - T 2 ) - T 2 log, I- 1 + Lj L i or W = (T! T 2 )( i + - 1 ) T 2 log e ~r heat units per Ib. . (2) *!/ 1 2 Now total heat supplied = H x /i 2 = L x + h h^ = L! + T a T, /. efficiency = . i- LI - (3) LI + TI 12 If the steam be initially wet, and of dryness fraction x lt the work done per pound becomes or W = ( T i ~ T 2 )(i + ^) - T 2 log^ 1 heat units . . (5) - 1 ! 7 'TI ART. 57] THEORY OF THE STEAM ENGINE 77 If the Fahrenheit units are used, W will be in B.Th.U. ; with Centigrade units W will be expressed in C.H.U. 56. Derivation of the Adiabatic Equation from this Result. Let i Ib. of water be raised from any temperature T 2 to T 1} and let the fraction x- be evaporated at Tj. Then Heat taken in = h h% + x{L . . . . (i) By expanding adiabatically back to T 2 and then condensing, the work done by equation 4, Art. 55, is Subtracting the work done from the heat supplied we have s rp T r- Heat rejected=/z 1 -// 2 +^L 1 - (h -A 2 -T 2 log f ^ - ^j .......... (.) J 2 M Now all this rejection must occur during the final condensation, since the expansion is adiabatic, and the amount of heat so rejected = x^L^ where x 2 = dryness fraction after expansion to T 2 . A result already obtained by a shorter method in Art. 46. 57. Temperature-Entropy Diagram for the Rankine-Clausius Cycle. Commencing with one pound of water at temperature T 2 , repre- sented by the point a on the diagram (Fig. 33), the water is heated to TI along the line ab t the gain of entropy being T T fg=C p loge - = loge =r (C p for water being assumed unity) X 2 A 2 The water at temperature T x is now turned into steam at T 1 along the line be, the gain of entropy be or gk being ~r 3 where L a is the latent heat at temperature Tj_. The steam then expands adiabatically along cd, down to the original temperature T 2 , and then condensation follows along da back to the original condition of water at temperature T 2 . The work done during the cycle is represented by the area abed and the total amount of heat supplied is represented by the area fabck The efficiency of the cycle is therefore Heat converted into work _ area abed Heat supplied ~~ area fabck 78 THE THEORY OF HEAT ENGINES The area abed = area.fa&g area/# -|- area Ibcd CHAP. iv. = C P( T I - T 2) - T 2 X loge Y + TJT X (T! - T 2 ) and taking C p = i = (T 1 -T 2 )(i + ti)-T 2 loge^l ....(!) a result already obtained in Art. 55. T b me '/ r r V 1 \ K / \ 5 <^/ \e 1 T n d\ e 1 f 8 Entropy < f > . FIG. 33. T0 diagram for Rankine cycle. The area abtk = area/0% + area gbck = L 1 -\-T 1 T 2 (as in Art. 55) . . The efficiency is therefore which may also be written (3) (3A) If the steam is initially wet, having dryness fraction -r- , the adiabatic expansion will take place along mn, and the work done per pound of steam will be represented by the area abmn ART. 57] THEORY OF THE STEAM ENGINE 79 and the total amount of heat supplied will be represented by the area fabmh the area abmn = area/^ area /#/+ area Ibmn = (T! - T 2 ) - T 2 X log, + (Ti - T a ) = (T! - T 2 )(i + ) - T 2 log e ^ (as in Art. 55) (4) the area^^// = area^-f area ^Li + Ti-T,. . ......... (5) The efficiency is therefore . ( 6 ) It will be seen from the diagram (Fig. 33) that the engine working on the Rankine-Clausius cycle does more work per pound of steam than the Carnot engine, the excess being represented by the area abL To do this greater amount of work, however, the engine has to take in a proportionately greater amount of heat represented by the area/a^, and is therefore less efficient, which fact has already been shown in Art. 53. \i now the entropy line ce for dry saturated steam be drawn, the dry- ness fraction of the steam at any point of the expansion can be found as in Art. 46, the dryness fraction at the end of the expansion being . For any other working fluid for which the specific heat of the liquid is s equation (i) becomes Work done per pound = (Tj - T 2 )(s + ^] - sT 2 log, ^ . (7) \ 1 1/ 1 2 and heat supplied = L x + j(T x T 2 ) ...... (8) The efficiency will therefore be -*. kg. With the engine working on Carnot's cycle the final stage of opera- tions is adiabatic compression from T 2 to T x . This is represented in Fig. 33 by the line $, and it is obvious that the compression must be commenced when the steam has a dryness fraction . Similarly the fraction represents the dryness fraction at any stage " r " of the adiabatic compression. A comparison between the indicator (pv) diagram, and the tempera- ture-entropy diagram for the Rankine cycle is given in Fig. 34. The curve BC represents the adiabatic expansion, and the dryness fraction of 8o THE THEORY OF HEAT ENGINES [CHAP. iv. the steam at any point of the expansion may also be shown on the " pv " diagram as follows : From B draw the curve for saturated steam BR /z/Te = const, (see Art. 70). Then at any point M on the expansion curve the dryness fraction is (neglecting the volume of the water) GM gm x = and on the entropy diagram x =~ Suppose now the toe of the indicator diagram is cut off at E (it being uneconomical to expand below a certain pressure, as explained in Art. 53). Entropy Volume FIG. 34. pv and T diagram for the Rankine cycle. This results in a sudden drop in pressure EF at constant volume. To show this loss, due to incomplete expansion, on the temperature-entropy diagram, take any point P on the/z; diagram, and find a point "/" on the T diagram such that = .^~ . If this be repeated for several points between E and F, the curve " epf" is obtained on the T< diagram, the shaded area " epfc n representing the heat wasted by the incomplete expansion. 58. The effect of using Superheated Steam. Let the steam be superheated to a temperature T 3 . Fig. 35 shows the T diagram. The adiabatic expansion is now represented by the vertical line " de" and il the ratio of expansion is large enough (or what amounts to the same thing if the temperature of the superheat is not too high) the line " de n will cut the saturated steam line at some point/, at which point the steam will be just dry and saturated. Further expansion down to e results in wet steam of dryness ->. The total amount of work done per pound of superheated steam is represented by the area abcde ART. 58] THEORY OF THE STEAM ENGINE 81 T 3 the increase of work done due to superheating is represented by the area gcde The extra amount of heat which is supplied to obtain this increase of work done is represented by the area mcdn and the efficiency of this con- version of heat into work is given by area gcde area mcdn which is greater than the effi- ciency using saturated steam, namely area abcg area habcm because the additional heat " mcdn " is received at a much higher temperature than that at which the other portion of the heat is received. Algebraic Expression for the Efficiency of an Engine using Superheated Steam and working on the Rankine- Clausius Cycle. The total 3 k Entropy n o FIG. 35. T diagram for engine using super- heated steam on the Rankine cycle. work done per pound of superheated steam is represented by the area dbcde = area habk area halk + area lbeg-\- area mcdn area mgen = (T! - T 2 ) - T 2 log + ( Tl - T 2 ) X The total amount of heat supplied is represented by the area habcdn = area habcm -|- area mcdn The efficiency is therefore ! - T 2 )(i + C lo - T (3) when C p is the specific heat of steam at constant pressure. 82 THE THEORY OF HEAT ENGINES [CHAP. iv. The efficiency is slightly greater than when saturated steam is used due to the widening of the temperature range. In actual practice, the use of superheated steam results in greater economy than the thermodynamic efficiency would lead one to expect, the advantage being rather in pre- venting initial condensation (see Chap. V.). The widening of the tem- perature range does not result (in the actual engine) in a corresponding increase of efficiency, for a greater part of the heat supplied is taken in at the temperature of saturation, and since its conversion into work depends upon the temperature at which it is taken in, it follows that the act of supplying a very much smaller quantity of heat afterwards, and so increas- ing the temperature range, will not materially increase the thermodynamic efficiency. 59. Engine in which the Steam is kept Dry and Saturated during Expansion. This is partly secured in practice by applying a steam jacket to the engine cylinder. Steam jacketing, while never super- ^C F E FIG. 36. -pv diagram for engine when steam is kept dry and saturated during the expansion. heating, tends to keep the steam dry and prevent condensation. The expansion curve, known as the saturation curve, follows the approximate law pv = constant. Fig. 36 shows the theoretical indicator diagram from which we see Work done during admission = area ABEF =A^i ...... . . . . . (i) (2) Work done during expansion from v to ART. 59] THEORY OF THE STEAM ENGINE Work expended during exhaust = area GCDF=/ 2 ^2 .'.net amount of work done by the steam = Again Heat received from the boiler = H x h^ Heat rejected to the condenser = H 2 h% Let HI = heat supplied by the steam jacket then J(H! H 2 + Hj) = i or HJ = and the efficiency is j (3) (4) (5) (6) (7) (8) Another method of considering this effect of the steam jacket is as follows : The temperature-entropy dia- gram for the ideal case is shown in Fig. 37, where, instead of the adiabatic ft ce" on the Rankine cycle, the expansion follows the saturation curve " cd" Hence the area " ced" represents the extra work done due to the heat sup- plied by the jacket. To find the Heat supplied by the Jacket and the Effi- ciency of the Cycle. Consider an element of the diagram xy of height dTT. The area of this strip represents the work done 8\V over the small change in tempera- ture dT. FIG. 37. T< diagram for engine when steam is kept dry and saturated during the expansion. Let Then Now W = work done during the cycle .-.W =/;; = al>T (Art. 35) m ^x =**! -~- J T 2 T T 3 84 THE THEORY OF HEAT ENGINES [CHAP. iv. Let Hj heat supplied at temperature T 1 Hi = by jacket H 2 = heat rejected at absolute temperature T 2 //j ;= sensible heat at T %2 = > T 2 then W=H 1 + H j -H 2 ............. (10) ..H) = W + H 2 H! = W-f (L 2 + /i 2 ) - (L x + /y = W + (h*-hi) + L 2 - LI Substituting for W, L 2 and Lj we have Hi - a log,^- 1 - b(\\ - T 2 ) + (/ ?2 - ^) + (- T 2 ) - (0 A 2 = * log ^ - ^(T! - T 2 ) + (// 2 - ^) + ^(T! - T 2 ) A 2 T (n) Regarding the sensible heat of the exhaust 7i 2 as returnable, we can write Nett heat supplied = H! + HI h 2 ^2+H T 2 +H j ...... (12) Substituting for Hj in (5) we have Nett heat supplied = L 1 + T 1 T 2 + a log ^ (T x T 2 ) A 2 T ^Li + rtloge^ ......... (13) X 2 W ,. Efficiency = . The numerical values of the constants " a " and " b " in (7) are a = 1437 and = 0*7 when Fahrenheit units are used, and a = 797 and ^=0*7 when Centigrade units are employed. EXAMPLE i. A steam engine using saturated steam works non- ex- pansively, the initial pressure being 100 Ibs. per square inch absolute (/=327'6 F.) and the final pressure 15 Ibs. per square inch absolute (/= 213 F.). Estimate its probable efficiency, given that the specific volume of steam at 100 Ibs. is 4-34 cubic feet, and the volume of i Ib. of water is 0*016 cubic foot. ART. 59] THEORY OF THE STEAM ENGINE 85 By Art. 54, equation (5), (A efficiency = - To find H! and h z ^=1082+0-3 X 3 2 7'6 = 1082 + 98-28 = 1180-28 B.Th.U. ,$2=213 32 = 181 B.Th.U. . 144(100- 15X4-34 0-016) :ienc ? = 778 X (1180-28-181) = 144X85X4^4 = O . o6g or 6 . g 778 X 999'28 N.B. Note that the Carnot efficiency working between the same limits of temperature would be 327-6 21* 114-6 3.7-6 + 461 = 788* = 0- ' or '4-5 per cent. EXAMPLE 2. If a steam engine works between the same pressures as in Example i, but with complete adiabatic expansion, i.e. on the Rankine cycle, estimate its efficiency, the clearance being neglected. The law of expansion is pv l ' im = constant. By (7) Art 53 efficiency = (A To find the ratio of expression " / " Since A^i 1 ' 135 = A^ 1 ' 135 .A_/^ 2 yi35_ "A V /. i'i35 log r = log 100 log 15 = 2-000 1-1761 = 0-8239 . 0-8239 '' logr== 7^ =:0 '7 2 59 .-.^=5-32 /. emciency ~~~, ( IOOX r 44X4'34 15X144X5-32 X4'34) (100 i5) I 44Xo-oi6 . ___ \J J __ ^ _ 778(1180-28 181) 8 '33 X 4'34(ioo 79'8)i44 1*36 X 144 778 X 999'28 (8-33 X 4'34 X 20-2 1-36)144^(730-23 1-36)144 778 X 999-28 778 X 999-28 104,060 ' The above result may be checked by the expression for the Rankine efficiency (Art. 55, equation 3) (Ti- efficiency = 86 THE THEORY OF HEAT ENGINES [CHAP. iv. Now L!= 1114 0-7 X 327-6= 1114 229-3 = 884-7 Tj = 327-6 + 461 = 788-6 T 2 = 213 + 461 = 674 Substituting these values we get efficiency 77 = 999-28 114-6 X 2-i2i 674 X 2-303 X (2-8968 2-8287) 999-28 243-06 674 X 2-303 X 0-0681 999-28 243-06 105-7 999-28 = 0-137 or J 3'7 P er cent< 999-28 Volume. V, cw.ft. FIG. 38. EXAMPLE 3. In a Stirling's engine, fitted with a perfect regenerator, the maximum pressure is 140 Ibs. per square inch absolute and the minimum 15 Ibs. per square inch absolute, the upper and lower temperatures being 650 F. and 60 F. A perfectly re- versible steam engine uses dry satu- rated steam between the same limits of pressure. Compare their efficiencies, and if the piston speed and stroke be the same in each engine, compare the diameters of the cylinders for equal power. Given temperature of saturation at 160 Ibs. abs. = 352 F., and at 15 Ibs. abs. = 213 F., and specific volume at 15 Ibs. abs. = 25-85. Stirling engine Efficiency = Steam engine Efficiency = 60 590 -T^ = ~ - = -53i or Per cent. or 17-1 per cent. 352-8 + 461 813-8 Since the piston speeds and strokes are the same, the ratio of the areas of the cylinders will be inversely as the ratio of the mean effective pressures for equal power. To find the mean effective pressures (P m ). Stirling engine : T x = 1 1 1 1 ; T 2 = 5 2 1. The indicator diagram is shown in Fig. 38. ART. 59] THEORY OF THE STEAM ENGINE 87 Let/ A = pressure at point A at the end of expansion Then / A =/ B X ^ = 1 5 X ^ = 3 r 9 8 ( sa 7 32 Ibs. per square inch) area of diagram _ r 15X4*37^1 log ^^ ^1(140 T 5X 4*37)^6 4'37 . z> 2 ^i "4*37^1 ^1 ^5 v5 / 0-6405 74-45 X 2-303 X 0-6405 = 109-818 = 3'37 3'37 lbs> per sq . inch 14-0 15 A Isothermal at 352-8 F. Isothermal at 213 F. <* Temperature T- g ^ & A a \ \ \ \ ^ i i i i i J ^""""""'^ X y / ED C F Entropy . FIG. 40. Volume v FIG. 39. Steam engine: Tj_ = 813-8 ; T 2 = 674. The indicator diagram is shown in Fig. 39 and the T< diagram in Fig. 40. The length of the diagram will be the volume at point C after adiabatic expansion. Let # 2 = dryness fraction after expansion. Then length of diagram = 25-85 X x^. To find x z . For adiabatic expansion we have the equation L x = 1114 0-7 X 352-8 = 1114 246-96 = 867-04 B.Th.U. L 2 = 1 1 14 0-7 X 213 = 1114 149-1 = 964-9 B.Th.U. 88 THE THEORY OF HEAT ENGINES [CHAP. iv. _. = 964-9 v 964-9 P> /" x z may also be found directly from the T< diagram, i.e. x 2 = ^ (Fig. 40). .*. neglecting the volume of the water, the volume after expansion is 25*85 X 0-866 = 22-38 cubic feet, Work done = area of T< diagram ABCD 144 867-04 778 X 139-8 X 22-38 X 144 = - g - - = 35'96 Ibs. per square inch absolute. Hence area of Stirling^ 35-96 ^ 1*073 area of steam 32-58 ~~ i t diameter of Stirling _ /- - _ 1*03 diameter of steam '* i EXAMPLE 4. An engine is fitted with steam jackets so that the steam remains dry and saturated throughout the expansion. If the initial temperature is 400 F. and the final back-pressure temperature is 126 F., calculate the heat supplied by the jacket per Ib. of steam, and the efficiency of the engine. T x = 400 -}- 461 = 861 T 2 = 126 + 461 = 587 H, = 1437 Jog,^ - ( T i - T 2) ( Art - 59> equation (n)) L 2 = 1437 X 2-303(2-9350 2-7686) (861 587) = 5507 274 ..^=276-7 B.Th.U. The efficiency by equation (14), Art. 59 = Now L! = 1 1 14 07 x 400 = 1 1 14 280 = 834 B.Th.U. Substituting we get 7 - 07 X 274 07-98 Efficiency = 5507 = ,o/'= ' 2 59 or 25*9 Per cent. 7 ART. 60] THEORY OF THE STEAM ENGINE 89 60. The Regenerative Steam Engine. It has been shown in Art. 57 that the efficiency of the Rankine cycle is always less than the ideal TI-T, of the reversible Carnot cycle. The efficiency of the steam- T, engine cycle may, however, be increased, and may then approximate to the ideal value of the perfectly reversible heat engine. The principle of regenerative feed-heating is exactly the same as that invented by Dr. Stirling and described in Art. 27. Heat is abstracted from the working steam at one or more stages during expansion by the feed water on its way to the boiler. In the ideal case the number of stages will be infinite, and the feed water will enter the boiler at the same temperature as the boiler steam, the temperature-entropy diagram being the same as that of the Stirling cycle with perfect re- / | generator (Fig. 8). / j The complete temperature- / | entropy diagram of the ideal / regenerative steam-engine cycle is shown in Fig. 41. During 7^ a( ^| ^ I/ the expansion from Tj to T 2 along cd, an amount of heat represented by the 'area dchg is abstracted from the working steam in order to heat the feed water on its way to the boiler. The heat necessary to raise the temperature of i pound of feed water from T 2 to Tj is repre- sented by the area abfe ; hence, since in the ideal case under consideration there are no losses, the areas dchgoxiA. abfe are equal. The area fbch represents the latent heat (L x ) taken in during evaporation at the constant FIG. 41.- f ^ h Entropy

FIG. 42. T diagram for compound regenerative cycle. T 3 . The nett amount of heat externally supplied will, there- fore, be represented by the area eabcdfl^ and the net amount of heat rejected to the condenser by the area eagl. The efficiency will therefore be area abcdfg area eabcdfl Let T 3 denote the temperature in the receiver. Consider i pound of steam passing completely through the engine, while w pound is abstracted from the receiver for feed heating. Then i-\-w pound passes through the high-pressure cylinder and i pound through the low-pressure cylinder. Further if H 2 denotes the heat rejected to the condenser, and R the heat taken from w pound of receiver steam and afterwards imparted to the feed water, we have For i pound of steam working between T 1 and T 2 receiving entropy and parting with the whole of it in returning to its original state T-i . Li Ho , , ART. 60] THEORY OF THE STEAM ENGINE 91 and for w pound working between T x and T 3 Also since i pound of the feed is heated from T 2 to T 3 by w pound of receiver steam R = T 3 -T 2 (4) R T, Hence from 1 (3) w = and from (4) w = - - ^r~ ....... (5) Also the work done will be W = heat received heat rejected = ( I + ^)(L 1 + T 1 -T 3 )-H 2 .... (6) Hence the efficiency will be T^-H, , . Substituting in (7) the value of H 2 from (2) we get (i + wXLi + T! - T,) - (8) which is always greater than the efficiency of the Rankine cycle, Art. 57, equation (3A). In the above theory it is assumed that the feed water abstracts as much heat from the receiver steam as it can, in order that its temperature may be raised from T 2 to T 3 . Actually the efficiency will be a little less than that obtained by substituting the value of w from (5) in (8), because the resulting temperature of the feed water will be less than T 3 . Triple-Expansion Engine. The regenerative principle may also be applied to the triple-expansion engine, the ideal cycle for which is shown on the temperature-entropy diagram (Fig. 43). cm is the curve of perfect regeneration, cd expansion in the high-pressure cylinder, fg expansion in the intermediate cylinder, and kn the expansion in the low-pressure cylinder. The gross amount of heat supplied is represented by the area eabch, and since an amount of heat represented by the area Ikgfdh is returned to the feed, the net amount of heat supplied is shown by the area eabcdfgkl. The work done is represented by the shaded area, and the efficiency is _ area dbcdfgkn ^ area eabcdfgkl THE THEORY OF HEAT ENGINES [CHAP. iv. Let /]_ = weight of steam taken from the first receiver, between the high-pressure and intermediate cylinders, and T x ' its temperature. Rj = heat supplied to the feed from w-^. w z = weight of steam taken from the second receiver between the intermediate and low-pressure cylinders, and T 2 ' its temperature. R 2 = heat supplied to the feed from w 2 . T, K. Q) Entropy < n FIG. 43. T diagram for triple-expansion regenerative cycle. Then for each pound of steam passing completely through the engine we have For i pound of steam working between T x and T 2 (9) For w pound working between T x and T/ LI - t -\-w ir -^ r -^, = o . . . . (10) ART. 60.] THEORY OF THE STEAM ENGINE 93 For a/2 working between 1\ and T 2 ' T 1 T ~R ' 2 log e ^,+ ^ 2 ^-.p 2 , = .... (n) 2 A l A 2 R 2 = T 2 '-T 2 (12) Ri = (i + W2)(Ti'-T 8 ') . . . (13) Hence from (n) and (12) From (10) and (13) The work done (shaded area) per pound of steam passing through the condenser, or per (i + #>i + a' 2 ) pound passing from the boiler through the high-pressure cylinder, is W = (i + Wi + z^XLi + T! - IV) - H 2 = (i + ^ 1 + W2 )(L 1 + T 1 -T 1 ')-T 2 (log e ^ + ^) (16) and the efficiency W ( i + Wl + ^(Li + T! - In practice this complete cycle is not carried out : Mr. Weir only takes steam irom the low-pressure receiver, in which case w^ = o and equations (14) and (17) reduce to the same as (5) and (8). EXAMPLE. In a compound steam engine, the admission pressure to the high-pressure cylinder is 150 pounds per square inch absolute, and the exhaust pressure in the low-pressure cylinder is 4 pounds per square inch absolute. The back pressure in the high-pressure cylinder and the admission pressure of the low-pressure cylinder is 45 pounds absolute. Steam for feed heating is drawn from the low-pressure steam chest. Assuming complete adiabatic expansion and no heat losses, estimate the efficiency of the engine, and the gain due to feed heating. From steam tables we find, at 150 pounds absolute Lj = 863 and T 1 = 358 + 460 = 818 ii 45 T 3 = 274 + 460 = 734 4 ii T 2 = 153 + 460 = 613 9 4 THE THEORY OF HEAT ENGINES From (5) [CHAP, iv _ = T 613 1 734 ._ _ i 0-8351 1-0550 and from (8) 0*1640 J w = ^= 0-1417 pound 1-1634 61-2 OI3 81 1-1417 X 947 = i 0761 = 0*239 or 23-9 per cent. Without feed heating and working on the Rankine cycle between tem- perature limits T! and T 2 the efficiency will be by (3 A), Art. 57, i-6i 3 (log.SJ+Hi) 863+ 818 613 i 0-772 = 0*228 or 2 2 '8 per cent. The increased efficiency due to feed heating is therefore 23-9 22-8 = 1*1 per cent., or the percentage saving is 100 X ^ = 4 '8 per cent. 61. The Expansion Curve to be assumed in estimating the probable Indicated Horse- AD = hyperbolic expansion, po const. L r _ _ AC = actual expansion. Power of Steam Engines. i? When dry saturated steam is AB = saturation curve 2>z/ 16 = const. AE = adiabatic expansion /to 1 ' 1 admitted into the cylinder the pressure will not remain constant up to the point of cut-off, but falls during admission due to wire-draw- ing and initial condensation. When cut-off occurs the pressure drops quickly for a portion of the stroke, due probably to further condensation, and then some of the previously con- densed steam re-evaporates during the latter portion of the stroke and so keeps the pressure up higher than if there had been no initial con- densation. The result is that the expansion curve follows neither the adiabatic nor the saturation curve, being less steep than either of them, approaching very closely to the iso- thermal for a gas, the curve being approximately a rectilinear hyperbola. Fig. 44 shows approximately the shapes of these three curves. The Volume v. FIG. 44. ART. 61] THEORY OF THE STEAM ENGINE 95 expansion curve, which is always taken for approximate calculations on the probable indicated horse-power, is the hyperbola, hyperbolic expansion (pv = constant) being assumed. Fig. 45 shows the assumed theoretical indicator diagram. FIG. 45. Theoretical indicator diagram. Let '! = initial pressure in Ibs. per square inch absolute. / 2 = pressure after expansion in Ibs. per square inch absolute. p^ = back pressure in Ibs. per square inch absolute. v = initial volume in cubic feet) , v 2 v z = final volume J then ^ = r r = ratio of expansion. Then work done is represented by the shaded area = work done during admission + work during expansion work during exhaust /. W = i44(/ 1 z' 1 -j-^z/j log e r fibV%) ft.-lbs (j) and mean effective pressure (2) (I + logg r) _ Equation (2) gives the theoretical mean effective pressure in terms of 9 6 THE THEORY OF HEAT ENGINES [CHAP. iv. the initial pressure, ratio of expansion, and the back pressure, all pressures being in Ibs. per square inch. In practice, since a sharp cut-off cannot be obtained, and also because of the rounding off of the diagram at release and the compression, the mean effective pressure is always less than the ideal given by (2) and may be written (3) where e = a constant less than unity, and called by Professor Unwin " the diagram factor." If L = length of stroke in feet, A = area of cylinder in square inches, N = number of working strokes per minute, then the indicated horse-power LH .P.=^AN ....... 33,000 Engine using Superheated Steam. In this case the law of the expansion curve being pv n = constant, we have Now the work done is also equal to p m X fl -i-.-l i- n 1 }-* (7) For superheated steam the value of "" is 1-3 and substituting this value in (3) will give p m ; for dry saturated steam n= 1*135, an d fo r steam which is kept dry during expansion n = ~ = 1-0646. 62. Effect of Clearance on the Mean Effective Pressure. In actual practice the piston does not come right up to the end of the cylinder at the end of the exhaust stroke, a small space being left to allow for the wear of the bearings and other mechanical reasons. In addition, there is the volume of the steam ports between the cylinder and valve face. This total volume is called the dearatice volume, and must be taken into account in all calculations on the curves of expansion. It constitutes a volume through which the piston does not sweep, but which is always filled with steam when admission occurs, and this steam in the clearance space forms part of the total steam which expands after cut-off. The ART. 62] THEORY OF THE STEAM ENGINE 97 theoretical indicator diagram as modified by the clearance is shown in Fig. 46. As will be seen, it results in a higher mean effective pressure, and a lower ratio of expansion than is obtained when the clearance is neglected. Let r = ratio of expansion when clearance is neglected. R = true ratio of expansion (allowing for clearance). c = fraction of piston displacement representing clearance, final volume v% -f- clearance clearance + Then R = initial volume i volume of piston displacement -j- clearance volume to point of cut-off -j- clearance i+er This value of R must be substituted for " r " in equation (3) and equation (7), Art. 61, when estimating the mean effec- tive pressure. To obtain an expression for the mean effective pres- sure in a single cylinder engine in terms of the initial pressure, p Ibs. per square inch absolute, the back pres- sure /& Ibs. per square inch absolute, the ratio of expan- sion neglecting clearance " r" the clearance volume being c times the piston displace- ment. W = shaded area (Fig. 46) Volume v. FIG. 46. Theoretical induction diagram modified by clearance. R (by (i)) A. = (3) 9 8 THE THEORY OF HEAT ENGINES [CHAP. iv. If the expansion follows the general law/^ n = constant, (6), Art. 61, becomes : w = pbl'z (4) /m = n i (5) EXAMPLE i. The diameter of a steam-engine cylinder is 6 inches and the stroke 12 inches. If the initial pressure is 100 Ibs. per square inch absolute and cut-off is stroke, find the mean effective pressure, neglecting clearance, the back pressure being 3 Ibs. per square inch absolute. If the clearance is ^ of the piston displacement, find the probable I.H.P. if there are 400 working strokes per minute (i.e. the engine is double-acting and runs at 200 revolutions per minute). Obviously r = 4. 100 pm ===(* + log e 4) 3 = 25(i +1*3862) -3 = 25 X 2-3862 3 = 59-65 3 = 56-65 Ibs. per sq. in. The actual ratio R = ' 2 $+ ' T = 35 = 8 i-f-o-i IOO i'i ~ 3 roo -1555 -3 = 68-3 = 65 Ibs. per square inch . IHP _ 65 X i X (07854 X 36) X 4QQ_ . ... 33,000 EXAMPLE 2. Find the diameter of the steam-engine cylinder needed to develop ioo I.H.P. when the piston speed is 600 feet per minute, initial steam pressure 150 Ibs. per square inch absolute, back pressure 15 Ibs. per square inch absolute, cut-off of stroke, clearance volume 8 per cent, of piston displacement, and the effect of early release, compression, etc., reduces the actual mean effective pressure to 90 per cent, of the theoretical. -A ((3) Art. 62) ART. 63] THEORY OF THE STEAM ENGINE 99 Here **= 5, ^ 0*08 'An = L1 +( J + ' 8 X 5) ^i - 15 = 3{ I + I '4log 6 3*857} J 5 = 30(1 + 1-4 X 2-303 X 0-5863) 15 = 20 X 2-89 15 = 57-815 = 42*8 Ibs. per square inch Let A = area of cylinder required then 600 X 42-8 X T 9 o X A = 100 X 33,000 ioo X 33,000 . .'. A = Q ^, = 1427 square inches 54 X 42-8 X 10 /. diameter = \/ I4 f - =13-27 say 13^ inches V 0-7854 63. Binary Vapour Engine. In this engine, invented by Du Tremblay about 1850, a combination of steam and a more volatile liquid (ether) was used as the working fluid. The heat in the exhaust steam from the steam-engine cylinder was utilised in evaporating ether, which in its turn did work in a separate cylinder. After being exhausted from the ether cylinder the ether was condensed in a surface condenser and the cycle repeated. By this means it was possible to have a net gain in efficiency, since more work was done per pound of steam than would have been possible in a steam-engine cylinder alone. It has already been mentioned in Art. 53 that it does not pay to expand below a certain pressure on account of the frictional resistance, etc., but at the low temperature of the exhaust steam, ether vapour exerts a consider- able pressure and enables the ether cylinder to be utilised with advantage. In order to get the best results with the binary engine it was found necessary to have a fairly high back pressure (about 7 pounds per square inch) in the steam cylinder, the result being that although the binary engine was more economical than the steam engine alone when working with this back pressure, yet in a modern well-designed steam engine with a low back pressure of about 2 pounds per square inch absolute, the gain would be a negligibly small quantity. In other words, an inefficient steam engine may, by the addition of an ether (or other volatile liquid) cylinder, be con- verted into an economical binary engine, but a binary engine will have a very little higher efficiency than the simpler modern steam engine. The thermodynamic principles of the binary vapour engine will be best illustrated by means of the following example. EXAMPLE. In a binary vapour engine using SO 2 as the more volatile liquid the initial temperature of the steam = 327 F., exhaust temperature = 100 F. The initial temperature of the SO 2 = 95 F., exhaust tempera- ture = 60 F. Given that the specific heat of liquid SO 2 = 0*4, and the latent heat of SO 2 = 176 o'27(/ 32) find (a) The efficiency of the steam engine alone. (b) The weight of SO 2 used in the vapour cylinder per pound of steam in the steam cylinder. (c) The combined efficiency of the two cylinders. ioo THE THEORY OF HEAT ENGINES [CHAP. iv. Assume both cylinders to work on the Rankine cycle, and that both steam and SO 2 are dry at admission into their respective cylinders. [L.U.] L! = 1114 07 X 327 = 1114 229 = 885 B.Th.U. () By (3), Art. 57. Efficiency of steam engine alone = 227 X 2-124 190*58 _ 292 III2 ~III2 = 0*262 or 26*2 per cent. (b) We first require the dryness fraction of the exhaust steam, or we may get .# 2 L 2 directly = 560(? + lg. Hi) = 560(1-124 + 0-340) = 560 X 1*464 = 820 B.Th.U. ?. This may also be found as follows. Since the expansion is adiabatic .# 2 L 2 = net neat supplied work done = 1 1 12 292 = 820 B.Th.U. as before. Heat available for evaporating SO 2 per pound of exhaust steam. = 820 + 5 = 825 B.Th.U. Heat of evaporation of SO 2 from 60 F. to 95 F. = 0-4(95 6 ) + J 76 0-27(95 32) = 14+ 176 17 = 173 B.Th.U. .. weight of S0 2 per pound of steam = = 4-767 pounds. (c) Work per pound of SO 2 by (7), Art. 57, = (95 - 6o)(o- 4 + g|)- 0-4 X 520 log e g = 35 X 0-686 0-4 X 520 X 2*3026 X 0-0283 = 24-01 13-55 = 10-46 B.Th.U. .'. work for 4-767 pounds of SO 2 = 10-46 X 47^7 = 49-86 B.Th.U. and total work done for each pound of steam = 292 -f- 49-86 = 34i-86B.Th.U. Sensible heat per pound of exhaust SO 2 = 0-4(60 32) = 1 1-2 B.Th.U. Sensible heat of 4-767 pounds = 11*2 X 4767 = 53-39 B.Th.U. ART. 63] THEORY OF THE STEAM ENGINE 101 .*. nett heat supplied per pound of steam to the binary engine = 885 + (327 - 32) - 53-39 = 1126-6 B.Th.U. Hence the combined efficiency will be, Total work done 341*86 Heat supplieT :== 7i^6*6 = '3<>3 EXAMPLES IV 1. Calculate the work done when I pound of steam expands adiabatically from 80 pounds per square inch absolute to 15 pounds per square inch absolute, (a) when the steam is originally dry saturated, and (b) when the dryness fraction is originally 0*9. Given, volume of I pound of dry steam at 80 pounds absolute = 5*47 cubic feet, and at 15 pounds absolute = 26^27 cubic feet. Use steam tables for any other data that may be required. 2. Steam 30 per cent, wet at 100 pounds per square inch absolute expands adiabatically to 20 pounds per square inch absolute. Find its wetness after expansion. If the expansion can be represented by pv n constant, find n. 3. A steam engine using saturated steam works non-expansive ly, the initial pressure being 120 pounds per square inch absolute, and the exhaust pressure 5 pounds per square inch absolute (/ = 162 '3 F.). Estimate the work done per pound of steam and the efficiency of the engine by calculation, and from a temperature-entropy chart, (a) when the steam is initially dry saturated, and (b) when the initial dryness fraction iso'8. [Given, volume of I pound of dry saturated steam at 120 pounds absolute = 3*726 cubic feet, and temperature = 341 F.]. 4. An engine using dry saturated steam works on the Rankine cycle between tempera- ture limits of 350 F. and 140 F. Estimate the work done per pound of steam and the efficiency of the cycle. 5. Solve Problem 4 if the initial dryness fraction of the steam is 0*85. 6. A steam engine requires 300 B.Th.U. per minute per horse-power when working between temperature limits of 390 F. and no F. What is the ratio of its thermal efficiency to that of an ideal engine working between the same temperature limits (a) on the Rankine cycle, (b} on the Carnot cycle ? 7. Estimate the pounds of steam required per hour per horse-power by an engine working on the Rankine cycle between temperatures of 330 F. and 210 F. 8. Superheated steam at 180 pounds per square inch absolute (/ = 373 F.), and temperature 520 F., expands adiabatically down to a pressure of 6 pounds per square inch absolute (t = 170 F.). Assuming that ihe Rankine cycle is followed by this steam, determine the weight of steam required per hour per horse-power, and the dryness of the steam after expansion (C^ = 0*5). 9. An engine is supplied with superheated steam at 120 pounds per square inch absolute (t = 341 F.) and exhausts at 4 pounds absolute (/ = 153 F.). Taking the mean specific heat of the steam as 0*5, find the superheat which must be given to the steam at the higher pressure in order that after adiabatic expansion to the lower pressure it may be just dry and saturated. An engine works on the Rankine cycle with that degree of superheat and between the above pressures. Find its thermal efficiency and the work done per pound of steam. 16. In a Stirling engine fitted with a perfect regenerator, the maximum pressure is 135 pounds per square inch absolute and the minimum 15 pounds per square inch absolute, the upper and lower temperatures being 600 F. and 80 F. A perfectly reversible steam engine uses dry saturated steam between the same limits of pressure : compare their efficiencies. If the piston speed and stroke be the same in each engine, compare the diameters of the cylinders for equal power. [Given, temperature of steam at 135 pounds absolute = 350 F., and at 15 pounds absolute = 213 F., and specific volume at 15 pounds absolute = 26*27 cubic feet.] 11. An engine is fitted with steam jackets so that the steam remains dry and saturated throughout the expansion. If the initial temperature is 400 F. and the final back- pressure temperature is 110 F., calculate the heat supplied by the jackets per pound of working steam and the efficiency of the engine. 12. Estimate the weight of steam required per hour per horse-power by an engine working between temperature limits of 200 C. and 60 C., (a) on the Rankine cycle, 102 THE THEORY OF HEAT ENGINES [CHAP. iv. (b) when by the use of steam jackets the steam remains dry and saturated throughout the expansion. 13. In a compound steam engine the admission pressure to the high-pressure cylinder is 170, pounds per square inch -absolute (L = 855 B.Th.U. and / = 368*5 F.), and the exhaust pressure ifj.ihe low-pressure cylinder is 2 pounds absolute (/ = 126 F.). The back pressure of the high-pressure cylinder and the admission pressure to the low-pressure cylinder i5C"pounds+) If the flow of heat has attained a steady condition, the difference between the amounts of heat flowing across the sections D and B must be lost in radiation, hence Heat lost in radiation per second from D to B (i) Let e= surface emissivity of the bar, and/ its perimeter, then Heat radiated per second from C to B = ^T8jc . . (2) Equating (i) and (2) gives /72T 1 ^2=#T ....... . (3) Next consider the flow of heat before a steady state has been attained. During this stage the difference between the heat flowing into and that flowing out of the layer DB of thickness 8x is expended in raising the temperature of the layer in addition to that lost in radiation. Let C be the thermal capacity per unit of volume of the material of the bar, and let the mean temperature of the layer DB increase by the amount 8T in time 8t, then Heat expended per second in raising the temperature of the layer . dl =c.A.a*.-^ and equation (3) becomes K.A.S = ^T + C.A.f ..... (4) If now the bar be assumed covered with a perfect non-conductor of h eat, or what amounts to the same thing, consider the heat to be flowing through a plate of infinite area, in which case e = o, equation (4) becomes K ART. 68] THE STEAM ENGINE 109 which may be written ,<**r (5) TT where k = the diffusivity of the material. (*> Equation (5) gives the connection between the temperature at different points in the thickness of the metal and the rate of change of temperature. Assuming a periodic change in temperature of the inside skin of the cylinder walls, the temperature changes at a point in the thickness of the walls will, after a time, attain a fixed character, and the mean temperature will remain steady. Let N be the uniform speed of the crank in revolu- tions per minute ; T the surface temperature of the walls at any instant ; TI the maximum temperature and T x the minimum temperature, i.e. the temperature range of the inside skin of the cylinder walls varies from Tj above to Tj below the mean temperature. Then T = T l cos = T! cos 2?rN/ ....... (6) where 6 is the angle turned through by the crank in / minutes. From (5) and (6) -^ cos (0 px) .-. = -/ny-M* cos (d - f^) + /ny-/** s i n (0 - putting x o, at the surface ~=/*T 1 (sin0--cos0) ctx .-. Q = KfJ,T 1 (cos6-sm6) ..... (11) Now Q has its maximum positive value when (cos 8 sin 6) is a maximum. i.e. when -TA (cos 9 sin 6) = o sin d cos 6 = or when cos 6 = sin 6 This happens when 6 = -- We next want the values of 6 between which heat is flowing ^ Q the metal. From (n) Q = o when cos 6 = sin 6, i.e. when 6 = - ^ _ and since Q is a maximum when 6 = - , it follows that heat must be flowing into the metal whilst 6 changes from to -. 1 Proc. Inst. C. E., vol. cxxxi. 1898. ART. 68] THE STEAM ENGINE in Now heat flow per minute Q = KjuT^cos 6 sin 6) ((n) above) O /. heat flow per unit angle = -4: and the heat flowing in per revolution is O K /. heat flow per unit angle = -4: = ^ pT^ (cos 6 sin 6) 27rN It I * s (cos 6 - sin B)dB J li 37 ^ 27TN VA/2"~V2/" V 2 " l ~" h /NTT 7 But u = /y/ -y-and K = 5*4, k= 1*2 .'. heat flowing in per revolution = -^ \/ ^ - TI 3-I4N From (12) we see that the heat absorbed by the cylinder walls per /iT square foot of surface per revolution of the crank is -^, where TI is half the actual temperature range of the metal surface and N the revolutions made by the crank per minute. The temperature of the cylinder walls does not necessarily follow that of the steam, as may be shown from the following example taken from Professor A. L. Mellanby's paper on the " Effects of Steam- Jacketing upon the Efficiency of a Horizontal Compound Steam-Engine." 1 In one trial (No. 95) the temperature range of the steam was from 357 F. to 247 K, i.e. 110 F. or T x above = 55 F. The speed of the engine was 60 revolutions per minute, and the surface available for con- densation at each end of the cylinder was 6 square feet, hence from (12) Heat absorbed per square foot per revolution = - L -~ B.Th.U. A/N and heat absorbed at each end per minute = 4 ^ X 60 X 6 = 10,200 B.Th.U. 1 Proc. Inst. Mech. Engrs., 1905, p. 544. ii2 THE THEORY OF HEAT ENGINES [CHAP. v. Taking the latent heat of steam at 357 F. as 860 B.Th.U. per pound, the steam which would be condensed per minute 10,200 __ ^ pounds 860 at each end. Hence the steam condensed per hour atv both ends if the walls followed the , , , same temperature variation as the == Ir 9 X 6o X 2 = M8 P ounds steam ) In the actual trial the missing quantity (Art. 71) at cut-off was only 753 pounds per hour, so that it is evident that in this case the cylinder walls did not have so great a range of temperature as the steam. If we assume that all this missing quantity (753 pounds per hour) was due to initial condensation, it follows that the maximum temperature range of the cylinder walls was 1428 whereas the temperature range of the steam was 110 F. It has been shown above that if the temperature range of the surface is known, the range at any depth within the metal can be found (eq. 8). Conversely, if the range at any depth within the metal can be measured, then, assuming a simple harmonic temperature-range at the surface, the surface range and also the heat absorbed per cycle can be easily calcu- lated. Messrs. Callender and Nicolson 1 found by experiment that the surface temperature, instead of following that of the steam, went only through a very small range. In one particular case, at 70 revolutions per minute, the temperature of the steam varied from 335 F. to 212 F., a range of 123 F., but the temperature range of the inside surface of the cylinder walls was only 7 F. 69. Indicated Weight of Steam. In order to estimate the steam consumption of an engine from the indicator diagram it is assumed that the steam in the cylinder is dry and saturated. Thus, in Fig. 52 AB represents the stroke volume and OA is made equal to the clearance volume to the same scale. Then at any point c after cut-off, the volume of steam in the cylinder as shown by the indicator diagram, i.e. the indicated volume, is represented by the length OC, equal to, say, v cubic feet. The absolute pressure of the steam at this point (p) is represented to scale by the length cC. From steam tables the volume of one pound of dry saturated steam at absolute pressure / is obtained ; suppose this is V cubic feet, then assuming dry steam at point c it is evident that the weight of steam present in the cylinder is indicated volume (cubic feet) OC v r-. 7 : pounds = or pounds volume per pound cubic feet r V V r 70. Saturation Curve applied to an Indicator Diagram Dryness of Steam. This curve represents the ideal expansion curve to be obtained in a steam-engine cylinder, being the curve that would be obtained if the whole contents of the cylinder were present throughout the stroke as dry saturated steam. In Fig. 53 OA represents the clearance 1 Proc. Inst. C. E., vol. cxxxi. 1898. ART. 70] THE STEAM ENGINE 113 volume to the same scale that AB represents the stroke volume. The A C B FIG. 52. Indicated weight of steam. total steam present in the cylinder during expansion is made up of two o A FIG. 53. Saturation curve. parts, namely, the weight of steam passing through the engine per stroke i ii4 THE THEORY OF HEAT ENGINES [CHAP. v. called the cylinder feed, and the weight of steam contained in the clearance space before admission, called the cushion steam. The cylinder feed is determined experimentally by running the engine for, say, one hour, condensing and then weighing the exhaust steam ; from the measured steam consumption and the number of strokes made, the cylinder feed in pounds per stroke is easily calculated. The weight of cushion steam is found from the indicator diagram as follows : Any convenient point such as C, Fig. 53, is selected on the compression curve and the pressure/! pounds per square inch absolute, and the indicated volume Vi cubic feet measured. Let Vj denote the volume in cubic feet of one pound of dry saturated steam at pressure / a , obtained from steam tables, then, assuming the cushion steam to be dry and saturated, weight of cushion steam w = v^ pounds i Let W = measured cylinder feed in pounds per stroke. FIG. 54. Dryness fraction of steam. Then, assuming no leakage past the valve or piston, the total weight of steam present in the cylinder during expansion will be w -\- W pounds The saturation curve is obtained by plotting on the indicator diagram with the help of steam tables, a curve for w -\-W pounds of steam as shown by SS in Fig. 53. Dryness Fraction of the Steam from an Indicator Diagram. - Having drawn the saturation curve as explained above, the dryness of the steam at any point during the expansion may be found as follows. At any point b (Fig. 54) the actual volume occupied by the steam in the cylinder is represented by the length ab j the volume which would be occupied by the steam if it were dry and saturated is represented by the length ac t hence, neglecting the volume of the water contained in the cylinder steam its dryness fraction will be represented by ab ac ART. 71] THE STEAM ENGINE 115 The dryness fraction may also be expressed as indicated weight total weight present in the cylinder indicated volume (ab cubic feet) X density (in pounds per cubic foot) ~ weight of cushion steam (w pounds)-]- weight of cylinder feed ( YV pounds) 71. Missing Quantity. It will usually be found in practice that the indicated weight of steam after cut-off is less than the measured steam consumption ; the difference is known as the missing quantity and is represented on the indicator diagram by the length be (Fig. 54). Owing to 90 80 70 20 13 12 8 II Cubic feet FIG. 55. re-evaporation during expansion the missing quantity is almost always less at release than at cut-off. EXAMPLE i. A calibrated indicator diagram is shown in Fig. 55. The measured steam consumption was 1344 pounds per hour at a speed of 200 revolutions per minute. The engine is double-acting and the card shown is an average diagram from both sides of the piston. Estimate the dryness fractions and missing quantities at cut-off and release, and deduce the interchange of heat per pound of steam between the steam and cylinder walls. From the figure we see that at the point C on the compression curve when the pressure is 20 pounds per square inch absolute, the indicated n6 THE THEORY OF HEAT ENGINES [CHAP. v. volume is 0-098 cubic foot. From steam tables the volume of i pound of dry saturated steam at this pressure is 20 cubic feet. 0*098 Hence weight of cushion steam = gQ = 0*0049 pound Cylinder feed per stroke = ~ , = 0-056 pound. .-. Total weight of steam present during expansion = 0-0049 + '56 = 0*0609 pound At cut-off (point A) the pressure is 76 pounds absolute, the indicated volume is 0*166 cubic foot, and the specific volume (from steam tables) 574 cubic feet per pound. If, therefore, the steam were dry and saturated its volume at cut-off would be 5*74 X 0*0609 = '349 cubic foot 0*166 Hence dryness fraction at cut-off = =0*475 or 47*5 P er cent - Indicated weight at cut-off= ; = 0*0289 pound .'. missing quantity = 0*0609 0*0289 = 0*032 pound per stroke = 0*032 X 2 X 200 X 60 = 768 pounds per hour At release (point B) the pressure is 30 pounds absolute, the indicated volume 0*575 cubic feet, and the specific volume (from steam tables) 137 cubic feet per pound. The indicated weight at release will therefore be = 0*042 pound Hence dryness fraction at release = , = 0*600 or 60 per cent. 0-0609 Missing quantity at release = 0*0609 0*0420 = 0-0189 pound per stroke = 0*0189 X 2 X 200 X 60 = 453 pounds per hour From the diagram the mean effective pressure during expansion from A to B is 36 pounds per square inch, hence the work done by 0*0609 Ib. of steam is Mean pressure (Ibs. per sq. ft) X change in volume (cub. ft.) = 36 X 144 ('575 0*166) = 2128 foot-pounds /. work done per pound of steam = 7 5 = 44'9 B.Th.U. 0*0609 X 770 At cut-off the pressure is 76 pounds absolute (L = 903, /= 308 F.). .'. heat per pound at cut-off = 308 32 -j- 0-475 X 903 = 276 + 429 = 705 B.Th.U. ART. 71] THE STEAM ENGINE 117 At release the pressure is 30 pounds absolute (L = 945, /= 250 F.). .*. heat per pound at cut-off = 250 32 -f- 0*690 X 945 = 218 + 652 == 870 B.Th.U. Let HI = heat receivedi per pound from the cylinder walls between cut-off and release, then assuming no heat losses 705 + Hi = 870 + 44-9 = 914-9 .'. Hf = 914-9 705 = 209-9 B.Th.U. EXAMPLE 2. Dry steam is admitted to an engine cylinder at 84 pounds per square inch absolute (/= 315 F., L= 898 B.Th.U., specific volume = 5*22 cub. ft.) and the condensation during admission is 25 per cent, of the whole steam supply. During expansion one-half of the heat absorbed by the cylinder walls during admission is returned to the steam at a uniform rate as the temperature falls. If the expansion be complete and the back pressure be 8 pounds per square inch absolute (/= 183 F., L = 988 B.Th.U., v = 47-3 cub. ft.), find the dryness fraction at the end of expansion. Also, assuming the exhaust steam homogeneous in quality, find its dryness fraction. Neglect clearance, heat losses due to radiation and conduction, and assume the specific heat of water to be constant and equal to unity. From steam tables we find the following : Pressure. Temperature, F. Latent heat. Specific volume. Entropy. Water. Evaporation. 84 315 183 898 9 88 5'22 47 '3 o'4579 0-2673 1-I58I I-5380 The temperature-entropy diagram is shown in Fig. 56. Since the condensation during admission is 25 per cent., it follows that the dryness BC" fraction of the steam at cut-off is 0*75 or ^p The dryness fraction at AD the end of expansion is represented by -r^ . ALr The heat absorbed per pound of steam by the cylinder walls during admission is the heat returned during expansion while the temperature falls 132 F. = ^5 =112-25 B.Th.U 2 .-. rate of heat return = II2 ' 2 5 132 = 0-85 B.Th.U. per F. fall in temperature /. 8H = o'8s8T SH . , Q 6T or = 8^ = 0-85 n8 THE THEORY OF HEAT ENGINES [CHAP. v. Hence total gain of entropy during expansion = FD (Fig. 56) /775 ?T = Lo' 8 5 T = 0-85 log. Jit = 0-85X0-187= 0-1589 units .-. AD = AH + HF + FD = (o'4579- 0-2673) + (o'75 X 1-581) + = 0-1906 + 0-8596 + 0-1589 = 1-2091 units 775 643 B FIG. 56. AD Hence dryness after expansion = AVjr _ I-2O9I = I-538 = 0-786 or 78*6 per cent. During exhaust the cylinder walls must return the remainder of the ART. 72] THE STEAM ENGINE 119 absorbed heat to the steam namely 112-25 B.Th.U. Hence the further rise in entropy DK during exhaust will be * != 0-1745 units and the dry ness fraction at the end of exhaust will be AK_ 1-2091 -f- ' I 745 AG~ 1-538 = 0*90 or 90 per cent. It should be noticed that the shaded area CDF represents the extra work done per pound of steam as the result of re-evaporation during expansion, and that the actual expansion line CD, which may also be called the re-evaporation line, shows at a glance how the expansion deviates from the adiabatic CF. 72. Method of Drawing the Temperature-Entropy from the Indicator Diagram. The method will be best illustrated by means of an example. Consider the diagram shown in Fig. 55. The maximum steam pressure at admission is 82 pounds per square inch absolute (/= 313 F.) and the back pressure during exhaust is 12 pounds absolute (/= 202 F.). The first step consists in finding the dryness fraction of the steam at any point on the expansion curve. This has been found to be 0*690 for the point of release B in Example i worked out in Art. 71. The point B may therefore be transferred directly on to the temperature-entropy chart. On the indicator diagram at B (Fig. 55) it has already been shown that the cylinder contained 0-0609 pound of steam, of dryness 0-692, and occupying a volume of 0-575 cubic foot. The temperature-entropy diagram, however, is drawn for i pound of steam. The actual volume occupied by i pound of steam of dryness 0*692 and pressure 30 pounds absolute is, neglecting the volume of water, 13-6 X 0*690 cubic feet The T< diagram will, therefore, be drawn for an engine which is 13*6 X 0-690 - ; - = 1 6 '3 6 times larger than the actual engine. If, now, any point on the indicator diagram be taken and the volume of the steam actually present in the cylinder be measured and then multi- plied by 16*36, the corresponding point on the T< diagram is completely determined. A series of points i, 2, 3, 4, etc., are next taken on the pv diagram at convenient pressures, and the pressure and volume read off for each point. Each volume is then multiplied by the factor 16*36 and the points plotted on the T(j) diagram. The results obtained are shown in the following table. It should be remembered that in the above method the effect of valve leakage is neglected, and it is assumed that the weight of steam present in the engine cylinder during expansion is a constant quantity. If the amount of leakage is known, an allowance should be made for it. For the method to be followed in such cases see the First Report of the Steam Engine Research Committee of the Institution of Mechanical Engineers. 1 1 Proc. I. Mech. E., 1905, p. 239. T2O THE THEORY OF HEAT ENGINES [CHAP. Point. Absolute pressure, Ibs. per sq. in. Volumes. > T0. I 82 0-055 0-90 2 80 0-100 1-64 A 7 6 0-166 2-71 3 70 0-184 3-01 4 60 O'22I 3'6i 5 50 0-268 438 6 4 0-374 6-12 7 33 O'5OO 8-18 B 30 0-575 9-40 8 20 0-575 9-40 9 15 0-566 9-27 10 12 0'200 3'27 ii 17 O'HO i -80 12 20 O'lOO 1-64 3 30 0-074 I-2I M 40 0-055 0-90 The temperature-entropy diagram is most conveniently drawn by laying a piece of tracing paper over the T chart and plotting the points directly on it. Thus, the position on the chart corresponding to point i is the intersection of the constant-pressure line for 82 pounds per square inch, with the constant volume line for 0*90 cubic foot, and similarly for all the other points. The complete tempera- ture-entropy diagram, corresponding to the indicator diagram shown in Fig. 55 is thus drawn, being shown in Fig. 57. When the temperature - entropy diagram is drawn in this, or any other way, the expansion line clearly shows the nature of the interchange of heat between the steam and the cylinder walls. Referring to Fig. 57 it will be seen that throughout the expansion from A to B there is a gain of entropy, and therefore the cylinder walls are restoring heat to the steam, and as re- evaporation continues, the dryness fraction of the steam increases as shown. 73. Boulvin's Method of draw- ing the Temperature - Entropy Diagram from the Indicator Dia- gram. Draw the axes OP, OT, OV, and OE (Fig. 58). Set off along OV a scale of volumes, in cubic feet for i pound of steam, making its length to represent at least the volume of i pound of steam at the lowest pressure on the indicator diagram. On OT set off a scale of temperature and along FiG. 57. T diagram for pv diagram shown in Fig. 55. ART. 73] THE STEAM ENGINE 121 OP a scale of pressure ; then by the aid of steam tables plot the tempera- ture-pressure curve for saturated steam as shown in the quadrant POT. The indicator diagram is next to be transferred to the quadrant POV. Draw in this quadrant the saturation curve for i pound of steam, using steam tables. On the actual indicator diagram draw the saturation curve for the actual weight of steam present during expansion by the method already explained in Art. 70. Next transfer the actual indicator diagram V H \ S i & 3\ - *k< SB sg cv ^\ 9 v.\ i i* FIG. 58. Boulvin's method. (Fig. 59) to the quadrant POV with the same relationship to the saturation curve already drawn therein as it has to its own saturation curve, plotting it directly to the new scale of pressures and volume. In the quadrant TOE draw the steam and water lines of the tempera- ture-entropy diagram for i pound of steam by the method of Art. 42, or transfer these lines directly from a temperature-entropy chart. Take any convenient point A on the saturation curve and draw the 122 THE THEORY OF HEAT ENGINES [CHAP. v. vertical line AB to the temperature-pressure curve, and the horizontal BF cutting the entropy lines in D and F. Project DG vertically to cut OE in G, and FH to cut the horizontal from A in H. Join HG ; from J draw a horizontal to cut HG in L, and from L draw a vertical to cut BC in K. Then K is a point on the T< diagram required. Repeat this process for different points on the indicator diagram, and in this manner the indicator diagram is transferred to the temperature-entropy diagram. Fig. 59 shows the actual indicator diagram, already considered by the other method, with its saturation curve, and Fig. 58 is drawn from it by the above graphical method. 74- Valve Leakage. In order to draw the saturation curve on an indicator diagram one of the assumptions made in Art. 70 was that the engine cylinder contained a constant weight of steam during expansion, namely, the sum of the cushion steam and cylinder feed. On this assump- 90 80 70 60 50 40 30 20 10 / -2 3 '4 '5 -6 FlG. 59. Actual indicator diagram. tion it is obvious that the whole of the missing quantity (Art. 71) is due to condensation. If, however, steam leaks past the valve or engine piston or both, the theory given in Arts. 70 and 72 is not true, since there is no longer a constant weight of steam present in the cylinder after cut-off. Neglecting leakage past the piston and even assuming no leakage past the valve after cut-off, it is evident that, although there will be a constant weight of steam present during expansion, the theory previously given will require modifica- tion, 1 and in order to estimate the dryness fraction of the steam at any point on the expansion curve, the total steam present must be made up of cushion steam, cylinder feed, and leakage steam. If leakage does take place, then the missing quantity will be due both to condensation and leakage. Referring to Fig. 54, which represents an indicator diagram having drawn on it the saturation curve for the cushion 1 See the First Report of the Steam Engine Research Committee, Proc. I. Mech. E., March, 1905, p. 239. ART. 74] THE STEAM ENGINE 123 steam plus cylinder feed, i.e. when valve leakage is neglected, it will be seen that if leakage occurs the length of ab will be unaltered, but the length of be, which represents the missing quantity, will be increased, and, more- over, if leakage continues throughout the expansion, the weight of steam present will be a continuously increasing quantity from cut-off to release. Great diversity of opinion exists amongst various authorities as to the existence of an appreciable valve leakage in modern steam engines. If it be accepted as true that all valves leak when under working conditions, it is evident that in order to reduce the missing quantity, and therefore obtain greater economy, more attention should be paid to the design of valves instead of endeavouring to reduce the clearance volume to the smallest possible value. There is a considerable amount of evidence that some valves at any rate do leak, such leakage having been measured by such authorities as Messrs. Callendar and Nicolson, Professor Capper, Captain Sankey and others; a brief review of some of the results obtained will be instructive. Professor Callendar and Nicolson l give their opinion of the way in which leakage takes place through slide valves in the following words : " So long as the valve is stationary, the oil film may suffice to make a perfectly tight joint ; but as soon as it begins to move, the oil film becomes broken up and partly dissipated. Water is being continually condensed on the colder parts of the surface exposed by the motion of the valve. This water works its way through, and breaks up the oil-film under the combined influence of the pressure and the motion. The continual evaporation taking place in the exhaust tends to maintain the leaking fluid in the state of water. The exhaust steam from the cylinder has the same tendency. ... It is not improbable that the quantity of water which can leak through a given crack under the given difference of pressure, may be from twenty to fifty times greater than the quantity of steam which can leak under similar conditions. This agrees with well- known facts in regard to leakage, and explains how it is that the leakage in the form of water is so great. . . . An explanation is thus furnished of a possible form of leakage, indirectly due to condensation and re- evaporation, so many times greater than steam leakage, which, alone, engineers have been in the habit of contemplating, that it might well claim attention on its own merits, apart from the very limited number of valves on which it has hitherto been possible to make direct experiments. " The analysis of a large number of observations, in addition to the few made by the authors, leads to the conclusion that all valves leak more or less when in motion, and that in many cases the greater part of the missing quantity is to be attributed to leakage of this description. What- ever the precise manner in which the leak takes place, it appears to be nearly proportional to the difference of pressure and to be in most cases independent of the speed. In any case it appears probable that the leakage is connected in some way with the condensation taking place on the valve surfaces. If so, it may evidently be greatly reduced, if not entirely cured, by jacketing, or otherwise heating the valve seat, to minimize the condensation. " These views have an important bearing on the design of valves. For low-speed engines, separate steam and exhaust-valves should possess 1 See Proc. Inst. C. E., 1897-8, vol. cxxxi. p. 179. i2 4 THE THEORY OF HEAT ENGINES [CHAP. v. advantages over the ordinary slide valve. The superiority of the com- pound engine would also appear to be partly due to the great reduction of possible leakage." From their results with slide valves, Messrs. Callendar and Nicolson state the law of leakage in pounds per hour as Diff. of press, on the two sides of the valve X perim. of steam port _ - _ x O'O2 Mean overlap Professor Capper measured the leakage by blocking up the steam ports of the slide valve, driving the engine at different speeds by external power with steam admitted to the steam chest at varying pressures, and con- densing and weighing the steam which leaked past the valve. The con- clusions he arrived at may be briefly summed up as follows l : Effect Of Steam Jacket. The leakage when the barrel of the cylinder was warmed by means of a steam jacket was considerably less than when unjacketed. Effect of Lubrication. There is a distinct reduction in leakage when the sliding surfaces are well lubricated over the corresponding leakage with scant lubrication. Effect Of Pressure. When the valve is stationary, and in mid- position, the leakage is approximately proportional to the steam pressure. When the engine is running, however, the leakage does not increase so rapidly as the pressure, and the higher the speed the larger does this divergence become ; hence, much of the leakage must be in the form of moisture condensed on the valve face and re-evaporated as it passes over into the exhaust. It will be noted that this confirms Callendar and Nicolson's conclusions, although they found that the leakage was practically independent of speed. Effect of Wire- drawing. The dry steam used was wire-drawn between the main steam pipe and the steam chest, and it was found that in all cases there was a sensible reduction in leakage as a result of super- heating the steam and therefore reducing the amount of condensation. Effect of Speed. The leakage was consistently less at 250 revolu- tions per minute than at 50 revolutions per minute. This may be partly due to a more perfect spreading of the oil film at the higher speeds, or it may be due to the reduced time allowed for condensation and re- evaporation, or to a combination of the two. This clearly shows that in these experiments, leakage was not produced by a lifting of the valve from its seat when working, but by a more or less steady flow of moisture or steam between the valve and the slide face. Effect of Overlap. The persistent and considerable difference between the leakage when the engine is stationary with the valve in mid- position and when it is running, point strongly to the conclusion that the amount of overlap and its variation has an important effect on the leakage. As already mentioned, Callendar and Nicolson give the mean rate of leakage as Difference of pressure X perimeter Mean overlap 1 First Report of the Steam Engine Research (Committee, Proc. Inst. Mech. E., March, 1905. ART. 74] THE STEAM ENGINE 125 Professor Capper finds that the leakage is not directly proportional to the pressure for all speeds, and states that there is reason to doubt whether it is exactly inversely proportional to the overlap ; assuming, however, that it is so, then he finds that C is not constant although its mean value is o'02, which is identical with that found by Messrs. Callendar and Nicolson. Leakage of Piston-Valves and Piston-Rings. 1 Captain H. Riall Sankey measured the leakage past the piston valve and rings of a Willan's engine. He found the above conclusions for a slide valve to be substantially correct for a piston valve, although the value of the constant C was 0*003 instead of 0*02. Warping of the Valve. A slide valve is a rather intricate casting which when cold may be true and steam-tight, but when heated under working conditions may become distorted and so allow steam to leak past. It is well known that a truly cylindrical shape is the least susceptible to warping, and that therefore the probable warping of a piston valve could be less than that of a slide valve. There is some evidence to show that warping may be wholly or partially the cause of leakage by the fact that the value of C found by Captain Sankey was 0*003 as against 0*02 for a flat valve. From tests on a piston valve Mr. H. Denzil Lobley 2 found that the leakage was very small, and in the particular valve tested would account for a negligible proportion of the missing quantity. Professor Mellanby 3 also brings forward considerable evidence to show that valve leakage may account for a large proportion of the missing quantity. For instance, in Trial No. 95, which has already been quoted in Art. 68, the highest temperature of the steam was 357 F. and the mean temperature was 284 F. The mean temperature of the cylinder wall was 335 F., hence the maximum temperature range of the walls could not be more than 2(357 335) = 44 F. Using the method of Art. 68 (p. in) it would appear that this temperature range at a speed of 60 revolutions per minute would result in 555 pounds being the maximum amount of steam that could be condensed per hour. Now the actual missing quantity near cut-off was, for this trial, 753 pounds, hence it would appear that the remaining 753 555 or 198 pounds represents the minimum amount of valve leakage per hour. It will also be noted that in this trial the mean temperature of the cylinder walls (335 F.) is considerably higher than that of the steam (284 F.). , He also found that in all trials, both jacketed and unjacketed, the apparent re-evaporation during expansion in the high-pressure cylinder was less when jacketed than when unjacketed. It is difficult to see how this can be so unless the view be accepted that the missing quantity is largely due to leakage and that the steam in the cylinder is dry before leakage takes place. The experiments enumerated above have been made on a few valves only ; hence one is justified in saying, that before a definite law of valve 1 See discussion on First Report of Steam Engine Research Committee, Proc. I. Mech. E., March, 1905, p. 275. 2 The Engineer, Feb. 9, 1912. 3 See " Effect of Steam-Jacketing of a Compound Engine," Proc. I. Mech. E., June, 1905. 126 THE THEORY OF HEAT ENGINES [CHAP. v. leakage can be stated it will be necessary to make numerous tests of different types of valves constructed of cast-iron and other materials under various conditions of working. In a physical sense it is difficult to see how all valves can leak, considering the present state of perfection attained in modern workshop practice, or that valves should leak more than stuffing boxes, etc. The elucidation of the vexed problem of valve leakage is beset with enormous difficulties, and because a certain valve may be found to leak in any particular engine it does not necessarily follow that the same type of valve will leak when fitted to another engine working under similar or under different conditions. For further information on this subject the reader is referred to a paper by Professor A. L. Mellanby on "Surface Condensation in Steam Cylinders," read before the Institution of Engineers and Shipbuilders in Scotland, on December 21, 1911. 75. The most Economical Ratio of Expansion. Consider first of all the theoretical indicator diagram in which clearance is neglected and the expansion assumed to be hyperbolic (Art. 61). The work done per cubic foot of steam admitted to the engine cylinder is, by Art. 61, W= i44{/ 1 (i -f- loge r) r.pb] foot-pounds . . . . (i) where p = initial pressure in pounds per square inch, absolute r = ratio of expansion The value of r which makes this a maximum is obtained by putting - = o, hence dr ^W_/i_ dr ~ r ~* b ~ or r=^ (2) The problem is not so simple as this, however, in the actual engine. In the first place, there will be less work done on account of initial condensation. Let w represent this loss due to condensation, then the work done per cubic foot of steam may be written W = i44/i( z + lo g r ) mr-pb w (3) dW p\ dw = 144 144/6 -7- = o for a maximum dr r dr dw ( p-\ \ / x or r- = 1441 pb) (4) dr V r For non-condensing engines Mr. P. W. Willans finds that the best ratio of expansion is given by r = -~ for a simple engine and r = for a compound engine. 1 Taking ? = -- and /& = 17 for a non-con- densing engine and substituting in (4) we find -^=144(25 17) = 8 X 144 = 1 Proc. Inst. C. E., 1887-8, vol. xciii. ART. 76] THE STEAM ENGINE 127 Integrating we have w= 1 1 527- + a constant ...... (5) This is the equation to a straight line, or in other words, on this theory the loss due to condensation is a linear function of the ratio of expansion, the effect of condensation being to increase the back pressure by 8 pounds per. square inch. In addition to the loss due to condensation there will be a further reduction in the work available as a result of engine friction. Suppose friction to be equivalent to a back pressure of /"pounds pei square inch, and writing c for the equivalent back pressure due to condensation, we have the net work available per cubic foot of steam, W = 144^(1 + loge r) I44K pb +/) i44(^ + A) . (6) rfVf 144/1 / r\ r = ^ i44(/6 +/) 144^ =0 for a maximum - (A +/)-'= r= On the above method of reasoning it would therefore appear that the most economical ratio of expansion to adopt for a non-condensing engine is equal to the initial steam pressure divided by a constant ; as already mentioned, Mr. Willans adopted 25 as the value of this constant for the particular engine he experimented on. In the case of a condensing engine, however, he says I : "It is hardly possible to define, for a con- densing engine, the best ratio of expansion. This question must remain one for each particular engine builder to answer. The useful work obtainable from the steam depends so greatly on the condenser back pressure and on the friction of each individual engine that it is impossible to give any general rule." 76. The Steam Jacket. The theory of the steam engine when the steam is maintained dry and saturated throughout the expansion has already been given in Art. 59 ; we will now see how close the actual engine, when fitted with steam jacket, approximates to the ideal case there considered. When dry saturated steam is admitted to an unjacketed cylinder there is always a certain amount of initial condensation (Art. 67) and the steam will be wet at cut-off. In order to prevent initial condensation when using saturated steam, it is essential that the temperature of the cylinder walls should not be below that of the entering steam, and this condition may be obtained by the application of a suitable steam jacket. Let abed (Fig. 60) represent the temperature-entropy diagram with unjacketed cylinder walls, the dryness fraction at cut-off being -r and at release -7 the expansion cd being adiabatic. If now a steam jacket be fitted to the cylinder and the steam maintained dry up to cut-off after which the expan- sion is again adiabatic, the heat supplied to the working steam by the jacket will be represented by the area cekh^ and the extra amount of work 1 Proc. Inst. C. E., 1892-3, vol. cxiv. 128 THE THEORY OF HEAT ENGINES [CHAP v. done by the area cegd. If the effect of the jacket is such that the steam is maintained dry throughout the expansion, the total heat supplied by the jacket will be represented by the area ceflh, and the extra work done by the area cefd, the result in either case being an increased efficiency (Art. 59). See also Fig. 56, and the example worked out in Art. 71. From the published results obtained with steam engines working with and without jackets 1 it appears that the increased efficiency resulting from the use of steam jackets varies from 3 per cent, to 25 per cent, depending on the type of engine. Speaking generally, it appears that jackets are more useful for slow- speed engines than for high-speed engines, and are also useful for simple and compound engines, but their efficiency is doubtful if they are ap- plied to triple or quad- ruple expansion engines. It has also been found f in practice that the more economical an engine is, apart from jacketing, the less advantage is gained by the application of a jacket. If an unjacketed engine uses say 13 pounds of steam per hour per I.H.P. with an initial pressure of about 1 20 pounds per /i fc / square inch, the value where L = stroke in feet, A = area of low-pressure cylinder in square inches, and N = number of working strokes per minute. The value of e depends upon the type of engine, i.e. whether simple or 1 Proceedings Inst. Mech. E. 1905, p. 300. 