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THERMODYNAMICS, ABRIDGED 
 
THERMODYNAMICS, ABRIDGED 
 
 BASED ON "APPLIED THERMODYNAMICS FOR 
 ENGINEERS" BY THE SAME AUTHOR 
 
 BY 
 
 WILLIAM D. ENNIS, M.E. 
 
 MEM. AM. SOC. M.E.J FELIiOW A. A. A. a., CONSULTING ENGINEER: VICE PRESIDENT OV THE 
 
 TECHNICAL ADVISORY CORPORATION OP NEW YORK: FORMERLY PROFESSOR OF 
 
 MECHANICAL AND MARINE ENGINEERING IN THE POST GRADUATE 
 
 DEPARTMENT OF THE UNITED STATES NAVAL ACADEMY 
 
 SECOI^D EDITION— CORJREC TED 
 
 73 ILLUSTRATIONS 
 
 NEW YORK 
 
 D. VAN NOSTRAND COMPANY 
 
 Eight Waeeen Steeet 
 
 1922 
 H 
 
Copyright 1920, 1922, by 
 D. VAN NOSTBAND COMPANY 
 
 PRINTED IN THE UNITED STATES OF AMERICA 
 
PREFACE 
 
 (WHICH STUDENTS SHOULD READ) 
 
 This brief statement of fundamental principles has been 
 prepared especially for the use of midshipmen in the United 
 States Naval Academy. The work was undertaken at the 
 request of Commander J. O. Richardson, U. S. N., Head of the 
 Department of Marine Engineering and Naval Construction, 
 and has been carried on in close cooperation with officers of that 
 Department, particularly Comdr. W. L. Friedell, U. S. N., 
 Lt. Comdr. T. W. Johnson, Professor of Ma|thematics, U. S. N., 
 and Lt. Comdr. H. W. Boynton, U. S. N. 
 
 Thermodynamics is difficult, but worth while. To some 
 extent, it has been simplified by planning the problems for easy 
 solution. The table preceding Chapter II will be found useful 
 for exponential expressions. The solution of many problems 
 is necessary in order that a real grasp of the subject may be 
 attained. All problems should be solved with the slide rule. 
 This implies that answers will be absolutely reliable only with 
 respect to two significant figures, the third figure being estimated. 
 An error which may be as high as 1 per cent, is therefore allow- 
 able. The answers given have been obtained by slide rule, 
 and are subject to this error. Other errors may occasionally be 
 found during a first year's use of the book. The student's 
 answer mat be right, therefore, even when it disagrees with the 
 answer in the book. 
 
 One per cent, accuracy is good enough for almost all practical 
 engineering calculations. We rarely know the strength of a 
 material with any greater exactness. Even the dimensions of 
 structiu-al parts are subject to a comparable error. Good 
 engineering is largely a matter of not straining at gnats and 
 swallowing camels. 
 
 If there is a "royal road" to any kind of learning, thermo- 
 dynamics will be found to be the royal road to real compre- 
 hension of steam machinery. 
 
NUMERICAL CONSTANTS AND COMMONER 
 
 SYMBOLS 
 
 1.34 hp. = 1 kw. 746 watts = 1 hp. 
 
 2.3026 = log. X -T- logio X. 778 ft. lb. = 1 B.t.u. 
 
 5.403 = 778 -=- 144. 1544 = PVm -=- wT, Art. 12. 
 
 32.17 = acceleration of gravity. 2545 B.t.u. per hr. = 1 hp. 
 
 42.42 B.t.u. per min. = 1 hp. 33 000 ft. lb. per min. = 1 hp. 
 
 1 980 000 ft. lb. per hr. = 1 hp. 
 
 B.t.u. = British thermal unit(s): Art. 3. 
 c = clearance: Arts. 7, 39. 
 C = compression ratio: Arts. 38, 149. 
 D = displacement: cu. ft: Arts. 7, 42. 
 e = eflSciency: Arts. 34, 80. 
 = base of Napierian system of logarithms. 
 Br, = volumetric efficiency: Art. 42. 
 E = T + I = internal energy of vapor: Art. 63. 
 F. = Fahrenheit. 
 
 F = factor of evaporation: Art. 73. 
 / = diagram factor: Art. 104. 
 g = 32.17 = acceleration of gravity. 
 h = heat of 1 lb. of boiling liquid: Art. 63. 
 H = total heat of 1 lb. of dry vapor: Art. 65. 
 H' = total heat of 1 lb. of superheated vapor: Art. 83. 
 S, H = heat received by a substance: Art. 2. 
 
 SH = net amount of heat received or rejected in a cycle: Art. 21. 
 hp. = horse power, 
 ihp., Ihp. = indicated horse power. 
 
 I = disgregation work: Art. 2. 
 /o = ice melting effect per ihp.-hr. : Art. 59. 
 kw. = kilowatt (s). 
 fc = specific heat at constant pressure: Art. 14. 
 I = specific heat at constant volume: Art. 14. 
 L = stroke of an engine, ft. : Art. 7. 
 = latent heat of vaporization: Art. 64. 
 log = logarithm to the base 10. 
 loge = logarithm to the base e = 2.3026 log. 
 m = molecular weight: Art. 12. 
 n = polytropic exponent : Art. 24. 
 
 vii 
 
viii Numerical Constants and Commoner Symbols 
 
 n = entropy or change of entropy: Art. 68. 
 riw = entropy of 1 lb. of liquid: Art. 71. 
 rie = entropy of vaporization: Art. 71. 
 n, = total entropy of dry vapor: Art. 71. 
 n' = total entropy of superheated vapor: Art. 83. 
 N = r.p.m.: Art. 7. 
 
 p = absolute pressure, lb. per sq. in. (= gauge pressure + 14.696). 
 P = absolute pressure, lb. per aq. ft. = 144 p. 
 Pm = mean effective pressure: Arts. 7, 39. 
 Pu = intercooler pressure: Art. 45. 
 Q = the quantity of heat in 1 lb. of a vapor: Art. 87. 
 B=PV -r wT: Art. 12. 
 
 r = internal latent heat of vaporization: Art. 65. 
 = ratio of expansion: Arts. 106, 162. 
 r.p.m. = revolutions per minute, 
 s = specific heat: Arts. 3, 27. 
 t = temperature Fahrenheit. 
 T = t + 460 = absolute temperature: Art. 11. 
 T = temperature effect of heat: Art. 2. 
 t' = temperature of superheated vapor: Art. 83. 
 u, U = velocity, ft. per sec: Arts. 56, 133. 
 V = volume of 1 lb. of a substance. 
 v' = volume of 1 lb. of a superheated vapor: Art. 83. 
 v„ = volume of 1 lb. of liquid: Art. 65. 
 V = volume of w lb. of a substance. 
 w = weight, lb. 
 Wf W = external mechanical work done: Arts. 2, 26. 
 STF = net external work done in a cycle: Art. 21. 
 X = dryness fraction: Art. 74. 
 
 y = k ^l = ratio of specific heats (Art. 17) : also adiabatic exponent 
 (Art. 28). 
 
CONTENTS 
 
 Numerical Constants and Commoner Symbols vii 
 
 Chapter I: Preliminary — General Principles. Values 
 
 of Exponentials in Chapters II and III . . 1 
 Chaptek II: Permanent Gases. Laws and Processes . . 13 
 Chapter III: Air Engines, Air Compressors, Air Refrig- 
 eration 46 
 
 Chapter IV: Steam and Other Vapors 76 
 
 Chapter V: Efficiency and Power of Steam Engines . . 118 
 
 Chapter VI: Vapor Refrigeration 138 
 
 Chapter VII: Heat Transfer Appliances 153 
 
 Chapter VIII: Fluid Flow and the Steam Turbine 171 
 
 Chapter IX: Internal Combustion Engines 203 
 
THERMODYNAMICS, ABRIDGED 
 
 CHAPTER I 
 PRELIMINARY— GENERAL PRINCIPLES 
 
 1. Subject-Matter. Thermodynamics treats of heat as a 
 source of motive power. All artificial motive powers (not 
 excluding windmills, tide motors and muscular effort) originate 
 from heat. Heat is not matter, but energy. It is measurable 
 in foot-pounds. Heat may be regarded as due to molecular 
 motion. Properly speaking. Thermodynamics is not concerned 
 with phenomena of heat transmission, which belong in the field 
 of general physics. Neither does it traverse the domain of 
 thermochemistry, which deals with heats of formation without 
 relation to mechanical effects. Thus our subject is from one 
 standpoint merely a subdivision of the subject of heat, which 
 latter constitutes one of the departments of physics. More 
 broadly viewed, however. Thermodynamics deals with molecular 
 mechanics while general physics (including practical astronomy) 
 is concerned with mass mechanics. 
 
 Questions. How does muscular energy originate from heat? 
 Show that the action of a windmill is due to heat. Show that 
 the force which propels a steamship is derived from the heat 
 of the sun. 
 
 2. Effects of Heat. Thermodynamics ignores electrical and 
 chemical effects and (for the present) considers only three results 
 following the application of heat to a substance : 
 
 (1) Increase of Temperature. This is the most familiar effect. 
 
 If heat results from molecular motion, temperature may be 
 
 regarded as due to molecular velocity. The molecules are 
 
 1 
 
2 Thermodynamics, Abridged 
 
 incessantly vibrating. The faster they move, the higher is the 
 temperature. Temperature is measured by the thermometer 
 and may be expressed in Fahrenheit or Centigrade units. Ice 
 melts, normally, at 0° C. or 32° F.; water boils under standard 
 conditions at atmospheric pressure at 100° C. or 212° F. A range 
 of 100° C. means the same thing as a range of 180° F. Hence 
 1 degree range of Centigrade temperatmre = 9/5 degree range of 
 Fahrenheit temperature, and 
 
 Tjr = 9/5 Tc + 32, (1) 
 
 !rc = 5/9(r^- 32). (2) 
 
 Note that the Fahrenheit freezing point, 32°, enters into both 
 equations. 
 
 (2) Disgregation Work. The molecules of fluids exert forces 
 upon one another. When heat is supplied, the molecules may 
 take up new positions, i.e., assume a new configuration. This 
 requires that energy be expended to destroy the original equi- 
 librium. Potential energy is stored up in establishing a new 
 condition of equilibrium. The expenditure of energy necessary 
 for rearranging the molecules is called disgregation work. Dis- 
 gregation work occurs, for example, when ice is melted by heat. 
 
 (3) External Mechanical Work. In general, bodies expand 
 when heated. The work done by reason of such expansion 
 against resisting objects is called external work and is the 
 effect of heat with which we are primarily concerned. 
 
 The action of heat may then be summed up thus: the heat, H, 
 received by the substance is utilized, (a) in raising its temperature, 
 (6) in rearranging its molecules, and (c) in performing mechanical 
 work. In symbols, 
 
 S== T+I+ W, (3) 
 
 where T = heat expended in changing temperature, 
 
 I = heat expended in doing disgregation work, 
 W = heat expended in doing external work. 
 
 All terms must be expressed in the same unit, which may be 
 
Pkeliminaey — General Principles 3 
 
 either a heat unit or a work unit. Signs may be either + or — - 
 Thus if heat is emitted, the sign of fl" is — . The sign of T 
 is — when the temperature falls. The sign of IF is — if work 
 is CONSUMED, i.e., if work is done on the substance instead of 
 BY it. When a substance expands, TF is +. When it is com- 
 pressed, TF is — . Note that T is not the temperature, but the 
 quantity of heat consumed in changing the temperature. It is 
 not even the change of temperature. The sign of J is + if 
 molecular forces have been overcome, as when a separation of 
 mutually attractive molecules is effected. Its sign is — in the 
 reverse instance, as when such molecules are brought closer 
 together. 
 
 Prob. 1. At what temperature do the Centigrade and Fahren- 
 heit thermometers read alike? (Ans., — 40°.) 
 
 Prob. 2. What is the (algebraic) amount of work done in 
 rearranging the molecules if the heat received is 400, the heat 
 represented by a rise of temperature 100 and the mechanical 
 work done 120? (Ans., + 180.) 
 
 Prob. 3. If TF = - 20, J = + 30, T = + 10, find H. 
 (Ans., -F 20.) 
 
 Prob. 4. When 1 lb. of water at the boiling point is converted 
 into steam, 970.4 units of heat must be supplied. The mechani- 
 cal work done by the expansion of water into steam is 72.8 units. 
 Compute Tand I. (Ans., T = 0, J = + 897.6.) 
 
 3. Definitions and Laws. Two bodies are at the same temper- 
 ature when there is no tendency toward a transfer of heat from 
 one to the other. When a heat-transfer occurs, heat always 
 passes from a high-temperature body to one of lower temperature. 
 The transfer tends to continue until the two temperatures are 
 equal. 
 
 The heat unit (British thermal unit, B.t.u.) is the quantity 
 of heat required to raise the temperature of 1 lb. of water 1° F. 
 Since this quantity is somewhat variable according to the initial 
 temperature, we will use the mean B.t.u., defined as 1/180 of 
 
4 Thermodynamics, Abeidged 
 
 that quantity of heat consumed in eaising 1 lb. of water 
 FROM 32° F. TO 212° F. 
 
 The CALORIE is the quantity of heat required to raise 1 kilo- 
 gram (2.2046 lb.) of water 1° C. (9/5° F.). Its value is 
 
 2.2046 X I = 3.9683 B.t.u. 
 
 
 
 The SPECIFIC HEAT of a substance is the quantity of heat 
 (number of heat units) necessary to increase the temperature 
 of 1 lb. of that substance 1° F. By definition, then, the mean 
 specific heat of water from 32° F. to 212° F. is 1.0. The specific 
 heat of a gas is not an absolute constant (see Art. 14). 
 
 When II B.t.u. are supplied to w lb. of a substance having 
 a specific heat s, 
 
 H = ^(^2 - Ti), (4) 
 
 where T2 and Ti are the final and initial temperatures respec- 
 tively. 
 
 When several fluids are mixed, some will lose heat and others 
 will gain heat. The total loss (including loss of heat by any 
 surrounding bodies) must equal the total gain. In other words, 
 the algebraic sum of heat losses is zero : 
 Loss by first body + loss by second body 
 
 + loss by third body • • • + radiation "loss" = 
 
 wMTa - Tm) + wMTb - TJ 
 
 + wMTc - T,n) ■■■ + Hj,= 0. (5) 
 
 Here Wa, Wb, Wc, denote weights of the several fluids in lb., 
 Say Sb, Sc, their respective specific heats, 
 Ta, Tb, Tc, their respective initial temperatures, 
 Tm, their common final temperature, 
 
 Hb, the heat contributed by surrounding bodies (which 
 will be negative if Tm exceeds the surrounding temper- 
 ature). 
 Prob. 5. How many (mean) B.t.u. are required to heat 4 lb. 
 of water from 32° to 212°? (Ans., 720.) 
 
Preliminary — General Principles 5 
 
 Prob. 6. If 1 meter = 39.37 in., how many kilogram-meters 
 are equivalent to 1 ft. lb.? (Ans. 0.1383.) 
 
 Prob. 7. If the specific heat is 0.10, how many calories are 
 consumed in heating 10 lb. 100° F.? (Ans., 25.2.) 
 
 Prob. 8. When 1 lb. of water is heated from 100° to 200°, 
 99.97 B.t.u. are expended. What is the mean specific heat of 
 water for this range? (Ans., 0.9997.) 
 
 Prob. 9. The specific heat is 0.3, the weight 3^ lb. How 
 many B.t.u. are consumed if the temperature rises 500° F.? 
 
 (Ans., 500 B.t.u.) 
 
 Prob. 10. A mixture is made of three liquids, the weights 
 being 5, 3 and 22 lb., the corresponding specific heats 1.0, 0.3 
 and 0.12 and the final temperatiu-e 189°. The original tempera- 
 tures were 200°, 110° and 220°. How much heat was lost by 
 radiation during the mixing? (Ans., Hs = — 65.7 B.t.u.) 
 
 4. Mechanical Equivalent of Heat. The " first law " of 
 Thermodynamics is. 
 
 Heat and mechanical energy are mutually 
 convertible, in the ratio of 778 ft. lb. (777.5 more 
 nearly) to 1 B.T.U. 
 This law has been established by direct experiment (expending 
 work in agitating a fluid and measuring the heating accomplished) 
 and from abundant inference. 1 B.t.u. = 778 ft. lb., just as 
 1 yard = 3 ft. We do not always realize 778 ft. lb. of useful 
 work from 1 B.t.u., because heat does not equal W. but T -\- I 
 + W (Equation 3). The number 778 is called the mechanical 
 equivalent of heat. 
 
 In a heat engine test, each pound weight of steam entered the 
 cylinder containing 1125 B.t.u. Leaving the cylinder, each 
 pound contained 1000 B.t.u. Then 125 B.t.u. disappeared. 
 The engine developed 150 horsepower = 150 X 33 000 or 
 4 950 000 ft. lb. per min. It used 55 lb. of steam per min., 
 or 55 X 125 = 6875 B.t.u., if we assume the heat which dis- 
 appeared to have been " used." Then 6875 B.t.u. = 4 950 000 
 
6 Thermodynamics, Abridged 
 
 ft. lb. or 1 B.t.u. = 720 ft. lb. This is as close an approximation 
 to the true value as might be expected from this type of experi- 
 ment. 
 
 Prob. 11. How many kilogram-meters are equivalent to 1 
 calorie (see Prob. 6). (Ans., 427.) 
 
 Prob. 12. 1 lb. of coal contains 14 000 B.t.u. If this could 
 be fully utilized, how much coal would be burned per min. to 
 develop 1 hp.? (Ans., 0.00303 lb.) 
 
 Prob. 13a. If a 100 hp. engine operates so as to cause the 
 disappearance of 4250 B.t.u. per min., what is the probable value 
 of the mechanical equivalent of heat? (Ans., 778.) 
 
 5. EflBiciency. Efficiency is in general effect -f- cause or 
 OUTPUT -7- INPUT. In Thermodynamics, efficiency is mechani- 
 cal WORK done -j- TOTAL HEAT RECEIVED. Um'ts must be alike. 
 A HORSE POWER (hp.) is 33 000 ft. lb. per min. or 1 980 000 ft. 
 lb. per hr. It is also 42.42 B.t.u. per min. or 2545 B.t.u. per hr. 
 At 100 per cent, plant efficiency, an engine of 1 hp. would con- 
 sume 2545 B.t.u. (in THE coal) per hr. At 10 per cent, efficiency 
 it would consume 25 450 B.t.u. The efficiency of a mechanical 
 device (a chain hoist, for example) may be increased indefinitely 
 by fine fitting, ball bearings, etc., with 100 per cent, as the 
 ultimate limit. Efficiencies of heat engines have a much lower 
 (though for given conditions, a perfectly definite) limit. The 
 actual efficiency of a given engine will always be below this 
 limit and will vary with the conditions according to laws which 
 Thermodynamics aims to discover. Thermodynamics shows 
 how big to make an engine and how efficient it is likely to be, 
 and why. 
 
 Prob. 13b. A ship uses 2 lb. of coal per hr. per indicated hp. 
 The heat value of the coal is 12 725 B.t.u. per lb. What is 
 the efficiency from fuel to cylinder? (Ans., 0.10, or 10 per 
 cent.) 
 
 Prob. 14. An oil engine has an efficiency of 0.13, or 13 per 
 cent. It develops 100 hp. using oil of 19 000 B.t.u. per lb. 
 What weight of oil will it consume per hr.? (Ans., 102 lb.) 
 
Pkeliminary — General Principles 7 
 
 6. Friction: Traction: Plant Efficiency. Thermodynamics is 
 concerned essentially with the conversion of heat in steam to 
 WORK IN the cylinder of the engine. The corresponding 
 efficiency, Et, is called the thermal efficiency. Mechanical 
 friction makes the work at the shaft or " brake " less than 
 that done in the cylinder. The ratio of shaft hp. to cylinder 
 hp. is Eif, the mechanical efficiency. The efficiency from 
 steam to shaft is E^ X E^. The efficiency of a series of con- 
 versions is always the product of the efficiencies of the individual 
 conversions : E = EiY. E^Y, Ez ■ • • . In a turbine, there is no 
 "cylinder" and the only efficiency determinable is the ratio of 
 work at the shaft to heat supplied by the steam. 
 
 If the resistance of a ship or railway train or other propelled 
 object is R lb., the actual hp. expended in propulsion is 
 
 RV _RVm_RVk 
 550 ~ 375 ~ 326 ' 
 
 where V, Vm and Vs are speeds in ft. per sec, miles per hr. and 
 knots per hr., respectively. 
 
 Prob. 15. The shaft hp. is 80, the horse power lost in friction 
 is 20 and 2 545 000 B.t.u. are consumed per hr. Find Et 
 and En. (Ans., 0.10, 0.80.) 
 
 Prob. 16. In an electric central station, 2 lb. of coal (14 000 
 B.t.u. per lb.) are used per kw. hr. (1 kw. = 1.34 hp.). Effi- 
 ciencies are as follows: fuel to steam, 0.75: steam to work in 
 cylinder, 0.25 : shaft to switchboard output, 0.8. Find efficiency 
 from cylinder to shaft. (Ans., 0.81.) 
 
 Prob. 17. The battleship Pennsylvania has a resistance of 
 254 000 lb. at 20 knots. If Em = 0.80 and the efficiency of the 
 propellors is 0.70, what horse power is developed in the cylinders? 
 
 (Ans., 27 700.) 
 
 7. Engine Action. Fig. 1 shows the piston P moving in the 
 cylinder C and communicating via the piston rod R with the 
 external mechanism. Inlet and outlet passages, / and E, are 
 provided with valves, not shown. The piston being in its ex- 
 
8 
 
 Thehmodynamics, Abridged 
 
 treme left-hand position (i.e., as far to the left as the external 
 mechanism will permit), the distance a is called the linear 
 CLEARANCE. The volume then included between the left face 
 
 Extreme right liand 
 position of piston ^^ 
 
 I 
 
 R- 
 
 ^^2^ 
 
 
 J 
 
 Fig. 1. Action of Engine. 
 
 of the piston and the closed faces of the valves is the clearance 
 VOLUME, Vc. The piston now makes one full stroke to the right. 
 The length of stroke, or distance moved, is determined by the 
 external mechanism. If the whole volume swept through by 
 the piston in such a stroke is B, then the clearance (proportion 
 of clearance) is c = v^jD. If d = piston diameter, L = piston 
 stroke, D = (ir/4)(Z^L. The quantity D is called the displace- 
 ment PER STROKE. 
 
Preliminary — General Principles 9 
 
 Most steam engines have the right-hand end of the cylinder 
 closed by a head with inlet and outlet passages. The rod then 
 passes through a stuflBng box in the head, not shown. The 
 engine is then called double-acting, steam being used on both 
 sides of the piston. Most gas and oil engines are single-acting. 
 In single-acting engines, the right-hand end of the cylinder is 
 left open. This gives a better opportunity for cooling the piston 
 and piston rod without circulating cold water through them. 
 In double-acting engines there is of course a clearance at each 
 end of the cylinder. 
 
 The indicator depicts on paper the pressure existing at 
 various points (piston positions) during the stroke. The lower 
 part of Fig. 1 shows a record of this kind, called an indicator 
 diagram. Ordinates represent pressures on the left-hand side 
 of the piston, abscissas represent piston movements. The upper 
 line 123 represents the pressures applied while the piston is 
 moving to the right. The lower line 341 represents resisting 
 pressures during the return (left-hand) stroke. The diagram 
 shows what occm"s on the left-hand side, only, of the piston. 
 It records two strokes. Under constant conditions, subsequent 
 records would simply duplicate this one. 
 
 Abscissas (which represent piston movements) also represent 
 volumes (to another scale), since the cylinder diameter is the 
 same at all points. The volume swept through = area of 
 piston X distance piston has moved. The vertical axis (of zero 
 volume) falls directly below the inner face of the left-hand cyl- 
 inder head only when the valve faces are flush with that face. 
 This is a condition which rarely if ever occurs. The ov axis is 
 located at zero absolute pressure, i.e., 14.696 lb. per sq. in. below 
 the pressure of the atmosphere. Obviously c = a -j- 6. In a 
 double-acting engine, a symmetrically disposed oppositely placed 
 diagram represents the action on the right-hand side of the 
 piston. 
 
 The average ordinate of the indicator diagram (= area -j- b, 
 with proper consideration of scale) is pm, the mean effective 
 PRESSURE, usually taken in lb. per sq. in. It is the average 
 
10 Thermodynamics, Abeidged 
 
 pressure exerted on one side of the piston during one stroke, the 
 pressure during the succeeding stroke being regarded as zero. 
 In a double-acting engine, however, there is an approximately 
 equal average pressure on the other side of the piston during 
 this succeeding stroke: hence the average pressure pm is main- 
 tained continuously. li d = piston diameter in inches, the 
 average total pressure on the piston (disregarding the reduction 
 of area on one side by the rod) is (7r/4)d% lb. If i = stroke 
 in ft., A = piston area in sq. in. = (7r/4)d^ and the engine 
 makes N r.p.m., the pressure p^ acts through a distance of L ft. 
 per stroke, 2L ft. per revolution and 2LN ft. per min., and the 
 cylinder horse power is 
 
 , ,^, 2LN wpmLiPN 
 
 ihp. = (7r/4)ci% X 33-QQ5 - 66 000 
 
 p^AS _ 2pmND _ 2pJLAN 
 33 000 12 X 33 000 33 000 ' 
 
 (6) 
 
 where S = piston speed = 2LN and D = (ir/4)d^i X 12. 
 
 The OVERLOAD CAPACITY of an engine is (maximum power 
 
 — NORMAL power) -t- NORMAL POWER. If pm is the Same at all 
 
 powers, 
 
 N' - N 
 overload capacity= — ^ — , (7) 
 
 where N = normal speed and N' = maximum speed. If N is 
 constant for all powers, 
 
 overload capacity = , (8) 
 
 Pm 
 
 where pm is the normal value and pm' the maximum value. In 
 most engines, pm varies. In marine engines, both pm and N vary. 
 The mean effective pressure fully determines the power of an 
 engine of given size and speed. 
 
 Prob. 18. In an engine 10 in. diameter by 20 in. stroke the 
 linear clearance is J in. Find the displacement per stroke in 
 cu. in. If the valve faces are flush with the inner face of the 
 
Preliminaky — General Principles U 
 
 cylinder head, what is the clearance volume? The proportion of 
 clearance? (Ans., 1570.8: 19.635 cu. in.: 0.0125 or 1| per cent.) 
 
 Prob. 19. The clearance volume of a 10 by 20 (diameter by 
 stroke) in. engine is found to hold 58.905 cu. in. of water. What 
 is its clearance? (Ans., 3| per cent.) 
 
 Prob. 20. The indicator diagram from a 10 by 20 in. engine 
 is 3.1416 in. long. What displacement is represented by 1 in. 
 of diagram length? (Ans., 500 cu. in.) If the clearance of 
 this engine is 5 per cent., what will be the dimension a, in Fig. 1 ? 
 
 (Ans., 0.15708 in.) 
 
 Prob. 21. The area of an indicator diagram is 3 sq. in. and 
 its length is 4 in. If the vertical scale is such that 1 in. repre- 
 sents 40 lb. per sq. in., what is the mean effective pressure? 
 
 (Ans., 30 lb. per sq. in.) 
 
 Prob. 22. In a 10 by 20 in. double-acting engine at 100 r.p.m., 
 with pm = 30, find the piston speed and ihp. 
 
 (Ans., 333^ ft. per min., 23.8 hp.) 
 
 Prob. 23. A constant-speed 10 by 20 in. engine has a normal 
 Pm = 30 and a maximum pm = 60. What is its overload 
 capacity? (Ans., 1.0 or 100 per cent.) 
 
 Prob. 24. At 100 r.p.m., p„ = 30. At 300 r.p.m., p„ = 24. 
 Find overload capacity if 100 r.p.m. is the normal speed. 
 
 (Ans., 140 per cent.) 
 
 Prob. 2S. An engine rated at 100 hp. has an overload capacity 
 of 210 per cent. What is the maximum hp. it can develop? 
 
 (Ans., 310 hp.) 
 
12 Thermodynamics, Abridged 
 
 VALUES OF EXPONENTIALS IN CHAPTERS II AND III 
 
 
 
 J. 
 
 n - 1 
 
 
 
 n 
 
 n 
 
 n 
 
 
 
 1.3 
 
 0.76t 
 
 0.231 
 
 
 
 1.35 
 
 0.74 
 
 0.259 
 
 
 
 1.4 
 
 0.714 
 
 0.286 
 
 
 
 1.406 
 
 0.713 
 
 0.289 
 
 
 0.355»-2»i = 0.787 
 
 
 3.5»" = 
 
 2.52 
 
 o.g"" 
 
 = 0.925 
 
 
 40.259 = 
 
 1.43 
 
 (¥)'•' 
 
 = 6 
 
 
 40.289 = 
 
 1.49 
 
 2W 
 
 = 2.24 
 
 
 40.74 _ 
 
 2.78 
 
 31.406 
 
 = 4.68 
 
 
 \Q-ll.m = 
 
 0.514 
 
 3.04"'* 
 
 = 2.28 
 
 
 100.286 = 
 
 1.93 
 
 3.370.269 
 
 = 1.37 
 
 
 100.289 = 
 
 1.95 
 
 3.37'>'* 
 
 = 2.46 
 
 
 100.714 = 
 
 5.18 
 
 3.390-269 
 
 = 1.38 
 
 
 100.74 
 
 5.50 
 
 3.390'* 
 
 = 2.47 
 
 
 11.080.259 = 
 
 1.86 
 
 3.50.259 
 
 = 1.38 
 
 
 11.08»'* = 
 
 5.93 
 
CHAPTER II 
 PERMANENT GASES. LAWS AND PROCESSES 
 
 General Statement 
 
 A clear understanding of the laws governing the behavior of 
 permanent gases is absolutely essential to a proper comprehen- 
 sion of steam machinery. Although those laws may seem for 
 the next few pages to have little bearing on the heat-power 
 equipment of a ship, their application will soon be shown. The 
 few paragraphs which follow are as important in steam engine- 
 ering as the multiplication table is in arithmetic. 
 
 8. Gases. Thermodynamics deals, generally, with fluids 
 rather than with solids, and (more narrowly) considers vapors 
 and gases rather than liquids. A vapor is formed when a liquid 
 evaporates. If the temperature of a vapor is greatly increased 
 it becomes what we call a gas. Steam at 1200° is a gas. Air 
 exists in the liquid form at about — 300° F. It might be con- 
 sidered a vapor at — 250° F. In ordinary human experience, 
 it is a gas. The present discussion deals with the "permanent" 
 gases. Dry air, hydrogen, nitrogen, oxygen and a few of the 
 compound gases are of this variety. Steam, NH3, CO2 and SO2J 
 on the other hand, are to be regarded as vapors. They will be 
 examined later. In many thermodynamic calculations, great 
 simplicity results (without any serious sacrifice of accuracy) 
 from considering the permanent gases to be "perfect," i.e., to 
 conform exactly to certain physical laws which in reality are 
 only approximated. 
 
 The properties of gases which are chiefly to be considered are 
 the PRESSURE, VOLUME and temperature. By pressure {■p, lb. 
 per sq. in.) we are to understand the uniform absolute (not 
 gauge) pressure exerted per unit of surface by the gas on sur- 
 rounding objects or by surrounding objects on the gas. The 
 
 13 
 
14 Thermodynamics, Abridged 
 
 volume, «, is taken in cu. ft. The symbol t represents the uniform 
 Fahrenheit temperature of the gas. 
 9. Boyle's Law. If the temperature of a gas is kept 
 
 CONSTANT, THE PRESSURE VARIES INVERSELY AS THE VOLUME. 
 
 
 
 \ 
 
 
 
 
 
 
 \ 
 
 \ 
 
 \, 
 
 
 
 
 \ 
 
 \ 
 
 
 V. 
 
 =5t- 
 
 
 ^ 
 
 s. 
 
 S. 
 
 \ 
 
 ^<C 
 
 
 
 V, 
 
 
 
 /^A-, 
 ^^^9 
 
 
 
 
 
 ^^^^i 
 
 Fig. 2. Boyle's Law. 
 
 £i = !? or 'pivi = p2»2 or yv = const, when t is const. (9) 
 
 This is the equation of an equilateral hyperbola, referred to its 
 
 asymptotes. Whenp= 14.696 
 and t = 32°, v for 1 lb. of air 
 is 12.387: hence at this tem- 
 perature and for 1 lb. of air 
 •pv is ALWAYS equal to 14.696 
 X 12.387 = 182.04. Fig. 2 
 represents Boyle's law, the 
 various curves being plotted 
 for various values of the 
 constant. 
 
 Note that at sea level 
 and normal barometer, p = 
 
 GAUGE PRESSURE + 14.696. 
 
 For ordinary calculations, 14.7 is near enough. 
 
 Prob. 26. What is the volume of 5 lb. of air at 32° F. if the 
 pressure is 10 atmospheres? (Ans., 6.1935 cu. ft.) 
 
 Prob. 27. If the volume of 1 lb. of hydrogen is 100 cu. ft. 
 when its pressure is 5.304 lb. by gauge, what wUl be the volume 
 of 4 lb. of the same gas at the same temperature when its gauge 
 pressure is 15.304 lb.? (Ans., 266| cu. ft.) 
 
 Prob. 28. The gauge pressure on a gas tank is 85.304 lb., 
 and its contents weigh 50 lb. What weight will it hold of the 
 same gas at the same temperature under a gauge pressure of 
 185.304 lb.? (Ans., 100 lb.) 
 
 10. Charles' Laws, (a) If the volume of a gas is kept 
 
 CONSTANT, EQUAL INCREilENTS OF PRESSURE ACCOMPANY EQUAL 
 
Permanent Gases — Laws and Processes 
 
 15 
 
 INCREMENTS OF TEMPERATURE. Fig. 3 illustrates this. At 
 constant volume, the relation between pressure and temperature 
 is represented by a straight line. The equa- 
 tion is, p = at -\- b, where a and b are con- 
 stants. 
 
 (6) If THE PRESSURE OF A GAS IS KEPT 
 CONSTANT, EQUAL INCREMENTS OF VOLUME 
 ACCOMPANY EQUAL INCREMENTS OF TEMPER- 
 ATURE. Again there is a straight line law: 
 V = ait+ 6i: Fig. 4. 
 
 Prob. 29. The pressure in a closed tank 
 is 200 lb. (absolute) at 32°, 400 lb. at 524° 
 
 Fig. 3. Charles' 
 Law. (Constant Vol- 
 ume.) 
 
 What will it be at 
 
 -Absolute Zero 
 
 Fig. 4. Charles' 
 (Constant Pressure.) 
 
 Law. 
 
 770°? (Ans., 500 lb.) 
 
 Prob. 30. The absolute pressure of 
 the gas in a balloon is 15 lb. at 40° 
 and 30 lb. at 540°. If we disregard any 
 increase in volume of the balloon, what 
 is the pressure at 90°? (Ans., 16.5 lb.) 
 
 11. Absolute Zero. In Fig. 4, sup- 
 pose the gas to be cooled indefinitely 
 while conforming with the straight line 
 law. At some very low temperature, the volume will then become 
 zero. Experiments indicate a contraction of 1/492 of the original 
 volume per degree of cooling if the gas is originally at 32° F. 
 Hence zero volume would be reached after 492° of cooling, or 
 at a temperature of — 460° F. This temperature (more accur- 
 ately — 459.6° F.) is called the absolute zero. Experiments 
 on the reduction of pressure at constant volume indicate the same 
 point for the absolute zero. Hence Charles' laws may be written. 
 
 when v is constant 
 
 — = ^ when p is constant 
 
 
 (10) 
 
 Here T = t -{■ 460 and denotes (as always hereafter) the 
 absolute temperature. 
 
16 Thekmodtnamics, Abridged 
 
 No actual gas conforms exactly with the laws of Boyle and 
 Charles. Zero volume would never be actually realized. The 
 laws are close approximations for the behavior of "permanent" 
 gases under normal conditions. 
 
 Prob. 31. What is the location of the absolute zero on the 
 Centigrade scale? (Ans., — 273° C.) 
 
 Prob. 32. If an automobile tire did not expand, what would 
 be the percentage increase in absolute pressure when the tempera- 
 ture rose from 40° to 90° F.? (Ans., 10 per cent.) 
 
 Prob. 33. The pressiu-e in a closed tank is 200 lb. absolute 
 when t = 32°. What is the pressure at 524°? Note Prob. 29. 
 
 (Ans., 400 lb.) 
 
 12. The Perfect Gas: Its Law. Conceive of a gas which con- 
 formed rigorously with the laws of Boyle and Charles (Arts. 9, 
 10, 11). Such a hypothetical substance is called the perfect 
 GAS. In many calculations, for dry air and its constituents and 
 for hydrogen, these gases are treated as perfect, with a close 
 approach to correctness. The behavior of the perfect gas may 
 
 be expressed by a single equation 
 which combines the laws already given. 
 In Fig. 5, let the points 1 and 2 rep- 
 resent any two states of a given weight 
 of gas. Draw an equilateral hj^jcr- 
 
 FiG. 5. Law of a Perfect ^°^^ ^"^ ^^^ ^ vertical line 3-2, inter- 
 Gas, secting at some point 3. Along 1-3, 
 
 t is constant and piVi = pzVs. Along 
 2-3, V is constant and p^/ps = TijT^, or pa = 2)2(^3/ To). Then 
 (since V3 = V2), piVi = p^vs = piViiT^lT^) or (since T^ = Ti), 
 
 PiVi P2V2 pv 
 
 "jT = ^— or ™- = constant. 
 
 If pvjT = const., then pv = const, when T is const. — Boyle's 
 law. If piiiilTi = p^tilTi, then when p is const. Vi/vi = Ti/Ti 
 — Charles' law. 
 For air, when p = 14.696 and t = 32°, T = 492° and v per lb. 
 
Permanent Gases — Laws and Pbocesses 17 
 
 is 12.387. Usual units employed are P = 144p, the pressure in 
 lb. per sq. ft. Then PvjT = (144 X 14.696 X 12.387) -^ 492 
 = 53.36 or Pd = 53.36 T. For w lb. having a volume of V cu. ft., 
 PV = 53.36wr. For any gas, 
 
 PV = wRT, (11) 
 
 in which R is some constant. This is the equation of the 
 
 PERFECT GAS. 
 
 P = 14Ap = absolute pressure, lb. per sq. ft., 
 
 V = volume, cu. ft., 
 
 w = weight, lb., 
 
 T = absolute temperature = i + 460. 
 
 Equation (11) includes three variables. It is therefore the 
 equation of a surface. A perspective representation of this 
 surface appears in Fig. 5a. Any plane perpendicular to the T 
 axis cuts an equilateral hyperbola (like psxv) in the surface, 
 the plane being a locus of constant temperature and the hyperbola 
 representing Boyle's law. Charles' law for constant pressure is 
 indicated by the straight lines map, Is, cut in the surface by 
 planes perpendicular to the OP axis. If planes are drawn 
 perpendicular to the OV axis, the intersections with the surface 
 are straight lines xy, vben, which represent the other form of 
 Charles' law. 
 