2 For further particulars on the use of superheated steam the reader is referred to the following papers: Institution of Marine Engineers, "Marine Engines and Super- heated Steam," by Mr. A. F. White, in the Marine Engineer and Naval Architect^ Dec. 1909. Institution of Naval Architects, " Superheaters in Marine Boilers," by Mr. Harold E. Yarrow, read March 28, 1912, and reproduced in Engineering of April 5, 1912. Proceedings Inst. C. E., vol. cxxviii., " Superheated Steam Engine Trials," by Prof. W. Ripper. 132 THE THEORY OF HEAT ENGINES [CHAP. v. compound, jacketed or unjacketed, high-speed or low-speed, etc. The Kw. output Stop valve Vacuum at Set. of generator coupled to Load at test. steam press. Lbs. per engine. Inches of Date of Test. engine. sq. inch. Mercury. A 208 Full 155 26 gin. 1904. B 2 2O . j 175 2 5 ov. 1902. C 308 > 190 25 Dec. 1902. D 362 J 162 2 5 -8 Feb. 1903. E 500 | ISO 26 Mar. 1904. F, F 2 700 5 80 1 190 192 27 27 Jan. 1905. Feb. 1905. G 1456 Full I8 3 20 July & Aug. 1903. 50 100 150' 200 250" 300 350 4-00 Superheat/ Degrees Fahr. above Saturated Steam Temperature/ FIG. 63. Non-jacketed quick-revolution triple-expansion condensing engines, using superheated. (Experiments on Messrs. Belliss and Morcom's engines.) difficulty experienced in estimating the diagram factor for a new engine lies in the uncertainty of the probable value of the back pressure. For ART. 78] THE STEAM ENGINE 133 the same condenser pressure, the actual back pressure in the engine (Engine A of Fig. 63), showing effect of superheat on steam-consumption at varying loads. The percentage figures indicate the increase in Ibs. of water per B.H.P.-hour over full load consumption. 20 5000 19 4-500 11 500 -j 1 c 1 1 x: 1 i vi , - ~ 1 1Q Q 50 100 150 200 250 300 B. H. R FIG. 64. Non-jacketed quick-revolution triple-expansion engine. cylinder will not be a constant quality ; it will usually vary with the speed 134 THE THEORY OF HEAT ENGINES [CHAP. v. of the engine, the initial pressure, and the ratio of expansion. If the initial pressure or speed of the engine, or both, be increased,, it is evident that more steam will be passed through the engine, the back pressure will usually rise, and e is at once affected. Mr. C. H. Wingfield 1 has shown that if a quantity which he calls the virtual back pressure is substituted for the actual back pressure, this virtual back pressure, which can be found from trials of similar engines, is not only independent of the speed and number of expansions, but the diagram factor e is less affected by the expansions, and scarcely at all by the speed. He applies his method to the results obtained by Mr. P. W. Willans in his condensing engine trials. 2 The virtual back pressure is obtained by plotting the actual mean effective pressure, and the theoretical mean pressure as shown in Fig. 65. The intercept of the resulting straight line on the vertical axis gives this virtual back pressure. Actual Mean Effective Pressures. 12-6 i Theoretical Mean Effective Pressures. FIG. 65. Virtual back pressure. With a constant number of expansions (/= 4-82) he finds the virtual back pressure to be 12-6 pounds per square inch, absolute, and the diagram factor e to vary from 0*69 at 400 revolutions per minute to 072 at 200 revolutions per minute. The slight reduction of e with increase of speed is in all probability due to the influence of wire-drawing. With a constant speed of 400 revolutions per minute, the virtual back pressure was 3 Ibs. per square inch absolute, and with the expansions used by Mr. Willans the values of the diagram factor are Expansions. 4-82 lO'OO 15-55 0-69 077 0-83 1 Engineering, Oct. 20, 1 893. 2 Proceedings Inst. C. E., vol. cxiv. 1892-1893. ART. 79] THE STEAM ENGINE 135 The values of the diagram factor obtained by Prof. A. L. Mellanby on a compound engine 1 are shown in Fig. 66 for the engine when both jacketed and unjacketed, together with the actual and theoretical mean effective pressures. This diagram shows very clearly how the diagram factor varies with the number of expansions. From the results obtained in practice, 2 it appears that the probable values of e for various engines are : for a compound marine engine, about 07 ; for a triple expansion marine engine about 0*635 ; for a horizontal Corliss engine without jackets, from 076 to 0*86. For a locomotive it is 16 20 Number of expansions FIG. 66. Diagram factor. 24 difficult to fix any value, but at a high piston speed of from 700 to 800 feet per minute, e appears to be about 0-63, and at slow speeds of about 170 feet per minute about o'8. In all cases, for a given engine, the lower the speed the greater is the value of the diagram factor; this is no doubt due to there being less wire- drawing and more time for initial condensation to take place, the subse- quent re-evaporation during expansion increasing the mean pressure at the lower speeds, and to the lower compression pressure. 79- Steam Consumption the Willans Law. It may be shown 1 Proceedings Inst. M. E., June, 1905, p. 535. 2 See a paper by C. H. Innes, M.A., Practical Engineer ; June 17, 1892. 136 THE THEORY OF HEAT ENGINES [CHAP. v. from theoretical considerations that, provided the ratio of expansion remains constant, i.e. if the engine governs by throttling, the steam consumption is a linear function of the indicated horse-power. Let W = steam consumption in pounds per hour, and P the indicated horse-power, then using the same notation as before 4- log Since r remains constant, this may be written 33,000 Let #>! be the weight of i cubic foot of steam at absolute pressure p 6oALN , x then W = -- w-i ...... (2) 1447- If w 1 and/! be plotted on squared paper using steam tables (p. 480), it will be found that approximately ^1 = where a and j3 are constants. Substituting for w in (2) we get Putting (i) in (3) gives 6oALN/ . j3 ^ \ , 6oALN 33, = - a + C A> + ~^ ' CLAN or W = a + P (4) where a and b are constants having the values a = (a + -/ft) and 1447- c _ 6oALN 33,000 1447- 'CLAN' This equation W = a + b? is known as the Willans' straight line law, and is found to be sensibly true for all actual engines working with a constant ratio of expansion, as will be evident from the curves given in Fig. 64, and in Mr. Willans' papers already referred to. EXAMPLES V I. Estimate the work done per cubic foot of steam in the following cases : (a) When there is no clearance and no compression. (b) When the clearance is 0-5 cubic foot and with no compression. (c) When the clearance is O'5 cubic foot and the compression pressure 50 pounds per square inch absolute. EX. V.] THE STEAM ENGINE 137 (d) When the clearance is 0*5 cubic foot and the compression equal to the initial steam pressure. In each case assume an initial steam pressure of 100 pounds per square inch absolute and a ratio of expansion of 4. 2. The following particulars are obtained from an indicator diagram taken from the high-pressure cylinder of a compound steam engine fitted with Corliss valves : Cut off ^ stroke ; at a point on the compression curve the pressure was 50 Ibs. abs., and indicated volume 4 cubic feet. Pressure at a point on expansion curve just after cut-off =155 Ibs. abs., and indicated volume = 7*2 cubic feet. Pressure at \ stroke on expansion curve =112 Ibs. abs. Pressure at release = 62 Ibs. abs., and indicated volume = 17*5 cubic feet. The diameter of the cylinder is 28 inches and stroke 4 feet, the clearance volume being 7 per cent, of the stroke volume and the cylinder feed 2*58 pounds per stroke with 150 working strokes per minute. Estimate the dryness fraction and missing quantity in pounds per hour (a) at cut off, () at stroke, (c] at release, given p V 155 112 62 So 2-92 3-98 6'95 8-51 3. Dry steam is admitted to an engine cylinder at 60 pounds per square inch absolute, and the condensation during admission is 20 per cent, of the whole steam supply. During expansion three-quarters of the heat absorbed by the cylinder walls during admission is returned to the steam at a uniform rate as the temperature falls. If the expansion be complete and the back pressure be 4 pounds per square inch absolute, find the dryness fraction at the end of expansion, and, assuming the exhaust steam to be homogeneous in quality, find its dryness fraction at the end of the exhaust stroke. Neglect clearance and all heat losses, and assume the specific heat of water to be constant and equal to unity. Use the steam tables given on p. 480. 4. A steam engine cylinder is 33^ inches diameter and the piston has a stroke of 3 feet 3 inches. The engine develops 600 I.H.P. at 100 revolutions per minute. Assuming a diagram factor of 0*82 what is the ratio of expansion if the initial steam pressure is 155 pounds per square inch absolute and the back pressure 2 pounds per square inch absolute ? CHAPTER VI COMPOUND EXPANSION 80. Advantages of Compound Expansion. In order to obtain economical results with the high boiler pressures used in modern practice it is necessary to work with a large number of expansions (Art. 75). Several disadvantages attend the use of a single cylinder for this purpose. In the first place, a very early cut-off would be required ; and in order to allow for the low release pressure desirable, the volume of the cylinder would have to be large enough to accommodate the large volume occupied by the steam at this pressure. The temperature range of the steam and the loss due to initial condensation would be excessively large (Art. 75). Also since the mean effective pressure on the piston would be a small portion of the initial steam pressure, the diameter of the cylinder would have to be made very large to develop the power required ; and further, all the working parts of the engine would have to be made strong enough to withstand the high initial pressure. The combined effect would be an engine of excessive size and weight (and therefore cost) which would be uneconomical in working for two reasons, namely, the high steam con- sumption produced by the excessively large initial condensation, and the comparatively low mechanical efficiency resulting from the large frictional resistance unavoidable with heavy moving parts. In addition to the above disadvantages the variation of the turning movement on the crankshaft would be very great, and in order to reduce the cyclic variation in speed a very heavy fly-wheel would be necessary (Art. 243), which in its turn would again increase the weight of the engine and the frictional resistance at the main bearings and tend to lower the mechanical efficiency still further. By employing compound or multiple expansion, in which the expansion is carried out successively in two or more cylinders, the initial condensa- tion, and therefore the steam consumption, is reduced, and further, the range of stress on the different pistons is diminished and a more uniform turning moment obtained on the crankshaft. In the compound engine the expansion takes place in two stages. The high-pressure steam is admitted into the high-pressure cylinder, and after cut-off expands through a certain ratio and is then exhausted from the high-pressure cylinder into the larger low-pressure cylinder in which the expansion is completed. In the triple expansion engine the expansion is carried out in three stages, the successive cylinders being known as the high-pressure, intermediate pressure, and low-pressure cylinders respec- tively, whilst in the quadruple expansion engines (used in conjunction with the highest boiler pressures) four cylinders are commonly used, the 138 ART. Si"] COMPOUND EXPANSION 139 high-pressure and low-pressure with two intermediate cylinders. In all cases the temperature range of the cylinder walls is reduced, and only the low-pressure cylinder is ever in communication with the low temperature of the condenser. By employing multiple expansion in this manner the initial condensa- tion is reduced in two ways, the temperature range being reduced to a practical value in each cylinder and also on account of re-evaporation during exhaust, the steam as it enters the successive cylinders is drier than it otherwise would be ; further, by a suitable arrangement of crank angles in conjunction with the smaller range of pressure in each cylinder a more even turning moment is obtained on the crankshaft (see Art. 238). 81. Compound Engines without an Intermediate Receiver. In this type of engine the steam is exhausted from the high-pressure cylinder directly into the low-pressure cylinder, the two cylinders remain- ing in communication throughout each stroke, there being continuous FIG. 67. Compound diagram without receiver. expansion in the low-pressure cylinder without an independent cut-off. In this type the cranks must be set either o or 180 apart; in the former case both the high-pressure and the low-pressure pistons are attached to a common piston rod and therefore both move together, this type of engine being called the tandem compound. When the cranks are 180 apart the cylinders are arranged side by side, each having its own piston driving a crank. In the case of a vertical engine, when the high-pressure piston is moving downwards, the low-pressure piston is travelling upwards and vice versci, and the steam is exhausted from below the high-pressure piston to the under side of the low-pressure piston where the expansion is com- pleted. On the next (upward) stroke of the high-pressure the steam is exhausted from the top of the high-pressure piston to the top of the low- pressure and so on. The theoretical indicator diagram for this type of engine, commonly called the Woolf type, is shown in Fig. 67. ABCD represents the diagram for the high-pressure cylinder, in which OA is the absolute admission 140 THE THEORY OF HEAT ENGINES [CHAP. vi. pressure; cut-off takes place at B and the steam expands down to C followed by exhaust, CD, from the high-pressure cylinder directly into the low-pressure cylinder where the expansion is completed. EFGH repre- sents the diagram for the low-pressure cylinder in which the expansion EF continues down to the required release pressure at F followed by exhaust GH. The release pressure in the high-pressure cylinder, i.e. the pressure at C, is the same as the initial pressure (at E) in the low-pressure cylinder and at the end of the exhaust stroke in the high-pressure the pressure is of necessity the same as the release pressure! in the low-pressure cylinder, i.e. the pressure at D is the same as at F. The two diagrams may be combined by drawing a number of horizontal lines, such as JM between E and D, and making LM equal to JK. The combined diagram ABF'G'H represents the equivalent theoretical indicator diagram which would be obtained by carrying out the expansion in a single cylinder of the same size as the low-pressure cylinder and is equal in area to the sum of the H K 0' H' FIG. 68. Compound diagram with receiver. areas of the high-pressure and low-pressure diagrams. The mean effective pressure obtained from the combined diagram ABF'G'H is called the mean effective pressure referred to the low-pressure cylinder and in the theoretical diagram shown is equal to ^(i+log e r) /& where/! is the initial absolute pressure OA, pb the absolute back pressure OH, and r the ratio of expansion -^-g- or -^g 82. Compound Engines with an Intermediate Receiver. In this type of engine a receiver is usually fitted between the high-pressure and the low-pressure cylinders. In many cases a separate receiver is not fitted, the steam pipe between the two cylinders answering the same purpose. Fig. 68 shows the compound diagrams of a tandem engine of the receiver type, ABODE being the high-pressure and EFGHK the low-pressure ART. 82] COMPOUND EXPANSION 141 diagram. After expansion, BC, in the high- pressure cylinder, the steam is exhausted into the receiver, and during the first portion of the high- pressure exhaust stroke CD, steam is admitted into the low-pressure cylinder from the receiver along the line EF. Steam is then cut-off in the low-pressure cylinder at some point F and the expansion is continued along FG in this cylinder independently of the high-pressure cylinder. After cut-off takes place in the low-pressure cylinder the exhaust stroke in the high-pressure cylinder is completed by compressing the remaining steam into the receiver along DE, until at the end of the stroke the pressure in the receiver (at E) is the same as the release pressure (at C) in the high-pressure cylinder. In order that the initial pressure (OE) in the low-pressure cylinder may be the same as the release pressure in the high-pressure cylinder, the cut- off in the low-pressure cylinder must occur at a certain fixed point, or in other words, the point D must be so chosen that at the end of compression FIG. 69. into the receiver the pressure is the same as at C ; in such cases there is evidently no drop in pressure between release in the high-pressure cylinder and the receiver. Since the two cylinders are never in direct communication it is evident that the cranks may be set at any required phase angle (usually 90), and further the temperature range in the high-pressure cylinder, and therefore the initial condensation, are both less than in the case of the Woolf type of engine. The greater the volume of the receiver the less will be the variation in the back pressure of the high-pressure cylinder, i.e. the more nearly will CDE approach to a horizontal straight line. In what follows the receiver volume will be assumed infinitely large, in which case the back pressure in the high-pressure and the admission pressure in the low-pressure cylinder will be constant, the combined diagram taking the form shown in Fig. 69, in which ABCD is the high-pressure and ADEFG the low- pressure diagram, the receiver pressure remaining constant and equal to OA. 142 THE THEORY OF HEAT ENGINES I[CHAP. vi. 83. Ratio of Cylinder Volumes. With a given ratio of expansion let ABODE (Fig. 70) represent the equivalent indicator diagram for one cylinder of the same volume as the low-pressure cylinder ; its area will represent the work done by a certain weight of steam and will be equal to the sum of the work done in the high- and low-pressure cylinders. If the total work done is to be equally divided between the two cylinders, the area of the high-pressure diagram ABFG must be made equal to that of the low-pressure diagram GFCDE, and the ratio of cylinder volumes will be Volume of low-pressure cylinder ED Volume of high-pressure cylinder GF FIG. 70. If equal distribution of work is not required, it is evident that the ratio may have a large number of values depending upon the position of the line GF. When designing an engine the size of the low-pressure cylinder is first estimated (Art. 88), and then a suitable volume ratio, obtained from practical experience gained with similar engines, is taken in order to fix the size of the high-pressure cylinder, and the point of cut-off in the high- pressure cylinder is then adjusted in order to obtain the ratio of expansion required. Let L be the ratio of the low-pressure to the high-pressure cylinder, R the total ratio of expansion required, then the ratio of expansion in the high-pressure cylinder must be R 84. Effect of varying the Point of Cut-off in the High- ART. 86] COMPOUND EXPANSION 143 pressure Cylinder on the Distribution of Work Cut-off Governing. The effect of varying the point of cut-off in the high- pressure cylinder and keeping both the cut-off in the low-pressure cylinder and the speed of the engine constant, is to vary the total power developed by the engine. By making the high-pressure cut-oft' later, more steam is supplied per stroke and the mean effective pressure referred to the low- pressure cylinder is increased, which results in an increased power being developed. The increase of power is not, however, equally distributed between the two cylinders. Neglecting clearance and assuming hyperbolic expansion, the release pressure in the high-pressure cylinder (and therefore the pressure in the receiver for no drop) will be , where /j is the initial pressure and r the ratio of expansion in the high-pressure cylinder. Now as the cut-off is made later, r decreases, hence, the back pressure in the high-pressure cylinder (which is equal to the receiver pressure) increases. The initial pressure in the low-pressure cylinder is equal to the receiver pressure, and since the number of expansions in this cylinder is not altered, it is evident that the mean effective pressure is largely increased, since its back pressure remains constant and a greater proportion of the increased power will be developed in the low-pressure than in the high-pressure cylinder. If the engine is governed by varying the point of cut-off in the high- pressure cylinder it will, when running on light load, have an early cut-off, and the number of expansions so obtained in the high-pressure cylinder will be sufficient to ensure a greater proportion of the total power being developed in the high-pressure cylinder ; when running light, the power developed in the low-pressure cylinder may be practically nothing. 85. Effect of Throttling the Steam to the High-pressure Cylinder on the Distribution of Work Throttle Governing. Consider the combined indicator diagram shown in Fig. 71, in which abed represents the work done in the high-pressure cylinder and adefg the work done in the low-pressure cylinder. Suppose that in order to meet a reduced load on the engine the steam is throttled down to the pressure ok, the cut-off in each cylinder remaining unaltered. Then ahkdviill represent the high-pressure diagram and adefg the low-pressure diagram ; the work done in the high-pressure cylinder (area hkda) will be less than before, but the work done in the low-pressure cylinder (area adefg) will be unaltered. If, therefore, the engine is governed by throttling the power developed in the low-pressure cylinder will remain practically constant, and when running on light load the power developed in the high-pressure cylinder may be very small. This is the converse to what happens with cut-off governing on the high-pressure cylinder. 86. Effect of varying the Cut-off in the Low-pressure Cylinder on the Distribution of Work. If the total number of expansions remains the same, i.e. with constant cut-off in the high-pressure cylinder, the total amount of work done per pound of steam remains the same, and the effect of varying the point of cut-off in the low-pressure cylinder is merely to alter the distribution of work between the two cylinders. Let abcde (Fig. 72) represent the theoretical indicator diagram referred to the low-pressure cylinder ; for any particular cut-off g, in the low-pressure cylinder, the area afgde represents the work done in that 144 THE THEORY OF HEAT ENGINES [CHAP. vi. cylinder, and the area/^ the work done in the high-pressure cylinder. FIG. 71. Effect of throttling. a FIG. 72. Effect of varying the cut-off in low-pressure cylinder. If now the cut-off in the low-pressure be made later, at h, the work ART. 87] COMPOUND EXPANSION 145 done in this cylinder will be reduced by the area fghk, and that done in the high-pressure cylinder will be increased by the same amount kbch will now be the high-pressure diagram and akhde the low-pressure diagram. Conversely, if the cut-off is made earlier, at /, the work done in the low-pressure cylinder will be increased to the area amide, but the work done in the high-pressure cylinder will be reduced to the area mbcl. In order to increase the work done in the low-pressure cylinder, therefore, the cut-off should be made earlier ; and vice versd. Making the cut-off take place later in the low-pressure cylinder has the effect of increasing the mean pressure in the receiver, and therefore the back pressure in the high-pressure cylinder, which results in less work being done in the high-pressure, and more in the low-pressure cylinder. Further, if there is a drop in pressure between release in the high-pressure cylinder and the receiver, it will be reduced, and there will be some cut-off at which the drop will be entirely eliminated. It will be evident, therefore, that by choosing a suitable ratio of cylinder volumes and cut-off in the low- pressure cylinder, it is possible to have equal distribution of work between the two cylinders and no drop. 87. Initial Loads on the High -pressure and Low-pressure Pistons. The initial load on the piston is the difference between the total forces exerted by the steam on the two sides of the piston. Thus, on the high-pressure piston the initial load will be (neglecting the area of the piston rod) Area of high-pressure piston X (initial steam pressure receiver pressure) and on the low-pressure piston Area of low-pressure piston X (receiver pressure back pressure) It will frequently happen that if equal distribution of work is allowed between the cylinders, the initial loads will be unequal and vice versd. When designing an engine it is an advantage to have the initial loads equal, in which case the distribution of work may be unequal, as will be evident from the following example. The ratio of cylinder volumes should be so chosen that both the work done in the two cylinders and the initial loads on the pistons are approximately equal. Suppose the initial steam pressure in the high-pressure cylinder is 100 pounds per square inch absolute, the total number of expansions 10, back pressure 4 pounds per square inch absolute, and the ratio of cylinder volumes 3. Let x be the receiver pressure in pounds per square inch absolute. Then, since the area of the low-pressure cylinder will be 3 times that of the high-pressure cylinder for equal strokes, for equal initial loads we have 100 ^ = 3(^ 4) .-. x = 22 pounds per square inch absolute Assuming complete hyperbolic expansion, the ratio of expansion in the high-pressure cylinder will be 100 146 THE THEORY OF HEAT ENGINES [CHAP. vi. and in the low-pressure cylinder 10 The mean effective pressure in the high-pressure cylinder will be ^( I + 1 Se4'55)-22 = 22(l -f- I*5l5l) 22 = 33*33 pounds per square inch The mean effective pressure in the low-pressure cylinder will be 22 2-2) 4 = 10(1+07885) -4 = 13*88 pounds per square inch Hence the ratio Work done in high-pressure cylinder 33*33 Work done in low-pressure cylinder 3 X 13*88 1-24 EXAMPLE i. In a two cylinder compound engine, the admission pressure to the high-pressure cylinder is 105 pounds absolute, cut-off 0*6 stroke. The release pressure in the low-pressure cylinder is 12 pounds absolute and the condenser pressure 3 pounds absolute. If the initial loads on the pistons are equal and the curve of expansion is pv 1 ' 2 = constant, estimate the cylinder volume ratio, the mean pressure in the receiver, the point of cut-off in the low-pressure cylinder, and the ratio of the work done in the two cylinders. (L.U.) Let R = total ratio of expansion, then, assuming a continuous expansion curve, 105 X i = 12 X R 1 * 2 from which R = 6*095 /. cylinder ratio = 6*095 X 0*6 = 3*657 Let x *= mean receiver pressure in pounds per square inch absolute. For equal initial loads, 105 ^=3*657(^3) x = 24*9 pounds absolute Let n = ratio of expansion in low-pressure cylinder, then 24-9 X i = 12 X n l ' z from which n = 1*838 .'. cut-off in low-pressure cylinder = . g g = 0-544 of the stroke. Let/ 2 = the absolute release pressure in high-pressure cylinder ' 2 from which / 2 = 56*87 pounds per square inch absolute The theoretical indicator diagram is shown in Fig. 73, in which abcde ART. 87] COMPOUND EXPANSION 147 is the high-pressure and afghk the low-pressure diagram. It should be noticed that there is a drop in pressure between the higrupressure release FIG. 73. and the mean receiver pressure of amount 56*87 24*9 = 31 '97 pounds per square inch. To find the Distribution of Work between the Cylinders. The mean effective pressure in the high-pressure cylinder is 105 X 105 X 1 56-87 X 7 O'O 1*2 I 24-9 105 + 5 1 24-9 r66 = 68-8 pounds per square inch The mean effective pressure in the low-pressure cylinder is 24-9 X i 12 X 1-838 1*2 I 24-9 X i + 1-838 24-9 + 14-2 M _ _, *j 1-838 = 18*25 pounds per square inch Hence, work done in high-pressure cylinder _ 68-8 work done in low-pressure cylinder 3*657 X 18-25 = 1-03 148 THE THEORY OF HEAT ENGINES [CHAP. vi. In this example, the approximately equal distribution of work and the equal initial loads are only obtained by the " drop " between the high- pressure release and the receiver. 88. Method of estimating the Cylinder Dimensions in order to develop a Given Power. The usual method is as follows : The ratio of expansion is first decided, and then, assuming hyperbolic expan- sion, the mean effective pressure referred to the low-pressure cylinder is calculated, a suitable diagram factor (Art. 78) being employed. The mean piston speed is next decided upon, and the area of the low-pressure piston calculated. A cylinder volume ratio is next selected, and the areas of the high-pressure and intermediate cylinders estimated. The stroke of the several pistons, invariably being equal, is next decided with reference to the mean piston speed and the revolutions per minute at which the engine is to run. The following table gives average values of cylinder ratios, etc. for different types of engines : Type of engine. Initial pressure, Ibs. per sq. in. (gauge). Total number of expansions. Cut-off in high- pressure cylinder. Ratio of cylinder volumes. High presssure cylinder = i. Compound engines Light engines . Heavy engines IOO-I4O 90-100 P 0-5-07 0-5-0-7 i : 3-2-3-8 4-4-6 Triple expansion 150-220 9-5-I2 0-6-0-7 1:2-6: 6-8 to i : 2-2 : 72 Quadruple expansion 190-220 10-13 0-65-0-72 i : 2 : 4 : 8 to 1:2-2: 4-4 : 9-2 EXAMPLE i. Determine the cylinder diameters of a horizontal com- pound steam engine with trip-gear to develop 600 indicated horse-power under the following conditions: Pressure in steam chest 155 pounds per square inch absolute, vacuum 26 inches, number of expansions 12, diagram factor 0-82, piston speed 650 feet per minute, cut-off in high-pressure cylinder \ stroke. Determine also the point of cut-off in the low-pressure cylinder, and compare the work done in the two cylinders when the initial loads are approximately equal. The cylinder ratio will be L = = = 4. Hence if A 2 and A x denote the area (in square inches) of the low- pressure and high-pressure cylinders respectively, A - - = 4 the strokes being made equal A i Referred to the low-pressure cylinder = 0-82 x 43 = 35 '3 pounds per square inch ART. 88] COMPOUND EXPANSION 149 Hence A 2 X35'3X 33,000 A 2 = 600 X 33, 35*3 X 650 = 86 1 square inches Since A 2 = 4A 1 , it follows that J 2 = 2d 1 , hence the diameter of the high-pressure cylinder will be 16*5 inches. Let x = mean receiver pressure in pounds per square inch absolute. I SS~ x = 4(x2) .: x = 32-6 pounds per square inch. If n is the number of expansions in the low-pressure cylinder, 32-6 X i=^X n n = 2'$ .'. cut-off in low-pressure is - - or 0-4 of the stroke 2 '5 The mean effective pressure in the high-pressure cylinder will be 0-82^(1 +log 3) -32-6J = o-82pp-X 2-097 32-6 j =-0-82 X757 = 62 pounds per square inch. The mean effective pressure in the low-pressure cylinder will be 0-82^(1 + log. 2-5) -4} = 0-82 X 21 = 17-3 pounds per square inch Hence, work done in high-pressure cylinder 62 work done in low-pressure cylinder 4X17*3 i'n EXAMPLE 2. Estimate the diameters of the cylinders required for a quadruple expansion marine engine to develop 12,000 I.H.P. with a piston speed of 960 feet per minute. Pressure in steam chest 210 pounds per square inch gauge, number of expansions 14. Assume a ratio of low- pressure to high-pressure of 9, and a diagram factor 0^65 . Find also the point of cut-off in the high-pressure cylinder. Assuming a vacuum of 26 inches as in Example i, i.e. a back-pressure 150 THE THEORY OF HEAT ENGINES [CHAP. vi. of 2 pounds per square inch absolute, the mean effective pressure referred to the low-pressure cylinder will be 2 = 0-65 X 5 6< 5 = 367 pounds per square inch Hence, if A be the area of the low-pressure cylinder, A X 367 X 960= 12,000 X 33, 12,000 X 33> 000 367 X 960 = 11,240 square inches. ' d= V 4= IJ 9' 6 inches and diameter of high-pressure cylinder 119-6 = f- = 39'5 mches o Taking a ratio of cylinder volumes of 1:2-1: 4-4 : 9 Diameter of ist intermediate cylinder = 39-5 X \/2'i = 58 inches. i> 2nd =39'5 X \/4 7 4=83 The cut-off in the high-pressure cylinder will be ratio of low-pressure to high-pressure 9 total number of expansions = ^4 89. The combination of Indicator Diagrams from a Com- pound Engine. The indicator diagrams of a large horizontal compound engine developing 1415 I.H.P. are shown in Figs. 74 and 75. In order to combine these diagrams an average indicator diagram must be first constructed for each cylinder. The average diagram shown in Fig. 76 represents the mean diagram for both sides of the high-pressure piston; similarly Fig. 77 represents the mean diagram for both sides of the low- pressure piston. The saturation curves SS and S'S' are then drawn one on each diagram by the method already explained in Art. 70. Next set off any convenient distance AB, Fig. 78 (say 10 inches), to represent the piston displacement of the low-pressure piston (i.e. area of cylinder X stroke), and oh. to represent to the same scale the clearance volume then choosing a convenient scale of pressures re-plot the mean diagram from the low-pressure cylinder together with its saturation curve S'S', all volumes being plotted from the ordinate through point o. Next, to the same scale of volumes, set off CD to represent the clearance volume , of the high-pressure cylinder, and DE the stroke volume of that cylinder ; then re-plot the mean diagram from the high-pressure cylinder together with its saturation curve SS as shown in Fig. 78. In the combined diagram drawn in Fig. 78 the saturation curves SS and S'S' do not form one continuous curve. This is because the total ART. 89] COMPOUND EXPANSION Ubs 180 140 100 60 20 Lbs 40 30 20 10 Atmospheric Atmospheric Atmos. FIG. 74. FIG. 75. 80 FIG. 76. Line Line Line 152 THE THEORY OF HEAT ENGINES [CHAP. vi. weight of steam present in the high-pressure cylinder during expansion is not the same as that in the low-pressure cylinder, the difference being due to unequal weights of cushion steam in the two cylinders. A single saturation curve can only be drawn on the com- bined diagram when the same weight of cushion steam is present in each cylinder. As a rule the weight of cushion steam is less in the low-pressure than in the high- pressure cylinder, and this causes the saturation curve of the low-pressure to fall inside that of the high-pressure cylinder, as is shown in Fig. 78. 20 14-7 to FIG. 78. EXAMPLES VI 1. In a two-cylinder compound engine, the admission pressure to the high-pressure cylinder is 80 pounds per square inch absolute, cut-off 0*5 stroke. The release pressure in the low-pressure cylinder is 8 pounds per square inch absolute and the condenser pressure 2 pounds absolute. Assuming hyperbolic expansion and equal initial loads on the pistons, estimate the ratio of cylinder volumes, the mean pressure in the receiver, the point of cut-off in the low-pressure cylinder, and the ratio of the work done in the two cylinders. 2. Solve Question I if, instead of hyperbolic expansion, the law of expansion is ^t/1'15 = const. 3. Determine the cylinder diameters of a horizontal compound steam engine to develop 500 indicated horse-power under the following conditions : Pressure in steam chest 140 pounds per square inch absolute, vacuum 26 inches, number of expansions 10, diagram factor o'8o, piston speed 600 feet per minute, cut-off in high-pressure cylinder O'35 stroke. Determine also the point of cut-off in the low-pressure cylinder and compare the work done in the two cylinders when the initial loads are equal. 4. In a two-cylinder compound engine the ratio of cylinder volumes is 5 and the total number of expansions is 10. The initial steam pressure is 100 pounds per square inch absolute and the back pressure 4 pounds per square inch absolute. Assuming con- tinuous hyperbolic expansion and equal distribution of work between the two cylinders, compare the initial loads on the pistons. 5. A three-cylinder triple expansion engine is required to develop 5 indicated horse-power at 90 revolutions per minute under the following conditions : Pressure in high-pressure steam chest 200 pounds per square inch absolute, cut-off in high-pressure cylinder 07 stroke, average piston speed 720 feet per minute, vacuum 28 inches. Using a cylinder ratio of I : 3 : 7*5 and a diagram factor 0-63, determine the dimensions of the cylinders. 6. If the initial loads on the pistons are equal estimate the mean receiver pressures for the engine in Question 5. CHAPTER VII MECHANICAL REFRIGERATION 90. Types of Mechanical Refrigerating Machines. The function of a refrigerating machine is to produce and maintain a low temperature. This is done by employing some substance which is capable of absorbing heat, and by a suitable arrangement rendering it possible to continue withdrawing heat from the body to be kept cool as fast as it flows in; the heat which is taken from the cold body is then transferred to another body which is at a higher temperature. This process requires external assistance in the shape of expenditure of heat in the form of mechanical work, since by the second law of thermodynamics it is impossible for a self-acting machine to convey heat from one body to another at a higher temperature (see Art. 25). Mechanical refrigerating machines may be divided into two classes, namely, (i) those which use air as the working substance, and (2) the compression type which employs a vapour which in the saturated state exhibits a high pressure at low temperatures. Cold Air Machines. In machines of this type, air is compressed to about 50 pounds per square inch and is then passed through a cooler the tubes of which are surrounded by water. In passing through the cooler the heat of compression is removed, after which the air is then passed through an interchanger. This interchanger consists essentially of a series of tubes surrounded by the cold air returning from the cold chamber, and in it the compressed air is further cooled, and any moisture it may contain is deposited as snow. From the interchanger the chilled compressed air passes to an expansion cylinder in which it expands, doing work and helping to drive the machine. The air is still further cooled by the expansion, after which it is exhausted and led away to the cold room. Vapour Compression Machines. The vapours used in modern machines of this type are ammonia, carbon-dioxide and occasionally sulphur-dioxide. Wet vapour is drawn into the compressor cylinder and after being com- pressed, is cooled and condensed, after which it expands (in practice through an expansion valve without doing useful work) to a low tempera- ture and then passes on through a series of pipes immersed in brine. The brine is thereby cooled to a low temperature and is itself used for refrigerating purposes. This type of machine is the one most frequently adopted in practice. pi. Coefficient of Performance. From what has been said in the previous Art., it is evident that a refrigerating machine is simply a heat pump which pumps up heat from a low to a higher temperature and has to be driven by mechanical means. The most economical machine 154 THE THEORY OF HEAT ENGINES [CHAP. vn. will be the one which extracts the greatest amount of heat from the cold body for a given expenditure of mechanical work. The ratio heat extracted work expended is known as the coefficient of performance, both quantities being reckoned in the same units. It was shown in Art. 24 that the ideal heat engine works on a reversible cycle, taking in heat at a temperature T 1? and rejecting heat at a lower temperature T 2 . If H x be the amount of heat taken in at temperature T 1 and H 2 the amount rejected at temperature T 2 , then the work done is Hj H 2 and the efficiency is If now, such an engine be reversed, it would take H 2 units of heat from the cold body at temperature T 2 and H x H 2 units of heat being supplied to drive the machine, H x units would be delivered from the machine at the higher temperature T x . The coefficient of performance would therefore be heat extracted _ H 2 work expended H 92. The Cold Air Machine. Consider an engine working on the ideal Carnot cycle reversed. Starting at point c the cycle is in a clockwise V FIG. 79. 5* FIG. 80. direction cdob (Figs. 79 and 80). The engine will take in a quantity of heat H 2 at temperature T 2 and will reject a larger quantity Hj at tempera- ture Tj, the work required to drive the machine being represented by the area cdob ; in fact, the reversed perfect engine will act like a heat pump, withdrawing heat at a temperature T 2 and delivering it at a higher temperature T 1} the coefficient of performance is H 2 MECHANICAL REFRIGERATION area cdef area abfe area dcef ART. 93] or referring to Fig 80 + 2 d) "T^Ts From (i) we see that the smaller the range of temperature (Tj T 2 ) the greater will be the co-efficient of performance and the greater the amount of refrigeration for a given expenditure of mechanical work. The Coo/er A FIG. 81. above expression only refers to the ideal reversed engine ; no refrigerating machine used in practice works on this cycle, but the commonest cold air machine in use works on the cycle of the reversed Joule engine. 93. The Reversed Joule Engine or the Bell-Coleman a Refrigerating Machine. The most important develop- ment of the Joule hot air engine, described in Art. 29, .has been in its reversed form, which has been developed as the Bell- Coleman refrigerating machine. A diagrammatic arrangement of the machine is shown in Fig. 81, whilst Fig. 82 represents the indicator diagram. The cycle is as follows : (1) The pump cylinder C takes in air from the cold V chamber R at the lower tempera- FIG. 82. ture T 2 during its suction stroke fc. (2) During the first portion of the return stroke the air is compressed 156 THE THEORY OF HEAT ENGINES [CHAP. vn. adiabatically up to the pressure existing in the cooler A, the temperature rising above that in A. This portion of the cycle is shown by cb (Fig. 82). (3) The pump C then delivers air into A by completing its stroke from b to e; the temperature falls to that of A and the air rejects to A a quantity of heat H such that where T& and T a denote the temperatures at b and a respectively. While these operations take place in the compressor cylinder C the following take place in the expansion cylinder E : During (i) and (2), E takes in an equal quantity of air from A at temperature T a and expands it adiabatically down to the pressure in R as shown by ea and ad respectively (Fig. 82). At the end of expansion the temperature T d is lower than that of the cold chamber R. During (3). After expansion in E the chilled air is discharged into R during the return stroke df. The work done on the air per cycle in the compressor cylinder is represented by the area /<;&?, and the work done by the air in the expansion cylinder is represented by the area eadf, hence the net amount of work expended in driving the machine is given by area eadf area deb a The net amount of heat extracted from the cold chamber per pound of air is and since the ratio of expansion in E is the same as the ratio of compres- sion in C Tb d\ T T ......... \ x / id AC hence the coefficient of performance is heat extracted H 2 work expended H^ H (T 6 T) (T T d ) rys = - ^=- by (i) above . ... (2) -l *d T This expression is less than the T J* T Q f equation (i), Art. 92, for the same reason that Joule's engine is less efficient than Carnot's, i.e. all the heat is not taken in at the temperature T 2 and all is not rejected at the temperature T x ; in other words, T a Td is greater than the difference of temperature T x T 2 of Art. 92. The actual coefficient of performance obtained by this machine is very much lower than that of a vapour compression machine and varies from i to f, this low value being due to (i) The necessity of working with a wide range of temperature (Ta Td) since air is a poor conductor and absorber of heat. ART. 94] MECHANICAL REFRIGERATION 157 (2) The large amount of air friction in the cylinders, particularly with large machines. A difficulty experienced with this type of machine was the presence of moisture in the air coming from the cold chamber. At the end of expansion this moisture was deposited as snow which had a tendency to choke up the valves. Mr. Lightfoot got over this difficulty by employing compound expansion. In the first expansion cylinder the temperature was reduced to about 35 F. and the moisture deposited and then drained away. The most usual method, however, is to employ an interchanger as mentioned in Art. go. 1 94. Reversed Heat Engine as a Warming Machine. A machine of the Bell-Coleman type may be used for this purpose, as was first pointed out by Lord Kelvin in 1852.2 The machine would take in air from the atmosphere, expand it down to a lower temperature and pressure and then allow its temperature to rise again by contact with the external air, after which the air would be compressed back again to atmospheric pressure, and its temperature thereby raised preparatory to being delivered into the room to be warmed. Let H 2 = heat taken from the atmosphere at temperature T 2 , H x = heat delivered to the room at temperature T 1} W = work expended in heat units. Then ^v= H ^ = T ^ T (i) When the range of temperature Tj T 2 is small, Hj may be many times greater than W, i.e. a large amount of heat may be raised through a small range of temperature with little expenditure of mechanical work. EXAMPLE i. If the compression pressure of a reversed Joule heat engine be 45 pounds per square inch gauge and the suction pressure 15 pounds per square inch absolute, find the lowest temperature produced in the engine, the air being cooled at the highest temperature by circulating water at the temperature of the atmosphere, which is 60 F. The/z> diagram is shown in Fig. 82. T a = 460 + 60 = 520 absolute, and T^ is the lowest temperature. v-i Now =? = \f~ } by (6), Art. 1 1 * d \A*' 1-4-1 + 15 \ r4 -("?) .-. TV,= , 4 7 5 2 4 T 52O T d = 350 absolute or 350 460 = 110 F. 1 For details of this and other types of refrigerating machines, see "The Mechanical Production of Cold," by Sir J. Ewing, Cambridge University Press. 2 Proc. Phil. Soc. of Glasgow, vol. iii. p. 269, or Collected Papers^ vol. i. p. 515. 158 THE THEORY OF HEAT ENGINES [CHAP. vn. EXAMPLE 2. Find the least horse-power of a perfect reversed heat engine that will make 900 Ibs. of ice per hour at 27 F. from water at 70 F. Take the latent heat of ice as 142 B.Th.U. per pound and the specific heat as 0*5. T 2 4604-27 487 Coefficient of performance = , r * = - = = 11*33. L 1 1 2 70 7 43 Heat extracted from i pound of water at 70 F. to produce i pound of ice at 27 F. will be (70 32) + 142 + 0-5(32 2-7) = 38 + 142 + 2-5 = 182-5 B.Th.U. Hence the least horse-power will be 182-5 X 900 __ 2545 X n'33 ~ 57 ' EXAMPLE 3. An oil engine uses Russolene of calorific value 20,000 B.Th.U. per pound. If it drives a reversed heat engine which takes in air at 40 F. and delivers it at 55 F., how much heat will be given to the air per brake horse-power hour if the reversed heat engine works at 80 per cent. of the ideal performance ? From (i), Art. 94 H^W.^^VX Tj T 2 100 Now W = i B.H.P. hour = 2545 B.Th.U. = 69,900 B.Th.U. A modern oil engine would use, say, 0-5 pound of oil per B.H.P. hour containing about 10,000 B.Th.U. When this oil is used to drive an oil engine which in turn drives a reversed heat engine as above, 69,900 B.Th.U. are given to the air, whilst if it were burned directly to warm the air, only 10,000 B.Th.U. would be available, i.e. about one-seventh as much. This does not, of course, mean that 69,900 10,000 or 59,900 B.Th.U. are created, but that the horse-power hour (2545 B.Th.U.) is converted into heat in the reversed engine and the remaining 69,900 2545 or 67,355 B.Th.U. are merely raised in temperature. 95. The Vapour Compression Machine. This type of refrigerating machine is shown diagrammatically in Fig. 83 and the indicator diagram is shown in Fig. 84. Starting with volatile liquid in the refrigerator R the cycle is as follows : (1) The valve Q is opened and the refrigerant drawn into the cylinder C as a wet vapour at temperature T 2 . This portion of the cycle is represented by dc in Fig. 84. (2) The vapour is compressed adiabatically to temperature T x along cb. Some of the heat of compression is absorbed in evaporating the liquid and at b the vapour should be just dry and saturated. (3) The vapour is next discharged through the valve U into the condenser A at constant pressure and at constant temperature T^. The ART. 95 ] MECHANICAL REFRIGERATION vapour is cooled and condensed in A by cooling water and gives up its latent heat. This portion of the cycle is represented by ba. Refrigerator FIG. 83. (4) In all machines used in practice, the liquid is then expanded freely through the reducing valve V back to its original state in R as represented by ad (Fig. 84). It might have been expanded adiabatically in a motor cylinder and thus help to drive the machine along of, and by so doing save an amount of work represented by the shaded area afd. FIG. 85. The theory of this type of refrigerating machine is more conveniently explained by reference to the temperature-entropy diagram shown in Fig. 85. The adiabatic compression of the wet vapour is shown by cb, then follows condensation at temperature T a along ba ; ae represents the cooling of the liquid to T 2 in the cooler and ec its evaporation in the refrigerator. It is evident that when the expansion valve is used, the cycle is for all practical purposes the Rankine cycle reversed. (Cp. dcba 160 THE THEORY OF HEAT ENGINES [CHAP. vn. Fig. 84 and ecba Fig. 85 with Figs. 30 and 33.) We will now consider the theory of the cycle with both wet and dry compression, with both an expansion valve and an expansion cylinder. 96. Vapour Compression Machine working without Super- heating and with an Expansion Cylinder. Consider the tempera- ture-entropy diagram shown in Fig. 85. The cycle is (1) dc, pump suction the vapour taking up its latent heat at T 2 . (2) cb, adiabatic compression until just dry and saturated at Tj. (3) ba, isothermal compression at Tj the vapour being condensed and giving up its latent heat to the cooling water. (4) ad, adiabatic expansion doing useful work in a motor cylinder and helping to drive the machine. The heat extracted from the refrigerant is represented by the area dcmn, and the work done in driving the machine by the area dcba, hence the coefficient of performance will be heat extracted _ area dcmn work done ~~ area dcba This may be put in algebraic form as follows : Let s denote the specific heat of the liquid and Lj its latent heat of evaporation at T 1} then ed=s loge -~ and ab = =r A 2 AI hence, area dcmn = * X T 2 " *! and area dcba = ~ (T x T 2 ) T 1 T .*. coefficient of performance = T 1 = - T^T (*) VT T \ 1 i"~ *2 fr(T x - T 2 ) Note that this is the same expression as (i), Art. 92. Case when the Vapour is Wet at the end of Compression. In this case compression starts at point k (Fig. 86) and finishes at some point h, the dryness fraction at the end of compression being ah ****& Heat extracted per pound = area dkln _ *|5i v T -- /N A 2 *l work done = area dkha *ii T ~7p A A 2 T 1 and coefficient of performance = . i - = = ^7=- as before ~ In this case, although the coefficient of performance is the same as ART. 97] MECHANICAL REFRIGERATION 161 before, there is less refrigeration per pound of the liquid, and therefore more of the liquid will be required for the same amount of refrigeration. h 6 "\ \ \ 97- Vapour Compression Machine working without Super- heating and with an Expansion Valve. This is the usual case in practice, and instead of adiabatic expansion in a motor cylinder we have free, unresisted expansion through the expansion valve, the expansion terminating at some point / (Fig. 85). We have here for all practical purposes the Rankine cycle reversed, the indicator diagram being as shown in Fig. 84, namely dcba. Assuming no gain or loss of heat when passing through the expansion valve, the heat contained by i pound of the stuff will be the same before and after throttling (cp. Arts. 38 and 47). The heat before expansion is represented by the area eanp (Fig. 85), and that after expansion by the area efrp. Hence these two areas must be equal, or since the area ednp is common, the area ead must equal the area dfrn. The net amount of refrigeration per pound, i.e. the heat extracted, will be represented by the area/<:wr, and the work done by the area eabc, hence the coefficient of performance will be area fcmr area eabc Suppose that by the use of an expansion cylinder, w represents the amount of work done per pound in the cylinder. This is represented by the area ead'm Fig. 85 and afd in Fig. 84; and since area dfrn (Fig. 85) is equal to area ead y it follows that the net amount of refrigeration is less than the area dcmn by an amount w. Hence if H denotes the amount of refrigeration per pound and W the work done when an expansion cylinder is employed, the coefficient of performance when using an expansion valve will be heat extracted H w work done ~ W-j-o/ TT which is obviously less than ^7. M i6 2 THE THEORY OF HEAT ENGINES [CHAP. vn. Using the same notation as in Art. 96, this result may be expressed algebraically as follows : Area eanp area efrp = s(T 1 T 2 ) . ..r_ Now ec = ed + _ T 2 T 1 ~ T 2 Heat extracted = we&fcmr = T 2 xft Also, work done == area eabc = MQ&peabm area ecmp = area pean + area ^w area edpn area . . ... (3) (This is the same as the work done on the Rankine cycle, cp. (7), Art. 57, p. 77-) heat extracted Coefficient of performance = wor k done ^ / v Vapour is Wet at the end of Compression. Here the compression commences at point k (Fig. 86) and finishes at some point h, the dryness fraction at the end of compression being *!=4 ab Heat extracted = area fklr = ^T 2 log 6 J + ^- 1 .T 2 -.(T 1 -T 2 ). (5) 12 A! work done = area eahk l-T 2 .,log e (6) . T 2 - s(T t - T 2 ) Coefficient of performance = - 1 - 2. - T" ( 7 ) ART. 98] MECHANICAL REFRIGERATION 163 98. Vapour Compression Machine working with Superheat- ing and with Expansion Cylinder. Consider the case in which the compressor draws in dry saturated vapour from the refrigerator. The temperature-entropy diagram is shown in Fig. 87. The compression now commences at point g and ends at some point ^, on the superheat FIG. 87. curve corresponding to the pressure at saturation temperature T 1} where the temperature is, say, T 3 . Heat extracted = area dgsn = area dcmn -j- area cgsm where Cp is the specific heat of the vapour at constant pressure (assumed constant). Work done = area ddbhg = area dabc -j- area bhsm area cgsm = tl (Tj - T 2 ) + C*>(T 3 - TJ - T, . C, log, . . (2) Coefficient of performance = (3) 164 THE THEORY OF HEAT ENGINES [CHAP. vn. To calculate T 3 we may proceed as follows : In passing along the superheat curve from b to k, the gain of entropy is <2r=C P log e I? .......... (4) also f gtg ec L% Ti Li = ^-Hog 1 - x ...... (5) 1 2 A 2 A l Equating (4) and (5) we have Ci log. = -* !<*,- .... (6) A l 1 2 X 2 - 1 ! an equation from which T 3 may be calculated. 99. Vapour Compression Machine working with Super- heating and with Expansion Valve. The expansion now ends at some point/ (Fig. 87) as in Art. 97. Heat extracted = area^r Now e and < = &=2 hence fg=egef L 2 ff hence heat extracted =fg X T 2 = L 2 -XT 1 -T 2 ) ...... (i) Work done = area edbhg = area ead -\- area . . (2)' (Cp. (i), Art. 58) Coefficient of performance EXAMPLE i. In an ammonia refrigerating machine the temperature in the refrigerator is 14 F., and after compression 86 F. In the cooler the vapour is condensed at 86 F. and then passes through an expansion valve into the refrigerator. Estimate the coefficient of performance when the vapour at the end of compression is, (a) just dry and saturated, (b) 90 per cent. dry. Take the latent heat of ammonia at 86 F. as 490-5 B.Th.U. per pound and the specific heat of the liquid as ri2. ART. 99] MECHANICAL REFRIGERATION 165 (a) By (2), Art. 97 T T heat extracted per pound = sT 2 log, = + -^ - T 2 j(Tj T 2 ) 1 2 1 l here T x = 86 -f 460 = 546 absolute T 2 = 14 + 460 = 474 absolute L! = 490-5 /.heat extracted = ri2 X 474 X log e ~ h~ 49 5 * 474 1-12(546 474) 474 54 & = 75' 8 + 4 2 5'65 80^64 = 420-09 B.Th.U. By (3), Art. 97 work done = (T - T 2 )s + - T 2 . s log 1 = 72 X 2-02 75-08 = 70-36 B.Th.U. 42O'OQ .-. coefficient of performance = = 5-97 (fi) By (5), Art. 97 heat extracted = /T 2 Iog 6 ^ + ^ . T 2 - s^ - T 2 ) X 2 L l = 75' 8 + '9 X 425*65 8o>6 4 - 377-52 B.Th.U. By (6), Art. 97 work done = (T x - T a )(j + -^ - T 2 . s \og f ^ = 72(1-12 + 0-81) 75-08 = 63-88 B.Th.U. .-. coefficient of performance = 7^-^ = 5-90 03-88 EXAMPLE 2. Solve the problem of Example i when an expansion cylinder is used instead of an expansion valve. (a) By Art. 96 Heat abstracted = = 474 = 425-65 B.Th.U. Work done =^(T 1 - T 2 ) = 49 ' 5 5 J 7 " =64-69 B.Th.U. Coefficient of performance = , , = 6-58 64-69 (b) Heat extracted = 0-9 x 425*65 = 383-08 B.Th.U. work done = 0-9 X 64-69 = 58-22 B.Th.U. coefficient of performance = ^-^ =6-58. 166 THE THEORY OF HEAT ENGINES [CHAP. vn. Note. By Art. 96 the coefficient of performance in both cases is T simply T _^ T and there is no necessity to work out separately the heat extracted and the work done in each case. It is done here in order to allow for comparison with the other cases (see p. 167). EXAMPLE 3. Consider the machine taken in Examples i and 2, but let the vapour be just dry and saturated when compression begins. Estimate the coefficient of performance. (Take Q> = 0*508 and L 2 = 577*4 B.Th.U. per pound.) The temperature after compression (T 3 ) is given by (6), Art. 98. 0.508 X 2 '33 fcfc- ,- X ,303 T ri 7 lo foo^= 1-218 0-158 0*898 = 0*162 o' 162 .-. log 10 T 3 Iog 10 546 = Y^~ = ' I 384 from which T 3 = 750-9 absolute. With expansion valve heat extracted = L 2 s( r l\ T 2 ) (by (i), Art. 99) = 577*4 i'" X 72 = 496-76 B.Th.U. work done by (3), Art. 98 ^) -T 2 . s Io g6 g + C,( T 3 - T x - T 2 log, / 75'9 = 70-36 + 0-508(^750-9 546 474 log e g = 70-36 + 27-33 = 97-69 B.Th.U. 496*76 .-. coefficient of performance = ~^T^T = 5' 8 With expansion cylinder heat extracted = - 2 + T 2 .Cp log, (by (i), Art. 98) = 425-65 + 7676 = 502-41 B.Th.U. ART. 99] MECHANICAL REFRIGERATION 167 work done = ^p~ (M ^2) ~h Cp(T 3 Tj) T 2 . Cp log e (by (2), Art. 98) = ~ r~7 h 0-508(780-9 546) 474 X 0-508 log e = 64-69-}- 27-33 = 92-02 B.Th.U. .'. coefficient of performance = = 5*45 92*02 For convenience in comparison the results obtained in the above examples are tabulated below. State at end of compression. Heat abstracted B.Th.U. per pound. Work done B.TH.U. per pound. Coefficient of performance. Expansion valve . . > > Dry and saturated 90 per cent, dry Superheated to 750-9 F. 420-09 377-52 496-7 6 70-36 63-88 97^9 5'97 5'9o 5-08 Expansion cylinder . i > > > > Dry and saturated 90 per cent, dry Superheated to 750-9 F. 425'65 383-08 502-4I 64-69 58-22 92*02 6-58 6-58 5'45 It will be seen that the coefficient of performance is greatest when the refrigerating agent used is just dry and saturated at the end of compression. The effect of allowing superheating to take place during compression increases the amount of refrigeration at the expense of a greatly increased amount of work done in driving the compression, the result being a reduced coefficient of performance. The heat of compression also raises the temperature of the compressor walls, and on the entrance of the next charge of cold vapour, heat is absorbed by it and the vapour expands. The result is that a smaller charge is taken in and there is less refrigera- tion per cycle, although, as shown in the table, the refrigeration per pound is greater. EXAMPLE 4. An ammonia compression refrigerating machine has to do an amount of refrigeration equal to the production of 25 tons of ice per 24 hours from and at 32 F. If the temperature limits in the compressor are 75 F. and 5 F., calculate the horse-power of the compressor on the assumption that the cycle is a perfect one. T The coefficient of performance of a perfect machine = ~ ^-=~ 1 68 THE THEORY OF HEAT ENGINES [CHAP. vn. Taking the latent heat of ice as 142 B.Th.U. per pound, we have Heat to be extracted per minute = Now w = .'. work done per minute, W = * 42 24 X 60 = 5560 B.Th.U. 5-687 X 778 foot-pounds and horse-power = ^ = 22-4 H.P. 5' 68 7 X 33,000 Note. An actual machine will have a coefficient of performance of from 60 to 70 per cent, of the ideal. Hence the B. H.P. of the engine driving the above compression would have to be about = 37 B.H.P. O'O ioo. Choice of a Refrigerating Ag;ent. In deciding upon the liquid to use in a vapour compression machine the most important pro- perties to be considered are, the specific volume of the vapour, the latent heat of evaporation, and the specific heat of the liquid. The ideal liquid will have a very high latent heat and a low specific heat, whilst the vapour will have a low specific volume at a moderately low pressure. None of the liquids in use possess all these properties, and in practice the number is restricted to one of three, viz. carbon dioxide (CO 2 )j ammonia (NH 3 ), and sulphur dioxide (SO 2 ). The accompanying table shows a comparison between these three agents taken for a lower temperature limit of 5 F. 1 Agent. Pressure, Ibs. per sq. in. abs. P at 5 F. Latent heat, B.Th.U. per Ib. at 5 F. Specific vol. cu. ft. per Ib. at 5 F. Possible refrigera- tion per cubic foot, L V Relative size of compression, a to CO 2 NH 3 S0 2 334 33-67 II'76 115-25 582-10 (best) 169-74 0-267 8'39 6-49 43 1 -6 (best) 69'3 26-1 I (best) 6-26 16-60 From the above table it is evident that SO 2 is inferior to both CO 2 and NHg, and that the CO 2 machine is the smallest and has a much greater possible refrigeration per cubic foot of vapour. Further comparisons, how- ever, are necessary. If the higher temperature limit be taken as 86F. reference to tables will show that at this temperature the pressure of CO 2 vapour is 1040 pounds per square inch, of NH 3 180 pounds per square inch, and of SO 2 67 pounds per square inch ; hence, although the CO 2 machine is the smallest, its compression pressure is very high (particularly in the tropics, where the maximum temperature may exceed 86 F.), and this necessitates greater attention to mechanical details. In the case of NH 3 , the range of pressure (from 33*67 to 180) is moderate and more con- venient for general use, but at the same time its use precludes the employ- ment of brass and copper for any of the working parts of the machine ; 1 The values of/, L, and V are taken from the tables given in "Technical Thermo- dynamics," by Dr. Zeuner, A. Constable & Co. ART. lOo] MECHANICAL REFRIGERATION 169 fortunately, however, this agent has no action on iron. CO 2 , on the other hand, has no chemical action upon either iron, copper or brass. As already mentioned in Art. 97, practically all vapour compression machines use an expansion valve not an expansion cylinder. The resulting loss of refrigeration consists in the amount of heat carried with the liquid into the refrigerator as it flows through the valve. This, as shown in Art. 98, is represented by the area eanp of Fig. 85, being equal to s(Ti T 2 ) ; the greater the specific heat of the liquid the greater will be this loss. The following table shows this loss for the three agents under discussion, the temperature limits being taken for this purpose as 86 F. and 14 F. : Heat carried Latent heat at Possible Proportionate Agent. over in liquid 86 F. refrigeration. (^i ^2) L L co a 54'45 110-65 J6-20 0-49 NH 3 79'56 577'4 7 '94 0-137 S0 2 23-213 168-18 0-138 From this table it will be seen that the proportionate loss in the case of CO 2 is very much greater than with the other two agents, and of the three substances ammonia is the best. It should also be pointed out that leakages are more easily detected in the case of NH 3 and SO 2 on account of their smell. EXAMPLES VII 1. By means of a reversed perfect heat engine, ice at 32 F. is to be made from water at 67 F., the temperature of the brine or freezing mixture being 12 F. How many pounds of ice at 32 F. can be made per I.H.P. hour? (Latent heat of ice 142 B.Th.U. per pound.) 2. If the compression pressure in a Bell-Coleman refrigerating machine is 60 pounds per square inch gauge and the suction pressure 15 pounds absolute, find the lowest temperature produced in the machine if the air after compression is cooled to 60 F. What is the coefficient of performance, and how much ice can be made from and at 32 F. per I.H.P. hour? 3. Find the least horse-power of a perfect reversed heat engine that will make 1200 pounds of ice per hour at 25 F. from water at 60 F. (Take specific heat of ice as 0*5 and latent heat 142 B.Th.U. per pound.) 4. In an ammonia refrigerating machine the temperature in the refrigerator is 15 F. and after compression 90 F. In the cooler the vapour is condensed at 90 F. and then passes through an expansion valve. Calculate the coefficient of performance when the vapour at the end of adiabatic compression is (a) just dry and saturated ; (b) 85 per cent, dry. Take the specific heat of liquid ammonia as ri, and the latent heat of vaporisa- tion as 5660-8^ F. 5. Solve Problem 4 when an expansion cylinder is used instead of an expansion valve. 6. If in Problem 4^the ammonia is just dry and saturated at the beginning of com- pression, estimate the coefficient of performance (a) when an expansion valve is used, and (b) when an expansion cylinder is used. (Assume C p = 0*508.) 7. A vapour compression machine has to produce 50 tons of ice per day of 24 hours at 28 F. from water at 50 F. If the temperature limits in the compressor are 80 F. and 10 F., calculate the horse-power of the compressor on the assumption that the cycle is a perfect one. 8. The following are approximate expressions for the entropy of ammonia liquid and 170 THE THEORY OF HEAT ENGINES [CHAP. vn. dry saturated vapour : liquid, O'ooi84(/ 32) ; vapour, I'I58 O'ooi92(^ 32), /being the temperature on the Fahrenheit scale : obtain corresponding expressions of the form a + btc> t c being the temperature on the Centigrade scale. Draw the 6 chart between temperatures of 14 F. and 77 F. (- 10 C. and 25 C.). Find the coefficient of per- formance of a refrigerator working on a reversed Rankine cycle between these limits, the vapour being 5 per cent, wet at the end of compression. If the actual performance is o'6 of the amount in the above ideal case, calculate the pounds of ice produced per horse- power hour from water at the freezing-point. Latent heat of ice, 144 B.Th.U. (SoC.H.U.)- (L.U.) CHAPTER VIII FLOW OF STEAM THROUGH ORIFICES AND NOZZLES ioi. Adiabatic Flow through an Orifice. Assume that the orifice has well-rounded edges so that the least cross-sectional area of the jet of steam is the same as the area of the orifice, and further, that there is no frictional resistance offered to the flow. Let p and / 2 denote the pressure on the two sides of the orifice, i.e. let the steam flow from rest, through the orifice from a vessel in which the pressure is maintained at p into a vessel where the pressure is maintained at / 2 . Instead of the steam doing work on, say, an engine piston, it does work on itself in generating kinetic energy ; the kinetic energy per pound of steam issuing from the orifice will, in heat units, be equal to the difference between the heat contents before and after expansion to the lower pressure / 2 . Let V be the velocity of the steam jet in feet per second, then V 2 kinetic energy = = k + -^i L i (^2 + ^2^2) ( l ) where x and x 2 are the initial and final dryness fractions respectively. Now x 2 = g(^l + I g e II) . . . (3), Art. 46, and assuming the specific heat of water to be constant and equal to unity, we have V2 - = h l hz T 2 log. ... (2) hence the kinetic energy is the same as the work done on the Rankine cycle, and the velocity of the steam is The velocity may also be expressed in terms of pressures, since approximately V 2 n . . . (see (6), Art. 53) 171 172 THE THEORY OF HEAT ENGINES [CHAP. vm. Now the expansion follows the law/z> n = constant, hence by Art. 12 n-l (Pi\~^\ 2) f In (5) /! and / 2 denote the initial and final pressures respectively in pounds per square foot, and v^ the specific volume of the steam in cubic feet at pressure p<. If the steam be initially dry saturated the value of n for adiabatic expansion is 1*135 (Art. 50). 102. Weight of Steam discharged per Second. Let A be the area of the orifice in square feet and ^ 2 the volume per pound at pressure / 2 tnen tne weight discharged per second will be (i) Now v v . r - hence W = Eq. (2) will give the number of pounds of steam discharged per second through an orifice A square feet in area when the fall in pressure is from any given value from p to / 2 pounds per square foot. Maximum Discharge. The discharge will be a maximum when the expression in (2) under the root sign is a maximum, i.e. when is a maximum - i - -ru- d ( n n \ , p* This occurs when I a a ) = o, where a =^? da \ / 4> t ART. 102] ORIFICES AND NOZZLES J73 - !_ 2 n / I\ n *>. when -a li + -)a =0 TZ \ n' ( 2 V" 1 / \ or when a = ^ . i j - (3) Inserting (3) in (2) we have for maximum discharge 2 2 Substituting : 35 &ves^= 0-577 from (3) and taking - as 32-2 Pi - (5) or where Pj is the initial pressure in pounds per square inch. The above value of , namely, 0-577, is for steam which is initially dry and saturated. In any other case the value to take for n may be estimated from Zeuner's equation (Art. 50). If this be done for a number of initial pn dryness fractions, varying from i to 075, it will be found that the ratio will not differ materially from 0-58, and that therefore this ratio may be taken for all cases of saturated steam likely to be used in practice. n When ~ = ( } n the maximum velocity will be, from (5) Art. 101, For steam initially dry n 1-135, and taking as 32-2 this reduces to V = 5'85A/7^. ....... (8) or if P! is the initial pressure in pounds per square inch / . . ..-. (9) 174 THE THEORY OF HEAT ENGINES [CHAP. vm. The expression for maximum flow given in (5) may also be deduced from (8) as follows : = A. 5-85^/^x0-380 -A. 5-85 X 0-615 103. Flow Of Superheated Steam. The Mollier diagram affords the most convenient means of estimating the velocity of flow, but in its absence the velocity may be calculated by the same method as Art. 102, using equation (4) in which n = 1-3 for superheated steam. Since the kinetic energy per pound of steam is the same as the work done on the Rankine cycle, equation (i), Art. 58, may be used for estimating the velocity, in which case -T 1 )-T 1 log.^+9 1 , log.^). (i) n For a maximum discharge = ( 7 ) A \*+i/ Substituting n = 1*3 for superheated steam, we have 12 2 \0'3 The maximum velocity will be the same as that given by (7), Art. 102, viz. Substituting n = 1-3 and taking as 32*2, this becomes ~ ....... (3) where p is the initial pressure in pounds per square foot, or if Pj is the initial pressure in pounds per square inch> V= 72-36 VT^ ..'