 Values of R for permanent gases are found as follows. Put a 
 gas under "standard" conditions of pressure (14.696 lb. per sq. 
 in. absolute) and temperature (32° F., or 491.6° absolute) and 
 measure the volume of 1 lb. Call this volume Vo. Then 
 
 P PV 144 X 14.696t;o ^_ 
 ^ = ^ = 1 X 491.6 = ^-^^^'o- 
 
 At any fixed condition, however, the volumes of gases are 
 
 INVERSELY PROPORTIONAL TO THEIR MOLECULAR WEIGHTS, Or 
 
 Vo = c -V- m, where c is a number which has the same value for 
 all gases, and m = molecular weight. The value of c may be 
 determined from the volume and molecular weight of any gas. 
 
18 
 
 Thermodynamics, Abridged 
 P 
 
 Fig. So. Thermodynamic Surface for a Perfect Gas. 
 
 Thus, for hydrogen (molecular weight = 2.016), «o is known to be 
 1 -f- 0.00562. Then c (based on standard conditions, as above 
 defined) is 2.016 -^ 0.00562 = 358.7. For any permanent gas, 
 »o = 358.7 -r- m. The value of R for any gas is then (i.305 
 X 358.7) -^ m = 1544 ^ m. For hydrogen, R = 1544 -i- 2.016 
 = 766. The expanded form of the perfect gas equation is then 
 
 IBUrcT ^^2^ 
 
 Pi- 
 
 rn 
 
 The perfect gas is one having inert molecules: i.e., molecules 
 which neither attract nor repel one another. Permanent gases 
 approach this condition. 
 
Permanent Gases — Laws and Processes 19 
 
 Prob. 34. An automobile tire carries 100 lb. gage pressure 
 at 40° F. and expands 5 per cent, when heated from 40° to 90° F. 
 What is the pressure after heating? (Ans., 105.3 lb., gage.) 
 
 Prob. 35. A balloon filled with hydrogen at 15 lb. per sq. in. 
 absolute pressure has a volume of 766 000 cu. ft. at 80° F. 
 What weight of gas does it hold? What weight of air at 14.696 
 lb. pressure and 70° F. does it displace? When the volume 
 increases 10 per cent, and the temperature of the gas is 90° 
 what is the pressure? 
 
 (Ans., 4000 lb.; 57 400 lb.; 13.9 lb. per sq. in. absolute.) 
 
 Prob. 36. What is the temperature of air at 1000 atmospheres 
 pressure (i.e., P = 1000 X 14.696 X 144) when it weighs 80.75 
 lb. per cu. ft.? (Ans., 32° F.) 
 
 Prob. 37. What is the ratio of sizes of tanks for an oxy- 
 hydrogen light if both tanks have like pressures and tempera- 
 tures? Note 2H2 + 02 = 2H2O. Take molecular weight of 
 oxygen at 32. (Ans., the oxygen tank is 50.4 per cent, the size 
 of the hydrogen tank.) 
 
 Prob. 38. A tank holds 560.4 lb. of nitrogen (molecular 
 weight 28.02) at a pressure of 100 000 lb. per sq. ft. and a tempera- 
 ture of 40° F. What is its volume? (Ans., 154.4 cu. ft.) 
 
 Prob. 39. A 44 cu. ft. tank holds 10 lb. of gas at 794° F. 
 and at a pressure of 10 000 lb. per sq. ft. What is the molecular 
 weight of the gas? (Ans., 44.) 
 
 Prob. 40. In the previous problem, after using gas from the 
 tank the pressure is found to be 4400 lb. per sq. ft. and the 
 temperature 140° F. What weight of gas has been removed? 
 
 (Ans., 0.8 lb.) 
 
 13. Mixtures of Gases: Dalton's Law. Excluding chemical 
 actions, the pressure of a mixture of gaseous bodies is the 
 
 SUM OF the pressures AT WHICH THE VARIOUS CONSTITUENTS 
 WOULD EXIST IF EACH IN TURN ALONE OCCUPIED THE SAME 
 SPACE AS THE MIXTURE AT THE SAME TEMPERATURE. Let the 
 
20 Thermodynamics, Abridged 
 
 mixture condition be P, V, T, and let subscripts denote con- 
 stituents. Then Dalton's law is 
 
 r - Fl+ Fi--- - y -f- y 
 
 T „ , RTw ,,^, 
 
 = y {RlWl + R2W2 • ■ •) = -p^ (13) 
 
 where w = Wi+ Wi- ■ ■ and R refers to the whole mixture. The 
 equivalent value of R in the equation PV = wRT is thus 
 
 {RiWi + R2W2 • • •) -^ {wi + Wi ■■ •)• 
 
 Prob. 41. What are the values of R for oxygen and nitrogen? 
 
 (Ans., 48.25, 55.1.) 
 
 Prob. 42. Considering air to be 3 parts oxygen to 10 of 
 nitrogen, by weight, what should be the value of R for air? 
 
 (Ans., 53.5.) 
 
 Prob. 43. In the preceding, what part of standard atmos- 
 pheric pressure (14.696 lb. per sq. in.) is due to the oxygen? 
 (Ans., the pressure due to oxygen is 26.3 per cent, of that due to 
 nitrogen and the oxygen pressure is 3.06 lb. per sq. in.) 
 
 14. Specific Heats of Gases. A gas may absorb heat while 
 remaining at constant temperature (Art. 20), the specific heat 
 being then infinite; for specific heat is heat divided by change 
 of temperature: or it may change in temperature without 
 absorbing heat (Art. 28), when the specific heat is zero. That 
 this must sometimes be the case is evident from the fundamental 
 equation, ^= T+I+ W. If ffis + and(IF+ J)is-t- and 
 greater than H, T must be negative. We cannot assign numeri- 
 cal values for the specific heats of gases unless at the same time 
 we stipulate the type of process with which such specific heats 
 are realized. 
 
 Two standard processes are clearly recognized. The gas may 
 be kept at constant pressure. The corresponding specific 
 heat, called k (value 0.2375 for air), will then be fairly constant 
 over ordinary temperatures. Second, the gas may be kept at 
 
Permanent Gases — Laws and Processes 21 
 
 CONSTANT volume. In this case, the (reasonably) constant 
 specific heat is called I, the value for air being 0.1689. At very 
 high temperatures, k and I are rapidly variable. 
 
 15. Difference of Specific Heats. When a gas is kept at con- 
 stant VOLUME, it can do no mechanical work, for there is no 
 such work possible without mass-motion. If heated at constant 
 volume, energy will be stored up: so will it be stored when the 
 gas is raised in temperature at constant pressure. If at constant 
 pressure mechanical work is done, then the difference between 
 k and I is due wholly to the performance of work in the one case 
 and not in the other, and k will be greater than I. 
 
 16. Work Done at Constant Pressure. Mechanical work is 
 done when a force acts over a distance. If the force is in lb. 
 and the distance in ft., the work, in ft. lb., is force X distance. 
 Take the case of a gas, at the pressure Pi, having a free (movable) 
 surface of A sq. ft. Let expansion occur in a direction perpen- 
 dicular to such surface, and let the linear movement in this 
 direction be L ft. The volume of the gas must increase during 
 the movement, by an amount equal to the space swept through 
 by the moving surface. This space is AL cu. ft. If Fa and Vi 
 are final and initial volumes in cu. ft., then AL = V2 — Vj and 
 W = PiAL = PiiVi — Vi) ft. lb. Thus the external me- 
 chanical work, in ft. lb., done by a gas kept at constant 
 pressure is equal to the pressure (lb. per sq. ft.) multi- 
 plied BY THE increase IN VOLUME (CU. FT.). 
 
 Suppose the initial and final temperatures to have been Ti 
 and T2. Then by Charles' law 
 
 YJ-Ii E^ _ 1 _ ?^ _ 1 ^2- Vi ^ T2- Ti 
 and 
 
 Hence 
 
 F2-Fi = ^\r2- Ti). 
 
 W = P^iV2 - Fi) = ^ {T2 - Ti). 
 
22 Thermodynamics, Abridged 
 
 But PiVi = wRTi, hence 
 
 Then W = ivR{T2 - T{). Then 
 
 which we may put equal to kwiTi — Ti) since the process is 
 one of constant pressure. Note that T2 and Ti are tempera- 
 tures, while T is the heat consumed in changing temperature, 
 as in Art. 2. 
 
 For heating at constant volume, on the other hand, W = 
 and 11= T -\- I = lw{Ti— Ti). This expresses the value of 
 T-{- 1 for ANT process from a state 1 to a state 2. The value of 
 T + J depends on the two states only, and not on what happened 
 in moving from one to the other of them. See Art. 18. Apply- 
 ing this value of T + -T to a process at constant pressure, 
 
 kw{T2 - Ti) = twin - Ti) + ^ (^2 - T{) 
 or 
 
 ^-/ = 4. (14) 
 
 Thus the quantity R in the perfect gas equation is simply 
 THE difference OF THE TWO SPECIFIC HEATS expressed in ft. lb. 
 It is the external work done in heating 1 lb. of gas 1° at constant 
 pressure. 
 
 Prob. 44. Given k for hydrogen = 3.409, find I. 
 
 (Ans., 2.424 B.t.u.) 
 
 Prob. 45. What value of R is indicated for air, with k and I 
 as given? (Ans., 53.37.) 
 
 Prob. 46. 1 lb. of air at 69° F. is placed in a vertical cylinder. 
 A piston without weight rests above this air. There is no 
 friction, so that the pressure on the air is that of the standard 
 atmosphere above, or 14.696 X 144 = 21 16.2 lb. per sq. ft. 
 
Permanent Gases — -Laws and Processes 23 
 
 The cylinder cross-sectional area is 1 sq. ft. State the pressure, 
 volume and temperature of the air in the cylinder, using standard 
 units. (Ans., P = 2116.2, V = 13.34, T = 529.) What will 
 be the volume after heating this air 1°? The mechanical work 
 done during such heating? (Ans., 13.365 cu. ft.: 53 ft. lb.) 
 
 17. Ratio of Specific Heats. The ratio of k to I, called y, is 
 of very great importance in Thermodynamics. Its value for 
 air is 0.2375 -4- 0.1689 = 1.406. For all of the permanent gases, 
 its value is close to 1.4. The following relations should be noted: 
 
 "^ ' 778' I ^' 778(2/ -1)' 
 
 ^ = 7-78(^^ ^=778^^' E=778Z(,-1). 
 
 Prob. 47. Find y for hydrogen. (Ans. 1.406.) 
 
 Prob. 48. If for nitrogen k = 0.2438, make two estimates of 
 the probable value of I. (Ans., taking R = 55.1, / = 0.173. 
 Taking y = 1.4, I = 0.174.) 
 
 18. Joule's Law. It has been demonstrated experimentally 
 that with perfect gases the amount of disgregation work is so 
 small that it may be ignored. For the permanent gases, 
 
 UNDER NORMAL, CONDITIONS AND WITH SUFFICIENT ACCURACY 
 FOR MOST ENGINEERING APPLICATIONS, Z = and H = T -{- W. 
 
 This refers to Equation (3). In a perfect gas, I would be 
 absolutely zero (Joule's law). 
 
 It follows from Joule's law that H = T during a constant 
 volume (zero work) operation. A constant volume process may 
 therefore be employed to measure the T effect per unit of heat 
 supplied. It is found that H is I B.t.u. per lb. per degree of 
 temperature change in such a process. Then for any process 
 whatever, H = lw{T2 — Ti) -{- W^ where W is expressed in 
 B.t.u. This principle underlies much of what follows. For a 
 given weight of a given gas, the W effect depends on the process 
 and may range from zero upward : but the T effect is determined 
 solely by the temperature change, no matter what the process is 
 or how the change is effected. 
 
24 
 
 Thermodynamics, Abridged 
 
 Prob. 49. The temperature of 8 lb. of air is raised 125° while 
 mechanical work equivalent to 100 B.t.u. is done by the air. 
 How much heat is supplied? (Ans., 268.9 B.t.u.) 
 
 Prob. SO. By what type of process can the temperature of a 
 gas be increased at the least expenditure of heat? (Constant 
 volume process.) What is this expenditure for 100 lb. of hydro- 
 gen heated 10°? (Ans., 2424 B.t.u.) 
 
 Prob. 51. In an air compressor, 778 000 ft. lb. of work are 
 done ON the air, which meanwhile rises 1000° in temperature. 
 If 5 lb. of air are present, state values of T, I, W and H. 
 
 (Ans., r = + 844.5, I = 0, PF = - 1000, H= - 155.5.) 
 
 Thermodynamic Properties of Permanent Gases 
 
 Gas 
 
 Symbol 
 
 Boiling 
 
 Point at 
 
 Atmospheric 
 
 Pressure, 
 
 Deg. Fabr. 
 
 Molecular 
 Weight 
 
 R 
 
 Tc 
 
 ( 
 
 y 
 
 Hehum 
 
 Hydrogen 
 
 Nitrogen 
 
 Oxygen 
 
 Air 
 
 Methane 
 
 Carbon monoxide . 
 
 He 
 H, 
 
 Ni, 
 O2 
 
 CH4 
 CO 
 
 -450 
 -412 
 -318 
 -297 
 -314 
 -265 
 -313 
 
 4.0 
 2.016 
 
 28.02 
 32 
 
 16.032 
 28 
 
 386 
 
 766 
 55.1 
 48.25 
 53.36 
 96.31 
 55.14 
 
 1.25 
 
 3.409 
 
 0.2438 
 
 0.2175 
 
 0.2375 
 
 0.5929 
 
 0.2425 
 
 0.75 
 
 2.424 
 
 0.173 
 
 0.1555 
 
 0.1689 
 
 0.4698 
 
 0.1716 
 
 1.667 
 
 1.406 
 
 1.409 
 
 1.399 
 
 1.406 
 
 1.26 
 
 1.413 
 
 Note: R = ?^ and I 
 m 
 
 ■778" 
 
 19. Graphical Representation of External Work. It has been 
 shown that the external work done at constant pressure is equal 
 to the pressure multiplied by the increase of volume. In Fig. 6, 
 let ab represent a constant-pressure process. Horizontal dist- 
 ances represent volumes. Vertical distances represent pressures. 
 Then W = Pab{Vt - 1\) = the area abdc. If the volume 
 DECREASES, as when the process is from b to a, the work is 
 represented by the same area, but it is work done on the gas, 
 and has a negative sign. 
 
 In Fig. 6, consider any path, ef. Take a short portion, gh, 
 of this path. When gh is infinitely short, it may be considered 
 horizontal. Then Wg^ = ghkj. The work done along the whole 
 
Permanent Gases — Laws and Processes 
 
 25 
 
 path ef is the sum of such areas as ghkj, or is the area efml. If 
 the path is from left to right (e to/), Wis +; otherwise, it is — . 
 The following theorem may be stated : 
 
 /7 
 
 a Peon s/, b 
 
 I I 
 
 i 
 
 f 
 
 w I 
 
 I 
 
 P 
 
 Vconst 
 
 h^p^M 
 
 I i^r^l 
 
 
 1 
 
 c c/ / J k m s r ' 
 
 Fig. 6. Areas Representing External Work. 
 
 On a PV diagram, the value or W fob any process is 
 
 REPRESENTED BY THE AREA BELOW THE LINE REPRESENTING 
 THAT PROCESS. 
 
 For a constant-volume process, no, Fig. 6, there is no sub- 
 tended area and hence W = 0, as already inferred. 
 
 20. Isothermal Process. An isothermal operation is one in 
 
 which THE TEMPERATURE REMAINS CONSTANT. The gaS must 
 
 then conform to Boyle's law and the process is represented (Fig. 6) 
 by an equilateral hyperbola pq. The area below this curve, 
 pqrs, represents W. The magnitude of this area can be obtained 
 by integration, the equation of the curve being PpVp = PgVq. 
 It is 
 
 IF=PpFplog«^ft.lb., 
 
 (15) 
 
26 Thermodynamics, Abeidged 
 
 where loge = 2.303 logio- The symbol log (without a subscript) 
 means logio- 
 
 In Fig. 5a, psxv, ah, mlyn, are isothermaJs. 
 
 For an isothermal process, the fundamental equation, H = T 
 + I + W, becomes H = W: for T = for an isothermal and 
 Z = f or a permanent gas. To obtain H in B.t.u., divide W 
 by 778. 
 
 Prob. 52. .If 10 lb. of air at standard atmospheric pressure 
 and 32° F. are heated at constant pressure to 524° F., find 
 H and W. (Ans., H = 1168.5 B.t.u., W = 262 134 ft. lb.) 
 
 Prob. 53. In the previous problem, show that numerically 
 H - (PF/778) = IwiTi - Ti). 
 
 Prob. 54. If 10 lb. of air at standard atmospheric pressure 
 and 32° F. are expanded isothermally to twice the original 
 volume, find W and H. (Ans., F' = 181 730 ft. lb.: F = 233.6 
 B.t.u.) 
 
 Prob. 55. If 10 lb. of air are cooled at constant volume from 
 524° to 32°, find W and H. (Ans., W= 0,H = - 830.99 B.t.u.) 
 
 21. Cycle. Most thermodynamic operations consist of a series 
 of processes which restore the fluid to its original condition. 
 Such a series constitutes a cycle, which is represented by a 
 closed figure on a PV diagram. For an entire cycle, the funda- 
 mental equation H = T -\- I -]- W becomes H = W: because 
 r = when the original temperature is regained and 1=0 for a 
 permanent gas. Hence, 
 
 In a closed cycle, the net absorption of heat 
 equals the net external work done. 
 SZT and 2 IF must be expressed in the same units, either heat 
 units or foot-pounds. This refers to a clockwise cycle. If the 
 cycle is counter-clockwise, the law holds, but ZH and STF are 
 both negative. The area of the cycle itself is the algebraic sum 
 of subtended areas under the paths which compose it: hence the 
 area of the cycle represents "SH and STF. 
 
 The area of a closed cycle on a PV diagram 
 
 REPRESENTS THE NET AMOUNT OF HEAT ABSORBED OR 
 
Permanent Gases — Laws and Processes 
 
 27 
 
 EMITTED AND ALSO THE NET AMOUNT OP EXTERNAL 
 WORK DONE BY OR ON THE SUBSTANCE: both in ft. lb. 
 
 Prob. S6. In the cycle abc of Fig. 7, for 10 lb. of air, find the 
 external work done along each path and the heat absorbed along 
 each path and show that SJT = SPF. See Probs. 52-55. (Ans., 
 Hob = 1168.5, Hie = - 830.99, Hoa = - 233.6, Sff = 103.9 
 B.t.u. = 80 400 ft. lb.: Wab = 262 134, Wic = 0, W,a = - 
 181 730, SPF = 80 400 ft. lb.) 
 
 22. Efficiency of Cycle. The object of a thermodynamic oper- 
 ation is the performance of some net amount of external work. 
 This is the result of the reception of heat. Since efficiency 
 = effect -i- cause, 
 
 The efficiency of a cycle is the net amount of 
 heat absorbed or external work done (area in 
 b.t.u.) , divided by the whole amount of heat 
 received. 
 In other words, it is SJ? divided by the sum of all of the H terms 
 having + signs. In Fig. 7, it is ahc in B.t.u. divided by Hob- 
 
 Prob. 57. Find the effici- 
 ency in Fig. 7. 
 (Ans., 104 -5- 1168.5 = 0.089.) 
 
 23. Clearance, ^OT, Displace- 
 ment. If such a cycle as 
 that of Fig. 7 were actually 
 employed, the displacement 
 
 (per 10 LB. OF AIR USED) 
 
 would be Fo - F„ = 123.87 
 cu. ft. The clearance (Art. 
 7) would be Va ^ {Vc - Va) 
 = 1.0, or 100 per cent. The 
 
 MEAN EFFECTIVE PRESSURE 
 
 (Art. 7) would be pm = abc 
 
 ^ {Vo - Va) = (104 X 778) -^ 123.87 = 653 lb. per sq. ft. or 
 
 4.54 lb. per sq. in. If 100 lb. of air were used per min., the engine 
 
 would develop (100/10) X (80 400/33 000) = 24.5 hp. (see Prob. 
 
 56). 
 
 p 
 
 
 
 
 a 
 
 b 
 
 albx- 
 
 1 ^ 
 
 c 
 
 £$ 
 ^ 
 
 ^ 
 ^ 
 
 Fig. 7. Problem. 
 
28 
 
 Thermodynamics, Abridged 
 
 Prob. 58. In the cycle of Fig. 8, find clearance, pm, displace- 
 ment and hp. if 50 lb. of air are used per min. and the cycle 
 represents 1 lb. of air. State the abso- 
 lute temperature at each point. (Ans., 
 clearance = 0.333: displacement = 26.68 
 cu. ft. per lb. of air or 1334 cu. ft. per 
 min., pm = 20.83 lb. per sq. in., hp. = 
 121.273, Ta = 500°, Ta = 1000°, n = 
 4000°, To = 2000°.) 
 
 P 
 6000 
 
 3000 
 
 1 
 
 on 
 
 00 
 
 o6 
 
 CM 
 
 Fig. 8. Problem. 
 
 ing as P decreases. 
 
 24. General Case of Gas Process: the 
 Polytropic. Practically all heat engine 
 processes may be represented on the PV 
 diagram by smooth curves, V increas- 
 Fig. 9 shows several such curves. An 
 equation which represents them is PT^" = const, (the exponent 
 applying to V only). When several such curves (called poly- 
 tropic curves) pass through a given point, like a, the value of 
 n indicates the steepness of the curve. Equality of n values 
 does not imply coincident curves. For n = 1, for example, one 
 curve is plotted through a and another through b. 
 For n = 0, PV = const., or P = const. — the process is one 
 
 of CONSTANT PRESSURE. 
 
 For n = 1, PV = const. — the process is isothermal. 
 ¥ovn= 00 , PV" = 00 , P'l'^V = const., V = const. - the 
 operation is at constant voliime. 
 
 25. Relations of Properties in a Polytropic Process. Consider 
 a case of polytropic {PV" = const.) expansion from 1 to 2. 
 Then PiFi" = PiVi". The substance being a gas and the 
 weight constant, the perfect gas equation may be written PiT'ir2 
 = P2V2T1. Dividing the second of these equations by the first, 
 
 T 
 
 -(Rr- 
 
 (16) 
 
 Extracting the n root of the first equation, and then dividing 
 the second, by it, and transposing. 
 
Permanent Gases - 
 
 
 Laws and Processes 
 
 (n-l)/7i 
 
 29 
 
 (17) 
 
 When 71 = 0, Equation (16) reduces to Charles' law for con- 
 stant pressure. When n- = oo , Equation (17) reduces to Charles' 
 law for constant volume. 
 
 «3 
 
 
 n 
 
 =0 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 |\ 
 
 
 b 
 
 
 
 n=o 
 
 
 V 
 
 ^ 
 
 
 \ 
 
 \A^ 
 
 
 
 \\ 
 
 \ 
 
 ^^^^s 
 
 
 
 
 
 \\ 
 
 ^ 
 
 n 
 
 ■■oo 
 
 
 ''i/ 
 
 
 
 \^ 
 
 'o. 
 
 ^"v^ 
 
 ^ 
 
 
 
 n- 
 
 ■OD 
 
 V 
 
 ^^.^ 
 
 
 
 
 V 
 
 Fig. 9. Polytropic Processes. 
 
 Thus suppose fi = 492, Pi = 2116.2, V^ = 12.387. Let 
 n = 1.4, Vi = 123.87. Then fa = 492 X 10-°S log Ti = 2.692 
 - 0.4 = 2.292, ^2 = 195.9. Also (since P2/P1 = (F1/F2)"), 
 Pa = 2116.2 X (1/10)*-', log Pa = 3.3256 - 1.4 = 1.9256, P2 
 
30 Thermodynamics, Abridged 
 
 = 84.26. As a check, from PiViT^ = P^V^Ti, T^ = (84.26 
 X 123.87 X 492) -H (12.387 X 2116.2) = 195.9. 
 
 Prob. 59. Given Pi = 10 000, Fi = 10, Fz = 50, n = 2. 
 The substance is 4 lb. of nitrogen. Find Ti, T^, P^. 
 
 (Ans., Ti = 453.72, Ta = 90.74, P^ = 400.) 
 
 26. External Work, Polytropic Process. The area under a 
 curve passing through a high-pressure point a and a low-pressure 
 point b, if the equation of the curve is PaVa' = PbVb^, is by- 
 integration, 
 
 ^^PaV^-PbV,^ (18) 
 
 71—1 
 
 This gives the work in ft. lb., which is to be regarded as + if the 
 process is from a to 6 (left to right). If the reverse, PT is — . 
 For n = \, the work is indeterminate by this method, and 
 Equation (15) should be used. 
 
 Prob. 60. In the problem immediately preceding, how much 
 external work is done? (Ans., TF = + 80 000 ft. lb.) 
 
 27. Heat Absorbed, Polytropic. In any gas process, M == T 
 + W. For all processes, T = lw{Ti, — Ti). For a polytropic, 
 
 n — 1 
 From Equation (11), PiVi = wRTi and P2V2 = wRTi. Hence 
 
 '^=7-78S^(=^^-^^)^-*- 
 Hence for a polytropic. 
 
 Noting that 
 R 
 
 778 
 
 = k-l, 
 
 H=.iT2- TO {l -'--{), 
 
 and putting k = ly, 
 
 ^ = ^{^l-^)iT2-T^). (19) 
 
Permanent Gases — Laws and Processes 31 
 
 But for any process, if the specific heat is s, H = sw{T2 — Ti). 
 Hence for a polytropic, s = l{n — y)l{n — 1). The value of s 
 for a given gas depends on n alone. The polytropic is then a 
 curve of constant specific heat. This is why the polytropic 
 is so important. Practically all permanent gas processes are 
 represented by polytropic curves. So also are many vapor 
 processes. Many practical applications will be found in Chapter 
 III. For n = 0, s = ly = k, as is to be expected. Iin> 1 < y, 
 s is negative, i.e., heat is emitted when the temperature 
 RISES or absorbed when the temperature falls. This actually 
 occurs in every air compressor. During compression, the 
 temperature of the air rises although the jacket water is con- 
 tinually abstracting heat from the air. The reason is that we 
 are doing external work on the air faster than we are absorbing 
 heat from it. For positive values of s, a rise of temperature 
 indicates reception of heat. When ?i = 1, 5 = oo ; a condition* 
 necessary for an isothermal process, in which no finite supply 
 of heat can ever change the temperature. 
 
 Prob. 61. In Prob. 59, what is the specific heat? (Ans., 
 • 0.1022.) How much heat is absorbed or emitted? (Ans., 
 JT = - 148 B.t.u.) What is the value oi TinH = T + I + W^ 
 (Ans., - 251 B.t.u.) 
 
 28. Adiabatic. An adiabatic process is one during which no 
 heat is received or emitted by the fluid in question. Under 
 adiabatic conditions, H=T+I-{-W=0: but T and W are 
 both finite. (If the substance is a permanent gas, / = 0.) If we 
 assume that the adiabatic may be represented by an equation 
 in the form PV"^ = const., n being undetermined as yet, 
 
 PY _ p 7 
 ^=-^ - W(^f=^-(^^-^^)' 
 
 where the process is from state i to state 2. Since PiFi = wRTi. 
 P2V2 = WRT2, this leads to I = i?/778(ra - 1). In Art. 17, it 
 was shown that I = R/778iy — 1). Hence for an adiabatic 
 PROCESS n = y. The equation of the adiabatic is PiVi'" = P2F2" 
 and for such process, following Arts. 25, 26, 
 
32 
 
 Thermodynamics, Abridged 
 PiFi - P2V2 
 
 W = 
 
 y-l 
 
 During an adiabatic expansion, the temperature falls. In Fig. 
 ha, ah is an isothermal, ae an adiabatic. The two curves pro- 
 jected to the PV plane appear as zk, zg. 
 
 Prob. 62. Air expands adiabatically from Pi = 10 000, Vi 
 = 10 to V2 = 30. Find P2 and W12. 
 
 (Ans., P2 = 2140, Wn = 88 100 ft. lb.) 
 
 29. Summary. The following table summarizes what has 
 been learned concerning permanent gas processes. The values 
 of H are in B.t.u.: those of TTare in ft. lb. if P is taken in lb. 
 
 Pennanent Gases — Processes 
 
 Name 
 
 Horizontal . . . 
 Vertical 
 
 Isothennal . 
 Adiabatic. . 
 Polytropic. . 
 
 Constant 
 Property 
 
 Pressure 
 Volume 
 
 Temperature 
 Entropy' 
 Specific heat 
 
 kw{Ti - TO 
 lw{T2-T,) 
 
 PiFxlog,^ 
 
 778 
 
 
 sw{Tt-Ti) 
 
 w 
 
 Pu(V2 - 7i) 
 
 
 r 1 
 P171-P2F2 
 
 k 
 I 
 
 y -1 
 
 n- I 
 
 n-y 
 n-l 
 
 Remarks 
 
 F2/F1 = T^ITi 
 
 PilPy = T,/Ti 
 
 PiV, =p,r, 
 
 \Tr\vJ =[pj'' 
 
 * The meaning of this term may be disregarded for the present. 
 
 per sq. ft. Only the last line of the table need be emphasized. 
 With the exceptions in heavy type, all of the other necessary 
 relations may be derived from those of the polytropic by sub- 
 stituting the proper value for n. 
 
 30. Application. A cycle will now be worked out in detail to 
 illustrate these formulas and to show how a complete check of 
 
Permanent Gases — Laws and Processes 
 
 33 
 
 the computations is possible. In Fig. 10, data are as follows: 
 
 Clockwise cycle, 2 lb. of nitrogen 
 
 Pi = P2 = 140 000, pr 
 
 P,V2^ = PsFa^ 
 
 ]\ = Fb = 0.5, 
 
 T2 = 1270.32, 
 
 Ps = 80 000, 
 
 P,J\ = P4F4, 
 
 Pi = 20 000, F3 = 3, 
 
 P4F4''- = P3F3", with n 
 
 unknown. 
 
 The unknown properties 
 are first found: 
 
 / 
 
 1 1 1 
 .Constant Pressure 
 
 ^i' 1 1 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 —5 
 
 1 
 
 
 
 
 
 \ 
 
 r 
 
 
 
 
 
 
 %, 
 
 
 \^ 
 
 
 
 
 
 ^ 
 
 ^ 
 
 ?^^ 
 
 3 
 
 
 
 
 4 
 
 
 V 
 
 T, 
 
 PiVi 
 wR 
 
 • Fig. 10. Cycle for Computation. 
 140 000 X 0.5 
 
 2 X 55.1 
 
 = 635.16, 
 
 T,= 7',-=635Xj^oooo=363 = 
 
 T 1 970 
 
 F,= F,j-;-0.6X-^=1.0, 
 
 /|7„\ 1.409 /1\1.409 
 
 P3=P2(^) =140000(3] ' 
 
 -'(fty 
 
 = 1270 X S-"-*"', 
 
 log P3 = 5.1461 - 0.6723 = 4.4738, P3 = 29 770, 
 Jog fa = 3.1040 - 0.1951 = 2.9089, T^ = 810.5. 
 
34 Thermodynamics, Abridged 
 
 Also, to check. 
 
 Ti = -^ = I .'" r, = 810.5. 
 
 P3F3 29 770 X 3 
 wR ~ 2X 55.1 
 
 To find the value of n for the process 4 — 3 : 
 P4F4" = P3F3", log Pi + n log Vi = log Pz + n log F3, 
 
 _ log P 3 — log f 4 
 ""logF4-logF3' 
 
 4,4738-4,3010 ^ _ 
 "" ~ 0.3010 - 0.4771 "•^"^• 
 
 The negative sign is unusual: note the slope of the curve. 
 It does not affect the method of procedure. The specific heat 
 for the path 4 — 3 is 
 
 543 = ^'^ = 0.173 X ^m? = + 0.2087. 
 n — 1 — i.ysi 
 
 We now compute the heat quantities: 
 
 7/12 = wh{Ti - Ti) = 2 X 0.2438(1270 - 635) = 309.7 (+) 
 
 ^23= 
 
 Hu = ws{Ti - Ts) = 2 X 0.2087(363 - 810.5) = 186.8 (-) 
 
 F4 
 _ ^sF5 loge y^ ^ 80 000 X 0.5 X 2.3 log 4 „, ^ , , 
 /f46- 773 - 773 - 71.L (-) 
 
 ^61 = wl{T-, - n) = 2 X 0.173(635 - 363) = 94.1 (+) 
 
 2F = 146 B.t.u. 
 
 The external work quantities are : 
 
 ^12 = Pl2(F2 - Fl) 
 
 = 140 000(1.0 - 0.5) = 70 000 (+) 
 P2F2 - P3F3 
 
 TF23 = 
 
 2/-1 
 (140 000X1.0)^- (29 770X3) ^^^3^^^^^^ 
 
Permanent Gases — Laws and Processes 35 
 
 P3F3 - P4F4 
 
 ^ (29 770X3) -^(20 000X2 )^ 24 900 (-) 
 
 y i 
 
 = 80 000 X 0.5 X 2.3 log 4 = 55 400 (-) 
 
 W,i= 
 
 S IF = 113 600 ft. lb. 
 
 = 146 B.t.u. 
 
 31. Isodiabatics are poly tropic curves not coincident, but 
 having the same value of n. In Fig. 9, the lines M and N are 
 isodiabatics. The following theorems will be found later to 
 save much time in computation with many common forms of 
 cycle : 
 
 1. Let a pair of isodiabatics, M and N, Fig. 11, be cut by lines 
 of constant pressure, Pi and P2. Then 
 
 Dividing the first of these equations by the second, 
 
 PeVe^ P,V„- 
 
 or since Pe = P/ and Pg = Ph, 
 
 PfVf- PhVh^ 
 
 V T. T_ 
 
 (20) 
 
 y e y g ■* e ■* fl 
 
 Vf Vh T, Th' 
 
 the equality of temperature ratios arising from Charles' law. 
 
 2. Let a pair of isodiabatics, M and N, be cut by lines of 
 constant volume, Vi and F2. Then 
 
 PjVj^ = PtVr, PkVk'' = Pr^Vm^, 
 PkVk^ PmVm" Pk Pm Tj, r^' ^^ ' 
 
 Charles' law being again applied. 
 
36 
 P 
 
 Thermodynamics, Abridged 
 
 \ 
 \ 
 
 \\ 
 
 
 
 ■V 
 
 -p, 
 
 \ \ 
 
 
 \ 
 
 
 \ \ 
 
 
 V, \ 
 
 ^ 
 
 \ \ 
 
 
 
 \h 
 
 
 
 
 
 ^\ ' 
 
 k 
 
 \ 
 
 
 \ 
 
 \\ 
 
 N 
 
 
 \^ 
 
 \\ 
 
 \ 
 
 
 > 
 
 \\ 
 
 
 \ 
 
 
 iA 
 
 L 
 
 \, 
 
 
 
 \ 
 
 1 ^. 
 
 '< 
 
 
 ^ 
 
 ^^- 
 
 
 t: 
 
 Fig. 11. Isodiabatics. 
 
 3. Let a pair of isodiabatics, M and iV, be cut by lines of 
 constant temperature, Ti and T^. Then 
 
 n \Pj ~Ta~\Pa) 
 
 Pb Pd' Pc Pd Va n' 
 
 Boyle's law being applied in this instance. 
 
 (22) 
 
Permanent Gases — Laws and Processes 37 
 
 Nearly all heat engines work in cycles made up of one or more 
 pairs of isodiabatics (see, for example, Art. 38). 
 
 Prob. 63. In Fig. 11, given 4 = 40° F., tj = 1040° F., 
 Pk = 20, Pi = 27.3, find Py and P™. (Ans., Py = 60, P™ = 9.1.) 
 
 Prob. 64. In Fig. 11, if Ve = 15, V, = 30, t, = 904° F., 
 Vn = 36, tg = 260° F., find F„, Tk and Te. 
 
 (Ans., F„ = 18, Th = 1440, Te = 682.) 
 
 Prob. 65. In Fig. 11, Pa = 10, Pb = 26.5, Va = 74, F. = 69, 
 T, = 750, Ta = 575. Find Vi, Pc, P., Fa. 
 
 (Ans., Vi = 27.9, Po = 14, Pa = 37.1, F,, = 26.) 
 
 32. Imperfect Gases: Application to a Gun. An imperfect 
 gas may be conceived as a perfect gas through which there are 
 distributed incompressible particles. Let the volume of such 
 particles in 1 lb. of gas be a cu. ft. Then the remainder of the gas 
 conforms to Equation (11), or 
 
 PiV - wa) = wRT. (23) 
 
 This is the Van der Waals equation for imperfect gases. The 
 gas liberated in the powder chamber of a gun furnishes a good 
 example of an imperfect gas. For powder gases, the value of a 
 is approximately 0.001«o, where vq = specific volume at 32° F. 
 and normal barometer. Hence 
 
 '^ = 0-001^« = lS^=«-0°«232P. 
 
 Prob. 66. Find a when R = 46. (Ans., 0.0107.) 
 
 Prob. 67. The powder chamber of a gun has a volume of 
 8.5 cu. ft. The weight of powder is 300 lb., the temperature 
 1000° F. before the projectile moves. Assuming R for powder 
 gases = 46, find the powder pressure at the moment the pro- 
 jectile begins to move. Compute first by Equation (11) and 
 afterward by Equation (23). (Ans., p = 16 400 if the gas is 
 perfect: p = 26 500 by the Van der Waals equation.) 
 
38 Thermodynamics, Abridged 
 
 Prob. 68. If the gun in the foregoing problems is 12-in., 
 50 ft. long from seat to muzzle, what is the whole volume of gas 
 behind the projectile when it emerges from the muzzle? 
 
 (Ans., 47.77 cu. ft.) 
 
 Prob. 69. If maximum pressure occurs when the projectile 
 has traveled half way from seat to muzzle, what is the volume 
 of gases behind the projectile then? (Ans., 28.13 cu. ft.) If 
 maximum pressure occurs when the projectile has moved 19.6 ft., 
 what is the volume at that point? (Ans. 23.88 cu. ft.) 
 
 Prob. 70. In the second case of the foregoing problem, 
 assume that a plot of pressure against volume would be a straight 
 line from atmospheric pressure and 8.5 cu. ft. volume to 30 000 
 lb. per sq. in. pressure at 23.88 cu. ft. volume. How much 
 external work is done up to the point of maximum pressure? 
 
 (Ans., 33 200 000 ft. lb.) 
 
 Prob. 71. In the foregoing, the further expansion of the 
 powder gases is polytropic, with n — 7/6. What is the pressure 
 at the muzzle? (Ans., 13 400 lb. per sq. in.) How much 
 external work is done from point of maximum pressure to 
 muzzle? (Ans., 66 000 000 ft. lb.) 
 
 Prob. 72. In all of the foregoing (Probs. 70, 71) if the powder 
 contains 1620 B.t.u. per lb., what is the thermodynamic efficiency 
 of the gun? (Ans., 0.26.) 
 