...-,. (4) ART. 104] ORIFICES AND NOZZLES The maximum flow will be given by Substituting n = 1-3 for superheated steam this reduces to = 6-o 3 AX 0-627 \/~ where/j is in pounds per square foot, or W = 45-43 (5) (6) where Pj is in pounds per sgtiare inch. 104. Flow through Nozzles. In the case of an orifice, when / 2 is greater than 0-5 8/j, the velocity of the issuing steam jet may be com- puted from either (3) or (5), Art. 101, and the discharge per second from (2), Art. 102. If the discharge takes place into a chamber in which the pressure is less than o'58/j, the jet of steam will have to expand further until its pressure is reduced to that of the receiving chamber; the pressure in the jet itself, however, as it issues from the orifice will be for all prac- tical purposes equal to 0*5 8/j. This further expansion results in a gain of velocity and therefore of kinetic energy, provided the jet is allowed to assume its natural shape. In turbine work the steam is usually allowed to expand through a larger range of pressure than is given by the ratio ^=0-58, and in order that the gain of velocity may be continuous down to the final back pres- sure the nozzle is made divergent, as shown in Fig. 88. The inlet end of the nozzle, known as the throat, has rounded edges, and from the throat to the discharge end the nozzle is made conical. The length of the nozzle should be such that for given inlet and outlet pressures the steam completely fills it, no energy being wasted by means FIG. iy6 THE THEORY OF HEAT ENGINES [CHAP. vm. of eddies. If the nozzle is made too short, eddies will be formed ; if, on the other hand, it is too long, the frictional resistance becomes too large. The best length is decided upon as the result of practical experience. The diameters of the throat and discharge end are first decided, and a taper of from i in 20 to about i in 12 allowed in order to fix up the length of the nozzle. 105. Design of Nozzles. In order to design a nozzle to pass a given weight of steam per second with given inlet and outlet pressures the first thing to do is to find the area of the throat of the nozzle, using (5) or (6), Art. 1 02, namely The area of the discharge end (A 2 ) of the nozzle must next be found. This may be done by using (2), Art. 102, which gives W n or the velocity at discharge may be found by either of the methods given above and then the area estimated by using (i), Art. 102. Choosing a suitable taper, say i in 1 2, the length of the nozzle from the throat to the discharge end is The ratio between the two areas is, from (i) and (2) A! = <35 3-6 Q-I55 / (_ " V W W After A x has been calculated the use of (3) is less laborious than (2) for estimating the value of A 2 . 106. Use of the Mollier Diagram. The Mollier Diagram, described in Art. 48, forms a very simple and convenient means of estimating the velocity on the assumption that the flow is frictionless and adiabatic. The " heat drop," or kinetic energy, and the velocity can be read off directly on the vertical scales arranged on the left-hand side of the diagram. The use of the diagram for this purpose will be best illustrated by means of the following examples. EXAMPLE i. Dry steam in a vessel expands through a nozzle from a pressure of 200 pounds down to 140 pounds per square inch absolute. ART. 106] ORIFICES AND NOZZLES 177 suming the flow the steam jet. By (i), Art. 101 Assuming the flow to be frictionless and adiabatic, estimate the velocity of the steam jet. V 2 kinetic energy j = fo + L x ) (h 2 + * 2 L 2 ) The dryness fraction x 2 will be found to be 0-974, and using steam tables, V 2 .'.= (1198*1) -(324-6 + 0-974 X 867-6) = 1198-1 1168-6 = 29-5 B.Th.U. .-. V = A/64-4X-778X 29-5 = 1215 feet per second Using (2), Art. 101, we have = 29X2 28-45 = 29-55 B.Th.U. which agrees very closely with the value obtained above. Using (5), Art. 101, we have (since / 2 is greater than o'58/j) / ~ f ~, TO"-135. v=V 64-4 >< - - >< '44 X = ^ 64-4 X r X 28,800 X 2-2 9 {i 0-960} = 1207 feet per second By means of the Mollier Diagram the velocity is read off directly and found to be 1200 feet per second. EXAMPLE 2. Dry steam at a pressure of 200 pounds per square inch absolute is to expand down to 5 pounds absolute. Determine the principal dimensions of the nozzle if the discharge is 60 pounds per minute. By (6), Art. 102, the area of the throat I /2-2G l i=Z^ x V 7^ TO = 0-00247 square foot or 0-355 square inch By (3), Art. 105, 0-155 =Mr hence A 2 = 6'8i X 0*355 = 2*417 square inches N i 7 8 THE THEORY OF HEAT ENGINES [CHAP. vm. Alternative Solution. From the Mollier Diagram we find the heat drop to be 253 B.Th.U., the velocity at discharge to be 3560 feet per second, and the dryness fraction 0-815. From steam tables the volume of i pound of dry steam at 5 pounds absolute is 73*33 cubic feet, hence the volume per pound of steam as discharged is (neglecting the volume of the water) o'8i5 X 73'33 cubic feet hence 3560 A 2 = 0-815 X 73'33 * _ 0-815 X 73'33 2 ~ ~l5^~ = 0*01678 square foot or 2-416 square inches 107. Effect of Friction on the Flow of Steam. In the preceding Articles it has been assumed that the flow is frictionless as well as adiabatic. Actually, there is a frictional resistance between the steam and the sides of the nozzle and also an internal resistance between the particles of steam themselves. In the theory already considered, the kinetic energy of the steam is represented by the area abed on the temperature-entropy diagram (Fig. 89). The actual kinetic energy is less than this amount because of the energy wasted in overcoming frictional resistance. The energy so lost is absorbed, '* as heat, by the steam, with the result that the steam is drier on leaving the nozzle than it would have been if the flow were friction - less ; or expressed in other words, the steam contains more heat on leaving the nozzle and less kinetic energy than in the ideal case when the frictional resistance is nil. At the end of expansion the steam is in the condition repre- sented by the point e (Fig. 89), the expansion being represented by some such approximately straight line as te. The total kinetic energy generated (including that utilised in overcoming friction) is represented by the area abee, the amount lost in friction by the area feeg, hence the net amount of kinetic energy available is equal to area abee area = area abed f FIG. 89. The actual position of the point e is difficult to decide, and in practice when allowing for frictional resistance de is usually made equal to about \dk. EXAMPLE. An impulse turbine of the de Laval type is to develop 250 H.P. with a probable consumption of 15*5 pounds of steam per H.P. hour, the initial pressure being 180 pounds and the exhaust 2 pounds per ART. 107] ORIFICES AND NOZZLES 179 square inch absolute. Taking the diameter at the throat of each nozzle as J inch, find the number of nozzles required. Assuming that 12 per cent, of the heat drop is lost in the diverging part of the nozzle, find the diameter at the exit of the nozzle and the quality of the steam, which is to be fully expanded as it leaves the nozzle. (L.U.) By (6), Art. 102, the weight discharged per second from each nozzle will be W = dr2 W = 3^ x -X - U/ = 0-1213 pound 144 4 V 4 / V 2-53 Total steam required per second = ~4 -^ = 1*076 pounds 60 X 60 Number of nozzles = = 8'8. say o nozzles 0-1213 If the expansion were frictionless and adiabatic from 180 pounds to 2 pounds absolute, the kinetic energy of the jet at exit from the nozzle would be . (5), Art. 101 Since 12 per cent, of this heat drop is lost, the kinetic energy will be 88 per cent, of the above, namely V2 1-13 C / 2 \i-is == 0-88 X - X i44 X 180 X 2*53 1 1 ( ~ - ) 2g o<1 3 ' \IoO / = 202,700 foot-pounds, or 260 B.Th.U. and the velocity V = A/64'4 X 202,700 =3610 feet per second From steam tables (p. 480) we find the total heat of dry saturated steam at 180 pounds absolute is 1196-4 B.Th.U. Hence the total heat per pound after expansion to 2 pounds absolute is h z + -* 2 L 2 = 1 196-4 260 = 936-4 Substituting for h^ and L 2 from steam tables 94 + x 2 X 102 1 = 936-4 x z = 0-825 Volume per pound = 0*825 X 173*5 cubic feet r i . 0*1213 X 0*825 X i73'5 /. area of nozzle at exit = , 3 '-^- 3 - 3610 = 0*00481 square foot = 0*693 square inch .*. diameter of nozzle at exit = \/ ?*- = 0*939 inch Tf O*7 O K A. i8o THE THEORY OF HEAT ENGINES [CHAP. vm. 108. Theory of the Injector. The action of an injector will be understood by reference to Fig. 90. The steam used for working the injector expands through the conical nozzle A, issuing therefrom with a high velocity, and coming into contact with cold water flowing in from the feed tank E, is condensed in the convergent combining tube or cone B. The resulting jet of water enters the divergent delivery tube or cone C, and at its smallest cross-section is moving with its maximum velocity. The kinetic energy of the jet of water is then converted into pressure energy in its passage along the delivery tube, its pressure increasing as its velocity decreases, until on leaving the tube the pressure is greater than the boiler pressure and the water enters the boiler. An outlet is provided at D through which any excess of water may overflow when starting. Overflow FIG. 90. The velocity of the issuing steam jet may be estimated by either of the methods already given. Let V be the velocity of the steam jet in feet per second,/! the initial steam pressure, and p z the pressure in the jet just outside the steam nozzle where contact occurs between the steam and the entering water, then V may be calculated from either of the following equations : V= V / 2^J{(T 1 -T 2 )(i +^_T 2 log e J| (see (3), Art. 101) or V = Art. 101) It is usual to assume p z = o'6j> lt or if the ratio for maximum discharge be assumed, namely / 2 = '577/i then V = 5-85 or (see (8), Art 102) where P x is the initial pressure in pounds per square inch. ART. 108] ORIFICES AND NOZZLES 181 Weight of Water per Pound of Steam. Let W be the number of pounds of water drawn from the feed tank per pound of steam, h the head of water in feet on the injector (see Fig. 90), p the boiler pressure in pounds per square foot, and H the height, in feet, of the boiler feed check valve above the delivery cone of the injector, then neglecting losses the least velocity of the jet Vj entering the delivery cone will be given by ^L^ + H ........ (i) 2g 62-4 In actual practice/ in (i) should be taken about 20 per cent, greater than the absolute boiler pressure to ensure that the injector works properly. The velocity with which the water will flow into the injector under the head of h feet will be the momentum of W pounds of this water will be W 77 -V* g * the momentum of i pound of steam moving with velocity V will be y g and the momentum of the resulting jet Hence, equating the momentum before to that after combining we have V W . == ^ + T"V2^ g g & or V. If the water is not supplied under pressure to the injector but the feed tank is h feet below the injector, as is the case of injectors of the lifting type, equation (2) becomes y -yy \v _ In most cases the term .. . <\/gh is so small that it may be neg- lected. By substituting the value of V ; - obtained from (i) in (2) the value of W may be found. Estimation of the Feed Temperature. -Let t be the temperature of the water in the feed tank and / 3 the temperature of the delivery from the injector, i.e. the feed temperature to the boiler. Then per pound of steam used _ Kinetic energy of the jet =- - 1 ~- in heat units heat gained by W pounds of water = W(/ 3 /) heat lost by i pound of steam = x^Li + (/i 4) 182 THE THEORY OF HEAT ENGINES [CHAP. vm. equating the heat lost by the steam to the heat gained by the water, we get .. . . (3) from which / 3 may be estimated. The kinetic energy of the jet is usually so small in comparison with the other items in (3) that for practical purposes it may be neglected. Area of Steam Nozzle. The dryness fraction, jv 2 , of the steam at pressure / 2 ( or '6 /i) is found from a temperature-entropy diagram, or by calculation from *2 = ? r i + 1 g< (see (3), Art. 46) i, 2 \ Ij 1 2 / Let w pounds be the weight of steam used per second and v% the specific volume at pressure / 2 , then, neglecting the volume of the water it contains, its volume will be w X and the area of the steam nozzle = - % ... (4) Area of Water-discharge Orifice. The quantity of water drawn from the feed tank per second will be w X W pounds, or w.W - - cubic feet 02*4 hence, w.W-\-w (5) area of the discharge end of the combining nozzle = fi ' I0p. Types of Injectors. Injectors may be divided into two classes, "non-automatic" and "automatic," or self-acting injectors. Non-automatic injectors do not restart automatically if for any reason the discharge is interrupted ; and further, both the steam and water must be independently regulated by hand when starting in the first instance, or when restarting in the event of any discontinuity in the discharge. Non-automatic Injectors are further divided into two classes, namely (a) Non-automatic Lifting Injectors which lift their feed water and usually have adjustable cones, that is to say, their cones can be adjusted to suit a great variety of conditions and a wider range of steam pressures than non-adjusting injectors. (b) Non-automatic Non-lifting Injectors which do not lift, but require their feed water to flow to them under pressure. Automatic or Self-acting Injectors, on the other hand, lift their feed water if required, and start working as soon as steam and water are turned on, without any manipulation. They will also restart instantaneously and automatically should the jet be accidentally broken, 1 EXAMPLE i. Calculate the area of the orifices of a live steam injector to take 1000 gallons of water per hour from the feed tank to a boiler, the absolute steam pressure being 165 pounds per square inch. The steam supplied to the injector may be assumed dry, the pressure in the steam 1 For the description of various types of injectors, see the author's "Steam Boilers " (Edward Arnold). ART. 109] ORIFICES AND NOZZLES 183 orifice o'6 of the boiler pressure, and the temperature of the water in the feed or suction tank 60 F. The pressure in the steam jet = 0*6 X 165 = 99 pounds absolute. From steam tables we find p t V L H T 165 99 366 327 2753 4'47 856-8 886-6 1195-0 1186-2 826 7 8 7 From (3), Art. 101, v / 64-4 x -j 420 feet pe fraction 7R\^&26 P 78'7H T I , 1 *Qn 1~~ f = i The dryness 7A^ U 77A T o v. \ o2 r second _ 787 /8 5 6-8 6/ ' / **p7J 8 2 6\ >g < 7T 7 ) 886'6\ 8 6 = 0-963 (Note. Both V and x% could be obtained directly from a Mollier Diagram.) From (i), Art. 108, we have, neglecting H V j = 1-2 X 165 X 144 64*4 62*4 from which V^ = 1 7 1 feet per second Neglecting the second term on. the right-hand side of (2), Art. 108, we have 1420 *7*--*w+i 1420171 W == = 7*30 pounds The feed water drawn from the feed tank per hour is 10,000 pounds = ^ pounds per second, hence the weight of steam used per second is 10,000 3600 x 7 - 3 oP unds and by (4), Art. 108, the area of the steam nozzle will be 10,000 0*963 X 4'47 X 3600 X 7'3 1420 = 0*00115 square feet = 0*00115 X 144 = o* 1 66 square inch The discharge from the injector per hour will be . 10,000 10,000 -j 7'3 = 11,700 pounds 11,700 = ^ pounds per second 3600 p i8 4 THE THEORY OF HEAT ENGINES [CHAP. vm. and by (5), Art. 108, the area of the water discharge orifice will be 3600 X 62-4 X 171 = 0-000304 square foot = 0*0438 square inch The feed temperature will be by (3), Art. 108 from which / 3 = 199 F. EXAMPLE 2. Calculate the diameter of the orifices for an injector to deliver 1200 gallons of water per hour into a boiler containing steam at 60 pounds per square inch absolute. The steam supplied to the injector may be assumed dry, the pressure in the steam orifice o'6 of the absolute boiler pressure, the temperature of water in the suction tank 100 F., and the temperature of the feed water 180 F. (L.U.) Taking the necessary data from steam tables we have = 1400 feet per second The dryness fraction *, = Jl(2!2 + b 753\ ' V / 9377 V 753 = 0-970 The feed temperature is here given as 180 F., the weight of water per pound of steam may therefore be estimated as follows. From (3), Art 108, we have (neglecting the kinetic energy of the jet) 914-9 -f- 293 180 = W(i8o 100) W= 12-84 pounds The velocity of the jet, neglecting the second term on the right-hand side of (2), Art. 108, is Vj = -~g- = 101 feet per second Assuming that the water drawn from the suction feed tank is 12,000 pounds per hour, the weight of steam used per second is 12,000 pounds 12-84 X 3600^ and by (4), Art. 108, the area of the steam orifice will be 12.000 0-970 X 11-58 ^ 12-84 X 3600 * 1400 = 0*295 square inch hence the diameter = ^_ = . 6l3 i ncn ART. 109! ORIFICES AND NOZZLES 185 The discharge from the injector per hour will be 12,000 = I2 >934 pounds _ I2 >934 p 0un( j s p er second 3600 and by (5), Art. 108, the area of the water discharge orifice will be 12,934 x 3600 X 62-4 X ioi = 0-0821 square inch hence the diameter = A /^ 5 = 0-323 inch v 07854 The above areas have been designed for a given feed-water temperature and the injector may or may not work. The pressure of the water at the feed check valve will be from (i), Art. 108, ioi X ioi __/ X 144 64-4 62-4 ioi X ioi X 62-4 p= - -.. - = 68'6 pounds per square inch absolute 144 X 64-4 Since the absolute boiler pressure is 60 pounds per square inch it is evident that the injector will work against this pressure. EXAMPLES VIII 1. Boiler steam of dryness fraction 0*98 and pressure 150 pounds per square inch absolute expands through a nozzle down to a pressure of 100 pounds absolute. As- suming the flow to be frictionless and adiabatic, estimate the velocity and dryness fraction of the steam jet. 2. Dry steam at a pressure of 180 pounds per square inch absolute expands through a properly designed nozzle down to a pressure of 3 pounds absolute. Determine the areas of the throat and discharge end of the nozzle to discharge 3000 pounds per hour, and state the dryness fraction at these places on the assumption that the flow is friction- less and adiabatic. 3. Superheated steam at a pressure of 200 pounds absolute and with 100 F. super- heat (volume per pound = 2'68 cubic feet) expands through a nozzle down to 15 pounds absolute. Determine the principal dimensions of the nozzle to discharge 3600 pounds per hour, and state the condition of the steam in the throat and at the discharge end. Assume frictionless and adiabatic flow. 4. An exhaust steam injector is to be used for feeding a locomotive boiler in which the steam pressure is 200 pounds absolute. If the pressure of the exhaust steam for working the injector is 17 pounds absolute and its dryness fraction is 0^85, estimate the weight of water that can be pumped per pound of steam, the area of the steam and of the water discharge orifice, and the feed temperature, if the weight of water taken from the feed tank is 10,000 pounds per hour at a temperature of 50 F. CHAPTER IX THEORY OF THE STEAM TURBINE 1 10. Function of a Steam Turbine. The steam turbine is a machine which converts heat energy into kinetic energy, and the kinetic energy into a form in which it can do mechanical work. Kinetic energy is generated in a series of nozzles or vanes, and the steam, issuing there- from at a high velocity, impinges on another series of moving vanes or buckets and gives up some of its kinetic energy to them. In the ideal turbine the steam would leave the moving vanes with zero velocity FIG. 91. (relative to the turbine casing) and all its kinetic energy would be changed into work done on the rotating vanes. The turbine is enabled to do this by changing the momentum of the steam in its passage through the moving vanes, the change of momentum per second constituting the driving force on these vanes. Let V be the absolute velocity of the steam as it impinges on the moving blades, and v the absolute velocity as it leaves them, then the change of momentum per second of i pound of steam in the direction of motion of the blades is V cos a cos . _. (see Fig. 91) 186 ART. no] THEORY OF THE STEAM TURBINE 187 If W pounds of steam are supplied per second the driving force on the vanes will be W (V cos a + v cos j8) And if V 6 be the velocity of the vanes, the work done per second by W pounds of steam will be W (V cos a -f- v cos P) . V& o (I) or, the work done per pound of steam per second will be U= -(Vcosa o The velocity diagram is shown in Fig. 92. AB represents the absolute velocity V at an angle a with the direction of motion of the vanes, and V& FIG. 92. the velocity of the blades. AC will therefore represent the velocity of the steam (R) at entrance relative to the blades ; let its direction be inclined 6 to V&, as shown. The absolute velocity v of the steam leaving the blades is represented by CD at angle j3, whilst CE will represent the velocity of the steam (r) relative to the blades at exit ; let it be inclined to V&, as shown. Now the work done per pound of steam per second is U = (Vcosa + cosj3) <> and since BD is equal and parallel to CE, CD cos j3 + CB = CE cos .'. CD cos ]8 = CE cos < CB But AB cos a = CB + AC cos hence AB cos a + CD cos ]8 = AC cos -f CE cos or, V cos a-\-v cos j8 = R cos 6 -j- r cos i88 THE THEORY OF HEAT ENGINES [CHAP. ix. Hence (2) may be written = -(Rcos0 + rcos<) ( 3 ) in. Impulse and Reaction Turbines. In impulse turbines the fall of pressure and generation of kinetic energy takes place in a set, or sets, of fixed nozzles or blades only. In reaction turbines the fall of pressure and generation of kinetic energy takes place partly in the fixed and partly in the moving blades. During the first portion of its motion through the moving blades kinetic energy is taken from the steam, and consequently its velocity is reduced ; but in the latter portion the velocity is increased by a suitable arrangement of the blades, and therefore genera- tion of kinetic energy takes place whilst work is being done on the moving blades. A hard-'and-fast line of distinction cannot be drawn between the two types, and many modern turbines are a combination of both the purely impulse and the purely reaction types. 112. Single Stage Turbines. The De Laval turbine is the best Moving Blades Axis J of Rotation FIG. 93. known and most extensively used single-stage turbine. It is a pure impulse turbine, and in it the steam is expanded in nozzles from the initial down to the exhaust pressure before being directed on to a ring of moving blades fixed on the circumference of a rotating wheel, as shown diagrammatically in Fig. 93. The velocity diagram is shown in Fig. 94. Neglecting all losses, the work done on the moving blades per pound of steam will be equal to the difference between the initial and final kinetic energy of the steam, viz. and the efficiency of the blades or turbine wheel will be V2 Z/2 The Speed of the Wheel to give Maximum Efficiency. Case I. Suppose V and a to be fixed and 6 = (/>, then using the same notation as in Art. no, we have, since r= R, ART. 112] THEORY OF THE STEAM TURBINE 189 or, Since CD2 = CB 2 + CE2 2 . CB . CE cos < = CB2 + AC 2 2 . CB . AC cos . Z/2 = V 6 2 + R2 _ 2 . V 6 . R COS 6 . . = V 6 2 -f & 2 V b r cos AC cos e = FC (2) may be written CD 2 = CB 2 + AC 2 2 . CB . FC (2) (3) FIG. 94. The efficiency will be a maximum when CD or v is a minimum, i.e. when CB . FC is a maximum. This occurs when CB = FC, or when Also since V 6 = R cos 6 (3A) = cos a -cos a 2 Hence v is a minimum when and since V& = R cos 6 = cos a FB = 2 FC tan 6 = 2 tan a (4) (5) In practice a is small (in the De Laval turbine it is about 20), and therefore cos a will be very nearly equal to unity, hence the maximum efficiency will be obtained when the velocity of the blades is approximately equal to half the velocity of the steam jet. Efficiency of the Blades The work done per pound of steam per second is by (3), Art. no V = (R cos + r cos ) o V& = (Rcos0 + R cos0) 2 V& . R cos = - (6) 190 THE THEORY OF HEAT ENGINES [CHAP. ix. Substituting for R from (3A) U - v and since by (4) V& = cos a V 2 U = i-cos2a (7) By (3) above z/ 2 = V 6 2 + R 2 - 2 . V 6 . R cos V; 2 hence v* = V & 2 + ^ 2 . V 6 2 cos^ a = V 6 2 (sec 2 i) = V& 2 . tan 2 /. z> = V 6 tan (8) From (4) and (5) we have V V& = cos a, tan 6 = 2 tan a V Hence v = cos a . 2 tan a 2 = V sin a (9) Neglecting all losses, the efficiency is V2 z;2 E = V 2 V 2 V 2 sin 2 a y 2 = i sin 2 a /. E = cos 2 a (10) The efficiency might also be deduced as follows ; it will be work done on the blades ~ initial kinetic energy of the steam Effect of Friction. Frictional and eddy losses will tend to make r less than R. On the other hand, any kinetic energy that may be developed by the steam in its passage through the moving blades will tend to make r greater than R. Hence r may be either less than R, equal to R, or greater than R. In general terms, therefore, we may assume that in actual practice r = R, where k is a constant. Equation (3) will therefore become 2 V^ . R cos B .... (n) ART. 112] THEORY OF THE STEAM TURBINE 191 As before, this will be a minimum when V V& = R cos 6 = - cos a Equation (3), Art. no, will become U=- 6 (Rcos0 + Rcos0) f (i+) ...... (12) Equation (n) becomes, since V&= R cos tan 2 _f_ j _ 2 k) 0+(/& 1)2} ..... (13) Substituting for V& in (13), from (4) and (5) v* = JV2-cos2 a{ 2 4 t an2 a + (k i) 2 } = 2 V 2 sin2 a + JV 2 cos2 a(^ i) 2 . . . (14) The efficiency of the blades will now be (is) When k= i, equations (12), (14), and (15), reduce to (7), (9), and (10) respectively. Case II. Suppose V is fixed in magnitude only, and being" fixed but not equal. Then neglecting frictional resistance CD 2 = CB 2 + AC 2 2 . CB . AC cos z; 2 = V 6 2-fR2_ 2 .v 6 .Rcos0 . . . . (16) Jn this case v will be a minimum when CB . AC is a maximum, i.e. when AC = CB, and the triangle ABC is isosceles. Hence = 2 a And from (16) and (17) we have V 2 V 2 2COS 2 V V 6 = V & = sec a= -sec- .... (17) = 2 . sec 2 - 2 . sec 2 - . cos ct> 4242 V 2 = sec 2 -(2 2 cos (/>) ..... . -:.-. . (18) 192 THE THEORY OF HEAT ENGINES [CHAP. ix. Now the work done per pound of steam per second is U = ^(R cos 6 + r cos ) 6 Substituting for V& and R from (17) V 2 6 U = sec 2 -(cos + cos <) ..... (19) The efficiency of the blades will be = J sec 2 -(cos 9 + cos <) _ cos 6 + cos c/> 2 COS 2 ? 2 cos 6 + cos d> . i+cos0 ........ (20) Hence in this case for maximum efficiency we must have V 9 V& = sec- ......... (17 above) V 2 6 V= . sec 2 -(cos 6 4- cos 6) . . . (19 above) 4f 2 , , cos 6 + cos , E = - ...... (20 above) i -J- cos 6 Effect of Friction. Writing r = &R the above expressions will be found to reduce to V 6 y b = - sec- as before, see (17) 2 2 V 2 U = . sec 2 -(cos0+ cos() ..... (21) i -\-cos6 Effect of Vane Speed on the Efficiency. Assuming that r= R, we have from Fig. 94 z/ 2 = R 2 _f-V 6 2 2 RV 6 cos^ ..... (23) arid R 2 = V 2 + V 6 2 _ 2 VV 6 cos a ..... (24) (24) in (23) gives __ z/ 2 = V 2 + V 6 2 2 V V b cos a + V 6 2 2 V b cos ^y/V8 + V 6 2 2 VV 6 cos a = V 2 + 2 V 6 2 2 VV & cos a 2V 6 cos fv/ v2 + v & 2 2V V& cos a _ s a y b cos ^y2 _|_ y & 2 _ 2 yy 6 cos a | V 2 ART. 112] THEORY OF THE STEAM TURBINE Writing V& = V where ^\/i 2n cos a + or E = 2cos a -- cos \/i 2 cos a n z n (25) Taking a = = 20 the following table shows the value of E for various values of n : 0'2 03 0*604 0789 0'4 '5 0-913 0-966 0-6 07 0-949 0-880 0-8 0-9 0-784 0-666 ro 0-540 The above values of n and E are shown plotted in Fig. 94A, from which it will be seen that the maximum efficiency (neglecting friction) of o | ) ) | I | | |||) The total expansion from ' ' ' J ' ' ' ' ' ' ' : the initial to the exhaust Fixed ^*^^^^^*^^^^ pressure is divided up into ^O Ss "^ s O s O s O s ^ _ a number of stages, each of Manna ^H^ \\\\\\\A\\ which is made up of a ring ///////// / _ of nozzles and a ring of T y V V V V moving blades, being essen- XXoOsVsQ^ tially a De Laval turbine Movmd w * tn a sma ^ drop m P res " sure. Each ring of moving FIG. 95. Rateau blading. wheels is mounted on a separate wheel, which is keyed to the rotor shaft, and the steam only exerts one effort on the moving blades in each stage. By this means the speed of the turbine is reduced, there being from twenty to thirty stages in the Rateau and about ten in the Zoelly turbine. For a given range of steam pressure, therefore, the Rateau runs at a lower speed than the Zoelly turbine. The number of stages is decided with reference to the total pressure drop in such a manner that the necessary velocity of the steam is obtained, which, with a reasonable speed of wheel, will give a satisfactory blade efficiency. M. Zoelly 1 selects a ratio of pressure fall per stage of = 173 /2 as giving the best combination of efficiency and practical constructional economy, where p = the initial steam-pressure, and / 2 tne ^ na ^ steam- pressure in the same stage. With an initial steam-pressure of 10 atmospheres absolute and an exhaust pressure of o* i atmosphere absolute this pressure ratio per stage requires nine stages, giving a steam velocity of about 1600 feet per second, and requiring a peripheral speed of the blades of about 700 feet per second. 1 See Proc. Inst. Mech. E., p. 686, July, 1911, ART. 113] THEORY OF THE STEAM TURBINE 195 By using from twenty to thirty stages in the Rateau turbine, depending upon the total pressure drop, the speed of the moving blades is reduced to about 350 feet per second. 1 The velocity diagram for both the Rateau and the Zoelly turbines is shown in Fig. 96. It consists of a series of diagrams, each similar to that Moving FIG. 96. Velocity diagram for the Rateau turbine. of the De Laval type ; ab represents the absolute velocity of the steam leaving the first set of fixed vanes or nozzles, cb the velocity of the tips of the first set of moving blades, and ac the velocity of the steam relative to the moving blades. The angle (6) of the blades at inlet must therefore be parallel to ac t in order that the steam may enter them without shock. The velocity of the steam relative to the moving blades at exit is represented by cd, and the absolute velocity leaving the moving blades by ce, being the angle of the moving blades at exit. The steam leaving the first set of moving blades with absolute velocity ce next impinges on the second set of fixed blades, and is guided by them in the direction ef, the fall of pressure being such that the absolute velocity Centering the second set of moving blades is the same as ab. The same 1 See Proc. Inst. Mech. Eng., June, 1904, p. 737. 196 THE THEORY OF HEAT ENGINES [CHAP. ix. action is now repeated, and efg is the triangle of velocities at entrance to the second set of moving blades, and ghk the triangle of velocities at exit. The entrance angle of the second set of fixed blades is j8, the exit angle being y. 114. Turbines with one or more Stages, each compounded for Velocity. The Curtis turbine is the best known example of this type, a diagrammatic sketch of the arrangement of nozzles and blades being shown in Fig. 97. In the example illustrated two stages are shown, in each STEAM CHEST STATIONARY BLADES TIONARY BLADES MOVING BLAGES NOZZLE. DIAPHRAGM MOVING BLADES STATIONARY BLADES MOVING BLADES MOVING BLADES mDWDWDnDDWDDWDDW i I I 1 I I FIG. 97. Arrangement of nozzles and blades'of Curtis turbine. of which there are one ring of nozzles, three rings of moving and two rings of stationary blades. The velocity diagram for one stage is shown in Fig. 98. Using the same notation as before, 6 and denote the angles of the moving blades at entrance and exit respectively, whilst j8 and y denote the corresponding angles for the fixed blades. Fig. 98 has been drawn on the assumption that the relative velocity of the steam at entrance and exit of each ring of blades is the same, i.e. r = R (see Art. 112), and also that the velocity of the tips of all the moving blades is equal, i.e. If the absolute velocity of the steam entering the first set of moving blades be denoted by V, and the absolute velocity on leaving the last set ART. 115] THEORY OF THE STEAM TURBINE 197 of moving blades' by v, then, as in Art. 112, the efficiency of the stage will be (neglecting losses) V2 - z;2 V2 ac- cd. ce=ef. Moving Fixed Am-mn, Moving Fixed Moving n FIG. 98. Velocity diagram for one stage of Curtis turbine. 115. Multi-Stage Reaction Turbines The Parsons Tur- bine. 1 Although commonly called a reaction turbine, this turbine works partly by impulse and partly by reaction. The arrangement of blading is shown diagrammatically in Fig. 99. The steam expands through the first ring of fixed blades, from which it is directed on to a ring of moving blades^ 1 For a detailed description of this and other types of turbines see "The Steam Turbine," by R. N. Neilson. Longmans. 198 THE THEORY OF HEAT ENGINES [CHAP. ix. then through another ring of fixed blades and another of moving blades, and so on alternately right along from the high-pressure to the exhaust end of the turbine. Kinetic energy is developed in the fixed blades and also in the moving blades whilst energy is being taken up by them from the expanding steam. Each pair of fixed and moving rings of blades consti- tute a stage, and since a large number of stages are employed in this type FIG. 99. Arrangement of blading for Parsons turbine. of turbine the velocity of the steam (and therefore of the moving blades) is not very high. The velocity diagram for two stages is shown in Fig. 100; db repre- sents the absolute velocity (V) of the steam leaving one of the fixed blades, cb the velocity of the next moving blade, and ac the relative velocity (R) with which the steam enters the moving blades. The expansion being con- tinued in the ring of moving blades the relative velocity of the steam on leaving (r) is given by cd, and its absolute velocity (v) by ce. In the next ring of fixed blades the velocity is increased to ef t which is the same as ab or V. With velocity V the steam enters the next ring of moving blades, and the action is repeated in successive stages, the pressure drop being the same in each ring of fixed and moving blades. In actual practice the velocity V does not remain the same in each stage, but increases continuously as the steam flows through the turbine, on account of the increasing specific volume of the steam as its pressure falls. In order to keep the velocity V as uniform as possible provision has to be made to accommodate the increasing volume of the steam. The moving rings of blades are fixed into the circumference of the rotor and project radially outwards. The rings of fixed blades are fixed into the outer casing, and project radially inwards between the rings of moving blades. The area through which the steam passes is that of the annular ring between the rotor and outer casing, minus the area of cross section of the blades, and in order to increase this area the length of the blade is increased at certain points along the length of the rotor and casing, the diameter increasing in steps towards the exhaust end, the lengths of the blades in the several rings of blades in each step being the same. The angle of the blades at exit is of very great importance. Referring to Fig. 100, it will be seen that the smaller the exit angle from the fixed blades (y), the greater will be the velocity of the steam (V) as it leaves ; also, the smaller the exit angle of the moving blades (), the smaller is the absolute velocity of the steam (v) leaving them, and consequently the less is the kinetic energy remaining in the steam as it leaves the moving blades. ART. us] THEORY OF THE STEAM TURBINE 199 It would appear, therefore, that the smaller these angles are made the better, but a limit is imposed by practical considerations. The smaller the angle the less is the width ab (Fig. 101) of the steam passage between adjacent P Moving Moving CCk FIG. 100. Velocity diagram for Parsons turbine. blades which makes the thickness of the blade / greater in proportion to ab, resulting in an increased tendency to form eddies and a consequent reduc- tion in the available energy of the steam. In addition, very small exit angles have the defect of increasing the length of the blade pas- sages, and therefore the frictional resistance. In turbines of this type it is usual to make the exit FIG. 101. angle about 20 and to make the moving and fixed blades exactly alike. At the low-pressure end of the turbine, the exit angles of the last few rings of blades is generally 2OO THE THEORY OF HEAT ENGINES [CHAP. ix. < J, increased in order to provide for an increase in the steam area without the use of excessively long blades. Such blades are called " semi-wing " or f full-wing," depending upon the outlet angle. 1 EXAMPLE i. The steam chest pressure in a De Laval turbine is 200 pounds per square inch absolute, and the exhaust pressure 5 pounds per square inch absolute, the steam being initially dry and saturated. The peripheral speed of the blades is 1200 feet per second, and the nozzle is inclined 20 to the direction of motion of the blades. Estimate the angles of the blades, the work done on the blades per second per pound of steam, the ab- solute velocity of the steam at discharge from the blades, and the efficiency of the tur- bine. Neglect frictional re- sistances and assume adia- batic flow. From the Mollier Dia- gram we find the heat drop in the nozzle to be 253 B.Th.U. and the velocity at entrance to the wheel A/778 X 253 X 64-4 = 3560 feet per second The velocity diagram is shown in Fig. 102. On setting this out to scale, is found to be 30, and AC = 2470 feet per second. The De Laval blades are made symmetrical with equal inlet and outlet angles, hence, making 6 = and CE = AC, the absolute velocity at discharge is found to be 1530 feet per second. The work done per second per pound of steam is by (3), Art. no, U = 1(2470 cos 30+ 2470 cos 30) _ 2400 x 2470 X o'866 32-2 = 159,400 foot-pounds The efficiency will be = 0-815 FIG. 102. The value of the inlet and outlet angles and the absolute velocity at discharge may be obtained by calculation as follows : R 2 = (I200) 2 + (356o) 2 2 X 1200 X 3560 COS 2O = io 6 (i'44 + 12-6736 8-0288) = 106 X 6-0848 /. R = 2466 feet per second, which agrees with the result found above. 1 For a detailed discussion of blade angles see " Steam Turbine Design," by J. Morrow. Edward Arnold. ART. 115] THEORY OF THE STEAM TURBINE 201 sin 6 _ sin 20 Also ~ 2466 2466 .-. = sin- 1 0-4935 = 29-6 and iP = (1200)2 -f (2466)2 2 X 1200 X 2466 cos 29-6 = 106(1-44 -j- 6-0848 5-1486) = 106 X 2-3562 .-. v = 1535 feet per second as above EXAMPLE 2. If the steam consumption of the above turbine is 3600 pounds per hour, estimate the horse-power of the turbine. From the above Example i, the work done per second is 3 x i59)4oo foot-pounds 60 x 60 and the horse-power = I59)4 Q = 290 H.P. 550 EXAMPLE 3. In the turbine given in Example i,find the blade angles work done per second per pound of steam, the speed of the wheel, and the efficiency, if the efficiency is to be the greatest possible. This is an example on Case i, Art. 112. Here a = 20, V = 3560 feet per second, and 9 = . By (4), Art. 112, V 6 = ^~ cos 20 = 1780 X 0-9397 = 1673 feet per second By (5), Art. 112, tan 8 = 2 tan 20 = 2 X 0-364 .. = tan- 1 0-728 36 nearly = 173,800 foot-pounds By (10), Art. 112, the efficiency = (cos 2o) 2 = (o'9397) 2 = 0-883 ,/' EXAMPLE 4. An impulse turbine of the Curtis type has two stages with one set of nozzles and three rotating and two stationary sets of blades. Superheated steam is expanded adiabatically in the first set of nozzles from a pressure of 180 pounds per square inch absolute and temperature 473 F. to 24 pounds per square inch absolute. The nozzles are inclined at an angle of 20 to the plane of rotation of the running blades, which have a peripheral speed of 450 feet per second. Determine the angles of the three moving and two stationary sets of blades in the first stage, and give the absolute velocity of the steam as it leaves the last set of running blades. Referring to steam tables (p. 480), we find that the temperature of saturation at 180 pounds absolute is 373 F. Hence, the steam is super- heated, 473 - 373 = 100 F. 202 THE THEORY OF HEAT ENGINES [CHAP. ix. By means of the Mollier Diagram, the velocity of the steam issuing from the nozzles is found to be 2840 feet per second. The velocity diagram has been drawn for three different conditions, in Moving Fixed Moving Fixed Mov/njj n o FIG. 103. each of which the effect of friction has been neglected. In the first case taken (Fig. 103), the moving blades have all been made symmetrical, the inlet and outlet angles being the same for each ring. In Fig. 103, which is self-explanatory, the angle at inlet (6) has been made equal to the angle ART. 115] THEORY OF THE STEAM TURBINE 203 at exit (<), and all the absolute velocities are represented by means of a thick line. The results obtained are as follows : Moving blades 6 = = 24 for each ring, ist ring. 2nd ring. Fixed blades . . Inlet angle fa = 30 Outlet angle 7! = i8'6 Inlet angle 2 = 35 Outlet angle B 2 = 15 Moving " Movintf 68 " 204 THE THEORY j OF HEAT ENGINES [CHAP. ix. The absolute velocity of the steam leaving the last ring of moving blades is mo = 375 feet per second. Second Case. In this case 6 has been made equal to (/> in the same ring of moving blades, and j8 equal to y in the same ring of fixed blades, but the angles in one ring differ from those in another (see Fig. 104). The results obtained are ist ring. and ring. 3rd ring. Moving blades . . e l = , i.e. the blades in the three rings of moving blades are symmetrical, but the entrance and exit angles are different. The magnitude of the outlet angle has been fixed at 20, ART. 1 1 6] THEORY OF THE STEAM TURBINE 205 and the velocity diagram drawn accordingly (see Fig. 105). The results obtained are : ist ring. 2nd ring. 3rd ring. Moving blades . Inlet angle 6 t = 24 Outlet angle fa = 20 6, = 2 4 2 = 20 3 = 24 3 - 20 Fixed blades . . Inlet angle ft l = 24 Outlet angle y t = 19 0, = 27'5' 7 2 = 24 The absolute velocity of the steam leaving the last ring of moying blades is mo = 340 feet per second. 116. Losses in Steam Turbines. The chief losses which occur may be divided into those produced by (a) Steam leakage. (b) Heat leakage. (c) Frictional resistance. (a) Steam Leakage. In all types of steam turbines some of the steam leaks past the tips of the moving blades without doing work on them. The proportion of available energy of this leakage steam, which is wholly lost, depends upon the type of turbine. In the De Laval turbine the available energy in the amount of steam which does not enter the moving blades is practically lost, since it passes directly from the nozzle into the turbine casing without doing useful work. In the Rateau and Zoelly turbines (Art. 113) each running wheel is mounted in a separate compartment, and any steam that leaks out between any ring of nozzles and the next set of rotating blades is enclosed in a separate chamber and its kinetic energy is there converted back into heat energy, which may be utilised again in the next stage. Also, since the kinetic energy of this leakage steam is only a comparatively small fraction of its total available energy (there being a large number of stages), the total loss due to leakage is comparatively small in spite of the large number of stages. In the case of the Curtis turbine (Art. 114), steam, on leaving the nozzle, may leak past the first set of moving blades, but (unlike the de Laval turbine) all its kinetic energy is not lost, because, since it passes through the remaining rings of fixed and moving blades at constant pres- sure, some of the kinetic energy is available for extraction by the moving blades, and although some leakage may occur between adjacent rings of fixed and moving blades, the net loss is less than if these rings were absent. The facilities for leakage are very much greater in the Parsons turbine. Unlike the Rateau and Zoelly turbines, all the rings of moving blades are fixed on a common spindle running in one chamber between rows of fixed blades, and, further, there is a continuous expansion and fall of pressure along the turbine (Art. 115). The difference of pressure on the two sides of each ring of moving blades increases the tendency to leakage, particu- larly at the high-pressure end of the turbine, since at this end of the turbine the necessary height of the blades is small because of the high density of the steam and the leakage area, i.e. the annular ring between the tips of the moving blades and the outer casing which is necessary for clearance, 2O6 THE THEORY OF HEAT ENGINES [CHAP. ix. is a greater proportion of the total area through the blades than at the low-pressure end where the blade height is much greater. (b) Heat Leakage. One of the advantages possessed by a turbine over a reciprocating engine consists in the absence of initial condensation when once the turbine has settled down into working conditions. In the case of the Parsons turbine, for example, at any point on the length of the rotor the temperature remains constant. There will, however, be a steady flow of heat from the high-pressure towards the low-pressure end. The amount of heat which thus leaks through to the exhaust does not all repre- sent a dead loss, because part of it is available for drying the steam, and thereby reducing the frictional resistance between the moving blades and the steam. In addition to this leakage there will be the loss due to radia- tion and conduction by the metallic casing. The general effect of heat leakage on the efficiency of the turbine is best seen with reference to the temperature-entropy diagram. The diagram with no heat loss is represented by abed in Fig. 106, this FIG. 106. Effect of heat leakage. being the diagram for the Rankine cycle already considered (Art. 57). If the steam loses heat throughout the expansion the final dryness fraction will be less than if the expansion were adiabatic, and the expansion line will terminate at some point e on the left of represents the final dryness fraction. Some of the steam will therefore be condensed at temperatures varying from T l to T 2 , and assuming the loss of entropy during expansion to be proportional to the fall of temperature the expansion line will be represented by ce instead of the adiabatic cd. The temperature entropy diagram with heat leakage is therefore represented by abce, this area representing the amount of heat converted into work, whilst the area ced represents the loss due to heat leakage. ART. 116] THEORY OF THE STEAM TURBINE 207 The heat supplied is denoted by the zxv&gabch and the overall efficiency of the turbine will be Area abce Area gabch (c) Frictional Resistance. Of the frictional resistance in a steam turbine, mechanical friction at the bearings is usually very small, the principal resistances being those between the steam and the blades. The eifect of friction on the flow of steam through the fixed nozzles has already been discussed in Art. 107. The effect in the moving blades is to make the relative velocity at exit less than at entrance to the blades (r = R), and its effect on the efficiency of the blades has been dealt with in Art. 112. The other losses which occur are due to the frictional resistance offered to the rotation of the turbine wheel or wheels in the steam, and the loss due to shock at entrance to the moving blades. The frictional resistance between the running wheel or wheels and the steam may in some cases be very great. Its magnitude depends upon the density of the steam in which the wheel rotates, upon the quality of the steam and upon the speed of the wheel. From results of tests carried out on De Laval turbines it appears that for a given speed of wheel and quality of steam the frictional resistance is proportional to the density of the steam. If dry saturated steam be used it will be wet after expansion, and the presence of this moisture will greatly increase the friction. The greater economy which results from the use of superheated steam is doubtless due to the reduced frictional resistance resulting from the absence of water in the steam. For a given quality and pressure of steam the resistance increases rapidly with increasing speed; with wheels of the De Laval type the frictional loss appears to be proportional to about the cube of the speed, 1 and the fifth power of the diameter, i.e. the wheel friction for this type is proportional to * 3 \ Y ^ * F * !*. i\ 7^-^** _u4 ^ >i'^ FIG. 116. Substituting (3) in W E 2) we have and efficie logf r , ^ 4 (3) The ratio of the volumes of the cylinders will be 129. Case VI. Three-Stage Compression with Spray Injection. The law of the compression curve will be pv n = constant, where n = 1*2 about. e^ work done per Ib. of air will be and efficiency 7] = ( 2 ) 224 THE THEORY OF HEAT ENGINES [CHAP. x. EXAMPLE i. An air compressor is required to compress from one to ten atmopsheres. Estimate the amount of work which must be expended for each Ib. of air compressed in each of the above cases. Calculate also the efficiency in each case, the initial temperature being 60 Fahr. Case I. Simple compressor with adiabatic compression. By equation (4), Art. 124 1-4-1 , xr T iu r 53 >l8 X i'4 X 52i[Yio\ 1-4 Work per Ib. of air = - - I I i 1- i L\ i / 1-4 X i- f ,. lbs . By equation (6), Art. 124, loge 10 2-303 7] = ,., , = - - = 0706 or 7o'6 per cent. 3-SX 0-931 I Case II. Simple compressor with spray injection. By equation (i), Art. 125, Work per lb . = = 53'i8 X 521 X 6 X 0-466 =77,110 ft.-lbs. By equation (2), Art. 125, __ loge io 2-303 oj? 1-2 X (101-2 i) 7J ~ o ~ 6 X 0-466 = 0-823 or 82-3 per cent. Case III. Two-stage adiabatic compression. By equation (5), Art. 126, W = -] = " X ' 5 " X ' '' X 0-39 = 75,63 .*.. 0*4 By equation (6), Art. 126, = !?!L = ' 8 43, or 84*3 cT 4 X ' 39 Case IV. Two-stage compression with spray injection. By equation (i), Art. 127, ... s.ri8 X S ai X 4[ IO F!_ ,] = S3 . l8 x ,. x x Q . 2I2 O'2 = 70,470 ft.-lbs. By equation (2), Art. 127, 12 X 0-212 ART. 129] THEORY OF AIR COMPRESSORS AND MOTORS 225 Case V. Three-stage compression with adiabatic compression. By equation (4), Art. 128, 53-18 X 5*1 X4-r J = J = SJ . I8 x 52I x IQ 04 = 71,550 ft. -Ibs. By equation (5), Art. 128, 2*303 77=- z = 0-891, or 89*1 per cent. 10-5 X 0-246 Case VI. Three-stage compression with spray injection. By equation (i), Art. 129, w = 53-18 Xj|2i X 3*[JI_ t ] = 5yi8 x 52I x i = 67,810 ft.-lbs. By equation (2), Art. 129, 2-203 18 X 0-136 94 ' EXAMPLE 2. Calculate the size of the cylinder required for a double- acting air compressor of thirty-five indicated horse-power working as a simple compressor with spray injection, the law of compression being ^zA-2 constant, the ratio of compression being from 15 Ibs. per square inch absolute to 120 Ibs. per square inch absolute. Revs, per min. = no, and average piston speed 560 ft. per min. Neglect clearance. Now, work done per Ib. of air = p^v^ Pi v i .*. Mean effective pressure = p 2 ~ />i + . . (i) Also '93i = 07526 , = 5-657 Substituting this value of in equation (i), we have 120 Mean effective pressure = - J ^ i e x 5' 6 57 5^57 02 = (21-21 15) x ^=37-26 Ibs. per sq. in. 226 THE THEORY OF HEAT ENGINES [CHAP. x. Let A = area of cylinder in square inches then 37-26 X 560 X A = 35 X 33> 5,37 inches /. Diameter = /v/ 55 37 = 8-40 inches V 07854 Since revs, per min. = no, and there are 220 working strokes per minute (the compressor being double-acting), the stroke is = 2 -5 45 feet = 2 feet 6J inches 130. Effect of Clearance. The clearance in the compressor cylinder should be as small as possible since its effect is to reduce the efficiency. This is shown in Fig. 117. During the suction stroke the 1 ^Clearance Stroke Po/urne FIG. 117. compressed air in the clearance space of expands to volume og t and less atmospheric air is drawn into the cylinder, the volume being only gh. The result is that the compressor must make a greater number of strokes to deal with the same volume of air, with a resulting greater loss by friction. 131. Air Motors. On leaving the compressors the air is delivered by the mains to the motors, in which it may do work by expansion in a variety of ways. The air may be allowed to expand adiabatically, or it may be warmed during expansion by spray injection, or it may be expanded in two or more stages, being warmed in intermediate receivers of large enough capacity. The best method, both from a thermodynamic and a practical point of view, is to pass the air through a " preheater " or heating stove, and to commence the expansion in the motors with the air at a con- veniently high temperature, and to work adiabatically. Occasionally, in very large motors, the air is heated up twice, being exhausted from the ART. 132] THEORY OF AIR COMPRESSORS AND MOTORS 227 high-pressure cylinder at a pressure of two or three atmospheres, and passing through a heater is expanded in the low-pressure cylinder down to atmospheric pressure. The theory of the different types of motors may be considered as follows : 132. Case I. Simple Adiabatic Expansion. Let the air enter the motor at pressure / m , volume v m , and temperature T m , the suffixes indicating the state of the air in the mains at the motors. Fig. 118 repre- sents the indicator diagram. H * i_ Then, Now, Hence, work done per Ib. W = *- (p m Vm f-l = T m (compare (i), Art. 124) w = If the expansion is isothermal W = RT m lo ge ^' . . Pz and the efficiency of the simple adiabatic motor is (I) (2) (3) (4) (5) (6) log. 228 THE THEORY OF HEAT ENGINES [CHAP. x. 133. Case II. Simple Motor, with Spray Injection. In this case the law of the expansion curve would be pv n = constant, where n varies from 1*25 to 1*35, and the work per Ib. would be 134. Case III. Two-Stage Adiabatic Expansion. In this type of motor the air is heated in an intermediate receiver to approximately the same temperature as it entered the high-pressure cylinder from the mains, which will be the temperature of the atmosphere if no preheater be used. The indicator diagram is shown in Fig. 119. f ^^-/sothermal at T m \ \ \ \ FIG. 119. Work done per Ib. in high-pressure cylinder is y-l In the low-pressure cylinder Hence total work done is W 1 -f- W 2 ; W = W, + W 2 = RT. . _*_[, _() If p 2 = Vpmfy, which is the condition for maximum work or for equal power developed in each cylinder y-l W = RT OT . -J2L [i -() 2y ] (compare (5), Art. 126) , . (4) ART. 135] THEORY OF AIR COMPRESSORS AND MOTORS 229 In this case r = -p, and the efficiency is given by 1 r\rffi>> (5) 135. Case IV. In this type let each Ib. of air entering from the T l i FIG. 120. mains in the state p m v^ m be heated at constant pressure in a heater to temperature T, so that its volume becomes JL The indicator diagram is shown in Fig. 120. Assuming adiabatic expansion, the work done per Ib. in the high- pressure cylinder will be Wl -* y i /-i (i) (2) y i p Now, let the air on its way to the low-pressure cylinder pass through another heater which raises its temperature to T again. Assuming adiabatic expansion down to atmospheric pressure the work done in the low-pressure cylinder will be W 2 = i .-i y (3) 230 Again, taking p 2 = THE THEORY OF HEAT ENGINES [CHAP. x. , we have the total work done y (4) Comparing this expression with (4), Art. 134, it will be seen that the work done depends upon the initial temperature of the air on entering the high-pressure cylinder. 136. Case V. Three-Stage Adiabatic Expansion Motor without Preheater but with Intermediate Heaters. In this case the air enters the high-pressure cylinder at the temperature of the mains T m , and after expansion is heated up to T m again, and then passes into the intermediate cylinder. Exhausted from the intermediate cylinder it is again heated to T m , and then expands adiabatically in the low-pressure cylinder down to atmospheric pressure. The indicator diagram is shown in Fig. 121. FIG. 121. Reasoning in exactly the same way as in the case of the three-stage compressor, Art. 12 8, the work done per Ib. of air reduces to the expression where p e = exhaust pressure and p m = pressure of mains. efficiency And the ART. 138] THEORY OF AIR COMPRESSORS AND MOTORS 231 9 = y 137. Case VI. This case is similar to Case V. (Fig. 121), only a preheater is used in addition to the intermediate heaters, and in a similar manner to Case IV., Art. 135, the work done per Ib. of air is w = The amount of preheating to be used depends upon the size of the motor and the desired temperature of exhaust. If the air enters without preheating it is exhausted at a low temperature (from 10 F. to 25 F.), in which state it may be used for cold storage or other similar purposes. If this low temperature is allowed there may be trouble caused by the deposition of snow in the cylinder, which may clog the valves. If, how- ever, a considerable amount of preheating be used, the exhaust temperature may be above the atmospheric temperature, and with a large motor enough warm fresh air may be obtained for heating and ventilation purposes in the winter. 138. Efficiency of Compressors. From tests made on the installa- tion in Paris it appears that the mechanical efficiency of air compressors is about 86 per cent. The thermodynamic efficiency of compressors (taking the isothermal compressor to have an efficiency of 100 per cent.) is about 85 per cent, for single-stage and 92 per cent, for double-stage compressors, with spray injection. The loss in the mains for a five-mile transmission, due to leakage and fall of pressure, may be put at 3*8 per cent., so that the efficiency of the mains is about 96*2 per cent. The thermodynamic efficiency of a simple adiabatic motor without preheater is about 77 per cent. ; if fitted with a preheater the efficiency becomes about 85 per cent. In the case of a two-stage adiabatic motor the efficiency is about 90 per cent, without a preheater, and no per cent, to 113 per cent, if a preheater is used in conjunction with an intermediate heater. EXAMPLES X 1. In a simple compressed air installation the compressor draws in air from the atmosphere, and compresses it adiabatically to twelve atmospheres. The temperature of the atmosphere is 60 F. In the mains the air is cooled at constant pressure to its original temperature, and is then delivered to the motor, where, after cut-off, it expands adiabatically to atmospheric pressure. Estimate the efficiency. 2. Estimate the B.H.P. of the engine required to drive an air compressor that takes in 260 cubic feet of air per minute at 60 F. and at atmospheric pressure, and compresses it adiabatically in one stage to ten atmospheres. Assume the mechanical efficiency of the compressor to be 86 per cent, and neglect all losses due to clearance, etc. 232 THE THEORY OF HEAT ENGINES [CHAP. x. 3. An engine is supplied with compressed air at 90 pounds per square inch absolute and at 65 F. The air is expanded, according to the law /z/ 1 ' 3 = const., down to 15 pounds absolute, and then exhausted at that pressure. Determine the pounds of air that will be used per hour per I.H.P., and calculate the temperature of the air at the end of expansion. Neglect losses due to clearance, etc. (L.U.) 4. An ideal air compressor works between pressures of one and twelve atmospheres, with adiabatic compression, the initial temperature being I7C. Estimate the amount of work which must be expended per pound of air compressed in the following cases : (a) Single-stage compression. (l>) Two-stage compression. (c) Three-stage compression. In (b) and (c] how much heat must be extracted in the intercoolers per pound of air compressed ? (Q> = 0*2375). 5. Solve Problem 4, if by spray injection the law of the compression curves is pv^-'' i = const. 6. Calculate the size of cylinder required for a double-acting air compressor of 50 I.H.P., working as a simple compressor, at 120 revolutions per minute, the law of compression being pv*'* const, and the ratio of compression pressures being from 15 pounds per square inch absolute to 150 pounds absolute. Take an average piston speed of 600 feet per minute. CHAPTER XI COMBUSTION COMBUSTION is a chemical combination of the inflammable constituents of a fuel with oxygen, the process resulting in the evolution of heat. The combustible elements in all fuels whether solid, liquid, or gaseous are carbon, hydrogen, and sulphur ; of these, the sulphur is of minor import- ance in contributing to the heating value, because only small quantities are present, also, the injurious sulphurous acid, which it forms on burning, makes it undesirable to use a fuel which contains much sulphur, particularly for steam-raising purposes. The following table gives the approximate atomic and molecular weights of the various substances to which reference will be made in this chapter : Atomic weight. Molecular weight. Hydrogen H 2 .... I 2 Oxygen O 2 16 32 Nitrogen N 2 . . ... H is Carbon C . . . 12 _ Sulphur S . . . 32 Water H 2 O . . . 18 Carbon monoxide CO 28 Carbon dioxide CO 2 . 44 Sulphur dioxide SO 2 64 Marsh gas CH 4 . 16 139. Combustion of Hydrogen. The process is represented by the following equation, the molecular weight being written under each symbol : (i) From (i) we see that i pound of hydrogen burning to steam (H 2 O) requires 8 pounds of oxygen and yields 9 pounds of steam. In the process of combustion i pound of hydrogen gives out about 60,930 B.Th.U. (33. 8 3 C.H.U.), but it forms 9 pounds of steam which absorbs about 9X9708730 B.Th.U., leaving in round figures 52,200 B.Th.U. (29,000 C.H.U.) available for useful purposes. Now, in actual practice, the steam formed as a result of the combustion of hydrogen in a fuel passes off as steam in the flue gases leaving a boiler or in the exhaust gases from an internal combustion engine, hence the effective calorific value of hydrogen must be taken as 52,200 B.Th.U. (29,000 C.H.U.) per pound. 233 234 THE THEORY OF HEAT ENGINES [CHAP. xi. This figure is known as the lower or effective calorific value, whilst 60,930 B.Th.U. per pound is called the higher or gross calorific value. The lower calorific value is, therefore, the higher calorific value minus the latent heat of the steam formed during combustion. 140. Combustion of Carbon. When carbon burns completely to CO 2 the process is represented by the equation C + 2 = C0 2 (i) 12 + 32 = 44 From (i) we see that i pound of carbon burning completely to CO 2 requires ff = 2-66 pounds of oxygen and forms ff = 3-66 pounds of CO 2 . In the process of combustion i pound of carbon when burning to CO 2 gives out in round figures 14,540 B.Th.U. (8080 C.H.U.). When i pound of carbon burns incompletely to CO it only gives out about 4400 B.Th.U. the action being represented by the equation. 2 C + O 2 = 2CO 2 (2) From the above it is evident that incomplete combustion results in a loss of heat equal to 14,540 4400 = 10,140 B.Th.U. per pound of carbon so burned. Hence the importance of having no CO in the flue gases given off from a steam boiler furnace or in the exhaust gases from an internal combustion engine. Also, if CO 2 at a high temperature comes in contact with incandescent carbon it is reduced to the lower oxide CO according to the equation C0 2 + C = 2CO (3) This occurs with a thick fire (see also Art. 152), and if the loss due to the presence of CO in the flue gases is to be prevented, air must be admitted above the fire to burn the CO to CO 2 . 141. Combustion of Sulphur. The action is represented by the equation S + 2 = S0 2 (i) 32 +32 = 64 From (i) we see that i pound of sulphur burning to SO 2 requires i pound of oxygen and forms 2 pounds of SO 2 . In the process of combustion i pound of sulphur gives out about 4000 B.Th.U. 142. Minimum Quantity of Air required for the Complete Combustion of i Pound of Solid or Liquid Fuel. Let i pound of the fuel contain C pound of carbon H pound of hydrogen O pound of oxygen S pound of sulphur. In Art. 140 we have seen that i pound of 'carbon requires 2 '66 pounds of oxygen for its complete combustion. From Art. 139, i pound of hydrogen requires 8 pounds of oxygen, and from Art. 141 i pound of sulphur requires i pound of oxygen. Hence, since air contains 23 per cent, by weight of oxygen, i pound of oxygen is contained in ^ 3 - = 4-35 pounds of air and i pound of carbon requires 2*66 X 4*35 = n'6 pounds of air i pound of hydrogen requires 8 X 4'35 = 34'8 pounds of air i pound of sulphur requires i X 4'35 = 4*35 pounds of air ART. 143] COMBUSTION 235 Therefore i pound of the fuel will require i r6 X C + 34'8 X H + 4'35 X S pounds of air . . ( i) EXAMPLE i. A sample of Russolene oil contains 86 per cent, by weight of carbon and 14 per cent, by weight of hydrogen; estimate the minimum quantity of air required for the complete combustion of i pound of the oil. Minimum quantity = ir6 X o'86 + 34-8 X 0-14 = 9'97 +4'87 = 14*84 pounds In actual practice about i times this quantity of air would be supplied in order to ensure complete combustion. EXAMPLE 2. The analysis of a sample of Nixon's navigation steam coal by weight is C 87*8 per cent., H 4*10 per cent., the remainder being ash, etc. If 18 pounds of air are supplied per pound of coal and the combustion is complete, estimate the composition of the products by weight. Total weight of products per pound of coal = 18 -f- combustible in i pound of coal = 18*919 pounds. Hence, Weight of CO 2 per pound of coal = 0-878 X ff = 3*219 H 2 O = 0-041 X 9 = '3 6 9 N 2 = i8x $0 = i3'9 6 Total = 17-548 pounds /. weight of oxygen (by difference) = 18-919 17*548 = 1*371 pounds Hence the products of combustion will consist of Carbon dioxide (CO 2 ) = --^^- = 0-170 = 17-0 per cent. 10-919 Steam (H 2 O) =^| = 0-019 = ri 9 Nitrogen (N 2 ) =^^= 0738 = 73'8o Oxygen (O 2 ) = = 0*072 = 7-20 143- Minimum Quantity of Air required for the Complete Combustion of i Cubic Foot of Gaseous Fuel The method of calculation will be best illustrated by means of an example. Taking a sample of coal gas whose volumetric analysis is hydrogen (H 2 ) = 46*0 per cent, marsh gas (CH 4 ) = 39-5 per cent., carbon monoxide (CO) = 7-5 per cent., nitrogen N( 2 ) = 0-5 per cent., water vapour (H 2 O) = 2*0 per cent., the method of procedure is as follows : Hydrogen 2H 2 + O 2 = 2H 2 O ..... , . (i) i.e. 2 cub. ft. of hydrogen -j- 1 cub. ft. of oxygen give 2 cub. ft. of water vapour ''46 -i-' 2 3 0-46 236 THE THEORY OF HEAT ENGINES [CHAP. xi. Marsh Gas CH 4 -f 2 O 2 = CO 2 + 2H 2 O ..... (2) i.e. i cub. ft. of marsh gas-|-2 cub. ft. of oxygen give i cub. ft. CO 2 -j-2 cub. ft. water vapour ' '395 ,. +o'79 0-395 cub. ft. CO 2 -f 0-79 cub. ft. water vapour Carbon Monoxide 2CO + O 2 =2CO 2 ....... (3) i.e. 2 cub. ft. of CO + i cub. ft. of oxygen give 2 cub. ft. of CO 2 >i +'375 . '075 Hence i cubic foot of coal gas requires 0*23 -f- 079 + '375 = I>0 575 cubic feet of oxygen for complete combustion. Now air contains 21 per cent, by volume of oxygen, hence i cubic foot of the gas will require r 575 X 1 r = 5-03 cub. ft. of air for its complete combustion. The above might be put in algebraic form as fellows : From (i) i cub. ft. of H requires 0-5 cub. ft. of O = 0-5 X \ = 2-38 cub. ft. of air From (2) i cub. ft. of CH 4 requires 2 cub. ft. of O = 2 X ^ = 9*52 cub. ft. of air From (3) i cub. ft. of CO requires 0-5 cub. ft. of O = 0-5 X^r = 2 '3 8 cub - ft - air Hence i cub. ft. of the gas containing these combustible constituents will require 2-38 H + 9-52 CH 4 + 2-38 CO cubic ft. Composition of the Products of Combustion. The products will consist of CO 2 o'395 + '75 = '47 cubic ft. H 2 O = 0-46 + 0794-0-02 = 1-27 N 2 = 0-005 + 5'3 X $5 = 3'98 Total = 5-72 Before combustion commenced the volume of the mixture of gas and air was i -f- 5*03 = 6*03 cubic feet, hence the contraction in volume after combustion is 6-03 5-72 = 0*31 cubit foot or ^~ X 100 = 5*1 per cent. Calculation of the Quantity of Air supplied per Cubic Foot of Gas from the analysis of the Exhaust Gases leaving a Gas Engine. The calculation requires the analysis of the gas and also that of the exhaust gases. In the example taken above the minimum quantity of air required per cubic foot of gas has been estimated as 5*03 cubic feet, the total volume of the products of combustion when the steam remains a vapour is 5-72 cubic feet. If, however, the steam had been condensed, the volume of the products (neglecting the very small volume of the resulting water) would be 5-72 1*27 = 4-45 cubic feet Now, when the exhaust gases are analysed the steam will be condensed, hence if x cubic feet be the amount of air supplied per cubic foot of gas in ART. 143] COMBUSTION 237 excess of 5 '03 cubic feet, the total volume of the actual products of combustion from i cubic foot of gas will be x + 4*45 cubic feet Let O 2 be the percentage of oxygen (by volume) present in the ex- haust gases, then the excess quantity of air which will contain this volume of oxygen will be O 2 X ^r cubic feet volume of air left in actual products of combustion _ x total volume of actual products of combustion ~ x -{- 4*45 Hence or #(21 O 2 ) = or expressing it in words excess air supplied (cubic feet) = volume of products when the minimum theoretical quantity of air is supplied (the steam formed being condensed) percentage of oxygen in exhaust gases 21 percentage of oxygen in exhaust gases EXAMPLE. With the above sample of coal gas the exhaust gas analysis gave 10 per cent, by volume of oxygen. Estimate the volume of air supplied per cubic foot of gas, and the contraction in volume after combustion. From (4) excess air x = - - = 4*045 cubic feet 21 10 =5'3 + 4-45 =9'75 cubic feet Volume of mixture before combustion = i + 9'75 = I0 '75 cubic feet Volume of mixture after combustion j = = g cubic feet assuming water vapour condensed j .-. contraction due to combustion= 10-075 8-495 = 1-580 cubic feet X ioo = 15-6 per cent. 10-075 In the engine cylinder the water vapour will not be condensed, and the actual volume of the products of combustion will be * + 572 =4*045 + 572 = 9765 cubic feet /.contraction in engine cylinder j = _ 6 = Q cubic foot due to combustion j = 3I x ioo = 3-07 per cent. 10-075 238 THE THEORY OF HEAT ENGINES [CHAP. xi. 144. Calculation of the Quantity of Air supplied per Pound of Fuel from the Analysis of the Flue Gases leaving a Boiler. Let C = percentage of carbon in the fuel by weight. CO 2 = percentage of carbon dioxide in the flue gases by weight. CO = percentage of carbon monoxide in the flue gases by weight. N = percentage of nitrogen in the flue gases, by weight. O = percentage of oxygen in the flue gases by weight. Considering 100 parts by weight of flue gases there are N parts of nitrogen in it. Now air contains 77 per cent, by weight of nitrogen. .'. N parts of nitrogen are supplied with ^ X N parts of air. From Art. 140, it will be seen that in the 100 parts of flue gases there are (CO 2 X Jf) + (CO X if) parts of carbon Hence the ratio of air to carbon is (C0 2 X If) + (CO X Jf) ' Now i pound of the fuel contains pounds of carbon. IOO Hence the weight of air supplied per pound of fuel burned is N X which reduces to If CO 2 , CO, and N represent the percentages of carbon dioxide, carbon monoxide, and nitrogen respectively in the flue gases by volume, then the weight of air supplied per pound of fuel burned is given by _ N X ffi X 28 _ C_ (C0 2 X J| X 44) + (CO X if X 28) X loo P ' where 28 = molecular weight of N 2 and also of CO and 44 = molecular weight of CO 2 The above expression will be found to reduce to ; ;. . __E__ XC pounds . ... . (3) Loss of Heat by the Flue Gases escaping up the Chimney. Let .y = mean specific heat of the flue gases (about 0-238), i.e. the mean heat required to raise unit weight of the mixed gases one degree in temperature, W = weight in pounds of the flue gases per pound of fuel burned, /i = temperature of the flue gases leaving the boilers (Fahrenheit), / 2 = temperature of the air in the boiler house (Fahrenheit), then assuming that there is no preliminary heating of the air supply the heat carried away in the flue gases per pound of fuel is W X j(/i / 2 ) British Thermal Units ART. 145] COMBUSTION 239 It may here be mentioned that the air supply should as far as possible be free from moisture, and further that it conduces to economy if it be heated, provided that such preheating is obtained at little cost. Any water vapour present in the air when it passes through the incandescent fuel will become dissociated into its constituent hydrogen and oxygen, and then will again re-combine when they leave the furnace ; but they will leave at the higher temperature of the chimney, and the net result is that some heat has been lost in raising the temperature of the water vapour, or expressed in other words, the mean specific heat of the flue gases is higher than if the air had been dry. 145. Calculation of the Mean Specific Heat of the Flue Gases leaving a Boiler. The calculation requires the analysis of the fuel and also the analysis of the flue gases. The method will be best illustrated by means of a numerical example. Taking the following analyses : Analysis of Fuel by Weight. C 87*30 per cent. H 078 Ash 8-27 Other matters 2*36 Analysis of Dry Flue Gases by Volume. CO 2 9*88 per cent. CO 0-05 2 9-82 N 2 80-25 The first step in the calculation is to convert the analysis of the dry flue gases by volume (the gases are always analysed by volume) into the analysis by weight. By multiplying each of the volume proportions by the corresponding molecular weight, adding the products so obtained, and then dividing each separate product by the sum of all the products, the proportion by weight of each gas present may be obtained. By ., Molecular. By Constituent, volume. X weight weight. CO 2 = 0-0988 X 44 = 4'348 = 0-145 = U*5 per cent. CO = 0*0005 X 28 = 0-014 = 0-0005 = 0*05 O 2 = 0-0982 X 3 2 = 3'i42 = 0-1055 = 10-55 N 2 =0-8025X28 = 22-470 = 0-749 = 74-90 Total . . 