 Prob. 73. Ignoring frictional and other losses, the total 
 mechanical work becomes \MU^ where M = mass of projectile 
 (= lbs. weight -f- 32.17) and U = muzzle velocity, ft. per sec. 
 Find muzzle velocity if the projectile weighs 800 lb. 
 
 (Ans., 2820 ft.) 
 
 Prob. 74. In the preceding problems, how long would the 
 gun need to be in order to reduce the muzzle pressure to that of 
 the atmosphere? (Ans., about 4 miles.) 
 
 Prob. 7S. Under the conditions adopted above, find the 
 temperature (by the Van der Waals equation) at the point of 
 maximum pressure and at the muzzle. (Ans., 6500° : 6200° : abs.") 
 
Permanent Gases — Laws and Processes 
 
 39 
 
 Fig. 12. Heat Absorbed, Any Process. 
 = 0. Hence Ha> alone is repre 
 
 33. Graphical Representation of Heat Absorbed. In Fig. 12, 
 
 let ah represent any permanent gas process. Draw two adi- 
 
 abatics through a and h. At „ 
 
 an indefinite distance to the 
 
 right the two adiabatics will 
 
 meet. Designate the meeting- 
 point as M. Then the figure 
 
 ahM is a cycle and the area 
 
 ahM, as already shown, rep- 
 resents the net amount of 
 
 heat absorbed or rejected in 
 
 that cycle. There are only 
 
 three processes included in 
 
 the cycle: Ma, ah and hM. 
 
 We know by definition of an 
 
 adiabatic that Hma = 0, HiM 
 
 sented by the area ahM: or 
 
 The heat absorbed or emitted along any path 
 IS represented by the area included between 
 that path and two adiabatics passing through its 
 extremities and prolonged indefinitely to the 
 
 RIGHT. 
 
 If, as viewed from a point lying between the adiabatics, the 
 path is from left to right, H is + - Otherwise, H is — . Along 
 an adiabatic, by this theorem, H = 0. 
 
 34. Carnot Cycle. In Fig. 13, a cycle ahdc is formed by the 
 combination of a pair of isothermals, ah and cd, with a pair of 
 adiabatics, ac and hd. As in all cycles, the efficiency is net 
 HEAT -i- GROSS HEAT. The net heat is the area ahdc. The 
 "gross" or plus heat is made up of four possible terms: but of 
 these Hac = Hid = 0, the paths in question being adiabatic: 
 while Hdc is eliminated because for a clockwise cycle its value is 
 negative. Hence the efficiency is 
 
 ahdc Hab — Hdc 
 
 e = 
 
 mahM 
 
 H. 
 
 db 
 
40 
 
 Thekmodynamics, Abridged 
 
 PaVa loge ^ - PcVa loge y 
 PaVa loge pT 
 
 From Equation (22), VhjVa = Vd/Vc: hence 
 
 PaVa - PoV, Ta - Tc Tl - T^ 
 
 •ta ' a 
 
 Ta 
 
 i\f 
 
 V(b 
 
 v^. 
 
 (24) 
 
 
 -m 
 
 Fig. 13. Carnot Cycle. 
 
 where Tx and Ta are the highest and lowest (absolute) tempera- 
 tures of the cycle. 
 This value is realized in a Carnot cycle, no matter 
 
 WHAT GAS IS used. 
 
 Between assigned limiting temperatures, no possible 
 
 CYCLIC process CAN HAVE AN EFFICIENCY EXCEEDING THAT OF 
 
 THE Carnot cycle. In Fig. 13, if the proposed superior cycle 
 
Permanent Gases — Laws and Processes 41 
 
 were abde, the point e would be below the lower limiting tenapera- 
 ture. If it were afdc, the point / would lie above the upper 
 limiting temperature. Neither condition is allowable under the 
 stipulation made. An allowable modified cycle would be abgc. 
 As compared with the Carnot, this gives a reduced amount of 
 external work represented by the area cgd: but the gross heat 
 absorbed is still mabM and the efficiency is hence reduced. 
 Consider next the cycle hbdc. It gives less work than the 
 Carnot, represented by the area abh: but it saves heat, the 
 gross heat being now vihbM. The saving of heat is exactly 
 equal to the loss of work. Then the efficiency of the new cycle is 
 
 abdc — abh 
 
 mabM — abh ' 
 
 Since obdcjviabM < 1.0, the efficiency of the new cycle is less 
 than that of the Carnot cycle. 
 
 If some such form as abjdc were proposed, it would be sur- 
 passed in efficiency by a broadened-out Carnot cycle ahlc, and 
 hence by the original cycle abdc, which lies between the same 
 temperature limits as aklc. 
 
 No heat engine can have an efficiency exceeding 
 
 {Ti — T2)jTi WHERE Ti AND Ti ARE THE HIGHEST AND LOWEST 
 ABSOLUTE TEMPERATURES OF THE FLUID IN THE ENGINE. 
 
 Equally definite, but lower, limits of efficiency will presently 
 be established for various specific types of engine. 
 
 Prob. 76. A Carnot cycle gives, area = 100, gross heat = 200. 
 A competitive cycle hbdc is formed, with abh = 20. State the 
 efficiency of each cycle. (Ans., 0.50 and 0.444.) 
 
 Prob. 77. What is the least amount of coal (12 725 B.t.u. 
 per lb.) per hour per horse power used for an engine using steam 
 at 340° F. and exhausting to a condenser at 140° F.? 
 
 (Ans., 0.8 lb.) 
 
 Prob. 78. The maximum temperature in an oil engine is 
 3500° abs., the exhaust temperature 1000° abs. The engine 
 develops 190 hp. and uses oil of 20 000 B.t.u. per lb. What is 
 the least possible oil consumption per hr.? (Ans., 33.8 lb.) 
 
42 Thermodynamics, Abridged 
 
 35. Carnot Cycle Reversed. In Fig. 13, suppose the cycle to 
 be worked counter-clockwise. Begin at c. Suppose the gas to 
 be very cold — say at 0° F. Let it pass through a room, receiving 
 heat (area mcdM) but not increasing in temperature. The 
 point d is reached. Then compress the gas adiabatically to b. 
 Its temperature will rise — say to 150° F. Then pass it across 
 coils through which cold water is circulating. Heat will be 
 abstracted (area mabM) but the temperature may be kept con- 
 stant by adjusting the rates of flow of air and water. The heat 
 
 EMITTED ALONG ha IS THE HEAT THAT WAS ABSORBED ALONG cd 
 
 (plus something). Then expand the gas adiabatically along ac. 
 the temperature falling and the initial condition being restored, 
 
 We have now done work on the gas, or expended power. 
 For what purpose? In order that heat might be taken up by 
 the gas at 0° and discharged by it at 150°. What good object 
 does this serve? 
 
 Suppose it is desired to make ice. Water is available at 50°. 
 Our gas, at 0°, can cool and freeze this water. For the opera- 
 tion to continue, it must then get rid of the heat which it has 
 taken from the water. By expending power we raise the gas 
 temperature without adding heat and the heat taken from 
 the water frozen can then be transferred to the circulating water 
 in the coils of the "condenser " and so discharged from the system. 
 
 This is the operation of a refrigerating machine. In the 
 Carnot cycle used for refrigeration, all changes of temperature 
 would take place without any flow of heat, and all transfers of 
 heat would take place without changes of temperature. Actual 
 refrigerating machines do not realize these conditions. 
 
 This is unfortunate. The efficiency of the Carnot cycle used 
 for refrigeration is 
 
 mcdM _ T2 
 
 ^ ~ mabM - mcdM ~ Ti - f^ ' ^^^^ 
 
 A reversal of the argument of Art. 34 shows that this is the 
 maximum possible value. No actual refrigerating machine can 
 surpass this eSiciency. 
 
 Efficiency is effect divided by cause. The effect, in refrigera- 
 
Peemanent Gases — Laws and Processes 43 
 
 tion, is the absorption of heat by the fluid at low temperature. 
 The cause is the mechanical work done. The part played by 
 the circulating water being disregarded, as in Equation (25), 
 the efficiency may exceed 100 per cent. It is a maximum when 
 the temperature range is least. With engines, on the other 
 hand, efficiency is a maximum when the temperature range is 
 greatest. 
 
 Prob. 79. What would be the percentage saving of coal, in 
 Prob. 77, if the exhaust temperature could be reduced to 40° F.? 
 
 (Ans., 33| per cent.) 
 
 Prob. 80. With the gas temperatures specified in Art. 35, 
 what is the efficiency of refrigeration in the Carnot cycle? 
 (Ans., 3.07 or 307 per cent.) What would it be if heat were 
 rejected at 50° F.? (Ans., 920 per cent.) 
 
 Prob. 81. Under the gas temperatures specified in Art. 35, 
 the heat absorbed by the gas (useful refrigeration) is 307 000 
 B.t.u. per hr. What amount of mechanical work must be done 
 on the gas? (Ans., 100 000 B.t.u. per hr.) What engine horse 
 power is ideally required? (Ans., 39.3.) 
 
 36. Availability of Heat. Recapitulating: for efficient power 
 generation in a Carnot cycle, the heat should be supplied at a 
 high temperature and discharged at a low temperature. The 
 latter temperature should be low, but is more or less fixed by 
 surrounding conditions. The supply temperature is the one 
 which is controllable. It should be high for high efficiency of 
 the Carnot cycle. No actual engine can exceed the efficiency 
 of the Carnot cycle. Anything that keeps the latter efficiency 
 low is bound to keep the former efficiency low, also. Hence 
 for all engines, a high supply temperature is desirable. Any- 
 thing that lowers temperature lowers the potential efficiency. 
 A fall of temperature implies a loss of possible power. The 
 availability of heat is determined by its temperature. A foot 
 pound is 1/778 of a heat unit: but a much smaller fraction of a 
 heat unit is actually realized in mechanical work by actual 
 
44 Thermodynamics, Abridged 
 
 engines. Low fluid supply temperatures make the fraction 
 particularly small. Steam at 600° F. is worth potentially about 
 twice as much as steam at 300° F. It does not cost twice as 
 much. 
 
 Whenever the temperature falls without the simultaneous 
 performance of mechanical work there is a loss of availability 
 of the heat which can never be recovered. This principle is 
 sometimes referred to as the second law of Thermodynamics. 
 
 37. Carnot Cycle Impracticable. An engine should be efficient 
 and powerful. The latter condition renders it necessary that 
 the mean effective pressure, pm, should be fairly high. This 
 should be accomplished without excessive maximum pressures. 
 In the Carnot cycle. Fig. 13, 
 
 nearly, since for any perfect gas y is approximately 1.4. Thus 
 if Ti is twice T2 (which would give an efficiency of 0.50), Pb 
 would be 11.3Pd. Pa is still greater. Suppose we narrow the 
 Carnot cycle by moving ac toward bd. Ultimately the two 
 curves coincide, the enclosed area becomes zero, pm = and still 
 the maximum pressure is 11.3 times the lowest pressure for the 
 temperature range stated. 
 For a finite area, the mean effective pressure is 
 
 area ZTF / ^ , p^Vj - yaV a 
 
 p™ = vT^Y^ = vT^Va = V^"^" i°g^ v+ y-i 
 
 - PcV. loge-pF y_ { j -5- (Fd - FJ 
 
 = j (PaVa - PcVo) l"g^^+ y-1 
 
 Pb y 
 
 . PcVo - PdVd 
 
 -^ (Yd - Va). 
 
 Since pcVc = PdVd and pbVb = PaVa, the last two terms within 
 the bracket disappear. Then 
 
Peemanent Gases — Laws and Processes 45 
 
 [T. - T.)wR log.^ {T, - T,) log. , . „ . 
 
 \Vd Pa) Vd Pa 
 
 For nlTi = 0.5, y = 1.4, pd = 15, pa = 300, Ti = 1000, this 
 gives 
 
 500 X 2.3 log (20 X 0.5'-^) „ ^ ,^ 
 
 ^- = 500_1000 = ^-^ ^^- P"' '^- ^°- 
 
 15 300 
 
 Thus, although the efficiency is 50 per cent., the mean effective 
 pressure is only 9| lb. per sq. in., in spite of the fact that the 
 maximum pressure is 300 lb. per sq. in. This would never do. 
 It would require a very large cylinder in order to develop even a 
 small amount of power. A steam engine, not working in the 
 Carnot cycle, might easily give p^ = 100 when the maximum 
 pressure was 300 lb. It would thus be more than ten times as 
 powerful as the Carnot engine. This would compensate for its 
 lower efficiency. 
 
 Prob. 82. A Carnot cycle has an efficiency of 0.4. Find the 
 lowest possible value of Pa/Pd, using y = 1.4. (Ans., 5.98.) 
 What is then the value of pm? (Ans., 0.) 
 
CHAPTER III 
 
 AIR ENGINES, AIR COMPRESSORS, 
 AIR REFRIGERATION 
 
 Ideal Cycle 
 
 38. Joule Cycle. A cycle underlying the operation of several 
 types of air machinery is shown in Fig. 14, diagram abed. It is 
 
 bounded by a pair of adia- 
 batics and a pair of constant- 
 pressure lines (hence by two 
 pairs of isodiabatics — Art. 
 31). The EFFICIENCY is 
 
 l^H/Hab, Hdc being negative 
 and HdaSind Htc being both 
 zero (compare Art. 34). 
 
 H 
 
 e = 
 
 ab 
 
 H, 
 
 cd 
 
 = 1 - 
 
 1 
 
 Hab 
 Hcd 
 
 Hab 
 
 kw{Tc - Td) 
 
 Fig. 14. Joule Cycle, 
 and k = specific heat at constant pressure. From Equation (20), 
 
 kwin- To)' 
 where w = weight of fluid 
 
 rm J m I rp -»-, 
 
 d la id la 
 
 To- Td 
 
 n- Ta 
 
 , Td Ta 
 
 .= l-y=- 
 
 Ta' 
 
 ■ Td 
 
 (27) 
 
 This looks like the expression for efficiency of the Carnot cycle 
 (Art. 34), which is, however, in terms of the limiting tempera- 
 tures. For Fig. 14, the Carnot eflBiciency is (fi — Td)lTi, 
 which is HIGHER than the Joule efiiciency. 
 
 46 
 
Air Engines, Aie Compressors, Air Refrigeration 47 
 
 The COMPRESSION RATIO (ratio of pressures during the adiabatic 
 compression da which closes the cycle) being called C, we have 
 
 P / T \w(!/-i) 
 (^ = rr[rj ' e = 1 - C^'-y^i". (28) 
 
 Prob. 83. Find the efficiency of a Joule cycle with a com- 
 pression ratio of 10 when y = 1.406. (Ans., 0.486.) If the 
 lowest temperature is 500° abs., and y = 1.406 what is the 
 temperature at the end of adiabatic compression? (Ans., 972° 
 abs.; 
 
 Prob. 84. If Ti, Fig. 14, is 1500° abs., what is the Carnot 
 efficiency in the foregoing problem? (Ans., 0.667.) 
 
 39. Mean Effective Pressure. The exponent y of the curves 
 be and ad is simply a special value of the polytropic exponent n. 
 The mean effective pressure (average ordinate) is 
 
 abed 
 
 Pa{Vb-Va) + ^_ ^ Pc{V-Vd) - ^_ ^ 
 
 (Arts. 16, 26) 
 
 n 
 n 
 
 ^ {pbVt - paVa - PcVo + PdVd) -^ (Vc - Va) 
 
 n 
 
 = ^1 {MVb - Va) + PdiVa - F.)} -f- (F. - Va). 
 
 TV x 
 
 Now clearance = c= Va-^ {Vc— V a) ■ hcncc (Vo — Va) 
 = VJc. Also pblpd = C. 
 
 cnpa C{Vb - Va) + Vd - Vc 
 
 ^"^ ~ Jl - 1 ■ Va 
 
 cnpd ( riY± n j-Yl Yj\ 
 
 - ^i^^^V Va" ^ ^ Va Va)- 
 
 Substitute 
 
48 
 
 Thermodynamics, Abridged 
 
 Va~ ' Va C 
 
 Va~ V/Va C ■ 
 
 Pm 
 
 cnpd 
 
 L j (J(.n-l)ln 
 
 C+1 
 
 - C+ C'l 
 
 "-^) 
 
 npd 
 
 {(c + l)(C("-i)/« - 1) + ciC" - C)]. (29) 
 
 n- 1 
 
 This holds whether n = y ov not. 
 
 Prob. 85. In Fig. 14, pd = normal atmospheric pressure, 
 C = 10, Ta = 500, Tb = 1500, n = 1.4. Find T, (Ans., 965), 
 To (Ans., 778), VdVa (Ans., 1.56), Fd/F„ (Ans., 5.18), F./Fa 
 (Ans., 8.05), c (Ans., 0.142), p™ (Ans., 19.6). 
 
 Compressed Air 
 40. Air Machinery. This section deals only with those air 
 machines in which the pressure changes are of sufficient magni- 
 tude to introduce noticeable 
 temperature effects. It ex- 
 cludes such devices as cen- 
 trifugal fans and blowers. It 
 considers the air as pure and 
 dry and treats the fluid as a 
 perfect gas. Properties of 
 moist air are to be considered 
 later. 
 
 41. Compressor. A com- 
 pressor is the reverse of an 
 engine. Work is done on 
 the fluid (air) by the supply 
 of power from a steam cylin- 
 der, an electric motor, or 
 some other suitable source. Fig. 15 represents the essential 
 parts of a compressor cylinder C, with piston, inlet valve a, dis- 
 charge valve h and receiver D. Suppose the piston to be at the 
 extreme left, and both valves to be closed. It is moved toward 
 
 Fig. 15. Air Compressor. 
 
Air Engines, Air Compressors, Air Refrigeration 49 
 
 the right. At a suitable point early in this stroke, the inlet valve 
 is caused to open. The piston suction draws air into the cylinder 
 throughout the rest of the stroke. At the end of the stroke, the 
 inlet valve is closed. The piston now moves to the left, com- 
 pressing the air. When the pressure has been sufficiently in- 
 creased, the discharge valve opens. The high-pressure air is 
 then discharged to the receiver and pipe line throughout the re- 
 mainder of this left-hand stroke. When the stroke is completed, 
 the discharge valve is closed. The air left in the clearance space 
 expands, during the first part of the succeeding right-hand 
 stroke. When its pressure has been reduced to that of the at- 
 mosphere, the inlet valve opens and the cycle is repeated. 
 
 Practically all compressors are double-acting. 
 
 In Fig. 14, the inlet valve opens at d. Air is drawn in along dc. 
 The pressure along dc is a pound or so below that of the atmos- 
 phere. The inlet valve closes at c and the air is compressed as 
 along cb. If the compression were infinitely rapid, the com- 
 pression path would be an adiabatic, for there would be no time 
 for heat to escape. If compression were infinitely slow, the 
 curve would be an isothermal, because any increase of temper- 
 ture would be offset by conduction of heat. Power is saved in 
 the latter case. The lower the value of n the less is the area of 
 the cycle (which is counter-clockwise) and the less is the 
 power consumed. The value of n depends on the piston speed 
 and the thoroughness of cooling by the water jacket which sur- 
 rounds the cylinder. (Some small compressors are air-cooled 
 only.) Usually, n is between 1.3 and 1.35. The discharge 
 valve opens at b. The pressure along ba is a pound or so higher 
 than the receiver pressure and does not vary much, because of 
 the relatively large receiving capacity into which the air is dis- 
 charged. The discharge valve closes at a. The clearance air 
 then expands as along ad, the value of n for this curve being 
 practically the same as for cb. 
 
 42. Compressor Power Consumption and Capacity. Equation 
 (29) gives a very close approximation to the mean effective 
 
50 Thermodynamics, Abridged 
 
 pressure of an actual compressor if a proper value is selected for n. 
 The value of pd may be taken in good machines at 14 lb. per 
 sq. in. Clearance averages 0.04. C is determined by the dis- 
 charge pressure desired. If S = piston speed, ft. per min., 
 N = r.p.m., d = piston diameter, in., D = displacement, cu. 
 ft. per min., the indicated horse power of the compressor cylinder 
 (compare Art. 7) is (double-acting) 
 
 ^ TT pmd^S ^ 14 4ff„Z) 
 inp. - 4 • 33 000 ~ 33 000 
 
 and the stroke in ft. is S -r- 2N. Values of S are from 400 to 
 800. The primary horse power required to drive the compressor 
 is the above power divided by the mechanical efficiency of the 
 compressor and its driver. 
 
 The IDEAL action might be considered as realized when there 
 was no clearance, when pd = 14.696 lb. per sq. in., and when 
 compression (path eg) was at constant temperature. Then 
 
 D=Vc-Ve= Vc, 
 
 SJF = Weg + W,, - Wcf 
 
 = P„{V, - F.) + P,V, loge ^ - P.{Vo - Vf). 
 
 But Ve = 0, F/ = 0, P,V, = PcVc. 
 
 Sr = P,Vc log. J^ = 144 X 14.696 X 2.3026D log ( J^gg) , 
 
 Ihp- = 33000 = 0-148i> (log p, - 1.167), (30) 
 
 Pm = J44^ = 33.84 (log p„ - 1.167). (31) 
 
 Here (since D is taken as displacement per minute) Fig. 14 is 
 to be regarded as a summation of all cycles realized in one minute 
 of time. 
 
 In the ideal cycle, the volume of air drawn into the cylinder 
 at atmospheric pressure is F^ = D. In an actual cycle the 
 volume is always less than D. Suppose in the cycle abed the 
 
Air Engines, Air Compressors, Air Refrigeration 51 
 
 line hj to represent normal atmospheric pressure. The whole 
 volume of air compressed, measured at this pressure, is Vj. Part 
 of this is clearance air; the volume of which, also measured at 
 atmospheric pressure, is Vh. Then the volume drawn in and 
 compressed, measured at atmospheric pressure, is Vj — Vh- 
 
 Free air is air at normal atmospheric pressure. 
 
 Compressor capacity is measured by the volume of free air 
 drawn in and compressed per minute (symbol F). 
 
 VoLiTMETRic EFFICIENCY (not an efficiency in the thermo- 
 dynamic sense) is the ratio of free air drawn in and compressed, 
 to the displacement. 
 
 e, = FID. (32) 
 
 For the ideal cycle fcge, Fig. 14, F = D and e^ = 1.0. For the 
 actual cycle dcba, 
 
 =(^+i)(ii^6r-Ki4^6y'"=(^3) 
 
 v(-^T=v.'-^(-^^T 
 
 '^'=V14.696/ " c \UmQj ' 
 
 V = V { ^'^ I 
 cft *^"V 14.696/ ■ 
 
 High volumetric efficiency (which leads to large capacity for a 
 given size and speed), results when the clearance is low, the 
 valve ports and passages of ample area, the piston speed low 
 and the value of n low. It is usually between 0.70 and 0.90. 
 
 Prob. 86. An air compressor of 2000 cu. ft. capacity has an 
 intake pressure of 13.226 lb. per sq. in., absolute, 4 per cent, 
 clearance, a discharge pressure of 132.26 lb. gauge. Find pm if 
 n = 1.35. (Ans., 35.2 lb.) 
 
 Prob. 87. In the previous problem, what is the volumetric 
 efficiency? (Ans., 0.742.) 
 
 Prob. 88. In the foregoing, find displacement and Ihp. 
 (Ans., 2700 cu. ft. per min., 415 hp.) 
 
 By = 
 
 F 
 D~ 
 
 Vc- 
 
 Vh 
 
 Va" 
 
 for. 
 
 
 
 
 Fc- 
 
 -Va 
 
 Va 
 C ' 
 
 Vj 
 
52 Thermodynamics, Abridged 
 
 Prob. 89. The compressor cylinder is directly driven by a 
 steam cylinder and the steam cylinder develops 532 Ihp. What 
 is the mechanical efficiency? (Ans., 0.78.) 
 
 Prob. 90. What would be the displacement of an ideal com- 
 pressor giving the diagram /c^'e, Fig. 14? (Ans., 2000 cu. ft. per 
 min.) What would be its Ihp.? (Ans., 296.) 
 
 Prob. 91. In the actual compressor (Prob. 88), the piston 
 speed is 800 ft. per min. and the rotative speed 100 r.p.m. Find 
 cylinder diameter and piston stroke. 
 
 (Ans., diameter = 24.9 in., stroke = 4 ft.) 
 
 43. Cooling. Low values of n for the compression curve are 
 realized as a result of jacket cooling. The limit of gain arises 
 from the slow rate of heat transfer from air through the cast 
 iron walls of the cylinder (Chapter VII). Below this limit, the 
 heat lost by the air, in B.t.u. per min. is weight X specific 
 
 HEAT X TEMPERATURE CHANGE 
 
 Pc(l + c)T) 
 53.36 n 
 
 0.1689"- '•*« 
 
 n-n(g)-}. 
 
 n — \ 
 This may be equated to the heat gained by the water 
 
 = W»(<2 — ti), 
 
 in all of which Wa and Ww are weights of air compressed and 
 water circulated, lb. per min., «„ = specific heat of air for a 
 polytropic path having the exponent n, h and U are outlet and 
 inlet temperatures of water at the jacket. Normal values are 
 Tc = 660 (it cancels out anyway), h = 130, <i = 60 or 70. 
 
 Prob. 92. In the problems of Art. 42, find «„, Wa and w„ if 
 T, = 660, h = 70°, U = 130°. (Ans., *„ = - 0.027, w„ = 152, 
 Wy, = 39.) What is the temperature at 6? (Ans., 767° F.) 
 
 44. Two-Stage Compressor. Compression in successive stages 
 saves power. In Fig. 16, compression begins at c. ct is an 
 adiabatic, cz a polytropic, cs an isothermal. The last is the 
 
Air Engines, Air Compressors, Air Refrigeration 53 
 
 preferred process, but cannot be realized because heat can be 
 transmitted only slowly through the cylinder walls. 
 
 Compress from c to w adiabatically, or slightly better than 
 that if possible, then discharge the air to an external cooler which 
 reduces its temperature to v. Again compress and cool {vw, wx). 
 With an infinite number of steps and complete cooling at each 
 
 jj^Normal 
 
 Afmospheric 
 c Pressure 
 
 Fig. 16. Two-Stage Compressor. 
 
 step, the isothermal path is realized, and power is saved. A 
 number of cylinders and intercoolers are required for multi- 
 stage compression,. The intercooler is a cylindrical shell filled 
 with small tubes through which cold water circulates, while the 
 air surrounds the tubes, on its way from one cylinder to the next. 
 The diagrams ahci and ABCD show the action of a two-stage 
 machine, with polytropic compression in each cylinder. With 
 good intercooling, tc = U. The gross saving of work, as com- 
 
54 Thermodynamics, Abridged 
 
 pared with single stage compression, is represented by the area 
 Bzbr. To offset this, partially, there is half the duplicated 
 hatched area due to friction of air passing through the intercooler, 
 which makes the suction pressure of the second cylinder a pound 
 or so less than the discharge pressure of the first, and thus some- 
 what decreases the saving. 
 
 Multi-stage operation saves power, though increasing first 
 cost and complication. The saving is greatest if the discharge 
 pressure is high. The higher this pressure, the greater the 
 number of stages desirable. Two stages are common in shop 
 practice, with discharge pressures around 100 lb. gauge. 
 
 45. Power Required in Two-Stage Compression. The least 
 total power is consumed when the intercooler pressure is 
 
 The two cylinders will then (if tc = Q require equal amounts 
 of power. The suction pressure pc, with properly designed valve 
 passages, may be taken at po — 0.5, and the discharge pressure, 
 Pa, at 2^0 + 0.5. Equation (29) gives the .mean effective pressure 
 of either cylinder, but the following points should be observed: 
 First (low-pressure) cylinder, clearance = c; = Va/ivc — Va), 
 
 C = Palpd- 
 
 Second (high-pressure) cylinder, clearance = ca = Va/(.vc—va), 
 C = PaIpd, Pd should read po- The equations of Art. 42 may 
 then be used to obtain the Ihp. of each cylinder, for a given 
 displacement, D, of the Low-pressure cylinder. With perfect 
 intercooling, as shown in Fig. 16, the high-pressure displace- 
 ment is D X PcIpc- 
 
 Thus, take the conditions of Prob. 86. 
 
 Po = V13.226 X 146.96 = 44.1. 
 
 Then pc = 43.6, p^ = 44.6. For the first cylinder, the mean 
 effective pressure is 
 
 13.226?i r 
 P™' = n - 1 I ^^' + 1) 
 
 1(1372) -i| 
 
 mr-m)}] 
 
 + c, 
 
AiK Engines, Air Compbessoks, Aie Refrigeration 55 
 For the second cylinder, it is 
 
 f /147.0 V'" /" 147.0 \ n 
 The displacement of the second cylinder is 
 
 The horse power consumed in the two cylinders is 
 
 lUDi / 13.2 \ 
 
 33 000 V ^"'"^ 43.6 ^""V" 
 
 46. Capacity, Two-Stage. The volumetric eificiency e^i of the 
 first cylinder is given by Equation (33) and Di — F -¥■ e^u It is 
 higher than the volumetric efficiency of the single-stage machine 
 with the same extreme pressure limits {^a and pc) and the same 
 clearance. 
 
 Prob. 93. In Art. 45, with n = 1.35, Ci = Ch = 0.04, find 
 Pmi and p„,h- (Ans., p^i = 18, p^h = 58.) 
 
 Prob. 94. In the foregoing, what is the ratio of displacements 
 of the two cylinders? (Ans., DJD„ = 3.3.) 
 
 Prob. 95. In the foregoing, what is the volumetric efficiency 
 of the first cylinder? Compare Prob. 87. (Ans., 0.87.) 
 
 Prob. 96. If i^ = 2000, E^i = 0.87, find 2)^, D„, in the fore- 
 going. (Ans., D^ = 2300, D„ = 700.) 
 
 Prob. 97. Find the horse power consumed by the whole 
 machine, noting that the cylinders require practically equal 
 powers. (Ans., 356 hp.) How much power is saved over single 
 stage operation as computed in Prob. 88? (Ans., 59 hp.) 
 
 Prob. 98. With S = 800, r.p.m. = 100, state cylinder dimen- 
 sions. (Ans., both strokes are 4 ft.: diameters are 23 and 12.65 
 in.) 
 
56 Thermodynamics, Abbidged 
 
 47. Heat Removed in Two-Stage Compression. The method 
 of calculation is much the same as for single-stage machines, 
 given in Art. 43. In Art. 45, and Probs. 93-98, assume Tc = 660. 
 Then Ti = TdpblpcY^-'-^"' = 660(44.6/13.226)°-2« = 904. Assume 
 Tc = 660, as should easily be the case with reasonably good 
 intercooling. Then Tb = TcipB/pcY"-^''"' = 904, also. The air 
 temperature is reduced 904 — 660 = 244 degrees in the inter- 
 cooler. It is increased 244° in each cylinder. The weight of 
 air in the low-pressure cylinder is Pdl + ci)Di -h 53.36 Tc 
 = (13.226 X 144 X 1.04 X 2300) -^ (53.36 X 660) = 130 lb. per 
 min. The weight in the second cylinder is practically the same 
 if its clearance is the same. The weight in the intercooler may 
 differ somewhat, since the intercooler does not cool the clearance 
 air: this difference will be disregarded. The passage of air 
 through the intercooler may be regarded as taking place at 
 constant pressure. The specific heat there is consequently 
 k = 0.2375. The specific heat of air during compression in the 
 cylinders is 
 
 .= Z^ = 0.1689 X=^=- 0.027. 
 n — 1 0.35 
 
 Then the heat removed from the air is its weight multiplied by 
 the sum of the products of specific heats and temperature ranges, 
 or 
 
 130{ (0.2375 X 244) + (2 X 0.027 X 244) } = 9250 B.t.u. per min. 
 This must be removed by water supplied at the jackets and 
 intercooler. It considerably exceeds the heat removed in single- 
 stage compression, where there is no intercooler. 
 
 Prob. 99. How much heat was removed in Prob. 92? 
 
 (Ans., 2330 B.t.u. per min.) 
 
 Prob. 100. In Art. 47, 150.4 lb. of water are delivered to the 
 intercooler every minute. How much will be the rise of tempera- 
 ture of this water? (Ans., 50°.) 
 
 Prob. 101. In Art. 47, the cooling water enters the jackets 
 at 70° and leaves at 156.5°. How much water must be supplied 
 at each jacket? (Ans., 9.9 lb. per min.) 
 
Air Engines, Air Compressors, Air Refrigeration 57 
 
 48. Multi-Stage Compression in General. For four stages, 
 the intercooler pressures are determined by an extension of the 
 method of Art. 45. There will be three intercoolers. The 
 pressure in the middle one, P3, will be a mean proportional 
 between the suction pressure. Pi, of the lowest pressure cylinder 
 and the discharge pressure, Ps, of the highest pressure cylinder. 
 The pressure in the first intercooler, P2, will be VP1P3. That 
 in the third will be P4 = VP3P5. The method may be further 
 extended for eight stages. For three stages, with successive 
 suction pressures of Pi, P2 and P3 and a final discharge pressure 
 of Pi, the condition for least power consumption and (with 
 intercooling sufficient to restore the pre-compression temperature 
 in each case) for equal power consumption by all three cylinders is 
 
 P3=>/P^, Pi=UKP?. 
 
 Prob. 102. A four-stage compressor operates between pres- 
 sures of 14 and 250 lb. absolute. What are the best intercooler 
 pressures? (Ans., 29, 59, 122.) 
 
 Prob. 103. What would be the best intercooler pressures in a 
 three-stage compressor for the same conditions? 
 
 (Ans., 37 and 96 lb.) 
 
 49. Vacuum Pump. The vacuum pump connected with a 
 steam engine or turbine condenser is an air compressor having 
 a suction pressure between 0.5 and 2.0 lb. absolute and a dis- 
 charge pressure around 15 or 16 lb. absolute. Because of the 
 presence of water in the cylinder, the value of n is low. 
 
 50. Air Engine. The most common example of an engine 
 using compressed air is the pneumatic tool. Suppose some of 
 the air compressed, in Prob. 86, to be used in an engine. The 
 essential parts of an engine are shown in Fig. 17, the upper 
 diagram indicating the cylinder, C; piston, P; piston-rod, R; 
 inlet passage, I, from the compressor; and outlet passage, E, 
 to the atmosphere. These passages have valves, not shown. 
 
 The piston is at the left-hand end of its stroke, position Pi. 
 The inlet valve is open. Hence there is a pressure on the piston. 
 
58 
 
 Thermodynamics, Abridged 
 
 tending to force it toward the right. The compressor discharge 
 pressure (Prob. 86) was 146.96 lb. absolute. The pressure in 
 the engine cylinder will be somewhat less: we will take it at 
 145.53 lb., as indicated. In the lower diagram, the point a 
 represents the state of the fluid, Va being the clearance volume. 
 
 ^ ?,Y^\^\v^v^^^w vv^^vv^^^vm\\^^^^^ 
 
 1^ 
 
 '& 
 
 ^ ^^k^^^W^^^^^^4,'^m'^^^^^^^^^^^^^^ ^^^^^^^ 
 
 /e./7- 
 
 c W 
 
 i/ji 
 
 1/5 
 
 '-0.07D 
 
 D ^ 
 
 Fig. 17. Air Engine. 
 
 The piston moves toward the right, until the position Pa is 
 reached. The inlet valve is then closed (point h on the lower 
 diagram). The piston will continue to move to the right, 
 impelled by the expansive force of the air, giving the process 
 curve he. The equation of this curve is ptFt" = pcVc", where n 
 is usually around 1.3 to 1.35. High speed or a heat-insulated 
 condition of the cylinder tend to increase the value of n and thus 
 to decrease work and power. The temperature falls along be. 
 
 Expansion terminates at c, when the discharge valve opens, 
 air rushes out of the cylinder through the outlet E and the 
 
Air Engines, Air Compressors, Air Refrigeration 59 
 
 pressure falls, the piston having reached the end of its stroke 
 (path cd). The farther the expansion is carried, the more work 
 will be obtained from the expansion of a given quantity of air. 
 On the other hand, long expansion leads to low mean effective 
 pressure and hence to large size of engine for a given power. We 
 will assume cut-off (the closing of the inlet valve at b) to occur 
 at one-quarter stroke. (The diagram is not drawn to scale.) 
 
 Now on a volume scale, Vd — Va = D, the displacement per 
 stroke. If clearance be taken at 7 per cent., Vn. = 0.07D. If 
 cut-off is at quarter stroke, ab = Z)/4 or 0.25Z) and Vb = 0.25D 
 + 0.07Z) = 0.32D. The pressure at c is ^(.(Fs/F.)"- With 
 n = 1.3, this gives pc = 145.53(0.32/1.07)1-' = 30.3. 
 
 The piston is now in the position P3. It moves to the left 
 (by the inertia of the reciprocating parts, or if the engine is 
 double-acting, under the action also of high-pressure air on its 
 right-hand side). Throughout the first part of the left-hand 
 return stroke, the pressure remains about constant and a little 
 above that of the atmosphere. (It must be somewhat above, 
 else air would not flow out of the cylinder.) We will take the 
 pressure at 16.17 lb. 
 
 At some position P4, before the return stroke is completed, the 
 outlet valve is closed. The comparatively small amount of air 
 then remaining in the cylinder is compressed (path ef) during the 
 remainder of the return stroke. The point e, at which the exhaust 
 valve closes is called the point of compression (strictly, " be- 
 ginning " of compression). The object of compression is to 
 cushion the action of the engine. We will assume compression 
 to begin at 8/10 the return stroke. Hence Ve = 0.27D. 
 
 The path ef follows the law p/Vf" = peVe^, the value of n 
 being again about 1.8. Then 
 
 ,, = p. (f;)"= 16.17(^-1)"= 93.5. 
 
 The compression curve terminates at /, the piston having then 
 again reached its extreme left-hand position. The inlet valve 
 immediately opens and the pressure increases, giving the path fa 
 before the piston starts to move again. 
 
60 Thermodynamics, Abridged 
 
 51. Air Engine Design. Still referring to Fig. 17, the work 
 done in the cylinder by the air is, since W/a = and Wed = 0, 
 
 •EW = W^ + Wic - Wie - Wef 
 
 = i44Z> j paiVb - Va) H — :;;^zr^ — 
 
 f , , (145.53X0.32)-(30.3X1.07) 
 = 144Z) (145.53 X 0.25) + ^ -^ 
 
 (93.5 X 0.07) - (16.17 X 0.27) \ 
 — (16.17 X 0.8) Q-g 
 
 = 9120Z) ft. lb. per revolution, on each side of a double- 
 acting piston. 
 Suppose 10 hp. to be required, the piston speed to be 300 ft. 
 per min. and the piston diameter to equal the stroke. If N 
 = r.p.m., S = piston speed, d = dia. in in., L = stroke in ft., 
 
 IT d^ d^ 
 
 91202) X 2i\r = 10 X 33 000, D = 
 
 -4: 
 
 4 1728 2200' 
 ^ = 2T=¥' 9120 X^-^^^X^f =330 000, 
 
 330 000X2200 ,^. ^ ,^ ,„ ^onoi-. 
 9120 X 12 X 300 = ^-^ ^"•' i = 4-7 ^ 12 = 0.392 ft., 
 
 N = 1800 -^ 4.7 = 383 r.p.m., D = ~^ = 0.0472 cu. ft. 
 
 52. Air Consumption. The volume of fresh air present at b, 
 Fig. 17, when the supply is cut ofif, is Vb = 0.32Z). Some of this 
 is clearance air. How much? Not Va, for some of the air at a 
 is fresh air, admitted during the operation fa. We know that 
 the volume of clearance air alone was Va = F/, at the pressure 
 Pf. What is it at the pressure jSa? It is reasonable to assume a 
 continued compression of clearance air by incoming air, along 
 the path fx, which is the path ef produced. Then the volume 
 Vx is that of clearance air present at the pressure pa, when the 
 t otal volume of air received is Vt. The value of Vx is 
 
Air Engines, Aib Compeessoes, Air Refeigeeation 61 
 
 Hence the volume of fresh air supplied to the cylinder by the 
 compressor is Vb- V^ = 0.27D = 0.27 X 0.0472 = 0.0127 cu. 
 ft. per stroke or 0.0127 X 383 X 2 = 9.7 cu. ft. per min. This 
 is air at 145.53 lb. pressure. Measured as free air, its volume 
 would be 9.7 X 145.53/14.696 = 96 cu. ft. per min. 
 
 The AIR RATE, which measures the economy of the engine, is 
 
 the VOLUME OF FREE AIR USED PEE HOUE PER HP. Its valuC for 
 
 our conditions is 
 
 96 X 60 
 
 10 
 
 = 576 cu. ft. 
 
 53. Comparisoii with Ideal Conditions. The ideal air engine 
 cycle would have no clearance, expansion would be "complete" 
 (i.e., cut-off would occur so early that the points c and d would 
 coincide), the exhaust pressure would be atmospheric and n would 
 equal 1. Cycle egcf, Fig. 14, will represent the action if p/c 
 = 14.696. Then the fresh air = eg, free air = eg X Pg/pc = fc, 
 and the work of the cycle is 
 
 PeiVg - Ve) + PaVg log. p^ - Pf{V. - Vf) 
 
 = 144 X 14.696F X 2.3 log^^ = 48602^ log ^^, 
 
 F denoting the volume of free air used to produce this work. 
 If s cycles are produced per hr., the free air consumption is Fs 
 cu. ft. per hr. and the hp. is 
 
 i8&OFs , p, 
 .log: 
 
 60 X 33 000 ^14.696' 
 so that the ideal air rate is 
 
 Fs 1 980 000 407 
 
 hp. 4860 (log pg - 1.167) logpg - 1.167 " ^^^^ 
 
 For our conditions, pg = 145.53, log pg = 2.163, ideal air rate 
 = 407/0.996 = 409 cu. ft. The "relative efficiency" of our 
 engine, or efficiency as compared with that attainable under 
 ideal conditions, is 409 -i- 576 = 0.71. (This is not a true 
 efficiency, but only the ratio of two efficiencies.) 
 
62 Thermodynamics, Abridged 
 
 The mean effective pressure of our actual engine is 
 
 f^ 9120 „„„„ 
 Vm = ^ = 3^ = 63.3 lb. per sq. in. 
 
 That of the ideal engine is given by Equation (31) and is for our 
 conditions 33.6 lb. per sq. in. If both engines have the same 
 dimensions and speed, the powers are in the same ratios as the 
 mean effective pressures. The actual engine develops 10 hp. 
 The ideal engine will therefore develop only 
 
 ||xi0 = 5.3hp. 
 
 Its lower mean effective pressure is due to its "complete" expan- 
 sion, which makes the pV diagram depart more noticeably from 
 a rectangle. 
 
 54. Preheat. Reference has been made to the fall of tempera- 
 ture which occurs along he, Fig. 17. This is important. The 
 air may cool below 32° F. Any moisture then present congeals 
 and may obstruct the action of the engine. This condition may 
 be avoided by using very little expansion, as in pneumatic tools. 
 Another method is to heat the air, close to the engine, in a sort 
 of hot air furnace or preheater. An interesting result follows 
 the use of the preheater. 
 
 Assume in Fig. 17 that Ti = 560° and that by employing a 
 preheater this temperature can be raised to 840°. Then, with- 
 out any change in the quantity of air admitted to the cylinder, 
 its volume at cut-off becomes 
 
 T 84.0 
 
 V, = v/f^= 0.32Z) X ^ = 0.48Z), 
 
 by Charles' law. The air now expands along gh, the law being, 
 such that 
 
 .. = P.(^;y=145.53(5f)'"=61.6. 
 Added work is gained, equal to the area hghc, the value of which is 
 
Air Engines, Air Compressors, Air Refrigeration 63 
 PoV„ - PhVh PtVt - PcVc 
 
 = PoiV, - n) + 
 
 n — I n — I 
 
 = 144D I (145.53 X 0.16) + (145-53 X 0.48) - (51.6 X 1.07) 
 
 _ (145.53 X 0.32) - (30.3 X 1.07) 
 0.3 
 
 = 144 X 0.0472 X 26.3 = 178 ft. lb. per cycle. 
 
 This is equivalent (there being 766 cycles per min.) to 4.1 hp., 
 added to the engine output without any increase in size 
 
 OR AIR consumption. 
 
 What does this cost? The preheater must supply enough 
 heat to warm all the air in the cylinder (both fresh air and 
 clearance air) from 560° to 840° absolute. The specific heat may 
 be taken as at constant pressure, k = 0.2375. The weight of 
 air is PiVi -^ RTt = (144 X 145.53 X 0.32 X 0.0472) -=- (53.36 
 X 560) = 0.01056 lb. per cycle. The preheat per cycle is then 
 0.01056 X 0.2375 X 280 = 0.7 B.t.u. or 545 ft. lb., and the 
 efiiciency of the preheating operation is 178 -;- 545 = 0.326. 
 The importance of this will now be shown. 
 
 55. Efficiency of Compressed Air Plant. In Probs. 86 and 88, 
 it was found that a 2000 cu. ft. (free air capacity per minute) 
 single-stage compressor discharging at 132.26 lb. gauge pressure 
 (146.96 lb. absolute) required 415 indicated hp. in the air cylinder. 
 Prob. 89 gives the corresponding hp. of the steam cylinder which 
 drives the compressor as 532. Prob. 90 shows that the ideal 
 compressor would have required only 296 hp., measured in its 
 own cylinder. 
 
 If we start with 532 Ihp. steam, we lose at once 532 — 415 
 = 117 hp., in friction of compressor mechanism. We then lose 
 415 — 296 =119 hp. in the compressor cylinder due to departure 
 from ideal conditions. The " relative efficiency " or ratio of 
 actual and ideal efficiencies of this cylinder is 296 -r- 415 = 0.713. 
 
 The compressed air reaches the engine (Art. 52) at a pressure 
 somewhat less (145.53 lb. absolute) than that at which it was 
 discharged from the compressor, and also at a much reduced 
 
64 Thermodynamics, Abridged 
 
 temperature. Having defined the ideal engine and compressor 
 cycles as we have, the only loss chargeable to pipe line trans- 
 mission is the pressure loss, which may be estimated as follows. 
 
 Art. 53 shows that the ideal air rate for our engine is 409 cu. ft. 
 of free air per Ihp. hr. If we installed a suflBcient number of 
 engines to use all the air furnished by the compressor, the ideal 
 Ihp. of all of these air engine cylinders would be (2000 X 60) 
 -^ 409 = 293. The pipe line loss is then 296 - 293 = 3 hp., 
 or about 1 per cent. If the compressor discharge and engine 
 inlet pressures had been the same, this loss would have been zero. 
 
 By Art. 52, the air rate of the actual engine is 576. Hence 
 the total Ihp. available in actual engine cylinders is (2000 X 60) 
 -V- 676 = 208. The " relative " efficiency of our engine is as 
 already stated, 208 -^ 293 = 0.71. 
 
 The efficiency of the whole conversion process, from Ihp. of 
 steam engine to Ihp. of air engine, is then 208 -i- 532 = 0.391. 
 This is not a thermodynamic efficiency covering the conversion 
 of heat to work, but merely an efficiency of conversion from 
 work in one form to work in another form. A compressed air 
 plant is merely a method for power distribution. A further loss 
 occurs in converting the Ihp. of the air engine cylinder to shaft 
 hp., this being due to mechanical friction of the air engine. The 
 thermodynamic process would start with the heat in the steam 
 entering the steam cylinder of the engine which drives the 
 compressor. As the efficiency of the steam engine under average 
 conditions is around 0.10, the efficiency from steam to Ihp. 
 of air engine is about 0.10 X 0.391 = 0.0391: less than 4 per 
 cent. 
 
 The preheater " boosts " the air engine output to a moderate 
 extent, and the efficiency of its part of the process has been 
 shown, in one illustrative case, to be 32.6 per cent. Thus its 
 effect is to improve the efficiency of the whole system, although 
 it is not installed primarily for that purpose. Compressed air 
 is unique among transmission systems, in that supplementary 
 energy can be supplied at the receiving end with highly beneficial 
 influence on efficiency. 
 
AiK Engines, Air Compkessobs, Air Refrigeration 65 
 
 56. Tests. Observed air rates in well-designed engines are 
 from 477 to 1000 cu. ft. of free air per hour per brake hp. Small 
 engines without expansion use around 2000 cu. ft. 
 
 The probable fall of pressure in the pipe line may be estimated 
 from Unwin's formula: 
 
 j>2 and pi are pressures at end and beginning of line, lb. per 
 
 sq. in., absolute, 
 / = 0.00435 for 6 in. pipe, 0.004 for 8 in. pipe and 0.003 for 
 
 pipe 12 in. or larger, 
 
 u = velocity of air in pipe, ft. per sec. 
 
 _ volume _ ( F_ 14.7 \ tt 
 = "^i^ ~ V60^"^/"^4'*' 
 
 G — length of pipe line, ft., 
 
 T = absolute temperature of air, 
 
 d = internal diameter of pipe, ft. 
 
 Prob. 104. A 12-in. pipe line 5000 ft. long carries 10 000 cu. 
 ft. of free air per min., the initial pressure being 147 lb. per sq. in. 
 Find the volume and velocity per sec. of air flowing through the 
 pipe. (Ans., vol. = 16.67 cu. ft., u = 21.25 ft.) 
 
 Prob. 105. If the temperature of this air is 40° F., find the 
 pressure at the end of the pipe line. (Ans., 145 lb. per sq. in.) 
 
 Prob. 106. If T„ Fig. 17, is 840°, find k. (Ans., 201° F.) 
 
 Air Refrigeration 
 57. Joule Cycle Reversed. The fall of temperature which 
 occurs during expansion in the cylinder of an air engine may be 
 (and has been) employed for refrigeration. In Fig. 14, suppose 
 the Joule cycle abed to be worked counter clockwise. Heat is 
 absorbed from d to c, the fluid is compressed (the temperature 
 rising) from c to b, heat is emitted from 6 to a and the fluid is 
 expanded (the temperature falling) from a to d. (Compare Art. 
 35.) The operation dc is a refrigeration of surrounding bodies. 
 
66 Thermodynamics, Abridged 
 
 During this operation, the fluid is circulating through pipes 
 placed in a brine tank or room to be cooled. 
 
 Refrigeration being the aim in view, the efficiency must be 
 regarded as 
 
 ^~i:W~ Hba- Hdc kin- Ta)-HTc- Ti) 
 
 n- Ta- Tc+ Ta' 
 Since the curves be and ad are isodiabatics, 
 
 rp rp rp rp 'F V 'P 
 
 ib £c ji5 __ 1 I_^ „ 1 ih — I a _ i_a 
 
 rp T? f nn J 'P 'P T' 
 
 i a -id J- a J- d i c -' I S Id 
 
 rfp rp \ rp rp rp rp T'T 
 
 b~ Jo~ Jci" -trf _ ib — 1 a . _ _/^ _ . _ 1 a — I d 
 
 rp rp T' V T* "^ T* 
 
 ■Ic ~ J^ d ic — J- d Id id 
 
 and 
 
 which may of course exceed unity, as in Art. 35. 
 
 The comparable efficiency of the Carnot cycle would be based 
 on the range of temperatures during the refrigerating 
 PROCESS ALONE, i.e., along dc, and would be Td-^ {Tc— Td). 
 This exceeds the value by Equation (36). 
 
 Prob. 107. In a Joule cycle for refrigeration, the compression 
 ratio (C = pjpd, Fig. 14) is 4. Find the efficiency it y = 1.406. 
 
 (Ans., 2.04.) 
 
 Prob. 108. If in Prob. 107 and Fig. 14, Ta = 420, Tc = 571, 
 find Ta, Tb and the efficiency of refrigeration in the Carnot 
 cycle. (Ans., Ta = 628°, Tt = 855°, ec = 2.78.) 
 
 58. Ideal Air Machine. In Fig. 14, the diagram fcbe may be 
 assumed to represent the action of a compressor, eadf that of an 
 engine. The area abed represents the difference between the 
 power absorbed by the compressor and that generated by the 
 engine. In an ice machine, power equivalent to this area, plus 
 the power necessary to overcome friction of mechanism, must 
 
Air Engines, Aik Compressors, Air Refrigeration 67 
 
 be provided from a separate source, such as a steam engine. 
 Let subscripts c, e> and ^ refer to compressor, air engine and 
 steam engine, respectively. Then for 1 lb. of air, 
 
 abed = 778{Hah - Hu) = 778{fc(n - Ta) -k{T,- Ta)], 
 
 Wc-Ws _ abed _Wc_ 
 
 W^ = W,a + Wad - Waf = feiVa - F.) 
 
 PdiVd - Vf) 
 
 
 ' y-1 ' 
 
 = {VaV. - 
 
 P^V,)^^ = ^^{Ta-Td), 
 
 Wc , , 778k{Ti- Ta- To+ Td) , ,, 
 
 w,-^ 1 
 
 Ry{Ta-Td) '" ^" 
 
 or since 
 
 
 
 R = l(y-l)X778, 
 
 Wc 
 
 w^- 
 
 n- Te r. n Dc 
 
 ' ^ Ta- Td Td'Vd D^ 
 
 Also, since W ^ 
 
 = Wc- W^, 
 
 where D = displacement. 
 
 Prob. 109. Under the conditions of Prob. 108, what is the 
 ratio of air engine power to compressor power? Of steam engine 
 power to compressor power? (Ans., 0.735, 0.265.) 
 
 59. Ideal Power and Units. The heat absorbed along dc, in 
 Fig. 14, for the temperatures assigned in Prob. 108, is 
 
 H=h{T,- Td) = 0.2375 X 151 = 35.9 B.t.u. per lb. of air. 
 
 The unit of refrigerating capacity is "ice-melting effect." 
 If water exists at 32° F., then under purely ideal conditions, the 
 abstraction of 144 B.t.u. from 1 lb. of this water will convert it 
 to ice. Conversely, the addition of 144 B.t.u. to 1 lb. of ice 
 
68 , Thermodynamics, Abridged 
 
 will liquefy it. The quantity 144 is the latent heat of fusion 
 of ice. The "ice-melting effect" of a refrigerating machine, in 
 lb., is the number of B.t.u. it absorbs along dc, Fig. 14, divided 
 BY 144. Suppose the ice-melting effect required is 113 lb. per 
 HR. Then the refrigeration to be done per min. is (113 X 144) 
 H- 60 = 271 B.t.u. If this is to be done when, as above, 35.9 
 B.t.u. are removed per lb. of air, the weight of air to be circu- 
 lated per min. is 271 -J- 35.9 = 7.54 lb. 
 
 Efficiency is refrigeration divided by the heat equivalent of 
 power expended. Its value here (Prob. 107) is 2.04. Hence 
 since 1 hp. = 42.42 B.t.u. per min., the steam cylinder hp. 
 necessary is (271 -^ 2.04) -^ 42.42 = 3.16 and (from Prob. 109) 
 the air cylinder will develop (0.735/0.265) X 3.16 = 8.8 hp. 
 while the compressor cylinder will consume 8.8 + 3.16 = 11.96 
 hp. In symbols, let Q, = B.t.u. of refrigeration done per minute. 
 Then steam cylinder hp. = Q -5- 42.42e, ice-melting effect 
 = Q -T- 144 lb. per min., lb. of air circulated per min. = Q 
 ■i-k{Tc- Ta) = Wa. 
 
 Heat must be removed from the air along ha, Fig. 14, in an 
 external "cooler" or "condenser." The heat removed per min. 
 is g = wJi{Tb — T^. (It should be noted that steam cylinder 
 hp. = {q— Q) -^ 42.42, if the curves he and ad are adiabatic.) 
 
 The TONNAGE of a machine is the ice-melting effect per 24 hr., 
 expressed in tons of 2000 lb. If T = tonnage, 
 
 60QX24 _ Q 
 2000 X 144 200 " 
 
 A unit of economical performance is the ice-melting effect 
 
 PER HR. PER IhP. OF STEAM CYLINDER. Call this /q- Then 
 
 60Q Q 
 
 ^» = 1^^4212^ =17-6- (37) 
 
 Prob. 110. From Prob. 108, find q if 7.54 lb. of air are circu- 
 lated per min. (Ans., 407 B.t.u.) 
 
 Prob. 111. What is the Ihp. of the steam cylinder computed 
 from q = 407 and Q = 271? (Ans., 3.2.) 
 
AiK Engines, Aik Compressors, Air Refrigeration 69 
 
 Prob. 112. For Q = 271, what is the tonnage? (Ans., 1.355.) 
 Prob. 113. Find Jo for the conditions of Art. 59. (Ans., 35.9.) 
 
 C357°/: 
 
 Fig. 18. Dense Air Ice Machine. 
 
 60. Dense Air Ice Machine. Fig. 18 refers to the action of an 
 actual ice machine, which consists of compressor and air engine 
 (expander) cylinders side by side, with a steam cylinder. The 
 action of the steam cylinder is not shown by the diagram. The 
 
70 Thermodynamics, Abridged 
 
 compressor cycle is EBCG, the pressure limits noted being 
 approximately those of practice and the compression ratio 4, as 
 used for the Joule cycle, Prob. 107. The engine (expander) 
 cycle is HDAF, the pressure differences between GC and HD 
 and between AF and EB being somewhat exaggerated. The 
 curves are NOT adiabatic. We will take n = 1.35. The com- 
 pressor should be jacketed like any compressor. This reduces 
 the work of the condenser, which must cool the air from C to D. 
 The air engine cycle is given " complete " expansion, in order 
 that the fall of temperature may be a maximum. 
 
 The air leaving the engine goes to the brine tank or other 
 place to be cooled, then returning to the compressor. The 
 necessary value of <a is determined by the service. We will 
 assume i^ = - 40° F., whence Ta = 420. Tb will be higher. 
 If ice is to be made, it cannot be much above 0° F. (460). It 
 will exceed the cold room temperature somewhat, because the 
 air will receive heat from the atmosphere on its way from the 
 cold room to the compressor. 
 
 In a specific machine, the compressor and expander cylinders 
 were 5f by 10 and 4f by 10 inch and connected to the same shaft. 
 The ratio of displacements is then Dc/D^ = (5.75/4.75)^ = 1.46. 
 
 PaVa ^ VbVb Tb _ PbVb _ 65 Be _ 
 Ta Tb ' Ta VaVa 70' D^~ ^■^^' 
 
 both clearances being 10 per cent. Then Tb = 420 X 1.36 
 = 571 or 111° F. This is an unusually high value, but we will 
 adopt it, as in Prob. 108. 
 
 Prob. 114. Find refrigeration per lb. of air in Fig. 18. Note: 
 consider the pressure as constant during heating from A to B. 
 
 (Ans., 35.9 B.t.u.) 
 Prob. 115. Compute Td, To. (Ans., 582°, 817°.) 
 
 The value of Td (122° F.) is easily realized at the condenser: 
 which should in fact be designed to reduce this temperature to 
 80°-100° F. The heat to be carried away at the condenser is 
 0.2375(^(7 - Td) = 55.9 B.t.u. per lb. of air. 
 
Air Engines, Air Compressors, Air Refrigeration 71 
 
 The value of Be is (Tr/4) X (5.75)^ X (10/1728) = 0.15 cu. ft. 
 
 per stroke. The volume of air drawn into the compressor (and 
 
 therefore circulated through the system) per stroke is V b — V^ 
 
 = 0.82Dc = 0.123 cu. ft. (see Prob. 116). The weight of air 
 
 per stroke is (P^ X 0.123) -^ (53.36^^) = (65 X 144 X 0.123) 
 
 -r- (53.36 X 571) = 0.0377 lb. Assume N = 100, i.e., 200 single 
 
 strokes per minute, the machine to be double-acting. Then 
 
 the weight of air circulated per min. is Wa = 200 X 0.0377 
 
 = 7.54 lb. (Compare Prob. 110.) We then have 
 
 Q = 35.9 X 7.54 = 271 (see Art. 59 and Prob. 114), 
 
 q = 55.9 X 7.54 = 421. 
 
 In this actual engine, the steam cylinder hp. is not determined by 
 
 q — Q, since heat transfers occur along all the curved paths of 
 
 the cycles. The ihp. of each cylinder (compressor and expander) 
 
 is first worked out in detail. Applying Equation (29) to the 
 
 compressor, 
 
 1 qf; v fi'i 
 Pmc = Q3^ {1A{4P-''' - 1) + 0.10(4'>-'« - 4)} = 88.6 
 
 and to the engine (expander), 
 
 P^M = 33 {l.l(3.5''-^59 _ 1) + o.lO(3.5''-'« - 3.5)} = 88. 
 
 The corresponding horse powers are 
 
 ^, 2 X 88.6 X 10 X 0. 7854 X 5.75 X 5.75 X 100 ,^ ^ 
 
 Ihp.o = 12X33 000 = ^^•^- 
 
 /4 75 \^ CO 
 Ihp.. = Ihp.ox(5;75)x 33^6 =7.8. 
 
 (Powers at equal piston speeds are proportional to the products 
 of area and pm-) Then the net power to be derived from the 
 steam cylinder is 11.6 — 7.8 = 3.8. To this we may add 50 
 per cent, to cover friction losses, so that Ihp.^ = 5.7. These 
 results may be compared with those for the ideal cycle computed 
 in Art. 59. It should be noted that compression in the air 
 engine is here " complete," i.e., the curve FH, Fig. 18, con- 
 tinues until the air-supply pressure is attained. 
 6 
 
72 Thermodynamics, Abridged 
 
 The eiBciency is in this case 
 
 Q 271 
 
 = 1.13. 
 
 42.42 X 5.7 241 
 For the ideal cycle, it was 2.04. The tonnage is that of the 
 ideal cycle, 1.355. The ice-melting effect per Ihp.-hr. is 7o 
 = 17.6 X 1.13 = 19.8 lb. The value realized for this depends 
 mainly on the temperature range. In practice, the useful value 
 of Q is much reduced by radiation, and values of e around 0.5 
 or 0.6 or of 7o around 10 are common. 
 
 If the machine were to make ice from water at 100°, it would 
 have to cool the water well below the freezing point, first, — say 
 to 0°. This would require the extraction of approximately 100 
 B.t.u. per lb. Adding the latent heat of fusion, the refrigeration 
 necessary per lb. of ice made is 244 B.t.u. The machine con- 
 sidered could then make a maximum of 271/244 = 1.11 lb. of 
 ice per min. During the preliminary cooling, it could lower the 
 temperature of 271 lb. of water 1° per min. 
 
 Prob. 116. In Fig. 18, find V^ and Vb - V^. 
 
 (Ans., 0.28Dc and 0.82Dc.) 
 Prob. 117. In Fig. 18, find Vc- (Ans., 0.393Z)c.) 
 
 Prob. 118. In Art. 60, check the values of pmc and p^E- 
 Calculate Ihp.^ by the formula 2-p^LAN -^ 33 000. 
 
 Regenerative Hot-Air Engine Cycles 
 61. Remarks. As has been stated, the air engine of a com- 
 pressed air plant is not a thermal engine but only a part of a 
 work-conversion system. Another type of engine, which uses 
 high-temperature air, discharging it at a lower temperature, is 
 now to be considered. This is really a thermodynamic machine. 
 Such engines are called hot-air engines. They are not 
 internal combustion engines, for fuel does not enter the cylinder. 
 They differ from most engines (not from dense air ice machines) 
 in that a fixed mass of fiuid is used over and over again. The 
 use of the regenerator, common with hot air engines, makes 
 the ideal efficiency of their cycle equal to that of the Carnot 
 
Air Engines, Air Compression, Air Refrigeration 73 
 
 CYCLE. In this respect they are unique. The regenerator is a 
 closed vessel filled with wire gauze, glass or any substance of 
 large heat absorbing capacity. 
 
 62. Stirling Cycle. Fig. 19 shows the principle of operation 
 of the Stirling engine. The piston being at the extreme left. 
 
 Fig. 19. Hot Air Engine with Regenerator — Stirling Cycle. 
 
 position Pi, the state on the lower diagram is d. Valves V2 
 and Fe are closed, valves Vi and Vb are open. Hot air from the 
 regenerator flows into the cylinder, the temperature rising to Ta 
 and the pressure increasing as long as the piston does not move. 
 Valve Fb is now closed and Fe is opened. Heat from the furnace 
 
74 Thermodynamics, Abridged 
 
 causes the air in the cylinder to expand (path ab) and the piston 
 moves to the extreme right (position Pi). The supply of heat 
 is so controlled with relation to the piston speed that the path ab 
 is ISOTHERMAL. The valve Fi is now closed, Vz and Vi are opened 
 and Fs closed. Air iBows from the cylinder to the regenerator. 
 The pressure in the cylinder falls (path be), the piston remaining 
 stationary. As the air flows into and through the regenerator, 
 its temperature falls: but the inlet temperature at the regenerator 
 finally approximates Tb, while its outlet temperature approxi- 
 mates Tc. Valve Vt is now closed and F3 opened. Then the 
 piston makes a full stroke to the left, forcing air into the con- 
 denser. The water supply in the condenser coils is so controlled 
 as to withdraw heat without lowering the temperature. The 
 path cd is consequently isothermal. 
 
 Under purely ideal conditions, all heat discharged to the 
 regenerator along be is regained along da, so that this quantity 
 of heat may be disregarded as a debit against the engine. The 
 only heat chargeable is Hab, and the efficiency is 
 
 ji y 
 
 e = {Hab — Hcd) -^ Hab = — ^ 
 
 (Art. 34), which is the efficiency of the Carnot cycle. 
 
 The operation of the Ericsson hot-air engine was similar, 
 and its efficiency is the same, but the regenerative processes 
 were at constant pressure instead of at constant volume. 
 
 For Tab = 1000, Ted = 500, e = 0.50. Without the regen- 
 erator, the efficiency would have been 
 
 Hd.+ Hab 778;(j,^ _ y^) ^ ^y^ l^g F. 
 
 y a 
 (Ta-Td)l0geJ 
 
 7781 ' 
 
 {Ta- Td)+ TalogeJ 
 
 R 
 
 in which J = Vt/Va and (since ab and de are isodiabatics) also 
 
Air Engines, Air Compression, Air Refrigeration 75 
 
 = VclVd. For the temperatures above assumed, this becomes 
 
 _ 1150 log J log J 
 
 ^ ~ 1232 + 2300 log ./ ~ 1.07 + 2 log J * 
 
 Then e increases as J increases. 
 
 The mean effective pressure, with or without regenerator, is 
 
 Wg, - TFrfo _ R loge J{Ta - Tg) 
 
 ^^ 144(F„ - Vi) 144Fd(J - 1) • 
 
 For the assumed conditions, this is 
 
 _ 53.36 X 2.3 X 500 log J _ 426 log J 
 ^™ ~ 144Fd(J - 1) ~ Vi{J - 1) • 
 
 This decreases (for a fixed value of Vd) as J increases. Hence 
 low mean effective pressures accompany high efficiencies. 
 
 Hot-air engines with regenerators have given up to 25 per 
 cent, efficiency even in small sizes, and have used as little as 1.7 
 lb. of coal per Ihp. per hr., but the mean effective pressures are 
 low and the heat-transmitting surface at the furnace burns out 
 rapidly. 
 
 Prob. 119. In Art. 62, find efficiencies without regenerator 
 for J = 4, 5, 10, 20, oo . (Ans., 0.27, 0.28, 0.33, 0.36, 0.50.) 
 
 Prob. 120. Find VaVm in the above. 
 
 (Ans.j 86, 75, 47, 28, indeterminate.) 
 
CHAPTER IV 
 STEAM AND OTHER VAPORS 
 
 Vapor Properties and Tables 
 
 63. Heating of Liquid. Consider the fundamental equation, 
 H = T+ I+W, Art. 2, Equation (3). For liquids and their 
 vapors, / (the disgregation work) is not equal to zero. The 
 equation is usually written for such fluids, H = E -\- W-. where 
 E = T -\- I and comprises all effects of heat other than external 
 work, and may be called the internal work, or gain of internal 
 
 ENERGY. 
 
 When 1 lb. of liquid at 32° F. is heated, a rise of temperature 
 and (in general) an increase of volume occur. W is therefore 
 positive. Its value is very small. There are few cases, in the 
 instance of water, for example, where it exceeds 1 B.t.u. Hence 
 in the heating of water E is nearly equal to the whole amount 
 of heat supplied (see Art. 77). 
 
 The heating of the liquid may be continued until it begins to 
 boil. At atmospheric pressure, each liquid has its own definite 
 boiling point. At a lower pressure, it will boil at a lower tempera- 
 ture. At a higher pressure, it will boil at a higher temperature. 
 Fig. 20 shows the temperatures at which boiling occurs, for 
 various liquids, at various pressures. Water appears here 
 (with one exception) as having the highest boiling temperatures, 
 but there are many liquids less volatile than water. All curves 
 show that as the temperature increases, the pressure increases 
 more and more rapidly. Therefore if high temperatures are 
 desired in a steam engine (so as to attain high Carnot efficiency) 
 high pressures almost necessarily result. There is an exception 
 to this, however: see Art. 83. 
 
 The HEAT OF THE LIQUID is THE QUANTITY OF HEAT NECES- 
 SARY TO RAISE THE TEMPERATURE OF 1 LB. OF LIQUID FROM 
 
 76 
 
Steam and Other Vapoes 
 
 77 
 
 32° F. TO THE BOILING TEMPERATURE. Its Symbol is h. The 
 boiling temperature depends upon the pressure at which boiling 
 occurs. There is a definite temperature of boiling for every 
 
 3240 80 120 ISO 200 240 
 
 Pig. 20. Boiling Temperatures for Various Pressures. 
 
 pressure (Fig. 20) of a given liquid.* The temperature 32° F. 
 is arbitrarily taken as a starting point. The value of h for all 
 
 ' As tabulated on pages 82, 83, values of h consider the pressure at the 
 start to be that pressure at which the boiUng point is 32°, viz., 0.0886 lb. per 
 sq. in. The pressure then increases as the water is heated, until the boiling 
 s reached. 
 
78 Thermodynamics, Abridged 
 
 liquids is at this temperature, consequently, zero. For every 
 different pressure at which boiling occurs, there will be a different 
 value of h. 
 
 The boiling point of ammonia at atmospheric pressure is 
 — 27° F. This is below 32° F., hence the value of h at this 
 temperature is negative. li t = temperature at which boiling 
 occurs, s = mean specific heat of liquid from 32° to t°, h = 
 s{t — 32). The value of s is generally somewhat variable. 
 
 Prob. 121. Compute h for water when t = 212°. 
 
 (Ans., 180 B.t.u.) 
 
 Prob. 122. When water boils under 100 lb. absolute pressure, 
 its temperature is 327.8° and h = 298.3. Find s from 32° to 
 327.8°. (Ans., 1.008.) 
 
 Prob. 123. The mean specific heat of liquid ammonia from 
 5° to 32° is 1.052. Find h at 5°. (Ans., - 28.4 B.t.u.) 
 
 64. Vaporization. If more heat is added after the liquid 
 reaches its boiling temperature, vapor is formed. No vapor 
 can be formed until the liquid has reached boiling temperature. 
 During the formation of the vapor, it and the liquid are at the 
 same temperature. This temperature is that at which the liquid 
 boils at the existing pressure. The temperature of the vapor 
 cannot be increased above this as long as any liquid is present. 
 The removal of heat from a vapor (if kept at constant pressure) 
 does not lower its temperature, but liquefies it, or part of it. 
 
 A pound of liquid completely evaporated makes a pound of 
 vapor. The vapor has a definite specific volume, v. Its 
 temperature we have shown to be t. The whole operation of 
 converting the pound of already boiling liquid into vapor con- 
 sumes or requires the heat L, defined as the latent heat of 
 vaporization, all of which is supplied at constant tempera- 
 ture, t, of the fluid. 
 
 The kind of vapor thus formed is called saturated vapor 
 or DRY vapor. At a given pressure, saturated vapor has the 
 lowest possible temperature and specific volume of any vapor. 
 Wet vapor may be regarded as incompletely evaporated liquid. 
 
Steam and Other Vapors 79 
 
 That is, for example, from 1 lb. of liquid at the boiling point, 
 5 lb. of saturated vapor may have been formed, so that the 
 resulting fluid is a mixture of equal parts of vapor and boiling 
 liquid. The temperature and specific volume of the vapor in 
 this mixture will be t and v. Superheated vapor is formed by 
 adding still more heat to saturated vapor. This can only be 
 done aftet all of the liquid is evaporated. The temperature, t', 
 of superheated vapor is always higher than t: its volume, v', is 
 greater than v. It behaves like an imperfect gas. 
 
 Prob. 124. A saturated vapor exists at the pressure p = 100 
 and the temperature t = 327.8. The pressure is increased 
 while the temperature remains constant. What happens? 
 (Ans., the vapor must liquefy.) If the pressure is lowered 
 while the temperature remains constant, what happens? (Ans., 
 the vapor becomes superheated.) 
 
 65. Variation of L, In general, L decreases at high pressures. 
 This does not mean that it is easier to generate vapor at high 
 pressure, because as the pressure increases, the boiling tempera- 
 ture increases and hence h increases. The increase in h is more 
 than the decrease in L, in the case of steam. The total heat 
 necessary to generate dry saturated vapor from 1 lb. of liquid 
 at 32° is called the total heat, H, and is equal to h-\- L. This 
 quantity increases, for steam, as the pressure increases. It 
 always requires more heat to transform water (at a given initial 
 temperature) to dry steam, when the pressure is high. The 
 amount of heat required to vaporize liquid already at the 
 boiling point decreases with increase of pressure. This quan- 
 tity we call L. 
 
 At moderate pressures and for most vapors, L is greater than h. 
 Thus for water boiling at 212° F., L = 970.4 while h = 180. 
 As the pressure increases, h increases and L decreases. Finally a 
 pressure is reached at which i = 0. The temperature at which 
 boiling then occurs is called the critical temperature. This, 
 for steam, is 689° F. At the critical temperature, the liquid 
 (brought to the boiling point) passes to the condition of saturated 
 
80 Thermodynamics, Abridged 
 
 vapor without any further addition of heat. The liquid and 
 vapor states thus merge into each other. If a liquid is heated 
 above its critical temperature it becomes a gas. It cannot 
 again be liquefied by mere increase of pressure until its tempera- 
 ture has first been reduced below the critical point. This is 
 of importance in connection with carbon dioxide, which has a 
 critical temperature around 88° F. This substance is used as a 
 fluid in refrigerating plants on some vessels. The refrigerating 
 fluid must be condensed in one part of the operation. If CO2 
 is used, its temperature must then be brought below 88°. 
 Conditions may arise in tropical waters where this would be- 
 impossible. The CO2 refrigerating machine cannot then be 
 used. 
 
 In the equation, heat = E -\- W (Art. 63), as applied to 
 vaporization alone, how much of the heat of vaporization, L, 
 is an E effect, and how much a W effect? We are considering 
 vaporization alone: the liquid is already at the boiling point: 
 L = E -\- W. Of the heat added during vaporization, W units 
 are used in performing external work, and E units in performing 
 internal work, or increasing the internal energy. Take the case 
 of steam formed from water at 400°, and therefore (by the 
 steam table) at 247.1 lb. pressure. The specific volume of the 
 water at 400° is 0.0187. During vaporization, this is increased 
 to ?) = 1.872. The external work part of L is therefore W^^ 
 = pressure X increase of volume = 
 
 144 X 247.1(1.872 - 0.0187) 
 —^ = 84.2 B.t.u. 
 
 The E effect must be L — 84.2. The value of L given by the 
 steam table is 827.2. Hence the E effect is 743: much larger 
 than the W effect. The symbol r is used for the E effect in- 
 cluded in L. Since h is nearly all E effect (Art. 63), H = h+ L 
 consists approximately of an E effect equal* to h + r and a W 
 effect nearly equal to 144p(?) - v„) -7- 778, where Vy, = specific 
 volume of liquid at the boiling point. 
 * The EXACT value is given in Art. 77. 
 
Steam and Other Vapors 81-82 
 
 Prob. 125. Using the steam table, pages 228, 229, take values 
 of p, V, L, as given at 327.8° F., and check the value given for r. 
 Note v„ = 0.01774 from Fig. 22a. 
 
 66. The Steam Table. The steam table, given on pages 228- 
 229, is a typical vapor - property table. Similar tables exist 
 for other vapors like ammonia, SO2, CO2: but the properties of 
 many vapors are only imperfectly established. The basis or 
 "argument" of the table is either the pressure or the tempera- 
 ture. For a given value of the pressure, p, t is the Fahrenheit 
 temperature at which boiling occurs. The specific volume of 
 the vapor is v, which is less for high than for low pressures. The 
 heat quantities are for 1 lb. of fluid, originally liquid at 32° and 
 at 0.0886 lb. pressure. The meaning of h, L, H and r has been 
 explained. All tables have an elaborate experimental basis, 
 and for saturated steam, the differences between the various 
 existing tables are of small importance. The value of H here 
 used is based on the Davis formula, 
 
 H = 1150.3 + 0.3745(< - 212) - 0.00055(< - 212)^- (38) 
 
 Prob. 126. From the Davis formula, find H when t = 212°. 
 
 (Ans., 1150.3.) 
 
 Prob, 127. Compute H when t = 312°. (Ans., 1182.25.) 
 
 Prob. 128. From the table, what is the mean specific heat of 
 water between 32° and 312°? (Ans., 1.007.) 
 
 Prob. 129. What is the approximate value of E in the forma- 
 tion of 1 lb. of steam at 80 lb. pressure from water at 32° F.? 
 (Ans., 1101.8.) What is the value of W during vaporization? 
 (Ans., 80.5 B.t.u.) 
 
 Entropy 
 
 67. A New Diagram. Art. 33 gave a graphical method of 
 representing heat absorbed, of which we have not made much 
 use. The subtended area under a path on the PV diagram, 
 representing external work done, has on the other hand been 
 used repeatedly. The magnitude of this area for specific pro- 
 cesses is easily computed, because three of its boundaries are 
 
83-84 
 
 Thebmodynamics, Abridged 
 
 straight lines. We are now to develop a similarly simple area 
 for the representation of quantities of heat absorbed or emitted. 
 The new diagram may be conceived as having been produced 
 by a mechanism " interlocked " with the indicator, so that while 
 
 T 
 
 ^Absolute Zero 0.002 0.004 0.006 0.008 " 
 
 Fig. 21. Entropy Diagrams. 
 
 the pencil of the latter is describing lines the areas under which 
 represent external work, the imaginary pencil of the new diagram 
 will simultaneously sketch corresponding lines, areas under which 
 will represent heat. 
 