29*974 1*000 100*0 Therefore in 100 pounds of dry flue gases there are 14-5 xit+ o-o 5 XM = 3*95 + '02 = 3-97 pounds of carbon Hence the weight of dry flue gases per pound of dry fuel will be 100 X 6-0873 = 22-0 pounds 3*97 2 4 o THE THEORY OF HEAT ENGINES [CHAP. xi. But the 0-0078 pound of hydrogen in the fuel produces 0-0078 X 9 = 0*070 pound of steam, hence the actual weight of each constituent in the flue gases from i pound of dry fuel is Constituent. Weight in pounds per pound of dry fuel burned. Weight in pounds of each constituent in i pound of flue gases. (~*O 3-I90 _ CO 2 22-0 X 0-0005 = O'OII 22-0X0-1055= 2-32I 22-07 O'OII = o'ooo5 22-07 ^ = 0-1052 22-07 16-478 _ . 6 . 2 H/~ 22-07 0*07 ~~ O'OO'U 22-07 22-070 i -oooo The specific heats at constant pressure of the above gases may be taken as : CO 2 = 0-216 CO -0-245 O 2 = 0-218 N 2 = 0-244 H 2 = 0-480 Hence the mean specific heat of the flue gases is : (0-1445 X 0-216) + (0-0005 X 0-245) + (0-1052 X 0-218) + (07467 X 0*244) + (0-0031 X 0-48) = 0-238 146. Heat carried away by the Products of Combustion and Excess Air. In drawing up the Heat Balance for a boiler it is a common practice to divide the heat lost in the flue gases into two parts, viz. that carried away by the products of combustion, and that carried away by the excess air supplied. The method of doing this will be best illustrated by means of an example. EXAMPLE. The analysis of a certain oil fuel used in a boiler trial was C 86 per cent., H 14 per cent. The volumetric analysis of the flue gases was CO 2 9-4 per cent, CO i per cent., O 2 io'i per cent., N' 2 79-5 per cent. Estimate, per pound of fuel burned, the heat carried away by the products of combustion, and also by the excess air, the temperature of the flue gases being 600 F., and of the air supply 60 F. The total weight of flue gases per pound of fuel may first be found as follows : Using equation 3, Art. 144, we have N Air supplied per pound of fuel = T-^T ~ X C i = 19*92 pounds ART. 146] COMBUSTION 241 The total weight of gases = 19-92 + combustible in i pound of fuel =-19-92 + 1 20-92 pounds Minimum quantity of ahO theoretically required perX = ii'6C + 34*8H ((i), Art. 142) pound of fuel ) = n-6 X 0-86 + 34-8 X 0-14 = 14-84 pounds Excess air supplied = 19-92 14*84 = 5*08 pounds .-. weight of products of ^ combustion from i pound > =15-84 pounds = 20-92 5-08 of fuel j We now require the weight of each constituent in this 15*84 pounds of products of combustion. Converting the volumetric analysis of the gases into the analysis by weight, we have Con- By Molecular By stituent. volume. weight. weight. CO 2 = o- 094 X 44 = 4* 136 = O 138 = i3 8 per cent. CO = o- 010 X 28 = o- 280 = 009 = o 9 j) 2 = o* 101 X 32 = 3' 232 = O 108 = 10 8 > N 2 = o* 795 X 28 = 22* 260 = o 745 = 74 '5 ti Total . . 29-908 i-ooo loo-o We next require the proportion of carbon burned to CO 2 and CO respectively. In 13-8 parts of CO 2 there are 13-8 X ^ = 3*76 parts of carbon o'9 >, CO 0-9 X f =0*38 Total . . 4-14 7-76 .*. proportion of carbon burned to C0 2 =^~ 4-14 CO-^ 4 *4 Hence per pound of fuel burned Weight of CO 2 = X 0-86 X = 2-86 pounds 3 4'*4 Weight of CO = 7 - X 0-86 X ~ = 0-18 pound Weight of H 2 O formed = weight of hydrogen X 9 = 0-14 X 9 = 1*26 pounds Weight of N 2 (by difference) = 15-84 (2-86 + 0-18 + 1-26) = 11-54 pounds 242 THE THEORY OF HEAT ENGINES [CHAP. xi. We next require the proportion by weight of each constituent in the products of combustion. r-r, = = Total = 1*000 Hence, mean specific heat of the products of combustion = (0-181 X o-2i6)+(o-on X o'245)-j-(o-o79 X o'48)+(o'729 X 0-244) 0-2575 .-.heat carried away byproducts of combustion=i5'84X 0-2575 (60060) =2202 B.Th.U. Heat carried away by excess air =5*08 X 0-2375 (60060) =651 B.Th.U. 147. Calorific Value of Solid and Liquid Fuels. Several formulae have been given from time to time by means of which the calorific value of a fuel may be calculated from its chemical analysis. In Dulong's formula it is assumed that all the oxygen present in a fuel is already combined with one-eighth of its weight of hydrogen in the proportion to form water so that the hydrogen available for combustion is only H -g Using the same notation as in Art. 142, the lower calorific value according to this formula is 14,5400 + 52,200 ( H "3") + 4oS B.Th.U. per pound. or 8o8oC + 29,000 (H g J + 22208 C.H.U. in which the constants denote the calorific values of the various combustible elements. The above formula invariably gives a result too low as compared with the value obtained by direct calorimetric tests, probably because more hydrogen is free than is assumed. On the other hand, if all the hydrogen in a fuel is assumed available for combustion the calorific value obtained (when using the above constants) is usually too high. The above formula assumes that the amount of oxygen present in a fuel is known accurately. This is not so because it is always found by difference, being practically a fictitious figure which contains all the errors of the analysis. Also the same formula will not be equally correct for both solid and liquid fuels on account of the complex composition of the hydrocarbons in oil fuels, and in addition, it should be remembered that the calorific value of carbon depends upon its physical state, diamond, graphite, and amorphous carbon all being different. ART. 147] COMBUSTION 243 The Author found 1 that for many steam coals used in practice the formula 14,4000 + 52,200!! B.Th.U. per pound or 8 J oooC-f29,oooH C.H.U. gives results for the lower calorific value very closely in agreement with those obtained in the "Bomb" calorimeter, whilst for oil fuels the formula i3>5ooC + 52,2ooH B.Th.U. per pound or 7,5ooC + 2 9) oooH C.H.U. gives results practically the same as those given by the Bomb calorimeter. The calorific value of a solid or a liquid fuel can only be determined accurately by direct experiment in an approved calorimeter such as the Bomb calorimeter. The following tables give the calorific values of various solid and liquid fuels : COMPOSITION AND CALORIFIC VALUE OF VARIOUS SOLID FUELS. Description of fuel. Vfoisture, per cent. Carbon, per cent. Hydro- gen, per cent. Oxygen + nitro- gen, per cent. Sulphur, per cent. Higher calorific value. Calories, per gramme. B.Th.U. per pound. Wood (ordinary) . Wood (dried) .... Peat (poor) . 28-9 6*9 20-8 6-1 0-6 4-0 2 '20 I-I4 I'20 0'80 I'OO 7'30 3*40 4'80 36*4 47'4 40-8 53'2 78-4 75-42 84-10 84*34 81-30 81-50 83-80 87-80 67-90 91-50 86-40 88-40 4-6 5'6 3'3 5*5 6-18 4'93 S'SO 5'30 4-60 4-80 4-10 4-90 3-50 2-00 I-40 29*6 4 I-0 3''4 35'o 11-20 9-98 7*OO 6'oo 9-90 6"oo 5-10 5-00 16-00 3'40 2-20 3-27 0-4 2-18 o-55 0-78 I'20 1-20 I'4O I-30 0'60 o-35 3310 4480 3770 5490 7760 7500 7420 8300 8160 7970 8050 8580 722O 8460 7480 7600 5,960 8,060 6,790 9,880 I3>970 ^SS 13,370 14,870 14,690 14,35 14,490 15,450 13,000 15,220 i3,47o 13,600 Peat (air dried). Cannel coal (Wigan) . Cannel coal (Scotch) . . Yorkshire caking coal . Durham caking coal . Newcastle steam coal . Durham steam coal Welsh steam coal . Nixon's navigation steam) coal / Welsh anthracite . American ... Average English coke . 1 See Engineer ) Feb. 17, 1911. 244 THE THEORY OF HEAT ENGINES [CHAP. xi. COMPOSITION AND CALORIFIC VALUE OF VARIOUS LIQUID FUELS. l Higher calorific value. Description of fuel. Specific gravity. Carbon, per cent. Hydro- gen, per cent. Oxygen + nitro- gen, per cent. Sulphur, per cent. Calories, per B.Th.U. per gramme. pound. Russolene (H.Y.O.) . . O'SgO 85*95 I3'5 O 10,900 19,600 American kerosene . 0-780 85-05 14-40 11,160 2O,IOO American royal daylight . 0-797 85-70 I4-20 11,170 20,100 American petrol 80-58 I5-IO 4-31 O 1 1, 080 *995 Refined Russian petroleum \ (Baku) / 0-825 86-00 14*00 11,270 20,290 American crude petroleum Russian crude Caucasian \ (Novorossisk) . . . / I 86-90 84-90 I3-IO 11*63 1-46 10,910 io,330 19,650 18,600 Baku heavy oil . 86-70 12-94 10,800 19,450 Naphtha Russian crude .... 0-87I 75-57 86-90 10-57 I3-IO 3'9i O 9,250 10,830 16,650 19,500 Java crude 0*867 87*10 I2*7O o 10 650 IQ IQO Canadian crude 0-859 *-V A ^ 86-92 j.^ y \j I2-87 0-35 10,800 *y, *y^ j 19,450 Texas crude .... 0-947 86-62 II 80 0-63 10,520 18,960 Solar oil . O-8q6 86 -61 12 "6O o* to 10 780 10 4^O Coal oil . . . M ^y^ 0-9I7 83-20 II-87 u ju I-56 10,220 .7 9 TO 18,410 EXAMPLE i. In a boiler trial the fuel analysis, dry coal as burned was C 83 per cent., H 4 per cent., Q 8 per cent., ash, etc., 5 per cent. ; the volumetric analysis of the flue gases was CO 2 10 per cent, CO 1-7 per cent., O 8*1 per cent, N 80*2 per cent. The temperature of the flue gases was 600 F. and of the boiler house 80 F. Find (1) The proportion of C burned to CO, and the heat lost through imperfect combustion, expressing the latter as a percentage of the heat in the fuel. (2) The heat carried away in the flue gases per pound of fuel burned, the average specific heat being taken as 0*24. In the example worked out in Art. 146, the proportion of C burned to CO was obtained from the analysis of the gases by weight. The proportion may be obtained directly from the volumetric analysis as follows : (i) C in 10 parts by volume of CO 2 =ioXj|X molecular weight of CO 2 = ioX;|f X 44 = 1 20 parts Cm i -7 CO= 1 7 X||X molecular weight of CO i '7 X || X 28 = 20-4 parts Total carbon in gases = 120 -J- 20-4 = 140-4 parts Hence proportion of C burned to CO = = 0-1453 or 14*53 per cent. 140-4 and the heat lost through imperfect combustion per pound of coal = 0-83 X 0-1453(14,540 4400) = 1223 B.Th.U. 1 From a paper by the author on " The Calorific Values of Solid and Liquid Fuels," The Engineer, Feb. 17, 1911. ART. 147] COMBUSTION 245 ] = X 0*83 -f~ 52,200(0*04 0*01) = 12,068 = 13,634 B.Th.U. per pound. .-. proportion lost through imperfect) 1223 g g t combustion 5 13,634 X 83 (2) Air supplied per pound of fuel by (3),) Art. 144 ) ~ 80-2 = 17*25 pounds .. weight of flue gases per pound of fuel = 17*25 -j- 0-95 = 18*2 pounds Heat carried away in flue gases) ig . a (6oo _ } per pound of fuel burned ) = 18*2 X 0-24 X 520 = 2271 B.Th.U. EXAMPLE 2. In a trial of a Lancashire boiler with economiser the following results were obtained : Volumetric analyses of the flue gases on entering and leaving the economiser Entering. Leaving. CO, . . . CO ... o . . . . N . . . . 8 "3 per cent. 0*4 II-2 80- 1 6"2 per cent. 0'3 ,i 137 ,, 79'8 Total . . lOO'O lOO'O Temperatures of the flue gases on entering and leaving the economiser, 642 F. and 335 F. Temperatures of feed water on entering and leaving the economiser, 1 34 F. and 234F. Weight of feed water per hour, 7370 pounds. Weight of coal stoked per hour, 1000 pounds. Per pound of dry fuel stoked the carbon burned was 0735 pound, and the weight of the flue gases, including moisture, entering the economiser was found to be 22*5 pounds. The average specific heat of the gases may be taken as 0*25. Calculate, per pound of fuel stoked (a) The air leakage into the economiser. (b) The heat lost by the gases in passing through the economiser. (c) The heat gained by the feed water. (L.U.) (a) Total air per pound of fuel,) 79*8 by (3), Art. 144 3 = ^(6*2 + Total weight of gases leaving) economiser per pound of fuel] = 2 7'34 + i = 28*34 pounds X 73'5 = 2 7'34 pounds .*. air leakage into economiser = 28-34 22*5 = 5-84 pounds 246 THE THEORY OF HEAT ENGINES [CHAP. xi. (b) Heat lost by 22*5 pounds of) ,, gases j= 22-5X0-25(642 -335) =-1727 B.Th.U. (c) Feed water per pound of fuel = Jf = 7-37 pounds .-. heat gained by the water = 7*37(234 134) == 737 B.Th.U. N.B. The amount of air that leaks into the economiser is, in its passage through the economiser, raised in temperature from the tempera- ture of the boiler house to that of the gases leaving the economiser. The quantity of heat used for this purpose is of course wasted, and passes away in the flue gases, the efficiency of the economiser being thereby reduced. In order to prevent air leakage through the brickwork, the walls enclosing the economiser should be built with hard-pressed bricks and cement. Glazed bricks are better still, and their extra cost would soon be recovered by the economy in fuel resulting from their use. If ordinary brickwork is used, it should be coated on the outside with pitch or other viscous substance to reduce porosity. The practice of encasing econo- misers with wrought-iron panelling, having 2 inches of asbestos yarn between the inner and outer plates, is sometimes followed ; this, however, is an expensive remedy, but the heat saved, together with the greater facilities for access and -repair, provides a reasonable return for the extra outlay. 1 148. Boiler Draught. It is necessary to maintain a difference of pressure above and below the firegrate in order to supply the quantity of air required for combustion. This difference of pressure is known as the draught, and may be produced either by means of (a) A chimney (natural draught) ; (b) Steam jets (induced or forced draught) ; (c) Fans, which may either draw the gases from the flues (induced draught) or blow air under pressure into the ash-pit, or a closed boiler- room from which the air supply is drawn (forced draught). The theoretical velocity of the gases produced by a draught or difference of pressure above and below the firegrate is given by the equation V* = 2gl where / is the height of a column of air, measured in feet, corresponding to the draught pressure. Let h be the draught in inches of water. Then, since a head of i inch of water is equivalent to a pressure of 5*198 pounds per square foot, and one cubic foot of air at N.T.P. weighs 0-0807 pound /== 5-198* 0*0807 and v 2 = 2 X 32-2 X 0-0807 # 2 = 4148^ z/ = V4 1 48/2 feet per second . . . , . (i) 1 See the Author's ** Steam Boilers." Edward Arnold. ART. 148] COMBUSTION 247 The actual velocity of the gases will be less than that given by (i) because of the frictional resistance offered to their passage along the flues. Height of Chimney required to produce a Given Draught Let h = required draught in inches of water, H height of chimney above the firegrate in feet, Tj = absolute temperature inside the chimney (assumed constant), T 2 = absolute temperature outside the chimney n = number of pounds of air supplied per pound of fuel burned, A = cross-sectional area of chimney in square feet, then the difference between the weight of a column of external air equiva- lent in volume to the interior of the chimney, and the weight of the same volume of the gases which are in the chimney, is equal to the draught pressure, i.e. Weight of A X H cubic feet of external air weight of A X H cubic feet of gases in chimney = 5-198^ X A /AX HX 0-0807 vv -- --* X X -jj- x H X 0-0807 X 492(4- -^ LL '^-)= 5- VI g n 1 !/ 5-198^ _^1 T 2_ 0-0807 X 492 T! - (n + i)f 2 _ (-(- i)T 2 (2) In the above theory the variation in the density due to the slightly reduced pressure inside the chimney is neglected, and if in addition we neglect the increased density due to the products of combustion, and assume the contents of the chimney to consist of air at atmospheric pressure (3) and j = (^_ _J ( 4) In actual practice the effect of the frictional resistance offered to the passage of the air through the firebars, fire, flues, and chimney is to reduce the draught h below the value obtained from' equation (4) ; also, the temperature of the gases inside the chimney is diminishing for every foot of its height. 1 In other words, the height of chimney required to produce a given draught is greater than that given by equation (3). Induced and Forced Draught. With a plant properly designed for mechanical draught, greater economy in working may be obtained because complete combustion is possible with a lesser amount of air supplied per pound of fuel ; and further, the gases can be cooled down to a much lower temperature before turning them off up the chimney than if chimney draught is used, and in addition the draught is under better 1 For a more detailed account of practice see the Author's "Steam Boilers." 248 THE THEORY OF HEAT ENGINES [CHAP. xi. control. The B.H.P. of the engine driving a fan may be estimated as follows : Let V = volume of air or gases, in cubic feet, passing through the fan per minute, h = draught pressure in inches of water, 7j = efficiency of the fan, n = number of pounds of air supplied per pound of fuel burned, w = weight of fuel burned per second in pounds, then 33,000 X With forced dratight V/ = 6onwV X 492 where V = volume of i pound of air at N.T.P. (12-39 cu b- f eet )> T/ = absolute temperature of the air delivered by the fan. With induced draught there will be approximately (n -f- i)w pounds of flue gases delivered per second, and assuming the density to be the same as that of air H.P. of induced draught fan _ 6o(n -f i)/V . Tj _ n -f i T* ,,. ' H.P. of forced draught fan = 6onwV . T/ n X T/ U If air leakage through the brickwork be taken into account the ratio given by (6) will be further increased. The amount of leakage into the flues with induced draught will depend upon the intensity of the draught, and also upon the means taken to prevent it. With ordinary brickwork and no special precautions, the above ratio must be increased by about 20 per cent, with the result that the H.P. required for the induced draught fan will not be far from double that for the forced draught fan. The advantages of using a mechanically produced draught may be briefly summed up as follows : Increased evaporation per square foot of heating surface ; higher furnace temperatures and consequently increased efficiency ; small chimneys may be used ; cheaper coal can be economi- cally burned ; the easier regulation of the output of steam and the ability to meet sudden demands such as exist in electricity supply stations, rolling mills, warships, etc. no dependence need be placed on the weather. The chief disadvantage of the system lies in the heavy upkeep and repairs that must be expected owing to the increased duty which the boilers are called upon to perform. EXAMPLE. Compare the fan powers expended for induced and forced draught. Compare also the quantity of heat carried away by the flue gases per pound of fuel burned in the following case : Temp, of flue gases leaving boiler, F. Air supplied per pound of fuel (pound.s). Temp, of air in boiler-house, F. Induced draught . Forced draught Chimney draught . 350 350 600 18 18 24 62 62 62 ART. 149] COMBUSTION 249 Now temp, of air delivered by forced draught fan T/=62-j-46o=522abs. and ,, ,, induced T;=35o-|-46o=8io abs. - hv (6\ H ' P< for induced draught_ 18 + i 810 _ y * ' H.P. for forced draught 18 * 522 Allowing 20 per cent, increase for leakage into the flues, this becomes 1-64 X i '2 = i '968, i.e. approximately 2 Allowing 15 per cent, for leakage with chimney draught, the losses due to the heat carried away with the flue gases per pound of fuel are with chimney draught = 1*15 X 0*238 X 25 X (600 62) = 3700 B.Th.U. with induced draught = 1-2 X 0*238 X 19 X (350 62) = 1560 B.Th.U. with forced draught = 0*238 X 19 X (350 62) = 1300 B.Th.U. N.B. The actual saving obtained in practice would be less than that indicated by the above figures. The cost of the power expended in driving the fan would have to be taken into account. 149. Calorific Value of Gaseous Fuels. In calculating the calorific value of a gas, the method followed is similar to that given in Art. 147 for solid and liquid fuels. The calorific value per cubic foot is assumed to be the sum of the quantities of heat evolved by the com- bustion of each constituent gas. The following example will illustrate the method. EXAMPLE. 'Estimate the calorific value per cubic foot at N.T.P. of a sample of coal gas whose volumetric analysis is : H = 46*0 per cent., CH 4 = 39-5 per cent., CO = 7*5 per cent., N = 0*5 per cent., H 2 O = 2 per cent. The weight of C and H in i cubic foot might be calculated and then the calorific value estimated as in Art. 147 ; or the calorific value of CH 4 and CO might be found, from which that of the gas can be estimated. Adopting the latter method we have Since the density of any gaseous compound is proportional to half its molecular weight, and the density of hydrogen at N.T.P is 0*00559 pound per cubic foot, the weight of i cubic foot of any gas is equal to half its molecular weight X 0-00559 pound. Hence the density of CH 4 = - - X 0*00559 0*04472 pound per cubic foot and CO = - X 0*00559 = 0*07826 pound per cubic foot Lower calorific value of H = 0*00559 X 52,200 = 282 B.Th.U. per cub. ft. at N.T.P. Weight of C in i cubic foot of CH 4 = 0*04472 X yi '3354 pound i) H . = 0*04472 X i 4 6 = 0*01 18 pound 250 THE THEORY OF HEAT ENGINES [CHAP. xi. .-. heat evolved during combus-) tionofCH 4 j =0-03354X14,540+0-01118x52,200 = 487-6 + 583-6 = 1071 B.Th.U. per cub. ft. at N.T.P. But the heat of formation of CH 4 is 17,100 calories per gram molecule, or in B.Th.U. per cubic foot. This amount will therefore be absorbed when burning CH 4 in decomposing it into C and H. Hence the lower calorific value ofCH 4 v = 1071 in = 960 B.Th.U. per cubic foot at N.T.P. Weight of C in i cubic foot of CO = 0-07826 X Jf = 0*03354 pound .-. calorific value of CO = 0-03354 X 10,140 = 340 B.Th.U. per cubic foot at N.T.P. Hence lower calorific value of the sample of coal gas = 0-46 X 282 + 0-395 X 960 + 0*075 X 340 = 129-7 + 389-2 + 25-5 = 544 B.Th.U. per cubic foot at N.T.P. The calorific values as given by Professor F. W. Burstall of a few different combustible gases are given in the following table 1 : Gas. Calorific value in B.Th.U. per cubic foot. Higher value. Lower value. Carbon monoxide CO . Hydrogen . . H 2 . Methane . . . CH 4 . Olefiant gas . . C 2 H 4 . 338 344 1050 1670 338 295'5 952-9 Tetrylene . . C 4 H 8 . 3060 EXAMPLE. Using the above calorific values, estimate the calorific value of a producer gas of the following composition by volume : CO 2 7*66 per cent., CO 22-27 per cent, H 2 20*19 P er cent., CH 4 2778 per cent., O 2 0-04 per cent., N 2 47 '06 per cent. The tabular method is usually the most convenient for calculating the calorific value of a gas from its chemical analysis. Following the method explained above the results are as follows : 1 From the Third Report to the Gas Engine Research Committee, Proc. I. Mech. E., 1908, p. 26. ART. 150] COMBUSTION 25 1 Per- Heating value per cubic foot of each constituent. Heating value per cubic foot of mixture. (ja?. of gas. B.Th.U. B.Th.U. B.Th.U. B.Th.U. Higher Lower Higher Lower value. value. value. value. Carbon dioxide CO 2 7-66 _ _ _ _ Carbon monoxide CO 22-27 338 338 75-28 75^8 Hydrogen . . H 2 Methane . . . CH 4 20-19 2778 344 1,050 295*5 952-9 69-45 29-17 59-67 26-47 Oxygen . . . O 2 0-04 Nitrogen N 2 47-06 " ' " Totals .... 1 73 '9 l6l'4 The actual values as obtained by Junker's calorimeter were Higher value, 175*6 B.Th.U. per cubic foot. Lower value, 162-0 B.Th.U. per cubic foot. 150. Producer Gas, Theory. Producer gas is made by forcing or drawing air (with or without the addition of steam) through a deep bed of fuel in a closed producer. In a suction gas-producer the air is drawn through by the suction of the engine piston, and in all cases, when once the burning of the fuel inside the producer has been started, the partial combustion is maintained by the air supply, and a sufficiently high temperature kept up to decompose the steam and to effect other reactions. If there were a thin fire in the producer the carbon and other com- bustible elements of fuel would be completely oxidized and the products would be incombustible, and therefore useless as a fuel. In practice a considerable depth of fuel is used (several feet), and there will be an excess of incandescent carbon which will reduce the CO 2 formed at the bottom of the fire where the air enters to CO according to the equation CO 2 +C = 2CO . (i) Carbon monoxide may also be produced by the direct combination of carbon with oxygen, thus 2 C + 2 = 2 CO (2) Both these reactions may and probably do occur in an actual producer, and if gas is made by simply passing air through a thick fire the final result aimed at is to have the carbon oxidised to CO only. Each pound of carbon so burned gives out (as we have already seen) 4400 B.Th.U., the remaining 14,5404400 or 10,140 B.Th.U. being available in the carbon contained in the CO present in the producer gas. The 10,140 B.Th.U. is practically 70 per cent, of the heat in the carbon to start with, the remain- ing 30 per cent. (4400 B.Th.U.) being expended in keeping the fire alight and maintaining a high temperature ; this means that if there are no heat losses the efficiency of the producer will be 70 per cent. The 30 per cent, of the heat in the fuel need not, however, be entirely lost. A certain amount (say 8 per cent.) will be lost through radiation and conduction, but a large proportion will be carried away with the gas as 252 THE THEORY OF HEAT ENGINES [CHAP. xi. sensible heat, since the gas will leave the producer at a high temperature ; and if the gas can be used while it is hot not much of this sensible heat will be lost. The high temperature obtained in the producer with this sensible heat may be excessive and cause trouble in working by the forma- tion of clinker which will obstruct the air passages. To avoid the produc- tion of too high a temperature and also to increase the efficiency of the producer (i.e. to reduce the 30 per cent, loss) some of the sensible heat may be used to generate steam to be passed in the producer with the air. This steam may react on the carbon in the following ways : C + H 2 = CO + H 2 (3) C + 2H 2 = C0 2 + 2 H 2 (4) Both these reactions cause a large absorption of heat, the former (3) to the extent of 4300 B.Th.U. per pound of. carbon, and the latter to that of 2820 B.Th.U., so that each of them has the practical advantage of reducing the temperature of the producer, and in addition they increase the calorific value of the gas by the addition of hydrogen. At temperatures above 1832 F. (1000 C.) reaction (3) is more likely to occur; but at a temperature of 1112 F. (600 C.)and under, reaction (4) takes place ; while at temperatures between 1112 F. and 1832 F. the two take place simultaneously, the predominance of either being entirely a function of the temperature. 1 It is evident that since equation (3) gives - the richer gas and the greater absorption of heat, the best results will be obtained in practice when working at the highest temperature consistent with practical conditions. The proportion of steam which should be used to obtain gas of the highest calorific value can be determined theoretically as below, but this in practice is largely affected by the nature and composition of the fuel used and the size of the producer. It varies from 0-5 pound for large producers to 0-7 pound for small producers per pound of coal gasified. 2 The best thickness of fuel depends upon the size of the pieces, but for ordinary anthracite as used in producers, from 2 to 3 feet seems to give the best results according to Mr. Allcut, Mr. Dowson, and Dr. Bone and Dr. Wheeler.3 Theoretical Amount of Steam required per Pound of Fuel to give Maximum Efficiency. Assume that the reactions given in (2) and (3) are followed, 2 C + 2 = 2CO (2) 2 4+ 3 2 = 5 6 C + H 2 = CO + H 2 (3) 12+ l8 = 28-f-2 From (2), 24 pounds of carbon liberate 24 x 4400 = 105,600 B.Th.U. From (3), 12 pounds of carbon are required to dissociate 18 pounds of steam, the reaction absorbing 4300 X 12 = 51,600 B.Th.U. Now the H 2 O is supplied to the producer as water, not as steam, and assuming the temperature of supply to be 62 F., the amount of heat 1 See " Producer Gas," by Dowson and Larter, p. 55. 2 See the following papers : Proceedings Inst. Mech. Eng., 1911, "Gas Producers" by J. Emerson Dowson, and " Effect of Varying Proportions of Air and Steam on a Gas Producer," by E. A. Allcut. 3 Journal Iron and Steel Institute, No. III., 1908, p. 206. ART. 150] COMBUSTION 253 required to generate i pound of steam at atmospheric pressure from water at 62 F. will be (212 62) + 970 = 1120 B.Th.U. Now 1 8 pounds of steam are dissociated by 12 pounds of carbon, hence this reaction will absorb 51,600 + 18 X 1120 = 71,760 B.Th.U. /. amount of water required = 18 X - ' 6o = 26-48 pounds The amount of carbon required for this weight of H 2 O will be 24 pounds, which follow reaction (2), and 12 X g = 17*65 pounds, which follow reaction (3), giving a total of 41-65 pounds. .-. weight of water per pound of carbon = ^- = 0-63 pound If the reaction (4) is followed, C+2H 2 = C0 2 +2H 2 (4) 12 + 36 = 44-1-4 12 pounds of carbon are required to dissociate 36 pounds of steam the reaction absorbing 4300 X 12 = 51,600 B.Th.U.; adding to this the heat of evaporation of 36 pounds of steam from water at 62 F., the total amount absorbed will be 51,600 + 36 X 1120 = 91,920 B.Th.U. 10^,600 /. amount of water required = 36 X = 41*35 pounds The corresponding amount of carbon will be 24 + 12 X ^p = 3778 pounds .-. weight of water per pound of carbon = ^5 = 1*09 pounds O / / Analysis of the Producer Gas. Assuming reactions (2) and (3), 24 pounds of carbon follow (2) and 17*65 pounds follow (3), hence, From (2) 24 pounds of C yield 56 pounds of CO. Now the weight of i cubic foot of CO = 14 X 0*00559 = 0-07826 pounds (Art. 149). .-. 24 pounds of C yield QtQ 5 6 = 715 cubic feet of CO at N.T.P. From (3), 17*65 pounds of C yield 28 X ^ ^pounds of CO = I fx X o-o 7 7 8a6 = 5*6 cubic feet of CO at N.T.P. From (3) 17*65 pounds of C yield 2 X ^ X o * ==526 cubic feet of CO at N.T.P. From (2) the total amount of O supplied = 32 pounds = 357 cubic feet at N.T.P. 16 X 0-00559 254 THE THEORY OF HEAT ENGINES [CHAP. XL and since air contains 79 per cent, by volume of nitrogen, the amount of N supplied in the air with 357 cubic feet of O will be 357 x Jf = J 343 cubic feet at N.T.P. The total will therefore consist of CO = 715 + 526 = 1241 cubic feet = 39-9 per cent. H 2 = S 2 ^ = 16-9 Total = 31 10 loo-o The above analysis might be worked out more conveniently in C.G.S. system of units. Since the molecular weight of any gas occupies 22-25 litres (see Introduction, p. xii), From (2), 24 grams of C yield 2 X 22-25 44'5 litres of CO at N.T.P. From (3), 17-65 grams of C yield 22-25 X ^ =32-7 litres of CO at N.T.P. From(3), 17-65 grams of C yield 22-25 X ^ J 5 = 32'7 litres of H 2 at N.T.P. From (2), O supplied = 22-25 litres .-. N 2 supplied = 22-25 + I? = 83-7 litres of N 2 at N.T.P. The total therefore will consist of CO = 44-5 + 32-7 = 77-2 litres = 39-9 per cent. H 2= 32*7 ii = 16-9 N 2= 837 = 43-2 Total= 193-6 ,, = xoo'o ,, The lower calorific value of the gas per cubic foot at N.T.P. will therefore be CO = 0-399 X 340 = i35' 6 B.Th.U. H 2 = 0-169 X 282 = 47-6 Total = 183-2 B.Th.U. per cubic foot. EXAMPLES XI 1. The analysis by weight of a certain coal is C 80 per cent., H 2 5 per cent., S 0*5 per cent. : estimate the theoretical quantity of air required for the complete combustion of I pound of the coal. If 20 pounds of air are supplied per pound of coal and the combustion is complete, estimate the analysis of the flue gases by weight. 2. A producer gas has the following analysis by volume : H 2 18*73 P er cent., CO 25-07 per cent., CO 2 5-2 per cent., N 2 51 per cent. Estimate the minimum quantity of air required for the complete combustion of I cubic foot of the gas, the percentage contraction in volume after combustion, and the composition of the products of combustion. 3. The analysis (by weight) of the fuel used in a boiler trial was C 88 per cent., H 2 3'6 per cent., O 2 4'8 per cent., ash 3 '6 per cent., and the volumetric analysis of the dry flue gases was CO 2 10-9 per cent., CO 1*0 per cent., O 2 7-1 per cent., N 2 81 per cent. Estimate the mean specific heat of the flue gases, and the quantity of heat carried away by the flue gases per pound of fuel burned if the temperature of the gases is 550 F., and of the air in boiler house 50 F. 4. The flue gas analysis by volume in a boiler trial was CO 2 10-5 per cent., CO I per cent., O 2 8 per cent., N 2 8o'5 per cent., and the coal analysis as burned was C 82 per cent., H 2 4-2 per cent., O 2 4*8 per cent., other matters 9 per cent. Calculate the following ART. 150] COMBUSTION 255 items in the heat balance per pound of coal, the temperature of the flue gases being 600 F . and the temperature of the air supply 60 F : (a) Heat carried away by products of combustion, average specific heat 0*24. (t>) Heat carried away by excess air, average specific heat 0*238. (c) Heat lost by incomplete combustion. 5. In a boiler trial the fuel analysis, dry coal as burned, was C 85 per cent., H 2 4 per cent., O 2 7 per cent., ash, etc., 4 per cent., and the flue gas analysis by weight was CO 2 ii percent., CO 1-5 per cent., O 2 7't per cent., N 2 80-4 per cent. The temperature of the flue gases leaving the boiler was 600 F., and the boiler house temperature was 70 F. Estimate per pound of coal (a) The proportion of carbon burned to CO and the heat lost through this imperfect combustion, expressing the latter as a percentage of the available heat in the fuel. (b) The heat carried away in the flue gases per pound of coal burned, the mean specific heat of the flue gases being taken as 0*24. 6. In a trial on a Babcock and Wilcox boiler fitted with an economiser the following volumetric analyses of the gases entering and leaving the economiser were made : Leaving Entering CO 2 ... 7 '9 per cent. 8*3 per cent. CO ... nil nil ,, 2 ... 11-5 11-4 N 2 . . . 80-6 80-3 The temperatures of the flue gases on entering and leaving the economiser were 350 C. and 1 80 C. respectively. Temperatures of water on entering and leaving economiser were 15 C. and 115 C. Weight of feed water per hour 10,000 pounds, weight of coal used per hour 1060 pounds. Carbon in I pound of coal 0*8 pound. Assuming the average specific heat of the gases to be 0*25, estimate per pound of coal burned (a) The air leakage into the economiser. (b) The heat lost by the gases in passing through the economiser. (c) The heat gained by the feed water. 7. Estimate the minimum height of chimney required to produce a draft of | inch of water if 24 pounds of air are supplied per pound of fuel burned, the mean temperature of the gases inside the chimney being 600 F. and the temperature of the external air 80 F. 8. Estimate the B.H.P. of the engine required to drive a fan which maintains a draught of 2 inches of water under the following conditions for (a) Induced draught fan ; (b) Forced draught fan. Temperature of flue gases leaving the boiler in each case 400 F., and of the air in the boiler house 70 F. ; air supplied per pound of fuel in each case 1 8 pounds ; weight of coal burned per hour in each case 2000 pounds. Neglect the effect of air leakage through the brickwork and assume the efficiency of the fan to be 80 per cent. 9. A sample of coal gas has the following analysis by volume : H 2 46 per cent., marsh gas CH 4 39*5 per cent., olefiant gas C 2 H 4 2*53 per cent., tetrylene C 4 H 8 1^27 per cent., CO 7'5 per cent., N 2 0*5 per cent., water vapour H 2 O 2 per cent. Calculate (a) The volume of air required for the complete combustion of I cubic foot of the gas. (b} The higher calorific value in B.Th.U. per cubic foot. (c) The lower calorific value in B.Th.U. per cubic foot. Assume the calorific values of the above constituents the same as given in the table on page 250. 10. The gas used in a gas engine test was tested in a Junker calorimeter and the following results were obtained : Gas burned in calorimeter 2*13 cubic feet Pressure of gas supplied 2' i inches of water Barometer 29-92 inches of mercury Temperature of gas 53 F. (ii'7 C.) Weight of water heated by gas 50-3 pounds Temperature of water at inlet 47'6 F. (87 C.) Temperature of water at outlet 72-4 F. (22'4C.) Steam condensed during test 0*116 pound 256 THE THEORY OF HEAT ENGINES [CHAP. xi. Determine the higher and lower calorific values per cubic foot at the temperature of 60 F. (15*6 C.) and barometric pressure of 30 inches of mercury. [Specific gravity of mercury = 13*6.] 11. A sample of oil used in an oil engine trial was tested in a Mahler-Cook bomb calorimeter and the following results were obtained : Weight of oil taken = I '090 grams. Total weight of water, including water equivalent of calorimeter, 2800 grams. Corrected rise of temperature of the water = 4-26 C. Determine the higher calorific value of the oil. 12. Find the maximum efficiency of a suction gas producer, the composition of the gas produced, and the calorific value per cubic foot, assuming that the fuel is carbon and that only air is passed through the fuel ; given that one pound of hydrogen occupies 178-8 cubic feet ; that the calorific value of carbon monoxide is 342-4 B.Th.U. (190*2 C.H.U.) per cubic foot; and that the calorific value of I pound of carbon is 14,544 B.Th.U. (8080 C.H.U.). What is the effect of admitting steam in addition to the air (a) on the working ; (b) on the efficiency of the producer? (L.U.) 13. The volumetric analysis of a producer gas supplied to an engine is CO 2 7*66 per cent., CO 22-27 per cent., H 2 20-19 per cent., CH 4 2-778 per cent., N 2 47'i per cent. The exhaust gases contained 10 per cent, of oxygen by volume. Estimate the quantity of air actually supplied per cubic foot of gas and the contraction in volume in the engine cylinder due to combustion. CHAPTER XII HEAT TRANSMISSION 151. Transmission through Flat Plates. When the opposite sides of a flat plate of infinitely large area are maintained at two different constant temperatures the amount of heat conducted through the plate is H = k X *~ 2 B.Th.U. per square foot per hour . (i) x where ^ and / 2 denote the temperatures (F.) of the two sides of the plate, x the thickness of the plate in inches, and k the thermal conductivity of the plate, i.e. the number of B.Th.U. passing per hour through one square foot of a plate i inch in thickness when the sides are kept at a constant difference of temperature of i F. The value of this constant for wrought iron and mild steel is 450. Assuming the temperature of the gas side of a steel plate 0*5 inch thick to be, say, 1500 F. and that of the water side 400 F., the above formula will give 400 = 990,000 B.Th.U. per square foot per hour Applying this to a boiler heating surface, the equivalent evaporation from and at 212 F. will be 990,000 , 66 =1020 pounds per square foot per hour. Now in actual practice an average evaporation of only about 5 pounds is obtained, corresponding to a transmission of only about 5000 B.Th.U. per square foot per hour, which would be obtained if the difference in temperature of the two sides of the plate was about 5*5 F. The difficulty attending the use of this simple formula lies in our uncertainty as to the temperatures on the two sides of the plate, the temperature head across the plate being very much less than the difference between the temperatures of the hot gases and the water. From the results of numerous experiments Rankine gave the following empirical formula for the heat transmitted per square foot per hour : H = ^~ /2 ^ B.Th.U. per square foot per hour . . (2) where a is a constant varying from 160 to 200 depending upon the type of boiler, 4 and / 2 the temperatures of the gases and water respectively. 257 s 258 THE THEORY OF HEAT ENGINES [CHAP. xn. Taking a = 200, this formula applied to the above case gives TT (i5 ~ 200 = 6050 B.Th.U. per square foot per hour If now (i) be applied to this result it is evident that 6o5o=%^> or The temperature of the water side of the plate will not be very much above that of the water ; suppose it to be 430 F., then the temperature of the gas side of the plate will be about 437 F., say 440 F., according to the above result, and the drop in temperature between the hot gases and the plate will be 1500 400 = 1060 F. The results of experiments by Mr. Hudson * and Sir John Durston 2 confirm this conclusion and show that the actual temperature head across a boiler heating surface is but a very small proportion of the difference of temperature between the hot gases and the water. 152. Efficiency of Heating Surface. Consider the case in which the gases pass along inside a cylindrical tube of diameter d and length / which is surrounded by water. Let T x = absolute temperature of the gases entering the tube. T 2 = leaving 0= water. w = weight of gases passing along the tube per second. s = specific heat of the gases. Further, let T be the absolute temperature of the gases at a distance x from the inlet end of the tube (Fig. 122) assumed constant over a small length I e Water / v >- Gas Oullel T, k 1 ^ > Gas Inlet X H. \ FIG. 122. 8x of the tube, then, if Q denotes the amount of heat transmitted through the tube per square foot per second per degree of difference in temperature on the two sides, we have, assuming the heat transmitted to be proportional to the temperature difference on the two sides of the plate, ws8T == Q(T 0)77^8* (i) where ST = fall in temperature of the gases in moving 8x along the tube. 1 Engineer, vol. 70, p. 523, 1890. 2 Trans. Inst. Naval Arch., vol. 34, p. 130, 1898. ART. 152] HEAT TRANSMISSION 259 8T TF /i -- dx T 6 ws Integrating (2) we have ^ ' (3) T 2 _E*.J TV^0 = * " ....... (4) Available heat entering the tube per second = ws(T^ 9) leaving =ws(T 2 0) i - T 2 ) v efficiency = ttv (I 1 ! 6) T t To T 2 (4) in (6) gives efficiency = i e~ ~ws ' l (7) From (7) we see that for a given length of tube and a given weight of gases passing per second the efficiency decreases as the diameter of the tube decreases. Equation (7) may be expressed in terms of the velocity of the gases as follows : Let v = speed of the gases, p = density of the gases, ltd* then w = v X Xp 4 ' and . / = -- . / ws svdp hence efficiency = i e~sdp (8) For a given diameter and length of tube it is evident from (8) that the efficiency will decrease as -the velocity v increases. This is also evident from (7), since by decreasing the diameter the velocity will increase for a given weight of gases passing through the tube, This conclusion, how- ever, is not confirmed by experiment. On the contrary, the heat trans- mission is found to increase as the velocity increases, and small tubes are more efficient than large ones (see Arts. 156 and 158). 260 THE THEORY OF HEAT ENGINES [CHAP. xn. Case when the Heat transmitted is Proportional to the Square of the Difference of Temperature. In this case Let T! = difference of temperature between gases and water at entrance to the tube, T 2 = difference of temperature between gases and water leaving the tube, T = difference of temperature between the gases and water at a point x from the inlet end of the tube. We have wsST = QT2 . irdSx ....... (9) ws T 2 T! ws Writing the internal area of the tube irdl= A = -^ . (10) T 2 T! ws T 2 from which ^pr = Available heat entering the tube per second = wsT leaving =wsT 2 . ^(Ti T 2 ) .'. efficiency = * ^ = i _Tj (10) in (n) gives efficiency = i + (w X constant) d per second and efficiency = - ^ (13) or, if W pounds of coal are burned per second and the air supplied per pound of coal is constant where c is a constant If now the net amount of heat evolved per pound of fuel be constant and equal to H, then the total per second will be WH. ART. 153] HEAT TRANSMISSION 261 Let x be the equivalent evaporation from and at 212 F. per second, then from (13) approximately . . . . (14) If_y denotes the equivalent evaporation from and at 212 F. per pound of coal, then x H / x J = W == (i +cw)gio (I5) WH From (14) =(i+,W)* or /.From (15) y=^ = -~~-cx (16) Hence, if the above theory be accepted, the equivalent evaporation from and at 212 F. per pound of fuel is a linear function of the total equivalent evaporation per second or per hour. 153. Transmission through the Walls of a Thick Tube. Let / be the length of the tube, r and r 2 the internal and external radii respectively, Tj the temperature of the inside surface and T 2 the tempera- ture of the outside surface, then if Q = heat transmitted across i square foot of any surface per second per degree difference in temperature H = total heat transmitted across the inside surface of the tube per second Q = - K-g(Art. 68) .... , (i) Across the surface of radius r-\- x /. From (i) and (2) T_ H dx~ 27TK(r, 4- #)/ dx ~ 2?rK(r 1 + x)lt\ Integrating over the whole length of the tube, we have ( Tl ^_ H I [ dx JT, 27rK/r l Jr 2 -r l r 1 -{-M 262 THE THEORY OF HEAT ENGINES [CHAP. xn. . n _ (T t - T 2 )2 W r 1 K/ _ . 2 >i log,- where A is the internal area of the tube. 154. Effect of High Gas Speed upon Conduction. The formulae given above assume that the temperature of the gas side of the plate is the same as that of the gases, and further, no account is taken of the density of the gases nor of the velocity with which they sweep over the heating surface. The great drop in temperature between the hot gases and the plate (Art. 151) may be accounted for if we accept the existence of a film of gas which clings to the plate. Gases transmit heat chiefly by convection, being very bad conductors of heat, and evidently, the greater the thickness of this stationary gas film, the greater will be the drop in temperature between the gases flowing over the heating surface and the metal plate. Experiment shows that this gas film does exist, and also brings forward considerable evidence to show that a thin film of water clings to the water side of the heating surface. Professor Dalby l suggests that of the total temperature head between the hot gases and water, about 97 per cent, is required to overcome the resistance of the gas film, i per cent, to overcome the resistance of the plate, and 2 per cent, to overcome the resistance of the water film. From this it is evident that the material of which the heating surface is constructed makes very little difference to the transmission of heat ; and further, that the thickness of the plates has a negligibly small effect. The velocity of the gases flowing along a furnace or smoke tube will not be uniform at all points of the cross section of the tube, the flow near the centre of a tube of large diameter (such as the furnace tube of a Lancashire boiler) being of a turbulent nature, the velocity diminishing towards the surface. The greater the speed of the gases and the smaller the diameter of the tube the thinner will be the gas film adhering to the heating surface, and the higher will be the temperature on the gas side of the plate, resulting in an increased temperature head across the plate, and consequently in an increased amount of heat transmitted. Professor Osborne Reynolds in 1874 deduced from theoretical con- siderations that the amount of heat transmitted was a linear function of the speed of the fluids. He gave the following formula for the amount of heat passing from the gases to the heating surface : Q = (A 1 + B 1/ , 1/ x 1 )(T-^) ..... (i) where Q = B.Th.U. transmitted per square foot per second, P! = density of the gases in pounds per cubic foot, jzj = velocity of the gases in feet per second, T = temperature of the gases in F., 6 = mean temperature of the heating surface, A x and Bj = constants. 1 " Heat Transmission," Inst. Mech. Eng,, Oct., 1909, p. 939. ART. 155] HEAT TRANSMISSION 263 In 1897, Dr. T. E. Stanton showed that the above law held for water on opposite sides of a metal plate ; 1 the law may be written /) . (2) where p z = density of water in pounds per cubic foot, jLt 2 = velocity of the water in feet per second, / = temperature of the water in F., A 2 and B 2 = constants. The above law has been since verified by many notable experimenters, but for a detailed account of the researches which have been made on " Heat Transmission," the reader is referred to the paper by Professor Dalby 2 already mentioned. Mr. Hudson, 3 in 1890, gave the following expression for the heat transmitted through boiler tubes or flues per hour per square foot per degree Fahrenheit : 2 X B where 'T g = absolute temperature of the flue gases F. T s = steam F. v = velocity of the gases in feet per second, B = a constant. 155. Heat transmitted through a Cylindrical Tube in Terms of the Gas Temperature, and the Temperature of the Gas Side of the Tube. Let w\ = weight of gases flowing through the tube per second. s = specific heat of the gases. Q = heat transmitted per second per square foot per degree Fahrenheit difference of temperature of air and metal. d = diameter of tube. T! = temperature of gases at inlet end of tube. 6 l = metal T 2 = gases at outlet end of tube. 6 2 = metal Consider an element of the tube of length 8x distant x from the inlet end, where the temperature of the metal surface is 6 F., and over which the fall in temperature of the gases is ST, their temperature being T. Then ws8T = Q X M T Ofi* ..... (0 Now 0=0!^..* dx Assuming a uniform temperature gradient along the tube, this becomes 6=6 1 fx ........ (2) 1 Trans. Royal Society, vol. cxcix., 1897, pp. 67-88. 2 Proceedings Inst. Mech. Engineers, Oct., 1909. 3 Engineer, vol. 70, p. 523, 1890. 264 THE THEORY OF HEAT ENGINES [CHAP. xn. (2) in (i) gives ws8T = Q - ird{ T (0 ex _ST Q.W 8x ws .,_ = . T - (ei -) ... (3) s ws v 1 Qird . Writing P = ~ (3) becomes or + p T = P(0 1 -fl ; ) ......... (4) The integrating factor of (4) is = J(T -f- 0) = mean film temperature in F., Pi = density of gases in pounds per cubic foot, ju-j = speed of gases in feet per second, W-L = pounds of gas flowing per second, a = area of tube in square feet, m 1 = hydraulic mean depth of the tube in inches, area of tube in inches ~ perimeter of tube in inches 157. Estimation of the Temperature of the Gases on leaving the Boiler. 1 For roughly correct estimates Reynolds's Law may be written w ci where p = density of the gases in pounds per cubic foot, p, = speed of the gases in feet per second, a = area of cross section of the flues through which the gases pass (square feet), A = average difference between the gas and water temperatures in F., c = a supposed constant quantity. Applying this formula for the heat transmitted per second through an elementary length dx of a tube (or tubes) of bore d^ through which gas at temperature T is passing and around whose outside diameter d% water at constant temperature / is flowing, we have, using the notation of Arts. 154 and 156, v (i) (2) (3) From (i) and (2) r ^ ... (4) / (5) 1 See Professor J. T. Nicolson's paper on "Boiler Economics and the Use of High Gas Speeds," Trans. Inst. of Engineers and Shipbuilders in Scotland, 1911. ART. 158] HEAT TRANSMISSION 267 itutin (T From (i) and (3), substituting from (6) for T 6 r * T 2 dt or since = which may be written r S = /Mog f ....... (6) where S = area of heating surface in square feet, a cross-sectional area of flue in square feet, TI = furnace temperature, T 2 = chimney temperature, i.e. temperature of gases leaving the heating surface, / = steam temperature. Hence from (6) S S T 2 = /(l ~&) + 1V"^ = A + BTi .......... (7) from which it will be seen that, for a boiler of given " surface section ratio" (-J the chimney temperature is a linear function of the furnace temperature, increasing and diminishing therewith. 158. The most Efficient Rate of Combustion. 1 In all boilers the two chief sources of loss are furnace loss and chimney loss. The chief source of loss in the furnace is the blowing out of the fire of coal dust and small coal, which is carried right through the flues without being burned and goes away up the chimney. At low rates only very fine dust escapes this way and the action is probably unimportant j but with the fierce blast of a locomotive running at full speed there is a constant rain of quite large pieces of coal from the top of the funnel, and by far the greatest proportion of all the loss incurred in the furnace is due to this cause. 1 Professor Nicolson finds that the combined loss of heat due to 1 For the substance of this article the author is indebted to Professor Nicolson's paper on " Boiler Economics and the Use of High Gas Speeds," Trans. Inst. of Engineers and Shipbuilders in Scotland, 1911. 2 See papers by F. J. Brislee andL. H. Fry in Proc. Inst. Mech. Eng., 1908, pp. 237 and 269. 268 THE THEORY OF HEAT ENGINES [CHAP. xn. imperfect combustion and coal blown away amounts to 5oF, B.Th.U. per pound of coal, and that the heat actually generated in the fire per pound of coal is Qo = Q-5oF ....... (i) where Q = calorific value of the coal in B.Th.U. per pound, F = rate of firing in pounds of coal per square foot of grate per hour. Chimney Loss. The chimney loss per pound of coal is (Art. 144) WX j(/i /a) B.Th.U. where W = weight in pounds of the flue gases per pound of fuel burned, s = mean specific heat of the flue gases, /! = temperature ( F.) of the flue gases leaving the boiler, / 2 = temperature ( F.) of air in boiler house. Furnace Temperature. The heat generated in the fire per square foot of grate per hour is Q X F. This heat is disposed of in two ways (1) By radiation from the fire surface to the furnace plates. (2) By heat communicated to the products of combustion, their temperature being thereby raised from / 2 to Tp By Stefan and Boltzmann's law of radiation the quantity of heat radiated per hour per square foot of fire surface is R= 1600 (-^] B.Th.U ...... (2) IOOO, where T O = Tj -j- 460, the absolute temperature of the fire surface. The heat received by the furnace gases per square foot of grate per hour is W X s X F(T! / 2 ) B.Th.U. Therefore, the heat equation for one square foot of grate may be written (3) where / = absolute temperature of air supply = / 2 + 4^o. From the study of many boiler trials it appears that the amount of air supplied per pound of coal with good firing is A = p- + 9 pounds per pound of coal Hence the weight of the flue gases per pound of coal is A + i or W = ^?+io ....... (4) Substituting (4) in (3) the heat equation reduces to + iooo/ iooo = (Q-5oF) o>gi2F ..... IOOO From (5) the absolute furnace temperature T O , and therefore T 1} may be calculated for various values of F. If this be done and a curve be plotted connecting Tj and F, it will be seen that the furnace temperature, ART. 159] HEAT TRANSMISSION 269 Tj, as thus determined, rises very rapidly at first as F increases, reaches a maximum for F = 80, and then begins to fall. One ought to expect an increase of the furnace temperature with an increased rate of combustion, because less air is supplied per pound of coal, and there is, consequently, a smaller weight of products to be heated up. Recalling, however, the fact that the heat actually generated in the fire per pound of coal gets less and less, owing to imperfect combustion and coal blown away, it will readily appear that this source of loss will presently overtake the gain due to diminished air supply, the furnace temperature curve will rise less and less steeply, and will finally attain a maximum and afterwards fall. The chimney temperature f may now be found for the various values of TO (or T^) by using equation (7), Art. 157, and hence the chimney loss. By adding the chimney loss to the furnace loss the whole loss of heat in the boiler due to these two causes together for various rates of firing may be found. If this be done it will be found that the most efficient rate of firing occurs at between 25 and 30 pounds of coal per hour per square foot of grate area. For smaller values, the combustion is more perfect, but the larger chimney loss, due to the larger amount of air supplied, more than makes up for this. For higher rates the chimney loss gets less, but imperfect combustion and coal blown away out of the fire does away with this advantage. 159. Heat Transmission through Condenser Tube. In the usual arrangement of surface condenser the steam is condensed on the outside of a set of tubes through which the condensing water is circulating. The rate at which steam can be condensed in this way depends upon the temperature gradient through the tube walls and also upon the condition of the steam. The rate of condensation measured by Dr. Nicolson in a steam-engine cylinder was equal to 074 B.Th.U. per square foot of surface per second per degree difference of temperature between the steam and cylinder walls, but the conditions in an engine cylinder differ considerably from those in a surface condenser. Experiment shows that the rate of condensation depends upon (1) The dryness fraction and pressure of the steam, (2) The velocity of the steam over the tubes, (3) The amount of air present in the steam, (4) The velocity of the condensing water through the tubes. Considerable evidence exists to show that the .rate of condensation appears to increase according to a nearly linear law with increasing speed of flow both of the steam and condensing water. Mr. Jordan 1 records a rate equal to 1*26 B.Th.U. per second per square foot per degree difference of temperature due to the high speed of the steam and condensing water ; he also found that dry steam condensed more rapidly than wet steam, and further, that the rate of condensation was greater at 25 pounds per square inch absolute than at 40 pounds absolute. It should be pointed out, however, that the increased rate of heat transfer through the tubes, both for condenser and boiler tubes, due to speeds of flow greater than those commonly used in practice, would only be obtained at the expense of power for producing the circulation, and it must be a matter of experience to decide how far the rate of heat transmission may economically be increased. 1 Proc. Inst. Mech. Engineers, 1909, p. 998. 270 THE THEORY OF HEAT ENGINES [CHAP. xn. At the usual condenser pressures and temperatures used in practice, the presence of small quantities of air causes a rapid falling off in the rate of condensation, which becomes less rapid as the quantity of air is increased. For the results of actual tests and data the reader is referred to the following important papers : "Efficiency of Surface Condensers," R. L. Weighton, Inst. Naval Architects, April, 1906, and Engineering, April 13 and 20, 1906. " Efficiency and Design of Surface Condensers," T. E. Stanton, Proc. Inst. C.E., vol. cxxxvi., Part 2, 1898-9. " Surface Condensers for Steam Turbines," Engineering, December n, 1908. " Air in Relation to the Surface Condensation of Low-Pressure Steam," J. A. Smith, Victorian Inst. of Engineers, December, 1905, and Engineering^ March 23, 1906, p. 395. " Design of Surface Condensers," R. M. Neilson, Inst. of Engineers and Shipbuilders in Scotland, 1910. " Modern Condensing Systems/' A. E. Leigh Scanes, Inst. Mech. E., February 14, 1913. EXAMPLES XII I. In order to determine the amount of heat lost by radiation from a metal surface, a cast-iron bar of square section 4 inches X 4 inches was heated at one end. When a steady condition was attained the temperatures were read from thermometers placed at different distances along the bar. Obtain a formula by means of which the amount of heat radiated per square foot per degree difference of temperature between the tempera- ture of the bar and atmosphere can be calculated. Determine the actual amount of radiation from the figures given below : Distance from end in inches Temperature ( F.) . . o 235 6 171 12 21 97 '9 30 80-5 69-2 Conductivity of cast iron, 5-4 B.Th.U. per square foot per minute per F. per inch thick. Temperature of atmosphere, 59-5 F. 2. The furnace tube of a Lancashire boiler is 36 inches diameter ; coal burned on a grate 20 square feet in area, 400 pounds per hour ; air supplied per pound of coal, 24 pounds. Temperature of the gases leaving the fire 2200 F., temperature of the gases at the end of the tube 900 F., steam temperature 350 F. Estimate the number of B.Th.U. transmitted through the tube per square foot per hour (a) using Rankine's formula H = ( J ~ J , Art. 151 ; () using Jordan's formula (2), Art. 156 ; (c) using Nicolson's formula (4), Art. 156. 3. In a surface condenser with 6000 square feet of cooling surface it was found that t;8,ooo pounds of steam were condensed per hour. The average temperature of the steam was 132 F., and of the circulating water 80 F. The tubes were of brass 0*05 inch thick and of such a conductivity that 25 B.Th.U. could pass per minute through a plate I square foot in area I inch thick, per degree difference in temperature between the two surfaces. Calculate the temperatures of the metal on the steam and water sides of the tubes. Assume that when steam is in contact with a metal surface the rate of condensation is 074 (T /) B.Th.U. per square foot per second, where T is the temperature of the steam and t the temperature of the metal, and latent heat at 132 F. = 1020 B.Th.U. per pound. CHAPTER XIII THEORY OF THE GAS ENGINE 160. Internal Combustion Engines. General Considerations. The term " internal combustion engine " includes all types of gas, oil, and petrol engines; in this chapter the theory of the gas engine alone is considered, on the assumption that the specific heat of the gases remains constant^ the variable specific heat theory being investigated in Chapter XIV. It will be as well to briefly compare the internal combustion engine with the steam engine, as by that means it will be easy to understand why its thermodynamic efficiency is so much greater than that of the steam engine. In the steam engine the heat generated by the combustion of the fuel in the furnace passes through the metal plates or tubes of the boiler to the water, and so generates steam which is used in the engine cylinder to do work ; the working fluid is therefore entirely different from the products of combustion of the fuel. In the steam engine there are four different organs, namely : the furnace, boiler, engine cylinder, and condenser (if the engine is condensing). In the gas engine, however, there are only two organs, the gas producer and the engine cylinder, into which a mixture of gas, or oil vapour, and air in suitable proportions is admitted and exploded when the piston is beginning to move forward on its working stroke. Heat is developed as the result of the explosion, the gases expand, driving the piston forward and so doing useful work ; on the return of the piston the products of combustion are driven out of the cylinder and the operation is then repeated. An oil engine is essentially a special form of gas engine with a vaporiser or carburettor attached, wherein the oil is converted into a gas or vapour and then used in the cylinder in the same way as the gas in a gas engine. The maximum temperature reached in the engine cylinder is very high, so that the range of temperature T x to T 2 is large. The full thermo- dynamic advantages of this high initial temperature cannot be realised in practice, since, if the cylinder walls are allowed to reach this high tempera- ture, they would soon be destroyed and lubrication of the piston would be impossible, to say nothing of the risk of premature ignition of the incoming mixture of gas and air ; hence, the cylinder is water-jacketed to keep it cool. With large gas engines the difficulty is to keep the cylinder, piston and valves cool enough ; with steam engines, on the other hand, the cylinder should be kept hot to reduce the losses due to condensation of the steam. There are a great and increasing number of gas and oil engines on the market, for a detailed description of which the reader is referred to the Technical Press and to standard books on the subject. 1 1 See Robinson's "Gas and Petroleum Engines"; "The Internal Combustion Engine," H. E. Wimperis; or D. Clark's " Gas, Oil and Petrol Engines" (Longmans). 271 272 THE THEORY OF HEAT ENGINES [CHAP. xur. Internal combustion engines may be divided into three classes, namely 1. Engines in which the explosion takes place at constant volume without any previous compression. No modern engine works on this principle. 2. Engines in which the explosion takes place at constant volume with previous compression. 3. Engines in which the combustion takes place at constant pressure with previous compression. In the discussion which follows, ideal indicator diagrams and conditions are assumed which cannot be obtained in practice. Further, to obtain the greatest possible efficiency, it is assumed i. That the explosion is instantaneous and takes place at constant volume in the first two of the above classes, the specific heats C p and Cv remaining constant at all temperatures. Expansion pv y = const. FIG. 123. 2. That the expansion (and compression in those engines which use compression) is adiabatic, which necessitates the assumption that the cylinder walls and piston are perfect non-conductors of heat. 3. That the products of combustion expand right down to atmospheric pressure and are exhausted at that pressure. 4. That the explosive mixture of gas and air is admitted during the suction stroke at atmospheric pressure. 161. Engines in which the Explosion takes place at Constant Volume without Previous Compression. The ideal indicator diagram for this cycle is shown in Fig. 123. The mixture of gas and air is drawn in at atmospheric pressure along AB, then follows explosion at constant volume from B to C, followed by adiabatic expansion CD down to atmospheric pressure and exhaust DA at the same pressure. Let T! = initial absolute temperature of the charge at A or B, T 2 = maximum temperature reached at C, T 3 = temperature at D after expansion. ART. 162] THEORY OF THE GAS ENGINE 273 Then we have per pound of the gaseous mixture, Heat supplied by the explosion = H x = C V (T 2 T x ) . . . (i) Heat rejected = H 2 = C P (T 3 T x ) ... (2) Efficiency = 1 _C.(T 2 -T 1 )-C P (T 3 -T 1 )_ Ta-Tj C^TJJ-T!) y 'T 2 -T 1 No modern engine works on this cycle ; the Lenoir and Hugon engines were representatives of this type, the ideal indicator diagram being repre- sented by BCEF (Fig. 123). It will be noticed that there was a loss of work due to incomplete expansion represented in amount by the shaded area FED. The efficiency of this ideal Lenoir cycle may be obtained as follows : Let the conditions be at B A^i^i, at C A^g^i at E /4^T 4 , then we have Work done during adiabatic expansion from C to E y i and since pv = RT, this may be written y i Work done in driving out exhaust gases from F to B =A(*4 fr g) (5) Net amount of work done during the cycle = (4) (5) Heat supplied from B to C = C(T 2 Tj) Hence efficiency = 1 *1 T 2-Ti 162. Atmospheric Engine. The Otto and Langen engine worked on a modification of this cycle. Like the Lenoir engine this cycle is obsolete, but it will be instructive to briefly consider the cycle and to determine its ideal thermal efficiency. The mixture of gas and air is drawn in at atmospheric pressure along AB (Fig. 124), then follows explosion at constant volume from B to C and adiabatic expansion along CD, followed by isothermal compression DB and exhaust BA at atmospheric pressure. Let the conditions be at B A^iTj, at Cfov^Tfr a t D/a^V^s, then we have Work done on the piston during adiabatic expansion CD _A y i P\ y i 274 THE THEORY OF HEAT ENGINES [CHAP. xm. Work done by the piston on the gas during isothermal compression DB Net amount of work done during the cycle and efficiency - (3) i. .... (4) But - =^ ; since DB is an isothermal, .. =^, and ( J = ^ by (4), Art. n. ... | = (I?p = (I?)r-i sinc e T , = T, 'Adiabatic Expansion p ^ r = const. Substituting in (4) we get FIG. 124. efficiency = i A) A -A - (5) To T, (6) ART. 163] THEORY OF THE GAS ENGINE 275 163. Engine in which the Explosion takes place at Constant Volume with Previous Compression. The ideal cycle for this type of engine is shown in Fig. 125. The mixture of gas and air is drawn in at atmospheric pressure along the suction stroke AB and compressed adiabatically along BC during the return of the piston ; then follows explosion at constant volume from C to D and adiabatic expansion right down to atmospheric pressure at E, followed by exhaust at that pressure along EA. iiabat/c Compression p y y FIG. 125. Let the conditions be at B A^iTi* at 4 T 4 , then per pound of working fluid we hav at D Heat supplied at constant volume from C to D = C W (T 3 T 2 ) Heat rejected at atmospheric pressure from E to A = C P (T 4 Tj) Net amount of heat converted into work C y (T 3 T 2 ) at E (i) (2) T 3 -T (3) In practice, however, this cycle is not realised. For mechanical reasons it is usual to work with a ratio of expansion equal to the ratio of com- pression, the expansion being stopped at point F. This results in incom- plete expansion, the diagram being represented by BCDF, and the shaded area BFE represents the loss due to incomplete expansion. Compound gas engines have been proposed and constructed whereby this further expansion from F to E is carried out in a separate low-pressure cylinder ; but almost all modern engines work on the cycle represented by BCDF. This cycle is known as the Otto, or four-stroke constant volume cycle, and is discussed further in Art. 166. 2 7 6 THE THEORY OF HEAT ENGINES [CHAP. xin. Efficiency of the Otto Cycle. The ideal indicator diagram of this constant volume cycle is shown separately in Fig. 126 and the T^> diagram Ad ia bat ic Expansion pyy=const. AdiabaTic Compression pv y = const FIG. 126. FIG. I26A. in Fig. I26A. Let the conditions be at B/^Tj, at Cp z v^Y z , at at F/ 4 ^ 4 T 4 , then per pound of working fluid we have Heat supplied at contant volume from C to D = C r (T 3 T 2 ) (4) Heat rejected at constant volume from F to B = C(T 4 Tj) (5) ART. 163] THEORY OF THE GAS ENGINE Heat converted into work = heat received heat rejected Efficiency = 277 ~ T .) - C.(T 4 - T,) (6) Now, since the ratios of adiabatic expansion and compression are equal, 4 1 Qr * 4 .1 * 4 ~ * 1 /,_\ Trp ** .... \l/ 3 A 2 A 3 *2 ^3 A 2 Substituting (7) in (6) T TT Efficiency = i ? or i =^ (8) by (4), Art. n where ^= ^, the ratio of expansion or compression. Hence the efficiency may be written y-l (9) This efficiency is frequently called the Air Standard Cycle Efficiency rj where y is taken as 1-4, the value of -^ for air. Equation (9) also shows {^v that, on the assumption that the specific heat of the working fluid remains constant, increasing the ratio of compression and therefore increasing the pressure, i.e. the pressure at the end of the compression stroke, results in a higher efficiency. The actual value of y for the mixture in an internal combustion engine varies from 1-36 to 1-38. Within these limits the exact value taken for y does not materially affect the efficiency of the ideal engine, as will be evident from the following tabulated values : EFFICIENCIES OF AIR STANDARD CYCLE i v ( - j 3 I 7 10 0-36 o'43 0-47 0-52 o'SS 0-61 0-332 Q'399 o'45 0-49 0-51 0-28 0-38 0-42 0-44 The temperature-entropy diagram of this cycle is shown in Fig. 126 A. 2 7 8 THE THEORY OF HEAT ENGINES [CHAP. xm. 164. The Atkinson Cycle. The Atkinson cycle is a modification of the Otto cycle, and was an attempt to approach nearer to the ideal cycle shown by BCDE in Fig. 125. In this cycle the working fluid was allowed to expand adiabatically to double its volume before compression, so that the ratio of adiabatic expansion was twice the ratio of adiabatic compression. By this means the loss due to incomplete expansion was reduced although the pressure at the end of expansion was still considerably above atmospheric. The ideal diagram for this cycle is shown in Fig. 127. The mixture of gas and air was drawn in at atmospheric pressure along AB, then followed adiabatic compression BC, explosion at constant volume from C to D ; adiabatic expansion DE, the volume at E being twice that at B, and exhaust at atmospheric pressure from F to A. 'Adiabatic Expansion pv y * 'Const Adiabatic Compression p v y " const. FIG. 127. Let the conditions be at B /i^iT 1} at C fav^^ at D /s^V^s, at E l, at F/ 5 z> 6 T 5 , then per pound of working fluid we have Heat supplied at constant volume from C to D = C V (T 3 T 2 ) . (i) Heat rejected at constant volume from E to F = Ct,(T 4 T 5 ) Heat rejected at constant pressure from Fto B = Qp(T 5 T x ) Total heat rejected = G;(T 4 T 5 ) + Cp(T 5 Ti) ... (2) Heat converted into work = heat supplied heat rejected f~\ /rrt rp \ f* /rr\ np \ /-! /rp fy \ = Ct>( 13 Iz) {->v( A 4 it,) ^p( A 5 -I I/ ^3 - T 2 ) - C,(T 4 - T 6 ) - C,>(T 5 - Tj) Efficiency = (3) It is obvious that this ideal efficiency is greater than that of the Otto cycle, because, for the same quantity of heat supplied during the explosion at constant volume there is more work done on account of the increased expansion. ART. 165] THEORY OF THE GAS ENGINE 279 165. Engine in which the Combustion takes place at Constant Pressure with Previous Compression. The ideal/^ diagram for this cycle is shown in Fig. 128, and the T0 diagram in Fig. I28A. The mixture of gas and air is drawn in at atmospheric pressure during the suction stroke AB, and on the return of the piston is compressed adiabatically from B to C. Combustion then takes place at constant pressure along CD, followed by adiabatic expansion from D to E, the gases being finally exhausted at atmospheric pressure from E to A. FIG. 128. FIG. I28A. Let the conditions be, at B p^v{^^ at C p^v^Y^, at D / 3 ^ 3 T 3 , at E , then per pound of the working fluid we have Heat supplied at constant pressure from C to D = C^(T 3 T 2 ) (i) Heat rejected at constant pressure from E to B = Qp(T 4 T x ) (2) Heat converted into work = Q,(T 3 T 2 ) C P (T 4 T x ) (3) 3 - T, Also, since the expansion and compression are adiabatic, hence Hence (3) may be written T 3 T 2 Efficiency = i -~ or i =r lo 1? Tj 2 (4) 280 THE THEORY OF HEAT ENGINES [CHAP. xm. '%-% and (4) becomes y-l y-l =-(0 y-l (5) where r is the ratio of compression. No gas engine works on this cycle, but the Diesel oil engine approxi- mates to it (see Art. 192). EXAMPLE i. In an ideal Lenoir engine the temperature of the in- coming charge is 60 F. at 15 Ibs. per square inch absolute, and the maximum temperature of explosion is 2800 F. If the ratio of expansion is 2 and the initial volume one cubic foot, estimate the efficiency. (Take Here T x = 60 -f- 460 = 520 T 2 = 2800 -f 460 = 3260' To find T 4 , the temperature after adiabatic expansion, y-l 0-38 S^Cf!) =6) .