Steam and Other Vapors 85 
 
 The ordinate of the diagram is to be absolute temperature. 
 In Fig. 21, then, T is the ordinate and ahcd represents the heat 
 absorbed or emitted along ah, or briefly ahcd = Hob- If ah is 
 described from left to right, Hab is + : if from right to left, Hab 
 is — . Bearing in mind that subtended areas represent 
 HEAT and ordinates are absolute temperatures, answer the 
 following : 
 
 Prob. 130. What is the shape of an isothermal on this 
 diagram? 
 
 Prob. 131. What is the only possible shape of an adiabatic? 
 Why? 
 
 Prob. 132. What is the shape of a Carnot cycle? Sketch 
 and letter the areas representing Hjk and Him- Which is plus 
 and which minus if the cycle is clockwise? What area represents 
 the net external work done in the Carnot cycle? (SPF = 1,H 
 = heat received — heat rejected.) Efficiency is HiH divided by 
 all of the + heat. In this case, it is the quotient of what two 
 areas? Of what two lengths (in rectangles of equal width, areas 
 are proportional to lengths). State the efficiency in terms of 
 Tj^ and Tmi and prove the statement. 
 
 68. The Abscissa of the Diagram. Consider the heating of 1 
 lb. of water from 32° F. or 492° absolute. Heat it 1°. The 
 heat absorbed is 1.01 B.t.u. (the specific heat of water is not 
 exactly 1.0). Let o. Fig. 21, represent the condition of the 
 water before heating. A path representing the heating process 
 must be subtended by an area of 1.01 B.t.u. If we assume the 
 path to be a straight line, the subtended area is a trapezoid of 
 area 1.01 and parallel vertical sides of lengths 492 and 493 
 (absolute temperatures). Its horizontal width is therefore 
 AREA -^ mean height = 1.01 -^ 492.5 = 0.00205. The process 
 is represented by the line o-p: the coordinates of o being 492, 
 (the latter fixed as an arbitrary starting point) and those of p 
 being 493, 0.00205. 
 
 Heat this water another degree. The heat supplied is this 
 time 1.0 B.t.u., the mean temperature 493.5, the horizontal 
 
86 Thermodynamics, Abridged 
 
 width of the trapezoid 0.00203, and the coordinates of q are 494 
 and 0.00205 + 0.00203 = 0.00408. 
 
 Prob. 133. In heating from 34° to 35° F., and also from 35° 
 to 36° F., 1.01 B.t.u. are supplied. Locate points r and s, Fig. 
 21. (Ans., abscissas are 0.00613 and 0.00817.) 
 
 The whole operation from 32° to 36° is represented by os, 
 which is nearly a straight line. This method is approximate, 
 involving the assumption that each of the segments op, pq, qr, 
 rs, IS a straight line. The error due to this is exceedingly small. 
 
 Consider the process tu. The area under this curve repre- 
 sents Htu. Consider a very small portion of the curve, along 
 which the temperature may be assumed to be constant. Call 
 this temperature T. The narrow area under the small portion 
 of the curve is dH and the width of this area is dHjT. The 
 horizontal component or abscissa of the whole path tu is then 
 
 CdH , ,. ■ 
 
 ntu = flu — nt = \ ~m , between proper limits. 
 
 The dimension n is called the entropy. The only definition 
 OF ENTROPY IS J'idH/T). It is the abscissa of our diagram, 
 which is accordingly marked n. There is no necessary origin 
 for entropy. The line os puts it at 32° F. or 492° absolute, but 
 this need not always be the case. Only changes of entropy are 
 important. The symbol ritu or the quantity Uu — rii both mean 
 
 the CHANGE OF ENTROPY Or the HORIZONTAL COMPONENT OF THE 
 
 PATH from t to u. We must now establish specific values for 
 J'{dH/T) for various processes. 
 
 69. Change of Entropy at Constant or Definite Specific Heat. 
 
 For the path <m, 7i„ — n.( = J'{dH/T). Suppose this path to be 
 one along which the specific heat is constant = s. Then H 
 = siTi- Ti) for 1 lb. weight heated from Ti to T2 or dH = sdT. 
 Then 
 
 X"' sdT T 
 
 -Y = ■S(l0ge Tu - log, Tt) = S loge Y ■ 
 
 (39) 
 
 Note that Tu and Tt are absolute temperatures. Suppose the 
 specific heat to be variable: for example, si= a-\-hT. Then 
 
Steam and Other Vapors 87 
 
 n.-nt= j'"^+£bdT=alogeY+HT.- T,). 
 
 The case of water comes in neither of these classes. Its specific 
 heat is too irregularly variable to be expressed as a linear function 
 of the temperature. An approximation can be made, for water, 
 by taking s as the mean value over the temperatm-e range 
 specified, in Equation (39). 
 
 Thus for water heated from 32° to 36° F. the heat expenditure 
 is 4.03 B.t.u. Take s, then, at 4.03 -^ 4 = 1.0075. Then by 
 Equation (39), 
 
 496 
 W36 - ^32 = 1.0075 X 2.3026* log ^ = 0.0082. 
 
 This result is comparable with that reached in Prob. 133 and 
 used in plotting the point s, Fig. 21. It should be noted that 
 with a positive specific heat the path must slope upward to the 
 right and downward to the left: that low values of the specific 
 heat lead to steep paths: and that any path of constant specific 
 heat becomes steeper as it ascends. 
 
 Prob. 134. Given Z'„ = 23",. Find n,„ when s = 0.1689, 
 0.2375, 0.5, 1.0. (Ans., 0.117, 0.164, 0.347, 0.693.) 
 
 Prob. 135. "VSTiat is the gain of entropy in heating 1 lb. of 
 water from 32° F. to 312° F.? Note: take s = h -^ {t - 32). 
 
 (Ans., 0.4535.) 
 
 70. Entropy at Various Specific Heats. For an adiabatic, 
 » = and the path is vertical. The table in Art. 29 mentioned 
 the adiabatic as a path of constant entropy. For an isothermal, 
 5 = 00 and the path is horizontal. For all constant values of s 
 between and + <», the curve is a logarithmic curve. Sev- 
 eral paths are plotted in the small diagram in the upper right- 
 hand corner of Fig. 21. Equation (39) may be used to determine 
 either end-points or intermediate points of these paths. 
 
 Prob. 136. If s = 1, Ts = 500, r„ = 750, find utu- 
 
 (Ans., 0.406.) 
 * The use of the modulus permits employing a table of common logarithms 
 instead of Napierian logarithms. 
 
88 Thermodynamics, Abridged 
 
 71. Entropies of Vapor. The entropy of the liquid, tiw, is 
 
 the CHANGE OP entropy DUE TO THE HEATING OF 1 LB. OF LIQUID 
 
 FROM 32° F. TO THE BOILING POINT. It was Computed approx- 
 imately for water with t = 312° F. in Prob. 135. Similar com- 
 putations taken for successive short ranges of temperature give 
 all of the values of ?i„ which appear in the steam table, third 
 column from the last. 
 
 The ENTROPY OF VAPORIZATION, He, is the CHANGE OF ENTROPY 
 
 of 1 lb. of fluid which occurs during vaporization. Vaporiza- 
 tion takes place at constant temperature. A constant tempera- 
 ture process, in Fig. 21, is represented by a horizontal line. The 
 subtended area is a rectangle. The width of the rectangle is 
 the change of entropy. It is equal to the area divided by the 
 height or to the heat absorbed divided by the absolute tempera- 
 ture. Hence rie — L -i- {t -{- 460) . Values are given in the 
 column of the steam table next to the last. 
 
 The TOTAL ENTROPY, OR ENTROPY OF 1 LB. OF VAPOR, is Ks 
 
 = n„ + He. Values appear in the last column of the table for 
 steam. 
 
 Prob. 137. Compute ne and n^ for steam at 312° F. (See 
 Prob. 135. Ans., tie = 1.1665, n, = 1.62.) 
 
 72. Entropy Diagram for Steam. In Fig. 22, the ordinate 
 scale is converted into Fahrenheit temperature. As in Fig. 21, 
 the lower part of the diagram is omitted. All heat areas are 
 to be understood as carried down to — 460° F. Water is to be 
 heated from 32°. The state is a, taken on the vertical axis. 
 The path ac represents the heating of water from 32° F. to 
 150° F. The abscissa of this path (length ot) is (k„)i5o = 0.2149. 
 The water is then further heated to 327.8° F. along cb. The 
 abscissa of the whole path ah is os = (n,„)327.8 = 0.4743. The 
 gain of entropy in heating water from 150° to 327.8° is os — o 
 = is = 0.4743 - 0.2149 = 0.2594. The whole curve ah may 
 thus be plotted from the steam table as accurately and as com- 
 pletely as may be desired. 
 
Steam and Other Vapors 
 
 89 
 
 i 
 
 
 
 
 
 ^ 
 
 .%70W/0;Z^/77 
 
 ± 
 
 "JL 
 
 =_ 
 
 ' V 
 
 =1 
 
 rcf 
 
 ^.^ 
 
 
 
 300 
 
 
 
 
 
 7 
 
 
 
 
 
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 1 — 
 
 (^ 
 
 — 1 
 
 ^ — 
 
 -1— 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 4— 
 
 TO 
 
 Hi 
 
 
 
 i 
 
 
 
 
 
 
 
 
 
 
 A 
 
 
 
 
 
 -i 
 
 i 
 
 ' 
 
 \- 
 
 1 
 
 \ 
 
 1 
 
 
 
 
 1 
 
 
 
 / 
 
 t — 1- 
 
 
 
 
 
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 \ 
 
 t 
 
 
 
 Z5U 
 
 
 
 3>^ 
 
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 SI 
 
 \0^ 
 
 
 
 A- 
 
 200 
 
 
 y^ 
 
 
 F 
 
 i.=^ 
 
 
 
 hof; 
 
 bi 
 
 
 
 -= 
 
 s 
 
 ^% 
 
 
 h! — 
 
 
 
 
 
 
 A?-ti 
 
 /sIl 
 
 ^ 
 
 EZI^ 
 
 r 
 
 
 -1— 
 
 
 ^^x^ 
 
 ys^^w/-^ 
 
 A i' 
 
 -AH 
 
 ISO 
 
 
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 1 
 
 ~V"fl 
 
 ^ 
 
 ^ 
 
 
 ^ 
 
 
 
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 CIV, 
 
 
 A"^ 
 
 ^r 
 
 
 
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 vMr" fi— 
 
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 ^H 
 
 
 
 ^^i^ 
 
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 lUU 
 
 
 f- 
 
 k 
 
 
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 fVo/uf?^ 
 
 
 
 
 
 
 ^ 
 
 
 
 F*"- 
 
 
 / 
 
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 H 
 
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 ■ 
 
 
 
 
 
 
 
 
 -h 
 
 / 
 
 1^ 
 
 
 
 
 
 
 
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 bU 
 
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 P 
 
 
 
 
 
 ■■m 
 
 THisAredfei}fesentslBiu 
 
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 C 
 
 .8 
 
 1 
 
 
 
 1 
 
 1 
 
 1.4 
 
 1 1 
 
 '9 
 
 1 
 
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 / 
 
 /? 
 
 Fig. 22. Temperature-Entropy Diagram for Steam. 
 
 Prob. 138. What is the abscissa of that part of the curve oh 
 which terminates at 212° F.? (Ans., 0.3118.) How much 
 entropy is gained when 1 lb. of water is heated from 212° to 
 312°? (Ans., 0.1417.) 
 
 Suppose the pressure to be such that the water begins to boil 
 at the point h, Fig. 22. Here the temperature is 327.8° F- 
 Vaporization takes place at constant temperature, and is there- 
 fore represented by the horizontal line from h. At some poirit d, 
 on this line, vaporization will be completed. The length of the 
 line hd, represents the change of entropy during vaporization, or 
 briefly the entropy of vaporization. The symbol for this 
 quantity is Tie. Its value for the assumed condition is 
 
 W327.8 =hd= (L/TUlb = 1.1277. 
 
 If the pressure had been different, so that vaporization had 
 taken place at 150° F., the path representing vaporization would 
 have been ce. The dotted line de joining the end-points of 
 vaporization paths like ce and bd is a locus of completely vapor- 
 
 7 
 
90 Thermodynamics, Abridged 
 
 ized states. It represents every possible kind of dry steam. Its 
 abscissa at any point is n.,. The curves ab and ed meet at the 
 
 CRITICAL TEMPERATURE. 
 
 All heat quantities are measured above 32° F. as a starting 
 point. The area under ae {acgf, carried down to the — 460° F. 
 level) therefore represents the heat of the liquid at c, abbrevi- 
 ated h. The area under ce (cejg) similarly carried down, repre- 
 sents the heat consumed in converting liquid at c into dry vapor 
 at e, abbreviated as ice- The sum of these two areas represents 
 He, the total heat of dry steam at e, above 32° F. 
 
 Prob. 139. Locate d, where t = 327.8. 
 
 (Ans., T = 787.8, n = {n,)s27.s = 1-602.) 
 
 Any point on ab represents boiling water. Any point on de 
 represents completely evaporated (" dry ") steam. Any point 
 BETWEEN these two curves represents a mixture of steam and 
 boiling water, i.e., wet steam. If the point is close to de, the 
 mixture is nearly dry steam. If the point is close to ab, the 
 mixture is mostly water. If the state moves from a point near 
 ab toward de, water is being evaporated. If the movement is in 
 the reverse direction, steam is being condensed. 
 
 An adiabatic process is represented by a vertical line. If hot 
 water (as at b) expands to a lower pressure, adiabatically (path 
 bn), some of it will be vaporized, because the point n is farther 
 from ab than is the point b. This actually occurs when the blow- 
 off cock of a boiler is opened. If dry steam is expanded adiabat- 
 ically (path dp), some of it will condense. If very wet steam is 
 compressed adiabatically (path nb) it will be partially or wholly 
 liquefied. If nearly dry steam is sufficiently compressed adia- 
 batically (path pd) it will be made wholly dry. A point m, 
 half way between ab and de, represents steam " 50 per cent, 
 dry," i.e., it represents a mixture in which half the water has 
 been evaporated while the other half is still liquid. 
 
 Prob. 140. Locate the point u, Fig. 22, representing steam 
 50 per cent, dry at 150° F. (Ans., entropy at the point c = 
 0.2149, entropy at the point e = L8674, /i„ = 0.2149 + |(1.8674 
 - 0.2149) = 1.0412.) 
 
Steam and Other Vapors 91 
 
 The " constant dryness " curves mu, vw may be plotted by 
 computations made as in Prob. 140, or may be obtained graph- 
 ically by properly dividing horizontal distances between ab 
 and de. Thus bm = md, mv = vd, cu = ue, uw = we. 
 
 Suppose heat to be added when the state is d. There being 
 no liquid present, the temperature will rise; i.e., the path from d 
 will be UPWARD. Since heat is supplied, there must be a sub- 
 tended area. Hence the path will not be vertical. Since the 
 sign of the heat supplied is +, the path must be from left to 
 right: i.e., it will slope, so to speak, toward the northeast. If 
 the specific heat during this operation were constant, the path 
 would be a logarithmic curve, indicated by Equation (39), like dx. 
 The operation dx is one of superheating. Superheating implies 
 the addition of heat (no liquid being present), increase of temper- 
 ature and increase of entropy. All points lying to the right of de 
 represent superheated vapor. If superheated steam expands 
 adiabatically (path xy) it becomes less superheated, then dry and 
 finally wet. If dry steam (or nearly dry steam) is compressed 
 adiabatically (path er) it is superheated. The heat absorbed 
 during the superheating operation dx is represented by the area 
 under dx, carried down to the — 460° F. level. 
 
 73. Factor of Evaporation. In usual practice, we do not start 
 the formation of steam with water at 32°, but at some higher 
 temperature, to. The heat expenditure for warming from this 
 temperature to h is hi — ho B.t.u. The total expenditure of 
 heat in forming dry steam at ti° is then hi — ho -\- Li = Hi — ho 
 B.t.u. This varies with the pressure, y>i, at which evaporation 
 occurs, and with the temperature U at which water is supplied 
 in the first place. This temperature is called the feed-water 
 temperature. The generation of 1 lb. of dry steam from feed 
 water therefore represents a variable quantity of heat. 
 
 For the purpose of comparison, the standard condition to 
 which all others are referred is thus defined : the feed temperature 
 is 212° F. Hence ho = 180. Also, the steam pressure is normal 
 atmospheric pressure. Hence hi is also 180. Then the total 
 
92 Thermodynamics, Abridged 
 
 heat expenditure per lb. of steam formed is hi — ho+ Li = L212 
 = 970.4. This condition is described as from (a feed water 
 temperature of) and at (a steam pressure corresponding with) 
 212° F. In Fig. 22, the area under zA represents X212, the heat 
 expenditure necessary to generate 1 lb. of steam " from and at 
 212° F." The area under cbd represents the heat expenditure 
 when the feed temperature is tc and the steam pressure is the 
 pressure which corresponds with the temperature fed- The 
 quotient of the latter area by the former is called the factor of 
 EVAPORATION, which has the value 
 
 _ hi, — he -\- Lbd _ Hd — he 
 
 970.4 ~ 970.4 ' ^^^ 
 
 The factor of evaporation is the ratio of the heat actually em- 
 ployed to make steam, divided by the heat necessary from and 
 at 212° F. A high value of F means costly steam. 
 
 Some Equivalents 
 Mechanical Units Heat Units 
 
 1 ft. lb. = 1/778 B.t.u. 
 
 778 ft. lb. = 1 B.t.u. 
 
 754 971 ft. lb. = 1 lb. of water evaporated from and 
 
 at 212° F. = 970.4 B.t.u. 
 33 000 ft. lb. per min. = 1 hp. = 42.42 B.t.u. per min. 
 1 980 000 ft. lb. per hr. = 1 hp. = 2545 B.t.u. per hr. 
 
 Prob. 141. How many B.t.u. are required to heat 10 lb. of 
 water from 212° F. to 312° F.? (Ans., 1020 B.t.u.) 
 
 Prob. 142. How many B.t.u. are required to generate 1 lb. 
 of dry steam at 80 lb. pressure from feed water at 90° F. ? (Ans., 
 1124.3.) What is the factor of evaporation? (Ans., 1.16.) 
 
 Prob. 143. Find F when U = 80, pi = 200. (Ans., 1.18.) 
 Prob. 144. A boiler evaporates 10 lb. of water per lb. of 
 coal under the conditions of Prob. 142. Another boiler, using 
 the same coal, evaporates 9.9 lb. of water per lb. of coal under 
 the conditions of Prob. 143. Which is the better boiler? How 
 much better? (Ans., the second boiler is 1.27 per cent, better.) 
 
Steam and Other Vapors 93 
 
 Wet Vapor: Vapor Processes 
 74. Properties of Wet Vapor. If 1 lb. of a liquid is brought 
 to the boiling point t and the pressure p and then the portion x 
 lb. of this liquid is evaporated, the properties are: 
 
 h, B.t.u., heat of liquid; for all of the liquid was brought to the 
 
 boiling point: 
 Uw, entropy of liquid; for the substance moved along ah, Fig. 
 
 22, in the usual way: 
 i, the common Fahrenheit temperature of the liquid and the 
 
 vapor: 
 xL, the heat of vaporization; for only x lb. were vaporized: 
 xrie, the entropy of vaporization; for xLjT = xue'. 
 h + xL, the total heat, which is the sum of the heats of liquid 
 
 and of vaporization : 
 riw + XHe, the total entropy, which is the sum of the entropies 
 
 of liquid and of vaporization. 
 The expression for specific volume of wet vapor is less obvious. 
 The volume of 1 lb. of the liquid at the boiling point is »„. In 
 a MIXTURE weighing 1 lb., (1 — x) lb. of liquid are present. 
 There are also present a; lb. of dry vapor, the total volume of 
 which is XV. Then the volume of 1 lb. of the mixture is (1 — x)Vw 
 -{- XV = v„ -\- x{v — Vw) • This is very nearly equal to xv, and 
 may be so taken unless special note is made to the contrary. 
 The external work of vaporization was shown to be 
 
 W = ^{v-v^)B.t.n. 
 
 when DRY vapor was formed. The quantity in the parenthesis 
 is the INCREASE of volume from liquid to dry vapor. The 
 corresponding increase for wet vapor is x{v — »«). The value 
 of W for vaporization to wet vapor is consequently 
 
 144p 
 
 which is X times the value for dry vapor. The internal work 
 {E effect) in the vaporization to dry vapor was r = L — W. 
 For wet vapor it is xL — xW = x{L — W) = xr. 
 
94 Thermodynamics, Abridged 
 
 Recapitulation : 
 Properties that are the same 
 for wet vapor as for dry 
 
 dry vapor, L, Tie, r, 
 
 h, n^, p, t, 
 
 Properties multiplied by x 
 
 wet vapor, xL, xne, xr, 
 
 I dry vapor, v, H, ria, 
 
 Other properties j ^gt^^p^r, v„ + x(i; - v^), h+ xL, tu,+ xn„ 
 
 (approjdmately xv). 
 
 The quantity x is the dryness fraction or proportion of 
 DRYNESS. It is the weight of dry vapor in 1 lb. of the mixture 
 of dry vapor and liquid. 
 
 Prob. 145. From values given in the steam table at 100 lb. 
 pressure, compute all of the corresponding properties for steam 
 at this pressure if 0.90 dry. (Ans., p = 100, t = 327.8, h = 
 298.3, Uy, = 0.4743, xL = 799.2, xne = 1.0149, xr = 725.94, 
 Vy, + x{v - v„) = 3.9876 if «„ = 0.0177, h+xL= 1097.5, re„ 
 + xne = 1.4892.) 
 
 The factor of evaporation for a wet vapor, following Equa- 
 tion (40), is 
 
 xL+h-h ,.. 
 
 "~ 970.4 ' ^^ 
 
 where quantities without subscripts refer to the vapor condition 
 and ho to the feed water condition. 
 
 Prob. 146. What is the factor of evaporation for the steam 
 in Prob. 145, if that steam was generated from feed water at 
 90° F.? (Ans., 1.07.) 
 
 75. Volumes on the Tn Diagram. Suppose it to be required 
 that we plot in Fig. 22 the locus of states where the volume is 
 25 cu. ft. Write 
 
 25 = Du, + x{v — «„), or briefly, xv. 
 
 The values of «„ in the following are from Fig. 22a. Those 
 of V, n„ and Ue are from the steam table : 
 
Steam and Other Vapors 
 
 95 
 
 For t II 50 75 100 125 150 175 200 215 
 =.1702 743 350.8 178.4 96.9 55.7 33.6 25.35 
 •v^ = 0.01602 0.01603 001603 0.01623 0.01634 0.01648 0.01663 0.01673 
 
 25-«„' 
 
 x = 
 
 V—Vy, 
 
 = 0.015 
 
 0.034 
 
 0.071 
 
 0.14 
 
 0.258 
 
 0.45 
 
 0.743 
 
 0.987 
 
 Tie 
 
 = 2.0865 
 
 1.9631 
 
 1.8605 
 
 1.7475 
 
 1.6525 
 
 1.5645 
 
 1.4824 
 
 1.4354 
 
 xn. 
 
 = 0.031 
 
 0.0661 
 
 0.132 
 
 0.245 
 
 0.426 
 
 0.702 
 
 1.103 
 
 1.42 
 
 riu, 
 
 = 0.0361 
 
 0.0840 
 
 0.1295 
 
 0.1730 
 
 0.2149 
 
 0.255 
 
 0.2937 
 
 0.3163 
 
 n„-{-xn. 
 
 = 0.0671 
 
 0.1501 
 
 0.2615 
 
 0.4180 
 
 0.6409 
 
 0.957 
 
 1.3967 
 
 1.7363 
 
 Values in the last line, plotted against those in the first line, 
 give the constant volume curve of 25 eu. ft. in Fig. 22. Curves 
 
 ^200 
 |]00 
 
 0.016 
 
 0.020 
 
 0.017 0.018 0.019 
 Specific Volume Vw 
 
 Fig. 22a. Specific Volume of Water, 
 
 for less than 25 cu. ft. lie above this: curves for larger volumes 
 lie below it. The maximum volume on the diagram is toward 
 the "south-east" corner. 
 
 Prob. 147. Find the entropy of steam at 50° F. when its 
 volume is 100 cu. ft. (Ans., 0.158.) Find also at 75°, 100°, 125°. 
 (Ans., 0.349, 0.658, 1.152.) 
 
 76. Constant Heat Content. The total heat of wet steam is 
 
 h + xL. This may have any value between rather wide limits, 
 
 depending on p and x. Thus it may be 1000 B.t.u. when p = 100 
 
 if a; = (1000 - h)lL = 701.7/888 = 0.791. Then n„ + xn, = 
 
 0.4743 + (0.791 X 1.1277) = 1.366. From values found in this 
 
 * Note that it is sufficiently accurate to take x = 25/t). This would not 
 be a safe procedure, however, if we were plotting a line of very small volume. 
 
96 Thermodynamics, Abridged 
 
 way, the curve of constant heat content = 1000 B.t.u. is plotted 
 in Fig. 22. 
 
 Prob. 148. What are the dryness and entropy of steam at 
 atmospheric pressure when the heat content is 1000 B.t.u.? 
 (Ans., X = 0.845, n^ + xn^ = 1.53.) 
 
 Prob. 149. If the steam in Prob. 148 emits 500 B.t.u. without 
 change of pressure, what is then its dryness? (Ans., 0.33.) 
 If it then emits 406 B.t.u. more, what does it become? (Ans., 
 water at 126.15° F., if the pressure is allowed to fall after con- 
 densation begins: within a small fraction of 1° of this tempera- 
 ture if the pressure is kept at that of the atmosphere.) 
 
 Prob. 150. A boiler generates from feed water at 90°, 20 000 
 lb. of steam per hr. at 200 lb. pressure, 0.99 dry. The coal has 
 a heat value of 11 316.7 B.t.u. per lb. and the efficiency of the 
 boiler is 0.75. What weight of coal is burned per hr.? (Ans., 
 2667 lb.) 
 
 77. Work Done in Heating Liquid. The heat absorbed in the 
 operation ah, Fig. 22, is h per lb. of liquid. The external work 
 done during this heating, and included in h, is very small and 
 may often be neglected. The method of computing it should 
 be known. 
 
 A pound of water has been heated from 32° F. and a pressure 
 of 0.0886 lb. per sq. in. (Art. 63), until its pressure is that cor- 
 responding with th, or 100 lb. per sq. in. The external work is 
 that due to this increase of pressure. It is equivalent to the 
 work of LIFTING the pound of water to a height represented by 
 the difference of the final and initial pressures.* What would be 
 this height? Suppose the water to be delivered into the base of a. 
 vertical tube of 1 sq. ft. area. If the height to the top of the 
 
 *When we lift water 255 ft., its pressure (as measured on a gage placed at the 
 
 14.7 
 original level) is approximately ^xTi ^ 255= 110 lb., if the temperature is around 
 
 60". (Atmospheric pressure is equivalent to a head of 34 ft. of water.) Conversely, 
 the application of 110 lb. pressure is equivalent to lifting the water 255 ft. The 
 more complicated but more correct statement in the text is due to the fact that the 
 water in question is at a temperature of 327° after it is lifted and at a tempera- 
 ture of 32° before lifting occurs. 
 
Steam and Other Vapors 97 
 
 tube (where the water flows off) is / ft., and the density of the 
 water is d, the hydrostatic pressure at the base of the tube is 
 simply the weight of water, or fd Ibi per sq. ft. Or, inversely, 
 the height in ft. is the pressure per sq. ft. divided by the density. 
 At 100 lb. pressure per sq. in., the density of water is 56.4 lb. 
 per cu. ft. The external work of our process is equivalent to 
 lifting 1 lb. of water to a height of (substantially) 100 X 144 
 -i- 56.4 = 255 ft., or is 255 ft. lb. In general symbols. 
 
 External Work = W, = ^-^^^^ = ^ B.t.u., (42) 
 
 for Vw is the reciprocal of the density. The gain of internal 
 energy during the heating of the liquid is h — (p»„/5.403). The 
 external work done in heating 1 lb. of liquid from c to & is iWk)b 
 — (Wk)c. Values of Vm are given in Fig. 22a. 
 
 Prob. 151. Find W^ at the critical temperature, where 
 Vy, = 0.049. (Ans., 26.7 B.t.u.) 
 
 78. Vaporization or Condensation. As shown in rAt. 74, the 
 heat absorbed in partial vaporization is xL B.t.u. per lb. For 
 complete vaporization, put x = 1.0. Vaporization (partial) at 
 constant pressure is represented by such a line as bv, Fig. 22. 
 Condensation of steam initially at the point v would be repre- 
 sented by the reverse line vb. The heat emitted is (xL)„, and 
 its sign is negative. Note that the isothermal and the constant 
 pressure paths coincide when the fluid consists of a mixture of 
 liquid and vapor. The external work done (Art. 74) is 144 
 px{v — Vw) -J- 778 B.t.u. This is + for vaporization and a; = 1 
 if vaporization is complete. It is — for condensation. The 
 internal work is + a;r for vaporization and — xr for condensation. 
 
 Prob. 152. How many B.t.u. are emitted when 1000 lb. of 
 steam at 1 lb. pressure, 0.70 dry, are condensed and cooled to 
 90°? (Ans., 736 020 B.t.u.) What is the exact loss of internal 
 energy? Take d„ = 0.01614 at 1 lb. and 0.0161 at 90°. (Ans., 
 692 829 B.t.u.) 
 
 79. Adiabatic Process. No heat is absorbed or emitted if the 
 process is adiabatic. The external work is computed by writing 
 
98 Thermodynamics, Abridged 
 
 as the law of a vapor adiabatic, initial entropy = final entropy : 
 or, (Fig. 22) 
 
 ub = nc, 
 (uw + xne)s = (Wu, + xne)c- (43) 
 
 This is one of the most useful equations of our subject. Either 
 value of X may be 1.0. For an adiabatic, 
 
 H=0 = E+W, W= - E= Eb- Ec, 
 
 '^=(»-5^3+-),-(»-^+"X- <«' 
 
 This is very nearly equal to {h+xr)B— ih+xr)c, and in 
 ordinary calculations should be so taken. Do not use this 
 approximation, however, in Probs. 153-156. 
 
 The work is + when the path is downward and — when it 
 
 is UPWARD. 
 
 On a pv diagram, the vapor adiabatic is a smooth curve which 
 may be represented approximately by the equation PbVb" 
 = PcVc", but the value of a depends on the initial dryness: 
 a = 1.035 + 0.1a;B. Then vc = VBiVBlvcf" and 
 
 „, Pb'Bb — Pcvc 
 W = 
 
 = iA^{v^^^^^ytAh. 
 
 a- 1 
 
 Prob. 153. Steam at 250 lb. pressure, 0.99 dry, expands 
 adiabatically to 1 lb. pressure. What is the dryness after 
 expansion? (Ans., 0.7515.) What was the specific volume 
 before expansion? (Ans., 1.832 if u„ = 0.0187.) After expan- 
 sion? (Ans., 250.) 
 
 Prob. 154. In Prob. 153, find p»„/5.403 at the initial condi- 
 tion. (Ans., 0.87 B.t.u.) At the final condition, when ««, 
 = 0.01614. (Ans., 0.003 B.t.u.) What was the internal energy 
 before expansion? (Ans., 1108.91.) After expansion? (Ans., 
 800.93.) What was the external work done during expansion, 
 by Equation (44)? (Ans., 307.98 B.t.u.) 
 
 Prob. 155. What is the value of a in the exponential approx- 
 imate formula for this process? (Ans., 1.134.) Of »c? (Ans., 
 239.) Of the external work? (Ans., 302 B.t.u.) 
 
Steam and Other Vapors 
 
 99 
 
 Prob. 1S6. Consider the adiabatic expansion of Prob. 153 
 as BC, Fig. 22. How much heat was absorbed, and how much 
 external work was done, during the operation c6? (Ans., heat 
 = 305.4 B.t.u., work = 0.867 B.t.u.) Along 6B? (Ans., heat 
 = 818.04 B.t.u., work = 83.46 B.t.u.) Along Ccl (Ans., heat 
 = - 777.5 B.t.u., work = - 46.37 B.t.u.) Show that for the 
 
 B cIka . 
 
 Fig. 225. Vapor Paths and Cycles. 
 
 closed figure chBC, Sff = SJF. (Ans., ^E = 305.4 + 818.04 
 - 777.5 = 345.94. Also SPF = 0.867 + 83.46 + 307.98 - 46.37 
 = 345.94 B.t.u.) 
 
 Ammonia is compressed adiabatically from 0° to 
 n^= - 0.0709, 
 
 Prob. 157. 
 90°. At 0°, 
 
 n, = 1.245. At 90°, n, = 
 1.0219. What must be the 
 initial dryness if the fluid is 
 to be exactly dry (.r = 1.0) 
 after compression? (Ans., 
 0.878.) 
 
 80. IdealVapor Cycle, Com- 
 plete Expansion : " Rankine " 
 
 Cycle. The cycle c6£C, Fig. ^^^^^ ideal Cycles for Vapors. 
 23, represents the ideal opera- 
 tion of a STEAM POWER PLANT running condensing. Water is 
 
 b 
 
 .B 
 
 
 G 
 
 E 
 
 
 H 
 
 F^~^--~~~^ 
 
 c 
 
100 Thermodynamics, Abridged 
 
 heated along cb, the pressure and temperature increasing and the 
 volume increasing slightly. Vaporization occurs along bB, adia- 
 batic expansion along BC and condensation along Cc. The corre- 
 sponding diagram is similarly lettered in Fig. 225. We have 
 seen how to compute the work and heat for each path of this 
 cycle. The following is a briefer method of computing efficiency 
 and mean effective pressure. Referring to Fig. 225, the effi- 
 ciency is 
 
 e = cbBC ^ gcbBD 
 = (fabBD - facCD) H- (fabBD - facg) 
 = {Hb - He) H- {Hb - h), 
 
 where H indicates the total heat above 32° of steam either 
 WET OR DRY. This is best written in the specific form, 
 
 e= {{h + xL)i - ih + xLh} ^ {{h + xL)i - }h\, (45) 
 
 in which subscripts i and 2 refer to high-pressure and low-pressure 
 conditions, respectively. This efficiency is always less than that 
 of the Carnot cycle between the same extreme temperatures. 
 The value of X\ is usually close to 1.0. That of Xi is computed 
 from Equation (43) : (ra„ + xn^x = (7i„ -\- xn^i. 
 
 The mean effective pressure is the work in ft. lb. divided by 
 {144 X iaac -■».)}, Fig. 23. This divisor becomes a;2(»2 - «<.), 
 or with an error practically never exceeding 1/10 of 1 per cent, 
 for steam, simply x^Vi., where vt is the tabular volume of dry steam 
 at the low-pressure condition. Then 
 
 iji + xi)i — (A -f- xL)i 
 
 Prob. 158. Find the efficiency of a Rankine steam cycle in 
 which the extreme pressures are 250 and 1 lb. and the dryness at 
 the beginning of expansion is 0.99. (Ans., 0.308.) How much 
 net work is done? (Ans., 345.94 B.t.u.: compare Prob. 156.) 
 What is the mean effective pressure? (Ans., 7.5 lb.: mean 
 effective pressures are very low in this type of cycle.) 
 
 81. Incomplete Expansion. During the later stages of expan- 
 sion. Fig. 23, very little work is derived from the fluid in pro- 
 
Steam and Other Vapors 101 
 
 portion to the increased displacement. The cylinder of the 
 engine will therefore be large and costly for its power. (This 
 does not apply to turbines.) In usual engine practice, expansion 
 is terminated at some point E, before the pressure has been re- 
 duced to that at which the steam is to be condensed. A constant- 
 volume DKOP, EF, then occurs. See Fig. 225 also. The cycle 
 is cbBEF. Divide it into two areas GbBE and GEFc by a hori- 
 zontal line. The lower area is on the pv diagram very nearly a 
 rectangle, and its magnitude in B.t.u. may be taken as 
 
 Then the efficiency is 
 
 {h + a;i)i — hi 
 
 144 
 {h + xL)i- {h + xL)e + 1^^ v^ipE - Vi) 
 
 {h+xL)y-h2 ' ^ ' 
 
 where v^ is the volume of wet vapor at the point g, to be com- 
 puted by first finding the entropy and dryness. 
 
 Thus under the limiting pressures and initial dryness of Prob. 
 158, let Pb = 10. Then at this pressure n^ = 0.2832, ne = 
 1.5042. Equating the entropies at B and E, 
 
 0.5676 + (0.99 X 0.9600) = 0.2832 + 1.50420;^, 
 Xe ?= 0.819. 
 
 Then, very nearly, v^ = 0.819 X 38.38 = 31.4, the value 38.38 
 being that for dry steam at 10 lb. pressure. Then the area 
 GEFc = (144/778) X 31.4 X 9 = 52.1 B.t.u. The value of 
 (A + a;i)s is 161.1 + (0.819 X 982) = 967 B.t.u. The area 
 hBEG is 375.2 + (0.99 X 826.3) - 967 = 226.24 B.t.u. Then 
 'LW = 52.1 + 226.24 = 278.34 B.t.u. and e = 278.34 -^ (1193.24 
 - 69.8) = 0.247. This is less than the result in Prob. 158. 
 If expansion had terminated sooner, i.e., if p^ had been greater 
 than 10, the efficiency would have been still less. 
 
 Prob. 159. Find the efficiency as above, when ps = 20. 
 
 (Ans., 0.215.) 
 
102 Thermodynamics, Abridged 
 
 Prob. 160. Find the efiiciency as above in the limiting case, 
 when Vs = 250, cycle cbBH. (Ans., 0.074.) 
 
 The mean effective pressure with the cycle cbBEF is, very 
 nearly, 7787:W ^ 144d^, or 
 
 r .^o (h + xL)i - (h+xL)^ , 5.4032TF ^^^^ 
 Pn. = 5.403^^^= '- ^-^ ^+ V^-P2= — . (48) 
 
 ^E ^E 
 
 For the foregoing conditions, this becomes 48.1. It may be as 
 high as desired, up to the value pi — pi. Incomplete expansion 
 reduces efficiency, but increases mean effective pressure. 
 
 Prob. 161. Find prn in Probs. 159, 160. (Ans., 78 and 249 lb.) 
 