-. T 4 = 3260 X (i)' 38 = 2505 absolute. Now I 520 Using equation (7), Art. 161, we have T 2 - T 4 -A . r - efficiency = (3260 2505) 15 X -f-^-f X r Substituting efficiency = 3260-520 7^6 I"?7 L3 2JL 0-275 or 2 7'5 P er cent - 2740 N.B. Note that the Carnot efficiency would be = _ = o . g or g 3260 3260 EXAMPLE 2. In an ideal Otto and Langen engine the initial and explosion temperatures are the same as in Example i. Assuming y = 1-38, estimate its efficiency. ART. 165] THEORY OF THE GAS ENGINE 281 By equation (6), Art. 162, Efficiency = i Now T 2 =326o and T = 2o 0-38 X 5* X log. :. efficiency = i 3260 520 2-63 0-38 X 520 X 2-303 lo g 2740 . 0-38 X 5 20 X 2-303 X 2-0948 .*. efficiency = i L = i 0-348 = 0-652 EXAMPLE 3. In an engine working on the Otto cycle the following temperatures were measured: Temperature of suction = 210 F., at end of compression 590 F., maximum temperature of explosion 2093 F., at end of expansion 1594 F. Estimate the ratio of compression and the ideal efficiency. Assume the law of compression to be /z/ 1 ' 445 = const. (The above data is taken from the Ay trial given in the second report of the Gas Engine Research Committee, published in the Proceedings of the Institution of Mechanical Engineers, October-December, 1901.) Let T x = temperature of compression = 590 -}- 460 = 1050, T 2 = suction =210-1-460 = 670, T i ~ Then = T 2 .-. 0-445 lg f=log T! log T 2 = 3-0216 2-8267 = 0-1949 0-1949 "' gr= = ' 4379== log 2 ' 741 .-. ratio of expansion and compression =2741 AX 7 " 1 efficiency = i ^ - 1 0-445 I \ = i 0-638 =0-362 or 36-2 per cent. This result may be checked by the shorter method, Art. 163, equation 8. T Efficiency = i - L z 670 = i = i 0-638 = 0-362 as before 1050 282 THE THEORY OF HEAT ENGINES [CHAP. xm. EXAMPLE 4. Suppose the above engine worked on the Atkinson cycle with the same maximum and suction temperatures : estimate its efficiency, assuming the temperature of exhaust to be 400 F. We must first find T 4 , the temperature after expansion. T __ __ yO-445^ w here T 3 = maximum temperature of explosion. Now r= 2 X 2741 = 5-482 T - -2l = 2 93 "^ 46 - 2 553 5-482 ' 445 "sW 445 log T 4 = log 2553 0-445 lo S 5'482 = 3-4072 *445 X 0-7390 == 3-4072 0-3288 = 3-0784 .-. T 4 = 1198 absolute = 738 F. Hence efficiency = i ^ 4 ~ T j^bffi 5 ~ T i> (Art. 164, equation 3) A 3 *2 Substituting we get Efficiency = i - ("98 - 860) + 1-445(860 - 670) 2553 105 ==I 337 + i'445 = 190 337 + 274'55 1503 1503 611-55 --== i 0-406 =0-594 or 59-4 per cent. 166. Otto Cycle. Dr. Otto first made a gas engine to work on the cycle patented by Beau de Rochae, and many modern engines work on this, or at any rate a modification of this, cycle, which may be represented thus Direction of stroke. Name of stroke. One revolution -j One revolution | ~' charging compression explosion and expansion exhaust The charge of gas and air is first drawn into the cylinder on the outward stroke of the piston, and on the return stroke is compressed, ignition takes place just when the piston is moving on its next outward stroke, and on the return stroke the products of combustion are driven out of the cylinder and the cycle is repeated. The indicator diagram is shown diagrammatically in Fig. 129. It will be noticed that during the charging stroke the pressure is below atmospheric (this is exaggerated in the diagram for clearness), and during exhaust is above atmospheric. The area of this loop of the diagram represents the work done on the gases during the cycle. It will be seen that a working stroke takes place every two revolutions, or every four strokes when the engine is on full load and missing no explosions, hence the cycle is frequently spoken of as the four-stroke cycle. At the end of each exhaust stroke there remains in the clearance space ART. 1 6 8] THEORY OF THE GAS ENGINE 283 a volume of approximately one-third of the piston displacement filled with products of combustion largely composed of CO 2 and N 2 , which are both inert and non-combustible gases. The next charge of gas and air mixes with these inert gases, and, as a result, the explosion will not be so effective as it might be. It is for this reason that in many modern engines some arrangement is adopted to remove these products of combustion before the next charge is admitted into the cylinder. (See Art. 169.) 167. After Burning. In discussing the ideal condi- tions, one assumption made Charging (Art. 1 60) was that the ex- plosion was instantaneous and FIG. 129. at constant volume. Actually it is found that the maximum pressure and temperature reached at the end of the explosion are much less than they would be if the explosion were instantaneous. According to Mr. Dugald Clerk l only about 65 per cent, of the potential heat of combustion is required to produce the maxi- mum temperature reached, the remaining 35 per cent, of the heat being evolved by slow combustion during the expansion. This phenomenon is known as " after burning," and it is quite possible that in many cases the products which are discharged into the exhaust contain some incompletely burnt fuel. To explain this phenomenon of after burning, it has been suggested that the extremely high temperature reached in the first stage of the explosion prevents the chemical union which takes place during com- bustion from being completed, just as in the same way the products of combustion would be dissociated or split up into their constituent elements, until the fall of temperature by expansion and the cooling by the cylinder walls allows the process of union to continue. Against this theory is the fact that " after burning " is more pronounced with weak mixtures when the maximum temperature reached is much lower. 168. Effect of the Strength of the Mixture. The minimum theoretical quantity of air required for complete combustion of one cubic foot of average coal gas is about 5 cubic feet (see Art. 143), but more than this is always supplied. The limits in practice seem to vary from about 6 to 14 cubic feet of air per cubic foot of gas, the average being from about 8 to 10 cubic feet at full load. Professor Burstall obtained the most economical result with from 9'48 to io'8 cubic feet. 2 Dr. Otto laid great stress upon the desirability of having a stratified mixture of gases in the cylinder with a portion rich in gas near the place of ignition, but Mr. Clerk's experiments are conclusive against this. The only case in which this might be an advantage would be when dealing with a very weak 1 See Proceedings of I. C. E., 1882 and 1886. 2 See the " First Report of Gas Engine Research Committee " in Proc. Inst. of Mech. Engineers, April, 1898. 284 THE THEORY OF HEAT ENGINES [CHAP. xin. mixture, when a small quantity rich in fuel near the ignition point would tend to start combustion of the rest. Professor Burstall also found that, within limits, weaker mixtures give higher efficiency than stronger ones. Professor Hopkinson 1 also found the same result from a series of experiments in which the proportion of air to gas was varied from 9-5 to i (the weakest mixture) to 7-5 to i (the strongest mixture). Within that range he found that the efficiency decreased steadily as the strength of mixture increased, the efficiency with the weakest mixture being 4-5 per cent, greater than that obtained with the strongest mixture. He also found that when using a mixture containing 8 '5 per cent, of coal gas, the actual thermal efficiency was 0-87 of the ideal efficiency, but when the proportion of coal gas was increased to n per cent., the actual was reduced to 0*83 of the ideal efficiency ; the weaker mixture, in addition to giving a higher ideal efficiency, came nearer in practice to realising that ideal. This may be due to the fact that the percentage of heat lost to the cylinder walls during expansion is less with weak mixtures than with stronger ones, the difference being sufficient to counterbalance the reverse effect of the more rapid combustion of the stronger mixtures. Causes of Higher Efficiency with Weaker Mixtures. That the efficiency will be higher with weaker mixtures may be deduced from the fact that the specific heat of the working gases increases with the temperature. The work done in the gas-engine cycle is mainly determined by the rise of pressure which occurs on explosion ; and in the same engine the area of the indicator diagram with different mixtures is nearly proportional to this rise. If the specific heat of the working substance were constant, the rise of temperature and pressure due to the explosion would be proportional to the heat supply, and the efficiency would, therefore, be constant. Since, however, the specific heat is greater at high temperatures, the rise of temperature and pressure on explosion increases in a less ratio than the heat supply, and the efficiency therefore diminishes as the supply of heat is increased. This theory, however, does not explain why the weaker mixtures, in addition to giving a higher ideal efficiency, come nearer in practice to realising that ideal. When the mean effective pressure in an engine cylinder is increased by using a stronger mixture, the effect of the increased temperature of com- bustion is to greatly increase the amount of heat radiated from the flame to the cylinder walls, since the heat radiated is proportional to the fourth power of the absolute temperature (Art. 185). The jacket loss and the metal temperature are thereby increased in a much greater proportion than the fuel consumption, and the efficiency is therefore diminished. It is possible that with a strong mixture the percentage of gas present in all parts of the cylinder is not the same, and that some of the gas has not got next to it the requisite amount of oxygen for combustion, the result being that some of the gas is not burned. On the other hand, with a weaker mixture there is the great probability that all the molecules of gas are surrounded with sufficient oxygen to ensure combustion. If this assumption is justifiable, it offers another explanation why weak mixtures are more economical than strong ones ; Professor Hopkinson found, how- ever, that the percentage of unburnt gas in the exhaust did not depend 1 " Effect of Mixture, Strength, and Scavenging upon Thermal Efficiency," Proc. Inst. Mech. Engineers, April, 1908. ART. 169] THEORY OF THE GAS ENGINE 285 upon the strength of the mixture, and that, moreover, he considered it very improbable that more than i per cent, of the fuel is ever unburnt at the end of the expansion stroke. In large gas engines the use of strong mixtures is prohibited by the shock of the explosion, such engines always using producer gas or blast- furnace gas of very much lower heating value than coal gas ; it is an advantage therefore that these weaker mixtures result in higher efficiency. In order that these weak mixtures may be explosive it is necessary to have a high compression of the charge before ignition. Theoretically, the higher the compression, the higher will be the efficiency ((9) Art. 163), but, according to Professor Burstall, there is a limit with the coal gas he used, about 175 pounds per square inch, beyond which increased compression does not result in increased efficiency. 1 The advantages of high over low com- pression before ignition may be briefly summed up as follows : 1. The same weight of gas is compressed into a smaller volume and is therefore exposed to less cooling surface. 2. When ignition takes place the rate of combustion is more rapid under the higher pressure, so that the time of exposure to the smaller cooling surface is less than for slow combustion. 3. The maximum pressure and temperature is reached more quickly and the fall of pressure is more rapid than with the slower combustion of a low-pressure charge. 4. By high compression a weak diluted mixture is made combustible which will not burn at low pressure and temperature. 5. The power and efficiency are largely increased by high compression up to a certain limit of maximum pressure (about 600 pounds per square inch). 169. Scavenging. To get rid of the burnt products of combustion mentioned in Art. 168 various methods have been tried. The obsolete Griffin, Linford, and Beck engines worked on a six-stroke cycle as follows : One revolution {- Charging^ One revolution * n and e *P ansion One revolution and scavenging It will be seen that this cycle is the Otto cycle with two extra strokes added in which air only is drawn in during the suction stroke and on the return stroke is exhausted, driving out the burnt products of combustion of the previous explosion. The great drawback to this cycle is that there is only one working stroke every three revolutions when the engine is on full load and missing no explosions, which results in a large engine for a given power, and an irregular turning moment on the crankshaft necessitating the use of a very heavy flywheel effect if steady running is to be obtained. A great advantage is that the cylinder is well cooled. If an indicator diagram be taken of the charging and exhaust strokes 1 "Third Report of Gas Engine Research Committee," Proc, Inst. Mech. Eng., 1908. 286 THE THEORY OF HEAT ENGINES [CHAP. xm. with a light spring it will frequently be found that the exhaust shows as a wavy line, the pressure falling at some point such as A (Fig. 130) below atmospheric. Mr. Atkinson took advantage of this and opened the air valve at this point A on the exhaust stroke, where a partial vacuum is formed, and closed the exhaust valve late as shown at B, Fig. 131, which shows the valve setting for scavenging. Theoretically, this would have the advantage of sweeping the burnt products of combustion away, but in practice the method was not successful and has been practically abandoned. Suction FIG. 130. The valve setting necessary was adopted by Messrs. Crossleys, and required an exhaust pipe about 65 feet long and a special adjustment of the silencer, which, unfortunately, got out of order with a varying load on the engine. There was introduced, in addition, a higher compression of the charge than usual, and it was doubtless this, rather than the scavenging, which gave the good result obtained with the engine. The best device is undoubtedly the positive scavenging used on the modern Premier gas engine. In this engine air is compressed in a separate Scavenging 'Exhaust opens J.me of Stroke Air Opens FIG. 131. pump and forced through the engine cylinder near the end of the exhaust stroke. The approximate valve setting necessary is shown in Fig. 132. The cycle is as follows : During the charging stroke air is sucked in by the pump, and in the compression stroke the air, as well as the charge in the engine cylinder, is compressed. The explosion then occurs, and towards the end of the working stroke the exhaust valve opens early. During the compression stroke, just as the piston is turning, the gas valve closes, then the air valve ART. 170] THEORY OF THE GAS ENGINE 287 closes. Towards the end of the exhaust stroke the air valve is opened, and in this last portion of the exhaust stroke the compressed air sweeps across and clears out the burnt products of the previous explosion. LlneofSTroke Gas Closes Air doses Air opens FIG. 132. Fig. 133 shows the valve setting for the ordinary Otto cycle and should be compared with Fig. 132. Gas Closes Line of Stroke : Air Closes FIG. 133. 170. Ignition in Gas Engines. In all modern engines the charge is fired either by means of an ignition tube or by an electric spark. The ignition tube consists of a small closed tube of iron or porcelain kept at a bright red heat by means of an external gas flame. It is screwed into the end of the combustion chamber, and the open end is in communica- tion with the engine cylinder. In many cases a timing valve is used with the tube, which opens a communication between the cylinder and the inside of the tube at the correct time for the explosion to occur ; in other cases communication is always established and the charge is fired when the compression raises the pressure high enough to cause combustion. On large engines the ignition tubes are usually fitted in pairs, so that if 288 THE THEORY OF HEAT ENGINES [CHAP. xm. one bursts the other can be used while the damaged one is repaired ; by this the engine is kept running, no stoppage being required for the renewal of the damaged tube. Electric ignition is rapidly replacing tube ignition, the most common method being to use a magneto machine mounted on the engine cylinder, contact being made and broken between the sparking points by means of a rod or lever actuated from the cam shaft of the engine. 171. The Atkinson Cycle Engine. From theoretical considera- tions the greatest drawback to the Otto cycle consists in the fact that the ratio of expansion is the same as the ratio of compression. Mr. Atkinson surmounted this difficulty by so arranging the mechanism that he obtained a ratio of expansion twice as great as the ratio of compression, which of course resulted in greater efficiency (Art. 164). With his arrangement he was also able to obtain a working stroke every revolution as against every two revolutions in the Otto cycle engine ; this resulted in a more even turning moment, but mechanical difficulties caused the engine to be abandoned. 172. The Clerk or Two- Stroke Cycle. In this cycle, an indicator diagram of which is shown in Fig. 134, there are only two strokes, the Ibs. per o" - 355 Avg. Press. - 4?7 Us. per a' 121 R&vs. B.H.P. 605 -170 ATM. FIG. 134. Two-stroke cycle. working stroke and the compression stroke. Towards the end of the working stroke the piston uncovers the exhaust ports in the cylinder walls ; air is then admitted under pressure from b to a and drives out the products of combustion. The explosive mixture of gas and air is next admitted from a to b and compressed on the return of the piston, and fired as the piston starts to move on its next stroke. By this means one explosion is obtained every revolution of the engine shaft when using one single-acting cylinder, as against one explosion every two revolutions in the four-stroke or Otto cycle. If the cylinder is double-acting there will be an explosion on each side of the piston, every revolution resulting in two working strokes per revolution. The two-stroke has many advantages over the four-stroke cycle. A single-acting cylinder of a certain size arranged as a two-stroke will give approximately twice the power of a cylinder of the same size arranged as a four-stroke, provided that the revolutions per minute are the same. Also an engine of a certain size arranged as a two-stroke will develop the same power as an engine with the same sized cylinders arranged as a four-stroke at half the number of revolutions. This is an important consideration in ART. 173] THEORY OF THE GAS ENGINE 289 marine engines of small power, as a slow-running and more efficient pro- peller can be adopted with engines of the same weight and cost as the four-stroke type. The two-stroke engine will run equally well in either direction with adjustment of the ignition. In a four-stroke engine special reversing gear is necessary for the inlet and exhaust valves. In addition to these advan- tages the two-stroke engine has a more uniform turning moment than the four-stroke, because there is an impulse every revolution ; hence a lighter flywheel may be used. Also, the cutting out of an explosion by the governor will have a less disturbing effect on the angular velocity than is the case of the four-stroke engine. In the two-stroke engine the hot exhaust gases do not pass through a valve but through ports in the cylinder wall, whereas in the four-stroke engine the gases are exhausted through a mechanically operated exhaust valve, which in engines of even moderate size has to be water cooled. A disadvantage of the two-stroke engine is its slightly lower efficiency. During a portion of the time taken by the explosive mixture to enter the cylinder the exhaust ports are open, and the probability is that some of the gas enters the exhaust pipe, and is therefore wasted, resulting in a larger gas consumption This, however, is a minor point, because most two-stroke engines are of large size and use cheap fuel gas, such as the waste gas from blast furnaces, etc. 173. Governing". The methods of governing gas engines may be divided into two classes. In the quality method the volume of the charge remains constant and sufficient to fill the engine cylinder as nearly as possible at atmospheric pressure, but the proportion of gas and air in the mixture is varied according to the load on the engine. In the quantity method the proportion of gas and air in the mixture is kept constant, but the volume of the charge is varied either by closing the admission valve before the end of the suction stroke, or by throttling ; in either case the charge is only sufficient to partly fill the cylinder at atmospheric pressure. The quality method may be subdivided into hit and miss governing ; variable gas admission, the gas being admitted uniformly throughout the suction stroke ; and variable gas admission caused by opening the gas valve earlier or later during the suction stroke, but always closing it at the end of the stroke. The variation in the strength of the mixture obtained with this method of governing results in explosions of varying intensity, as will be evident on referring to Fig. 135. The best method to use depends to a large extent on the size and type of the engine, also upon the kind of gas used. The quality method is preferable with engines having a considerable weight of reciprocating parts attached to one connecting rod, because under these conditions the inertia of the reciprocating parts should be cushioned by the compression pressure (which in this method is constant), otherwise shock may be caused similar to that experienced in a steam engine working with insufficient compression or lead. The quantity method results in a variable compression and is specially suitable for engines with comparatively light reciprocating parts running at a moderate speed, as under these circumstances the reduced compression pressure may be sufficient to take up the inertia forces without shock at the time of ignition. Variable gas admission caused by opening the gas valve earlier or u 2 9 o THE THEORY OF HEAT ENGINES [CHAP. xm. later during the suction stroke, but always closing at the end of this stroke, is a most satisfactory method for large engines. By admitting the gas and air in this way, when working at light loads air only is first drawn in, the gas being admitted towards the end of the suction stroke ; the result is that the portion of the charge next to the piston contains only a small pro- portion of gas, whilst that portion near the ignition point has sufficient to form an explosive mixture. This method of governing has come rapidly into use of recent years. Quality governing by admitting the gas in varying quantities con- tinuously throughout the suction stroke is not generally adopted, although some very large engines have been governed in this way. If the charge is either too rich or too weak it will burn slowly, the combustion con- tinuing throughout the working and exhaust strokes until the commence- ment of the next suction stroke when the next charge will be ignited as it enters the cylinders, the explosion resulting in burnt products being driven through the open admission valve into the gas and air mains. This may so weaken the next one or two charges drawn into the cylinder that their Card with sharp explosion owing to rich Card with late firing owing to weak . mixture. mixture. FIG. 135. working strokes are spoiled. This method can be successfully used when the engine is working at, or nearly at, full load ; but for engines which have to govern throughout the full range from full load to no load it is not satisfactory. Hit and Miss Governing. In this method the governor allows a full charge of gas to be admitted into the cylinder at normal speed, and cuts oif the gas when the speed exceeds a certain limit, making the engine miss one or two explosions. Only air then enters the cylinder, is com- pressed, expanded and exhausted, by so doing cooling the cylinder and sweeping out the products of combustion. The next explosion is more violent and the turning moment on the crankshaft is very irregular. To keep the speed steady it is necessary to put on heavy flywheels and to run at high speeds on account of the fluctuation in the violence and number of explosions. As regards gas consumption this is the most economical method. A small block having a V groove cut in it is suspended between the end of the gas valve spindle and the striker which is driven from the cam ART. 174] THEORY OF THE GAS ENGINE 291 shaft of the engine. The end of this striker forms a knife-edge which fits into the V groove in the block when the engine speed is not above its normal value and so opens the gas valve. A very small movement of the block results in the striker missing the block on its next stroke and the gas valve is not opened. The block is moved by the governor, and for very close governing the governor decides whether there is to be a " hit " or a " miss " by a very small movement of the block, in some cases the thickness of the knife-edge. The governor itself, therefore, always governs when in one position, and it does not matter if it is very far from being isochronous (Art. 256) ; also, as it has very little work to do, the parts to move being very light and having very little resistance to overcome, a very small governor will effectually control a large engine. Unfortunately, hit and miss governing is scarcely suitable for large engines, owing to the inadvisability of opening a large and therefore heavy gas valve by means of a knife-edge, and to the uneven turning movement which necessitates the use of very heavy flywheels to keep the speed as uniform as desired. Quantity Governing. The quantity method of governing may be divided into throttle governing and cut- off governing. The latter method frequently necessitates some form of trip arrangement which requires a heavy and powerful governing gear, resulting in noisy working and con- siderable wear and tear. The most common methods are to govern by throttling the mixture in the inlet pipe to the engine Cylinder, or else to vary the lift of the admission valve. One objection to all kinds of quantity governing is that very strong springs are required on the admission and exhaust valves, as frequently the pressure towards the end of the suction stroke is as low as 9 pounds per square inch below atmospheric. Gas engine valves invariably open inwards, and a large valve requires a strong spring to prevent its being opened by a suction of 9 pounds ; on the Continent the throttle system is usually used. It should be remembered that practically all large gas engines work on producer gas which is of low heating value and slow to ignite. With weak mixtures combustion is impossible unless the compression pressure is fairly high ; if, therefore, the engine is governed on the quantity method it is important that the compression should be high enough with a full load charge to ensure its being high enough for a low load charge. 174. Study of the Indicator Diagram. Determination of the Laws of the Expansion and Compression Curves from the Indicator Diagram. The expanding gases follow the law pv n = constant ; the problem under discussion is to find the value of the index " n " from the indicator diagram. Let pv n = c Then taking logs we have log p -\-n\Qgv = log c = constant This is the equation of a straight line. If, therefore, this equation be plotted on squared paper and a straight line be drawn lying evenly between the plotted points, the errors of measurement will be eliminated and the value of " n " found. The method will be best illustrated by means of an actual example. 292 THE THEORY OF HEAT ENGINES [CHAP. xm. Fig. 136 shows an average indicator diagram taken from a 25 B.H.P. Campbell gas engine working on suction gas. Diagrams were taken every 10 minutes during a trial lasting 6 hours, from which the average diagram shown in the Fig. 136 was plotted on squared paper. The cylinder was 9-5 inches diameter and the stroke 19 inches, there- fore the piston displacement or stroke volume is 07854 X (9'5) 2 X 19 = i34 6 '7 cubic inches The clearance volume was 272 cubic inches. .44 O .300 .200 .100 \^Expansion Curve pv 1 ' 292 - const. on Curre pr 1 ' 38 * const. In plotting the diagram (Fig. 136) the length AB representing the stroke was made 5 inches, hence i inch represents = 269-34 cubic 3 inches. Set off the vertical OP representing zero volume so that AO represents the clearance volume of 272 cubic inches, i.e. AO = -,. or i -oi inch. The volume of the gas is proportional to the length of cylinder it occupies, hence we may write pin constant where /represents the length on the diagram measured from OP. Next divide the diagram into any number of parts by the ordinates as shown and tabulate the absolute pressures "/ " corresponding to the lengths " /." The results obtained for the expansion curve are as follows : ART. 174] THEORY OF THE GAS ENGINE 293 Pounds per square inch absolute. inches. log/). log/. 273 i'59 2-4362 0*2014 I8 5 2-09 2-2672 0-3201 145 2-1614 0-4133 1 13*5 3-09 2*055 0-4900 93 3'59 1*9685 0-5551 78-5 4-09 1-8949 0-6117 68 4'59 1-8325 0-6618 60 1-7782 0-7067 53 5'59 17243 0-7474 Next plot log/ and log /as shown in Fig. 137 ; the slope of the line 2-4 2-3 22 21 < ^ 2-0 s 1-9 1-8 /'7 \ \ \ \ \ \ \ Sf \ \ 4- \ \ \ \ V \ \ \ \ \ V O-2 0-3 O4 O-5 0-6 O7 FIG. 137. 294 THE THEORY OF HEAT ENGINES [CHAP. xm. so obtained gives the value of " nT Reading from this line we find that for an increase in log/ of (2^43 1 70) or 0*73, the corresponding increase in log /is (0*765 0*20) or 0-565 ' 73 =r 292 hence n = '5 6 5 The law of the expansion curve is therefore /zA-292 constant. In exactly the same way the law of the compression curve is found to be ^z/i'38 = constant. Now this engine was working with a mixture for which y may be taken as 1*38. It is obvious, therefore, that the expansion curve lies above the adiabatic/z^ 1 ' 38 = constant, hence on our assumption that the specific heat of the working gases remains constant,' the gas must be receiving heat during the expansion (Art. 14). But we know that the gases are losing heat to the water jacket, hence if their specific heat remains constant, the only conclusion to be drawn is that combustion takes place throughout the greater part of the expansion stroke, and that more heat is being supplied to the gas during the expansion than is lost to the cylinder walls, that is to say, we have after burning, which has been discussed in Art. 167. The following is a rough method of determining the laws of the expansion and compression curves when the clearance volume is not known. 175. Approximate Method of finding the Clearance Volume and the Index "n" of "pv" from the Indicator Diagram. Let FIG. 138. Fig. 138 represent the expansion curve. On a smooth portion of the curve take any two points i and 3, and choose an intermediate point 2 such that / 2 V/i/'s, or the pressure at 2 is the geometric mean of the pressures at i and 3 Let c clearance (to be found) p l v l = pressure and volume at point i A^2 = > 2 /3 7 '3 = 3 Then, since / 2 = rffaf* .-. == = constant = k say ..... (i) ART. 175] THEORY OF THE GAS ENGINE 295 If " v " is the indicated volume, then Law of curve is p(v -f- c) = constant = b say 1 _!_ or ? = < + "/"" ...... (3) Writing m = - for convenience, we have the volumes at points 1,2, and 3 are Vi = c+b^p^- ...... (4) V2 = c+tiy 1 -m ...... (5) V B = c + b^pz- ...... (6) From (5) and (6) we have * 8 ; 2 = * m (/3~ w A~ w ) ..... (7) From (4) and (5) we have Hence from (7) and (8) But by (i) V^ = A--A- (9) and / 2 = ^j>i .*. Substituting these values of/ 3 and/ 2 in (9) we have ^3 ^2^~ w A~ m A~ m 1 (10) Taking logs we have ^ 2 ^1 ^ = lQg * ....... (II) ^' Equation (n) gives "n" in terms of k = ~=?~ and z/ 3 , v& and z' l5 all of which are known. i To find the Clearance. Plot v vertically and/" horizontally, then by equation (3) above the intercept on the axis of "y " will give the clearance 296 THE THEORY OF HEAT ENGINES [CHAP. xm. " c" as shown in Fig. 139, equation (3) being the equation to a straight line. The Author has tested this method repeatedly on diagrams taken on various gas and oil engines, and found that by careful working the clearance volume can be obtained within three per cent, of the actual measured volume; it is not very reliable, how- ever, on small scale diagrams on account of the errors of observation of the pressures and volumes, and as only three points are taken these errors will not be eliminated as well as in the usual method of Art. 174, where any number can be taken (the more the better). 176. Rate of Heat Reception and Rejection from the Indi- cator Diagram. In Art. 14 it has been shown that the rate of heat . dIL . reception r- is given by dv -/I FJG. 139. y i If y and n are known can be calculated. EXAMPLE. Let the law of the expansion curve be/zA' 479 = 145-5, and the law of the compression curve /z/ 1 ' 304 = 39*36, and let y= 1-37 for the products of combustion, and y = 1*385 for the mixture. Then/i?r the expansion curve dv ~ P 0-37 0-37 This shows that the gas is losing heat throughout the expansion, since - is negative. dv Similarly for the compression curve x 1-385 1-304 _ _ 0-385 Therefore the gas is losing heat throughout the compression as well as throughout the expansion. 177. Method of Calculating the Temperature at any point of the Indicator Diagram. The following method, 1 due to Professor John Perry, is based on the assumption that the specific heats of the gas are constant, and that the gas follows the law : P = R = Cp CT? for a perfect gas or See Phil. Mag., July, 1884. ART. 177] THEORY OF THE GAS ENGINE 297 Knowing the pressure, volume and temperature at any point on the diagram, and the pressure and volume at another point from the diagram, the temperature at that point is easily calculated. The value of ^ is found from the point on the diagram at the end of the charging stroke, just before compression begins. EXAMPLE. The diagram shown in Fig. 140 was taken from a 20 B.H.P. Campbell gas engine in the Engineering Laboratory of the University College, Nottingham. Given, diameter of cylinder 9^ inches, stroke 19 inches, clearance volume 272 cubic inches, barometric pres- sure 15 Ibs. per square inch, temperature at beginning of compression 180 F., C p = 0-24, C v = 0-17, law of expansion curve /z/i-292 ^ = const., calculate : (a) the temperatures at FlG< A and B ; (b) the heat given to or taken from the gases between A and B ; (c) the rate of heat reception between A and B. (a) Clearance volume = 272 cubic inches = 0-157 cubic foot 1720 Piston displacement = - w J/ X = 0779 cubic foot 144 12 .-. Volume at beginning j = + = 6 w f of compression ) From the diagram z; A = 0*157 + 0-023 = 0-180 cubic foot /A = 475 + 15 = 49 Ibs. per square inch absolute ?; B = 0*157 + 0-719 = 0-876 cubic foot *^ B = 50 -j- 15 = 65 Ibs. per square inch absolute _ ' T~~ T A . 15 x 144 X 0-936 490 X 144 X 0-18 1 80 + 460 T A 15 x 0-936 65 x 0-876 640 T B .*. calculated temperature (maximum) at A = 3565 F. B=2i 3 8F. (b) The law of the expansion curve by the method of Art. 174 is found to be /fli'292 = constant. 298 THE THEORY OF HEAT ENGINES [CHAP. xm. By equation (5), Art. 14, the heat added during expansion is v % H = work done X y-i T ^, . , C O*24 In this example y = - = -= 1-41 and work done from A to B =~ foot-pounds n i 490 X 0-18 65 X 0*876 ~~~~~ X 144 1-292 i = 88-20 56-94 x = 31-26 X 144 0-292 0*292 = 15,430 foot-pounds = 19-85 B.Th.U. .*. heat added = 19-85 X 778 1-202 ' . 2 2 Hence heat supplied as during expansion from A to B = 7-75 B.Th.U. dYL 1*41 1-202 (c) Rate of heat reception -7- = p X -- ; - = 0-405^. 2^77 The value of the constant^- can also be inferred from the weight of the gas in the cylinder as follows : Let V g = volume of gas (in cubic feet) admitted per stroke at absolute temperature T^, and absolute pressure P,;. V volume of air (in cubic feet) admitted per stroke at absolute temperature T a and absolute pressure P a . y e == volume of the exhaust gases (in cubic feet) remaining in the cylinder at the end of the exhaust stroke at absolute temperature T ei and absolute pressure P e . Reducing these to N.T.P., i.e. 492 F. absolute, and 2116 pounds per square foot, the total volume in the cylinder at N.T.P. will be g a If w g , w a , We be the densities of the gas, air, and exhaust gases in pounds per cubic foot at N.T.P., the total weight of the contents of the cylinder will be or snce T = 492 and P = 2ii6 ART. 178] THEORY OF THE GAS ENGINE 299 Hence, the volume of i pound of the cylinder contents will be w TO ~~0'2 33 W or 4-2927/0 (3) (4) Hence, for the contents of the cylinder we have pV = 4^92 VQ X T (5) from which the temperature at any point in the cycle can be estimated. The value of T g may be taken as the mean temperature of the exhaust gases which can be mea- n sured directly; the weight w e may be calculated from the proportion of air and gas used. 178. Temperature- Entropy Diagram for the Ideal Otto Cycle. The temperature - entropy diagram corresponding to the indicator diagram (Fig. 126) is shown in Fig. 141. C The line BC, Fig. 141, repre- Q sents the adiabatic compres- Expansion sion, the rise in temperature being BC. The temperature at the instant compression begins, namely at B, may be approximately taken as the H FIG. 141. maximum temperature of the jacket water leaving the cylinder, hence points B and C are fixed on the T diagram. The explosion at constant volume gives the gain of entropy along CD, the curve CD being plotted from the equation 2 fa = C T ^r (Art. 1 6, equation (n)) assuming C v to be constant. In the case of the ideal diagram when the gases expand down to atmospheric pressure, the line DE represents the fall in temperature during the adiabatic expansion, while the line EB represents the constant-pressure atmospheric line plotted from the equation #1 = Q, = C P e ^ (Art. 16, equation (12)) For the ideal pv diagram (expansion down to atmospheric pressure) the T< diagram is therefore BCDE, Fig. 141, and heat produced by the explosion = area FCDG heat rejected to exhaust = area FBEG work done = area FCDG area FBEG = area BCDE . area BCDE Hence efficiency = 3 oo THE THEORY OF HEAT ENGINES [CHAP. xm. When the ratios of expansion and compression are equal, as in the Otto cycle, DH represents adiabatic expansion, DH being the fall of temperature during the expansion, and HB represents the exhaust stroke, while . area BCDH efficiency = ^r-^ ^ area FCDG The actual T< diagram corresponding to an indicator diagram taken from an engine is represented by the shaded area in Fig. 141, since, as already mentioned, the maximum temperature reached in the cylinder falls short of the calculated temperature of combustion, and also the expansion is not adiabatic. 179. Method of drawing the Temperature-Entropy Diagram from the Indicator Diagram. An indicator diagram taken from a 500 4-00 $300 200 100 Df\E 3( 0-157 Volume (cubic feet) FIG. 142. 0-936 25 B.H.P. Campbell gas engine working on coal gas is shown in Fig. 142. The law of the expansion curve as determined by the method of Art. 174 is/zA' 29 = constant, and of the compression curve /z; 1 ' 38 = constant. The temperature at the end of the suction stroke at B is assumed to be 100 C. or 212 F., i.e. 212 + 460 = 672 absolute, and Cp = 0-28, C = 0-201. The probable temperature of the working substance at different points on the indicator diagram is calculated by the method of Art. 177. Starting at point B, where the pressure is 15 pounds per square inch absolute, temperature 672 absolute, and volume 0*936 cubic foot, we have pv T 15 x 0-936 672 = constant = 0*0208 ART. 179] THEORY OF THE GAS ENGINE 301 and the temperature at any point where the absolute pressure in pounds per square inch is/' and volume is v' cubic feet is given by P'v' T' =-^ 5 absolute O-O206 The calculated temperatures at different points on the indicator diagram are shown in the table on p. 302. The entropy of the mixture at point B, reckoned from 32 F., i.e. 492 absolute, is or 0-28 X 2-303 Iog 10 = 0-087 um 't . / r ty L Y i = C v log e ^ (Art. 16 (n)) The results obtained are shown in the table below ; plotting columns 4 and 6 gives the temperature-entropy diagram shown in Fig. 143. 302 THE THEORY OF HEAT ENGINES [CHAP. xm. The above is a long and rather intricate method of drawing the temperature-entropy diagram from the pressure-volume diagram, but is given here because it is worked from first principles only. For practical work Captain H. Riall Sankey's method is preferable. 1 Point on /z/ diagram. Pressure, Ibs. per sq. in. absolute. Volume, cubic feet. Absolute temperature, Change in entropy. Entropy reckoned from 32 F. " 1 B 15 0-936 6 7 2 0*0868 | 1 i 30 0-5IO 735 0-0004 \ o 0864 ' n g "% 2 6 7 0*320 1030 O*OO22 2 o * 0846 O E> /-i CJ "S U I6 5 0*157 1245 0*0032 / J 0*0836 3 240 0-157 1811 0-0749 \ 0^585 rt 3 c 300 0*157 2264 0'120 o P> 0*2036 C/3 +J g "3.1 "> X trc J 400 0-157 3072 0-181 1 0*264 Wg c 8 D 490 0*160 4240 0-246 I rt to 0*329 E 490 o* 190 4476 0*0151 0-344 SJ / 6 | a 330 0*234 3712 0*0013 \ (W 0-345 u c 8 ? 180 0-388 3357 0*0019 0*346 V 8 105 0*621 3134 0*0024 I 0-346 M w M 9 80 0*790 3038 0*0026 / o'347 E / F 65 0*936 2924 0*0029 > o 344 **J2 1 6 > ( 10 rt g \ 40 0*936 1800 0*019 0*240 in B 15 0-936 672 0*260 / ~ 0-087 180. Losses in Gas Engines. The total amount of heat supplied during the explosion may be accounted for in three parts, namely 1. Heat converted into work. 2 . Heat carried away in the exhaust gases. 3. Heat lost during expansion. For the methods by which the above are measured the reader is referred to Chapter XVI. ; for the present we are only concerned with the problem of reducing the losses and thereby increasing the efficiency. Almost all modern gas engines are of the constant volume type discussed in Art. 163, 1 Proc. Inst. Mech. Eng., May, 1906. A direct graphical method will also be found in Proc. Inst. Mech. Eng., Feb. 1908. ART. 180] THEORY OF THE GAS ENGINE 303 in which the ratios of compression and expansion are equal, and con- sidering this type of engine, we have the unavoidable loss in the exhaust gases resulting from incomplete expansion. As we have already seen (Art. 164), Mr. Atkinson reduced this loss by carrying the expansion down to a much lower pressure, but mechanical difficulties caused this engine to be abandoned. The same result would also be obtained by employing compound expansion, in which the gases from the high-pressure or explosion cylinder are exhausted into a larger low-pressure cylinder, exactly in the same way as the expansion of steam is carried out in a compound steam engine. With the modern engine, therefore, the loss due to exhausting with high pressure at release cannot be materially reduced. The very high temperature attained during the explosion results in a very rapid extraction of heat by the cylinder walls. The higher the temperature the more rapid is this loss to the water-jacket through the cylinder walls. Any suitable method, therefore, which has for its object the reduction of the maximum temperature reached in the engine cylinder, without decreasing the mean pressure during the cycle, should be beneficial in reducing this loss and in raising the efficiency. Two methods have been tried in practice, the water injection method and the super-com- pression method. The Water Injection Method. This method, adopted by Messrs. Crossley, consists in injecting a fine spray of water into the engine cylinder during the suction stroke. The water entering in this manner with the air supply does not form a water film on the cylinder walls, but is distributed throughout the cylinder contents in the form of a mist, only a very small quantity of water being necessary. When the mixture explodes the water mist is evaporated into steam, and its latent heat so absorbed prevents the temperature rising too high. The Super-Compression Method. This method, due to Mr. Dugald Clerk, 1 may be described in his own words as follows : " Some time ago it appeared to me possible to reduce the maximum temperature by increasing the charge-weight per stroke given to an engine. I had experimented with two engines, one having a 7 -inch cylinder, 15 inches stroke, and the other a lo-inch cylinder, 18 inches stroke. These engines, which are of the ordinary standard four-stroke type, are allowed to take in the usual charge of gas and air ; then at the end of the stroke a further charge of air or inert fluid is added to increase the pressure in the cylinder to 7 or 8 pounds per square inch above atmospheric before the return of the piston. A small part of the return stroke is, however, made before the pressure can be materially increased, as the added charge takes some time to fill the cylinder. This has the effect of increasing the charge weight present in the cylinder by about 40 per cent, and of increasing the pressure of compression without, how- ever, increasing the temperature of compression. Indeed, in both experi- ments the temperature of compression was diminished. As the charge present is constant so far as gas is concerned, the maximum temperature capable of being produced is much reduced. The maximum temperature shown by the diagrams taken by me from these two engines is about 1200 C. Experiments were made, and it was found that the heat flow 1 James Forest Lectures, Inst. of Civil Engineers (1904). 304 THE THEORY OF HEAT ENGINES [CHAP. xm. was reduced to about two-thirds, and further that the mean available pressure was increased about 20 per cent." It is very fortunate that a reduction in the maximum temperature reached in the engine cylinder results in a higher efficiency, because a lower temperature also results in greater freedom from over-heating and the cracking of cylinders and pistons due to the temperature stresses set up in the material. The higher efficiency thereby obtained results in increased reliability in working, which is more important to the manufacturer than economy of fuel. 181. The Standard Cycle to be used for comparing the Per- formances of Internal Combustion Engines. There is great diversity of opinion upon the best cycle to be used. It has been shown definitely (Chap. XIV.) that the specific heats of gases are not constant, but increase at the high temperatures attained in the engine cylinder. This being so, it can hardly be considered fair to compare the engine with the /jVy-l air standard efficiency given in Art. 163, namely i \- ) . It is equally unfair to compare the engine with the Carnot cycle, in which the efficiency T T* is 1 r p (Art. 21). An unavoidable cause of loss of efficiency lies in the fact that all the heat is not supplied to the working substance at the maximum temperature, but some is supplied between the temperature at the end of compression and T lf the maximum temperature of combustion. As suggested by Professor J. A. Ewing, 1 it seems, therefore, a fairer com- parison to take the ideal standard of performance as that of an engine in which combustion takes place between two defined temperatures and in which the action is reversible in all other respects. (Compare the ideal Rankine cycle for the steam engine, Art. 55.) The ideal efficiency, taking into account the variable specific heat of the working substance, is considered separately in Art. 190. Let T be the absolute temperature of the gases before ignition, i.e. at the end of compression, T 2 the absolute temperature of exhaust, T! the maximum temperature at which, in the ideal case, com- bustion is supposed to be complete, then, assuming the specific heat to remain constant, if SH be a small quantity of heat supplied during combustion over a small range of tempera- ture 8T or T T 2 , the greatest amount of work that can be done per pound of working substance will be or T- and the total work done will be or a(T 1 -T )-C,.T S! log 6 li . * ... (i) x o 1 "The Steam Engine and other Heat Engines," 2nd edition, p. 434. ART. 182] THEORY OF THE GAS ENGINE 305 The total quantity of heat supplied during combustion will be Hence the ideal efficiency will be Heat converted into work Heat supplied T ) The Committee of the Institution of Civil Engineers on the Efficiency of Internal Combustion Engines in their Report l recommend the con- tinued use of the air standard efficiency on account of its simplicity, recognising as they do the uncertainty attached to the values of the specific heats of the gases at high temperatures. (See also Art. 190.) 182. Theories advanced to explain how it is that the Expan- sion Curve on the Gas Engine Indicator Diagram frequently lies above the Adiabatic. Considering the great amount of heat taken from the gases by the water jacket during expansion, it might be expected that the expansion curve will always be below the adiabatic. In many cases, particularly in modern engines in which the maximum explosion tempera- ture is very high (the example worked out in Art. 177, for instance), the curve actually is less steep and lies above the adiabatic. Various theories have been advanced to explain this phenomenon, the following being the principal ones. 1. Action of the Cylinder Walls. It has been suggested that the cooling action of the cylinder walls is sufficient to cause slow combustion. It is very doubtful, however, that this is the case ; but admitting it to be true, this action of the cylinder walls would have the effect of keeping the tem- perature of combustion below the theoretical temperature expected, and also, the combustion not being instantaneous, but gradual, will cause the pressure to be kept up during the expansion on account of the heat supplied during the combustion taking place on expansion. 2. After Burning. Admitting that "after burning" or retarded com- bustion takes place (as discussed in Art. 167), the effect would be the same as the above, only much more pronounced. 3. The Specific Heat of the Gas is not Constant. This is probably the best explanation of the phenomenon. It is well known that the specific heat increases with increasing temperature, and applying this fact to the study of the curve the phenomenon is explained. Fig. 144 shows an indicator diagram on which the adiabatic i has been drawn on the assumption of constant specific heat, and the adiabatic 2 on the assumption of variable specific heat. It will be noticed that the actual expansion curve which follows the law /zA' 461 = constant lies very close to the adiabatic for constant specific heat (in very many cases it lies well above this adiabatic, as before men- tioned). The expansion curve, however, lies very much below the adiabatic 1 Proc. Inst. C. E., vol. clxiii., 1905-1906, p. 241. 306 THE THEORY OF HEAT ENGINES [CHAP. xm. drawn on the assumption of variable specific heat, and on this hypothesis the gases must be losing heat during expansion. The diagram shown in Fig. 144 is taken from the Second Report of the Gas Engine Research Committee, 1 and in that report it is proved that the assumption of variable specific heat shows that in all cases the expansion curve lies below the adiabatic, as would be expected. This theory explains the phenomenon without admitting that after burning or retarded com- bustion takes place. The theory of the gas engine on the assumption of variable specific heat is given in Art. 190. 46% 20 |<- Clearance FIG. 144. 183. Heat Transmission through the Cylinder Walls. 2 The rate of heat-flow from the gas to any part of the walls at any instant depends upon the then temperature density and motion of the gas, and also upon the temperature and condition of the wall surface. It will therefore differ widely at different points of the working stroke, and by far the greater part of the heat-flow will occur in a very short time immediately after ignition and pass into the surface of the combustion chamber and piston and valves ; comparatively little will pass into the barrel of the cylinder, since it is not uncovered by the piston until the density and temperature of the gas have fallen. Mr. Dugald Clerk 3 has found that the average heat-flow 1 Proc. Inst. Mech. Eng., 1901, p. 1031. 2 See also the " Fifth Report of the Gaseous Explosions Committee,' B. A., 1912. 3 Proceedings Royal Soc., A., vol. Ixxvii. (1906), p. 500. ART. 183] THEORY OF THE GAS ENGINE 307 per square foot per second in the first three-tenths of the working stroke is three times that of the average over the whole stroke for equal temperature differences, and he calculates that the actual rate of heat- flow in the first three-tenths of the stroke is six times that of the wtule stroke in ordinary gas engines working at full load. In order that the heat may be conducted away through the walls at the required rate there must be a certain temperature gradient in the metal (Art. 151) and a corresponding mean surface temperature. The cyclic variation in temperature above and below the mean (Art. 68) will not be very large with a clean metallic surface; Professor Coker found a total cyclic variation of 7 C. at a depth of 0*015 inch in the wall of the combus- tion chamber of an engine running at 240 revolutions per minute. The chief problem in the design of large gas (and oil) engines is to prevent the temperature of the walls from rising too high and causing pre-ignitions, and also to prevent the metal from getting overstrained by unequal expan- sion, particularly at places where the metal wall is thick, as for example at the head of an ordinary flat-faced piston ; in this case the central portion will get very much hotter than the edge, which will therefore be put in tension, and to minimise these evils it is essential to cool the piston by water circulation. Radiation from the Gas. 1 The law of radiation given by Stefan and Boltzmann (Art. 158) has been confirmed for gaseous explosions by W. T. David, 2 who found that the rate of loss from this cause varies roughly as the fourth power of the absolute temperature. A large propor- tion of the heat loss through the walls is radiated directly from the hot gas, and it is evident that the rapid reduction in temperature consequent upon expansion results in comparatively little radiant heat reaching the barrel of the cylinder (see above). "An important practical consequence of radiation is the greatly increased loss of heat which occurs when the mean pressure in an engine is increased by increasing the strength of the mixture ; the jacket loss and the metal temperatures are raised in a much greater proportion than the fuel consumption and the efficiency is diminished. In very large engines this sets a fairly sharp limit to the possible output, which is, as a rule, con- siderably less than the maximum of which the engine would be capable if it were given all the fuel it could take. If the load be in excess of this limit the engine overheats rapidly in consequence of the greatly increased heat-flow." 3 The size of the cylinder will also have an effect on radiation, the greater the depth of hot gas (up to a certain value) the greater will be the amount of radiant heat reaching the walls. For this reason the difficulty of de- signing and working large engines is not only due to the greater thickness of metal required but also to the greater flow of heat. Effect of the Density of the Gas. For a given temperature, an increase in density of the gas results in a greater heat-flow to the walls. "The most important practical question connected with the relation between density and heat-loss is the effect of degree of compression on the working and efficiency of gas engines. To put the matter in its simplest 1 See "Third Report of Gaseous Explosions Committee " of the B. A. 2 Phil. Trans. Roy. Soc., A., vol. ccxi. p. 375. 3 " Fifth Report of Gaseous Explosions Committee," loc. cit. 3 o8 THE THEORY OF HEAT ENGINES [CHAP. xm. form we may suppose that the engine has a cylindrical combustion space and flat-headed piston, so that the enclosure containing the gas at the moment of firing is a cylinder. The length of this cylinder will in most cases be a fraction of the diameter, the ratio of diameter to length being of the same order as the compression ratio of the engine. The problem, then, is to determine how the amount and distribution of heat-loss to the walls is altered when the compression ratio of the engine is changed, say, by lengthening the connecting rod. In the ordinary case of a fairly high compression ratio, the effect of this alteration will be to reduce the length of the cylindrical combustion space without changing its diameter, and to keep the mass of gas confined therein substantially constant so that the density goes up in inverse proportion to the length of the space. At the same time there will be a small rise in the temperature of the fired mixture consequent on the higher temperature before firing. This, however, would not be very much, amounting to about 100 C. for an increase in com- pression ratio from 4 to 6. " The average heat-loss per square foot to the surface will increase, but not in proportion to the density. On the other hand, the area over which that loss is distributed is reduced, but again in a considerably less pro- portion than the density. For instance, with an engine of equal stroke bore ratio, having a cylindrical combustion chamber, the result of increasing the compression ratio from 4 to 6 will be to reduce the surface of the combustion chamber by nearly 16 per cent. The density is, of course, increased 50 per cent., and if the heat-loss increases in a greater ratio than the square root of the density, which is almost certainly the case, the effect of this increase of compression would be to increase the total heat-loss, and therefore to diminish the efficiency of the engine relative to the air standard. This in the case supposed would not, of course, lead to any reduction in actual efficiency, because the greater heat loss would be more than counter- balanced by the increase in the efficiency due to increased expansion. But it is clear that if the process were carried sufficiently far the abso- lute efficiency might also be reduced. Some approach to this state of things was found by Burstall when the compression exceeded about 7 1 (Art. 168). " The conclusion gained from practical experience, that there is a point beyond which it will not pay to increase the compression in the gas engine, is therefore in full accord with the results of laboratory experiments on the relation between density and heat-flow. Not only is there a point beyond which increasing compression is not followed by an increase in efficiency, but before that point is reached the flow of heat per unit area is increased to an amount at which trouble will begin to arise on account of the difficulty of cooling. It is sometimes supposed that the difficulties which arise from pre-ignition when the compression is increased too far are due in some way to the rise of temperature of the gas consequent on the high adiabatic compression. It is very improbable, however, that this has much to do with the matter. The real cause of pre-ignition is the over- heating of some part of the interior surface of the metal or of a deposit thereon, due to excessive heat-flow following an increased density. If the metal could be kept clean and cool, compression could be carried to very much higher values than are now used in practice without any danger of 1 "Third Report of Gas Engine Research Committee," Proc* I. Mech. E., 1908. ART. 183] THEORY OF THE GAS ENGINE 309 pre-ignition. The effect of increasing density on heat-loss is, however, a matter on which further experimental evidence is needed." Effect of Turbulence. During the suction stroke of an engine working on the ordinary four-stroke cycle, or during the charging stroke in the two- stroke cycle, when the charge is admitted into the engine cylinder under pressure from a pump, the mixture of gas and air enters with a high velocity, and the resulting eddying or tubulent motion will persist for some time, so that at the moment of explosion there may be still a good deal of turbulence. In consequence of this motion convection will go on more rapidly and the rate of heat-flow increase. Mr. Dugald Clerk has found that in the compression and expansion of air or CO 2 without firing, the engine being simply motored round, the rate of heat loss at a given Ordinary ignition a to b takes 0-037 second ; trapped ignition on third compression ; 1 a' to b' takes 0^092 second ; mixture in both cases i volume gas, 9^3 volumes air z other gases. FIG. 145. line and temperature is greater in the first compression after drawing in the charge than in subsequent compressions when the turbulence has died away. Mr. Clerk also found that the result of damping down the turbulence was to retard the rate of combustion of the gas. Figs. 145 and 146, which are reproduced by permission of the Council from the Fifth Report of the Gaseous Explosions Committee of the British Association, show two of his indicator diagrams which were taken with an optical indicator from an engine of 9 inches diameter cylinder and 17 inch stroke running at 180 revolutions per minute. The engine was fitted with two electric igniters ; one operating at the charge inlet valve at the back of the combustion chamber and the other operating at the side of the cylinder close to the piston. In Fig. 145 the back ignition was used, and in Fig. 146 the side igniter was in operation. The charge was drawn into the cylinder in the 3 io THE THEORY OF HEAT ENGINES [CHAP. xm. ordinary way ; the valves were then tripped the charge being compressed and expanded for two revolutions before firing. b' Ordinary ignition a to b takes 0-033 second ; trapped ignition on third compression ; line a' to b< takes 0*078 second ; mixture in both cases i volume gas, 9-3 volumes air and other gases. FlG. 146. EXAMPLES XIII I- A gas engine works on the ideal Otto cycle with adiabatic expansion and com- pression, receiving and rejecting heat at constant volume. The clearance volume is 272 cubic inches, the cylinder is 9-5 inches diameter, and the stroke 19 inches. At the end of the suction stroke the pressure is 13 pounds per square inch absolute, and the tem- perature of the charge is iooC. Estimate the ideal "air standard" efficiency, and the temperature and pressure at the end of the compression stroke (y = 1*4)- 2. Calculate (a) the ideal efficiency of a gas engine working on the Otto cycle when the compression pressure is 135 pounds per square inch above atmospheric, and (l>) of an oil engine working on the same cycle with a compression pressure of 65 pounds per square inch above atmospheric. Assume the pressure during the suction stroke to be 14 pounds per square inch absolute in each case, and the expansion and compression to be adiabatic with y = 1*38. 3. The engine in Question I receives 0*0836 cubic foot of gas per suction stroke at oC., and 14*7 pounds per square inch absolute, the density of the gas being 0*03 pound per cubic foot, and its calorific value 550 B.Th.U. per cubic foot. The strength of the mixture is I of gas to JO of air by volume, and its temperature at the end of the suction stroke is 100 C., the pressure being 14*7 pounds per square inch absolute. Find (a) Weight of the charge drawn in (take weight of I cubic foot of air at N.T.P. as 0-0807 pound). () Pressure and temperature at the end of compression (take 7 = 1*38). (c) Rise of temperature during explosion (neglect jacket loss and take C v = 0*18) (d) Pressure at the end of the explosion. (e) Temperature and pressure at the end of expansion. ( /) Efficiency of the cycle. ART. 183] THEORY OF THE GAS ENGINE 4. The following figures are taken from the expansion curve of a 100 B.H.P. Hornsby- Stockport gas engine indicator diagram having a clearance volume of 0*6035 cubic foot, and piston displacement 3*1528 cubic feet, the volume v representing the total volume of the gases : v (cubic foot) . 0-6035 07610 0-9188 1-234 1-549 I-865 2' 1 80 2-495 2-810 3-126 3'44i / pounds per square inch abso- lute . . 400 279 229 155 121 92 72 60 5o 42 35 Estimate the law of the expansion curve, i 5. The following data was obtained from a gas engine indicator diagram. Cylinder volume including clearance = 0*54 cubic foot. Atmospheric pressure 15 pounds per square inch absolute j temperature at beginning of compression stroke 150 F. ; pressure at beginning of compression stroke 15 pounds per square inch absolute ; law of expansion curve pv 1 ' 51 = constant. At a point A on the expansion curve the pressure was 245 pounds per square inch absolute, and volume 0*127 cubic foot. At another point B on the curve the pressure was 40 pounds per square inch absolute and volume 0-505 cubic foot. Estimate the temperature at A and B, and find the heat given to or taken from the gases between these two points. (Assume 7 = 1-39.) CHAPTER XIV THEORY OF THE INTERNAL COMBUSTION ENGINE AS- SUMING THE SPECIFIC HEAT A LINEAR FUNCTION OF THE TEMPERATURE 184. Explosion at Constant Volume. If a combustible mixture of gas and air be ignited in a closed vessel, the temperature and pressure of the mixture will rapidly rise, it being possible to calculate their values if certain assumptions are made. Assuming no loss of heat to take place from the vessel and that the specific heat of the gases remains constant, if H represents the number of units of heat evolved during the combustion, W the total weight of the contents of the vessel, and C v their specific heat at constant volume, then H = C, X W X (T! - T 2 ) where T x and T 2 denote the absolute temperatures after and before combustion respectively. The only unknown in this equation is T 1? the absolute temperature reached as a result of combustion. Assuming further that/z^ = RT, the resulting absolute pressure (/) may also be calculated when T is known. It is found that the maximum pressure calculated in this way is always very much greater than the actual pressure observed during experiments ; in most cases the calculated pressure is about twice as high as the actual pressure .observed. Several explanations have been offered from time to time to explain this discrepancy, the chief of which are : 1. The Dissociation Theory. It is well known that when definite chemical compounds such as CO 2 and H 2 O are heated to a high enough temperature, they are reduced to simple gases, and in so doing absorb heat. Bunsen thought that this dissociation or decomposition by heat at very high temperatures accompanied by the absorption of heat, reduced the temperature attained by the combustion of an explosive mixture. Accepting this theory one would expect that the higher the temperature reached with rich mixtures the greater would be the loss of pressure, and that with weak mixtures when lower temperatures and pressures are obtained, the less would be this loss. Examination of the results of experi- ments, however, show that this is not so, the loss of pressure being practically the same with rich as with weaker mixtures. Evidently, then, this theory does not of itself account for what happens. 2. Cooling Theory. This theory assumes that during combustion a 312 ART. 185] VARIABLE SPECIFIC HEAT THEORY 313 temperature is soon reached above which heat is conducted away by the metal walls more rapidly than is generated by combustion, with the result that the greatest pressure .reached is far less than it would be if no heat were lost. If this were the sole explanation, the loss of pressure would not be constant but would vary with the size and shape of the vessel in which combustion takes place, and this is not found to be the case. 3. Increasing Specific Heat Theory. Mallard and Le Chateleir con- cluded from the results of their experiments that the reduction of the explosion pressure and temperature might be produced by the continued increase of specific heat of the gases at very high temperatures. The objection to this theory as the sole explanation is the same as that to the dissociation theory already mentioned. In 1906 l Dugald Clerk carried out a series of experiments in a gas engine cylinder, and came to the conclusion that the apparent specific heat of the working fluid of the internal combustion engine (consisting chiefly of a mixture of N 2 , CO 2 , H 2 O, and O 2 ), when calculated from the first ^ of the stroke, undoubtedly increases between the observed temperatures 300 C. and 1500 C., but tends towards a limiting value at 1500 C.; and further, that the apparent change in specific heat is not entirely due to a real change in specific heat, but requires in addition, continuing com- bustion after the maximum pressure is reached, i.e. after burning, to account for all the facts. 4. After-Burning Theory. As mentioned above, Dugald Clerk sug- gested that combustion was not instantaneous, and that it continued long after the maximum pressure was reached. In a gas engine this would mean that combustion continued right through the working stroke, and that unburnt gas would pass away in the exhaust. Experiment has not proved conclusively that this is so, and it is unusual to find in a good engine more than a very small percentage of unburnt gas in the exhaust. Professor Hopkinson 2 came to the conclusion that even in the weakest mixtures, combustion, when once initiated at any point, is almost instan- taneously complete, and that the specific heat of the products is very much greater at high than at low temperatures. The discrepancy is doubtless due not merely to any one of the above four theories, but to all of them acting together. 185. Internal Energy of Gases at High Temperatures. 3 The physical properties of a gas in chemical equilibrium are completely specified when we know (1) The relation between the pressure and volume at constant tem- perature. (2) The internal energy per unit volume as a function of the temperature and the density. The internal energy of a gas per unit of mass is usually defined as Ct,(T T ) (Chap. I., Art. 6) where T is some standard temperature from which energies are reckoned, and C w the mean specific heat at constant volume between T and T . In all the cases met with in gas engine practice the first of the above relations is, for all practical purposes, 1 Trans. Roy. Soc. 2 Trans. Roy. Soc., 1906. 3 From the " First Report of the Gaseous Explosion Committee of the British Association." 314 THE THEORY OF HEAT ENGINES [CHAP. xiv. given by Boyle's Law, viz. pv = constant. 