 82. Ideal Steam Consumption. The area of the cycle, or the 
 numerator of the efficiency expression, represents the ideal work 
 obtainable per lb. of steam, liW (B.t.u.). Since 1 hp. = (33 000 
 X 60) -f- 778 = 2545 B.t.u. per hr., the ideal steam rate (lb. 
 of steam per hr. per Ihp. under best conditions) is 2545 -^ 2TF. 
 For Prob. 158, this rate is 2545 h- 345.94 = 7.33 lb. The heat 
 supplied being Ai + xiLi — h^, an engine having an efiiciency 
 of 100 per cent, would use 2545 -j- {hi + XiLi — h^) lb. of steam 
 per hr. per Ihp. For the conditions of Prob. 158, this is 2545 -j- 
 1123.44 = 2.26 lb. But this ideal engine has an efficiency of 
 only 0.308. Hence its ideal steam rate is 2.26 ^ 0.308 = 7.33 lb., 
 as before. The actual steam consumption will be greater than 
 7.33, under the assigned conditions. 
 
 The ideal heat rate is the heat consumed in B.t.u. per 
 MiN. PER Ihp. At 100 per cent, efficiency, it would be 33 000 
 -^ 778 = 42.42. At an efficiency e it is 42.42 -^ e. In Prob. 
 158, it is 42.42 -f- 0.308 = 138 B.t.u. 
 
 Prob. 162. Find ideal steam rate and ideal heat rate in 
 Art. 81. (Ans., 9.13 lb., 171 B.t.u.) 
 
 Prob. 163. Find these ideal rates for Prob. 160. 
 
 (Ans., 30.2 lb., 570 B.t.u.) 
 
 Prob. 164. Find the ideal rates in Prob. 159. 
 
 (Ans., 10.4 lb., 195 B.t.u.) 
 
Steam and Other Vapoes 
 
 103 
 
 Fig. 24 shows the variation in efficiency of such cycles as 
 cbBEF for steam initially dry under the initial (ps) and back 
 (Pp) pressures common in condensing and non-condensing 
 practice. 
 
 2 3 4 5 6 
 
 Ratio of Expansion Vp-f-Vg 
 for non-condensing engines 
 
 Fig. 24. Ideal Efficiencies with Incomplete Expansion: Steam Initially Dry. 
 
104 Thermodynamics, Abridged 
 
 Prob. 165. From Fig. 24, state the heat rate and steam rate 
 for an ideal cycle using dry steam at 215 lb. pressure, condensing 
 at 2 lb. absolute pressure, when the ratio of volumes during 
 adiabatic expansion is 30. (Ans., e = 0.265, heat rate = 160, 
 steam rate = 8.7.) 
 
 Superheated Vapor 
 
 83. Properties. Superheating almost always occurs .at con- 
 stant pressure. The pressure is that at which the saturated 
 vapor was formed. Vapor must be dry-saturated before it can 
 be superheated. The specific heat at constant pressure is 
 highly variable, both with pressure and with temperature. 
 Extensive tables are therefore necessary for mean values of h, 
 the specific heat, or of W , the total heat. \i t = saturation 
 temperature, t' = actual vapor temperature, both at the pressure 
 TP, the " SUPERHEAT " (amount or number of degrees of super- 
 heat) ist' — t and the total heat is 
 
 H' = H+Ht'-t). 
 
 The volume is given quite closely for steam by the rather 
 formidable equation 
 
 pv' = 0.5962(f' + 460) - p(l + 0.00142^) ^^|^|^^- 0.0833 ) . 
 
 The gain of entropy of a vapor during superheating may be 
 expressed as k loge (t' + 460) /(< + 460), and the total entropy is 
 
 T' 
 n' = n,+ k loge Y - 
 
 The external work done during superheating is 
 
 ]^' = ^(.'-«)B.t.u. 
 
 The whole amount of external work done during the formation 
 of superheated steam from water at 32° and 0.0886 lb. pressure 
 is (see Arts. 77, 78), 
 
 yow , PJi} - Vw) , p{v' - v) _ ^' 
 5.403^ 5.403 "*" 5.4031 ~ 5.403 *•"• 
 
Steam and Other Vapors 
 
 105 
 
 The gain of internal energy during this operation is then 
 
 H' - 
 
 5.403 • 
 
 Prob. 166. For steam at 250 lb. pressure, superheated 200°, 
 h = 0.562. Find t', H', v', n', W and the gain of internal energy 
 from liquid at 32°. (Ans., 601.1° F.: H' = 1313.9: / = 2.471: 
 n' = 1.646, W = 28.7, gain of energy = 1199.7.) 
 
 The factor of evaporation for superheated steam is 
 
 F' = ifl' - h) ^ 970.4, 
 
 following Equation (40). 
 
 Prob. 167. What is the factor of evaporation for the steam 
 in Prob. 166, if formed from water at 90°? (Ans., 1.29.) 
 
 The following is a greatly abridged sample of a superheated 
 
 steam table. 
 
 Properties of Superheated Steam 
 
 (Condensed from Steam Tables and Diagrams, by Marks and Davis, with 
 the permission of the pubMshers, Messrs. Longmans, Green, & Co.) 
 
 SUPEBHBAT. */' 
 
 
 40 
 
 90 
 
 200 
 
 300 
 
 Absolute Pressure Lbs. per 
 Square Inch 
 
 1 
 
 v' 
 
 H' 
 
 [ n' 
 
 = 141.7 
 = 357.8 
 = 1122.6 
 = 2.0069 
 
 191.7 
 
 387.9 
 
 1145.3 
 
 2.0434 
 
 301.7 
 
 453.7 
 
 1195.6 
 
 2.1145 
 
 401.7 
 
 513.4 
 
 1241.5 
 
 2.1701 
 
 100 
 
 v' 
 H' 
 n' 
 
 = 367.8 
 =*4.72 
 = 1208.4 
 = 1.6294 
 
 417.8 
 
 5.07 
 
 1234.6 
 
 1.6600 
 
 627.8 
 
 5.80 
 
 1289.4 
 
 1.7188 
 
 627.8 
 
 6.44 
 
 1337.8 
 
 1.7656 
 
 140 
 
 v' 
 ]H' 
 ( n' 
 
 = 393.1 
 = 3.44 
 = 1215.8 
 = 1.6031 
 
 443.1 
 
 3.70 
 
 1242.8 
 
 1.6338 
 
 553.1 
 
 4.24 
 
 1298.2 
 
 1.6916 
 
 653.1 
 
 4.71 
 
 1346.9 
 
 1.7376 
 
 ISO 
 
 ( t' 
 I v' 
 Iff' 
 
 = 398.5 
 = 3.22 
 = 1217.3 
 = 1.5978 
 
 448.5 
 
 3.46 
 
 1244.4 
 
 1.6286 
 
 558.5 
 
 3.97 
 
 1300.0 
 
 1.6862 
 
 658.5 
 
 4.41 
 
 1348.8 
 
 1.7320 
 
 250 
 
 v' 
 H' 
 n' 
 
 = 441.0 
 = 1.98 
 = 1229.3 
 = 1.5593 
 
 491.0 
 
 2.14 
 
 1257.7 
 
 1.5900 
 
 601.0 
 
 2.47 
 
 1313.9 
 
 1.6460 
 
 701.0 
 
 2.75 
 
 1363.5 
 
 1.6905 
 
106 Thermodynamics, Abridged 
 
 84. Rankine Cycle with Superheat. Such a cycle is repre- 
 sented as cbdxJ, Fig. 225, or as cbBC, Fig. 23. There is no 
 change in direction of the constant pressure line on the pv 
 diagram. The eflBciency is 
 
 (H%-(h+xL)j . 
 
 Thus, take the conditions as p^ = 250, U = 601°, pc = 1. 
 First find the dryness at J: 
 
 (n™ + xne)j =n' = 1.646 (Prob. 166) 
 
 0.1327 + lM27xj = 1.646, Xj = 0.82. 
 Taking from Prob. 166 HJ = 1313.9, 
 
 1313.9 - {69.8 + (0.82 X 1034.6)] 
 
 e = 
 
 = 0.319. 
 
 1313.9 - 69.8 
 
 In Prob. 158, the eflSciency for the same pressures but with 
 SLIGHTLY WET steam was 0.308. Superheating improves 
 
 EFFICIENCY. 
 
 The mean effective pressure, following Equation (46), is 
 
 5.403ZTF 
 ^™- ixv)j • 
 Noting that SPF is the numerator of the expression for efficiency, 
 this becomes (5.403 X 397) -^ (0.82 X 333) = 7.84 for our 
 conditions. 
 
 Prob. 168. Find e and pm in Art. 84 if the superheat is 100° 
 instead of 200°. (Ans., e = 0.313, pm = 7.64.) 
 
 85. Incomplete Expansion. Following Art. 81, if expansion 
 terminates before the line of back pressure is reached, the work 
 of the superheated cycle is 
 
 in which subscripts 3 refer to the " terminal " condition, E, 
 Fig. 23. The cycle is cbBEF (cbdxKF in Fig. 22B). The 
 efficiency is 
 
 ' = K^^o (50) 
 
Steam and Other Vapors 107 
 
 and the mean effective pressure is 
 
 5.4032: TT- _^ 
 
 Vm = — ;; . (51) 
 
 Suppose that under the conditions of Art. 84 the terminal 
 pressure is Vk — V^— 10. Then 
 
 w/ = 1.646 = (n„ + xne)3 = 0.2832 + 1.5042a;3, 
 
 X3 = 0.906, (xL)3 = 0.906 X 982 = 891, 
 
 % = 0.906 X 38.38 = 34.9 = vz, 
 
 SPF = 1313.9 - 161.1 - 891 + -f^ = 320 B.t.u., 
 
 _ 320 _ n or;7 _ 5.403 X 320 _ 
 
 ® ~ 1313.9 - 69.8 ~ ^^'' ^" ~ 34.9 ~ ^ 
 The eflBciency exceeds that computed for saturated steam in 
 Art. 81. 
 
 Prob. 169. Find e and 'pm when the pressure limits are 250 
 and 1 lb., the initial superheat 100° and the terminal pressure 
 20 lb. (Ans., 0.223 and 79 lb.) 
 
 86. Cases Not Hitherto Considered. In Arts. 84 and 85, the 
 steam at the end of its adiabatic expansion was wet. In cases 
 where at this point it is still superheated, an exact solution is 
 not possible by any method thus far shown. We know in general 
 the terminal pressure and entropy. We have no way of finding 
 terminal volume, temperature, superheat, or heat content. 
 A graphical method for finding these properties will now be 
 developed. 
 
 In Fig. 25 the coordinates are total heat above 32° and total 
 entropy. Plot the " saturation curve " ah from values in the 
 steam table. Thus at 100 lb. pressure the coordinates are 
 1186.3 and 1.602: at 1 lb. they are 1104.4 and 1.9754. These 
 values establish the points h and n. 
 
 The points e and c are those at which the pressxu-es are 150 
 and 50 lb. Plot the constant pressure curves ef and cd. 
 This is done by finding the coordinates of points like /, for some 
 
108 
 
 Thermodynamics, Abridged. 
 
 assumed dryness. Thus, assume a dryness of 0.80. Then at / 
 the coordinates are, 330.2 + (0.8 X 863.2) = 1020.8, and 0.5142 
 + (0.8 X 1.055) = 1.3582. Taking other drynesses, additional 
 points may be located to establish the whole line ef. Plot cd in 
 the same way. Joining / and d gives a line of constant dryness. 
 
 1.6 1.7 
 ^ Entropy 
 
 Fig. 25. Development of a Heat Chart. 
 
 Prob. 170. Locate the point o on the line of 1 lb. pressure, 
 where the dryness is 0.80. (Ans., coordinates are 897.5 and 
 1.6069.) 
 
 The constant-pressure curves are now carried upward into 
 the superheated region, using the table in Art. 83, and lines of 
 constant superheat, gj, hk, are drawn. 
 
 Prob. 171. Find coordinates of points g and j. (Ans., 1244.4 
 and 1.6286: 1145.3 and 2.0434.) 
 
 87. Application of Total Heat Entropy Diagram. For com- 
 plete EXPANSION (Rankine) cycles. Equations (45) and (49) 
 may both be written 
 
Steam and Other Vapors 109 
 
 where subscripts i and 2 refer to conditions at the beginning 
 and end of adiabatic expansion, respectively, h stands for heat 
 of liquid and Q for total heat of vapor either wet, dry or super- 
 heated. Values of Qi and Q2, and also of the final dryness, xi, 
 may be read directly from a chart like Fig. 25. A more complete 
 chart appears in Fig. 26. Still larger scale charts, which may 
 be used for accurate work, will be found in recent steam tables 
 Uke those of Goodenough or Marks and Davis. The saving in 
 labor due to their use is very great. Thus suppose a cycle with 
 complete expansion to work between pressure limits of 150 lb. 
 and 1 lb., with an initial superheat of 200°. Locate A, Fig. 25 
 and read off Qi = 1300 at the left. Draw W vertically. Read 
 off, at I, interpolating between the 0.84 and 0.85 dryness curves, 
 Xi = 0.842, Q2 = 944. Then e = (1300 - 944) -^ (1300 - 69.8) 
 = 0.256. The value hi = 69.8 is from the steam table. Taking 
 the specific volume at 1 lb. pressure at 333, also from the steam 
 table, 
 
 1)2 = 0.842 X 333 = 280.4, p™ = ^'^^^^^^^ = 6.86. 
 
 This method is applicable whether the steam is initially wet, dry 
 or superheated. 
 
 Incomplete Expansion. The most difficult case is the one thus 
 far omitted, that in which the steam is superheated at the end 
 of expansion. Equations (47) and (50) may be written 
 
 n n I ^3(g3 - Vi) 
 _ yi - ^3 + 5.403 (53) 
 
 Qi — hi 
 
 where 3 is the terminal condition. If this condition is one of 
 WET steam, the volume v% is found from the dryness, as above. 
 But suppose steam at 150 lb. pressure, 200° superheated, to 
 expand to pa = 50 lb. pressure and then to be condensed at 
 1 lb. pressure. In Fig. 25, Qi = 1300, as before: draw hm 
 
110 
 
 Thermodynamics, Abridged 
 
 
Steam and Other Vapors 111 
 
 vertically. At m, Q3 = 1203, superheat = 54°. Then h = 281 
 + 54 = 335, and the equation in Art. 83 for the volume of super- 
 heated steam gives Vz = 9.26. Then 
 
 9 26 
 1300-1203 + ^-^3(50-1 ) ^^^ 
 
 1300 - 69.8 1230 
 
 and 
 
 181 X 5.403 ^_ 
 
 P"' = 9:26 =^°^- 
 
 Note the marked difference between these results and those of 
 the last paragraph. Incomplete expansion has reduced the 
 efficiency from 0.256 to 0.147 but has increased the mean 
 effective pressure from 6.86 to 106. A corresponding increase 
 would occur in the power of an engine of given size and speed. 
 
 Prob. 172. Check from Fig. 26 all drynesses and efficiencies 
 computed in Probs. 153, 158-160, 168, 169. 
 
 To avoid the labor of computing V3 from the long formula of 
 Art. 83, an auxiliary chart called the total heat-pressure 
 diagram is often used. This uses the quantities mentioned as 
 coordinates, and plots curves of constant dryness, superheat and 
 volume. Pressure and heat content being taken from Fig. 26, 
 volume may be read directly from the total heat-pressure 
 diagram. 
 
 Factors Affecting Ideal Engine Efficiency 
 
 88. Incomplete Expansion, Dry or With Superheat. Effi- 
 ciencies of various cycles with superheat are plotted in Fig. 27. 
 These may be compared with values in Fig. 24. 
 
 Back Pressure has of course a pronounced influence. This 
 explains the gain in economy by running condensing. A very 
 low ratio of expansion, running condensing, gives better efficiency 
 than a high ratio running non-condensing. A back pressure 
 around 25 lb. absolute is often used for small auxiliary engines 
 on ships. This is for the purpose of using the exhaust steam 
 from these engines advantageously in heating the boiler feed 
 water. Fig. 24 shows that it leads to low engine efficiency. It 
 
112 
 
 Thermodynamics, Abridged 
 
 pays, because practically all of the heat wasted by the engines 
 goes back to the boilers. 
 
 Ratio of Expansion-Condensing Engines 
 
 4 5 6 7. 
 
 Ratio of Expansion-Non-Condensing Engines 
 
 FiQ. 27. Ideal Efficiencies with Superheated Steam. 
 
Steam and Other Vapors 113 
 
 Ratio of Expansion has an especially important bearing on 
 eflBciency. Many land engines running at constant speed are 
 governed by varying this ratio. Hence the curves show how 
 eflBciency varies with output. The best ratio is not to be deter- 
 mined by thermodynamic considerations alone, but is established 
 from various practical and commercial factors (Arts. 95, 96). 
 At a given pressure, eflBciency increases with the ratio of expan- 
 sion, but mean effective pressure decreases. A ratio is finally 
 reached at which there is little or no further gain in eflBciency. 
 This limit is very soon attained if the initial pressure is low, and 
 is postponed if high superheat is used. 
 
 High Initial Pressure is important. At a given ratio of ex- 
 pansion, the eflBciency increases with the pressure: but for 
 very low ratios, high pressures are of little advantage. In con- 
 densing engines with superheat, high pressure is less important 
 than in non-condensing engines using superheat. 
 
 Superheat increases eflBciency at a given pressure or ratio of 
 expansion. In general, the gain of eflBciency is roughly pro- 
 portional to the amount of superheat. Superheat should not be 
 continued beyond the point where the terminal condition is 
 dry. Superheat gives high eflBciency at moderate ratios of 
 expansion and therefore permits of reasonably high mean 
 effective pressures along with good eflBciency. 
 
 Initial Dryness is not considered in these diagrams. Within 
 reasonable limits (0.95 to 1.0) it has practically no influence on 
 eflBciency. 
 
 All eflBciencies are below those of the Carnot cycle between the 
 same extreme temperature limits. 
 
 89. Vapors Other Than Steam. In the Carnot cycle, the 
 eflBciency is independent of the fluid used. In the actual steam 
 cycles here described, it depends on the properties of the fluid 
 (latent heat, specific heats). Various fluids have been used in 
 vapor engines. In simple condensing engines with low ratios 
 of expansion, a considerable gain is possible over the best results 
 with steam. A desirable vapor would be one whose pressure 
 
114 Thermodynamics, Abeidged 
 
 was less than that of steam at high temperatures and more at 
 low temperatures. The latter characteristic would make high 
 vacuums unnecessary. It should have a high latent heat of 
 vaporization and a low specific heat of liquid, since this, in 
 Fig. 225, would make the Rankine cycle approximate more 
 closely the Carnot. In the cycle cbBC of this diagram, the heat 
 supplied by the boiler is the area under cbB. That carried 
 away at the condenser is the area under Cc. The ratios of these 
 areas to the power developed depend on the properties of the 
 fluid used. Hence a suitable substitute fluid might decrease the 
 relative size of boiler or condenser necessary. The mean effective 
 pressure, and hence the size of cylinder, also depends on the 
 properties of the vapor. 
 
 Prob. 173. In Fig. 20, what vapor apparently has a higher 
 pressure than that of water at' low temperatures, and a lower 
 pressure at very high temperatures? 
 
 Mixtures of Air and Steam 
 
 90. Saturated Air. The word "saturated," as applied to air, 
 has a different meaning from that of the same word applied to 
 steam. Saturated steam is dry steam. Saturated air is the 
 wettest kind of air. Supersaturated air is air containing all 
 the moisture it can hold, in the presence of an additional body 
 of moisture. 
 
 When air and a sufBciency of moisture are mixed at the 
 temperature t, the mixture contains water vapor, i.e., steam, 
 at the corresponding pressure ps, which may be taken from the 
 steam table. The total pressure of the moist air is the sum of 
 the partial pressures of dry air and steam. If we are dealing 
 with normal barometer, the partial pressure of dry air is then 
 Pa = 14.696 — ps. The steam is saturated steam and the air 
 is saturated air. One cubic foot of pure dry air unmixed with 
 steam at the temperature t and normal barometer would weigh 
 (144 X 14.696) -J- (53.36 r) lb. One cubic foot of air in the 
 present saturated mixture weighs Wa = 144(14.696 — pa) -f- 
 (53.363') lb., which is less than the weight of dry air unmixed 
 
Steam and Other Vapors 115 
 
 with steam. The weight of steam intermixed with the 1 cu. ft. 
 of dry air in the mixture is Ws = 1/v, where v is the specific 
 volume from the steam table at t° F. By Dalton's law, if we 
 employ partial pressures ps and pa as above, 1 cu. ft. of mixture 
 contains both 1 cu. ft. of dry air and 1 cu. ft. of steam. 
 
 Prob. 174. What is the weight of a cubic foot of pure dry air 
 at normal atmospheric pressure and 40° F.? (Ans., 0.0794 lb.) 
 
 Prob. 175. The pressure of saturated steam at 40° F. is 
 0.1217 and its specific volume is then 2438. If air at 40° F. and 
 normal barometer is saturated with moisture, state the partial 
 pressures of steam and air. (Ans., 0.1217 and 14.5743.) 
 
 Prob. 176. In Prob. 175, state the weights per cu. ft. of air 
 and steam. (Ans., 0.0787 and 0.00041 lb.) What is the density 
 (weight per cu. ft.) of the mixture? (Ans., 0.07911 lb.) 
 
 91. Unsaturated Air. The " absolute humidity " is the ratio 
 of weight of steam to weight of air. In Prob. 176, it is 0.0052. 
 If there is insufficient moisture present to saturate the air, its 
 temperature will be t, but its pressure will be p/, less than ps- 
 Hence the moisture in such a mixture will be superheated 
 STEAM. Call its weight per cu. ft. w/. Then the relative 
 HUMIDITY is defined as Vh = pZ/ps = w/lws. The partial pres- 
 sure of air in the mixture, at normal barometer, is 14.696 — ps', 
 from which the weight of dry air in a cubic ft. of mixture may be 
 computed as before. The weight of steam will be rhWs- 
 
 Prob. 177. What is the absolute humidity of saturated air 
 at 212° F.? (Ans., 00.) 
 
 Prob. 178. At t = 200°, p, = 11.52, Ws = 0.02976. For a 
 relative humidity of 0.9, find for moist air the values of ps' 
 (Ans., 10.368), pa' (Ans., 4.328), Wa' (Ans., 0.026784), Wa' (Ans., 
 0.0177). 
 
 Prob. 179. As t increases, how will pa, Pa, Ws and Wa vary? 
 (Ans., Ps and Ws will increase, pa and Wa will decrease.) 
 
 92. Moist Air Processes. Following Equation (13), the value 
 of R for a mixture of air and steam is 
 
116 Thermodynamics, Abridged 
 
 RsWs' + 53.36w/ 85.8w/ + 53.36w/ 
 
 the value of R for steam being taken at 1544/18.016 = 85.8. 
 Rm will often differ notably from 53.36. , 
 
 When a fixed body of unsaturated moist air is cooled, the value 
 of Ws ddes not change, but that of Ws decreases. Hence the 
 relative humidity increases. If the process is carried sufficiently 
 far, the mixture will become saturated air. Beyond this point, 
 the air will be supersaturated and dew will be deposited. Sup- 
 pose Th = 0.60 at 100° and the mixture is cooled to 90°. The 
 original 1 cu. ft. becomes (460 + 90)/(460 + 100) = 0.982 
 cu. ft. At 90°, the density of saturated steam is 0.002131. The 
 weight of steam in 0.982 cu. ft. of mixture, if saturated, would 
 be {w,)w = 0.982 X 0.002131 = 0.00209. At 100°, the weight 
 present if saturated is 0.002851. This weight, multiplied by 
 the original relative humidity of 0.60, is still present after cooling 
 to 90°. Hence at 90° the relative humidity will have become 
 
 
 , ... . J.002851 X 0.60 ^„^ 
 '■" = '-' = 0:00209 =°-^2. 
 
 Prob. 180. Find the value of R„ in Prob. 176. (Ans., 53.5.) 
 
 Prob. 181. In Art. 92, what is the relative humidity if the 
 mixture is further cooled to 85°, where Ws = 0.001832? (Ans., 
 0.96.) 
 
 Pure steam at 102° F. has a pressure of 1 lb. absolute. Hence, 
 however low the temperature of condenser circulating water, 
 a low temperature of heat rejection in the steam engine (shown 
 in Art. 36 to be essential for highest potential efficiency) is 
 dependent on the maintenance of a good vacuum by the air 
 pump. By using a mixture of air with steam in an engine, a 
 rejection temperature as low as the circulating water will permit 
 might be attained along with any pressure or vacuum (above the 
 pressure given in the steam table for cooling water temperature) 
 which might be desired. We might have the steam and air 
 at 70° F. The partial pressure of the steam would then be 
 
Steam and Other Vapors 117 
 
 only 0.3626 lb. per sq. in., with saturated air. The total pressure 
 might however be 1 or 2 lb., a range of values easily maintained 
 by modern vacuum apparatus. 
 
 Prob. 182. A mixture of air and steam at 2 lb. pressure has a 
 temperature of 126.15° F. What would be the air pressure if 
 the air were saturated? (Ans., 0.) What are the steam and 
 air pressures if r^ = 0.5? (Ans., each 1 lb.) In the latter case, 
 how much is the steam superheated? (Ans., 24.32°.) In a cubic 
 foot of such mixture, state the weights of air and steam. (Ans., 
 0.00461 and 0.00289 lb.) The mixture is cooled at constant 
 pressure to 70° F. What will then be its volume? (Ans., 0.903 
 cu. ft.) What weight of steam could be held by this volume of 
 mixture, if saturated? (Ans., 0.001037 lb.) Is it saturated? 
 (Ans., super-saturated.) How much steam must have been con- 
 densed as dew? (Ans., 0.001853 lb.) What weights of water, 
 satvu-ated steam and dry air are now present? (Ans., 0.001853 
 lb. water, 0.001037 lb. dry steam, 0.00461 lb. air.) How much 
 heat above 32° is contained in the water and steam? (Ans., 
 1.2005 B.t.u.) 
 
CHAPTER V 
 EFFICIENCY AND POWER OF STEAM ENGINES 
 
 Efficiency 
 
 93. Prediction of Efficiency. Following Art. 82, if an engine 
 uses W^ lb. of steam per hr. while developing M Ihp., its actual 
 
 INDICATED THERMAL EFFICIENCY is 
 
 _ 2545 2545M 
 
 ]^(Qi->2) 
 
 where Qi = total heat above 32° of steam at the throttle condi- 
 tion (wet, dry or superheated) and hi = heat of liquid at the 
 pressure existing in the exhaust pipe. Let e denote the efficiency 
 of the Rankine (complete expansion) cycle under the conditions 
 Qi, hi. Then the relative efficiency, or ratio of actual and 
 ideal efficiencies, is 
 
 ^ie = ^o -^ e. (55) 
 
 The heat rate of the actual engine is 
 
 MM ^^^ ~ ^^^ ^-t-"- per Ibp. min. 
 The work done by it per lb. of steam is 
 
 2545 ^ 
 BaiQi - h) = ^^r-^r-^B.t.u. 
 
 Prob. 183. Under the conditions of Prob. 158, a certain 
 compound engine uses 13.3 lb. of steam per Ihp.-hr. Find the 
 actual efficiency, relative efficiency, heat rate and work per lb. 
 of steam. (Ans., 0.17: 0.55: 250: 190.) 
 
 94. Probable Values. A rough estimate of the probable 
 economy of a new engine for definite steam conditions may be 
 formed by using the following values of 6^: 
 
 118 
 
Efficiency and Power of Steam Engines 
 
 119 
 
 
 simple 
 
 Compound 
 
 Triple 
 Expansion 
 
 Saturated steam, non-condensing .... ... 
 
 0.6 
 0.4 
 0.65 
 
 0.65 
 
 0.5 
 
 0.65 
 
 
 Saturated steam, condensing 
 
 0.6 
 
 Highly superheated steam 
 
 
 
 
 Best Recorded Performances, Standard Engines 
 
 steam Superheated 
 Saturated Steam, Above 150" F., 
 
 Lb. Per Ihp.-br. B.t.u. per Ihp.-min. 
 
 Simple non-condensing 21 300 
 
 Simple condensing 16.3 226 
 
 Compound non-condensing 16 245 
 
 Compound condensing 12 200 
 
 Triple-expansion condensing lOJ 190 
 
 Records as good as these are rarely attained. The following 
 steam rates may be realized with saturated steam by engines in 
 good condition and operating at their most favorable loads: 
 
 single 
 Valve 
 
 Double 
 
 Valve* 
 
 Four 
 Valvet 
 
 Four Valve, 
 Releaslngt 
 
 Simple non-condensing . . . 
 
 Simple condensing 
 
 Compound non-condensing 
 Compound condensing . . . . 
 
 33 
 
 27 
 
 25i 
 
 20 
 
 30 
 23 
 23 
 
 17 
 
 29 
 21i 
 22 
 15 
 
 26 
 21i 
 22 
 15 
 
 These rates will be surpassed with very high steam pressures 
 or exceptionally good vacuum. 
 
 95. Indicator Diagram vs. Rankine Cycle. So far as efficiency 
 is concerned, the Rankine cycle, which represents the operation 
 of the whole power plant, establishes a limit for the engine 
 specifically. A still closer limit is found in the " incomplete 
 expansion " cycle, ABCDF, Fig. 28. The actual diagram from 
 the engine appears relatively as abceg. The rounding of corners 
 in the actual diagram is due to slow movements of valves. The 
 reduced admission pressure and augmented back pressure are 
 caused by friction of steam flowing through valve ports and 
 
 * Separate cut-off control. 
 
 t Admission, cut-off, release and compression independently controlled. 
 
 t Automatic disengagement causing very sharp cut-off. 
 
120 
 
 Thekmodynamics, Abridged 
 
 passages. Clearance exists as a necessary evil in the actual 
 engine. It is relatively large in engines of small size. 
 
 The operation of compeession, eg, is chiefly introduced for 
 its cushioning effect, to promote quiet and smooth running. 
 Compression is greatest in high-speed engines. From an eco- 
 nomical standpoint its duration should probably never exceed 
 10 per cent, of the stroke. 
 
 P 
 
 ^Pressure at ThrotHe g 
 
 Rath ofExpansion=%or^ 
 
 ^3 "b 
 
 H 
 
 "^X fPoinf of Compression 
 ^^e ^ Exhaust 
 
 ''-Pressure of Atmosphere or Condenser 
 
 Fig. 28. Action of Steam Engine. 
 
 The lines he and BC are unlike in slope and position. In 
 particular, the volume at h is much less than that at B. This is 
 due to CYLINDER CONDENSATION, an important factor in engine 
 economy. If the cylinder were covered with a perfect heat- 
 insulating material, there would nevertheless be a rapid alterna- 
 tion of temperature of its metal walls, according to the tempera- 
 ture of the steam in contact with them. The walls are heated 
 during admission and the early part of expansion and cooled 
 during the late expansion and exhaust. They abstract heat from 
 the incoming steam very rapidly, so that at the point of cut-off 
 the steam is often less than 0.60 dry. This accounts for the 
 
Efficiency and Power of Steam Engines 121 
 
 small volume at b. Late in the expansion stroke, the steam has 
 cooled off and the walls of the cylinder now supply it with heat. 
 This raises the right-hand end of the expansion curve so that 
 be is less " steep " than BC. The restoration never compen- 
 sates for the initial loss. The availability of heat for doing 
 work depends on its temperature (Art. 36). Low-temperature 
 heat is worth less than the same quantity of high-temperature 
 heat. 
 
 The curve be is therefore not adiabatic. It may be represented 
 by an equation pF" = const., a rough average value of a being 
 1.0. This does not make it isothermal, the saturated steam 
 isothermal being a line of constant peessuke. For non- 
 condensing unjacketed engines, a = (%+ 0.614) -r- 1.258, nearly. 
 
 Condensation is accentuated by high ratios of expansion, 
 which increase the magnitude and duration of the temperature 
 alternations. It is apt to be excessive at slow speeds. It is 
 large in small engines, because their ratio of surface to volume 
 is large. It is reduced in multiple-expansion engines (Art. 
 100). 
 
 Prob. 184. What steam consumption might be expected from 
 the tabular value of e^ in a simple condensing engine under the 
 conditions of Prob. 183? (Ans., 18.1.) How much better than 
 this was the performance of the best standard simple condensing 
 engine? (Ans., 10 per cent, better.) What is the chief reason 
 why an average good engine uses 21| lb.? (Ans., it uses a lower 
 steam pressure.) 
 
 Prob. 185. In Fig. 28, will the rounding of corners be most 
 pronounced at light loads or heavy loads? Will the differences 
 of pressure between ab and AB, ce and DF, be greater at slow 
 speeds or at high speeds? Why should compression be greatest 
 in high-speed engines? 
 
 Prob. 186. What is the value of % when that of a for the curve 
 
 6c, Fig. 28, is 1.0? (Ans., 0.644.) What then is the approximate 
 
 value of Vb -^ Fs? (Ans., 0.644.) What is a when xt = 1.0? 
 
 (Ans., 1.28.) 
 9 
 
122 Thermodynamics, Abridged 
 
 Prob. 187. Assume that condensation varies inversely with 
 speed. The condensation loss is 10 per cent, at 300 r.p.m. 
 What will it then be at 100 r.p.m? (Ans., 30 per cent.) 
 
 Prob. 188. The surface of a cylinder is irdL + {irl2)d? 
 = Td{L + (d/2)) and its volume is {■n:jA)d?L. The ratio of 
 surface to volume is therefore 4/d + 2/X. If the actual con- 
 densation loss is proportional to surface and the weight of steam 
 used is proportional to volume, state a ratio expressing relative 
 condensation loss per lb. of steam used. (Ans., 4/rf + 2/i.) 
 Compare condensation losses in 10 by 20 and 40 by 60 in. cyl- 
 inders, all other factors being disregarded. (Ans., the condensa- 
 tion loss per lb. of steam in the large cylinder is only 26.7 per 
 cent, of that in the small.) 
 
 96. Ratio of Expansion. Values used for ratio of expansion 
 in practice are less than those of highest ideal efficiency, because 
 of condensation. We are limited by condensation to com- 
 paratively low expansion ratios. Values from 3 to 5 are common 
 in simple engines, and up to 36 in compounds. Too high a ratio 
 leads to large size of cylinders and excessive cost {pm being low) 
 as well as to excessive condensation. In constant speed engines 
 the ratio of expansion decreases with increased power. The 
 efficiency increases with increase of power up to about the 
 designed load (Fig. 29). The engine should give its best effi- 
 ciency at the load usually carried. Fig. 29 gives several curves 
 of steam rate against output or ratio of expansion. All show 
 poor economy at high ratios (light loads). Practically all show 
 poor economy at very low ratios (heavy loads). The low points 
 of the curves are the best operating points. A flat curve is 
 particularly desirable where the load is variable. 
 
 97. Jackets. The steam jacket consists of a hollow space 
 surrounding the cylinder, filled with high-pressure steam. By 
 keeping the cylinder walls hot at all times, it prevents or dimin- 
 ishes condensation. The steam condensed in the jacket is of 
 course chargeable to the engine {Qj — h B.t.u. per lb., where 
 Qj = total heat above 32° of steam supplied, h = heat of liquid 
 
Efficiency and Power of Steam Engines 
 
 123 
 
 at the temperature of the jacket drip), but there is usually a 
 net gain. Values of e^ tabulated in Art. 94 may be increased 
 0.03 to 0.05 if jackets are employed. The jacket is typical of a 
 large class of devices which permit of the saving of fuel by the 
 expenditure of money. It is therefore to be used where fuel 
 
 34r 
 
 50 60 70 80 90 100 
 Percent of Rated hp. 
 High-* — Ratio of Expansion— >-Low 
 
 Fig. 29. Variation of Economy with Ratio of Expansion: Constant Speed 
 
 Engine. 
 
 saving is most important: that is, where fuel is costly, or the 
 space which it occupies is valuable, or the load steady. The 
 maximum percentage gain due to jacketing will be realized if 
 the ratio of expansion is high and the speed low. The reason is 
 that these conditions would lead to excessive cylinder condensa- 
 tion if jackets were not used. Compression may be especially 
 small in jacketed engines. 
 
 Prob. 189. In Fig. 29, state the per cent, of rated hp. at 
 which best economy is attained, in engines A and B. (Ans., 
 90 and 103.) 
 
 Prob. 190. The Rice and Sargent engine in Fig. 29 was rated 
 at 700 hp. What weight of steam would it use per hr. when 
 developing 490 hp.? (Ans., 6700 lb.) 
 
124 
 
 Thermodynamics, Abridged 
 
 Prob. 191. Which engine in Fig. 29 has the lowest steam rate 
 at normal load? Which at 70 per cent, of normal load? At 125 
 per cent.? 
 
 Prob. 192. Under the conditions of Prob. 158, an engine 
 uses 12 lb. of cylinder steam and 1 lb. of jacket steam, both per 
 Ihp.-hr. The jacket supply is dry steam at 240 lb. pressure and 
 its drain temperature is 386°. How many B.t.u. per hr. are 
 chargeable as cylinder steam? (Ans., 13481.2.) As jacket 
 steam? (Ans., 841.7.) Total? (Ans., 14322.9.) What is the 
 actual efficiency? (Ans., 0.178.) The steam rate? (Ans., 13 
 lb.) The relative efficiency? (Ans., 0.578.) 
 
 98. Superheat. The thermodynamic advantage of superheat 
 is suggested by Fig. 27. In practice, the steam which expands 
 in the cylinder is usually not superheated, unless superheat 
 
 exceeding 100° is provided at 
 the boiler. The reason is 
 that the cold cylinder walls 
 cool the steam during admis- 
 sion. Superheat below 100° 
 or 150° simply mitigates con- 
 densation. Higher superheat 
 gives the additional advantage 
 of a more efficient cycle. In 
 Fig. 28, if ABCDF is the ideal 
 cycle and abceg the actual 
 cycle realized with saturated 
 steam, superheat (even slight 
 superheat) moves the point b 
 toward L and gives an ex- 
 pansion curve like LM, " flat- 
 ter " than the curve be. Since 
 with superheat we should use a larger ratio of expansion, the 
 actual curve, representing the behavior of reduced weight of 
 steam, lies like b'c'. The mean effective pressure is then less 
 when superheat is used. Fig. 30 shows the effect of low super- 
 
 1 
 
 0) 
 Q_ 
 
 2.1 
 2.0 
 
 
 
 V^ 
 
 
 
 
 
 
 
 
 
 
 J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 r 
 
 ^ 
 
 
 
 
 
 
 
 
 
 \ 
 
 h? 
 
 ^ 
 
 
 
 
 J 
 
 
 
 
 
 \ 
 
 ^\l 
 
 ^ 
 
 u 
 
 u 
 
 / 
 
 
 
 
 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 ^7 
 
 y 
 
 
 
 
 k 
 
 
 
 
 v 
 
 ^\ 
 
 H, 
 
 
 c\^; 
 
 — 
 
 
 
 
 
 \ 
 
 f^ 
 
 I' 
 
 k 
 
 ^ 
 
 Y 
 
 5 1.7 
 
 -Ql.6 
 
 '■^ 10 40 60 80 100 
 Percentage of Full Load 
 
 Fig. 30. Effect of Low Superheat 
 (72° to 92°) on Overall Economy. H. 
 M. S. Brittania. 
 
Efficiency and Power of Steam Engines 125 
 
 
 \ 
 
 
 
 
 
 
 
 
 248 
 
 \ 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 c 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 im 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 '(■224 
 
 3 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 m2l6 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 S 
 
 
 
 208 
 
 
 
 
 
 
 s 
 
 
 
 
 
 
 
 
 
 ^ 
 
 
 ?nn 
 
 
 
 
 
 
 
 
 '^ 
 
 50 100 150 200 250 300 350 400 
 Superheat, deg. F. 
 
 Fig. 31. Relation Between Efficiency 
 and Superheat. Compound Condens- 
 ing Engine. 
 
 heat. Fig. 31 gives average results at low and high superheat 
 from tests of 40 famous engines. Superheat exceeding 200° 
 may be expected to produce 
 a saving of 20 per cent, in 
 simple engines and 16 per 
 cent, in compounds. The 
 simple engine with super- 
 heated steam is often as 
 economical as the compound 
 with saturated steam. The 
 conditions which make su- 
 perheat desirable are the 
 same as those which favor 
 the use of jackets (Art. 97). 
 The two should not be used 
 together. Low compressions 
 should be used when superheat is employed. 
 
 Prob. 193. From Fig. 31, what is the gain by using 400° 
 superheat as compared with 100°? (Ans., 15 per cent.) 
 
 Prob. 194. In Fig. 30, what was the load of best economy? 
 (Ans., 70 per cent, of full load.) With saturated steam, how 
 much more total coal is used per hr. at full power than at best 
 economy? (Ans., 54 per cent.) Note that in warships engines 
 are designed for best economy at reduced power and speed. 
 
 99. Speed. If an engine duplicated its own indicator diagram 
 continuously, the mean effective pressure would be constant, 
 and the ihp. would be directly proportional to the r.p.m. In 
 single-valve engines with fixed cut-off, the valve action at high 
 speeds is inferior. Hence the amount of steam used tends to 
 increase somewhat more rapidly than the ihp. This is just about 
 offset by the favorable influence of high speed on cylinder con- 
 densation. The higher the speed, the less time is available 
 for transfer of heat. From this standpoint, the steam consump- 
 tion should increase a little less rapidly than the Ihp., as the 
 speed increases. As a final result, when the total hourly steam 
 
126 
 
 Thekmodynamics, Abridged 
 
 consumption is plotted against Ihp., the resulting curve is (for 
 
 engines of this type) a straight line. The steam rate (lb. per 
 
 Ihp.-hr.) is therefore best 
 (lowest) at high speeds. 
 
 In marine engines, with re- 
 versing valve gears, cut-off is 
 delayed at high speed, which 
 is the full-power condition. 
 The power therefore increases 
 faster than the speed (Fig. 
 32). The most economical 
 point of cut-off, and of the 
 ratio of expansion, are fixed 
 at maximum speed for mer- 
 chant vessels and at reduced 
 cruising speed for warships. 
 For the former, the steam 
 rate per Ihp. hr. should de- 
 crease as the speed increases. 
 
 For the latter, it should vary somewhat like the coal rate of 
 
 Fig. 30. 
 
 Prob. 195. An auxiliary engine uses 1730 lb. of steam per hr. 
 when developing 100 hp. and 11 730 lb. when developing 500 hp. 
 The curve of hourly steam against hp. is a straight line. What 
 is the steam rate when developing 300 hp.? (Ans., 22.43 lb. 
 per hp.-hr.) 
 
 Prob. 196. In Fig. 32, compare mean effective pressures at 
 90 and 120 r.p.m. (Ans., pm at the low speed is 51 per cent, of 
 that at the high speed.) 
 
 100. Multiple Expansion. By partially expanding the steam 
 in one cylinder and then conducting it to another, the alternations 
 of temperature in any one cylinder are reduced. Hence con- 
 densation is diminished,* and a much larger total ratio of ex- 
 
 *The temperature range in any cylinder is halved, but the total size 
 (surface) of cylinders is not doubled. 
 
 IZO 130 
 
 Fig. 32. Indicated Power of an Ar- 
 mored Cruiser at Various Speeds. 
 
Efficiency and Power of Steam Engines 127 
 
 pansion may be utilized with advantage. These ratios range up 
 to 30 or more in compound (two-stage) and triple expansion 
 engines. They are rarely over 5 in simple engines. There is a 
 corresponding gain in economy. The compound will have a 
 steam rate 20 to 30 per cent, lower than the simple engine. 
 The triple may be slightly better still. Four stages (quadruple 
 expansion) are scarcely to be considered for naval service. The 
 number of cylinders is often greater than the number of expansive 
 stages, in order to avoid the use of very large low pressure 
 cylinders. Compound engines usually have small compression. 
 They should always have high initial pressure and should run 
 condensing. The reheater is a vessel placed between the 
 cylinders. Steam passes through it on the way from one cylinder 
 to the next. It is equipped with coils containing high-pressure 
 steam. This dries and often superheats the steam which is on 
 its way to the second stage. The heat supplied is computed in 
 the same way as the heatsupplied at a jacket. The argument 
 in favor of the use of a reheater is the same as that for the use 
 of superheat in any cylinder. 
 
 Prob. 197. From the last table in Art. 94, state the per- 
 centage^ gain by compounding in four valve engines, (a) con- 
 densing, (6) non-condensing. (Ans., (a), 30: (6) 24.) 
 
 101. Quality of Exhaust Steam. The heat supplied to an 
 engine, per lb. of steam, is Qi — hi: Art. 93. The work done is 
 
 2545 2545>Ohp. 
 TTT _^ ^ = s? B.t.u. per lb. oi steam. 
 
 The heat rejected is the difference of these two quantities. 
 It is equal (ignoring radiation) to {xL)^, where the subscript ^ 
 refers to the condition of steam in the exhaust pipe. Then 
 
 2545 
 {xL)e= Qi-hi- ^^^ j^ . (56) 
 
 These relations are important in condenser design. The con- 
 denser may however absorb more heat than that indicated by 
 Equation (56), because it may cool the condensation. 
 
128 Thermodtnamics, Abridged 
 
 Prob. 198. In Prob. 183, how much of the heat in each lb. 
 of steam is rejected to the condenser? What is the dryness of 
 the steam in the exhaust pipe? (Ans., 933.44 B.t.u.: 0.90.) 
 If the condenser cools the condensed steam to 90°, how much 
 additional heat does it remove? (Ans., 11.8 B.t.u. per lb.) 
 
 102. Desirable Back Pressure. Even from a purely thermo- 
 dynamic standpoint there is a limit to the condenser vacuum 
 
 desirable in simple engines. 
 In Fig. 33, fhcde is an in- 
 complete expansion cycle. 
 Consider the effect of lower- 
 ing the back pressure, so 
 that the cycle becomes 
 abcdg. The gain of work is 
 afeg. The increased cost of 
 heat is hf — ha- The value 
 of afeg is from the pV dia- 
 gram, practically, («e/5.403) 
 
 Fig. 33. Desirable Back Pressure. (Pf - Va) B.t.U. Now for 
 
 engines of any importance, Vc 
 is never greater than 4. This volume is the volume of dry (or 
 nearly dry) steam at the throttle pressure. In simple engines, the 
 ratio of expansion is practically never over 5. Now Ve is equal 
 to Vc multiplied by the ratio of expansion. Hence a maximum 
 value of Ve is 20, and a maximum value of afeg is S.7{pf — pa), 
 in simple engines. If pf = 2, pa = 1, afeg = 3.7 B.t.u. But, 
 then, hf = 94, Ji^ = 69.8, heat cost = 94 - 69.8 = 24.2 B.t.u. 
 The efficiency at which we have gained the additional area afeg is 
 then 3.7 -^ 24.2 = 0.152. This may easily be less than the 
 efficiency of the original cycle fhcde: hence the lower back 
 pressure (improved vacuum) will lower the efficiency of the 
 whole cycle. Simple engines should operate at 2 lb. back 
 pressure (26 in. vacuum). With compound engines, the ratio 
 of expansion is much greater and a better vacuum is desirable. 
 
 Prob. 199. A compound engine uses dry steam at 250 lb. 
 pressure and a ratio of expansion of 30. Find v^. (Ans., 55.5.) 
 
 T . 
 
 €c 
 
 
 \ 
 \ 
 
 d\ 
 
 \ 
 
 ri 
 
 \ 
 
 n 
 
Efficiency and Power of Steam Engines 
 
 129 
 
 How much work is gained per lb. of steam by lowering the back 
 pressure from 2 lb. to 1 lb.? (Ans., 10.3 B.t.u.) At what 
 efficiency is this quantity of work derived? (Ans., 0.42.) 
 
 103. Results by Special Engines. The Uniflow engine has 
 unidirectional flow of steam, which is supposed to reduce con- 
 densation. There is no exhaust valve, the long piston uncovering 
 slots in the cylinder wall when near the end of its stroke. Fig. 34 
 
 Fig. 34. Uniflow Engine. 
 
 shows an engine with Corliss valves. Poppet valves are com- 
 monly used. The engine is not well adapted for running non- 
 condensing on account of the excessive compression which then 
 results. A simple unjacketed engine of this type has given the 
 remarkable steam rate of 9.06 lb. with 155 lb. pressure and 667° 
 steam temperatiu-e (implying considerable superheat) . 
 
 The binary vapor engine uses two fluids. Steam is used 
 in an ordinary cylinder. It then passes to a condenser, in which 
 the cooling fluid is not water, but a volatile liquid (ether or SO2). 
 This " binary fluid," evaporated in the steam condenser, is 
 used in a second cylinder and then discharged to a surface con- 
 denser employing water as a cooling agent. The maximum 
 efficiency is determined by the cooling water temperature and 
 is ideally somewhat higher than that attainable by steam alone 
 with present vacuum apparatus. Actual tests have given a heat 
 rate around 167 B.t.u. per Ihp.-min. 
 
 Regenerative engines have occasionally been built. Fig. 
 35 illustrates the action in four stages (quadruple expansion). 
 
130 
 
 Thermodynamics, Abridged 
 
 The path cd represents adiabatic expansion in the first cylinder. • 
 The steam then passes to a reheater (Art. 100) on its way to the 
 second cylinder. A portion of the steam is drawn off from 
 
 Fig. 35. Quadruple Expansion Regenerative Engine. 
 
 this reheater and employed to heat boiler feed water. We may 
 regard the heat withdrawn as the area under de, and the heat 
 transferred to the feed water as the (ideally equal) area under ga. 
 Similarly after expansion in the second cylinder, the heat under 
 hf becomes the heat under hg: and after a third expansion the 
 heat under Ij becomes the heat under pfc. With an infinite 
 number of stages, the equivalent expansion would be cq, the 
 heat under which would equal the heat under p6. This is a 
 REGENERATIVE CYCLE, like that in Art. 62, and the only heat 
 chargeable is that under he. The efficiency is ideally ( jTfc— Tp) j Tb, 
 that of the Carnot cycle. Actual tests gave heat rates as low as 
 169 B.t.u. per Ihp.-min. 
 
 Prob. 200. A Uniflow engine has 2 per cent, clearance and 
 opens its exhaust ports at 95 per cent, of the stroke. If the 
 compression curve is pv = const., find the pressure at the end 
 of the compression, the exhaust pressure being 1 lb. What is 
 it if the exhaust pressure is 15 lb.? (Ans., 48| lb., 727| lb.) 
 
Efficiency and Power of Steam Engines 131 
 
 Prob. 201. A Uniflow engine uses 9.5 lb. of steam per Ihp.-hr. 
 at 150 lb. pressure, superheated 300°. The back pressure is 
 0.505 lb. Find the heat rate. (Ans., 206 B.t.u. per Ihp.-min.) 
 Is any other value as low as this for a simple engine given in 
 this chapter? 
 
 Prob. 202. What is the ideal Caenot eflBciency for steam at 
 200 lb. initial pressure (dry) at 1 lb. back pressure? (Ans., 
 0.332.) If by using a binary vapor the rejection temperature is 
 reduced to 75° F., what is then the Carnot efficiency? (Ans., 
 0.363.) If the relative efficiency as compared with the Carnot 
 CYCLE is 0.45, what is then the heat rate? (Ans., 258 B.t.u. 
 per Ihp.-min.) 
 
 Prob. 203. Prove, in Fig. 35, that pbr = qcs. Prove that the 
 efficiency is that of the Carnot cycle. 
 
 Power or Capacity 
 104. Mean Effective Pressure, Simple Engine, Saturated 
 Steam. Methods of finding pm for ideal cycles have been given 
 in Arts. 80, 81, 84, 85, 87. The usual method is based on the 
 diagram ABCDF, Fig. 28, the curve BC being assumed to be, 
 PbVb = pcVc- Then 
 
 STF(ft.lb.) f jr , J, , Vc „ \ „ 
 
 P"^ = 1441^0 ^ yPBVB+PBVBl0gey--pDVD j^Vc 
 
 = PB-p^l l + l0ge|F^ I-PC (57) 
 
 in which pa = initial pressure, Pd = back pressure, Vc/Vb 
 = ratio of expansion. The maximum overload capacity (with 
 no cut-off) is 
 
 AHDF - ABCDF _ Pb-Pd- Pr. 
 
 ABCDF Pm • ^^^^ 
 
 Values of pm from Equation (57) are multiplied by a diagram 
 FACTOR, /, (sometimes called the " card factor ") to give the 
 probable mean effective pressure in the actual engine. Note that 
 / = abceg -^ ABCDF. Values of / range from 0.6 to 0.9, de- 
 
132 
 
 Thermodynamics, Abeidged 
 
 pending chiefly on the valve gear. Jackets increase / by 0.05 
 to 0.15. Its value is decreased by high ratios of expansion, 
 speeds above 225 r.p.m., excessive clearance, and inadequate 
 valve passages. 
 
 Piston speeds are usually between 500 and 1000 ft. per min. 
 The allowable r.p.m. depend chiefly on the valve gear. Long 
 strokes favor low clearances. See Art. 7. 
 
 Prob. 204. With an initial pressure of 115 lb., a back pressure 
 of 2 lb. and a ratio of expansion of 4.5, find pm and ideal overload 
 capacity. If / = 0.85, piston speed = 800, r.p.m. = 100, find 
 the cylinder diameter and stroke, the engine being rated at 500 
 Ihp. (Ans., j)m = 62, ideal overload capacity = 82 per cent., 
 diameter = 22.3 in., stroke = 48 in.) 
 
 105. Superheated Steam. For superheated steam, the expan- 
 sion curve BC, Fig. 28, is " steeper " than for saturated steam. 
 Its equation is PbVb" = PcVc", where a may have values about 
 as follows: 
 
 
 Superheat, "F. 
 
 Ratio of EJcpanslon 
 
 200 
 
 2S0 
 
 300 1 350 1 400 
 
 450 
 
 SCO 
 
 
 Values of Exponent a 
 
 8.0 
 
 1.05 
 
 1.07 
 
 1.09 
 
 1.11 
 
 1.13 
 
 1.15 
 
 1.18 
 
 7.0 
 
 1.06 
 
 1.08 
 
 1.10 
 
 1.12 
 
 1.14 
 
 1.16 
 
 1.19 
 
 6.0 
 
 1.07 
 
 1.09 
 
 1.11 
 
 1.14 
 
 1.16 
 
 1.18 
 
 1.21 
 
 5.0 
 
 1.08 
 
 1.10 
 
 1.12 
 
 1.15 
 
 1.17 
 
 1.19 
 
 1.22 
 
 4.0 
 
 1.09 
 
 1.11 
 
 1.13 
 
 1.16 
 
 1.18 
 
 1.20 
 
 1.23 
 
 3.0 
 
 1.10 
 
 1.12 
 
 1.15 
 
 1.17 
 
 1.20 
 
 1.22 
 
 1.25 
 
 The mean effective pressure is now 
 
 Pb- 
 
 Vb 
 
 a (Vb\ 1 
 
 (59) 
 
 This gives a lower value than Equation (57) — see Art. 98 — but 
 (as with jackets) the value of / is higher. With slight super- 
 heat, use Equation (57) for p„ but use high values of / in con- 
 nection therewith. 
 
Efficiency and Powee of Steam Engines 133 
 
 Prob. 205. Under the conditions of Prob. 204, what is the 
 value of pm for steam superheated 100°? (Ans., 62.) If the 
 steam is superheated 300°, what is the probable value of a? 
 Of p„? (Ans., a = 1.125, pm = 59.) 
 
 106. Compound Engine: Ratios. In the preliminary propor- 
 tioning of a compound engine, it is assumed that the combined 
 diagrams of the two stages will appear as ABCDF, Fig. 28, and 
 that the steam exhausted from the high-pressure cylinder along 
 JK enters the low pressure cylinders along KJ without fall of 
 pressure or change of volume. The curve BC is assumed to be 
 (with saturated steam) PbVb = PcVc- The ratio of cylinder 
 volumes (cylinder ratio) is C = Vc/Vj = pjjfc- For cyl- 
 inders of equal strokes (the usual case), C = Ai/Ah, where Ai 
 and Ah are the cross-sectional areas of low and high pressure 
 cylinders. But pc = Ps/r, where r = whole ratio of expansion. 
 Then C = rpjjpB- The receiver pressure (receiver between 
 the cylinders) is pj. The ratio of expansion in the high pressure 
 cylinder is r% = Vj/Vb = Pb/Pj- That in the low-pressure 
 cylinder is Vc/Vj = C. 
 
 There are five methods of establishing a value for pj. 
 
 (1) It may be assumed. 
 
 (2) A value may be assumed for C. Then pj = Cpb/t. 
 
 (3) The temperature ranges may be equalized in the two 
 stages. Then tB — tj = tj — to, tj = {Ib + <d)/2. Values of 
 Ib and to are those from the steam table, corresponding with the 
 pressures pa and po- The pressure pj is then the tabular 
 pressure for tj. 
 
 (4) Maximum total piston pressures may be equalized, i.e., 
 
 Pb — Pj= C{pj - Pd) 
 
 Pj= Pb- C{pj - Pd) 
 
 = Pb--- (Pj - Pd) 
 Pb 
 
 = pb^ -^ (Pb + rpj — rpn). (60) 
 
134 Thermodynamics, Abridged 
 
 This is the important method for marine practice. 
 (5) The two stages may develop equal power. Then ABJK 
 = KJCDF, or 
 
 ps ,- . „ - Vc 
 
 Vj 
 
 Vb 
 
 Vb 
 
 VbVb loge^ = VjVj + VjVj loge y- " VdV D 
 
 Vj 'J 
 
 loge Vb - loge Pj= 1 + loge C - r- 
 
 = 1 + loge Vj - loge Vb + loge r- r; 
 
 logeVj=-2[^oger + ^^r- ij— loge^ (61) 
 
 Prob. 206. Given initial and back pressures of 120 and 2 lb., 
 with r = 25, find pj under each of the following conditions: 
 (a) C = 4. (Ans., 19.2.) 
 (6) Temperature ranges equalized. (Ans., 22.) 
 
 (c) Maximum total pressures equalized. (Ans., 22.7.) 
 
 (d) Power division equalized. (Ans., 17.8.) 
 
 107. Approximate Dimensions. After having determined pj 
 by one of the foregoing methods, the mean effective pressures 
 in the high and low pressure cylinders are 
 
 ABJK pb, 
 
 Pmh = TT = — loge Th 
 
 KJCDF Pj,^ , , ^, 
 
 Pml = —y = ^ (1 + loge C) - po 
 
 (62) 
 
 The output of the engine is 
 hp.= 
 
 f^ r A \ A \ /'S^t ( Vmh I \ .„„^ 
 
 {PmhAh -r PmlAl) == go QQQ I ^ -r Pml j , (Od) 
 
 33 000 
 
 from which Ai is determined. The value of/ is about the same 
 as that in a simple engine using a ratio of expansion equal to Vr. 
 The quantity {pmh/C + pmi) is called the total mean effective 
 
 PRESSURE referred TO THE LOW-PRESSURE CYLINDER. ItS 
 
 value is the same as that given by Equation (57), applied to the 
 whole diagram ABCDF. Then 
 
Efficiency and Power of Steam Engines 135 
 
 ^ 33 000 
 
 ^ (1 + logo r)-pD\. (63a) 
 
 It thus appears that the size op the high pressure cylinder 
 IS without influence on the total power of the engine. 
 
 Prob. 207. The initial, back and receiver pressures are 120,2 
 and 17.8. The ratio of expansion is 25. The engine develops 
 500 Ihp. at 800 ft. piston speed and 100 r.p.m. Find cylinder 
 diameters and strokes. Use / = 0.9. (Ans., 20.8 and 40 in. 
 by 48 in.) 
 
 Prob. 208. In Prob. 207, what are the mean effective pressures 
 in each of the two cylinders? (Ans., 33.9 and 9.11.) Show that 
 the two cylinders do develop equal power. 
 
 Prob. 209. If in the foregoing two low pressure cylinders are 
 used, state dimensions. (Ans., all strokes 48 in.: diameters, 
 high pressure, 20.8 in. ; two low pressure, 28.5 in.) 
 
 Prob. 210. Given p^h = 35, pmi = 9, C = 4, in a compound 
 engine. What proportion of the total power is developed in each 
 cylinder? (Ans., high pressure 49.2 per cent., low pressure 50.8 
 per cent.) 
 
 108. Cylinder Ratio and Drop. The value of C has not been 
 thoroughly standardized. It should increase with r. High 
 values increase the initial steam pressure in the second stage. 
 The margin of overload capacity, obtainable by admitting 
 reduced pressure steam from the boilers to the low-pressure 
 cylinder, is thus decreased. Values up to 6 or 7 have given 
 highly efficient engines. The ideal diagram of Fig. 28 shows 
 terminal drop, CD, in the low pressure cylinder but none in 
 the high pressure cylinder. Drop (high pressure drop) is often 
 provided for. This reduces the value of / about 0.05. In 
 general practice, the mean effective pressure referred to the low 
 pressure cylinder {pmh/C + p^i) is kept around 1.2 + 0.12pB. 
 
 Prob. 211. Following the rule in Art. 108, what would have 
 been the diameter of the low pressure cylinder for 500 Ihp. at 
 800 ft. piston speed with/ = 0.9? (Ans., 43 in.) 
 
136 Thermodynamics, Abbidged 
 
 109. Compounds with Superheat. For superheat below 150°, 
 the foregoing method may be used, increasing / as when jackets 
 are employed. With high superheat, take the value of a in 
 Art. 105 which would be suitable for a ratio of expansion equal 
 to Vr in a simple engine. Then, following Equation (59) 
 
 Pja Pj ^ 
 
 P""' ~ C{a - 1) C»(a - 1) P"" 
 
 Vj ( 0- 1 \ 
 Vb fa 1 \ 
 
 (64) 
 
 In this case pj = pBiC/r)" and Vh = r/C. If the work of the 
 two cylinders is to be equally divided, it will be found that 
 
 The horse power is 
 
 , fSAi fpmh I \ 
 
 P- ~ 33 000 V C '*' P""^ J ' 
 
 where the quantity in the parenthesis has the value given by 
 Equation (59). 
 
 Prob. 212. For r = 25 at 300° superheat, what value of a 
 should be used? (Ans., 1.12.) If pe = 120, pn = 2, pj = 20, 
 find C. (Ans., 5.05.) Find the value of C for equality of work. 
 (Ans., 5.2.) What is then the value of pjt (Ans., 20.6.) 
 Under the latter condition, find r^, pmh, Pmi- (Ans., r^ = 4.8, 
 Pmh = 40|, pmi = 7.85.) What is the value of the total mean 
 effective pressure referred to the low pressure cylinder, as deter- 
 mined by Equation (59)? (Ans., 15.7.) 
 
 110. Triple Engines: General Case of Multiple Expansion. 
 
 For any multiple expansion engine using saturated steam (or 
 steam superheated less than about 150° F.), the size of the low 
 pressure cylinder is given by Equation (63a). The sizes of the 
 other cylinders, for zero drop, are then determined from assumed 
 
Efficiency and Power of Steam Engines 137 
 
 values of cylinder ratios or receiver pressures. Their sizes 
 influence efficiency but not power. 
 
 With high superheat, similarly, following Equation (59), 
 
 , fSAi /■pB a Vb _ \ 
 
 P' 33 000 V »• ' a - 1 r\a - 1) ^'^ ) 
 
 (65) 
 
 If h is the number of expansion stages, take a from Art. 105 for 
 a ratio of expansion of \r. 
 
 Prob. 213. The battleship Oklahoma uses triple engines at 
 280 lb. initial absolute pressure. Assuming 1 lb. back pressure, 
 and 10 000 Ihp. capacity at cruising speed when r = 30, with 
 / = 0.9, what would be the diameter of each of the two low 
 pressure cylinders at 1000 ft. piston speed, using saturated steam? 
 (Ans., 76 in.) 
 
 Prob. 214. The cylinder ratios in the Oklahoma are 1 : 2.84 : 
 9.93 (high — intermediate — low). Find the diameters of high 
 and intermediate cylinders under the conditions of Prob. 213. 
 (Ans., 34 and 64 in.) 
 
 Prob. 215. Under the conditions of Prob. 213, but with 
 steam superheated 300°, what is the probable value of a? (Ans., 
 1.15.) What is then the diameter of the low pressure cylinder? 
 (Ans., 84| in.) 
 
 10 
 
CHAPTER VI 
 VAPOR REFRIGERATION 
 
 111. Ideal Cycle. The ideal cycle in vapor refrigeration is a 
 Rankine cycle worked counter-clockwise (Fig. 36). Assume 
 the fluid to be at low temperature and pressure, somewhere on 
 the line eb, but nearer e than b, i.e., in the condition of a very 
 wet vapor. It is contained in coils in the room or tank to be 
 cooled, the contents of such room or tank being warmer than 
 the fluid. Heat will flow from such contents to the fluid. This 
 will not increase the temperature of the fluid, but will vaporize 
 whatever liquid it contains. 
 
 When the vaporization is nearly (but not quite) complete, 
 allow the fluid to flow out to a compressor. Here it is com- 
 pressed adiabatically until its temperature is above that of an 
 available supply of cooling water. The operation is indicated 
 by the path be. The pressure will rise, as well as the temperature. 
 The fluid now passes to the condenser, where it is subjected to 
 the cooling effect of water in the coils. This water condenses 
 the vapor (path cd) but in general cannot cool it much. The 
 final operation necessary is a reduction in temperature of the 
 resulting liquid. In air refrigerating machines (Arts. 57, 60) 
 the final operation took place in an expanding cylinder 
 With air, mere expansion from a high pressure to a lower pressure 
 would not produce cooling, unless at the same time work is 
 done. With a vapor, on the other hand, expansion alone will 
 reduce the temperature. Hence the final operation, which in 
 Fig. 36 reduces the temperature of the liquid from td to te, is 
 merely a reduction of pressure accomplished by allowing the 
 fluid to flow through an expansion valve. A high pressure 
 exists from the compressor outlet to the expansion valve inlet. 
 A low pressure exists from the expansion valve outlet through 
 the cold room to the compressor inlet. 
 
 138 
 
Vapor Refrigeration 
 
 139 
 
 112. Choice of Fluid. It is desirable that the high pressure 
 (necessarily the pressure corresponding with a normal cooling 
 
 d 
 
 -P 
 
 ' 
 
 c 
 r 
 
 .k 
 
 
 
 
 
 
 
 120 
 
 
 \ 
 
 \ 
 \ 
 
 
 
 
 
 
 
 80 
 
 \ 
 
 
 \ 
 
 \ 
 
 1 
 
 
 
 
 
 
 \ 
 
 
 
 \ 
 
 \ 
 
 
 
 
 
 
 40 
 
 
 \ 
 
 
 
 N 
 
 
 
 
 
 
 
 \ 
 
 \ 
 
 h 
 
 
 s 
 
 
 . 
 
 
 
 fii 
 
 
 
 
 
 
 
 
 V 
 
 
 
 
 
 
 
 
 
 
 
 
 b 
 
 1 23456789 10 
 
 f 
 
 i 
 
 0.2 
 
 0.8 
 
 0.4 0.6 
 Entropy 
 
 Fig. 36. Vapor Refrigeration with Ammonia. 
 
 J\ 
 
 1.2 
 
 water temperature of say 80° F.) be not too high. It is also 
 desirable that the low pressure (that corresponding with the 
 
140 
 
 Thermodynamics, Abridged 
 
 lowest temperature of the fluid, which for ice-making and general 
 service is about 0° F.), be not too low. The latter pressure 
 should preferably be above the pressure of the atmosphere. 
 If it is lower than this, any leakage will be inward. Leakage 
 cannot then be readily detected, and the fluid will be diluted if 
 leakage occurs. Fig. 20 shows that from the first standpoint the 
 best fluids are ether, ethyl chloride, sulphur dioxide and ammonia. 
 Carbon dioxide, at 80° F., gives a pressure of nearly 1000 lb. per 
 sq. in. From the second standpoint, ether, ethyl chloride and 
 SO2 are all unsatisfactory, giving pressures below that of the 
 atmosphere at 0° F. The fluids mostly used are SO2, ammonia 
 and CO2. The last is most popular in marine service, since it is 
 innoxious. In the following discussion, reference is made 
 particularly to ammonia. This is for two reasons. Its proper- 
 ties are fairly well known. It permits greater flexibility of 
 cycle, because there is in no case any approach to its critical 
 temperature (Art. 65). 
 
 113. Action in the Expansion Valve. A new type of vapor 
 process is here encountered, which will be more fully discussed 
 later (Chapter VIII). The liquid at d, Fig. 36, is converted to 
 fluid at the pressure pe without change of heat content. 
 Since ha > he, the fluid cannot be liquid at the new pressure. 
 It can only be a wet vapor. Its state is /, defined by heat at 
 / = HEAT AT d, or hf + {xL)f = hd- We must regard the process 
 as being represented by def. It is not an adiabatic process. 
 The refrigeration that can be performed by the fluid is then not 
 the area under eh, but only the area under /6. 
 
 Properties of Saturated Ammonia Vapor 
 
 Cfc,' 
 
 p 
 
 V 
 
 
 
 
 
 
 
 
 t ■ 
 
 Pressure, 
 
 Volume, 
 
 
 
 
 
 
 
 
 -^ss 
 
 Lb. per 
 
 Cu. Ft. 
 
 
 
 
 r 
 
 ««. 
 
 m. 
 
 n. 
 
 
 Sq. In. 
 
 per Lb. 
 
 
 
 
 
 
 
 
 
 
 29.95 
 
 9.190 
 
 -33.7 
 
 572.2 538.5 
 
 521.4 
 
 -0.0709 
 
 1.2450 
 
 1.1741 
 
 70 
 
 129.20 
 
 2.296 
 
 42.1 
 
 512.8 555.0 
 
 458.5 
 
 0.0813 
 
 0.9684 
 
 1.0497 
 
 90 
 
 181.80 
 
 1.650 
 
 65.3 
 
 493.5 558.9 
 
 438.9 
 
 0.1238 
 
 0.8981 
 
 1.0219 
 
 95^ 
 
 199.00 
 
 1.512 
 
 71.9 1 488.0 559.9 
 
 433.3 
 
 0.1356 
 
 0.8790 
 
 1.0146 
 
 (From Goodenough's " Properties of Steam and Ammonia") 
 
Vapoe Refeigeration 141 
 
 Prob. 216. Liquid ammonia at 70° expands to 0°. What is 
 the dryness of the resulting vapor? (Ans., 0.132.) What is 
 its entropy? (Ans., 0.0936.) 
 
 Prob. 217. This vapor in Prob. 216 is 0.87 dry when it leaves 
 the coal room. How much refrigeration has been done per lb. 
 of fluid? (Ans., 422 B.t.u.) 
 
 114. Ideal Power and EflS.ciency. In symbols, per lb. of fluid, 
 and in B.t.u., 
 
 Refrigeration done = heat under /6 = Lft,(xb — Xf) 
 
 Heat rejected at condenser = heat under cd — Lcd{xc) 
 
 Work expended = heat at c — heat at 6 = {Ji-\- xL)c — (A + xL)i, 
 
 T7ffl^;^^ Refrigeration Lfb{xb - x,) 
 
 Efiiciency = ^^^^ = ^^ ^ ^^^^ _ ^^ ^ ^^.^^ . (66) 
 
 The last may exceed unity. The value of Xc is found from 
 
 (n„ + xne)c = {riw + xne)h. 
 
 Efficiencies are much higher than with " dense air " machines 
 
 (Art. 60). The mean effective pressure is 
 
 work X 5.403 ^ . ^^^^ 
 
 Vm = ; i : see Equation (67). 
 
 maximum volume 
 
 Prob. 218. The fluid is ammonia. Temperature limits are 
 0° and 70°. The fluid is 0.87 dry at the beginning of com- 
 pression. Find the dryness at the end of compression. (Ans., 
 0.96.) 
 
 Prob. 219. In Prob. 218, how much heat is removed by the 
 circulating water at the condenser? (Ans., 494 B.t.u. per lb. 
 of fluid.) 
 
 Prob. 220. In Prob. 218, how much work is expended? 
 (Ans., 70.8 B.t.u. per lb. of fluid.) What is the ideal efficiency? 
 (Ans., 5.98.) 
 
 115. Capacity, Power and Size. If f = tonnage (Art. 59), 
 the necessary refrigeration is 200 T B.t.u. per min. Then the 
 weight of fluid to be circulated per minute is w = 2002" -^ ee- 
 FRiGERATiON PER LB. and the ideal hp. is (w/42.42) { (A + xL)c 
 
142 Thermodynamics, Abridged 
 
 — {h + xL)b}. The heat rejected at the condenser is wLcdXc 
 B.t.u. per min. A close approximation to the compressor dis- 
 placement may be made as follows: 
 
 The volume of fluid present at the beginning of compression 
 is Vb in the pv diagram. (Lettering on the pv diagram corre- 
 sponds with that on the Tn diagram.) This is substantially 
 (xv)b. Assume clearance = c and assume that the clearance 
 expansion curve is pgVg = phVh- Then Vh = cD{palph), where 
 D = displacement of compressor. The volume of fresh fluid 
 drawn in and compressed is vt — Vh = (1 + c)D — cD(pglph) 
 = D{1 + c{l — (pg/ph))}. The weight, w, is this quantity 
 divided by {xv)i, the specific volume of the wet vapor. The 
 weight being known, the displacement may be found. 
 
 Thus, suppose clearance to be 4 per cent., and Pglph = 5. 
 The volume compressed is Djl + 0.04(1 - 5)} = 0.842). Note 
 that 0.84 is the volumetric efficiency (Art. 42), suction fric- 
 tion being ignored. Suppose the vapor to be 0.90 dry at the 
 beginning of compression, when its temperatmre is 0°. Then 
 ixv)i, = 0.90 X 9.19 = 8.271. Suppose 100 lb. of fluid to be 
 circulated per min. Then the volume to be circulated is 827.1 
 cu. ft. = 0.84D whence D = 983 cu. ft. per min. 
 
 The equation given for mean effective pressure in Art. 114 
 may now be written 
 
 5.m{{h + xL%- {h+xL)b] 
 
 'P- = M. • (67) 
 
 Prob, 221. In Probs. 218-220, the capacity of the machine 
 is 100 tons. Find the weight of fluid per min. and the ideal hp. 
 (Ans., 47.4 lb., 79 hp.) 
 
 Prob. 222. In the foregoing, how much heat must be carried 
 away by the cooling water? (Ans., 23 400 B.t.u. per min.) 
 
 Prob. 223. In Prob. 221, the clearance is 5 per cent., the 
 piston speed 240 ft. per min., the compressor double-acting at 
 60 r.p.m. Find the diameter and stroke of the compressor 
 cylinder. (Ans., 18.6 and 24 in.) What is the mean effective 
 pressure? (Ans., 48 lb. per sq. in.) 
 
Vapor Refrigeration 143 
 
 Prob. 224. In Prob. 221, what is the ideal ice-melting effect 
 per Ihp.-hr.? See Art. 59. (Ans., 106.) 
 
 Actual Perfonnance. In the foregoing illustrative case, the 
 condenser temperature (70°) is lower than may in general be 
 safely assumed. Consequently the ideal efficiency is higher 
 than is to be expected in ordinary plants. A narrow temperature 
 range leads to efficiency in refrigeration. An ideal ice-melting 
 effect (Art. 59) of about 80 lb. per Ihp.-hr. is attainable between 
 limits of 0° and 90°. About 60 lb. would be considered a fairly 
 good result if actually attained in ordinary practice, with these 
 limits. There is thus implied a relative efficiency of 60/80 
 = 0.75. This is higher than the value of relative efficiency 
 attainable with steam engines, mainly because the temperatures 
 involved are closer to those of the surrounding atmosphere. 
 The horse power of the engine which drives the compressor is 
 greater than the horse power here computed, the ratio of the 
 two being the mechanical efficiency. Under the 0°-90° 
 limits, one horse power in the ammonia compressor cylinder 
 should give a trifle less than one ton of capacity. Under un- 
 favorable conditions, as much as 2 hp. per ton may be required 
 at the steam engine cylinder. (Capacity is ice-melting effect, 
 NOT ice-making capacity.) 
 
 Piston speeds range from 125 to 600 ft. per min., being low 
 in small machines. Mean effective pressures vary widely (30 
 to 90 lb.) with the assigned temperature limits. 
 
 In purchasing a compressor, the stated tonnage should be 
 checked against the displacement, for the temperature limits to 
 be employed. It is seldom safe to assume a condenser tempera- 
 ture below 90° for year-round shore service, but 75° may be 
 realized on ships in northern latitudes. A low temperature of 0° 
 will just about answer for ice-making with direct expansion 
 (ammonia circulated in coils in the freezing tank). A slightly 
 lower temperature still will accelerate the freezing. When the 
 brine system is used (ammonia cools the brine and brine does 
 the refrigeration) the possibly offensive fluid is kept away from 
 substances it might injure, but efficiency is sacrificed. The low- 
 
144 
 
 Thermodynamics, Abeidged 
 
 est ammonia temperature must be below the lowest brine 
 temperature, and the latter must be below the desired cold-room 
 
 30 
 
 u 
 
 s. 
 
 to 
 
 1 1 M 1 1 M N 1 I 
 
 Soecific Heaf Calcium -- 
 
 
 f\f^k < 
 
 ^^ ^^ oT-^ I 
 
 ^t- 'k--^ 
 
 sife -S4 
 
 ncfl S§*rt Tv 
 
 "•^__ --^^^ „ v^'^ 
 
 ^vfT ^°^y^ X 
 
 ^± iJg^ ,v^- 
 
 017 r^ '^^ 7^° ia 
 
 "•'2 s, / /'b^ ^j,A^ 
 
 s^'' -L^XW- 
 
 ^^ lmi&+- 
 
 me, SS^S1^\- 
 
 "■' - ^t&Wn 
 
 f^Y^ 
 
 1 \\p*\5pecific Heat 
 
 (\M\ y ^' '\r-rdSa It Brine 
 
 
 -IS^F'-i^,^/ -^%h 
 
 ^ t?2r\%. 
 
 nM /^ V^' 
 
 ,2 \-§ 
 
 - ^J - -W- 
 
 ^/x ^\ 
 
 nRft •\\^ i''-L Vo 
 
 ^T/^ ^1? 
 
 w-A'?:, Vi- 
 
 oXV.o^V'^ ^%' 
 
 092 - ^Mo^i'^ii- -^- 
 
 "•^^__ xw>.^\^Y/ ^ 
 
 ^'tW^•^^^v;C o 
 
 1&1^Ao\4l \_ 
 
 056 5iwA'^o<5- :1 
 
 ■ _K&^/ ^^ *vL 
 
 ?X*//.'?' 
 