2 It is usual also to make the further assumption that the product pv is proportional to the absolute temperature, i.e. pv = RT (Art. 4). If this latter assumption is true, then the internal energy is a function of the temperature only (Art. 6) ; if it is not true, then the relation between /, v t and T must be obtained from a knowledge of the internal energy, which will be a function of both temperature and density. So far as the present state of knowledge goes, the energy is to be expressed in terms of the absolute temperature only ; but an important part of future investigation must deal with its dependence on the density, either by direct measurement or by a determination of the relation between/, v t and T at high temperatures. The prediction of the temperature reached in combustion rests upon a knowledge of the energy function. For, subject to corrections for loss of heat, incomplete combustion, and work done while combustion proceeds, the thermal energy of the mixture of steam (H 2 O), CO 2 , etc., after com- bustion is equal to the chemical energy of the gases from which that mixture was formed. The chemical energy can be accurately inferred from the composition of the combustible gases, and, the thermal energy being thus known, the temperature can be calculated from a table of the energy function. The pressure or volume changes resulting from the combustion can be deduced from the temperature by the use of the/, v t T, relations, which again depend upon the energy function. A table of this function, i.e. the internal energy at high temperature, is therefore of very great importance. In order to predict the performance of a gas engine we must know the rise of temperature and pressure produced by the explosion. This is not only the principal factor in the mean pressure developed, it also determines in large measure the mechanical design of the engine and the necessary strength of its parts. In proceeding further to analyse the indicator diagram given by the engine with the object of accounting at each point for the heat which has been put in, a knowledge of this function is again required. The heat accounted for on the diagram is the work which has been done, plus the heat contained in the gas. The latter item can be calculated from the temperature, if the energy function be known. The balance unaccounted for, which it is usually the object of such investigations to find whether in the steam or the gas engine is the heat which has been lost to the walls or has been suppressed owing to incomplete combustion. In fact, the internal energy of the gases at high temperatures plays much the same part in the analysis of gas- engine phenomena as does the total heat of steam in investigating the working of the steam engine. From a table of internal energy it is possible to construct an ideal indicator diagram corresponding to the cycle of operations in use for any given combustible mixture on the assumption that the combustion is instantaneous and complete at the in-centre ; that there is no loss of heat in compression, explosion, or expansion ; and that during expansion the gases are at all times in thermal and chemical equilibrium. These conditions can never be completely realised in practice, but can in theory be approached asymptotically by improvements in design carried on 1 See Witkowski, Phil. Mag., vol. xli. (1896), p. 309. ART. i86] VARIABLE SPECIFIC HEAT THEORY 315 within certain defined limits namely, that the degree of compression and the nature of the mixture are to be unaltered. For example, the heat loss may be reduced by increasing the size of the engine and altering the nature of the cylinder walls, and the attainment of thermal and chemical equilibrium may be promoted by reducing the speed. Such an ideal cycle is, in fact, precisely analogous to the Rankine cycle of the steam engine, in that it takes account of the actual physical properties of the working substance, but leaves out of account such non-essential imperfections as heat loss due to the cylinder walls. It represents an ideal which the real engine may approach indefinitely but can never attain ; and the closeness of the approach is a true measure of the perfection of the engine. The ideal cycle which has hitherto been used in discussing the performance of gas engines is the well-known "air cycle" (Art. 163). This is based upon a special assumption as to the form of the energy functions namely, that it is a linear function at high, as it is known to be at low, temperatures. The specific heat of the working substance is taken to be constant and equal to 19 foot-pounds per cubic foot, or 4/8 calories per gramme molecule (see Introduction, p. xii). Recent researches, however, on the properties of gases at high temperatures have definitely shown that the assumption of constant specific heat is erroneous, and have given sufficient information about the magnitude of the error to show that it is of material importance. They have shown that the air cycle cannot be regarded as equivalent to the Rankine cycle in the steam engine, inasmuch as it does not take account of the properties of the actual working fluid, but postulates hypothetical fluid which has no real existence. It is as though in the theory of the steam engine the total heat of the steam were to be taken as equal to its latent heat, the sensible heat of the water being neglected. This assumption would lead to a simpler formula for the ideal efficiency for the steam engine, but would be erroneous in the same way and to about the same extent as the air cycle formula for the gas engine. If the sensible heat of the steam could be neglected, the Rankine cycle efficiency T . r|1 1 2 (Art. 55) would reduce ^i ~r * i 1 2 'p HP to the Carnot cycle efficiency ^rp -, for no heat would then be necessary to warm the water from the condenser to the boiler temperature and the whole process would become reversible. The closer approximation to the real cycle which is made by taking account of the actual properties of the working fluid in the steam engine, the total heat of the steam instead of only the latent heat, in the gas engine, the true value of the energy instead of that based upon the assumption of constant specific heat though it leads to some complication of formulae, gives compensating advantages of real practical value. It shows the engineer what are the limits to the improvements which can be effected by changes in design or increase in size, and it enables him to judge whether it is better that the lines of development should proceed in such directions or in the direction of radically modifying the cycle of operations. 186. Measurement of the Internal Energy or Specific Heats 316 THE THEORY OF HEAT ENGINES [CHAP. xiv. of Gases at High Temperatures. The experimental work done on this subject may be divided into three classes : (1) Constant-pressure experiments by Regnault, Wiedemann, Withowski, Lussana, Holborn and Austin, Holborn and Henning. The gas was at atmospheric pressure and heated from an external source in these experiments. (2) Constant-volume experiments by Mallard and Le Chatelier, Clerk, Langen, Petavel, Hopkinson, and Joly's determinations with the steam calorimeter. In these explosion experiments the gas is heated by internal combustion. (3) Experiments in which both volume and pressure are varied, the gas being heated by compression. The recent experiments of Clerk and the determinations of the velocity of sound in hot gas by Dixon and others belong to this class. (i) Constant-pressure experiments. The constant pressure experiments have been carried to a temperature of i4ooC. The gas under atmo- spheric pressure flows steadily through a heater and then through a calori- meter where it is cooled. The temperature just before entering and just after leaving the calorimeter and the quantity of heat evolved per gramme molecule of the gas are measured. This quantity of heat, less the work done in the contraction, which is i '98 times 1 the fall of temperature, is the change of internal energy corresponding to that fall. The values of the volumetric heat of air per gramme molecule found by various authori- ties between o and 100 C. are : Calories per gramme molecule. Foot-pounds per cubic foot. IQ'4. Regnault and Witkowski . Swann (at 100 C ). 5*O I9-2 IQ'8 Joly (steam calorimeter) . w 4-98 I 9 7 The Gaseous Explosions Committee of the British Association 2 consider that Swann's value is correct within i per cent. In the case of CO 2 the results obtained by Wiedemann and Regnault were : Wiedemann. Regnault. Increase of internal energy o to 100 C. . . 710 680 0t0200C. . . 1510 1490 ,, 100 to 200 C. . . 800 810 The result of these experiments may be summed up by saying that the volumetric heat of CO 2 at 100 C., taken as equal to the mean volumetric 1 The work done per degree fall in temperature is equal to pressure X change in volume 76 x 13*59 x 981 22,250 4I x I0 X -^rr~ = I '98 calories per gramme molecule. 2 See Second Report of this Committee, 1909, Winnipeg, p. 3. ART. 186] VARIABLE SPECIFIC HEAT THEORY 317 heat between o and 200 C., is between 7*45 and 7-55, and that its rate of increase with temperature is between 0*009 and 0*013, or > roughly, one six-hundredth part per C. For the volumetric heat of CO 2 the Explosions Committee consider Swann's values to be correct to within i per cent. They are as follows 1 : At 20 C. At 100 C. Specific heat at constant pressure (Cp) . Volumetric heat- Calories per gramme molecule (taken as 41 0'202 6'Q2 O'22I 7'76 Foot-pounds per cubic foot 27*4. 1O"7 Holborn and Austin carried the constant-pressure experiments for air and CO 2 up to 800 C. 2 The gas was heated electrically and the tempera- ture measured with a thermo-couple. Similar measurements on steam were made by Holborn and Henning, who subsequently carried the deter- minations for the three gases up to i4ooC. 3 The following table shows the volumetric heats (C.) of air, steam, and CO 2 at 100, 600, and 1 100 C. respectively. The values at 100 C. are derived from the experiments of Wiedemann and Regnault; those at 600 and uooC. are based on the specific heat values given by Holborn and Henning. The corresponding values of y are also shown (y = i + - - V 100 C. 600 C. IICO C. C. y- C. y. C. V. Air ... 4*9 1-404 5-2 1-38 575 i'345 Steam . 6-6 1-30 6-85 1-29 8-50 1-24 C0 2 . 7'5 1-26 9'95 I '20 II'IO ri8 These figures show that the volumetric heat of air increases by about 0*0009, that of steam by 0-0033, and that of CO 2 by 0-0036 per C. over the range 100 to nooC. There is no evidence that the rate of increase is other than constant in the case of air ; but there can be no doubt that the average rate of increase between 100 and nooC. in CO 2 is less than half the rate of increase between o and 200 C. as determined by Regnault and Wiedemann. There is also distinct evidence in these and in other experiments that the rate of increase of the specific heat of steam becomes greater as the temperature rises. Holborn and Henning give the mean value of C p between o and /C. for nitrogen and carbondioxide as 1 Ibid. 2 Wiss, " Abhandlungen der Phys. Techn. Reichsanstalt," 1905. 3 Ann. de Phys.) 23, 1907, p. 809. 318 THE THEORY OF HEAT ENGINES [CHAP. xiv. N 2 , C = 0-2350 -j- 0-000019^ (a straight line). CO 2 , C p p = o'20io -j- 0-0000742^ o'oooooooiS/ 2 (a slightly curved line). (2) Constant-volume Experiments. Extensive researches were carried out by Mallard and Le Chatelier and by Langen. The former experimenters used a cylindrical vessel 17 cm. X 17 cm., whereas Langen used a sphere , surface 40 cm. diameter. The ratio -, was 2*3 times as great in the first as in the second case. The temperatures reached in these explosion experiments vary from 1300 to 3000 C. Temperatures below 1500 C. are, however, obtained by the use of weak mixtures, involving slow burning and large cooling corrections, and but little reliance can be placed on the results. Langen made very few observations on mixtures giving lower temperatures than i5ooC., and takes that as the lower limit of the range of temperature to which his observations apply. The temperature 3000 C. is about that reached in the explosion of hydrogen and oxygen in their combining pro- portions. This is much above the mean temperature ordinarily reached in the gas engine, the upper limit of which may be put at about 2000 C., although 2500 C. or more is occasionally reached locally. The formulae given by Langen as representing the results of his observations are Air, C = 4'8 -j- o"ooo6/ calories per gramme molecule. CO 2 , C = 6*7 -j- o-oo26o/ ,, H 2 O, C = 5 > 9 + ' 2i5' where C is the mean thermal capacity over the range o to t C. Mallard and Le Chatelier, in 1887, published the following results : For CO 2 Cv= 0-1477 + 0*00017 6t H 2 O Ct, 0-3211 -j- 0-0002 19^ N 2 C v = 0-170 -j- 0-0000872^ O 2 Ct; = 0*1488 -J- 0-0000763^ The following table shows the internal energy of various gases. The energy at 800 C. and 1200 C. are based on Holborn and Henning's results, and Clerk's results are given for comparison. The results for the gas- engine mixture are plotted in Fig. 147, on which points obtained by Mallard and Le Chatelier's formula are also shown. 800 D C. 1200 c. 1600 C. 2000 C. Clerk. Holborn and Henning. Clerk. Holborn and Henning. Langen. Langen. Air JC7Q c 840 8,700 1 1, 5OO CO 2 6460 10,880 I7,OOO 2^,3OO H 2 . j 4670 7.930 14,400 19,900 Gas-engine mixture 4250 3840 6900 6,340 9,800 13,200 Ideal gas (C = 4-9). . 34, JO 54< X) 7,350 9,300 ART. 186] VARIABLE SPECIFIC HEAT THEORY To convert the above values to foot-pounds per cubic foot multiply by 3-96. The results in these units are shown plotted in Fig. 148. (3) Clerk's Experiments. 1 These cover about the same range of temperature as those of Holborn and Henning. The gas used was the product of an explosion in a gas engine, and therefore consisted of a mixture of CO 2 , steam, and air. It was first expanded in the ordinary course after the explosion, and was then heated by compression on the next in-stroke of the engine, the valves being kept closed for this purpose. On the next out-stroke the gas was again expanded, then compressed Energy, Calories per gramme molecule. N) * O * & / ^ / y Holbor Clerk Lang Mallat m ne & Henning .9 e I y 7 ^ 'en ... + .e ' v f ^ ~d & Le ChateJ/er. ( A y ^ t^ y^- < v<2 /. x^ X ^ \^ ^X S* ^ 2OO 4-OO 60O &OO IOOO '20O WOO I6OO /8OO 2OOO 2ZOO Temperature, Cerctiyrade. FIG. 147. again, and so on, the valves remaining closed and the engine running on its own momentum. An indicator diagram was taken of the whole opera- tion. The change of internal energy in any portion of a compression stroke, e.g. BC in Fig. 149, is equal to the work done less the heat lost to the cylinder walls ; in an expansion stroke (CD) it is the work done plus the heat lost. The loss of heat comes in as a correction on the work done and was estimated by a comparison of the compression line and the expan- sion line immediately following (CD). The calculation is based on the assumption that the total heat loss from the hot gases during any portion of a stroke is the same in expansion and compression if the mean tempera- ture be the same. In the first compression the temperature of the gas rose to about nooC. (at the point C, Fig. 149), during the first three-tenths of the following expansion stroke (CD), the temperature fell to about 700 C. The work done in this part of the expansion was measured and the heat 1 Pfoc. Roy. Soc., A., vol. Ixxvii. 3 20 THE THEORY OF HEAT ENGINES [CHAP. xiv. loss determined as above was added. Thus the change of internal energy corresponding to the temperature change nooto 700 C. was obtained. SO OOO 35 OOO 20000 5OOO y\& /' 2OO 4-OO 60O BOO /OOO /4-OO /6OO /GOO 200O 22OO Strode Temperature, Cenzigrcude. FIG. 148. The average volumetric heat over this range is, within the errors of experi- ment, equal to the volumetric heat at the mean temperature 900 C., which is by this method determined directly instead of by difference, as is neces- sarily the case when (as in Holborn and Henning's experiments) the whole internal energy change asso- ciated with complete cooling of the gas is measured. The following table shows the internal energy of the mixed gas with which Clerk ex- perimented, calculated from Holborn FIG. 149. and Henning's figures, together with the energy calculated from Clerk's values for the mean volumetric heat. The energies are reckoned from 100 C. and the energies of an ideal gas with constant volumetric heat 4*9 are added for comparison. ART. 187] VARIABLE SPECIFIC HEAT THEORY 321 Temperature C. Holborn and Henning. Clerk. Ideal gas. 400 1580 1720 1470 800 3840 4300 3430 1200 6285 7040 5390 It will be seen that Clerk's values are about 10 per cent, higher than the others. Corrections resulting from further experiments show that Clerk's values given above are about 3 per cent, too high. 1 The Probable Value of C v for a Gas Engine Mixture. For a mixture of i of gas to 9 of air, Burstall gives C v 0-178 + 0*105 From the straight part of Clerk's curve C v = 0*194 -f- 0*051 IOOO IOOO 187. Rate of Heat Reception with Variable Specific Heats. Now, But SH = C v = a and when Substituting for /= o C . - = y = ^ C v ft and R in (2), we have (3) v - (4) 1 " Second Report of Gaseous Explosions Committee," Section G, Winnipeg, 1909, P- 5- Y 322 THE THEORY OF HEAT ENGINES [CHAP. xiv. If the specific heats were constant, i.e. if fc = o, the above equation reduces to = {Py + v if\> ^ e same as ($)> Art. I 4' If the specific heats be expressed in terms of the absolute temperature, i.e. writing C P = a + T and C* = j8 -f- T, R = a ]8. This may be written reception when the expansion or compression follows the law = constant. Let /# w = ^r, say, then, as in Art. 14, v . -- = np dv Substituting this in (3) gives If the specific heats were constant, i.e. if k = o, this reduces to = ~ -- . /, the same as (4), Art. 14. 188. Adiabatic Expansion with Variable Specific Heats. For an adiabatic expansion =- = o, hence from (5), Art. 187, Dividing by pv we get Now, T=f dp . dv kv . kp P + ^+R-* + S ART. 189] VARIABLE SPECIFIC HEAT THEORY 323 Integrating, we get k j3 log^-f a log ^ + -(/^) = constant . ... (8) or j8 log / + a lg v + ^T = constant . . . . (9) Hence the adiabatic equation may be written *.!* ^z/V * = constant ....... (10) or pPv a e kT = constant ....... (n) 189. Reduction of Efficiency when the Working Fluid is changed for one having a Higher Specific Heat. Assuming the specific heat to remain constant, the efficiency will be given by E = i ( - ] wnere y = C?; Now since C p C^ = R TD y i =TT (equation (2), Art. 7) Differentiating with respect to Ct;, we have i-E dE R R(r-E) Or since \fV ^ R ' - (2) l*V"y V^V~ \A X (i) may also be written i-E, ) dCvt. vi E. . , . ' i> g log rj (3) EXAMPLE i. Find the fractional change of efficiency, assuming y=: and r = 5*96, corresponding to a i per cent, increase in C v . 324 THE THEORY OF HEAT ENGINES 0-4 CHAP. xiv. In this case 0*4 X 0-501 0-499 = 0*499 = 0-717 per cent. i. the efficiency would decrease 0*717 per cent, as the result of C v increasing i per cent. 190. Calculation of the Ideal Efficiency of a Gas Engine assuming that the Specific Heat is a Linear Function of the Temperature. The method will be best illustrated by means of an example. We will take the case of an engine working on the ordinary Otto cycle with "hit and miss" governing, rated to give 40 B.H.P. at a speed of 180 revolutions per minute. The following are particulars of the engine * : Cylinder diameter Stroke . . . . Compression space Compression ratio ii J inches 21 inches 407 cubic inches 6*37 inches Cambridge coal gas was used throughout as a. fuel. The following table gives its average compositions : Percentage by volume. O required for combustion Steam pro- duced. CO 2 pro- duced. H . . 47'2 2r6 47*2 CH 4 Heavy hydrocarbons CO 35'2 4*8 7'ic 70-4 22'6 r6 70-4 i6'o 35*2 14*4 7*15 N 5 -A O - 2S lOO'OO I2CX2 133'6 5675 The higher calorific value varies between 630 and 680 B.Th.U. per standard cubic foot, the lower value between 570 and 620. From anemometer -experiments it is found that, with a medium jacket temperature and with the engine exploding every time, the volume of mixed gas and air taken in is 0*85 of the stroke volume. If we assume a normal barometer (14*7 pounds per square inch absolute) and an outside tem- 1 The information in this article is taken by kind permission from Professor Hopkinson's paper on the "Thermal Efficiency of Gas Engines," Proc. Inst. Mech. Eng., April, 1908. ART. 190] VARIABLE SPECIFIC HEAT THEORY 325 perature of 15 C, the quantity taken in (reckoned in standard cubic feet) is 0-85 X fff = 0-805 of the stroke volume. Thisls mixed with the contents of the compression space and possibly also with some of the exhaust products which have backed in from the exhaust pipe. The volume of the compression space is o'lSj of the stroke volume ; the pressure of the gases is atmospheric and their temperature may be taken as that resulting from the nearly adiabatic expansion which took place at release. The release pressure is between 50 and 55 pounds per square inch absolute according to the strength of the mixture. We may assume 52 pounds. The volume at release is 0*90 times the total cylinder volume. Assuming that the suction temperature was 100 C., or 373 absolute, we may find the temperature (T) just before release, using the relation T! T 2 (52 X M4) X 0-90 (147 X 144) X i .*. T = 1190 absolute The expansion down to atmospheric pressure, which occurs very rapidly after release, reduces this temperature to 1-35-1 1-85 119 X \-~ assuming y = 1-35 0-26 = 860 absolute, or say 600 C. Assuming that this temperature does not change materially during the exhaust stroke, it follows that the contents of the compression space amount to 0-187 X dr^= '6 of the stroke volume, reckoned in standard cubic feet. 800 The total cylinder contents 'at the end of the suction stroke would therefore, at standard temperature and pressure, occupy 0^805 -j- o'o6 = 0-865 f tne stroke volume In what follows it is assumed that the gas has a calorific value of 600 B.Th.U. per standard cubic foot, the products of combustion being cooled to a temperature of 100 C. At the end of the suction stroke the valves are all closed and the cylinder is then full of the mixture of coal-gas and air (assumed to be dry) which has been drawn in plus pro- ducts of the previous explosion amounting to 7 per cent, of the whole, the temperature being 100 C. and the pressure 147 pounds per square inch absolute. The mixture is compressed adiabatically and is fired at the in- centre, the combustion being complete and instantaneous. The products of the combustion are then expanded without loss of heat to the out-centre, 326 THE THEORY OF HEAT ENGINES [CHAP. xiv. when the exhaust-valve is opened. The efficiency is calculated for two mixtures, of which the following are particulars : Volume of coal-gas taken per suction (cubic feet at external tempe- rature and pressure) Percentage of coal-gas in mixture drawn in Percentage of coal-gas in mixture in engine o'l 9-4 0-13 I2'2 Products of combustion of I cubic foot of the mixture. Steam , C0 2 . NandO Total A 0-125 0-053 0793 0-971 B 0-163 0^069 0732 0-964 The above analysis of the products of combustion is calculated from the average composition of the coal-gas. The internal energy curves (Fig. 150) have been calculated from these compositions, using the following values for specific heats : Temperature. 800 C. 1400 C. 1900 C. Air . IQ'Q 22'O H 2 O 26'O 3I'O 2 3 5 3Q"6 CO. 3T2 A\'A jy w 46'! The figures are the mean values of the specific heats at constant volume up to the temperature in question expressed in foot-pounds per standard cubic foot of the gas. Those at 800 and 1400 are the results of Holborn and Austin 1 and of Holborn and Henning 2 obtained by external heating at constant pressure. These are probably correct within 3 per cent. The figures at 1900 are from Langen's explosion experiments; 3 they are probably rather too high because of incomplete combustion and loss of heat, but they are the best available. The values are those given by Clerk 4 for mixed gases ; they are rather higher than those calculated from the above figures at 800 and 1400. It is most convenient to follow what happens to a standard cubic foot of the mixture in passing through the engine. The mixture contains 0-805 0-865 = 0*93 of its volume of gas and air, the rest being products of combustion. Starting at 100 C. or 373 absolute it is compressed adiabatically 6-37 times. The temperature after compression is 373 X (6'37)' 4 = 780 absolute. Rise in temperature during compression = 780 373 = 407 C. Work done during compression = 19 X 407 = 7 700 foot-pounds, 1 Researches of the Reichsanstalt^ vol. iv., 1905. 2 Annalen der Physik, vol. xxiii., 1907. 3 Zeitschrift des Vereines Deutsches Ingenieure, vol. xlvii., 1903. 4 " On the Limits of Thermal Efficiency in Internal Combustion Motors," Proc. Inst. Civ. Eng., vol. 169, p. 121. ART. 190] VARIABLE SPECIFIC HEAT THEORY 327 since the thermal capacity is constant and equal to 19 foot-pounds per cubic foot. This is the internal energy at the end of compression with either mixture. The pressure at the end of compression is 14*7 X (6'37) 1 ' 4 = 196 pounds per square inch absolute I. Strong Mixture (B). In this case 12-2 per cent, of the mixture drawn in is coal-gas. In the mixture as it exists in the engine (after mixing with the products of the previous explosion) the percentage of coal-gas is 12*2 X 0*93 = 1 1 '4 per cent. The heating value of the gas per standard cubic foot is therefore 0-114 X 600 X 778 = 53,000 foot-pounds Add to this the work of compression = 7,700 foot-pounds Internal energy after explosion = 60,700 foot-pounds After explosion the standard cubic foot of mixture becomes 0^964 cubic foot of products. .*. internal energy per cubic foot of products reckoned from 100 C. is 60,700 ^ = 63,000 foot-pounds From the curve (Fig. 150) the corresponding temperature is 2210 C. or 2480 absolute and the pressure is '964 X g Q X 196 = 600 pounds per square inch absolute. The expansion curve is computed by trial and error. We assume an expansion curve of the form pv n constant. The true adiabatic will not be of this form because the specific heat is not constant (Art. 188). But if n be so chosen that no heat is lost on the whole during expansion, the loss in the first portion being balanced by an equal gain in the second, we shall have a sufficiently close approximation to the real adiabatic. If we take n = 1*20 the temperature at the end of expansion is (6^20 = *7 1 3 absolute, or 1440 C. From the curve (Fig. 150) the internal energy at this temperature is read off to be 33,700 foot-pounds, and the loss of energy in expansion is therefore 63,000 33,700 = 29,300 foot-pounds The work area under this curve is most simply computed by noting that it is the adiabatic of a gas for which y is constant and equal to 1-20, and for which the specific heat is therefore n foot-pounds per pound ((2), Art. 7) = -~ - X 0-0807 (since i standard cubic foot of air weighs 0*0807 Ib.) = 387 foot-pounds per standard cubic foot. 328 THE THEORY OF HEAT ENGINES [CHAP. xiv. The fall of temperature during expansion is 2480 1713 = 767 C. .*. work done during expansion = 38*7 X 767 = 29,700 foot-pounds per standard cubic foot of products. This work done is slightly more than the loss of energy (29,300 foot- pounds), showing that along this assumed expansion line there must be some gain of heat on the whole. If the index 1-21 be tried, corresponding to an average specific heat of 36-9 foot-pounds per standard cubic foot, 70,000 60 ooo - / /'' 2 so.ooo - Duguld Clerk'* r &=< 12*1 %CoaL- fr*-^/t /"' 2, i o 40,000 - S\r ^/> s 9-4% a vL-gan ! S ^ > .-" .** *^ 2O.OOO - - \ <**** S' -< -V'J9J[A k) 10,000 . ^ .^ 1>2 = constant. As already pointed out, this is not an adiabatic expansion line, but possesses the property that ART. 191] VARIABLE SPECIFIC HEAT THEORY 329 no heat is lost or gained in the course of it, the loss of energy in the early parts being balanced by an equal gain in the latter parts. The true adiabatic for which n is an increasing quantity, at first greater and afterwards less than 1-20, will be at first above the assumed line, will cross it, and will finally be below it. The final temperature after true adiabatic expansion will be less than after the assumed expansion. Since no heat is lost to the walls (on the whole) in either case, it follows that the work area under the true adiabatic line must be slightly greater than under the line /z/ 1 ' 2 = constant. Thus the efficiency calculated above is a little lower than that of an engine using real adiabatic expansion, but it is easy to prove that the difference is appreciable. II. Weak Mixture (A). It is unnecessary to go through all the steps of the calculation with the weaker mixture. The following are the results which should be worked out by the reader following the above method : Compression work (as before) = 7,700} f oot .pounds per Heat in gas = 0-094 X 0-93 x 600 X 77 = 40,700] cubic foot of mix _ Internal energy after explosion = 48,400) ture from 100 C. Energy per cubic foot of products = ^. >4 I = 49,800 foot-pounds Corresponding temperature from curve (Fig. 150) = i94oC. = 2210 absolute. Assuming the expansion curve /z; 1 ' 24 = constant, the final temperature is 1418 absolute or ii45C. The energy (from the curve) is then 24,000 foot-pounds. Loss of energy = 25,803 Work area under expansion curve =32-3X792 = 25,600 Work of expansion per standard ) cubic foot of mixture \ = 2 S' 6o X '97S = *4,95 Net work = 17,250 Heat supply = 40,700 Efficiency = 42-4 per cent. The difference between the efficiencies with the two mixtures is mainly due to the fact that a greater rise of temperature, and therefore of pressure (in proportion to the fuel used) is obtained when exploding the weak than when exploding the strong mixture. The temperature rises are 1700 and 1430 respectively, and are in the ratio 1-19, but the amounts of fuel supplied are in the ratio of i -30. The pressure falls rather more rapidly in the adiabatic expansion of the weaker mixture, but the difference in this respect is not very material. The determining factor is the initial pressure produced by the explosion (see Art. 168). 191. By Wimperis's Formula. Mr. Wimperis 1 in his book on the internal combustion engine deduces the following expression for the ideal efficiency of a gas engine assuming the specific heat to be a linear function of the temperature (Ct, = j3 x -(- sT). 1 See " The Internal Combustion Engine," p. 85. Constable & Co. 330 THE THEORY OF HEAT ENGINES [CHAP. xiv. where 77 = air standard efficiency T 2 = maximum absolute temperature (Centigrade) T = suction temperature (absolute) and j8 x and s are the constants in the equation C v = P I + s1\ Using the value given by Dugald Clerk for an average working mixture for C V) namely C v = 0-194 + 0*051 - (Art. 186), this equation (i) becomes 1 .... (2) The air standard efficiency is 0'4 = i '477 = 0-523, or 52-3 per cent. Taking the same values for T 2 and T as in Art. 190, namely, T 2 = 248o absolute and T = 373 absolute, the efficiency becomes for the strong mixture E = 0-523) i yJjj (1 0-523 2480 + 373) } 7000 = o-523{i 0-2223} = 0-523 X 0-777 = 0-406, or 40-6 per cent. For the weak mixture, in which T 2 = 2210 absolute, E = 0-523} i 7^0 (i 0-523 2210 + 373)} =0-523 1- ^t 37 - 3 ! *( 7000 ) = o523{i 0-204} = 0-523 X 0-796 = 0-416, or 41*6 per cent. 1 At the absolute zero (T = o or / = - 273), C v = , Using C v = 0-194 + 0-051 where / = - 273 C. C. = o, 94 - = 0,80 /3, = O'lSO and JL _ '* 1 x JL.-^l = Ji- approximately 2j8 t ~ 2 X 0-18 A 1000 360 7000 F CHAPTER XV THEORY OF THE OIL ENGINE IQ2. The Diesel Engine. The Diesel engine has been developed, both as a four-stroke and a two-stroke cycle engine, exclusively as an oil engine using crude oil, and works on a cycle which approximates to the ideal constant pressure cycle already considered in Art. 165. The development of this engine is proceeding so rapidly that, although at present the four- stroke cycle engine may be regarded as the standard type for stationary Adiabatic Compression FIG. 151. plants up to about 600 horse-power, and the two-stroke for larger powers, it is probable that in the near future the double-acting two-stroke cycle may become the standard type for all powers. The ideal cycle considered in Art. 165 assumed complete adiabatic expansion down to atmospheric pressure. The Diesel engine differs from this cycle inasmuch as the expansion ends when the volume is the same as that at which compression begins, as shown in Fig. 151. 332 THE THEORY OF HEAT ENGINES [CHAP. xv. Let the conditions be at A p\v{\, at B / 2 7' 2 T 2 , at C / 3 ' 3 T 3 , and at D 4 T 4> then Heat received at constant pressure = Q/T 3 T 2 ) Heat rejected at constant volume = C/T 4 T x ) Heat converted into work = C/T 3 T 2 ) C V (T 4 Tj) TT- Q TT T) The mechanical efficiency is then the usual relation ,' ' ' The difference between the indicated and brake horse-power is known \\bodBloch. I FIG. 158. as the mechanical loss, and includes the negative loop of the indicator diagrams and also the negative work done when the engine takes no gas. The following method, devised by Mr. Morse, of obtaining the mechanical efficiency of a multi-cylinder engine eliminates the necessity of using an indicator. It consists in running the engine fully loaded with all the cylinders working. The load is put on by means of a Prony brake clamped to the flywheel or brake-wheel, carrying a dead weight which is partially supported by a spring balance. The spring balance reads the excess of the dead weight over the brake load in the usual manner, and small changes in the brake load may be very accurately read. In making the test, one cylinder is stopped from firing and the pressure on the brake blocks is reduced until the engine is again running at its normal speed. The reduction in brake load is then read off, and is approximately that ART. 200] TESTING OF INTERNAL COMBUSTION ENGINES 343 * corresponding to the indicated power of the cylinder which has been cut out. The other cylinders are treated in succession in this manner, and by adding the results the indicated horse-power of the engine is determined. 200. Fuel Consumption. The most reliable method of measuring the gas consumption of an engine is to pass the gas into a graduated gas- holder, from which it is drawn by the engine. This is more accurate than trusting to the readings of a gas meter, but either method may be used, depending upon the object of the trial. For a complete thermodynamic trial the gas-holder should be used, whilst for a commercial test the gas meter would probably be most suitable. The temperature and pressure of the gas should be taken, so that the volume used may be reduced to standard temperature and pressure, i.e. oC. or 32 F. and 760 mm. of mercury or 29-92 inches of mercury. The commercial test is often given in terms of cubic feet of gas per brake horse-power hour at 60 F. and standard atmospheric pressure, since this temperature is about the average at which the gas would be delivered to the engine in practice. A trial of one hour's duration should be ample, and if the engine has settled down to its working conditions, half an hour or even less should suffice to measure the gas consumption and indicated horse-power, etc. In the case of a gas suction plant the coal consumption per horse- power hour is required, and this will necessitate a trial under steady load of at least six hours' duration. The gas producer should be filled up at the commencement of the trial and weighed amounts of coal added at regular intervals, the producer being as full at the end as at the commencement of the trial. In the case of an oil engine trial the oil pump suction may be connected directly to the oil tank. This tank should have a narrow neck fitted with a gauge hook. At the commencement of the trial the point of the hook should be just at the surface of the oil, and the weight of oil which must be added during the trial so that the level of the oil is the same at the end will give the oil consumption during the trial. EXAMPLE i. The following particulars were obtained during a trial on a 25 brake horse-power Campbell gas engine in the heat engine laboratory of University College, Nottingham : Duration of trial, one hour; total revolutions of engine = 13,602 ; total number of explosions, 4620; nett load on brake, 277 Ibs.; mean effective pressure on piston, 106 Ibs. per sq. in.; gas consumption as registered by meter, 455-5 cubi c feet; lower calorific value = 592 B.Th.U. per cubic foot at N.T.P. ; pressure of gas, 771 mm. ; temperature of gas passing through meter, i5C. ; diameter of cylinder, 9^ inches; stroke, 19 inches; effective circumference of brake, 12*8 feet. Work out (a) the indicated horse-power; (b) brake horse-power; (c) mechanical efficiency ; ( C = ml, O 2 = 14 8 ^, N a = 80 9 , Temperature of exhaust 492 F,. Temperature of engine room 60 F. Cooling water per minute 43-0 Ibs. Inlet temperature 49 F. Outlet temperature i26'5F. The items to be worked out are per Ib. of oil : (1) Heat converted into work in engine cylinder. (2) Heat rejected to cooling water. (3) Heat rejected in exhaust. (4) Unaccounted for. (1) Heat converted into work. Oil per I.H.P. hour = ^f 126-6 .*. i Ib. of oil gives - = 3*119 I.H.P. hours. 40-55 = 3-119 X 2545 B.Th.U. = 7938 B.Th.U. (2) Heat rejected to cooling water per minute = 43(126-5 49) = 3332 B.Th.U. Oil used per minute = = 0*676 Ib. oo /. heat rejected to cooling water per Ib. of oil = ~, fi = 4929 B.Th.U. (3) Heat rejected in exhaust. To estimate this approximately we require the air supplied per Ib. of oil burned as follows. By equation (3), Art. 144, air supplied per pound of oil = X 85 = 48*2 pounds OO *^ T" O 346 THE THEORY OF HEAT ENGINES [CHAP. xvi. .-. weight of exhaust gases per pound of oil 49*2 pounds (approxi- mately). Assuming the specific heat of the exhaust gases to be 0-25, we have heat rejected to exhaust per pound of oil 49-2 x 0-25(492 60) = 53H B.Th.U. The heat balance will therefore be : B.Th.U. Per cent. Heat in I pound of oil ... . IQ, 3OO lOO'O Heat equivalent of I. H.P 7,0-38 41*12 Heat rejected to cooling water A Q2Q 2S" Total C0 2 0-337 H 2 O from 0*23 cubic foot of H 2 ..... 0-23 cubic foot 0*036 CH 4 ..... 0*072 Total H 2 N 2 from i cubic foot of gas ....... '433 cubic foot 1*34 cubic feet of air = 1-34 X T 7 73Q ,, dry air 8,460 entering gas 8,260 jacket water supply 109,345 Total . . 668,345 Heat carried away per hour above 32^., B.Th. U. By dry exhaust gases 231,250 ,, steam in exhaust gases 102,460 ,, jacket water 279,600 Unaccounted for, radiation, errors of observation, etc. 55,035 Total . . 668,345 EXAMPLES XVI 1. The following particulars were obtained from a trial of a four-stroke cycle oil engine : Duration of trial, 40 minutes ; oil used, 12-80 Ibs. ; total revolutions, 8142 ; jacket water, 738 Ibs. ; rise of temperature of jacket water, 74 F. ; mean effective pressure in cylinder, 96 Ibs. per square inch ; torque due to brake load, 786 Ibs. -feet ; lower calorific value of oil, 17,000 B.Th.U. per Ib. ; area of piston, 113 square inches; stroke, i8J inches. Find (a) the indicated and brake horse-powers ; (b) the oil used per I.H.P. and per B.H.P. per hour ; (c] the heat converted into indicated work per minute; (d) the heat rejected by the jacket water per minute ; (e) the heat lost by friction, exhaust gases, etc., per minute. 2. In a test on an oil engine running at half full load on Russolene the oil used was 6'8 Ibs. per hour, the I.H.P. was 8-33, jacket water, 935 Ibs. per hour, and its rise of temperature, 67-4 F. The total air sent into the engine per hour was 301-8 Ibs. The exhaust gases per hour consisted of 21-4 Ibs. of CO 2 , 8-05 Ibs. of H 2 O, and 279-15 Ibs. of air. The temperature of the air in the engine-room was 5IF., and of the exhaust gases 531 F. The lower calorific value of the oil was 18,000 B.Th.U. per Ib. Calculate : (a) Heat converted into work per hour ; (b) heat rejected in jacket water per hour ; (c) heat rejected in exhaust gases per hour ; (d) heat unaccounted for per hour. Given. Specific heat of CO 2 = o'2i6, of H 2 O = 0-480, of air 0*238. 3. The following particulars were obtained from trials of a four-stroke cycle oil engine: Cylinder diameter, 12 inches; stroke, i8J inches; dia. of brake wheel, ART. 204] TESTING OF INTERNAL COMBUSTION ENGINES 353 9 ft. of in. Draw to a base of B.H.P. curves showing mechanical efficiency and oil used per B.H.P. hour. Average revs, per min. . . . Brake load (Ibs.) 199 221; 202 IQC 203 ic6 204 117 204 60 205 o Mean effective pressure (Ibs. per so. in.) IO7 Q7 87 74 60 42 2O*7 16-3 I7'C IO'7 lO'O 0/C (L. U.) 4. The following data were obtained during a trial of a gas engine : Duration of trial, one hour. Total revolutions of engine 13,600 Total number of explosions Net load on brake Mean effective pressure on piston . . Gas consumption as registered by meter Lower calorific value of gas at N.T.P. Pressure of gas passing meter . . . Temperature of gas passing meter . . ,240 350 Ibs. 1 10 Ibs. per square inch 625 cubic feet 550 B.Th.U. per cubic foot 770 mm. 17 C. Diameter of cylinder, 9^ inches ; stroke, 19 inches ; effective circumference of brake wheel, I2'8 feet ; clearance volume, 272 cubic inches. Estimate : (a) Indicated horse- power ; (b) Brake horse-power ; (c) Mechanical efficiency ; (d) Thermal efficiency (on I.H.P.); (e) Overall efficiency; (/) Efficiency ratio referred to the "Air Standard" cycle engine. 5. The following data were obtained from a gas engine trial: Brake horse-power, 20-9 ; gas consumption per hour, 331 cubic feet at N.T.P. ; lower calorific value of gas, 561 B.Th.U. per cubic foot at N.T.P. ; higher calorific value, 622 B.Th.U. per cubic foot at N.T.P. ; volumetric analysis of the gas, per cent., CH 4 3373, C 2 H 4 474, H 2 41-29, CO 7-13, N 2 10-22, CO 2 2-62, O 2 0-27. Temperature of air in engine-room, 48-4 F. j difference between wet and dry bulb thermometers, 3 F. ; volume of air supplied per hour measured by anemometer, 3209 cubic feet ; atmospheric pressure, 14-58 Ibs. per square inch. Estimate: (a) Overall efficiency of the engine. (b) Heat carried into engine by the air supply per hour. ( f ) > g as 6. The analysis by weight of a sample of crude petroleum used in an oil engine gives C 85 per cent., H 2 13*5 per cent., incombustible matter 1*5 per cent. The volumetric analysis of the exhaust gases gives CO 2 7 per cent., O 2 11*3 per cent., and N 2 81-7 per cent. The engine uses 0*333 Ib. of oil per l.H.P. hour, and 14-8 Ibs. of water per I.H.P. per hour pass through the jacket. The rise in temperature of the jacket water is 52 C. The temperature of the air at the end of suction is 30 C., and the temperature of the exhaust is 384 C. The lower calorific value of the oil is 10,720 C.H.U. per Ib., and the specific heat of the exhaust gases 0*25. From the above data draw up a heat balance for the engine. (L. U.) 2 A CHAPTER XVII STEAM ENGINE AND BOILER TRIALS 205. Steam Engine Trials. In commercial tests it is generally sufficient to measure the steam consumption per I.H.P. or per B.H.P. hour, but for a complete thermodynamic trial it is necessary to measure the losses in addition, and also to draw up a heat account. The measure- ments necessary to determine the thermal efficiency and to draw up the heat account are : 1. Indicated horse-power. 2. Brake horse-power (if possible). 3. Rate of steam supply by weight, i.e. pounds of steam per hour. 4. Dryness fraction of the steam on the boiler side of the engine stop valve. Or 40. Temperature of the steam if superheated steam is used. 5. Pressure of the steam at the engine stop valve. 6. Temperature (or pressure) of the exhaust steam. 206. Indicated Horse-power. When proper precautions are taken it is possible to estimate the I.H.P. of a steam engine with great accuracy. It is essential for accurate results that the pressure inside the indicator cylinder should be the same as the pressure inside the engine cylinder at all points of the stroke, otherwise the diagram will not be a true representation of the variation in pressure on the engine piston. In order to ensure this, the connecting pipe between the indicator and engine cylinder should be as short and straight as possible and of large bore. The indicator should also be fixed vertically to reduce the possibility of water lodging in the indicator cylinder and connecting pipe. In the case of double-acting engines a separate indicator should be used for each end of the cylinder, and before taking a diagram the steam should be allowed to blow freely through in order to clear out as much as possible of the water which may have collected in the pipes. Considerable errors will jesult in the estimated mean pressure if the motion of the indicator drum be not a true representation of the motion of the engine piston. For this reason an accurate reducing motion should be used, 1 and the spring inside the drum should be sufficiently strong to prevent any backlash. The indicator spring should be tested for accuracy when under steam by direct comparison with a mercury column. When taking a diagram steam should be first blown through for about 1 See "Testing of Motive Power Engines," Royds. Longmans, Green, & Co. 354 ART. 206] STEAM ENGINE AND BOILER TRIALS 355 one minute, then the indicator cord coupled up to the reducing gear and the diagram taken, the pencil being pressed lightly against the paper for about twenty seconds. The atmospheric line should then be drawn and the indicator cord uncoupled ; the diagram may then be measured up for mean pressure by means of a planimeter or by the method of mean ordinates. The indicated horse-power is then calculated in the usual way. Let di = diameter of engine cylinder in inches. d% = piston rod Pi = mean effective pressure on the head end of the piston in pounds per square inch. p^ = mean effective pressure on the crank end of the piston in pounds per square inch. / = stroke in feet. n =5 revolutions per minute. Then the indicated horse-power will be On the head end I.H.P. = 33,000 On the crank end of the piston the effective area will be and the I.H.P. developed on the crank end will be A X -(4 B.ThU. and efficiency of boiler = - = 0*7653, or 76*53 per cent. From the above it will be seen that in this particular case the actual efficiency of the boiler is 76-53 per cent., or i per cent, higher than that obtained when the heat balance is drawn up on the total heat in i pound of dry fuel. Neglecting the water vapour in the air supply the efficiency will be 76*32 as against 75*62 per cent. EXAMPLES XVII 1. The diameter of a steam engine cylinder is 40 inches, and of the piston rod 5 inches, and the stroke is 5 feet. The mean effective pressure on the head end of the piston is 40 pounds per square inch, and on the crank end 42 pounds per square inch. If the speed of the engine is 120 revolutions per minute, what is the indicated horse- power ? 2. A locomotive has two double-acting cylinders supplied by steam, the admission pressure being 150 Ibs. per square inch absolute, and the exhaust pressure 18 Ibs. per square inch absolute. The cylinder diameters are 17 inches, stroke 26 inches, and the diameter of the driving wheels 6 feet. Find the tractive effort and the indicated horse- power when running at 40 miles an hour, the cut-off being 0*4. Allow a diagram factor of o - 9 and a mechanical efficiency of 80 per cent. ART. 214] STEAM ENGINE AND BOILER TRIALS . 369 3. In some trials with a triple expansion engine at constant boiler pressure and number of expansions, it was found that the steam used could be expressed by W = 678 + 7'i6R, and the mean effective pressure reduced to the L.P. cylinder by M.P. = 27-4 O'O42R, where W = Ibs. of steam per hour; M.P. = mean pressure in Ibs. per square inch ; R = revolutions per minute. Find the speed at which the steam used per hour per I.H.P. is a minimum. State the I.H.P. and the steam used per hour at this speed. Diameter of L.P. cylinder 23 inches, stroke 18 inches. (L. U.) 4. In a trial of a jacketed engine the steam chest pressure was 145 Ibs. per square inch absolute, the cylinder feed was 29 Ibs. per minute, and the jacket feed was 3'2 Ibs. per minute, the feed and jacket steam being 3 per cent. wet. The circulating water was 550 Ibs. per minute, inlet temperature 55 F., outlet temperature IO4'3F. The feed temperature was 125 F., and the I. II. P. no. Draw up a heat balance and find also the thermal efficiency. (L. U.) 5. The following data were obtained from a trial on a steam engine : Air pump discharge per hour 6417 Ibs. Weight of steam used in jackets per hour 1079 Ibs. Temperature of jacket drainage 352 F. Pressure of steam at boiler side of stop valve (Ibs. per sq. in. absolute) . 139*0 Moisture in steam ,, ,, ,, (dry saturated) . . . nil. Temperature of exhaust steam U9F. Indicated horse-power 494'3 Circulating water per hour 87,300 Ibs. Inlet temperature of circulating water 33 '2 F. Outlet 91-6 F. Draw up a heat account for this engine and in addition calculate (a) Steam consumption per I.H.P. hour. (b) Thermal efficiency of the engine. (c) Heat theoretically required per minute per I.H.P. by an engine working on the Rankine cycle between the above temperatures. (d) Efficiency ratio or coefficient of performance. 6. The flue gases from a boiler pass around the tubes of an economiser. The temperature of the gases entering the economiser is 3I5C., and on leaving it 149 C. The amount of feed water passing through the tubes is 90 pounds per minute ; the temperature of the water entering the economiser is 38 C., and on leaving H5C. If the boiler evaporates 10 pounds of water per pound of coal, find approximately the weight of air supplied per pound of coal burned. Assume that the specific heat of the gases is 0*25. 7. In a boiler trial 3600 pounds of coal were consumed in 24 hours. The weight of water evaporated was 28,800 pounds, mean steam pressure by gauge 95 pounds. The coal contained 3 per cent, of moisture and 3*9 per cent, of ash by analysis. Determine the efficiency of the boiler and the equivalent evaporation from and at 212 F. (i) per pound of dry coal, (2) per pound of combustible. Feed temperature 95 F., total heat of I pound of steam at no Ibs. per square inch absolute, 1184 B.Th.U., calorific value of one pound of the dry coal 13,000 B.Th.U. 8. The following data were obtained during a boiler trial : Feed water per hour 10,115 pounds Temperature of feed to boiler I74F- Steam pressure (Ibs. per square inch absolute) 170 Moisture in I pound of steam o'oig pound Coal fired per hour 1074 pounds Dry coal ,, 1054 ,, Calorific value of dried coal 14,000 B.Th.U. per pound Analysis of dried coal . C 88 %, H 2 3*6 %, ash 3*6 %, other matters 4'8 % Calorific value of ashes 900 B.Th.U. per pound Weight of ashes per hour 38 pounds Analysis of flue gases by volume CO 2 10*9 %, CO i'o %, O 2 7'l %, N 2 8ro % Temperature of flue gases leaving boiler 600 F. Temperature of air in boiler house 60 F. Draw up a heat account for this boiler. 2 B CHAPTER XVIII VALVE DIAGRAMS AND VALVE GEARS 215. Slide Valves. The functions of a slide valve are to admit steam to the engine cylinder j to cut off the steam ; to release the steam from the cylinder by placing it in direct communication with the exhaust pipe and to close the exhaust passage before the end of the exhaust stroke in order that the steam left in the cylinder may be compressed, and in the act of so doing " cushion " the reciprocating masses. The valve is shown diagrammatically in Fig. 161, which represents a longitudinal section through the steam-chest of an engine cylinder. The valve V is given a reciprocating motion by means of the eccentric rod R, '////////////////////////////////////////////t; f^mm^T^^^ R s f"***Ft ^ L a H f . ITI^^^^^^J ^i_>i^ V^!oO$v^SS^\ s v'v^S^^ Cyltnder FIG. 161. which is driven by an eccentric keyed on the crank-shaft of the engine. Steam is continuously supplied into the steam-chest S and admitted to the cylinder, when the valve uncovers the steam ports P from the outside of the valve. The inside of the valve is coupled up to the exhaust pipe E, and when the inside of the valve opens the ports P the cylinder is put into communication with the exhaust. Outside and Inside Laps. When the valve is placed in the middle of its stroke, the distance . 90 -|- B. Consider the motion of the valve from its mid position to the end of its stroke. Before the valve can open to steam it must travel a distance equal to the outside lap ; the remainder of its stroke will give the maximum opening of the valve to steam, provided that the width of the steam port is not less than this distance. Let r denote the " throw " of the eccentric or half the stroke of the valve, and s the maximum opening of the valve to steam, then half travel = outside lap + maximum opening to steam Similarly for the inside of the valve half travel = inside lap -|- maximum opening to exhaust (i) (2) where e denotes the maximum opening to exhaust. FIG. 162. Since the valve cannot open a greater distance than the full width of the port, it is evident that the distances r o and r i will give the 372 THE THEORY OF HEAT ENGINES [CHAP. xvm. maximum openings to steam and exhaust respectively only when they do not exceed the width of the port. The angle of advance may be obtained by the following simple construction : Draw a circle of radius ?, the diameter therefore being equal to the stroke of the valve. On a horizontal diameter set of ol (Fig. 162) equal to the outside lap o, and Id equal to the lead of the valve. Erect perpen- diculars at o and d and join ob. Then the angle abb will be the angle of advance (6) required, and the eccentric centre will lead the crank by the angle ebb 90 + 6. Since Fig. 162 is drawn to represent the path of the eccentric centre, and further, since in almost every case in practice the length of the eccentric rod driving the valve is great compared with the throw of the eccentric, the valve will describe a simple harmonic motion along cm of amplitude r, and for any position of the eccentric, such as b t the displace- ment of the valve from midstroke, when the crank is on the dead centre, will be given by od t or od = r sin d (3) If the crank has turned through an angle a from its dead centre, the general expression for the displacement of the valve will be (Fig. 163) x = r sin (a + 6) (4) FIG. 163. In the particular position shown in Fig. 162 x = od = outside lap + lead = r sin 6, or o + t=rsm8 . " (5) where / denotes the lead of the valve 216. Piston Displacement Curve. The piston will not execute a simple harmonic motion, because the connecting rod is not infinitely long. Draw AO (Fig. 164) to represent the line of stroke, and with centre O draw in the crank-pin circle, the diameter of which will represent the stroke of the engine to any convenient scale. Draw on the centre lines of the crank and connecting rod for any crank angle. With centre A and radius AC (the length of the connecting rod) draw the arc CP to cut the line of stroke in point P, then when the crank is at C the piston is evidently OP from the middle of its stroke. If this construction be repeated for a number of different positions of the crank, the piston displacement may be found for any crank angle. A more convenient construction is obtained by drawing in the arc DOE of radius equal to the length of the connecting rod AC, and whose centre lies at a point in the line of stroke OA. If now a line CC' be drawn ART. 216] VALVE DIAGRAMS AND VALVE GEARS 373 parallel to OA, CC', which is equal to OP, will represent the displacement of the piston from midstroke. Proceeding in this way and taking crank- FIG. 164. angles, say, every 30, the piston displacement curve may be drawn as shown in Fig. 165. An algebraic expression for the piston displacement may readily be found as follows : Let / = length of connecting rod. r = length of crank. n = ratio of length of connecting rod to length of crank = -. a = crank angle measured from the inner dead centre. (/> = inclination of connecting rod to line of stroke. Then referring to Fig. 164 But and OP = x = r cos a (/ / cos ) = r cos a /( i cos ) r sin a = / sin . , r . i . .-. sin = V i sin 2 (j> cos sin2 a sn a (2) in (i) gives x = r cos a /(i --- V 2 sin 2 a) = r cos a r ;z 2 374 THE THEORY OF HEAT ENGINES ^ [CHAP. xvm. Stroke FIG. 165. ART. 217] VALVE DIAGRAMS AND VALVE GEARS 375 Expanding y ri 2 - sin 2 a and using the first two terms only we have an approximate expression for x, namely x = r cos a /-j- r(n -- sin 2 a -f . .) r ;F= r cos a / -f- rn -- sin 2 a 211 = rcosa -- sin 2 a ......... (4) 2.11 and since sin 2 a = J(i cos 2 a) this becomes r . x = r cos a --- J(i cos 20) 2TI r <2. = r cos a -- (i cos 20) ....... (5) If the obliquity of the eccentric rod be not neglected (4) will give the displacement of the valve from midstroke instead of (4), Art. 215, in which case r will be the throw of the eccentric and / the length of the eccentric rod. 217. Rectangular Valve Diagram. For a given ratio of length of connecting rod to length of crank the piston displacement curve is first drawn either by the graphical method described in Art. 216, or plotted on, say, squared paper from equations (3) or (4), Art. 216. On the same diagram the valve displacement curve is plotted at the proper phase angle, as shown in Fig. 166. Since the eccentric leads the crank by an angle 90 -j- 0, it is evident that the valve reaches its maximum displacement at a point 90 -{-0 before the piston reaches its maximum displacement. Consider now the instroke, i.e. the stroke towards the crank shaft, the steam acting on the head end of the piston. Draw a line SS to the left of and parallel to AA, and distant from it the outside lap 0, to the same scale as the valve displacement curve. At point C, where SS cuts the valve displacement curve, the valve is closed and cut- off occurs. From C project horizontally to cut the piston displacement curve, and the fraction of the piston's stroke at which cut-off occurs can be read off directly. At the point B, where SS again cuts the valve displacement curve, admission occurs. The horizontal distances between SS and the valve displacement curve up to cut-off give the opening of the valve to steam, the distance when the crank angle is zero being the lead. For the exhaust side of the valve draw EE parallel to A A and distant from it the inside lap *'. At point R, where this line cuts the valve dis- placement curve, the valve is just on the point of opening to exhaust and release occurs. From R project horizontally to cut the piston displacement curve, and the fraction of the stroke at which release occurs can be read off directly. At point M, where EE again cuts the valve displacement curve, compression commences. The horizontal distances between EE and the piston displacement curve between R and M give the opening of the valve to exhaust. 376 THE THEORY OF HEAT ENGINES [CHAP. xvnr. EXAMPLE. Draw the rectangular valve diagram from the following data : Ratio of connecting rod to crank n = 4. Throw of eccentric r 2 inches. Angle of advance 6 = 30. Outside lap for both ends = 0-8 inch. Inside lap for both ends = 0-5 inch. * Draw the piston displacement curve on squared paper by the method of Art. 216, and divide the stroke into, say, ten equal parts for both the instroke and outstroke, as shown in Fig. 167. In drawing this curve the stroke has been made 5 inches, and therefore gives the piston displacement on a reduced scale. Next draw the valve displacement curve of amplitude ART. 217] VALVE DIAGRAMS AND VALVE GEARS 377 2 inches and go + 3 I20 fr m tne piston displacement curve as shown. Then draw the two lap lines. On the Instroke (on the head end of the piston). Admission takes place at point B near the end of the compression stroke at 0-995 of the stroke, the lead scaling o'2 inch. Cut-off takes place at point C at 0*83 of the "'"* Max, openintf Exhaust (Instroke) 0-99S B', Outstroke FIG. 167. stroke. Release takes place at R at 0-98 of the instroke, and compression commences at M at o'8 of the exhaust stroke. On the O^lt stroke (steam on the crank end of the piston]. Draw the two lap lines shown dotted. Admission occurs at B' at 0-995 f tne ret "rn stroke, the lead being 0*2 inch. Cut-off occurs at C' at 0-74 of the stroke. 378 THE THEORY OF HEAT ENGINES [CHAP. xvm. Release occurs at R' at 0-98 of the stroke, and compression commences at M' at 0-883 f the stroke. 218. Reuleaux Valve Diagram. To set out this diagram draw a circle whose diameter is equal to the travel of the valve. The diameter AB (Fig. 1 68) represents to scale the stroke of the engine, and the circle represents to scale the crank-pin circle, A and B being the inner and outer dead centres respectively. Draw another diameter db inclined 6 to AB, where 6 is the angle of advance of the eccentric. On AB (produced towards the left) as centre, and with a radius equal to the length of the connecting rod to the same scale that AB represents the stroke, draw the arc ED passing through the centre of the valve circle. The line db is called the valve displacement line, and for any position of the crank, such as C at a from the inner dead centre, the perpendicular CF drawn on to db gives the displacement of the valve from the middle of its stroke. This is at once evident from the fact that CF = OC sin L aOC = r sin (a + 0), which by (4), Art. 215, is equal to the displacement of the valve from midstroke. A horizontal line CP parallel to AB gives to scale the displacement of the piston from midstroke, as explained in Art. 216. Now draw a line cd parallel to ab and distant 0, the outside lap from it, and a line ef parallel to db and distant /, the inside lap from it ; cd is called the steam line, and ef the exhaust line. A perpendicular, such as CG, from any crank position C on to cd gives the opening of the valve to steam. The maximum opening to steam will evidently be HK. Similarly, per- pendiculars from various crank positions between / and e, and below / in this case to the left of the main valve centre, or in LH. Thread I? H. Thread 1 II II III If *i E _k\\\\\\ J c. Outside /ap of ^-Inside lap of Main Valve Main l/a/i/e FIG. 179. other words PO represents the displacement of the expansion valve relative to the main valve. Hence E v is called the centre of the virtual or equivalent eccentric, and when the relative displacement of the centres of the valves is equal to the distance between the outer edge of expansion Em. FIG. 1 80. FIG. 181. valve and the edge of the steam port in the main valve when the valves are in their mid-positions (i.e. the distance a of Fig. 179) then cut-off occurs, but whilst the relative displacement is less than this amount steam is being admitted. The expansion eccentric, E e , is placed directly opposite to the crank 390 THE THEORY OF HEAT ENGINES [CHAP. xvm. when the engine is intended to reverse, because, in reversing, the main valve alone is altered, and it is obvious that to secure equal grades of expansion for either direction of running, the expansion eccentric must be placed in the symmetrical position diametrically opposite to the crank. To find the events of the cycle we can treat E v just as an ordinary eccentric if we know its angle of advance A. The angle of advance (A) of the virtual eccentric may be found as follows : Let C be the position (Fig. 181) of the crank when on the inner dead centre. A Let off OE m to represent the throw of the main eccentric, the angle BOE TO being its angle of advance ; set off OE e diametrically opposite to OC to represent the expansion eccentric. Join E m E e and draw OE r parallel to and equal to E m E e . Then OE V is the throw of the virtual eccentric, and angle BOE its angle of advance A. Either the Reuleaux or the Zeuner valve diagrams may now be used to find the point of cut-off, etc. Using the Reuleaux diagram, draw a circle of radius equal to the throw of the virtual eccentric, and set out the displacement line bd (Fig. 182) at an angle A to AB. Draw the line cf parallel to db and 'f (Re-opening) FIG. 182. distant " a " from it, then c will be the position of the crank at cut-off, and when the relative displacement is again reduced to an amount " a " the expansion valve will reopen the main valve must have sufficient outside lap to ensure its being closed by this time or readmission to the engine cylinder will occur. The expansion valve is usually provided with a means (such as a right- and left-handed screw, Fig. 179) of altering the distance "0" and so making the cut-off early or late. The limit of earliness is imposed by the condition that the distance "a" must not be reduced below the amount necessary to give a fair steam opening, and the limit of lateness being imposed by the consideration that the main valve itself becomes closed at a position determined by its own outside lap. EXAMPLE i. Given the earliest and the latest cut-off ever required to find the values of " a " to be provided. ART. 225] VALVE DIAGRAMS AND VALVE GEARS 39i This problem may be conveniently solved as follows : Draw the main valve circle (Fig. 183) of diameter equal to the travel Radius FIG. 183. of the main valve. Find the position of the crank at the latest cut-off required and design the main valve to give the latest cut-off at b and suitable points of release, compression and admission. Next find the virtual eccentric and put in the circle with E v as radius. Join ob t cutting the circle E v in K 2 and draw the perpendicular K 2 M 2 on to the displacement line. The length of K 2 M 2 gives the value of " a " to be provided. Next find the position of the crank for the earliest cut-off. Suppose this is at K 1} then the perpendicular K 1 M 1 gives the value of " a " to be provided. If now the value of " a " be required for any other cut-off, all that is required is to find the crank position at cut-off (say A), and the length of the perpendicular AB gives the value of " a " required. EXAMPLE 2. In a Meyer valve gear the travel of the main valve is 4 inches, angle of advance 22^, lead J inch, and inside lap f inch. Find and state the piston positions at (a) main valve cut-off; (b) at release; (c) at compression; (d) steam port opening at o'i of stroke; (e) port opening to exhaust at end of working stroke ; (/) outside lap of main valve. If the travel of the expansion valve is 4 inches, its angle of advance 90, and the width of the port through the main valve is ij inches. Find the distance from the edge of the expansion valve to the outer edge of the port through the main valve to cut-off at 0*2 and at 0*5 stroke. In the former case find the width of the narrowest expansion plate that may be used. Connecting rod 3*5 cranks long. We will solve this problem by using the Reuleaux diagram and also by the rectangular diagram. 39 2 THE THEORY OF HEAT ENGINES [CHAP. xvm. Main Valve. The Reuleaux diagram is shown in Fig. 184 for the head end of the piston only, the results being : (a) Cut-off at 0-92 of the stroke. (b) Release at 0*99 (c) Compression at 0*89 of the stroke. Cut-off Admfssion Compress FIG. 184. (d) Opening to steam at o'i of stroke =1*12 inch. (e) Exhaust lead 0-4 inch. (/) Outside lap 0*52 inch. Radiuses FIG. 185. Expansion Valve. First find the virtual eccentric. This is shown in Fig. 184, OE r = 2*2 inches and A = 147. Then find the positions of the crank at 0*2, and at 0*5 of the stroke (see Fig. 185) and scale off the ART. 225] VALVE DIAGRAMS AND VALVE GEARS 393 perpendicular distances from these points to the displacement line ; it will be found that they are 0-48 inch and r62 inch respectively. From Fig. 185 the maximum opening to steam, or rather the throw of the equivalent eccentric minus 0*48 is 2*2 0*48 = 1*7 2 inch. Now during the time that the relative movement of the expansion plate over the main valve is 172 inches the port through the main valve is open, hence the width of the narrowest plate must be this distance plus the width of the port : i.e. i'72-f-i*5 = 3'22 inches. Rectangular Diagram. The piston and main valve curves are drawn in the usual way as explained in Art. 216. The expansion valve curve is 0-.99 092 -9 -8 Head End only , Instroke 5 * -3 2 "^ FIG. 186. then added as shown in Fig. 186. The results are the same as before, namely : Main Valve. (a) Cut-off at 0-92 of the stroke at A. (b) Release at 0-99 B. (c) Compression at 0*89 of the stroke at C. (d) Opening to steam at o-i of stroke = DE =1*12 inch. (e) Exhaust lead = FG = 0-4 inch. (/) Outside lap = HK = 0-52 inch. 394 THE THEORY OF HEAT ENGINES [CHAP. xvm. Expansion Valve. When the piston is at 0*2 of its instroke the main valve is the distance LM from its mid-position, and the expansion valve the distance NM from its mid-position. The distance of the centre of the expansion valve from the centre of the main valve, i.e. the lap "0," is therefore LM NM = LN = 0-48 inch (as before). Similarly when the piston is at 0*5 of its stroke the distance of the centre of the expansion valve from the centre of the main valve is OP =1-62 inch (as before) It should be noted that when a = O, i.e at the point a in Fig. 185, and at Q in Fig. 186, cut-off occurs at o-ii of the stroke. If the lap "a" be made positive, then cut-off could be obtained earlier than o'ii of the instroke. The various laps required for different cut-offs on the head and crank end of the piston are further shown in Fig. 186 and also tabulated below : Cut-off. Head end. Crank end. Difference. 0'2 0-48 0'95 0'47 0-3 0-98 I- 4 6 0-48 0-4 o'S 1-31 I'62 1-79 2-06 0-48 0-44 0-6 rpl 2'20 O'29 The right-hand column gives the differences between the laps required to give equal cut-offs at the stated fractions of the stroke. If the expansion plates be adjusted so that the laps are 0*98, and 1*46 for the head and crank end respectively, the cut-offs will be equal at 0-3 and 0*4 of the stroke, and nearly equal at 0*2 and 0*5 of the stroke. 226. Analytical Solution of Meyer Gear. The calculations for the main valve are the same as for the simple eccentric gear described in Art. 223. We are now concerned with the calculation of the lap "a" of the expansion plates required for a given cut-off. Let r = throw of main eccentric. 6 = angle of advance of main eccentric. x = displacement of main valve from its mid position, and r l9 l5 x lt these quantities for the expansion eccentric. Then if a denotes any crank angle measured from the inner dead centre x = r sin (a + 6) (4), Art. 215, or if and ^ denote the angular advance of the main and of the expansion eccentric respectively (i.e. = 90 + 0) then and Xi = TI cos (a -j- j) (2) Let 8 denote the relative displacement of the expansion plates from the central position of the main valve, then g__^, (3) = r cos (a + ) r cos (a -}- (/>). In this example r = r 1 = 2 inches, c/> = 120 and ^ = 180. .'. = 2[cos (a -f- 120) cos (a -j- 180)] 8 = 2 [cos (a + 120) + cos a]. 396 THE THEORY OF HEAT ENGINES [CHAP. xvm. The values of the laps 8 are shown in the following table : VALUES OF 5 (INCHES). Cut-off. Head end. Crank end. Difference. O'l 0-2 0'102 o - 6o I -00 - 0-438 - 0-98 - I-38 0-336 0-38 0-38 0-5 -1-60 -I-8 3 0-23 The right-hand column gives the differences between the laps required to give equal cut-off at the stated fractions of the stroke. If the expansion plates be adjusted so that the laps are o'6o and 0-98 for the head and crank end respectively, the cut-offs at o'i and 0*4 will be nearly equalized, but those at 0*5 will be unequal. The best setting will therefore be with the laps o'6o and 0-98. The reader should verify the above figures graphically, using, say, the the rectangular valve diagram. 227. Resolution of the Valve Displacement Curve into Two Components at Right Angles. Let OE be the position of a simple eccentric whose angular advance is (Fig. 187). The actual motion of the valve may be resolved into two com- , / '$ /! i c 01 1 i 1 1 1 B ponents, one of which is received from an imaginary eccentric OA 90 in ad- vance of the crank and the other from an imaginary eccentric OB 180 in FIG. 187. advance of the crank. Neglecting the effect due to the obliquity of the eccentric rods, the displacement curve obtained from the component eccentric OA will be a sine curve whose maximum ordinate is OA = r sin = Y say and the displacement for any crank angle a will be x = Y cos (a + 9) = Y sin a (i) (2) The displacement curve obtained from the component eccentric OB will be a cosine curve whose maximum ordinate is OB = r cos = X say ....... (3) The maximum X-component will evidently be OB = outside lap -f and the displacement for any crank angle a will be # 1 = X cos (a+ 1 80) = X cos a .... (4) The total displacement of the valve will be = r sin $ cos (a + 90) + r cos (/> cos (a + 180) (5) ART. 228] VALVE DIAGRAMS AND VALVE GEARS 397 Writing (/> = 6 + 9 where 6 is the angle of advance, (5) becomes Total displacement = /-(cos 6 sin a sin 6 cos a) == r sin (a + 0) (6) as given by (4), Art. 215. The reader should verify this graphically by plotting (2) and (4), and, adding the ordinates together, show that the resultant curve coincides with (6). 228. Link Motions. There are two forms of reversing gear used in practice, namely, link motions and radial valve gears. The underlying principle of both forms of motion may be explained briefly as follows : For forward running the valve receives its motion from the eccentric E (Fig. 1 88), set with a certain angle of advance 6. In order to reverse the FIG. 188. SM Eccentric for C conrra-clockwise rotation direction of rotation it is evident that the valve must be driven by a second eccentric E f having the same angle of advance. In a link motion two eccentrics are used, one for forward running and the other for reversed direction of rotation. In a radial valve gear either one or no eccentrics are used for imparting the required motion to the valve, but, in either case, for any position of the gear, a "virtual" or "equivalent" eccentric can be found which will give approximately the same motion to the valve as the actual gear itself. For any type of reversing gear suppose E (Fig. 188) denotes the virtual eccentric for full forward gear, and E' the virtual eccentric for full back- ward gear, then when shifting from full forward to full backward gear, the X component of the valve's motion remains unaltered, whilst the Y com- ponent decreases, passes through zero, and then becomes negative. When in midgear, i.e. at OB, the valve's motion is that due to the X component only, the half-travel of the valve being the outside lap + lead. A detailed analysis of the various reversing gears is beyond the scope of this book, and it is only proposed to give here an approximate solution to the most common types. 1 1 For a detailed analysis of these motions, see Prof. W. E. Dalby's book on " Valves and Valve Gear Mechanisms," Edward Arnold. 398 THE THEORY OF HEAT ENGINES [CHAP. xvni. 229. Stephenson's Link Motion. Fig. 189 shows the centre line diagram of this link motion with open rods. It consists essentially of a curved slotted link AB concave towards the crankshaft O and suspended by a lever from the " weigh-bar " shaft F in such a manner that it may be raised or lowered as required. A block R attached to the valve rod is free to slide in the slotted link AB, and the eccentric rods aA and B are pinned to the link as shown. When the link AB is lowered into one extreme position the block R is at the end of the slot close to A, and the FIG. 189. valve receives its motion principally from the forward eccentric a, this position being known as full forward gear. When the link AB is raised to its other extreme position, the valve receives its motion principally from the backward eccentric ^, this position being known as full backward gear. When the block R is half way between A and B (as in Fig. 189) the motion is in mid-gear; for forward running R may be in any position between A and mid-gear, and for backward running it may be in any position between B and mid-gear. By setting the link AB into any intermediate position the valve receives a motion very nearly the same as that which would be given by a single eccentric of shorter " throw " and of greater angular advance, and the effect is to give a distribution of steam in which cut-off is earlier. The mechanism thus serves to adjust the work done to the demand, in addition to giving reversal. Virtual or Equivalent Eccentric. -In full forward gear, when A is lowered to R (Fig. 189), the equivalent eccentric is oa, its angle of advance being L doa. In full backward gear, when B is raised to R, the equivalent eccentric is - I - . X r cos dba = r cos (y -f a + <) [ see Art- 239] (2) AJj "2C \AD=DB=c FIG. 195. Also, the displacement of B from its central position is r cos boe = rcos( a + &>) (3) The corresponding displacement of R relative to point A is -j-^ X r cos bbe r cos ( a + a>) [see Art. 239] (4) A.IJ 2C Hence the displacement (x) of R, and therefore of the valve, from its mid-position is the sum of (2) and (4), namely Expanding cos [(< + y) + a] and cos [(< + a)) a] (5) becomes cos (^ -f co) + r sin a| ^ sin (^ -j- a>) -- -sin ( + y) j . (6) ART. 232] VALVE DIAGRAMS AND VALVE GEARS 403 Expanding cos ( -f- y), cos ((/> -f- ^>), sin ( -f- &>), and sin (< -j- y), this becomes (V-f- , , c u . , . ^ x=rcos a{ -- (cos (f> cos y sin

cos a) sin

+ cos <6 sm w) 2C 2C X (sin < cos y -j- cos < sin yu ........... (7) Since y and cu are small we may write cos y = cos CD = i, and AR c u RB c + u , sm y = = 7 > and sin o> = ^ = * > hence we have C ^2 _ u z ) ( ?/ ) /. # = r cos a j cos < sin} . . (10) If now p be the " throw " of the equivalent eccentric and A its angular advance, we may write X = p cos A and Y = p sin A (i i) . Y and tanA = - . . (12) When u = c, i.e. when R is at A, X = r cos 0, and Y = r sin , the equivalent eccentric is therefore oa. When u = <:, i.e. when R is at B, the equivalent eccentric is ob. When u = o, i.e. in mid gear X = Y tan A = = o X .-. A = o or 1 80 according to the arrangement of the gear. Gooch Motion. The method is the same as for Stephenson's motion down to (7). Substituting in (7) sin y = sin a> = - , and cos y = cos a) = i it reduces to ( , c . .\ (u . u . .) x = r cos OA cos 9 -sm - sin f =p cos A and Y=r| - cos ^-| sin f =p sin A (14) 232. The Allan Link Motion. In this motion the slotted link AB (Fig. 196) is straight and is suspended from the weigh-bar shaft by the link AK. The arms HK and HG form part of the weigh-bar shaft and turn with it about the fixed fulcrum H. The valve rod is jointed at V and supported by the link FG. On turning KHG in a contra- 404 THE THEORY OF HEAT ENGINES [CHAP. xvm. clockwise direction by pulling the reversing lever L, the link AB is lowered and the block R simultaneously raised. The motion of AB from full forward to full backward gear is therefore much less than in the case of the Stephenson motion, and the weight of the link AB and the (Fixed Fulcrum) FIG. 196. eccentric rods can be made to balance that of the rod RV by giving suitable lengths to HK and HG. The solution to this motion is rather more intricate than either Stephenson's or Gooch's. 1 233. Hack worth's Radial Valve Gear. A centre line diagram of this gear is given in Fig. 197. A single eccentric E is used diametrically opposite to the crank C. The extremity F of the link FE is constrained to move along the guide bar or slotted link AB, and the valve V receives its motion from point G through the rod VG. The mean position of the link FE is arranged to be perpendicular to the line of stroke of the piston P, and the guide bar AB can be turned in a contra-clockwise direction (thus altering the angle 0), so that it may be made to occupy the dotted or any intermediate position. By this means the travel of the valve can be varied, and the engine reversed when AB has passed through the position DO, i.e. when changes from positive to negative. As the crank rotates the point G describes the full line oval in space. Hence to find the correct valve displacement, it is necessary to plot in the gear for a number of crank angles and scale off the displacement of V. When the bar AB is in, say, the dotted position, G describes the dotted oval and the engine runs reversed. Equivalent Eccentric. An approximate solution may be obtained by finding the equivalent eccentric which, driving through GV, will give an approximation to the actual motion of the valve. A simple graphical construction for this purpose is the following: Set off OC (Fig. 198) to represent the position and length of the crank when on the inner dead centre, and OE the position and " throw " of the eccentric. Draw OB 1 For further details, see Dalby's " Valves and Valve Gear Mechanisms." ART. 233] THE THEORY OF HEAT ENGINES 405 perpendicular to CE, and EB inclined 6 to OE. Divide OE in m z such that Om 2 FG and erect the perpendicular mm z . Then for this position of the guide bar Om represents the equivalent eccentric and L BOw its angle of FIG. 197. advance. If the angle L OEB (6) is the extreme value of the inclination of the slotted bar AB (Fig. 197), Om will be the equivalent eccentric for full forward gear, forward running being understood to be in the direction of the arrow in Fig. 197. For full backward gear set out L OEB' = L 9EB, then Om' will be the equivalent eccentric for full back- ward gear, its angle of advance f being L B'CW. When the motion is in mid- \ gear, i.e. when 6 = o, the equiva- lent eccentric is Om 2 , its angle of advance being BOw 2 = 9- For any intermediate position of the gear set off l_ O&fttf '. equal to the particular value of 0, and O/i' will be the equivalent eccentric. It should be noticed that for all positions of the gear, i.e. for all possible values of 6, the displacement of the valve from its mid-position when the crank is on the dead centre is equal to Om Zt or in other words is equal to the outside lap plus the lead of the valves. FIG. 198. 406 THE THEORY OF HEAT ENGINES [CHAP. xvm. 234. Marshall's Valve Gear. In this gear the eccentric OE (Fig. 199) is set at the same angle as the crank OC, and the point G describes an arc of a circle having L as centre. For any position of the gear L is a fixed centre. To reverse the engine, the centre L is swung over towards the left about the fixed centre M, so that L may occupy any position on an arc of a circle of radius ML and centre M. The motion is shown in full forward gear in Fig. 199, the dotted position ML' being for full backward gear. As the crank rotates, the point G swings to and fro along a small arc having L as centre, and the point F describes o i FIG. 199. the full line oval curve. When the centre L is shifted over into its other extreme position, the point F describes the dotted oval. The mean position of the link EF is perpendicular to OP (as in Hackworth's gear), and for all positions of the gear its inclination to OF is very small. Equivalent Eccentric. To obtain an approximate solution we may proceed as for the Hackworth gear. First find for any position of the gear the two extreme positions of the point G. Since the length of the arc described by G is small it may be replaced by a straight line DH, whose inclination 6 to OF is the same as the mean inclination of the arc. Set off OC (Fig. 200) to represent the crank, and OE the eccentric when the crank is on either dead centre (Fig. 200 shows it when the piston is at the bottom of its stroke, the crank being on the outer dead ART. 235] VALVE DIAGRAMS AND VALVE GEARS 407 centre). Draw OB perpendicular to OC, and EB making angle with EO. Produce EO to m 2 making OE EG and draw mm 2 perpendicular to Qm 2 . Then Om represents the equivalent eccentric, and L CO/0 its angular advance, or L ~BOm its angle of advance. When the motion is in midgear, 6 = o, and Qm 2 is the equivalent eccentric, its angle of advance being L ~BOm 2 = 90. For any inter- mediate position of the gear set off L OEm^ equal to the particular value of and Om^ will be the equivalent eccentric. When is negative, i.e. for reversed running, the construction is shown dotted. For all positions of the gear the displacement of the valve from its mid-position is Om 2 , which is equal to the lap plus the lead of the valve, being in this respect similar to the other link motions previously considered. 235. Joy's Valve Gear. A centre line diagram of this gear is shown in Fig. 201. For any position of the gear G! and G 2 are fixed axes, and FIG. 201. F describes an arc of a circle whose centre is at G 1? whilst H describes an arc whose centre is at G 2 . Points A, E, and D describe the oval curves shown. The valve V receives its motion from D through the rod DV, hence from the locus of D the corresponding valve displacements may be found. The engine is reversed by shifting the axis G 2 along the arc shown dotted. When reversed the path of E is unaltered, but the paths of H and D are inclined in the opposite direction to the vertical. Equivalent Eccentric. First find the two extreme positions of the point H ; since the length of the arc described by H is small, it may be replaced by a straight line (shown dotted in Fig. 201), whose inclination 6 to the vertical is the same as the average inclination of the arc. Set off OC (Fig. 202) to represent the position of the crank on the 408 THE THEORY OF HEAT ENGINES [CHAP. xvm. dead centre, i.e. when the piston is at the head end of the cylinder, make OE 1 equal to half the horizontal displacement of point E and OE 2 , equal to half the vertical displacement of E. Draw OB 90 in front of the crank, and make angle OE 2 B = 0. Produce E O to A, making EiO EH ,_. _ = _(Flg.20I> Join E X B and produce it to cut the perpendicular from A in point m. Then Om is the equivalent eccentric and COm its angular advance, or L BOw its angle of advance, A. For midgear 9 is zero, the equivalent FIG. 202. eccentric is OA, its angle of advance being 90. For any intermediate position of the gear, the equivalent eccentric may be found by repeating the above construction, the angle 6 depending upon the position of G 2 (Fig. 20I). 1 EXAMPLES XVIII. 1. The travel of a slide valve is 5 inches, lead at head end 0*25 inch, cut-off at 07, and release at 0*95 of the stroke for both ends. If the connecting rod is 4 cranks long, find (a) outside and inside laps ; (b) angle of advance j (c) maximum openings to steam ; (d) lead at crank end. 2. Neglecting the obliquity of the connecting rod, find the angle of advance, travel, and steam lap, so that cut-off may take place at 0*6 of the stroke, and so that the maximum opening to steam is I inch and the lead o'l inch. 3. Solve Question 2, if the connecting rod is 3*5 cranks long. 4. The travel of a slide valve is 4 inches, lead at both ends O'I25 inch, cut-off at head end O'6 stroke, release at both ends 0*96 stroke. Find : (a) outside and inside laps for both ends ; (b) cut-off on crank end ; (c) angle of advance. Connecting rod 4 cranks long. 5. In a steam engine the connecting rod is 4 cranks long, and the displacement of the slide valve is given by the equation x = 2'6 sin (a + 32) + O'2 sin (2a + 105) Draw a rectangular valve diagram and find the two outside laps, in order that cut-off may take place at 07 of the stroke at both ends of the cylinder. 6. An engine, with a cylinder 21 inches diameter and 36 inches stroke, runs at 72 revolutions per minute. It is fitted with an ordinary single-ported slide valve, and the point of cut-off is to be 0^64 stroke for both ends of the cylinder. The width of the ports (in plan) is 18*5 inches, and the steam speeds allowed are 120 feet per second for admission and 90 feet per second for exhaust. Draw the valve diagram and obtain the valve travel, maximum openings to steam and exhaust, leads, outside and inside laps, to give a reasonable indicator card. Connecting rod, 5 cranks. (L.U.) NOTE. Assume a lead (head end) of 0*25 inch. 7. In a Meyer valve gear the travel of the main valve is 3*5 inches, and its angle of advance 30. The travel of the expansion valve is 3*5 inches, and its angle of advance 90. Find the radius and angle of advance of the equivalent eccentric, and the " lap " of the expansion plate (a) for a cut-off at 0*2 stroke. Neglect the obliquity of the con- necting rod. 1 For further details, see Dalby's " Valve Gears and Valve Gear Mechanisms." ART. 235] VALVE DIAGRAMS AND VALVE GEARS 409 8. The following are data from an engine fitted with a Meyer valve : Connecting rod, 4 cranks long. Travel of main valve, 3*5 inches. Angle of advance of main valve, 30. Travel of expansion valve, 3' 5 inches. Angle of advance of expansion valve, 90. Find (using a graphical construction) the lap "a" of the expansion valve, so that cut-off may take place at 0*2, 0*3, 0*4, o'5, and O'6 on both sides of the piston, and deduce from these laps the best setting of the expansion valve. 9. Solve Question 8 analytically. 10. A Meyer valve gear is to be capable of varying the mean cut-off from O'l to O'6 of the stroke ; connecting rod, 4 cranks. The main valve is to give a maximum opening to steam of i'375 inches, and, if acting alone, would cut-off at 075 of the stroke at the head end of the cylinder. The lead of the main valve is to be 0*25 inch. Find : (a) Travel of main valve, its angle of advance, and outside lap. (b) If the travel of the expansion valve is I'l times that of the main valve, and its angle of advance is 90, find its lap " a " at the above fractions of the stroke. (c) If the width of the steam ports in the cylinder is 1*75 inches, find the width of the narrowest expansion plates required. 11. In a Stephenson's Link Motion the " throw " of each eccentric is 3 inches, angle of advance 20, length of slotted link 20 inches, length of each eccentric rod 60 inches. Find the equivalent eccentric (a) when in full gear ; (b) when in mid gear. 12. The displacement of the valve operated by a Stephenson Link Motion is approximately x X cos a Y sin a, in which X and Y are constants, having the values X = r(cos o* 1 1 sin ) Y = O'S$r sin r is the radius of eccentricity of each eccentric sheave, and $ is the angle between the crank and the eccentric radius of each sheave. Find values of r and which will secure a cut-off of 76 per cent, of the stroke, a lead of o'l inch, and a maximum opening for steam of I inch. Neglect the obliquity of the connecting rod. 13. In the Hackworth radial valve gear, shown in Fig. 197, OC = 1*3' j CP = 4*2' ; OE = 0-5' j OD = 3-45' ; 6 = 45 ; EF = 3-48' ; EG = 1-84' ; GV = 5-06'. Distance between strokes of P and V = i'83'. Plot the gear when the crank angle POC is 60, and find (a) Travel of valve V. (b) Angle of advance of equivalent eccentric actuating V. Find also the equivalent eccentric for mid gear, i e. when = o. 14. In the Marshall radial valve gear, shown in Fig. 199, OC = 1*2' ; CP = 4*78' ; OE = 0*25' ; OM = 1*5' (O and M being on the same horizontal line) ; inclination of ML to the vertical, 20 ; ML = LG = 2' ; EG = r& ; EF = 2-3'. Distance between strokes of P and V = 2^4' . Plot the gear when the crank angle POC is 135, and find the equivalent eccentric actuating V. 15. In the Joy valve gear, shown in Fig. 201, OC = i' i" ; CA = 4' i"; AB = 2 f - i" ; AE -- 7-25" ; EF = i' - of" ; GjF = 2' - 4" ; perpendicular distance of G! from BO = i' i|"; horizontal distance of G l from O = 6' 6". Distance between strokes of B and V = i' 2f " ; ED = 2' ; DH = 3" ; VD = 3' 7^" ; G 2 H = 2' ; perpendicular distance of G 2 from BO = 7" ; horizontal distance of G 2 from O = 6'. Find : (a) Horizontal displacement of E. (b) Vertical displacement of E. (<:) Half travel of valve and angle of advance of equivalent eccentric. CHAPTER XIX TWISTING MOMENT DIAGRAMS 236. Twisting Moment for any Crank Angle. Let P = effective force on the piston in pounds (obtained as explained in Art. 283). = thrust along the connecting rod in pounds. = crank angle measured from the inner dead centre. (f> = inclination of connecting rod to the line of stroke. r = length of crank in feet, i.e. half the stroke of the piston. Then, referring to Fig. 203 P A -X- FIG 203. Twisting moment T = Q X perpendicular distance OD. Now Q = P sec cf> .'. T = P,sec '.-. = P x OF pound feet . ; . . . . The twisting moment may also be expressed as follows : T = Q X OD = P sec X OC sin L OCD = P sec X r sin (0 + ) (I) cos < 410 pound feet (2) ART. 237] TWISTING MOMENT DIAGRAMS 411 After the value of P has been found for any crank angle 0, either (i) or (2) may be used in order to calculate the twisting moment. 237. Inertia of the Reciprocating- Parts Acceleration of the Piston. In order to find the effective force P, on the piston, we must know the acceleration of the piston for any position of the crank. This may be obtained analytically as follows : Let CD = angular velocity of the crank in radians per second (assumed constant). / = length of connecting rod. n = -. r x = displacement of the piston from the end of its stroke. Then, since AG = AC = / (Fig. 203) x = BE -j- EG, where CE is perpendicular to AO. = r(i cos0)+/(i cos<) ........ (i) Also CE = / sin < = r sin 6. .-. sin = - sin 6 = - sin 6 . . . . . . . . (2) and cos = <\/i sin 2 < = -\^ 2 sin 2 6 . . . (3) Substituting (3) in (i), we have x = r(i cos 0) + /(i - ^ 2 - sin2 0) . . . (4) Differentiating (4), we find the velocity of the piston v = -=- _dx _dQ_ dx_ dx ~~dt~~~dt~dQ~*~dQ , dx ( sin 6 cos 9 and - = or ( * , sin 26 \ , v a)r(sm9-{ . ; = 1 . ... (5^) Neglecting sin 2 6, an approximate expression for the velocity is, from (50), < n , sin 26 \ sin 9 -\ -- 1 ...... (6) 211 / Differentiating (5), we find the acceleration of the piston -^ or -^ at at*' ^__d^__dB_ dv _ dv d& ~~Tt~ Jt'de^^dd nrl - ^ X 2 r fl I ^ COS 2B + Sin4 a d .'. -j-= = co 2 r cos 9 -\ -- 2 4i2 THE THEORY OF HEAT ENGINES [CHAP. xix. Neglecting sin 2 6 and sin 4 0, an approximate expression is .... (8) which is accurate enough for practical purposes except when n is small. If the obliquity of the connecting rod be neglected the piston executes a simple harmonic motion, and its acceleration is o>V cos B = co 2 . OE. The approximate expressions (6) and (8) may also be deduced directly as follows : As before, x = BE + EG (Fig. 203) Now EG X (AG + AE) = (CE) 2 . Fr = But CE = r sin 0, and AG = AC = /, and writing AE = AG, we' have CE 2 r* sin 2 EG = ^ = 2AG 2/ r 2 /. x = r(i cos 0) -) - sin 2 approximately = r) (i cos 0) -| sin 2 [ approximately . . (9) C 2n j dx . . Q . sin 20 . . v = -= cor(sin -j ) as above at 2tl ' x i and = co 2 r(cos -4- -cos 20) as above. dt ' n Graphical Method. Klein's construction may be conveniently used for finding the acceleration of the piston. Draw the centre lines of crank OC and connecting rod AC, producing the centre line of the rod (if necessary) to cut the perpendicular to the line of stroke through O in point F (Fig. 204). With C as centre, and radius CF, describe a circle, and draw another circle on AC with AC as diameter. Draw the common chord to these circles, producing it if necessary to cut the line of stroke in point B. Then OB represents, to scale, the acceleration of the piston when the crank is at C. If the length of OC represents the centripetal acceleration (w z r) of the crank-pin, OB represents to the same scale the acceleration of the piston, or, in other words, the acceleration of the piston is o> 2 OB. The angular acceleration of the connecting rod is proportional to DB p\"D being equal to co 2 . AO Accelerations at the ends of the stroke. At the inner dead centre where = o, we find from either (7) or (8) Similarly at the outer dead centre where = 180 d^oc / i\ , . = o*(i--). ... . . . (ii) ART. 237] TWISTING MOMENT DIAGRAMS Position of the piston when its acceleration is zero. Equating (7) to zero, we find , # 2 cos 20 4- sin 4 9 cos 9 -\- - ^ = o (2 _ sin2 9)1 cos d(n 2 sin2 8)1 = ( 2 cos 26 + sin 4 d) Squaring both sides of this equation, we get COS 2 0(2 _ S in2 0)3 = (2 cos 20 + sin 4 0)2 which reduces to sin<5 2 sin 4 4 sin2 -f- ;* 4 = o . . . (12) a cubic equation for sin 2 0. The value of given by this equation is very nearly equal to that given by = tan" 1 n, which represents the crank FIG. 204. angle when ihe connecting rod is perpendicular to the crank, i.e. when the angle L AGO (Fig. 203) = 90. For example, when n = 4, the true value of from (12) is 76 43', whilst the approximate value = tan" 1 4 is 75 5 8 '; wnen = 8 (12) gives = 82 59', whilst tan" 1 8 = 82 52'. The error made in assuming the piston to have zero acceleration when the connecting rod and crank are at right angles, i.e. when = tan" 1 , is therefore negligibly small in practice. Taking this value of and substi- tuting in (9) we have the piston displacement = r\(i cos 0) + - sin 2 0> 414 THE THEORY OF HEAT ENGINES [CHAP. xix. (where s = the stroke of the piston), which may be written Eq. (14) will therefore give the position of the piston from the inner H FIG. 205. dead centre position when its acceleration is zero, quite accurately enough for all practical purposes. Acceleration curves. The curves shown in Fig. 205 have been drawn for n = 4, using the equations (8) and (14). The results of the calculations are shown in the following table : ART. 238] TWISTING MOMENT DIAGRAMS Crank angle e. Position of piston. Fraction of stroke X Values of (cos 6 + - cos 20). ~* r ' O I-250 30 0-083 0-991 60 0*296 0-375 7 6 0-438 O'OOO 90 0-562 0-250 120 0-796 - 0-625 150 0-980 - 0-741 180 I '000 - 0-750 The dotted lines in Fig. 205 show the acceleration curves on a stroke base for an infinitely long connecting rod, i.e. when = cu 2 /* cos 6. Vead tnd FIG. 206. These curves will also represent the inertia forces to another scale since W d*x Accelerating force = - where W = weight of reciprocating parts. 238. Method of drawing the Twisting Moment Diagram. We will take the case of a horizontal single-cylinder double-acting steam engine. The work may be conveniently carried out as follows : The first step is to draw from an indicator diagram the piston effort diagram, which consists of a curve of P (Art 236) on a stroke base. Let the indicator diagram be as shown in the upper part of Fig. 206. 416 THE THEORY OF HEAT ENGINES [CHAP. xix. Consider the instroke of the piston, i.e. the working stroke from the head end of the cylinder during which the piston is moving towards the crank-shaft. The curve abc gives the steam pressure on the head end piston at the various points of the stroke, and the curve def gives the pressure on the crank end of the piston during the same stroke. The effective driving pressure on the piston will evidently be the difference of the pressures on the two sides of the piston, i.e. at any point such as g the effective pressure will be gh kh =gk At the point m the effective pressure will be zero, and from m onwards to the end of the stroke it will be negative. A separate curve of effective pressure may be plotted vertically below the indicator diagram as shown, on which ordinates above the base line dc represent to scale the positive effective force on the piston and ordinates below the negative force. Next add the inertia line np as described in Art. 237. Produce the base line dc> and set out the crank-pin circle to the same scale, i.e. make M = = inclination of the connecting rod to the line of stroke. w = weight of the connecting rod. a) = angular velocity of the connecting rod. The first step is to obtain expressions for ^(a) Angular acceleration of the connecting rod. (b) Acceleration of the mass centre of the connecting rod 1 and - ] *^(c) Acceleration of the piston -,- (see Art. 237). ART. 239] TWISTING MOMENT DIAGRAMS (a) Angular Acceleration of the Connecting Rod. Now / sin (f) = r sin 6 .'. sin (f> = j sin 6 sm

cos ' dt n cos sn 2 /2 FIG. 210. Differentiating (2), we have i ) sin (3) (b) Acceleration of the Mass Centre of the Connecting Rod. From Fig. 210 and = / 2 sin (j) (4) dQ dy . T Q /..v 420 THE THEORY OF HEAT ENGINES [CHAP. xix. Also, from Fig. 210 s = x t z (icos) (8) = r(i cos 6} -j- ^(i cos (f>) / 2 (i cos (f>) = r(i cos 0) 4- (/ 4)(i cos <) = r(i cos ) -j- /i(i cos <) since / / -f- / 2 .-. s = r. l j(i - cos 9) + r. ^(i - cos 6) + /i^(i- cos <) = j{r(i cos 6) + /(i cos ()} -f- j . r(i cos 6) = j.x-\-- 2 .r(i cos6) (9) or s =-hA's horizontal displacement) + /(C's horizontal displace- ment) (9A) which is a well-known general principle. Differentiating (8), we have ds_ dx . , d$ Substituting for -j from (5), Art. 237, and for sin (f> and -j- from (2) above, we have ds ( . n . sin 6 cos 6 \ 4 to sin 6 cos dt / . ,, . sin 6 cos 6 \ / 2 . o> . sin cos . = o>Hsin^-f . _ =^}- . =r- . (n) r V^2 _ si n 2 0/ Jn sin2 Also, by differentiating (9), we have the general principle ds dx / = -j (A's velocity) + -j (Cs horizontal velocity) . . (i2A) Differentiating (10), we have Also by differentiating (12), we have the general principle = ^ (A's acceleration) -j- ~ (Cs horizontal acceleration) . (i4A) The resultant acceleration of the mass centre of the rod will be RT. 239] TWISTING MOMENT DIAGRAMS 421 Approximate effect of the inertia of the connecting rod on the Twisting Moment. In this approximation the angular motion of the rod is neglected. Dividing the rod into two parts of weight, w 2 at A and w at C, such that we have w% = \ . w and w = -, w I i Thrust along the rod Q = (P *~ . ^) sec \ ^ / where P = effective steam pressure on piston frictional resistance W cftx , in which W = weight of piston, piston rod, crosshead, and cross- o head pin. Resolving (15) perpendicular to the crank and multiplying by r, we have FIG. 211. Twisting moment in) direction of rotation j d z x '' (i) (i6A) The effect on the twisting moment of the inertia of the rod alone is found approximately by putting P = o in (16), which gives Twisting moment in) direction of rotation [ _ ^? . due to rod alone J S at * X in (0 + ) . (17) ( I7 A) Exact effect of the inertia of the connecting rod on the Twisting Moment. Referring to Fig. 211 422 THE THEORY OF HEAT ENGINES [CHAP. xix. Let X = horizontal force exerted by the crank on the rod. Y = vertical force exerted by the crank on the rod. Then, neglecting friction, we have the equations of motion < -{-X/sin< ..... (20) where IA = moment of inertia of the rod about the crosshead pin A = I a + ^ = L> + 4 s) A & in which k = radius of gyration of the rod about G. From (18) and (20) we have /cos 9 and the effect of the rod alone when P = o /cos ==__! +5.^. tan ,6 ....... (22) / cos

effect of the rod alone, taking account of its angular motion, is therefore from (24), Art. 239, -|..g. cos 0) --- y- -.sec < cos . . (2) The exact effect is therefore greater than the approximate effect by the amount The approximation will give the twisting moment accurately when 2 _ / 1 / 2j i n w hich case (3) is equal to zero. In almost all connecting rods 2 is less than /j/g, and the excess twisting moment given by (3) is positive when cos 6 is negative, i.e. when 6 is between 90 and 270, and negative when cos 6 is positive, i.e. when 6 is between o and 90 and between 270 and 360. The radius of gyration (k) of the rod may be conveniently found as follows : First find the position of the centre of gravity of the rod, / 2 , from the crosshead end ; then swing the rod as a pendulum about knife edges at the centre of the small end, and find experimentally the length (d) of a simple pendulum which has the same period. The point on the rod distant d from the centre of the small end is known as the centre of percussion, its distance from the centre of gravity of the rod being d / 2 . Then it may easily be shown that (4) *" 241. Kinetic Energy of the Connecting Rod. The kinetic energy of the rod when in any position is made up of two parts, namely 424 THE THEORY OF HEAT ENGINES [CHAP. xix. (1) The kinetic energy of translation = . (2) The kinetic energy of rotation = l( -Jj . The resultant velocity v of the mass centre of the rod is Hence (i) above is The kinetic energy due to rotation of the rod about its centre of gravity is where I = moment of inertia of the rod about an axis passing through its centre of gravity. Hence the total kinetic energy of the rod is the sum of (i) and (2), viz. : The twisting moment exerted on the crankshaft by the rod may also be found from its kinetic energy, since for a small angle 86 over which the twisting moment exerted by the rod is T, we have - 8E = T86 ' T 51 (2) 86 Differentiating (i) with respect to 6 we have ^ E T __ a)' dt d_ (dy^ * ^ di\dt) ~ "dt \dt _ ,^2 vgtdfl V/" 1 " dft ',#* ' dt' dt where k is the radius of gyration of the rod about its centre of mass. * 242. Graphical Method for Finding the Twisting Moment exerted by the Connecting Rod- The force required to give the mass centre of the rod its acceleration may be first found. It will be w - X acceleration of the mass centre ART. 242] TWISTING MOMENT DIAGRAMS 425 //d2s\ 2 /^2y\ 2 The construction for finding \/ (^2) + (j^jz) is snown in Fi g- 2I2 - Set out the centre lines of the rod and crank. Find the acceleration BO of the piston A by Klein's construction (Fig. 204). Join CB and draw GD parallel to AO to meet it in point D join DO. Then DO represents the FIG. 212. acceleration of the mass centre of the rod in magnitude and direction to the same scale as CO = a> 2 r, the centripetal acceleration of the crank pin, or acceleration of mass centre = w 2 DO and accelerating force on rod F = co 2 DO o Since the rod has an angular acceleration the force F will not pass through the centre of gravity G of the rod, but will give the couple about G, which is required to produce the angular acceleration of the rod. Find the point K such that AK is equal to thejength of a simple pendulum which has the same period as the connecting rod when it oscillates about an axis passing through A (K is called the centre of percussion). Draw K parallel to GD and join kO ; then /O gives the acceleration of point K. From K draw KL parallel to kQ to meet the line of stroke AO in point L. Then L is a point on the line of action of the force F, which must therefore pass through L and be parallel to DO. To find the turning moment exerted on the crankshaft, draw AE perpendicular to AO to meet the line of action of the force F in point E. Set off EH to represent the force F = eo 2 DO to any convenient scale, o 426 THE THEORY OF HEAT ENGINES [CHAP. xix. join EC and from H draw HM parallel to" AE. Then ME is the equivalent force acting on the crank pin, and the turning moment on the crankshaft is ME X perpendicular distance OR in a direction opposite to the direction of rotation. As shown in Fig. 212 6 is between o and 90, hence the twisting moment exerted on the crank- shaft by the rod is negative (cp. end of Art. 242)^ EXAMPLE i. The connecting rod of an engine weighs 100 pounds and is 4 feet long, the crank being i foot in length. The centre of gravity of the rod is 2-5 feet from the crosshead end. The diameter of the cylinder is 15 inches, trie engine runs at 240 revolutions per minute. If the weight of the reciprocating parts other than the connecting rod is 300 pounds, find the twisting moment when the crank angle is 30 from the inner dead centre, the steam pressure on the head end of the piston being 80 pounds per square inch, and on the crank end 20 pounds per square inch. Assume the rod to be equivalent to two masses, at the crank pin and crosshead pin respectively. Angular velocity of crankshaft co = -~ X 277 = 877 radians per second. Acceleration of piston when 6 = 30 is by (8), Art. 237 //2r ~ = (877)2 X i (cos 30 + i cos 60) = 64772 (0-8660 + 0*1250) = 625 feet per second per second. Portion of connecting rod to be assumed concentrated at the cross- head pin = L5 x ioo = 62*5 pounds 4 .'. total weight of reciprocating parts W= 300 + 62-5 = 362-5 pounds Accelerating force = ^~~ X 625 = 7040 pounds Effective force on piston? _ } g ( )2 due to steam pressure J v = 60 X 1767 = 10,600 pounds .*. P = 10,600 7040 = 3560 pounds and twisting moment by ((2), Art. 236) is Now sin (/> = - sin 6 ((2), Art. 237) , __ * _ 4 -~ (o-5) 2 = = 1-0080 3-968 ) = sec" 1 1 '0080= 7*2 1 For other graphical methods, see Dalby's " Balancing of Engines." Edward Arnold. ART. 242] TWISTING MOMENT DIAGRAMS 427 and sin (0 -j- (/>) = sin 37*2 = o'6o46 .'. T = 3560 X i X 1*0080 X '6046 = 2170 pound-feet EXAMPLE 2. The connecting rod of an engine weighs 1500 pounds, and is 6 feet long, and the crank is i foot 9 inches long. The centre of gravity of the rod is 4 feet from the crosshead end. The engine makes 100 revolutions per minute, and the connecting rod when oscillating about its small end swings in unison with a simple pendulum 5 feet long. Find the twisting moment exerted by the rod in the direction of rotation and the kinetic energy of the rod when the crank angle is 45 from the inner dead centre 100 I07T co = -j x 277 = - radians per second 60 3 First find the radius of gyration of the rod about its centre of gravity. & = 4 X (5 - 4) = 4 .*. k 2 feet Using the notation of Art. 239, we have 6 24 since n== - = =- 175 7 I007T 2 X 1750-707 / 7 \ (0-707 + --XoJ 987 X 1-7 5 - - X o 707 = 135 ft. per sec. per sec. dx / . Q . sin 20\ =a>Hsm0 + - -I dt 2n / IO7T / 7 \ = T * '75(0707+^) 31*416 X 1*75 - X 0-853 = 15-6 ft. per sec. 5 ds /i dx . 4 IO7T = 5-2 + 8-63 = 13-83 ft. per second = w . / 2 . T - cos ((6), Art. 239) I07T 1-75 = X 4 X -^ X 0707 = 8-63 ft. per second 428 THE THEORY OF HEAT ENGINES [CHAP. xix. 2 , A I007T 2 = 6 X I35 + 6 X ~ 9 ~ X I75 X '77 = 45 + 90 Z 3S ft. per sec. per sec. d gjcos 6 Tt= V ^^rg (H Art 23 9) _I07T 0707 _ _ (3'43) 2 + (0707) 2 16 X 3 X V _ 31-416 X 0707 ~ ~ 2<2 radians per second = I007r2 9 = 987 1075 x 0707 9 ' 38 = 22 radians per sec. per sec. The twisting moment exerted by the rod in the direction or rotation is T=- ^ jg (sin 0+ tan < cos ^ Now sin = - sin 6, cos < = -A/^ 2 - sin 20, tan = ./ 9 sm g . M n n V n 2 sin 2 V(3'43) 2 - (o' Substituting in the expression for T we have : 1500 C /' . 0707 3'36 .'. T = -- ^ X 1*75^135(0707 -4- ' ' . ^- 32-2 ^i ' r 3*36 S^ {i35Xo- 9I3 - 75 x 6 32-2 = 5750 pound-feet The reader should check this figure graphically. The kinetic energy of the rod is ART. 242] TWISTING MOMENT DIAGRAMS 429 + ( 8 ' 6 3) 2 ! + 1 X 4 X (2-2)2 1500 6000 = 7^77 X 265 + -x 4-84 04 4 6 4 4 = 6170 + 450 = 6620 foot-pounds EXAMPLE 3. In an inverted direct-acting engine the stroke is 2 feet, length of connecting rod 4 feet, cylinder 14 inches diameter, weight of reciprocating parts 300 pounds, and the speed 180 revolutions per minute. On the down stroke, at the commencement of the stroke the difference of pressures on the two sides of the piston was 40 pounds per square inch (acting downwards) ; at the end of the stroke the difference of pressure was 10 pounds per square inch (acting upwards). Find the effective pressure transmitted to the crank pin in these positions. If the steam pressures remained unaltered, at what speed would the engine have to run in order to make the effective pressure at the end of the stroke zero, and what would then be the effective pressure at the commencement of the stroke ? At the commencement of the stroke the accelerating force is W In this example r = i foot, co = 277 X ^ = 677, n = 4 .-. accelerating force = -=-7- X 36772 (i + |) = 4140 pounds area of piston = X (i4) 2 = 154 square inches 4 .*. accelerating force per sq. in. of piston area = ^~ = 26*8 Ibs.per sq. in. weight of reciprocating parts per sq. in. of piston area = fff = 1*95 Ibs.persq. in. .-. effective downward pressure transmitted to crank pin = 40 + rg$ 26*8 = 15 -15 Ibs.persq. in. or total force = 15*15 X 1 54 = 2 33 3 pounds. At the end of the stroke the accelerating force is W -- ((i i), Art. 237) = 2480 pounds = - = 1 6- 1 pounds per square inch of piston area .-. effective downward pressure exerted on crank pin =1*95 10 -j- 16*1 = 8*05 Ibs. per sq. in. or total force = 8-05 X 154 = 1240 pounds downwards. 430 THE THEORY OF HEAT ENGINES [CHAP. xix. For the effective pressure at the end of the stroke to be zero, the accelerating force must be equal to 10 1-95 = 8-05 pounds per square inch or total inertia force = 8-05 X 154 = 1240 pounds. Hence if a> is the angular velocity required we have f^XO^ 1 -- 1)=I2 4 1240x32-2 300 X 075 a>= Vi77 = 13*3 radius per second _3j3 _ I2 ^ revolutions per minute. 27T At the commencement of the down stroke Accelerating force = X 177 X (i + J) = 2165 pounds 2165 = - =13*4 pounds per square inch 154 /. effective pressure =i'9S + 40 *3'4 = 28*55 pounds per square inch or total force = 28-55 X 154 = 4400 pounds 243. Function of the Flywheel Cyclic Variation of Speed. Let Fig. 2 13 be the twisting moment diagram drawn as explained in Art. 238. D 180 360 FIG. 213. The area under this curve will represent to scale the work done on the piston, and should be compared with the work done as obtained from the indicator diagram. The resistance on the crankshaft may be assumed constant and be expressed as a resisting moment, being equal to the actual torque on the crankshaft plus the equivalent torque required to overcome engine friction. Find the mean height of the twisting moment diagram by measuring its area (using, say, a planimeter), and dividing by the length, and draw the line AB to represent this mean twisting moment. The accuracy of the work may be checked by calculating the mean twisting moment from the l.H.P. as follows : ART. 243] TWISTING MOMENT DIAGRAMS 431 Let T m = mean twisting moment in pound-feet, n = mean speed of shaft in revolutions per minute. Then I.H.P. = - 33,000 I.H.P. X 33,00 and l m = m 27772 Then at points C, D, E, and F the equivalent twisting moment exerted on the shaft is equal to the resisting moment and the speed is constant. From C to D, and also from E to F, the shaded areas above CD and EF represent the excess work done on the piston over that taken of the crank- shaft, and the engine will accelerate in speed from the minimum value at C or E to the maximum value at D or F. From D to E the work taken off the crankshaft is greater than that done on the shaft by the shaded area shown, and the speed will fall from the maximum value at D to the minimum value at E. The function of the flywheel is to absorb the excess energy from, say, C to D with a pre-determined rise in speed, and to restore it again from D to E with the same fall in speed. The flywheel must therefore be designed to give a pre-determined cyclic variation in speed. Let a) = mean speed in radians per second, eoj = maximum speed in radians per second, a> 2 = minimum ,, Then the excess energy to be stored by the flywheel is where I = moment of inertia of the wheel = / Now 0,= 2 a) 2 ) ........... (l) Let E = kinetic energy of the wheel when running at the mean speed co. Then E = JIco* ........ ( 3 ) (3) in (2) gives (4) V T/ or if q denotes the percentage variation in speed permissible COi COo ioo and , = (5) IOO or (6) 432 THE THEORY OF HEAT ENGINES [CHAP. xix. Also from (2) e=lo>*^ Q (7) EXAMPLE i. A gas engine working on the four-stroke cycle develops 15 I.H.P. at 250 revolutions per minute. Assuming one explosion every 2 revolutions, that the resistance is uniform and that the speed is not to vary more than i per cent, above or below the mean speed, calculate the weight of flywheel required if its radius of gyration is 2*5 feet. Take the fluctuation of energy as 075 of that developed during a working stroke. (L.U.) Excess energy to be stored by the wheel = 075 X * 5 X 33> 000 = 2970 foot-pounds By (7) e = la>*.^ ~\\7 / r 1 \S \ ^ 32*2 \ 60 / ioo W X 6-2S /25 X 77\2 2 2970 = ^ ^X( 5 ) X- 32-2 \ 30 / ioo f ,., .^ 2970 X 32*2 X 900 X ioo from which W-= -^ = 1 1 ia pounds 6-25 X 625772 X 2 EXAMPLE 2. A gas engine is provided with two flywheels each of weight 1 1 -5 cwts., and radius of gyration 1*87 feet. There is i working stroke in 2 revolutions. The diameter of the cylinder is 7*5 inches, stroke 9 inches and mean speed 250 revolutions per minute. The mean pressure during the firing stroke is 887 pounds per square inch, during the com- pression stroke 15*1 pounds per square inch, during the exhaust stroke 4'4 pounds per square inch and during the suction stroke, atmospheric. If the resistance be constant, find the percentage variation of speed of the engine. (L.U.) }= X ^^ X 7'5 2 ) X 075 = 887 X 44'i8 X 075 = 2939 foot-pounds. =< 1 S- I +4-4) X 44-18 X 0-75 = 646 foot-pounds net work done per cycle = 2939 646 = 2293 ft.-lbs. Excess energy during working stroke to be absorbed by the flywheels Moment of inertia of both flywheels together = 2939 - -F = 2366 ft.-lbs. X X (r8 7 )2 = 280 pound-feet units. 32*2 X 277 2577 . = -4- radians per second 60 3 ART. 243] TWISTING MOMENT DIAGRAMS 433 .'. Wi a) 2 _ 2366 2366X9 280 X 625^2 percentage fluctuation of speed = 1-23 per cent., or 0*66 per cent, above or below mean speed. EXAMPLES XIX 1. The connecting rod of a vertical engine weighs 210 pounds, and is 66 inches long, the stroke being 33 inches. The centre of gravity of the rod is 40 inches from the cross- head pin centre. The diameter of the cylinder is 20 inches, and the engine runs at 60 revolutions per minute. The combined weight of the piston, piston rod, crosshead, and crosshead pin is 600 pounds. Find the twisting moment when the crank is 60 from the inner dead centre and the piston is moving towards the crank shaft (a) neglecting the inertia of the reciprocating parts ; (b) allowing for inertia, and assuming the rod to be equivalent to two masses at the crank pin and crosshead pin . For this position of the crank assume the steam pressure on the head end of the piston to be 60 pounds per square inch and on the crank end 10 pounds per square inch. 2. A horizontal engine, stroke n inches, connecting rod 2 feet 9 inches between centres, runs at 300 revolutions per minute. The mass of the rod is 105 pounds, distance of e.g. from crosshead centre I '655 feet, and radius of gyration about the centre of gravity 1-075 feet - Find tne kinetic energy of the rod when the piston is moving towards the crank shaft and the crank angle is 30 with the inner dead centre, and the effect upon the turning moment. Find also the approximate effect on the turning moment when the rod is assumed to be equivalent to two masses at the crank pin and crosshead pin respectively. 3. A connecting rod weighs 105 pounds, distance between centres 33 inches, distance of e.g. from crosshead centre 20 inches, radius of gyration about the e.g. VrTJ feet. Find the error in the turning moment caused by the usual assumption, at any crank angle 0, when the engine is making 250 revolutions per minute and its stroke is ii inches. (L.U.) 4. A horizontal steam engine cylinder has a diameter of 20 inches, stroke 18 inches, connecting rod 3 feet 6 inches between centres, mass of reciprocating parts, including connecting rod, 280 pounds. Find how much the pressure per square inch of the cushion steam must exceed that on the other side of the piston at each end of the stroke, so as to relieve the crank pin brasses of all pressure when the engine is running at 350 revolutions per minute. (L.U.) 5. A compound steam engine develops 600 I.H.P.'at 90 revolutions per minute, and from the twisting moment diagram it is found that the fluctuation of energy is 20 per cent, of the energy exerted in one revolution. Find the moment of inertia of the flywheel required, in order that the fluctuation of speed may not exceed I per cent. 6. A single-acting Otto cycle gas engine develops 60 I.H.P. at 160 revolutions per minute, with 80 explosions per minute. The change in speed from the beginning to the end of the power stroke must not exceed 2 per cent, of the mean speed. Design a suit- able rim section for the flywheel, so that the hoop stress due to centrifugal force does not exceed 600 pounds per square inch. Neglect the effect of the arms, and assume that the work done during the power stroke is i^ times the work done per cycle. (L.U.) 7. A gas engine, running at a mean speed of 180 revolutions per minute, has a cylinder diameter of 20 inches and stroke 2 feet. The mean pressure during the forward stroke is 120 pounds per square inch gauge ; during exhaust stroke, 2 pounds per square inch gauge ; during the suction stroke, I pound below atmospheric ; and during the com- pression stroke, 40 pounds per square inch gauge. If the resistance on the crank shaft be constant, and the percentage cyclic variation of speed 1*5 per cent., find the moment of inertia of the flywheel required. 2 F CHAPTER XX BALANCING 244. Centrifugal Force. Let a body of weight W pounds revolve in a circle of radius r feet, with uniform angular velocity co radius per second. Then it may easily be shown that the body is subjected to an acceleration coV feet per second per second in a direction acting radially inwards towards the centre of the circle. The force acting on the body which gives to it this acceleration is commonly called the centripetal force, and must act in the same direction as the acceleration it produces. Hence this force acting on the body compelling it to move with uniform speed in a circle is equal to W o a)*r pounds g and it acts in a direction radially inwards. If, now, the body is connected to the axis of rotation by means of a string or a rigid link, the link will be put in tension, and will itself supply the inward force on the body, and will simultaneously exert an equal and opposite force on the axis of rotation called the centrifugal force. Any rotating weight carried at, say, the crank-pin of an engine will therefore put the crank arm in tension, and exert on the shaft a force which acts radially outwards, of magnitude W o) 2 r pounds ( i ) o This outward force on the shaft is sometimes called the dynamical load on the shaft due to the rotating weight, and may be expressed in terms of revolutions per minute (n) if so desired, since W , 2irn F = - coV and a> = -7 g 60 W /27m\ 2 IT* ... . 9 , . .-. F = . ( ) r = . W 2 /- = o-ooo34\V;7z 2 . . (2) g \ 60 / 900^- EXAMPLE. The crank arms and crank-pin of a crank-shaft are equiva- lent to a weight of 800 pounds at a radius of 12 inches. The crank-shaft is supported on bearings 5 feet apart from centre to centre, and the centre of the crank-pin is 2 feet from the left-hand bearing. If the diameter of the shaft is 9 inches, find the dynamical load on the shaft and on its bearings when the shaft is running at 300 revolutions per minute. Find also the loss in friction at each bearing if the coefficient of friction between shaft and bearing be taken as o'o6. 434 ART. 245] BALANCING 435 Total dynamical load on the shaft = = 0-00034 X 800 X i X (300)2 = 24,480 pounds Two-fifths of this load will evidently act on the right-hand bearing and three-fifths on the left-hand bearing, hence dynamical load on right-hand bearing = f X 24,480 = 9792 pounds left-hand = 24,480 9792 = 14,688 pounds H.P. lost in friction at right-hand bearing = 9792 X o'o6 X TT X9 X 300 12 33,000 H.P. lost in friction at left-hand bearing = 12-55 Xf = 18*82 H.P. 245. Dynamical Load on a Shaft due to any Number of Rotating Weights in the Same Plane. Suppose we have, say, five weights Wj, W 2 , W 3 , W 4 , and W 5 revolving in one plane with uniform angular velocity w at radii r lt r 2 , r 3) ? 4 , and r 5 respectively. Each weight will give rise to an outward pull on the shaft, and the total dynamical load will be the vectoral sum (W^ -f- W 2 ?2 + W 3 r 3 -j- W 4 ; 4 -f- W 5 r 5 ) OJ 2 we may therefore find the vectoral sum 27W> and, by multiplying by - find the total dynamical load in the shaft. EXAMPLE. Suppose we have five weights of 2, 3, 4, 5, and 6 pounds rotating at radii of 1*4, 2, 1-5, 175, and i foot respectively, the phase 2i wr=Z a -^ Jfo ^^<60* +~f* . w-^X. 90 X- ^s/ x^ tt' / < ^ 9 Wr-6 +e *? rrr-o v^ >x C7 Ja A-' Wr=6 M>=875 FIG. 214. angles between the radii being as shown in Fig. 214. Find the resultant dynamical load at a) radius per second. Z"W> is found by drawing the vector polygon shown in Fig. 214 to any 436 THE THEORY OF HEAT ENGINES [CHAP. xx. convenient scale. The sum is 2 '6 being represented by db in magnitude and direction ; the total dynamical load will therefore be CO 2 2 '6 X pounds 32-2^ 246. Method of Balancing any Number of Rotating Weights in one Plane. A set of rotating weights in one plane is balanced when the total dynamical load on the shaft is zero, i.e. when ZWr = o, and the force polygon closes. In the above example 2Wr = ab = 2 '6, hence in order to balance this system of rotating weights it is only necessary to add one balance weight W diametrically opposite to the resultant at such a radius r Q that W r is equal to 2*6. This is shown in Fig. 214, the radius r making an angle of 60 with the weight of 2 pounds. 247. Rotating Weights in more than one Plane. Let there be two equal and diametrically opposite weights W rotating at equal radii r, the distance between the planes of rotation being x (Fig. 215). In this W W Axis of Potation FIG. 215. case SWr = o, the force polygon being a straight line closing upon itself. There will, therefore, be no out of balance force, or dynamical load, on the axis of rotation, but the system will not be in perfect balance because there will be a couple exerted perpendicularly to either plane and proportional to W> X x. In order to neutralise this couple two rotating balance weights must be added, one in each of two different planes. This may conveniently be done by adding a balance weight W 1 diametrically opposite each of the rotating weights such that W^ = W>. The Resultant of a Number of Rotating Weights can be ieplacedby a Single Force in any Plane, together with a Couple perpendicular to it. Consider first, a single weight W rotating at radius r in the plane B (Fig. 216). The effect of this weight on any reference plane such as A, distant x from B, is to exert an equal and parallel force at A (proportional to W>) of W W magnitude / x w 2 and a couple of magnitude o) 2 rx (proportional to o o ART. 247] BALANCING 437 Wo:) tending to cause rotation about an axis through A perpendicular to the reference plane. t F=%rxa* > '** r ^ ' I \ i \ j \ i \ \ t \ \ 1 \ r \ +. ^ X * Axis of Rotation - >F j B i : i 1 \ \ 1 1 1 V \ \ I I i i \ V < t... ta*9>jM*> \ / / Reference Plane FIG. 216. Consider next a number of rotating weights, say five, in different planes and choose a reference plane to the left (or right) of these weights (Fig. 217). Find the value of Swr by drawing the force polygon Reference Plane Resultant Couple =AF^ FIG. 217. abcdef and find the resultant force which is proportional to Wr or of. Next find the resultant couple by drawing the couple polygon ABCDEF, each side of which is proportional to Wrx, i e. find the value of 438 THE THEORY OF HEAT ENGINES [CHAP. xx. The resultant couple will be represented by AF, and the combined effect of these five rotating weights will be a single force at O proportional to W> or of, and a couple proportional to Wrx or AF. If the five rotating weights be in perfect balance amongst themselves it is evident that both the force and the couple polygon will close, and there will be no unbalanced force and no unbalanced couple. 248. Method of Balancing any Number of Rotating Weights in different Planes. From Art. 247, it will be seen that any number N9I 500/6* /V2 600/bs. fi3 400lbs. N9I N93 FIG. 218. of rotating weights may be completely balanced by adding two rotating balance weights, one in each of any two different planes, the conditions required being that (1) The force polygon must close, i.e. 27W> = o. (2) The couple polygon must close, i.e. SWrx = o, the moments being taken about any plane. ART. 248] BALANCING 439 If the two planes, say A and B, in which the balance weights are to be placed are given, the method of procedure is as follows : Choose one of the planes, say A, as a reference plane, thereby elimi- nating the couple produced by the balance weight in this plane, and draw the couple polygon ; the closing line will represent the couple which must be exerted by the balance weight in the other plane B. This couple divided by the distance between the two planes will give the force, W>, required in the plane B. Repeat the construction using B as the reference plane, and find the value of W> required in the plane A. To test the accuracy of the work choose any other reference plane, and, having put on the balance weights in their proper angular position, draw a new couple polygon. If the work has been carried out accurately this polygon will close. The force polygon will evidently be the same for any plane. Rule for drawing the Couple Polygons. If all the rotating masses be on one side of the reference plane the vectors should be drawn radially cutwards from the axis of rotation. If some of the rotating weights are on one side and some on the other side of the reference plane, the vectors for one side should be drawn radially outwards, and for the other side radially inwards. For the Force Polygon all the vectors should be drawn radially outwards. EXAMPLE. Fig. 218 shows three rotating weights of 500, 600, and 400 pounds of radii of i, 1-5, and 1*25 feet respectively, balance weights required in the planes A and B. Find the 440 THE THEORY OF HEAT ENGINES [CHAP. xx. The values of Wr for the three weights are, 500, 900, and 500 respectively. To find the force required in plane B. Choose A as the reference plane and draw the couple polygon (taking moments about A), as shown at (a) Fig. 219. The closing line will be found to scale 6700. Hence the value of Wr for the balance weight in plane B is '- = 837. Next taking B as the reference plane draw the couple polygon (taking moments about B), as shown at (b) Fig. 219. The closing line will be found to scale 6530. Hence the value of Wr for the balance weight in plane A is - 816. The accuracy of the work is checked by drawing the force polygon, as shown at (c) Fig. 219. The angular position of the balance weights are shown in the lower portion of Fig. 218. A balance weight in plane B of 837 pounds at a radius of i foot, making 157 with the direction of No. 2, and a balance weight in plane A of 816 pounds at a radius of i foot, making 170 with No. 2, would therefore be suitable. 249. Balancing Reciprocating Weights assuming Simple Harmonic Motion. Primary Balancing. If the connecting rod be infinitely long, the piston moves with simple harmonic motion, its acceleration being where x is the displacement of the piston from mid-stroke. If r denotes the length of the crank, and 6 the crank angle measured from the dead centre, then x = r cos 6, and the acceleration is a> 2 r cos 6 The acceleration of the piston is therefore equal to the resolved part of the centripetal acceleration of the crank-pin (w 2 r) in the direction of the line of stroke. If W denotes the weight of the accelerating parts per cylinder (estimated as in Arts. 238 and 239), the inertia force for each reciprocating weight is for any crank angle afir cos g This force acts along the line of stroke, and is evidently equal to the resolved part of the centrifugal force of a weight W, at a radius r, in the direction of the line of stroke. If, therefore, a rotating weight W be attached to the crank- shaft diametrically opposite to the crank-pin, it will balance the inertia force of the reciprocating weight, and there will be no oscillations in the direction of the line of stroke. On the other hand, the addition of such a rotating weight will be to throw the engine out of balance in a plane perpendicular to the line of stroke, for there will always be the component of the centrifugal force of the rotating weight in the plane. It is therefore evident that a reciprocating weight cannot be perfectly balanced by means of a rotating weight; it can only be balanced by means of another reciprocating weight in the same plane, and always moving in the opposite direction, so that its inertia force balances that of the other reci- procating weight. In a vis-ti-vis engine in which the cylinders are on opposite sides of the crank, the connecting rods driving one common crank-pin, the reciprocating ART. 249] BALANCING 441 weights for each cylinder will be balanced amongst themselves if the con- necting rods are very long and the motion of the pistons simple harmonic. A two-cylinder engine with cylinders whose centre lines are x apart, and whose cranks are at 180, will be balanced if the reciprocating weights per cylinder are equal as regards forces, i.e. the force polygon will always close, being a straight line closing upon itself, but there will be an out of balance couple due to both rotating and reciprocating masses, as explained in Art. 247. A number of reciprocating weights in different planes cannot, therefore, 40 4-( 30 3' i ^o B c D 4'- 4 10' -> +- -4' > -> C"462 00 -T^rta** o\ 0=340 8=480 0=340 be perfectly balanced by rotating weights, but only by adding other recipro- cating weights driven by cranks at the correct angles. The general rules for primary balance, i.e. when the obliquity of the connecting rod is neg- lected and simple harmonic motion assumed, are exactly the same as those for rotating weights given in Art. 248, and by suitably arranging the crank angles the reciprocating weights may be 'arranged to balance themselves, provided there are not less than four of them. EXAMPLE i. In a four-crank engine the distances between the cylinder 442 THE THEORY OF HEAT ENGINES [CHAP. xx. centre lines are 4, 10, and 4 feet respectively, reckoned from left to right. Reading from left to right, the reciprocating weights per cylinder are, for crank A 340 pounds, crank B 480 pounds, crank C (not known), and crank D 340 pounds. Find the reciprocating weight required on crank C and the crank angles in order that they may mutually balance. Take C as the reference plane and draw the couple polygon, making it close (taking moments about C), as shown at (a), Fig. 220. Care must be taken in drawing the vectors in the right direction. A and B are on one side of the reference plane, and the vectors may be drawn outwards ; D is on the other side of the reference plane, and it must therefore act inwards. Suppose crank A be fixed vertically, then draw ab vertically to represent the couple for A equal to 340 X 14.= 4760; for crank B, with centre b and radius equal to the couple 480 X 10 = 4800, describe on arc; for crank D, with centre a and radius equal to the couple 340 X 4 = 1,360, describe another arc, cutting the first one in point c. Join be and ac> this will fix the directions of the cranks B and D ; draw OD parallel to ac and OB parallel to be, and scale off the crank angles. The crank angle for C and its reciprocating weight may next be found by drawing the force polygon as shown at (b), Fig. 220, drawing each vector radially outwards from the centre of the crank-shaft. Draw a'b' parallel to ab to represent 340 pounds for crank A, tic* parallel to ac to represent 340 pounds for crank D, c'd' parallel to be to represent 480 pounds. The closing line tfa 1 , which scales 482 pounds, gives the reci- procating weight required for crank C and its crank angle. From O draw parallel to d'd and scale of the crank angle. The accuracy of the work may be checked by taking any other reference plane and drawing a new couple polygon which will be found to close. 250. Balancing of Locomotives. The most common form of locomotive is the two-crank engine, with cranks 90 apart. It has been shown in Art. 249 that when the cranks are 180 apart, the rotating and re- ciprocating weights balance themselves as regards forces only ; when the cranks are at 90, however, there will be out of balance forces and couples unless some method is used to balance them. With the two-cylinder engine it is not convenient to add a reciprocating balance weight, and the usual prac- tice is to take two-thirds of the reciprocating weight per cylinder, and all of the out of balance rotating weight, and assume that the sum of these two acts at the crank-pin as a rotating weight. The method of balancing rotating weights is then followed as previously explained, the two balance weights being added to the wheels. FIG. 221. ART. 250] BALANCING 443 The two-crank engine may be conveniently treated analytically as follows : Let W be the total out of balance weight assumed concentrated at each crank-pin, and r the length of the crank, then referring to Fig. 221 for an inside cylinder engine, we can balance W at crank-pin A by two balance weights w and w 2 arranged diametrically opposite the crank A, such that W r w l - r + W 2 r 2 and Similarly the weight W at crank-pin B may be balanced by two weights w-i and w% arranged diametrically opposite the crank B. The two balance weights w and w may then be replaced by a single weight w Q at the same radius r, such that j i and tan = w l Similarly for the wheel nearest the crank B a balance weight will have the same effect as w z and w 2 , provided , 2 tan <6 = Y 0/2 The engine will, of course, only be very imperfectly balanced, since a little consideration will show that there will be for any position of the cranks (1) An unbalanced force in the line of stroke. (2) An unbalanced force in a vertical plane. (3) An unbalanced couple causing swaying from side to side. (4) An unbalanced couple which tends to cause oscillations in a vertical plane, in case the centre line of the driving axle is not in the same straight line as the drawbar pull. EXAMPLE i. Estimate the balance weights required for an inside cylinder uncoupled locomotive from the following data : Weight of reciprocating parts per cylinder = 600 pounds. rotating =700 Centre lines of cylinders 2 feet apart, length of cranks 13 inches, radius of balance weight circles 2 feet 3 inches. Balance all of the rotating and two-thirds of the reciprocating weights. Weight to be balanced per cylinder, assumed concentrated at the crank-pin : = 700 + f X 600 = 1 100 pounds Consider the left-hand crank. Taking moments about A (Fig. 222), we have 0/2 X 27 X 60 = 1 100 X 13 X 18 w 2 = 159 pounds 444 THE THEORY OF HEAT ENGINES [CHAP. xx. Taking moments about B, we have a>i X 27 X 60= noo X 13 X 42 w = 371 pounds WQ = v 37 1 2 -f- *59 2 = 43 pounds tan (j> = iff = 0-429 .*. < = 23 10' If the left-hand crank is leading, the required balance weights will therefore be, L.H. wheel 403 pounds, 113 io ; behind R.H. crank, or LH. R.H. 18 m* Left hand wheel, looking in direction of arrow. ^ ^Reference plane ^^ through A . Couple polygon Reference plane through B. FlG. 222. 156 50' in front of L.H. crank, R.H. wheel 403 pounds, 156 50' in front of R.H. crank shown dotted in Fig. 222. Graphical Solution. For balance weight on left-hand wheel, choose the right-hand wheel as a reference plane, and draw the couple polygon as shown in Fig. 222, i.e. for R.H. crank make a&= noo X 13X18=257,400, and for L.H. crank bc=. noo X 13 X 42 = 600,600; the closing line ca scales 654,900 and angle fca = = 23. Hence, if WQ is the balance weight required, ^o X 27 X 60 = 654,900 w Q = 404 pounds, agreeing with the calculated value. ART. 250] BALANCING 445 EXAMPLE 2. Find the position and amount of the balance weights required for the wheels of a coupled outside cylinder locomotive, the wheel centres being 5 feet, cylinder centres 7 feet 6 inches, and coupling rod centres 9 feet, cranks 13 inches long, radius of balance weights 3 feet. Weight of reciprocating parts per cylinder 500 pounds, and unbalanced rotating parts referred to the crank pin 200 pounds. The weight of each coupling rod is 600 pounds, and the coupling rod crank-pins are in line with the main crank-pins. Weight to be balanced per cylinder assumed concentrated at crank-pin = 200 -f- X 500 = 533 pounds at crank-pin = M2 = ^00 pounds at coupling rod crank-pin Driving wheels (see Fig. 223). In the outside cylinder engine the LH. R.H. 300 533 Q4" - IOQ" 90" - - 60' RH Balance Weight B \ Left hand wheel, looking indirection of arrow. FIG. 223. weights to be balanced are not all on the same side as the reference plane, hence, considering the left-hand crank, and taking moments about B x we have Wj (opposite L.H. crank), W x X 3 6 X 60 = (533 X 13 X 75) + (300 X 13 X 84) .*. Wj = 392 pounds. Taking moments about A, we have W 2 (in line with R.H. crank) since W 2 is on the opposite side of reference plane through A. W 2 X 36 X 60 = (533 X 13 X 15) + (30 X 13 X 24) W 2 = 9 1 '5 pounds tan = = 0-234 = 1 10 W Q = V(39 2 ) 2 + (9*'5) 2 = 403 pounds The position of the balance weight for each driving wheel is shown in Fig. 223. 446 THE THEORY OF HEAT ENGINES [CHAP. xx. Coupled wheels. Taking moments about B (Fig. 224), W x X 36 X 60 = 300 X 13 X 84 w i = 1 53 '5 pounds. Taking moments about B, W 2 X 36 X 60 = 300 X 13 X 24 ^2 = 43*3 pounds tan = = i 5 i6' w 'o = A/(43'3) 2 + ( I 53'5) 2 = i59'S pounds The position of the balance weight for each coupled wheel is shown in Fig. 224. The reader should check the above work graphically, taking L.H. R.H. 300 fc- h- 24-" ->- B 300 R.H.Balam Weight Left hand wheel, looking in direction of arrow. FIG. 224. care to follow the rule given in Art. 248 for drawing the vectors of the coupled polygon. 251. Secondary Balancing. The method of balancing the recipro- cating weights given in Art. 249 is only approximately correct when the connecting rod is not very long compared with the length of the crank. It has been shown in Art. 237 that the actual acceleration of the piston is given by cos o Q . coV)cos 04- C (2 sin* 0) or approximately by f cos 20 ) j cos -| ART. 251] BALANCING 447 For secondary balancing the latter expression is used, and the inertia force to be balanced is taken as W f . cos 26 f a . cos 2 \ { cos 6 + - - > I J Writing 6 = a>t and n = -, this becomes W( r ) a) 2 r I cos wt -f- - cos 2 co/ > (i) WC r 2 ) or jeu 2 r cos co/4~ ( 2Ct) ) 2 ^ cos 2 CD/? (2) The first term in (2) represents the projection OA (Fig. 225), on the line C OB FIG. 225. of stroke of the inertia force due to a weight W, concentrated at the crank pin C, and rotating at an angular velocity w. This is known as the primary force. The second term in (2) represents the projection OB (Fig. 225), on the line of stroke of the inertia force due to a weight W concentrated at an 7-2 imaginary crank D, of radius , and rotating at an angular velocity 2o> in 4/ the same plane as the main crank. This is known as the secondary force. For perfect secondary balancing the conditions will evidently be (1) The primary force polygon must close (2Wr = o). (2) The primary couple polygon must close (2Wrx o). r 2 (3) The secondary force polygon must close (27W . - = o). 4/ r 2 (4) The secondary couple polygon must close (27VV -,. x = o). In a two-crank engine with equal reciprocating weights and cranks at 1 80, the primary forces are balanced, but the primary couples and the secondary forces and couples cannot be balanced. A three-crank engine with equal reciprocating weights and cranks at 120 the primary and secondary forces may be balanced, but the primary and secondary couples cannot be balanced. A four-crank engine may be arranged to be balanced for primary and secondary forces and for primary couples, but not for secondary couples. A five- and six-cylinder engine may be completely balanced for both primary and secondary forces and couples. 1 1 For a complete discussion of secondary balancing, see Professor W. F. Dalby's book on " Balancing of Engines." Edward Arnold. 448 THE THEORY OF HEAT ENGINES [CHAP. xx. EXAMPLE i. In a three-crank engine the reciprocating weights per cylinder are each equal to i ton, and the length of the connecting rod is 4 cranks, the cranks being 120 apart, and i foot long. Estimate the out- of-balance primary and secondary couples if the distances between the cylinder centre lines are 3 feet, and the engine runs at 120 revolutions per minute. The primary and secondary force polygons will evidently close, since each of them will be an equilateral triangle. Out-of -balance Primary Couple. Taking I (Fig. 226) as a reference n o ; 120 Closure = 5-2 FIG. 226. plane, we have for crank II Wrx = i X 3 = 3> for crank m Wr# = 6, hence drawing db parallel to crank II to represent the couple 3, and be parallel to crank III to represent 6, we find the closing line ac scales 5-2. Now a) = -j~ X 27T = 477 radians per second .*. out-of-balance primary couple == (EWrx) o 32-2 _ feet Out-of-balance Secondary Couple. The crank angles for the imaginary secondary cranks are shown in Fig. 227.- Measured in order in a ART. 251] BALANCING 449 clockwise direction from No. I, the angle between Nos. I and II is 240, and between Nos. I and III 480. The radius of these cranks is = Afoot and their angular velocity 20) or STT radians per second. The out-of-balance couple may be most conveniently found by taking, say, I as a reference plane, and assuming unit radius for each secondary crank. This has been done in Fig. 227, and the closing line scales 5*2 (as before). The out-of-balance secondary couple will therefore be 32-2 X 16 EXAMPLE 2. A two-cylinder vertical engine has its cranks at i8o c The mass of the recipro- cating parts for each cylinder is 276 pounds, stroke 18 inches, con- necting rod 3 feet 9 inches, distance between / N \ \ centres of cylinders 3 feet, revolutions per ^^\^ 2 fS^ 7 /^L minute 354. Assuming D*""^ >m that the rotating masses Secondary Cranks are balanced, draw to a base of crank angles curves representing (i) the alternating force, (2) the alternating couple set up during a revolu- tion (L.U.). First draw the ac- celeration curve for one piston, using, say, Klein's construction (Art. 237). This will represent the accelerating force to scale. When the crank is on the inner dead centre 6 = o, the ac- celeration is ofirl i + - ), and the accelerating FIG. 227. W Closure =5 -2 force is o) 2 r(i -I ). g \ l */ 276 on inner dead centre F = ^- X = 3 il x X ffi-P 12 \ 45 900 3 6 X - X - = 10,600 pounds 2 G 450 THE THEORY OF HEAT ENGINES [CHAP. xx. The force scale may now be calibrated, since OA represents 10,600 pounds. Draw any line such as OB (Fig. 228) scaling io % 6 units, and FIG. 228. divide it into units from O. Join BA, and draw parallels, as shown, which will calibrate the curve for equal intervals of 1000 pounds. Next draw the curve for the other cylinder as shown in Fig. 229, add 360 FIG. 229. the ordinates of the two curves together, and obtain the resulting curve of out-of-balance forces shown. The alternating couple is W ^ X acceleration X distance between cylinders. ART. 251] BALANCING 451 The force curve (Fig. 228) will therefore represent the couple to a different scale. The maximum value of the couple is 10,600 X 3 = 31,800 pound-feet It should be noted that if the connecting rods were infinitely long, the force curves for the two cylinders would neutralise one another, and the resultant force for all positions of the cranks would be zero, i.e. the reciprocating weights would mutually balance as regards forces, but, as already stated, there would still be the out-of-balance couple. The reader should draw similar curves for the case in which the cranks are 90 apart. EXAMPLES XX 1. A single-cylinder gas engine of stroke 18 inches is fitted with two fly-wheels, one on each side of the crank. The distance between the planes of the wheels is 6 feet, the left-hand wheel being 2 feet from the centre of the crank pin. The rotating weight is 200 pounds at the crank pin and the reciprocating weight 300 pounds. Find the balance weights required on the wheels at I foot radius, all of the rotating and two-thirds of the reciprocating weights to be balanced. 2. Four weights of 2, 4, 6, and 8 pounds are rotating in the same plane at 120 revo- lutions per minute at radii of I, i'5, 2, and 1-75 feet respectively. Reckoning in order from the 2-pound weight the phase angles of the different radii are, between 2 and 4, 30, between 2 and 6, 90, and between 2 and 8, 210. Estimate the resultant dynamical load on the shaft, and the position and amount of the balance weight required at a radius of I foot. 3. Reckoning from the left in order, let BCD represent the cranks of a three-cylinder engine. The out-of-balance rotating weights are : at B 200 pounds, at C 400 pounds, and at D 250 pounds. The length of each crank is I foot. Find the balance weights required in the planes A (to the left of B) and E (to the right of D), given the following distances : between B and C 2 feet, C and D 2 feet, U and E 3 feet, and A and B 4 feet ; the crank angles are between B and C 120, between B and D 240. 4. In a four-crank engine the distances between the cylinders are all equal. Three reciprocating masses reckoned in succession from the left are i, 2, and 2 tons. Find the fourth reciprocating mass and the crank angles so that the reciprocating masses may mutually balance. (L.U.) 5. Reckoning from the left in order, let ABCDE represent the cranks of a five- cylinder engine. The reciprocating masses are : at A I ton, at B 2 tons, and at C 3 tons. The angle between A and B is 90, and between B and C (in order) 120. Find the reciprocating masses and the crank angles for D and E in order that they may mutually balance. 6. Find, from the following data, the magnitude and position of the balance weights for an inside cylinder uncoupled locomotive : radius of crank, 12 inches ; radius of balance weights, 33 inches ; weight of reciprocating parts of each cylinder, 550 pounds ; weight of rotating parts of each cylinder, 500 pounds ; cylinder centres, 25*5 inches ; wheel centres, 69 inches. All of the rotating and two-thirds of the reciprocating weights to be balanced. (L.U.) 7. From the following data of an outside cylinder uncoupled locomotive calculate the magnitude and position of the balance weights to balance all the rotating and two- thirds of the reciprocating parts : Weight of reciprocating parts for each cylinder 500 pounds, weight of rotating parts for each cylinder 600 pounds, cylinder centres 66 inches, wheel centres 56 inches, stroke 24 inches, radius of .e.g. of balance weight 30 inches. Find also the maximum variation in rail pressure when running at 30 miles per hour. Diameter of driving wheels 7 feet. (L.U.) 8. In a three-crank engine the reciprocating weights per cylinder are each equal to I ton and the connecting rod is 3 '5 cranks long. The distances between the cylinder centre lines are 4 feet, the engine runs at 180 revolutions per minute, and the cranks are 120 apart and I foot long. Calculate the out-of-balance primary and secondary couples. CHAPTER XXI GOVERNORS 252. Function of the Governor. The function of the governor is to control the speed of the engine over a long period, or in other words, to keep the speed from varying beyond desired limits when the load on the engine is changed. In the case of a steam engine the governor may be made to actuate a throttle valve in the main steam pipe (throttle govern- ing), or to alter the point of cut-off (cut-off governing). In an internal combustion engine the governor varies the amount of fuel admitted into the engine cylinder to suit the load on the engine. The governor can only work by a change of speed which alters the position of the governor balls, such alteration being utilised by a suitable gear in order to vary the amount of working fluid admitted to the cylinder. The cyclic variation in speed due to an uneven turning moment on the crankshaft is controlled solely by the flywheel, as already mentioned (Art. 243). 253. The Watt Governor. The early Watt governor is shown diagrammatically in Fig. 230. The vertical governor spindle is driven by the engine, thus causing the two balls to rotate in a circle. In order to make the balls rotate in a circle, a force must act radially inwards on each ball (Art. 244) of amount cuV; this force is supplied by the combined action of the weight of the ball (w Ibs.) and the tension (T Ibs.) of the supporting arm. Referring to Fig. 230, let the angular velocity of the ball be constant and equal to a) radians per second, then there are two forces acting on the ball, viz. its own weight w acting vertically downwards, and the tension T of the supporting arm. Let r = the radius of the ball circle in feet, h = height of the governor in feet. The sum or resultant of T and W is shown in the lower part of Fig. 230, where ab T and bc=- W ; the horizontal line ac represents the centripetal force F, i.e. the vectoral sum of T and W acting on the ball and making it W rotate with uniform speed in a circle ; its magnitude is evidently a> 2 r pounds. The two triangles ABC and dbc are similar, hence BO^fc CA ca h w _ r * . i.e. ""f r F = w 'fr W 452 ART. 253] GOVERNORS 453 h W9 P- or - = = A r WCO 2 f CO 2 ?" If n = revolutions per minute, then a) = -'-= and 60 30 , _^f_X 9 _ 3 2 * 2 X 900 i 772^2 772 2937 X 12 35240 . -- ~ (3) The height of the governor is therefore inversely proportional to the square of the angular velocity. FIG. 230. Let h = height when the angular velocity is coj and/& 2 = ,, ,, changes to Then from (2) above 454 THE THEORY OF HEAT ENGINES [CHAP. xxi. and the change in height is If, therefore o^ and cu 2 denote the two extreme speeds permissible, the change in height h h% must be sufficient to actuate the gear required. At high speeds the change in h or the movement of the governor sleeve rapidly falls off as the speed increases, as will be evident from the following : n. , = 3^o Change in h (inches). 60 lo 90 100 120 I4O 1 60 180 200 9*79 7*i93 5*507 4*35i 3*5 2 4 2-450 1-798 i*375 1-087 0-881 2-507 1-686 1-156 0-827 1-074 0-652 0-423 0-288 0-206 A change in speed from 180 to 200 revolutions per minute only changes the height (neglecting friction) o'2o6 inch, and, further, the height at the latter speed is only o'88i inch. On the other hand, a smaller change from 60 to 70 revolutions per minute changes the height 2-5 inches. It will be evident, therefore, that this type of governor is not suitable for high speeds, and, further, the power of the governor, or the work it can do on the governor sleeve, is very small, being only equal to 2w X extreme variation in h. 254. The Porter Governor. The power of a Watt governor is greatly increased by adding a dead weight to the governor sleeve as in the Porter governor (Fig. 231). Before any movement of the sleeve can occur, this dead load must be lifted, and this necessitates a higher speed of rotation of the balls in order to obtain the required lifting force at the sleeve. Let w = weight of each ball in pounds, W = dead load on the sleeve in pounds, , vertical movement of the sleeve vertical movement of the balls /W\ then each ball has to lift a total equivalent weight of w + ( \k pounds, and by the same reasoning as before, the centrifugal force F per ball must be , W N w 05 W + w (I) ART. 254] GOVERNORS 455 FIG. 231. 456 THE THEORY OF HEAT ENGINES [CHAP. xxi. In the Porter governor, as usually constructed, all the links are equal in length, and therefore k = 2, and w a) Expressing this in terms of revolutions per minute, we have , _ w -h W gX 900 (3) or in order to run at a height of h feet, the speed in revolutions per minute must be ,,2 ^937 a^ + W rl 7 n w An alternative method of treating the theory of this governor is to reduce it to a statical problem by inserting a hypothetical outward force on the ball, which we will call the controlling force, equal to the inward force F= afir. Taking moments about point C (Fig. 232) we have o W F Xy = wX ac-\ -- X be With equal links y = h, ac = r, and be = 2r .-. /i = wr + Wr or /i = (w + W)r and since F = -afir & as before. Effect of Friction. The effect of friction in the governor mechanism is usually measured as a force, say/pounds, at the sleeve, opposing motion of the sleeve. Then, if the speed increases and the height does not alter g But if the speed decreases and the height does not alter -/ g hence = I - ... (8) t> 2 W + W W -j- W ART. 254] GOVERNORS 457 Equation (8) gives the relation between the two extreme speeds and the mean speed which will not cause any change in the position of the governor balls on account of frictional resistance. EXAMPLE i. The balls of a Porter governor weigh 6 pounds each, and the central weight is 100 pounds. Calculate the height to which the balls will rise when running at 200 revolutions per minute if the resistance on the sleeve due to moving the governor gear is 5 pounds. How much must the speed decrease before the balls begin to descend ? 27T 2O7T co = 200 X 7 = - radians per second /. h= 1-323 feet When the balls start to descend we have the relation from (6) w co 2 2 III IOI g ~~2 CO 6 V" 6 i 2 ^ 1 1 1 l'. 101 and !$ ill IOI 2 = 190-8 revolutions per minute EXAMPLE 2.- The balls of a Porter governor weigh 4 pounds each, the load on the governor is 40 pounds, and the arms intersect on the axis. At what height will this governor run if it revolves at 240 revolutions per minute? If the speed of the balls is suddenly increased 2-5 per cent., what pull will be exerted on the gearing attached to the governor sleeve? II the friction of the regulating apparatus is equal to a dead load on the sleeve of 5 pounds, by how much must the speed increase before the balls begin to rise i? (L.U.) co = 240 X ^ = 8?r radians per second , 40 -j- 4 ^o- ~~$ X 647^2 = '55 8 f ot or 67 inches when the speed suddenly increases to 1*02560 h = 4 + 4 ^ g _ 44 + x g 4 *! be the speed at which the ball begins to rise, then , _ 44 g___ 44 + 5 = '- - 2/2 44 or = Vrii2 = 1*056 n n- L =n X 1*056 = 240 X 1*056 = 254 revolutions per minute, i.e. an increase of 5*6 per cent. EXAMPLE 3. In an ordinary Porter governor, suppose the frictional resistance reduced to the sleeve to be 15 pounds, dead load 100 pounds, and the weight of each ball 3 pounds. Find the change in position of the sleeve as the speed respectively rises 5 per cent, above or falls 5 per cent, below the normal speed which is 200 revolutions per minute. 277 At 210 revs. co = 210 X -^ = TTT radians per second 277 77 At 190 revs. co = 190 X 7-= 19- radians per second , r 277 2O77 At the normal speed of 200 revs. CD = 200 X 7- = radians per second. 60 3 If the governor rises through a speed from 200 to 210 revs. 5 X 9 .Q*nW "200 o = 2 oo feet 3 40077^ and /5 210 = IOO + 3 + I 3 _!__ = 2>594 feet 3 Hence, for the 5 per cent, increase in speed the change in height is 2*860 2-294 = 0*266 foot or 3*192 inches N.B. If, however, the speed was falling at the instant the balls were rotating at 200 revolutions per minute, an increase in the speed of 5 per cent, would not move the sleeve. If the governor /#//.$ through a speed from 200 to 190 revs. and ^ = . _,. 363 feet Hence, for the 5 per cent, decrease in speed the change in height is 2-363 2*133 = 0*23 foot or 276 inches The rise or fall of the balls may therefore be 3*19 inches and 2-76 inches respectively, or anything less according to the position of the above ART. 255] GOVERNORS 459 at 200 revolutions, which may vary from that due to a height of 2 '86 feet to that of 2 'i 3 3 feet. 255. Modified Proell Governor. Another form of loaded governor is shown diagrammatically in Fig. 233. A general statical theory of this governor may be given as follows : As before let w be the weight of each ball and W the weight of the w FIG. 233. FIG. 234. central load, then, considering one side of the governor only and taking moments about point P (Fig. 234) we have w 2 where F is the horizontal outward force known as the controlling force, which would have to be applied to the ball when at rest in order to main- tain it at radius r. Neglecting friction F will be numerically equal to the centrifugal force, i.e. t w W .'. ai 2 r X y = x -j- 2 wry 460 THE THEORY OF HEAT ENGINES [CHAP. xxi. or, expressing this in revolutions per minute (n) (Wx -f 2wa\ \ /27T\ 2 _ \ 60 / 7T 2 2 wry Wx -4- 2wa , . - ..... (3) For any position of the balls (3) will give the speed at which (neglect- ing friction) the governor must rotate. In the usual design of this governor the links AB and BC (Fig. 233) are equal and at the normal speed the arm carrying the ball is vertical. The actual movement of each ball on either side of this mean position is small and may without serious error be assumed horizontal. In this case, therefore, it is evident that a = r> and x = 2r and (2) becomes = (W . 2r + 2wr)g 2wry w y ** (4) The power of this governor may be expressed in the same terms as that of the Porter governor, namely, Weight of both balls X extreme vertical movement of balls (if any) + central load X extreme left of sleeve, or 2 (mean centrifugal force per ball X range through which this force acts) i.e. where o> = mean angular velocity, r = mean radius, r and ;- 2 the greatest and least radius of the ball circles respectively. 256. Stability and Sensitiveness of a Governor Hunting. A governor is said to be stable when it takes up a definite position for a definite speed, or in other words, for a definite change of speed a stable governor changes its position by a definite amount. For this condition to exist it is essential that the horizontal controlling force (F) on the balls must change more rapidly than the radius of the ball circle, since the centrifugal force is directly proportional to the radius. ART. 256] GOVERNORS 461 If a small change 8r in the radius produces a change F in the force, stability is assured providing 8F. 8r or or -7- r dr z>. the rate of change of the controlling force is greater than that of the radius. In the case of the unloaded Watt governor F = w . -, ( (i) Art. 253), r -7 ( (5) Art. 254). Now W and w - in the Porter governor F = are constant, and in both of these cases it is evident that h must decrease as r increases ; and that therefore F must increase faster than r, i.e. -T > and stability is assured. If in any case ^r = the governor would only have one speed of rotation for any position of the balls. Such a governor would be useless in practice because the most minute change in speed would cause the balls to fly from one extreme position to another. Such a governor is said to be isochronous. Sensitiveness. Throughout the complete range of the governor, or any B FIG. 235. part of the range there must be a definite change in the speed. If the change is small the governor is sensitive. The sensitiveness of a governor 462 THE THEORY OF HEAT ENGINES [CHAP, xxi. is therefore proportional to the sleeve lift for a given change in speed, and is usually denned as the ratio of the variation in speed to the mean speed, i.e. as where n and n 2 are the greatest and least speeds respectively and n the mean speed of the engine. An isochronous governor is infinitely sensitive since it has only one speed for equilibrium. Many governors are approximately isochronous, being at the same time stable and very sensitive, and designed so that F increases very little faster than r. The Watt governor may be made more sensitive by crossing the arms as shown in Fig. 235. In this case the point of suspension is virtually at O, the point of intersection of the arms, and any desired relation between the rates of variation of F and r may be obtained. In some position such as A the governor is stable since h decreases as r increases, whereas in another position such as B the governor is unstable since h will increase as r increases, and therefore -p- will be less than . There will always be some intermediate position in which -p- = and the governor will then be isochronous and infinitely sensitive. By suitably arranging the distance a and the length of the arms the path of the balls can be made a parabola, in which case h (the subnormal to the parabola) will be constant and the governor isochronous. Stability can be obtained by making the distance a a little less than that which gives isochronism, and to make the governor more powerful the sleeve may be loaded as is done in the case of the Porter governor. Hunting 1 is a state of forced vibration in the engine speed produced by a too sensitive governor. In deciding upon the degree of sensitiveness to adopt for the governor, due regard must be paid to the cyclic variation in speed which is allowed by the flywheel (Art. 243). If the cyclic variation is less than the sensitiveness of the governor^ - - j then; the governor is stable ; if, however, the cyclic variation is greater than the sensitiveness it will be evident that the governor will be continually oscillating in response to the cyclic changes in speed of the engine, giving rise to " hunting." In the case of an internal combustion engine the governor will not, as a rule, work equally well on all loads. The flywheel and governor will normally be designed for about full load on the engine, so that at the load usually maintained the cyclic variation in speed is less than the sensitiveness of the governor and hunting will not take place. On light load, however, the cyclic variation may be greater than the sensitiveness, and with a sensitive governor hunting will occur. From the above it will be evident that very sensitive governors are only required for the very close regulation in speed required for electric driving and similar work, in which case the cyclic variation in speed is necessarily very small and the sensitiveness of the governor can be made very low without causing the engine to hunt. In cases where the cyclic variation need not be very small a very sensitive governor is unnecessary, because ART. 257] GOVERNORS 463 the flywheel effect will be too small for the governor, or in other words, the cyclic variation may be greater than the sensitiveness and the governor will be continually hunting. In the case of a compound steam engine governed by throttling, if the governor is too sensitive it will be evident that, apart from the flywheel effect, a small increase in speed will cause the governor to close the throttle ; but this will take time, and before the closing is completed the engine must have stored a little excessive energy and more steam than is required will have passed the throttle. The extra amount of steam will continue expanding through the low-pressure cylinder after it has done work in the high- pressure cyclinder, with the result that the engine speed will accelerate for a time and the governor will close the throttle still further. After this extra steam has passed through the engine the speed will fall and the governor will open the throttle, but while it is doing so the speed will be falling with the result that the throttle will be opened further than is required, and in a short time the engine will again accelerate. A too sensitive governor will therefore alternately open and close the throttle too much and set up hunting. The inertia of the governor balls will also assist a too sensitive governor in producing hunting by making the balls overshoot their new position when the speed suddenly changes. This may be prevented and an approximately isochronous governor may be made as sensitive as required, without hunting, by fitting a dash pot to it which prevents a too sudden change in position of the governor sleeve, but has practically no effect when the change is gradual. 257. Spring Loaded Governors. Hartnell Governor. A small governor may be made very powerful and at the same time as sensitive as required by loading the sleeve by means of a strained spring instead of a dead weight. The Hartnell governor is shown in Fig. 236. A cap C fixed to the central spindle and driven by bevel wheels, carries bell crank levers L with a ball of weight w pounds at one end and a roller acting on a loose sleeve B at the other end of each lever. The cap is made hollow and contains a spring S which, when in compression, presses downwards on the sleeve B. As the engine speed in- creases the balls fly out until the centrifugal force is balanced by the compressive load on the spring. When the engine is running at its normal speed the FIG. 236. arms of the bell crank levers carrying the balls are usually arranged to be vertical and the total movement of the balls being small it is usual to assume that the balls move out horizontally in a radial direction. 464 THE THEORY OF HEAT ENGINES [CHAP. xxi. Let r be the radius of the ball circle at* speed co radians per second, and F the outward controlling force on each ball which, when the governor is at rest, will keep the balls at this radius, then, taking moments about the fulcrum and neglecting friction we have, for each ball, force on sleeve v balanced by the spring = F X - oc w 9 y = cu 2 r . - pounds and for both balls load = 2 - aj 2 r.- pounds If r and r z be the two extreme radii of the balls when running at speeds o^ and o> 2 respectively, the change in load on the spring will be w y 2 2 . 2 - - K>1 ay*) pounds The spring must be made of the required stiffness in order that when given a certain initial compression the above change in the load on it will allow the necessary movement of the governor sleeve. EXAMPLE i. A spring loaded governor of the type shown in Fig. 236 is provided with balls, each weighing 6 pounds. The arms of the bell crank levers carrying the balls are x = 4 inches and y = 5 inches. Neglecting the effect of the obliquity of the lever arms and the mass of the levers, find the pull exerted on the governor sleeve by the balls when the governor is driven at 300 revolutions per minute. Find the strength of the spring so that the governor will revolve steadily at the configuration shown in Fig. 236 and also in a configuration determined by a vertical movement of the sleeve of 0*2 of an inch when the speed increases 5 per cent. At 300 revs, per min. a) = 300 X g^= IOTT radians per second /. controlling force per ball F = ~^jX ioo7r2 x ^ = 61-3 pounds load on the sleeve = 2X6r3X|= i53'5 pounds For a vertical rise of 0-2 inch of the sleeve the radius of the ball circle will increase o'2 X f = 0*25 inch. Hence the radius when the speed increases 5 per cent., i.e. when a) = IOTT X 1*05 is 4' 2 5 inches and F= X (io'57r) 2 X ^^ = 71-8 pounds 32-2 12 /. load on sleeve = 2 X 71*8 X f = I79'5 pounds. Hence a 5 per cent, increase of speed increases the load on the sleeve, and therefore on the spring, by an amount i79'S i53' 2 5 = 26 ' 2 5 pounds this compresses the spring 0-2 inch, hence the stiffness of spring required is . = 131*25 pounds per inch compression. ART. 257] GOVERNORS 465 EXAMPLE 2. In the governor shown diagrammatically in Fig. 237 the strength of the spring is such that a force of 200 Ibs. is required to stretch it i inch. In the position shown the balls are against the stops and the initial compression of the spring is no pounds, and the weight of each ball is 5 pounds. Assuming that the balls move in a horizontal plane and that all control is effected by the spring, at what speed will the balls revolve at a radius of 5 inches ? Is the governor stable ? When the radius of the ball circle increases from 2-5 to 5 inches, the downward movement of the sleeve will be 2*5 X f = 1*25 inches This movement will put the spring in tension to the amount of 1*25 X 200 = 250 pounds The spring, however, is initially in compression to the amount of 1 10 pounds. Hence at a radius of 5 inches the tension of the spring is 250 no = 140 pounds The controlling force on each ball is therefore J X ^ = 35 pounds. .'. - - 5 - X cu2 x ~ = 35 32-2 12 ^35X32-002 25 w = 23-26 radians per second 23-26 X 60 n = = 222 revs, per mm. 277 An increase of i inch in the radius moves the sleeve 0-5 inch and changes the load on the spring 200 X 0-5 = TOO pounds. Hence the corresponding change in the controlling force per ball is ^ = 25 pounds. ...ff JS ' dr i At a radius of 5 inches F = 35 pounds. r 5 ' The governor is therefore stable since -=- > -. dr r EXAMPLE 3. What must be the initial load on the spring in Example 2 in order that the governor may be isochronous, and what will then be the equilibrium speed of the governor ? F d\? i* or isochronism ~ = - = 2 c r dr f| p .*. at the radius of 5 inches =25 F= 125 pounds 2 H 466 THE THEORY OF HEAT ENGINES [CHAP. xxi. This controlling force per ball will put a tension on the spring of 4 X 125 = 500 pounds It has been shown in the solution to Example 2 that the movement of the sleeve puts the spring in tension to the extent of 250 pounds, hence the initial tension of the spring must be 500 250 = 250 pounds. Let MI = equilibrium speed of the governor X Wi2 x - 5 - = 125, 32-2 12 0)^= 1931 a>i = 43 - 8 radains per second or = 420 revs, per minute A modification of the Hartnell governor is shown in Fig. 237. In this type the revolving weights, or balls, are carried on bell crank levers and rotate with the engine crankshaft, being tied to- gether by means of a spring or springs. The engine speed can be adjusted, while running, by varying the tension of the external spring B. When the ball radius increases due to the speed rising the bell crank levers move the sleeve C to the right and partially close the inlet valve through the lever D and valve rod A. In a well-designed governor of this type the sleeve C must be as light f 6" FIG. 237. FIG. 238. as possible, otherwise its inertia may cause hunting. The theory of this governor is very similar to that of the Hartnell previously given and will be best understood by studying the following numerical example. EXAMPLE. A spring loaded governor of the type shown in Fig. 237 is fitted with a spring which extends one inch for a pull of 30 pounds. The dimensions of the bell crank lever carrying the balls are shown in ART. 257] GOVERNORS 467 Fig. 238, the fulcrum being at point F. The weight of each ball is 5 pounds, and there is no tension on the spring when the balls are at a radius of 3 inches. Neglecting the controlling effect of the balls and arms, find the speed at which the governor will run when the balls are at a radius of 5 inches, and find also the force exerted on the sleeve if, when the balls are in that position, the speed increases 10 per cent. When the balls are at a radius of 5 inches the spring is stretched 10 6 = 4 inches. The controlling force on the balls will therefore be 4 X 30 = 120 pounds. 5_ V /.2 NX JL - 12 = 120 or 32-2 w = 43-2 radians per second, n = 413 revolutions per minute. If the radius remains unaltered but the speed increases 10 per cent, the controlling force will increase to 120 X (ri)2= 145-2 pounds, i.e. an increase of 25-2 pounds per ball. Taking moments about F (Fig. 238) we have Force on sleeve = 25*2 X pounds. |= 100-8 The Hartung Governor. In the gover- nors previously described the levers which transmit the movement of the balls to the sleeve also transmit the controlling force to the balls with the attendant frictional resistance which may be excessive in amount. One of the great advantages of the Hartung governor, other than its compactness, is that the levers merely trans- mit the motion, as will be evident from an inspection of Fig. 239. An outward move- ment of the weights A is resisted by the internal springs and the movement only is transmitted to the sleeve D by the levers C. The friction of the governor is therefore very small, and it may be made as powerful and as sensitive as desired. The Proell Governor. This governor FlG * 239< is shown in Fig. 240, a general statical theory being as follows : Let w be the weight of each ball, W the total load on the sleeve due to the compressed spring and F the controlling force on each ball. The outward horizontal force P on the end of the bell crank lever may be first found by taking moments about point C (Fig 241) which gives = FX 468 THE THEORY OF HEAT ENGINES [CHAP. xxi. W (a + A (-T-) Then taking moments about the pin A we have W X x = P X y . , FIG. 240. FIG. 241. 258. Curves of Controlling Force. It has been shown analytically in Art. 254 that the effect of friction is to increase the range of speed of the governor, and therefore to decrease its sensitiveness. The effect may also be very conveniently shown graphically, as first suggested by Mr. Hartnell, 1 by drawing a curve having the controlling force as ordinates, and the radius of the ball circle as abscissae. Such a curve for a Porter governor is shown in Fig. 242 at ab y in which, neglecting friction, the controlling force is given by . . F = ^J^ (5), Art. .54- f . , If now any convenient radius be chosen, such as on, and a perpendicular drawn, the distance nx may be set off equal to the centrifugal force W cuV, or o-ooo34^/?72 2 , where r = on, to the same scale as that of the the slope of the curve ab is greater than the slope of the straight line oe, then the governor is stable when in that position, since d F -f- is > -. Inspection of Fig. 242 will show that throughout the full range of the governor the slope of ab is greater than that of any such line as oe. The power of the governor will evidently be represented by the shaded area, since this is equal to the mean controlling force multiplied by the extreme radial movement of the balls. The effect of friction is shown in Fig. 243. Here cd is the controlling force curve, whilst the radius of the ball circle is increasing the friction, increasing the force to a greater value than w 470 THE THEORY OF HEAT ENGINES [CHAP. xxi. n Radius FIG. 243. r Radius FIG. 244. ART. 258] GOVERNORS As the radius decreases fe represents the controlling force, whilst ab represents the force, neglecting friction, as in Fig. 242. The power absorbed in friction is represented by the area cdfe, and the range of speed is increased from n^n 2 to nn. Spring Controlled Governors. The foregoing remarks apply, but now the curve becomes a straight line, since the force exerted by the spring is proportional to its deflection, and therefore to the radius of the ball circle. By varying the initial tension of the spring three conditions may be obtained, namely : JTj Tf (1) Stability as shown by ab (Fig. 244), since -^- > . dF F (2) Isochronism given by cd^ since ~T~ = - . d F (3) Instability given by ef t since ^- < ^. At any radius r, y gives the equilibrium speed of the stable governor from ab, and x gives the speed of isochronism. EXAMPLES XXI 1. In a simple unloaded Watt governor estimate the change in height when the speed increases from 100 to 120 revolutions per minute. 2. Prove that the height, in feet, of a loaded governor of the Porter type with equal arms and links is given by , _ W + w 0-816 fl ^ - where n is the speed in revolutions per second. If w = 4 pounds, W = 56 pounds, and the speed suddenly increases from 250 to 255 revolutions per minute, what will be the lifting force at the sleeve, neglecting friction? (L.U.) 3 The balls of a Porter governor weigh 5 pounds each, and the central load weighs 95 pounds. Calculate the height to which the balls will rise when running at 240 revolutions per minute, if the resistance at the sleeve due to moving the governor gear is 5 pounds. How much must the speed decrease to before the balls begin to de- scend, and what must be the ratio between the two extreme speeds which will not cause any change of position of the governor balls ? 4. The governor of a steam engine is of the Porter type, and runs at 1 80 revolutions per minute. If the height of the governor when running at its normal speed is 7*5 inches, what is the ratio between the load and the weight of one of the balls ? What would be the proportional increase of speed for this governor before it could come into operation, if the frictional resistance be assumed equal to o'l of the load acting on the sleeve? (L.U.) 5. In a spring-controlled governor, the curve of controlling force is a straight line ; FlG. 245. when the balls are 14 inches apart the controlling force is 240 pounds, and when 8 inches apart 120 pounds. At what speed will the governor run when the balls are 10 inches 472 THE THEORY OF HEAT ENGINES [CHAP. xxi. apart ? What initial tension would be required on the spring for isochronism, and what would then be the speed ? Each ball weighs 20 pounds. (L.U.) 6. A spring-controlled governor of the type shown in Fig. 245 is provided with balls each weighing 3 pounds. Neglecting the effect of the obliquity of the arms of the bell- cranks and the mass of the bell-crank levers, find the force exerted on the sleeve when the governor is running at 300 revolutions per minute. Find the strength of the spring, so that the governor will revolve steadily in the position shown in Fig. 245, and also in a position determined by a vertical movement of the sleeve of 0*2 inch when the speed increases 5 per cent. 7. Show that a simple Watt governor may be made nearly isochronous by crossing the arms. Draw for radii between 6 and 8 inches the controlling force curve of such a governor with crossed arms, each arm being 12 inches long, and pivoted at a point I '5 inches from the axis, and each ball weighing 10 pounds. Find from the curve the " power " of the governor, expressing it in foot-pounds. (L.U.) 8. Show for a loaded governor of the Porter type that when the central load rises and falls twice as fast as the balls s = */i + A Vi A where s = the difference between the highest and lowest speeds in any position due to friction, divided by the mean speed in that position. A = equivalent frictional resistance at the sleeve divided by the weight of a ball plus the load. (L.U.) 9. Find, for the position shown in Fig. 246, the speed at which the governor runs, FIG. 246. neglecting friction and the weight of the arms. Explain why the governor is more sensitive than if the balls were placed at the junction of the links. The weight of each ball is 20 pounds, and of the load 80 pounds. (L.U.) 10. In a loaded governor of the Porter type, with equal links 12 inches long, pivoted at the axis, the weight of each ball is 10 pounds, and of the load 60 pounds. When the ball radius is 7 inches, find the speed of revolution, allowing for the effect of the load and the centrifugal action of the links, which may be taken as of uniform section, and of weight 4 pounds each. (L.U.) ANSWERS TO EXAMPLES EXAMPLES I. (1) 8*47 cubic feet. (2) 524 F. ; 26170 foot-pounds ; 116-85 B.Th.U. (3) 1824 foot-pounds. (4) 12,440 foot-pounds. (5) 15,255 foot-pounds. (6) (a) 139-9 pounds per square inch ; (b) 23,900 foot-pounds. (7) (<*) 354 F. ; Heat expended = 39,600 foot-pounds ; work done = 10,800 foot-pounds ; (b} 1728 F. ; work done = loss of internal energy = 26,093 foot- pounds. (8) 347 F. ; 2913 F. (9) 56*54 pounds per square inch ; 634 F. (10) 667 F. ; 126,170 foot-pounds. (11) (a) 104 cubic feet; 778,500 foot-pounds ; (&) 57*4 cubic feet; 419,400 foot-pounds. (12) o'54/ ; +0-32^ ; 150 B.Th.U. per second. (13) (a) 6600 foot-pounds ; (b] 4750 foot-pounds; (c) 495 F. (14) 277 pounds per square inch; 85-23 B.Th.U. ; 0*0947 ranks. (15) 2704 pounds per square foot; 39720 foot-pounds; 51*05 B.Th.U.; 0*0947 ranks. (16) 0-0348 ranks. EXAMPLES II. (0 () '534; () 0-449- (2) (a) 0*562; (b) 0*172. (3) 0-588 ; 9-2. (4) 0-355. (5) 0-184- (6) 28*54 pounds per square inch. (7) 86*0 pounds per square inch ; 0*581. (8) 0*553. EXAMPLES III. (1) 4*33 cubic feet. (2) (a) H = 1191 B.Th.U. ; I = 1107*7 B.Th.U. ; (b} H = 933*2 B.TH.U. ; I = 874*9 B.Th.U. (3) A. B. C. (a) 10-98 (*) 7S'8i 10-45 72-I7 10*48 j pounds 72-32 f per cent. 473 474 THE THEORY OF HEAT ENGINES (4) 63 per cent. (5) 0*420 pound ; 0*900. (6) 0*987. (7) 0*467. (8) 0-968. (9) 0*920 (10) 0-987. (n) 1*693 ranks. (12) (a) o'86o ; (b) 0*688 ; (<:) 0*912 ; (d) 23 pounds per square inch absolute. (13) 90 pounds per square inch absolute. (14) Temperature = 414 F., superheat 201 F. ; gain of entropy 0*276 units. (15) Before expansion, temperature = 588 F., superheat = 260 F. After expansion, temperature = 286 F., superheat = 46 F. (16) 0*237 units. (17) Temperature = 247-9 F., superheat = 42 F. ; 0*2952 units. (18) Temperature = 298*1 F., superheat = 92*2 F. ; 0*30 units. (20) 0*2175 pound ; 0*0355 pound. EXAMPLES IV. (1) (a) 108*5 B.Th.U.; () 98*3 B.Th.U. (2) Dryness fraction 0*666 ; n no. (3) 00 79' 02 B.Th.U. ; 7*5 per cent. ; (b) 63-15 B.Th.U., 7*1 per cent. (4) 255*3 B.Th.U. ; 23*6 per cent. (5) 221*5 B.Th.U. ; 23*3 per cent. (6) (a) 0*483 ; (b) 0*428. (7) 17*63 pounds. (8) 9*92 pounds, 0*874. (9) 596 F. ; 26*2 per cent. ; 357-48 B.Th.U. , . .. , diameter of Stirling 1*03 (10) Stirling 49*0 per cent., steam 16*9 per cent. ; Y - ? s = diameter of steam I (n) 300-9 B.Th.U.; 0-272. (12) (a) 8-82 pounds ; (b} 7-77 pounds. (13) 27-6 per cent. ; without feed heating 26*4 per cent. (14) 26-8 per cent. 05) 00 3 1 '3 per cent. ; (b} 29*3 per cent. 16) (a) 13*50 per cent. ; (b) 13*74 per cent. 17) 80*65 pounds per square inch ; 85*1. 1 8) 8*12 inches. 19) 2*60. (20) (a) 26*4 per cent. ; (b) 31-2 per cent. EXAMPLES V. (1) (a} 19,967 foot-pounds; (&) 18,000 foot-pounds ; (c) 13,955 foot-pounds ; (d) 19,967 foot-pounds. (2) (b) 0-807, 53io ; (b} 0-804 ; 54 ' (0 ' 826 , 477. (3) 0-841 ; 0-887. (4) 12. EXAMPLES VI. O TT p i (0 5 ' 1 S pounds per square inch absolute ; ; pvr = ; TT p j (2) 3*702 ; 18*58 pounds per square inch absolute ; 0*480 ; ^ ' ' = - * TT T) y (3) L.P. = 31*42 inches, H.P. = 16*8 inches ; 0*428 stroke ; j-p-' = p- ANSWERS TO EXAMPLES 475 L.P. "172* (5) 3 1 '6 inches, 54*6 inches, 86*5 inches, stroke 4 feet. (6) 64*33 pounds per square inch ; 19*09 pounds per square inch. EXAMPLES VII. 123-3- '2) 1317 F. ; 1712; 30*6 pounds. (3) 5*90. (4) (a} 5745 W 5*66. (5) 00 6*33 ; (*) 6*33- (6) () 5-38 ; (*) 6-26. (7) 44-24- (8) Liquid $>= 0*003 31 2/c ; vapour =1-158-0-003456/0; 6-14; 65-1 pounds. EXAMPLES VIII. (1) 1260 feet per second ; 0-951. (2) Area of throat 0*329 square inch, x = 0-96 ; area of discharge end 3*069 square inches, x 0-802. (3) Area of throat 0*367 square inch, of discharge end ro66 square inch. Condition of steam, in throat superheat = 24 F., at discharge end x = 0-90. (4) 5'56 pounds ; steam orifice 1*134 square inches ; discharge orifice 0*0404 square inch ; feed temperature 200 F. EXAMPLES IX. (1) 29*5 ; 162,660 foot-pounds ; 1575 feet per second ; o'8o8. (2) 36 (nearly) ; 177,690 foot-pounds ; 1690 feet per second ; 0-883. (3) 28-8; 0789; 1 59-3 H.P. (4) 29-5 ; 154,540 foot-pounds ; 1345 feet per second ; 0-860. (5) Moving blades, inlet angle = exit angle = 24-5. Stationary blades, ist set, inlet angle = 34, exit angle = 17-3 ; 2nd set, inlet angle = 43, exit angle - 7. (6) 116-1 H.P. EXAMPLES X. 1) 49-2 per cent. 2) 6 3 *6. 3) 48-3 pounds; -ii3F. 4) (a] 100,655 foot-pounds ; () 83,410 foot-pounds ; 29*9 C.H.U. ; (c) 78,050 foot-pounds, in ist cooler 18*28 C.H.U., in 2nd cooler 18*28 C.H.U. (5) (a] 85,690 foot-pounds ; () 76,840 foot-pounds ; 15-9 C.H.U. ; (c) 74,166 foot-pounds in ist and 2nd coolers 10*2 C.H.U. (6) 9*12 inches diameter, 2 feet 6 inches stroke. EXAMPLES XI. (1) 11*04 pounds ; CO 2 14*06, H 2 O 2*15, N 2 73-87, O 2 = 9*92 per cent. (2) 1*0424 cubic foot; 10*6 per cent.; CO 2 16*60, H 2 O 10*21, N 2 73-19 per cent. (3) 0-2417; 2292 B.Th.U. (4) 00 1519 B.Th.U. ; (J) 852 B.Th.U. ; (c) 723 B.Th.U. (5) (*) I7'5 P er cent. ; 1502 B.Th.U. = 10-77 per cent. ; () 3214 B.Th.U. (6) (a} 1-28 pounds; () 1039 C.H.U. : (c} 943 C.H.U. (7) 93'5 feet. 476 THE THEORY OF HEAT ENGINES (8) (a) 5-39 B.H.P. ; (6) 3*148 B.H.P. (9) () 576 cubic feet; (b) 687-9; W 621-3. (10) 576-4 ; 5247 B.Th.U. per cubic foot. (11) 10,943 calories per gram, or 19,697 B.Th.U. per pound. (12) 70 per cent. ; CO 34-71, N 2 65-29 percent. ; 119 B.Th.U. per cubic foot. (13) 3-846 cubic feet ; r86 cubic feet. EXAMPLES XII. (2) 0) 7200 ; (0) 3064 ; (<:) 6100. (3) On steam side 128*3 F., n water side 127-98 F. EXAMPLES XIII. (1) 51 per cent. ; 488 C. ; 157-9 pounds per square inch absolute. (2) (a) 48 per cent. ; (b) 38' I per cent. (3) (#) 0-0393 pound: (ff) 158 pounds per square inch absolute and 488 C. ; (c) 3616; (d) 907 pounds per square inch absolute; (e) 1948 C. and 77-4 pounds per square inch absolute ; (/) 51 per cent. (4) /z/i-323 = 200 . 2< (5) At A, i88oF. ; at B, 1060 F. ; Heat taken from the gas 1271 foot- pounds, or 1-63 B.Th.U. . , EXAMPLES XVI. (1) O) I.H.P. = 50-9, B.H.P. = 20-45 ; () 0-37 pound, 0*63 pound ; (V) 2158 B.Th.U. ; (d} 1365 B.Th.U. ; (*) 1917 B.Th.U. (2) (a) 21,200 B.Th.U.; (b) 63,020 B.Th.U. : (c\ 35,960 B.Th.U. ; (d} 2220 B.Th.U. (4) (a) 39-0 ; (ff) 30-76 ; ( crank end = 1-29", head end = 1*29". V4J W \lnside lap, crank end = 0*8 1", head end = 0-65". () 0-477 stroke ; (c} 45. (5) At head end 1*59 inch, at crank end 1-31 inch. (6) Lead. Cut-off. Release. Compres- sion. Max. opening to steam. Max. opening to exhaust. Outside lap. Inside lap. Head end . 0'25 0*64 0-898 0-805 0*72 I'5 0-86 0'08 Crank end . 0-50 0*64 0-85 0-855 0'94 i'S 0-64 0'08 Travel = 3'i6 inches, angle of advance 46. (7) 175 inches ; 150 ; o'686 inch. (8) and (9) VALUES OF "8" INCHES, Cut-off. Head end. Crank end. Difference. 0'2 -o-53 -0-85 o'33 0-3 -0-87 I'2I o'33 0'4 0'5 -1-13 1-40 -i-45 r6o 0-31 0'20 0-6 -i'S9 -1-70 O'll (10) (a) travel = 5*55 inches, angle of advance 36 5, outside lap 1-42 inches. W) CUT-OFF AT O'l 0'2 0-3 0-4 o-5 0-6 Head end a = 0-04" 0-72" i -20" 1-65" 2'O2" 2'33' 4-35 inches. ANSWERS TO EXAMPLES 479 (u) {a) 3 inches, angle of advance 20 ; (6) i 6 inches, angle of advance 90. (12) r 2-83 inches ; $ = 102 - 36'. (13) (a) 8-4 inches; (b) 43- (14) r = 2*1 inches, angle of advance = 45. (15) (a) i foot 5 inches ; (6) 9 inches ; (c) 3", 20. EXAMPLES XIX. (r) (a) 21,910 pound-feet ; (b) 21,400 pound-feet. (2) 225 foot-pounds ; 199*5 pound-feet ; 221 pound-feet. x.x _, . sin e cos 6 (3) 5 ' 84 (is^nsp- (4) 33'86 and 21*90 pounds per square inch. (5) 49,490 pound-feet units. (6) I = 4776 pound-feet units, rim speed 78*68 feet per second. (7) 11,890 pound-feet units. EXAMPLES XX. (1) On left hand wheel 200 pounds ; on right hand wheel 100 pounds. (2) 46' i pounds ; 9-39 pounds, 301*6 from the 2 pound radius. (3) Plane A, 98*6 pounds at 269*5 to the 2 pound radius ; Plane B, 131*8 pounds at 347*5 to the 200 pound radius. (4) 4th mass = 172 tons ; Reckoning from the left, angle between ij and 2 = 146*5, between ij and 24 = 242, between ij and 1*72 = 40. (5) Crank D, mass = 2*88 tons, angle from A = 277 ; Crank E, mass = 2*72 tons, angle from A = 58. (6) R.H. wheel, 238 pounds, 155 behind R.H. crank; L.H. wheel, 238 pounds 155 leading L.H. crank. (7) With right crank leading, R.H. wheel, 409 pounds, 184*8 leading R.H. crank ; L.H. wheel, 409 pounds, 184*8 behind L.H. crank ; 3560 pounds. (8) 76-43 tons-feet ; 21*84 tons-feet. EXAMPLES XXI. (1) 1*074 inch. (2) 2*4 pounds. (3) 12*84 inches; 228*2 revs, per min. ; 1*051. (4) 5 '9 * 4 per cent. (5) 2 37 revs, per min. ; 40 pounds when r = 4" ; 265 revs, per min. (6) 76*62 pounds ; 65*6 pounds per inch of compression. (9) 35 revs, per min. (10) 162 revs, per min. PROPERTIES OF SATURATED STEAM SATURATED STEAM: PRESSURE TABLE. (Marks and Davis.) Pressure, IKc en Tempera- Sp. vol., _..L ft. Heat of fi__ i;,,.,:^ Latent heat of Total heat of Entropy. IDS* SC[. in. abs. ture, CUD. It. per Ib. .ne liquid. B.Th.U. evap., B.Th.U. steam, B.Th.U. Water. Evap. Steam. I IOI-83 333-0 69*8 1034*6 1104-4 0*1327 1-8427 1-9754 2 I26-I5 173-5 94*0 IO2I*O III5-0 0-1749 -7431 I*9l8o 4 153-01 90-5 120-9 10057 1126-5 0-2198 6416 1-8614 6 170-06 61*89 137-9 II33-7 0-2471 5814 1*8285 8 I82-86 47*27 ISO'S 988*2 II39-0 0-2673 5380 1-8053 10 193-22 161-1 982*0 II43-I 0-2832 5042 1*7874 12 201-96 32-36 169-9 976*6 1146-5 0-2967 4760 1*7727 H 209-55 28-02 I77-5 971*9 1149-4 0-3081 -4523 1-7604 16 216-3 24-79 184-4 967*6 II52-0 0*3183 '43 ll 1-7549 18 222*4 22-16 190-5 1154*2 0-3273 4127 1-7400 20 228-O 20-08 196-1 960-0 II56*2 0*3355 -3965 1-7320 22 233-1 18-37 201*3 956*7 1158-0 0-3430 38il 1*7241 24 237-8 16-93 206-1 953*5 1159-6 0*3499 3670 1*7169 26 242-2 15-72 2IO'6 950-6 Il6l"2 0-3564 -3542 1*7106 28 246-4 I4-67 214-8 947-8 II62-6 0*3623 3425 1*7048 30 250-3 I3-74 218-8 945; i 1163-9 0*3680 33" 1*6991 32 254-1 12*93 222-6 Il65*I 0-3733 3205 1-6938 34 257-6 12-22 226*2 940*1 1 166'3 0-3784 3107 1*6891 36 26l-O II-58 229-6 937-7 Il67*3 0-3832 3014 1*6846 38 264-2 iroi 232*9 935-5 1 1 68*4 0-3877 2925 1*6802 40 267-3 10-49 236*1 933-3 1169-4 0*3920 2841 1-6761 42 27O-2 10-02 239*1 931*2 1170-3 0-3962 2759 1-6721 44 273-1 9-59 242-0 1171*2 0-4002 2681 1-6683 46 275-8 9'20 244*8 927*2 1172-0 0*4040 2607 1-6647 48 278-5 8-84 1172-8 0-4077 2536 1*6613 50 281-0 8- S I 2 50* I 923*5 1173-6 0-4113 2468 1*6581 52 283-5 8-20 252-6 921*7 II74-3 0*4147 2402 I-6549 54 7-91 255-I 919*9 ii75-o 0*4180 2339 1-6519 56 288-2 7-65 2 57'5 918-2 H75'7 0*4212 2278 1-6490 290-5 7-40 259*8 916-5 1176-4 0-4242 2218 1-6460 60 292-7 7-17 262-1 914-9 1177*0 0*4272 2160 1-6432 62 294-9 6-95 264-3 9 J 3'3 1177*6 0*4302 2104 1-6406 64 297-0 6-75 266-4 911-8 1178-2 0-4330 -2050 1-6380 66 299-0 6*56 268*5 910-2 1178-8 0-4358 2007 1-6355 68 301-0 6-38 270*6 908-7 1179-3 0-4385 1946 1-6331 70 302-9 6*20 272-6 907-2 1179-8 0-4410 1896 1-6307 72 304-8 6*04 274-5 905*8 1180-4 0-4437 1848 1-6285 74 306*7 5-89 276*5 904*4 1180-9 0-4462 1801 1*6263 76 308-5 574 278-3 903-0 1181-4 0-4487 1755 1-6242 g 310-3 312-0 5-60 S'47 280*2 282*0 901*7 900-3 1181*8 1182*3 0-4511 0-4535 1710 1665 1-6221 1*6200 480 PROPERTIES OF SATURATED STEAM 481 Pressure, Ibs. sq. in. abs. Tempera- ture, F. Sp. vol., cub. ft. per Ib. Heat of ie liquid, B.Th.U. Latent heat of evap., B.Th.U. Total heat of steam, B.Th.U. Entropy. Water. Evap. Steam. 82 3I3*8 5*3* 283-8 899-0 Il82*8 0-4557 1*1623 1-6180 84 3!5'4 5*22 285-5 897*7 1183-2 0-4579 I*I58l 1*6160 86 W l 5-10 287-2 896-4 1183-6 0-4601 I-I540 1-6141 88 3187 5-00 288-9 895-2 1184*0 0-4623 I-I500 1*6123 90 320-3 4-89 290-5 893*9 1184*4 0-4644 1*1461 1-6105 92 321-8 4-79 292-1 892-7 1184*8 0-4664 1*1423 1-6087 94 96 323'4 324'9 4-69 4-60 293*7 295*3 89I-5 II85-2 1185-6 0*4684 0-4704 1*1385 1*1348 1-6069 1-6052 98 326-4 4'S 1 296-8 889*2 II86-0 0-4724 I'I3I2 1-6036 100 327-8 4*429 298-3 888-0 II86-3 0-4743 I-I277 1*6020 105 33 J *4 4-230 3O2-O 885-2 Il87*2 0-4789 I-II9I 1-5980 no 334*8 4-047 305*5 882-5 1188*0 0-4834 I-IIOS 1*5942 115 120 338-1 34 i '3 3*88o 3*726 309*0 3 I2 '3 879-8 877-2 1188-8 1189-6 0-4877 0*4919 1-1030 1-0954 1*5907 1-5873 J 25 344*4 3-583 3i5*5 874-7 1190-3 0-4959 I -0880 1*5839 I 3 347*4 3-45 2 318-6 872*3 1191-0 0*4998 1-0809 1*5807 "-35 350*3 3*33i 321*7 869-9 1191-6 0-5035 1*0742 i*5777 140 353*1 3*219 324*6 867-6 1192*2 0*5072 1*0675 ^5747 145 355-8 3-112 327*4 865*4 1192-8 0-5107 1*0612 i*57i9 150 358'5 3-012 330*2 863*2 I193-4 0*5142 1*0550 1-5692 '55 361-0 2*920 332*9 86 1*0 1194*0 0-5175 I -0489 1-5664 160 363*6 2-834 335*6 858-8 1194*5 0*5208 1*0431 I-5639 165 366-0 2-753 338-2 856-8 1195*0 0-5239 1*0376 i*56i5 170 368-5 2-675 340-7 854*7 ii95*4 0-5269 1*0321 i*5590 175 370-8 2-602 343*2 852-7 1195*9 0-5299 1*0268 I-5567 180 373*1 2-533 345-6 850-8 1196*4 0*5328 1*0215 1-5543 185 190 375*4 377*6 2-468 2-406 348-0 350-4 848*8 846-9 1196*8 H97-3 0-5356 0-5384 1-0164 1-0114 1-5520 I-5498 195 379-8 2-346 352-7 845-0 1197*7 0*5410 1*0066 1-5476 200 381-9 2-290 354-9 843-2 1198-1 0-5437 1*0019 i*5456 210 386-0 2-187 359-2 839-6 1198-8 0*5488 0*9928 1*5416 220 389-9 2*091 363*4 836-2 1199-6 0-5538 0-9841 1-5379 230 393*8 2-004 367*5 832-8 1200*2 0-5586 0-9758 1*5344 240 397*4 1-924 37i*4 829-5 I200'9 0-5633 0*9676 1-5309 250 401-1 1*850 375-2 826-3 I20I-5 0*5676 0*9600 1*5276 260 404*5 1*782 378-9 823-1 1 202- 1 0*5719 0-9525 i*5 2 44 270 407-9 1718 382-5 820-1 1202-6 0*5760 0-9454 1*5214 280 411-2 1-658 386-0 817-1 I2O3T 0*5800 0-9385 1*5185 290 414-4 1-602 389*4 814-2 1203*6 0*5840 0-9316 I*5I56 300 4I7-5 i*55i 392-7 811-3 I204T 0*5878 0*9251 1*5129 310 420-5 1*502 395-9 808*5 I204-5 0-5915 0*9187 1*5102 320 423*4 I-456 399-1 805-8 I204-9 0-5951 0*9125 1*5076 330 426-3 1-413 402-2 803-1 I205-3 0*5986 0-9065 1*5051 340 429-1 1-372 405-3 800-4 I205-7 0-6020 0*9006 1*5026 350 3 60 43i*9 434-6 i*334 1*298 408-2 411*2 797*8 795*3 I2O6- 1 1 206* 5 0-6053 0-6085 0-8949 0-8894 1*5002 1*4979 370 437*2 1*264 414-0 792-8 1206-8 0-6116 0-8840 i*4956 380 439*8 1*231 416-8 790-3 1207-1 0-6147 0-8788 i*4935 390 442-3 I'2OO 4i9'5 787-9 1207-4 0-6178 0-8737 i*49i5 400 444-8 1*17 422*0 786*0 1208-0 0*621 0*868 1*489 2 I THE THEORY OF HEAT ENGINES GLAISHER'S FACTORS FOR WET AND DRY BULB HYGROMETER. (From Ganot's ''Physics.") Dry bulb, temperature, F. Factor. Dry bulb, temperature, F. Factor. Below 24 8'5 34 to 35 2-8 24 to 25 25 26 6-9 6'5 35 40 40 45 2'5 2'2 26 27 6-1 45 50 2'I 27 28 5'6 5o 55 2'0 28 29 5'i 55 60 '9 29 30 4'6 60 65 8 30 3i 4'i 65 70 8 31 32 37 70 75 7 32 33 3*3 75 80 7 33 34 3*0 80 85 6 MATHEMATICAL TABLES 483 Angle. Chord. Sine. Tangent. Co- tangent. Cosine. De- grees. Radians. C 000 00 1 1-414 1-5708 90 1 2 3 4 0175 0349 0524 0698 017 035 052 070 0175 0349 0523 0698 0175 0349 0524 0699 67-2900 28-6363 19-0811 14-3007 9998 9994 9986 9976 1-402 1-389 1-377 1-364 1-5533 1-6359 1-5184 1-5010 89 88 87 86 5 0873 1047 1222 1396 1571 087 0872 0875 11-4301 9962 1-361 1-4835 85 6 7 8 9 105 122 140 157 1045 1219 1392 1564 1051 1228 1405 1584 9-5144 8-1443 7-1154 6-3138 9945 9925 9903 9877 1-338 1-325 1-312 1-299 1-4661 1-4486 1-4312 1-4137 84 83 82 81 10 1745 174 1736 1763 6-6713 9848 1-286 1 '3963 80 11 12 13 14 1920 2094 2269 2443 192 209 226 244 1908 2079 2250 2419 1944 2126 2309 2493 6 1446 4-1046 4-3315 4-0108 9816 9781 9744 9703 1-272 1-259 1-246 1-231 1-3788 1 3614 1 -3439 1-3265 7* 78 77 76 15 2618 261 2588 2679 3-7321 9659 1-218 1-3090 75 1 17 18 19 2793 2967 3142 3316 278 296 313 330 2756 2924 3090 3256 2867 3057 3249 3443 3-4874 3-2709 3-0777 2-9012 9613 9563 9511 9455 1-204 1-190 1-176 1-161 1-2915 1-2741 1*2666 1-2392 74 73 73 71 20 21 22 23 24 25 3491 347 3420 3640 2-7475 9397 1-147 1-2217 70 3665 3840 4014 4189 364 382 399 416 3584 3746 3907 4067 3839 4040 4245 4452 2-6051 2-4751 2-3559 2-2460 9336 9272 9205 9135 1-133 1-118 1-104 1089 1-2043 1-1868 1-1694 1-1619 69 68 67 66 4363 433 4226 4663 2-1445 9063 1-075 1-1346 65 26 27 28 29 30 4538 4712 4887 5061 450 467 484 501 4384 4540 4695 4848 4877 5095 6317 5543 2-0503 1-9626 1'8807 1-8040 8988 8910 8829 8746 1-060 1-046 1-030 1-015 1-1170 1-0996 1-0821 1-0647 64 63 62 61 5236 518 5000 5774 1-7321 8660 1-000 1-0472 60 31 32 33 34 35 5411 5585 5760 5934 534 551 568 685 5150 5299 5446 5592 6009 6249 6494 6746 1-6643 1-6003 1-5399 1-4826 8572 8480 387 8290 985 970 954 939 1 0297 1-0123 9948 9774 69 68 67 66 6109 601 5736 7002 1-4281 8192 923 9599 65 64 63 62 61 36 37 38 39 6283 6458 6632 6807 618 635 651 668 5878 6018 6157 6293 7265 7536 7813 8098 1-3764 1-3270 1-2799 1-2349 8090 7986 7880 7771 908 892 877 861 9425 9250 9076 8901 40 6981 684 6428 8391 1-1918 7660 845 8727 50 41 42 43 44 7156 7330 7505 7679 700 717 733 749 6561 6691 6820 6947 8693 9004 9325 9657 1-1504 1-1106 1 -0724 1-0355 7647 7431 7314 7193 829 813 797 781 8652 8378 8203 8029 49 48 47 46 45 7854 765 7071 1-0000 1-0000 7071 765 7854 45 Cosine. Co- tangent. Tangent. Sine. Chord. Radians. De- gree* Angle. 484 THE THEORY OF HEAT ENGINES 1 2 3 4 5 6 7 8 9 123 4 5 21 20 6789 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 9 13 17 4 8 12 16 25 30 34 38 24 28 32 37 11 12 0414 0792 0453 0828 0492 0864 0531 0899 0569 0934 0607 0969 0645 1004 0682 1038 0719 1072 0755 1106 4 8 12 15 4 7 11 15 3 7 11 14 3 7 10 14 19 19 18 17 23 27 31 35 22 26 30 33 21 25 28 32 20 24 27 31 13 14 1139 1461 1173 1492 1206 1523 1239 1553 1271 1584 1303 1614 1335 1644 1367 1673 1399 1703 1430 1732 3 7 10 13 3 7 10 12 3 6 9 12 3 6 9 12 16 16 15 15 14 14 20 23 26 30 19 22 25 29 18 21 24 28 17 20 23 '26 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 3 6 9 11 3 5 8 11 17 20 23 26 16 19 22 25 16 17 2041 2304 2068 2330 2095 2355 2122 2380 2148 2405 2175 2430 2201 2455 2227 2480 2253 2504 2279 2529 3 5 8 11 3 5 8 10 3 5 8 10 2 5 7 10 14 13 13 12 16 19 22 24 15 18 21 23 15 18 20 23 15 17 19 22 18 19 2553 2788 2577 2810 2601 2833 2625 2856 2648 2878 2672 2900 2695 2923 2718 2945 2742 2967 2765 2989 2579 2579 2479 2468 12 11 11 11 14 16 19 21 14 16 18 21 13 16 18 20 13 15 17 19 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2468 11 13 15 17 19 21 22 23 24 3222 3424 3617 3802 3243 3444 3636 3820 3263 3461 3655 3838 3284 3483 3674 3856 3304 3502 3692 3874 3324 3522 3711 3892 3345 3541 3729 3909 3365 3560 3747 3927 3385 3579 3766 3945 3404 3598 3784 3962 2468 2468 2467 2457 10 10 9 9 12 14 16 18 12 14 15 17 11 13 15 17 11 12 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2357 9 10 12 14 15 26 27 28 29 4150 4314 4472 4624 4166 4330 4487 4639 4183 4316 4502 4654 4200 4362 4518 4669 4216 4378 4533 4683 4232 4393 4548 4698 4249 4409 4564 4713 4265 4425 4579 4728 4281 4440 4594 4742 4298 4456 4609 4757 2357 2356 2356 1346 8 8 8 7 10 11 13 15 9 11 13 14 9 11 12 14 9 10 12 13 30 4771 4786 4800 4S14 4829 4843 4857 4871 4886 4900 1346 7 9 10 11 13 31 32 33 34 4914 5051 5185 5315 4928 5065 5198 5328 4942 5079 5211 5340 4955 5092 5224 5353 4969 5105 5237 5366 4983 5119 5250 5378 4997 5132 5263 5391 5011 5145 5276 5403 5024 5159 5289 5416 5038 5172 5302 5428 1346 1345 1345 1345 7 7 6 6 8 10 11 12 8 9 11 12 8 9 10 12 8 9 10 11 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1245 6 7 9 10 11 36 37 38 39 5563 5682 5798 5911 5575 5694 6809 5922 5587 5705 5821 5933 5599 5717 5832 5944 5611 5729 5843 5955 5623 5740 5855 5966 5635 5752 5866 5977 5647 5763 5877 5988 5658 5775 5888 5999 5670 5786 5899 6010 1245 1235 1235 1234 6 6 6 5 7 8 10 11 7 8 9 10 7 8 9 10 7 8 9 10 40 6021 6031 6042 6053 6064 6075 6085 6896 6107 6117 1234 5 6 8 9 10 41 42 43 44 6128 6232 6335 6435 6138 6243 6345 6444 6149 6253 6355 6454 6160 6263 6365 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6304 6405 6503 6212 6314 6415 6513 6222 6325 6425 6522 1234 1234 1234 1234 5 5 5 5 6789 6789 6789 6789 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1234 5 6789 46 47 48 49 6628 6721 6812 6902 6637 6730 6821 6911 6646 6739 6830 6920 6656 6749 6839 6928 6665 6758 6848 6937 6675 6767 6857 6946 6684 6776 6866 6955 6693 6785 6875 6964 6702 6794 6884 6972 6712 6803 6893 6981 1234 1234 1234 1234 6 5 4 4 6. 7 7 8 5678 5678 5678 5678 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1233 4 LOGARITHMS 485 1 2 3 4 5 6 7 8 9 123 4 5 6789 51 52 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 7118 7202 7284 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 7316 7396 1233 1223 1223 1223 4 4 4 4 4 5678 5677 5667 5667 5567 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1223 56 57 58 59 7482 7559 7634 7709 7490 7566 7642 7716 7497 7574 7649 7723 7505 7582 7657 7731 7513 7589 7664 7738 7520 7597 7672 7745 7528 7604 7679 7752 7536 7612 7686 7760 7543 7619 7694 7767 7551 7627 7701 7774 1223 1223 1123 1123 4 4 4 4 5567 55-67 4567 4567 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 1123 4 4566 61 62 63 64 7853 7924 7993 8062 7860 7931 8000 8069 7868 7938 8007 8075 7875 7945 8014 8082 7882 7952 8021 8089 7889 7959 8028 8096 7896 7966 8035 8102 7903 7973 8041 8109 7910 7980 8048 8116 7917 7987 8055 8122 1123 1123 1123 1123 4 3 3 3 4566 4566 4556 4556 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1123 3 4556 66 67 68 69 8195 8261 8325 8388 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 8344 8407 8222 8287 8351 8414 8228 8293 8357 8420 8235 8299 8363 8426 8241 8306 8370 8432 8248 8312 8376 8439 8254 8319 8382 8445 1123 1123 1123 1122 3 3 3 3 3 4556 4556 4456 4456 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1122 4456 71 72 73 74 8513 8573 8633 8692 8519 8579 8639 8698 8525 8585 8645 8704 8531 8591 8651 8710 8537 8597 8657 8716 8543 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 8681 8739 8567 8627 8686 8745 1122 1122 1122 1122 3 3 3 3 4455 4455 4455 4455 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1122 3 3 3 3 3455 76 77 78 79 8808 8865 8921 8976 8814 8871 8927 8982 8820 8876 8932 8987 8825 8882 8938 8993 8831 8887 8943 8998 8837 8893 8949 9004 8842 8899 8954 9009 8848 8904 8960 9015 8854 8910 8965 9020 8859 8915 8971 9025 1122 1122 1122 1122 3455 3445 3445 3445 3445 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1122 3 81 82 83 84 9085 9138 9191 9243 9090 9143 9196 9248 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9112 9165 9217 9269 9117 9170 9222 9274 9122 9175 9227 9279 9128 9180 9232 9284 9133 9186 9238 9289 1122 1122 1122 1122 3 3 3 3 3445 3445 3445 3445 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1122 3 3445 86 87 88 89 9345 9395 9445 9494 9350 9400 9450 9499 9355 9405 9455 9504 9360 9410 9460 9509 9365 9415 9465 9513 9370 9420 9469 9518 9375 9425 9474 9523 9380 9430 9479 9528 9385 9435 9484 9533 9390 9440 9489 9538 1122 0112 0112 0112 3 2 2 2 3445 3344 3344 3344 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0112 2 3344 91 92 93 94 95 9590 9638 9685 9731 9777 9595 9643 9689 9736 !_ 9782 9600 9647 9694 9741 9605 9652 9699 9745 9609 9657 9703 9750 9614 9661 9708 9754 9619 9666 9713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 0112 0112 0112 0112 2 2 2 2 3344 3344 3344 3344 9786 9791 9795 9800 9805 9809 9814 8818 0112 2 3344 3344 3344 3344 3334 96 97 98 99 9823 9868 9912 9956 9827 9872 9917 9961 9832 9877 9921 9965 9836 9881 9926 9969 9841 9886 9930 9974 9845 9890 9934 9978 9850 9894 9939 9983 9854 9899 9943 9987 9859 9903 9948 9991 9863 9908 9952 9996 0112 0112 0112 0112 2 2 2 2 2 I 2 4 86 THE THEORY OF HEAT ENGINES 1 2 3 4 5 6 7 8 9 1234 5 6789 00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 0011 i 1222 01 02 03 04 1023 1047 1072 1096 1026 1050 1074 1099 1028 1052 1076 1102 1030 1054 1079 1104 1033 1057 1081 1107 1035 1059 1084 1109 1038 1062 1086 1112 1040 1064 1089 1114 1042 1067 1091 1117 1045 1069 1094 1119 0011 i 1222 0011 0111 i i i i i i i 1222 222?, 2222 2222 2222 2223 2223 05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 0111 06 07 08 09 1148 1175 1202 1230 1151 1178 1205 1233 1153 1180 1208 1236 1156 1183 1211 1239 1X59 1186 1213 1242 1161 1189 1216 1245 1164 1191 1219 1247 1167 1194 1222 1250 1169 1197 1225 1253 1172 1199 1227 1256 0111 0111 0111 0111 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 0111 i 2223 11 12 13 14 1288 1318 1349 1380 1291 1321 1352 1384 1294 1324 1355 1387 1297 1327 1358 1390 1300 1330 1361 1393 1303 1334 1365 1396 1306 1337 1368 1400 1309 1340 1371 1403 1312 1343 1374 1406 1315 1346 1377 1409 0111 0111 0111 0111 2 2 2 2 2223 2223 2233 2233 15 1413 1416 1419 1422 1426 1423 1432 1435 1439 1442 0111 2 2233 16 17 18 19 1445 1479 1514 1549 1449 1483 1517 1552 1452 1486 1521 1556 1455 1489 1524 1560 1459 1493 1528 1563 1462 1496 1531 1567 1466 1500 1535 1570 1469 1503 1538 1574 1472 1507' 1542 1578 1476 1510 1545 1581 0111 0111 0111 0111 2 2 2 2 2233 2233 2233 2333 20 1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 0111 2 2333 21 22 23 24 1622 1660 1698 1738 1626 1663 1702 1742 1629 1667 1706 1746 1633 1671 1710 1750 1637 1675 1714 1754 1641 1679 1718 1758 1644 1683 1722 1762 1648 1687 1726 1766 1652 1690 1730 1770 1656 1694 1734 1774 0112 0112 0112 0112 2 2 2 2 2333 2333 2334 2334 25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 0112 2 2334 3334 3334 3344 3344 26 27 28 29 1820 1862 1905 1950 1824 1866 1910 1954 1828 1871 1914 1959 1832 1875 1919 1963 1837 1879 1923 1968 1841 1884 1928 1972 1845 1888 1932 1977 1849 1892 1936 1982 1854 1897 1941 1986 1858 1901 1945 1991 0112 0112 0112 0112 2 2 2 2 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 0112 2 3344 31 32 33 34 2042 2089 2138 2188 20461 2094 2143 2193 2051 2099 2148 2198 2056 2104 2153 2203 2061 2109 2158 2208 2065 2113 2163 2213 2070 2118 2168 2218 2075 2123 2173 2223 2080 2128 2178 2228 2084 2133 2183 2234 0112 0112 0112 1122 2 2 2 3 3344 3344 3344 3445 35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1122 3 3445 36 37 38 39 2291 2344 2399 2455 2296 2350 2404 2460 2301 2355 2410 2466 2307 2360 2415 2472 2312 2366 2421 2477 2317 2371 2427 2483 2323 2377 2432 2489 2328 2382 2438 2495 2333 2388 2443 2500 2339 2393 2449 2506 1122 1122 1122 1122 3 3 3 3 3445 3445 3445 3455 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1122 3 4455 41 42 43 44 2570 2630 2692 2754 2576 2636 2698 2761 2582 2642 2704 2767 2588 2649 2710 2773 2594 2655 2716 2780 2600 2661 2723 2786 2606 2667 2729 2793 2612 2673 2735 2799 2618 2679 2742 2805 2624 2685 2748 2812 1122 1122 1123 1123 3 3 3 3 4455 4456 4456 4450 45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 1123 3 4556 46 47 48 49 2884 2951 3020 3090 2891 2958 3027 3097 2897 2965 3034 3105 2904 2972 3041 3112 2911 2979 3048 3119 2917 2985 3055 3126 2924 2992 3062 3133 2931 2999 3069 3141 2938 3006 3076 3148 2944 3013 3083 3155 1123 1123 1123 1123 3 3 4 4 4556 4 5 5 G 4566 4566 ANTILOGARITHMS 487 1 2 3 4 5 6 7 8 9 1 234 5 6789 4567 50 51 52 53 54 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1123 4 3236 3311 3388 3467 3243 3319 3396 3475 3251 3327 3404 3483 3258 3334 3412 3491 3266 3342 3420 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3524 3296 3373 3451 3532 3304 3381 3459 3540 1223 1223 1223 1223 4 4 4 4 5567 5567 5667 5667 55 3548 3556 3565 3573 3581 3589 3597 3606 3614 36*2 1223 4 5677 56 57 58 59 3631 3715 3802 3890 3639 3724 3811 3899 3648 3733 3819 3908 3656 3741 3828 3917 3664 3750 3837 3926 3673 3758 3846 3936 3681 3767 3855 3945 3690 3776 3864 3954 3698 3784 3873 3963 3707 3793 3882 3972 1233 1233 1234 1234 4 4 4 5 5 5678 5678 5678 5678 6678 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1234 61 62 63 64 4074 4169 4266 4365 4033 41.78 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4121 4217 4315 4416 4130 4227 4325 4426 4140 4236 4335 4436 4150 4246 4345 4446 4159 4256 4355 4457 1234 1234 1234 1234 5 5 5 5 6789 6789 6789 6789 65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1234 5 6789 66 67 68 69 4571 4677 4786 4898 4581 4688 4797 4909 4592 4699 4808 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4955 4634 4742 4853 4966 4645 4753 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 1234 1234 1234 1235 5 5 6 6 6 7 9 10 7 8 9 10 7 8 9 10 7 8 9 10 70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1245 6 7 8 9 11 71 72 73 74 5129 5248 5370 5495 5140 5260 5383 5508 5152 5272 5395 5521 5164 5284 5408 6534 5176 5297 5420 5546 5188 5309 5433 5559 5200 5321 5445 5572 5212 5333 5458 5585 5224 5346 5470 5598 5236 5358 5483 5610 1245 1245 1345 1345 6 6 6 6 7 8 10 11 7 9 10 11 8 9 10 11 8 9 10 12 75 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1345 7 8 9 10 12 76 77 78 79 5754 5888 6026 6166 5768 5902 6039 6180 5781 5916 6053 6194- 5794 5929 6067 6209 5808 5943 6081 6223 5821 5957 6095 6237 5834 5970 6109 8252 5848 5984 6124 6266 5861 5998 6138 6281 5875 6012 6152 6295 1345 1345 1346 1346 7 7 7 7 8 9 11 12 8 10 11 12 8 10 11 13 9 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6412 6427 6442 1346 7 9 10 12 13 81 82 83 84 6457 6607 6761 6918 6471 6622 6776 6934 6486 6637 6792 6950 6501 6653 6808 6966 6516 6668 6823 6982 6531 6683 6839 6998 6546 6699 6855 7015 6561 6714 6871 7031 6577 6730 6887 7047 6592 6745 6902 7063 2356 2356 2356 2356 8 8 8 8 9 11 12 14 9 11 12 H 9 11 13 14 10 11 13 15 85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 2357 8 10 12 13 15 86 87 88 89 7244 7413 7586 7762 7261 7430 7603 7780 7278 7447 7621 7798 7295 7464 7638 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7516 7691 7870 7362 7534 7709 7889 7379 7551 7727 7907 7396 7568 7745 7925 2357 2357 2457 2457 8 9 9 9 10 12 13 15 10 12 14 16 11 12 14 16 11 13 14 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2467 9 9 10 10 10 11 13 15 17 91 92 93 94 8128 8318 8511 8710 8147 8337 853L 8730 8166 8356 8551 8750 8185 8375 8570 8770 8204 8395 8590 8790 8222 8414 8610 8810 8241 8433 8630 8831 8260 8453 8650 8851 8279 8472 8670 8872 8299 8492 8690 8892 2468 2468 2468 2468 11 13 15 17 12 14 15 17 12 14 16 18 12 14 16 18 95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2468 10 12 15 17 19 96 97 98 99 9120 9333 9550 9772 9141 9354 8572 9795 9162 9376 9594 9817 9183 9397 9616 9840 9204 9419 9638 9863 9226 9441 9661 9886 9247 9462 9683 9908 9268 9484 9705 9931 9290 9506 9727 9954 9311 9528 9750 9977 2468 2479 2479 2579 11 11 11 11 13 15 17 19 13 15 17 20 13 16 18 20 14 16 18 20 INDEX ( The numbers refer Absolute temperature, 2, 28 Acceleration of piston, 411 of connecting rod, 419 Actual indicator diagram, 103 Adiabatic expansion and compression, 6, 69, 322 equation, 57, 77 flow through orifice, 171 Advantages of compound expansion, 1 38 After burning, 283 Air, compressed, transmission of power by, 215 compressors, 219 , efficiency of, 231 motors, 226 for combustion of I pound of fuel, 234, 238 of i cubic foot fuel, 233, 236 Air-engine, Carnot, 24 , Stirling, 30 , Ericsson, 31 , Joule, 33 Air standard cycle, 277 Allan's link motion, 403 Analysis of flue gases, 362 Angle of advance of eccentric, 371 Atkinson cycle, 278, 288 Atmospheric engine, 273 Balancing rotating weights, 436 reciprocating weights, 440, 446 of locomotives, 442 Bell-Coleman refrigerating machine, 155 Bilgram's valve diagram, 382 Binary vapour engine, 99 Blade efficiency, 188, 192 Boiler draught, 246 heating surface, efficiency of, 258 Boulvin's method for 1> diagram, 120 Boyle's Law, 2 Brake horse-power, 341, 356 Calculation of mean specific heat of flue gases, 239 of dryness of steam after expansion, 57 Callendar, Prof. H., no, 123, 129 Calorific value of fuels, 242, 249, 361 Capper, Prof. F., 124 Carnot's cycle for gas, 24, 26 for steam engine, 70 principle, 26 Centrifugal governors, 452 Charles's law, 2 Choice of a refrigerating agent, 168 Chimney draught, 247 Clearance and cushioning, 96, 105 Co- efficient of performance, 153 Cold-air refrigerating machine, 154 Combination of indicator diagrams, 150 Combustion of hydrogen, 233 of carbon, 234 of sulphur, 234 , most efficient rate of, 267 Compound expansion, advantages of, 138 , indicator diagram, 139, 150 Compressed air, transmission of power by, 2I 5 Compressors, efficiency of, 231 Condensation, initial, 106 Conditions for maximum efficiency, 27 Connecting rod, inertia of, 418 , kinetic energy of, 423 Connection between /, z>, and T for a gas, 2,8 /, v, and / for steam, 41 Constant volume cycle, 275 pressure cycle, 279 volume lines, 56 Controlling force, curve of, 468 Curtis turbine, 196 Cut-off governing, 143 Cycle, Carnot's, 24, 26, 70 , Rankine, 73-80 , Otto, 276, 282 , constant volume, 275 , pressure, 279 Cycles of operation, 24, 272 489 49 THE THEORY OF HEAT ENGINES Cyclic variation in speed, 430 Cylinder dimensions, 148 feed, 1 14 volumes, ratio of, 142 walls, temperature of, 107 D De Laval turbine, 188 , losses in, 205 Diagram factors, 96, 131 of twisting moment, 415 Derivation of adiabatic equation for steam, Design of De Laval nozzle, 1 76 Diesel engine, 331 Draught for boiler furnace, 246 Dryness fraction, measurement of, 49, U2,357 , from indicator diagram, 1 14 Dynamical load on shaft, 435 Dynamometer rope, 342 Effect of pressure on turbines, 208 of superheat on turbines, 209 of vacuum on turbines, 209 of clearance, 96, 105, 226 of high gas speed heat transmission, 262 of strength of mixture, 283, 337 of turbulence, 309 of cylinder dimensions, 337 of density of charge, 307 Efficiency of a heat engine, i, 26 of a perfect engine, 24, 70 of gas engines, constant specific heat, 277 , variable specific heat, 324 maximum, conditions for, 27 of blades, 188, 192 Engine trials, steam, 354 , gas and oil, 339 Entropy, 15, 53, 6 1 and total heat, Mollier diagram, 63 temperature diagrams, 25, 30, 32, 35, 54, 89, 120 Equivalent eccentric, 388, 398, 401, 405, 406, 408 Ericsson s air engine, 31 Exhaust steam turbine, 212 Expansion valve, 388 Explosion at constant volume, 312 Feed water, measurement of, 362 First law of thermodynamics, I Flow of steam through orifices, 171 nozzles, 175 Flywheel, function of, 430 Four-stroke cycle, 276, 282 Forced draught, 247 Friction, effect of in jet, 178 governors, 470 Fuel consumption, 343 Fuels, calorific value of, 242, 249 Gain of entropy due to throttling, 61 Gases, permanent, laws of, 2 , specific heats of, 2, 315 Gas engine, theory of, Chaps. XIII. and XIV., pp. 271, 312 , losses in, 302 , expansion curves, 291, 305 , Atkinson, 278, 288 Gaseous fuels, 249 Generation of steam, 40 Glaisher's factor, 347, 482 Governing of gas engines, 289 of steam engines, 143 of steam turbines, 213 Governor, Watt, 452 , Porter, 454 , Proell, 459, 467 , Hartnell, 463 , Hartung, 467 , sensitiveness of, 461 , hunting of, 462 Gooch link motion, 400 Graphic representation of work done, 24 II Hackworth's radial valve gear, 404 Hartnell governor, 463 Hartung governor, 467 Heat drop, 176 mechanical, equivalent of, I carried away by flue gases, 238, 240 by exhaust gases, 348 into engine, 346 account for gas engine, 352 steam engine, 359 steam boiler, 365 reception, rate of, IO, 321 transmission, 257 through flat plates, 257 cylinder walls, 306 condenser tubes, 269 thick tube, 261 Heating surface, efficiency of, 258 Height of chimney, 247 High gas speeds, 262 Hornsby oil engine, 335 Ignition in gas engines, 287 in petrol engines, 336 Impulse turbines, 188 INDEX 491 Indicated horse power, 339, 354 weight of steam, 1 12 Indicator diagram, 95, 103, 291 , combination of, 150 , directions for taking, 339, 341, 354 , effect of friction on, 340 Induced draught, 247 Inertia of reciprocating parts, 411 of connecting rod, 418, 424 Initial load on piston, 145 condensation, 106 Injectors, types of, 182 , theory of, 180 Inside lap, 370 Internal combustion engines, 271, 312, 331 energy of a gas, 3,313,315 of steam, 43 Jacket, steam, 127 Joule's air engine, 33 reversed, 155 equivalent, I Joy's valve gear, 407 K Kelvin, statement of second law, 29 warming machine, 157 Kinetic energy of connecting rod, 423 Klein's construction, 413 Lap in valves, 370 Latent heat of steam, 43 Laval, de, turbine, 188 Laws of thermodynamics, I, 28 of permanent gases, I, 3 Lead in valves, 371 Leakage of valves, 122 Link motion, 397 , Allan, 403 , analytical solution of, 402 , Gooch, 400 , Stephenson, 398 Loaded governor, 454, 459, 467 Locomotive, balancing of, 442 Losses in gas engines, 303, 339 in steam engines, 104, 354 in steam turbines, 205 M Macfarlane Gray's construction, 399 Marshall's radial valve gear, 406 Maximum discharge through orifices, 173 Mean effective pressure, 95, 97 Mechanical equivalent of heat, I refrigeration, 153 Mellanby, Prof. A. L., in, 129 Method of drawing T diagram, 119, 120, 301 Method of twisting moment diagram, 415 Meyer expansion valve, 388 Missing quantity, 115 Mollier diagram, 63, 176 N Nicolson, Prof. J. T., no, 123, 129 Non-expansive engine, 74 Nozzles, design of, 176 Oil engines, 331 Otto cycle, 276, 282 Outside lap, 370 Oval valve diagram, 381 Parsons steam turbine, 197 Permanent gases, laws of, I, 2, 3 Petrol engines, 336 Piston displacement curve, 372 Porter governor, 454 Pressure, volume and temperature relations in a gas, 2, 8 in steam, 41 Primary balancing, 440 Producer gas, theory, 251 Proell governor, 459, 467 R Radial valve gear, Hackworth, 404 , Marshal, 406 , Joy, 407 Rankine cycle, 73, 75-80 Rankine's statement of the second law, 28 Rate of combustion, most efficient, 267 of heat reception, 10, 321 Rateau steam turbine^ 195 Ratio of cylinder volumes, 142 of expansion, most economical, 126 Reaction steam turbines, 188 Receiver engine, 140 Rectangular valve diagram, 375 Refrigerating machines, 153-169 , co-efficient of performance in, 153 , Bell-Coleman, 155 , vapour compression, 158 agent, choice of, 168 Regenerative steam engine, 89 Relation between C p and C v of a gas, 3 /, v and T of a gas, 2, 8 /, v and t of steam, 41 Reuleaux valve diagram, 378 Reynold's law of heat transmission, 262, 265 492 THE THEORY OF HEAT ENGINES Saturated steam, total heat of, 43 , internal energy of, 43 , specific volume of, 41, 44 , table of properties of, 480 Saturation curve on pu diagram, 1 12 Scavenging in gas engines, 285 Second law of thermodynamics, 28 Secondary balancing, 446 Single stage air compressor, 219 motor, 227 Slide valves, 370 Specific heats of gases, 3, 317 Steam, formation of, 40 , saturated, 43 , superheated, 45 , table of properties of, 480 engine, Carnot's cycle for, 70 , Rankine cycle for, 73 , dry saturated during expansion, 82 , regenerative, 89 , expansion curve, 94 jacket, 82, 127 turbines, 186 consumption, 135, 357 engine trials, 354 boiler trial, 360 Stephenson's link motion, 398 Stirling's air engine, 30 Superheated steam, 45 Temperature, absolute, 2, 28 Temperature-entropy diagram for steam, 54,58 from pv diagram, 1 19, 301 for gas engine, 299 Temperature range of cylinder walls, 107 Thermodynamics, laws of, I, 28 Throttling calorimeter, 50, 357 of steam, 46 Total heat in pressure diagram, 65 Transmission of power by compressed air, 215 of heat, 257 Trials of steam engines, 354 boilers, 360 of internal combustion engines, 339 Turbines, steam, 186 , impulse, 188 , reaction, 188 , single-stage, 188 , multi-stage, 194 Turbines, Curtis, 196 , De Laval, 188 , Parsons, 197 , Rateau, 195 , Zoelly, 195 , losses in, 205 , effect of pressure on efficiency of, 208 , of superheat efficiency of, 209 , of vacuum efficiency of, 209 , exhaust steam, 212 , governing of, 213 Twisting moment, 410 diagram, method of drawing, 415 U Unwin, Prof. W. .,96 Valve diagram, rectangular, 375 , Reuleaux, 378 , Zeuner, 379 , oval, 381 , Bilgram, 382 , choice of a, 383 gear, Joy's, 407 , analytical solution of, 386, 394 leakage, 122 displacement, components of, 396 , slide, 370 Vapour compression refrigerating machine, 158 Variable specific heat theory, 321 W Warming machine, 157 Watt governor, 452 Wet steam, 44 Willan's law, 135 Wire drawing, 46, 50, 104 Woolf engine, 139 Work done by expanding gas, 4, 5 during adiabatic expansion of steam, 69 Yarrow-Schlick-T weedy system of balanc- ing, 440 Z Zeuner valve diagram, 379 Zoelly turbine, 195 THE END PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES. UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. Fine schedule: 25 cents on first day overdue 50 cents on fourth day overdue One dollar on seventh day overdue. APR JUN 4 194 NOV 26 1 DEC 2. 194 APR 25 1950 APR I 7 1951 DEC 23 1 3RARY LD 21-100m-12,'46(A2012sl6)4120 YC 33310 969388 Engineering Library THE UNIVERSITY OF CALIFORNIA LIBRARY