 ^ISjt/'/iC 
 
 10 T-!'^?iir ...... _ _ 
 
 •10 1.1 1.2 
 
 
 
 en 
 
 
 
 w 
 
 70 
 
 
 ^ 
 
 lU 
 
 
 >> 
 
 
 
 .o 
 
 
 
 a> 
 
 
 
 ■c 
 
 10 
 
 u. 
 
 o 
 
 
 
 x: 
 
 
 -*-> 
 
 o 
 
 
 c 
 
 g 
 
 
 o 
 
 -^ 
 
 
 
 a. 
 
 ?4o 
 
 
 IS) 
 
 ID 
 
 
 c 
 
 o 
 
 
 ■ N 
 
 U 
 
 -10 
 
 0} 
 <1> 
 
 20-ii 
 
 
 L 
 
 (0 
 
 
 U. 
 
 to 
 
 ■20 
 
 
 16-H 
 
 o 
 u 
 
 t- 
 
 -30 
 
 12£2i 
 
 -40 8 
 
 Specific Gravity (Water- 1) 
 
 Fia. 37. Properties of Brines Plotted Against Specific Gravity. 
 
 temperature. Typical values for the three temperatures are 
 0°, 15°, 25°. If brine is used, the ammonia will work between 
 
Vapoe Refrigeration 145 
 
 0° and 90°. With direct expansion, it might have worked be- 
 tween 15° and 90°. The latter range gives a more efficient cycle. 
 Direct expansion is greatly to be preferred on economical grounds 
 for ice-making or very low temperature work. Brine circulation 
 gives opportunity for storage of " cold " and no harm is done 
 if the compressor has to be shut down for a short period. 
 
 In INDIRECT REFRIGERATION, air is used instead of brine. Air 
 is blown over coils carrying the refrigerating fluid, and the 
 chilled air then passes to the room to be cooled. 
 
 Fig. 37 gives the properties of the two types of brine used. 
 Calcium chloride permits of lower temperatures than salt brine, 
 and is less corrosive. 
 
 Prob. 225. In Prob. 224, the machine as, actually installed 
 developed 98 tons capacity and the compressor cylinder indi- 
 cated 107 hp. while the Ihp. of the steam engine driving the 
 compressor was 133.33. State the relative efficiency and the 
 mechanical efficiency. (Ans., 0.72, 0.80.) If the actual indi- 
 cated thermal efficiency of the steam engine was 0.10, what was 
 the efficiency from steam at the throttle to refrigeration? (Ans., 
 0.38.) 
 
 Prob. 226. What is the Carnot efficiency of refrigeration 
 between 0° and 90° F.? (Ans., 5.1.) Between 15° and 90°? 
 (Ans. 6.3.) 
 
 Prob. 227. The machine in Prob. 221 is used for indirect (air- 
 blast) refrigeration. The air is cooled at constant pressure from 
 70° to 10°. Ignoring radiation loss, what weight of air is circu- 
 lated per min.? (Ans., 1400 lb.) 
 
 Prob. 228. A calcium chloride solution has a freezing point 
 of - 20°. What is its strength? (Ans., 24.4 per cent.) What 
 is its specific gravity? (Ans., 1.217.) Its specific heat at 5° F.? 
 (Ans., 0.688.) If with a 100 ton machine its temperature rises 
 from — 15° to 25° F., what weight of solution must be pumped 
 per min.? (Ans., 726 lb.) 
 
 116. Tonnage Rating. The 100 ton machine of Prob. 223 has 
 a displacement of 454 cu. ft. per min., or 4.54 X 1728 = 7840 
 
146 Thermodynamics, Abridged 
 
 cu. in. per ton per min. This figure (the specific displacement) 
 depends on the clearance, the limiting temperatures and the 
 dryness at the beginning of compression. There is a tendency 
 to standardize 0° and 95j° as limits for rating. Under these 
 conditions a specific displacement proposed is 5.52 cu. ft. or 
 9570 cu. in. 
 
 Prob. 229. In Art. 116, if clearance is 5 per cent., and temper- 
 ature limits are 0° and 95|°, what value of % would give a specific 
 displacement of 9570? (Ans., 0.94.) 
 
 117. Dry Compression. In Prob. 229, the vapor is nearly dry 
 at the BEGINNING of compression. Inspection of Fig. 36 will 
 show that it is bound to become somewhat superheated during 
 adiabatic compression to 95|°. Such operation is described as 
 "dry compression," in distinction to "wet compression" hereto- 
 fore described. Wet compression leads to the more eflBcient 
 cycle, especially if Xh is so adjusted that x^ = 1.0. This is a 
 condition difiicult to maintain, and it is obviously not the 
 condition which gives maximum refrigeration per lb. of fluid. 
 Such maximum refrigeration is obtained (without allowing the 
 fluid temperature to rise in the cold room) when % = 1.0. In 
 order to maintain wet compression under this condition, some- 
 thing must be done to withdraw heat during compression. This 
 makes the path depart from the adiabatic, swinging it to the 
 left as it proceeds upward. Jackets can be used to accomplish 
 this result, as in air compressors. Multi-stage operation is 
 occasionally practiced. The injection of cold oil or of liquid 
 ammonia into the cylinder is frequently adopted. In any case, 
 the rate of flow of fluid should be increased as soon as the com- 
 pressor discharge becomes noticeably warm. This indicates 
 (with normal values of pc) the existence of some superheat. 
 Increasing the flow of fluid will decrease Xb and reduce or eliminate 
 the superheat. 
 
 Prob. 230. With 0° and 90° limits, find xt when Xc = 1.0. 
 (Ans., 0.88.) 
 
Vapor Refrigeration 147 
 
 Prob. 231. While steadily giving 100 tons of refrigeration, 
 the machine in Prob. 217 is speeded up so that the weight of 
 fluid circulated per min. is increased 10 per cent. What change 
 occurs in %? (Ans., it is reduced from 0.87 to 0.80.) 
 
 118. Ideal Cycle with Superheat. In ideal dry compression 
 without heat removal, we have such a cycle as ejkld, Fig. 36. 
 The point k is not located to scale. The line Ik is a line of 
 constant pressure. We now have, following Art. 114, 
 Refrigeration done = i/y(l — Xf), 
 Heat rejected at condenser = Hk — hd, 
 Work expended = Hk — Hj, 
 Lfj{l - Xf) 
 
 Efficiency = 
 
 Hk — Hj 
 
 5A0S{Hk' - Hj) 
 
 Mean effective pressure = Pm = 
 
 Vj 
 
 Volumetric efficiency as in Art. 115. 
 The value of x/ is found as before, in Art. 113. That of Hk is 
 best found from a total heat-entropy diagram. Fig. 38. The 
 entropy and pressure at k are known. The symbol Hj means 
 the total heat of dry vapor at j. 
 
 Thus take limits of 0° and 70°. From Prob. 216, Xf = 0.132. 
 Then 
 
 Refrigeration done = 572.2 (1 - 0.132) = 497, 
 
 Pressure at A; = 129.2, 
 
 Entropy at A; = (ns)oo = 1.1741, 
 
 Hk' from Fig. 38 = 630, 
 
 Heat rejected = 630 - 42.1 = 587.9, 
 
 Work = 630 - 538.5 = 91.5, 
 
 Efficiency = 497 -^ 91.5 = 5.45, 
 
 p,n = (5.403 X 91.5) -^ 9.19 = 54. 
 The efficiency is less than in Prob. 220 and the mean effective 
 pressure is necessarily greater. For 100 tons capacity, 40.25 lb. 
 of fluid must be circulated per min. This is of course less than 
 in Prob. 221. The volume of fluid per min. is 40.25 X 9.19 
 = 370 cu. ft., and with volumetric efficiency = 0.834 as in Prob. 
 
148 
 
 Thermodynamics, Abridged 
 
 o 
 
 O On 
 
 o 
 
 T— 1 (VI 
 
 
 I M \ 1 . I ^1 ' 
 
 "r_t5:^ 
 
 S, \ IliJ 
 
 
 1 ^ 
 
 \ V M \ Ik |1 
 
 ; \i^ .^ 
 
 k^AH 
 
 
 r \ 
 
 V M \ J \ \ IK 
 
 Vt^tVP^l 
 
 «*^ ^\i / 
 
 
 t \ 
 
 C Ok \ *^ K \^ 
 
 :^U^rx 
 
 
 
 tr\ _i_ \_ k_ 
 
 ^ \ ik \|\ \ 
 
 
 
 "S^^^ 
 
 K \i\f\i\ \ V i\ 
 
 I ' T ^if 
 
 <A V Xlo' 
 
 
 b ^S 
 
 V S IkN v V 1\ » 
 
 k ^ r^ J V T^ 
 
 
 
 ^ ^ 
 
 V kSix \ *^ Nl V K 
 
 \ ^-1-^ 
 
 
 
 1s s 
 
 ^ NKN k, \ K > 1 
 
 ^ Vi^ 1 * V ^ 
 
 J J 
 
 UO 
 
 ? ^ 
 
 A 'v 'l\ r I i\ tv 
 
 "S X V 1 
 
 \ 11 
 
 
 "^ ^ 
 
 i\ ^^ V\ V N r 
 
 ^iVo^ V K 
 
 K \J 
 
 r— 
 
 "^ 
 
 ■. l\ ^1 \ A \ \ ^s^'^ 
 
 5 B L S k-P 
 
 I w 
 
 
 ^ s 
 
 \ . \ ^llO w 1 3v 
 
 Cju^ k r xlIS 
 
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 i^ 
 
 s. K V u>j*^lP\ 
 
 ?a?N u"l' 
 
 i- T~ 
 
 
 t^ 
 
 ^ L V M^"?!! r^ 
 
 V 1 \ 
 
 
 
 ox J ^>P*Y^\jLSlJ 
 
 \ \l \ k M \ I V 
 
 r4 (-i-Ci- 
 
 
 
 ^ I Vo^\&^f^ 
 
 Ni \K \ \ I M 
 
 ^ N^ 
 
 
 
 '^ >\ VVAJ?S VNl 
 
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 t/y 
 
 r^j \r\ \ S. 
 
 ^ \ hNlM >i 1 \ 
 
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 VV C 
 
 v\\ vk\ N \ ^ 
 
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 ?^ \l 
 
 ^S. 0\\ N \. sjyv ^ 
 
 Pt^- c; 
 
 
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 N\v Nrt^ s. ' 1^ 
 
 AjJL.^ 
 
 
 cc 
 
 1 
 
 jxfA ^^r^^ 
 
 L Ct S. si 
 
 
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 5 
 
 jjSSj l^jS 
 
 -i - rS 
 
 
 ^ 
 
 o' 
 
 k V Vy A \ ^ \ V ^ 
 
 yp >^ 
 
 
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 fo' -3 
 
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 Q. 
 
 
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 J m 
 
 
 
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 rdfX ' > lN[ 
 
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 ^' 
 
 
 
 
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 tc 
 
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 H 
 
 
Vapor Refrigekation 
 
 149 
 
 223, the specific displacement is (3.7 X 1728) 4- 0.834 = 7700 
 cu. in., a value not notably different from that realized in wet 
 compression. 
 
 Prob. 232. Find the ideal eflBciency with dry compression 
 from 0° to 90°. (Ans., 4.12.) 
 
 119. Carbon Dioxide Machine. The temperature-entropy 
 diagram, Fig. 39, is plotted with lines of constant pressure and 
 constant superheat in the superheat region. Dry compression 
 
 
 
 
 
 90 , -^- ■« J sfl\/ ^i\ n V\lA^f 
 
 s/ 11 jW Y I y\^ll Yx\ 
 
 ^ ^ / / K/\ / 11 V \ A*^ / 
 
 
 
 ,r- / \ IV YA/A Iv^ f VT 
 
 c / ^"^ l\/<A ft t[^a M K ^ 
 
 £k« : :: : I" /:: : : r^CmKM'^'^P^'''^^^^' 
 
 ■SSO 7 _ ^NsIVyW DecfF 1 1 
 
 i2 -i ^ . \aV1V \\\\\ 
 
 * / ~ * V V \ w\ 
 
 
 S30 - / _;: MAlu 
 
 "S J nM\V 
 
 i_ V ^ y\ lu 
 
 o^m "^j , \A y 
 
 1+10 r^/ _ l1\J_ 
 
 
 L__ ^/ 1 1 / 
 
 -L 
 
 lO y t 
 
 ^ J 
 
 
 
 -005 0.05 0.10 0.15 020 0.25 0.30 035 
 Entropy 
 
 Fig. 39. Temperature-Entropy Diagram for Carbon Dioxide. 
 
 is common, but leads to low efficiency. Adiabatic compression 
 from a dry state at 0° to the pressure of 900 lb. (corresponding 
 roughly with a saturation temperature of 75°) would put the 
 final temperature far above 100°. In the tropics, the vapor 
 could not then be condensed. We must in practice in any 
 latitude either compress from an initially wet condition, or 
 remove heat during compression, or be content with com- 
 paratively low efficiency. 
 
150 Thermodynamics, Abridged 
 
 Properties of Saturated Carbon Dioxide Vapor 
 
 In 
 
 S 
 
 3 
 
 ii 
 
 
 
 
 
 
 
 
 1' 
 
 
 as 
 
 IS 
 
 A 
 
 L 
 
 If 
 
 r 
 
 ™» 
 
 ™. 
 
 ». 
 
 S 
 
 a 
 
 3 
 
 Is 
 
 
 
 
 
 
 
 
 
 
 313.5 
 
 0.2661 
 
 0.016298 
 
 -15.5 
 
 114.3 
 
 98.8 
 
 99.81 
 
 -0.03278 
 
 0.2494 
 
 0.21662 
 
 75 
 
 902.0 
 
 0.0708 
 
 0.022232 
 
 29.4 
 
 50.8 
 
 80.2 
 
 42.69 
 
 0.05718 
 
 0.09495 
 
 0.15213 
 
 In the first case, the method of analysis is precisely as in Arts. 
 113, 114. In calculating mean effective pressure and displace- 
 ment, however, it should be noted that the values of « and «„ 
 are too nearly the same to permit of the approximation hereto- 
 fore used (Arts. 74, 115) for the specific volume at the beginning 
 of compression. That volume is for vapor at 0° F., 0.90 dry, 
 for example, not 0.90 X 0.2661, but 0.90 (0.2661 - 0.016298) 
 + 0.016298 = 0.2418. 
 
 Prob. 233. With CO2 between 0° and 75°, the vapor is 
 exactly dry at the end of adiabatic compression. What was its 
 dryness at the beginning of compression? (Ans., 0.743.) 
 
 Prob. 234. In Prob. 233, what was the dryness at the expan- 
 sion valve outlet? (Ans., 0.39.) Compute the refrigeration per 
 lb. of fluid. (Ans., 40 B.t.u.) How much heat is rejected at 
 the condenser? (Ans., 51 B.t.u. per lb.) How much work is 
 done? (Ans., 10.5 B.t.u. per lb.) What is the ideal efiiciency? 
 (Ans., 3.83.) What is the specific volume at the beginning of 
 compression? (Ans., 0.202.) What is the volumetric efficiency 
 if clearance is 4 per cent.? (Ans., 0.9248.) What is the dis- 
 placement per lb. of fluid? (Ans., 0.218 cu. ft.) What is the 
 specific displacement — cu. in. per min. per ton — for the con- 
 ditions assigned? (Ans., 1870.) 
 
 120. Carbon Dioxide Cycle with Heat Removal. Suppose 
 heat to be removed during compression at such rate as to make 
 the vapor dry at both the beginning and end of compression. 
 The cycle is represented diagramatically (not to scale) by ejld, 
 
Vapor Refrigeration 151 
 
 Fig. 36. Assume the compression path jl to be a polytropic. 
 Then pi^i" = Pji>f, and for the 0°-75° limits, the exponent is 
 
 log pi — log pj log 902 — log 313.5 
 
 a = 
 
 log Vj - log vi log 0.2661 - log 0.0708 
 
 2.955 - 2.496 
 
 -0.5744+ 1.15 
 Then the external work done under this curve jl is 
 
 pm - PjVj (902 X 0.0708) - (313.5 X 0.2661) 
 
 = 0.798. 
 
 Wj 
 
 a- 1 - 0.202 
 
 144 
 X 7^ = 17.9 B.t.u. 
 
 The external work of the whole cycle is 
 
 ^ { pVw)d . Pl(.V - Vv,)l . „ Pj{v - Vw)j 
 
 5.403 "*" 5.403 '^ ^'-^ 5.403 
 
 = (F^)j. , ,7 Q _ Pj(« - ^w)j _ 902 X 0.0708 
 5.403"^ 5.403 5.403 
 
 313.5(0.2661-0.0163) 
 + ^^-^ 5.403 
 
 = 11.8 + 17.9 - 14.5 = 15.2 B.t.u. 
 
 The refrigeration done is Lj{l — xf). The heat rejection at 
 the condenser is Li. The specific volume at the beginning of 
 compression is the tabular volume Vj. Since I1H = liW, 
 
 Hi + Hij - Hj = 15.2 = 80.2 + Hij - 98.8 
 Hii = 33.8 B.t.u. per lb. of fluid. 
 
 This is the quantity of heat to be removed during compression. 
 
 Prob. 235. In the foregoing, find refrigeration and heat rejec- 
 tion per lb. of fluid, and the ideal efficiency. (Ans., 69.6, 50.8, 
 B.t.u.: efficiency = 4.59.) 
 
 Prob. 236. The machine in the foregoing is of 100 ton 
 capacity, double-acting, at 240 ft. piston speed and 5 per cent. 
 
152 
 
 Thermodynamics, Abeidged 
 
 clearance. Find cylinder diameter, ideal hp. and ideal ice- 
 melting effect per Ihp.-hr. (Ans., 8 in., 103 hp., 81 lb.) 
 
 Prob. 237. In Probs. 235-236, 33.8 B.t.u. must be removed 
 per lb. of fluid during compression. If this is accomplished by 
 injecting liquid at 0° into the cylinder, what weight of fluid 
 must be thus supplied per min., the injected liquid being com- 
 pletely vaporized at 75°? (Ans., 102 lb.) 
 
 121. Comments. This last type of cycle requires less power 
 for a given amount of refrigeration, and is consequently more 
 efficient, than the cycle with adiabatic compression: but it 
 requires jacketing or liquid injection. Carbon dioxide gives 
 lower thermal efficiencies than ammonia, but much higher 
 efficiencies than air. On account of its low specific volume (high 
 density) the displacement is less in proportion to refrigerating 
 capacity than is the case with ammonia. The very high pressures 
 reached limit the application of carbon dioxide to cylinders of 
 small diameter. Hence large machines rarely use this fluid. 
 Leakage is more probable than with ammonia, and is not easy 
 to detect. The fluid is usually cheaper than ammonia. For 
 extremely low temperatures, carbon dioxide can be used with- 
 out maintaining a vacuum between the compressor and the 
 expansion valve. 
 
 122. Sulphur Dioxide. This fluid is occasionally employed in 
 marine service. The following values are given by Zeuner: 
 
 Properties of Satiirated Sulphur Dioxide 
 
 la 
 II 
 
 a" 
 
 n ■ 
 
 K 
 
 ".d 
 
 >e 
 
 -4 
 6 
 77 
 86 
 95 
 
 9.272 
 11.756 
 56.386 
 66.359 
 77.630 
 
 8.051 
 6.486 
 1.445 
 1.220 
 1.036 
 
 -11.216 
 
 - 8.449 
 
 14.582 
 
 17.572 
 
 20.588 
 
 171 
 
 169.745 
 
 148.775 
 
 144.787 
 
 140.495 
 
 159.784 
 161.296 
 163.357 
 162.359 
 161.083 
 
 157.111 
 165.563 
 133.718 
 129.827 
 125.656 
 
 -0.0237 
 
 -0.0177 
 
 0.0284 
 
 0.0339 
 
 0.0394 
 
 0.3755 
 0.3655 
 0.2773 
 0.2655 
 0.2534 
 
 0.3518 
 0.3478 
 0.3057 
 0.2994 
 0.2928 
 
 Prob. 238. What property, in the above table, varies in an 
 unusual way? 
 
CHAPTER VII 
 HEAT TRANSFER APPLIANCES 
 
 123. Mixtures. Mixtures of simple substances have been 
 considered in Art. 3. For these, or for vapors, ignoring radia- 
 tion, 
 
 loss of heat = gain of heat, 
 
 Wa.{Qo! - Qa) = WbiQb' - Qb), 
 
 where w = weight in lb., Q = heat content in B.t.u., and primes 
 denote the high temperature conditions. In actual mixing, 
 tb' = ta'. i.e., the final temperatures of the two fluids are the 
 same. For simple substances, w{Q' — Q) is best written, 
 ws{t' — t), where s = specific heat. For vapors, 
 
 Q = A at the liquid condition, 
 
 Q = h-\- xL when wet, 
 
 Q = H when dry, 
 
 Q = H -{- A;(4up — 4at) when superheated. 
 
 Prob. 239. Ammonia at 199 lb. pressure, superheated 220°, 
 is cooled, condensed and further cooled until its temperature 
 is 70° F. See Fig. 38. How much heat is emitted per lb.? 
 (Ans., 598.9 B.t.u.) 
 
 Prob. 240. An injector is supplied with dry steam at 200 lb. 
 pressure and water at 90° F. The two fluids mix and the 
 resulting temperature is 170.06° F. What weight of water was 
 mixed with 1 lb. of steam? (Ans., 13.3 lb.)* 
 
 Prob. 241. In a jet condenser, 75.07 lb. of water are employed 
 per lb. of steam. Water enters at 80° F. Steam enters at 2 lb. 
 pressure, 0.70 dry. What is the temperature of the discharged 
 mixture? (Ans., 90° F.) 
 
 124. Temperatures in Surface Transfer. Transfers through 
 
 surfaces (like the metal wall of a condenser tube) are more 
 
 * See Art. 134. 
 11 153 
 
154 
 
 Thermodynamics, Abridged 
 
 s 
 
 Parallel Flow 
 
 f^ 
 
 \K. 
 
 u 
 
 complicated. There are four temperatures to be considered; 
 the inlet and outlet temperatures of both the heat-emitting and 
 the heat-receiving fluids. The rate of transfer depends not 
 
 only on the values of these 
 four temperatures, but also, 
 so to speak, on their ar- 
 rangement. Take the case 
 of an economizer — a group 
 of tubes containing feed 
 water, placed in the path of 
 the combustion gases from 
 a boiler to its stack. The wa- 
 ter moves vertically through 
 the tubes, but it also moves 
 (considering the whole sys- 
 tem of tubes) in a direction 
 which may be either toward 
 the boiler or away from it. 
 
 The two cases thus sug- 
 gested are represented dia- 
 grammatically in Fig. 40. 
 In parallel flow, the two 
 fluids move in the same gen- 
 eral direction. In counter 
 flow, they move in opposite 
 directions. In many devices, the direction of flow is uncertain 
 or indeterminate. Such is the case with a steam radiator, for 
 instance. 
 
 The rate of heat flow is generally proportional to the tempera- 
 ture difference, that is, to the difference between the temperatures 
 of hot and cold fluids. The total heat transfer is proportional 
 to the MEAN temperature difference, tm- This is the average 
 ordinate or vertical distance between the two curves of Fig. 40, 
 (upper diagram). It is not the arithmetical average tempera- 
 ture difference, which would be the average vertical distance 
 between the two dotted straight lines. For parallel flow, tm is 
 
 Counter Flow 
 
 Fig. "40. Two Cases of Heat Transfer. 
 
Heat Transfer Appliances 155 
 
 obviously less than the arithmetical average. Its value for 
 
 parallel flow or for cases where one fluid remains at constant 
 
 temperature, or for cases in which there is no determinate flow, 
 
 is given by 
 
 . id — tb — {tg — tb) ,„„. 
 
 im = T-> i (^^) 
 
 ta — tb 
 
 For counter flow, it is 
 
 , tg' — tb — {tg — tb) 
 
 tm = T-, 77 — , (68a) 
 
 1 ^a ~ tb 
 
 tg tb 
 
 symbols being as in Art. 123. The derivation of these formulas 
 furnishes an interesting exercise in calculus. 
 
 In parallel flow, fe' approaches ta as a maximum. In counter 
 flow, it approaches the higher value tj. Hence counter flow 
 permits of higher ultimate temperatures of the heat-receiving 
 fluid. On the other hand 4' actually becomes equal to td or tg 
 only when the surface is infinite. A finite difference of tempera- 
 ture always exists in practice at the outflowing point of the 
 colder fluid (i.e., the right-hand end for parallel flow and the 
 left-hand end for counter flow, in Fig. 40). With this difference 
 fixed, tm is greater for parallel flow. The quantity of heat trans- 
 mitted is proportional to tm- Hence with a fixed temperature 
 difference at the outflow of the cooler fluid, and when a given 
 quantity of heat is to be transmitted, parallel flow is more 
 economical of transmitting surface. 
 
 Prob. 242. One fluid cools from 100° to 60° while the other 
 is heated from 30° to 50°. What is the arithmetical average 
 temperature difference? (Ans., 40°.) State the value of tm for 
 parallel flow. (Ans., 30.9.) The direction of flow of the cooler 
 fluid is now reversed and its final temperature changes to 68°' 
 Find tm. (Ans., 30.9.) Compare the heat transfers in the two 
 cases. (Ans., they are equal.) Compare temperature differ- 
 ences at cold fluid outlet. (Ans., parallel flow, 10°, counter 
 flow 32°.) 
 
 Prob. 243. Should parallel flow or counter flow be used in 
 the condenser of a CO2 refrigerating machine? Why? 
 
156 Theemodtnamics, Abridged 
 
 125. Coefficient of Transmission. The general equation of 
 transmission is 
 
 WaiQa' - Qa) = W^CQft' - Qi) = pAtr., (69) 
 
 where weights are in lb. per he., A = surface in sq. ft., and 
 p = coefficient of transmission. Obviously, p indicates 
 
 the quantity of heat (B.t.U.) TRANSMITTED PEE HE. PEE 
 DEGEEE of mean TEMPEEATUEE difference by EACH SQ. FT. 
 
 OF SUEFACE. Fcw factors used in engineering are more variable 
 than p. It depends on the two fluids, their condition and purity : 
 and on the cleanliness and material of the transmitting surface. 
 It is highest for transfer between liquids and wet vapors, as in 
 condensers and feed water heaters: and lowest for transfer 
 Involving dry gases, as in superheaters and economizers. Besides 
 the factors already mentioned, p varies notably with some power 
 (0.3 to 0.6) of the velocity of fluid, or more strictly with the 
 MASS-FLOW (velocity X density). The values of p given in the 
 accompanying table are merely rough averages. 
 
 Prob. 244. A fan discharges 100 lb. of air per min. across 
 coils through which water is being pumped. The water cools 
 from 197.75° to 170.06°. The air is heated from 70° to 130°. 
 What weight of water is pumped per min.? What amount of 
 coil surface is necessary? Note: direction of flow is indeter- 
 minate, hence use Equation (68). (Ans., 51 lb., 113 sq. ft.) 
 
 Prob. 245. A feed water heater uses auxiliary exhaust steam, 
 0.70 dry at 25 lb. pressure, amounting to 10 000 lb. per hr. 
 Boiler feed water enters at 90°, its weight being 150 000 lb. per 
 hr. The condensed steam is 10° warmer than the outlet feed 
 water. What will be the outlet feed temperature? If counter 
 flow is used, find <„. and the necessary surface. (Ans., 140°, 
 274 sq. ft.) 
 
 Prob. 246. In Art. 47, 130 lb. of air are cooled 244° F., per 
 minute, in the intercooler. Assume indeterminate flow, water 
 entering at 70° and leaving at 110°. Find i„, and the necessary 
 intercooler surface if p = 3. (Ans., tm = 199°, surface = 755 
 sq. ft.) 
 

 
 
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158 Thermodynamics, Abridged 
 
 126. The Steam Boiler. If the potential heat (heat of com- 
 bustion) of the coal burned is Bi, the heat evolved in the furnace 
 is always some less quantity B2, the quotient B2IB1 being e^, 
 the FURNACE EFFICIENCY. The heat evolved, B2, is available 
 for transmission through the heating surface to the water and 
 steam. Only a part, £3, is so transmitted, the remainder 
 (Bi — £3) being lost up the stack. Then Bs/Bi = e, is the 
 SURFACE EFFICIENCY or transmissive efficiency and Bg/Bi = e^e, 
 is the efficiency of the whole boiler. 
 
 For uniform coal, Bi varies in direct proportion to the amount 
 of coal burned. Under good control, e^ is about constant. 
 Hence with such good control B2 is directly proportional to the 
 coal burned. The heat B2 exists in the form of a body of hot 
 gas, and is proportional to the weight and temperature of gas. 
 It is numerically equal to the product of weight of gas by its 
 mean specific heat and by . the difference between fire-room 
 temperature and furnace temperature. The fiu-nace tempera- 
 ture and the mean temperature difference between gas and 
 steam are scarcely affected by the weight of coal burned. The 
 weight of gas is (or should be, in good operation) directly pro- 
 portional to the weight of coal. 
 
 The heat transmitted is B3 = pAtm. In a given boiler, A is 
 constant. Under good operation, as above described, C is 
 constant. Hence £3 varies directly with p. Accept the mass- 
 fiow theory, i.e., p varies directly with the weight of gas, hence 
 with the weight of fuel. Then B3 varies du-ectly with the weight 
 of fuel. If Bp varies, B3 nevertheless varies directly with £2- 
 
 This analysis loses sight of the fact that part of the transfer in 
 a boiler is by radiant heat, with which the transmission varies 
 not as the first power, but as the fourth power of the mean 
 temperature difference. However, if t^ near the furnace is 
 constant, the radiant heat emission is constant. Hence the 
 BALANCE of heat transmitted is directly proportional to B2, or, 
 at constant furnace efficiency, to Bi. 
 
 This deduction is thoroughly confirmed by the straight line 
 law illustrated in Fig. 41 . The radiant heat emission is constant, 
 
Heat Transfer Appliances 
 
 159 
 
 and does not depend on the amount of coal burned. As a mathe- 
 matical fiction, it may be regarded as represented by the inter- 
 cept on the vertical axis {oa, for the Denver). The balance 
 of the heat is directly proportional to Bi. Hence e^ decreases 
 
 Fig. 41. 
 
 4000 8000 IZPOO 1^000 20,000 
 B.tu.Value of Coal Bumed=Bi 
 
 Navj' Department Tests on Water-Tube Boilers — Quantities are 
 per Square Foot of Heating Surface per Hour. 
 
 with forcing, but more and more slowly, and the limit of boiler 
 capacity is not easily reached. At very high rates of driving, 
 Bf. decreases. 
 
 If tests were carried down to zero weight of coal burned, the 
 lines could not, obviously, continue straight. This would imply 
 NEGATIVE radiant heat for the Nebraska. The imaginary 
 straight portion (o6 for the Denver) may, however, be utilized 
 in calculation. See Probs. 247-250. 
 
 A boiler "horse power" is equivalent to the transmission of 
 34.5 X 970.4 = 33 479 B.t.u. per hr., or the evaporation of 34| 
 lb. of water per hr. from and at 212° F. 
 
160 
 
 Thermodynamics, Abridged 
 
 Prob. 247. For the Denver, in Fig. 41, how much heat was 
 transmitted by pure radiation? (,Ans., 860 B.t.u. per sq. ft. of 
 heating surface per hr.) 
 
 Prob. 248. When 20 000 B.t.u. coal value was supplied per 
 sq. ft. of heating surface per hr., in the Denver, how much heat 
 was transferred by radiation? (Ans., 860 B.t.u.) By ordinary 
 transmission? (Ans., 11 800 B.t.u.) What was the boiler 
 efficiency? (Ans., 0.633.) What was the hp. per sq. ft. of 
 heating surface? (Ans., 0.377.) 
 
 Prob. 249. State the boiler efficiency and hp. per sq. ft. of 
 heating surface for the Denver when the coal equivalent is 
 JSi = 10 000 B.t.u. per hr. (Ans., 0.674 efficiency, 0.202 hp.) 
 
 Fig. 42. Babcock and Wilcox Stationary Boiler with Superheater. 
 
 Prob. 250. li t^= 1200, find p in Probs. 248, 249. Note: 
 p applies to ordinary transmission only. (Ans., 9.83 and 4.90.) 
 
 127. Superheaters. Superheaters (Fig. 42) are usually placed 
 between the first and second " passes " of a water tube boiler, 
 where the gas temperature is well above 1000°. The transfer 
 equation is 
 
Heat Transfer Appliances 161 
 
 Wsfcs(<s' — is) = Wgkg{tg' " tg) = fiAtm, (70) 
 
 where Ws = lb. of steam per hr. 
 
 ka = specific heat of superheated steam between t^ and <«' 
 
 at the given pressure. 
 ta = temperature of steam after superheating, 
 ts = temperature of saturated steam at the given pres- 
 sure, 
 Wg = Ih. of combustion gases passing the superheater per 
 
 hr., 
 kg = specific heat at constant pressure of combustion 
 
 gases, say 0.24, 
 tg' = temperature of gases approaching superheater, 
 tg = temperature of gases leaving superheater, 
 p = coefficient of transmission, around 3, 
 tm = mean temperature difference, 
 A = superheater surface, sq. ft. 
 In very good operation, Wg will vary directly with Wg, the ratio 
 Wglws being from 1.6 to 2.2. Hence 
 
 t/ - ts ^ 2 X 0.24 ^ 
 
 to' -tg~ 0.48 
 when Wgjwe = 2 and k, = 0.48 (rough average values). Thus 
 the rise in temperature of the steam is about equal to the fall in 
 temperature of the gas. As with boilers, p increases in direct 
 proportion when Wg increases: i.e., when the boiler is forced. 
 Under these conditions, the equations can be satisfied only 
 when the temperature ranges remain unchanged, because tm 
 cannot increase when either temperature range increases. In 
 practice, the ratios Wg/ws and p/wg both vary somewhat and the 
 amount of superheat varies in an irregular way with the rate of 
 driving. 
 
 Parallel flow should be used in superheaters to decrease 
 the necessary surface. 
 
 Prob. 251. A boiler of 1000 hp. uses feed water at 213° and 
 generates dry steam at 200 lb. pressure. What weight of steam 
 is produced per hr.? (Ans., 33 000 lb.) 
 
162 Thebmodynamics, Abridged 
 
 Prob. 252. In Prob. 251, it is required to superheat this steam 
 200°. Take h = 0.548. The boiler evaporates 8.5 lb. of water 
 per lb. of coal and the weight of gas is 17 lb. per lb. of coal. 
 The gas after passing the superheater is at 1000° F. What is its 
 temperature at entrance to the superheater? (Ans., 1230° F.) 
 What is the mean temperature difference? ^ (Ans., 610° F.) The 
 superheater surface? (Ans., 1970 sq. ft.) If the coal used 
 contains 11 267 B.t.u. per lb., what is the combined eflficiency 
 of boiler and superheater? (Ans., 0.85.) Note: efficiencies 
 above 0.80 are rarely attained in practice. 
 
 Prob. 2S3. In Prob. 252, 10 per cent, of the heat is trans- 
 mitted by pure radiation. Given t^ = 1200, what should be 
 the heating surface (not including superheating surface) of the 
 boiler? (Ans., 2500 sq. ft.) 
 
 128. Economizer. The definite equation is (see Art. 124), 
 
 w^it^' - ty,) = 0.24wj(i„' - Q = iAtm, (71) 
 
 where Wgjwv,, as with the superheater, tends to remain around 2 in 
 the best operation. Then the water is warmed about half as 
 many degrees as the gas is cooled. Countee flow gives the 
 highest final water temperatures, but economizer surface is 
 costly and parallel flow may therefore be used. 
 
 Prob. 254. Boiler feed water amounting to 10 000 lb. per hr. 
 is to be heated in an economizer. Water enters at 90°, gas at 
 500° and the ratio of weights of gas and water is 5.0. With 
 parallel flow, the water leaves at 234°. Find the outlet gas 
 temperature and the surface. What is the value of (<„ — U'), 
 Fig. 40 ? (Ans., tg = 380° F., A = 1400 sq. ft., (i„ - U)= 146. 
 
 Prob. 255. If w^ = 5454.54, t„ = 90, Wg = 5«)„, </ = 500, 
 with COUNTER FLOW, find the outlet gas temperature and the 
 surface when tj = 354°. (Ans., tg = 280°, A = 2130 sq. ft.) 
 Compare costs of installation and saving of heat in Probs. 254 
 and 255. (Ans., the temperature differences at water outlets 
 are the same; the quantities of heat transmitted are the same; 
 the counter flow economizer is 52 per cent, larger.) 
 
Heat Transfee Appliances 
 
 163 
 
 129. Surface Condenser, Feed Water Heater. In surface con- 
 densers (Fig. 43) and feed water heaters (Fig. 44), some of the 
 factors affecting the value of p have been established. For usual 
 tubes when clean, p = 400 Vw, where u = water velocity, ft. per 
 sec. Very badly fouled tubes will give half this value of p. 
 
 .Exhaust Steam 'T° °"^'^^ 
 
 '"'^^ WaterOui/eh \ 
 
 Tube 
 Sheet} 
 
 \ Wafer 
 Inlet 
 
 From 
 Met 
 
 Fig. 43. 
 
 -/- /• , / PathofC/rculatinqj 
 c 4.- A A iniei 'Sv'"-'der Water 
 
 Section A-A Dram ^AirPump Suction 
 
 Section B-B 
 
 Surface Condenser Showing Path of Water and Steam and General 
 Arrangement of Baffle Plates and Tubes. 
 
 Vulcanizing (coating) admiralty tubes decreases p nearly 85 
 per cent. If there is much air in the steam to be condensed, 
 p will be reduced. Air may enter through leaks in vacuum 
 piping. Fresh feed water also carries air. High water velocities 
 increase p and therefore decrease the necessary size of heater or 
 condenser. They also increase the power consumption of the 
 pump which handles the water. This last point is not of much 
 importance in condensers, where circulating water pressures are 
 low: but the water passing through feed water heaters is under 
 full boiler pressure and excess pressure should be avoided. With 
 high water velocities, excess pressures are necessary to overcome 
 friction. For feed water heaters, use m = 6 or 7. 
 For condenser or heater, with clean tubes of usual materials, 
 
 w^ilii + XiLi — hi) = w„(i4 - ta) = 400 Vm^4., (72) 
 
164 
 
 Thermodynamics, Abridged 
 
 \<-——uP-iZ 
 
 > 
 
Heat Transfer Appliances 165 
 
 where i and % are initial and final steam conditions, 3 and 4 initial 
 and final water conditions. Flow is usually indeterminate, hence 
 Equation (68) is used for 1^: this equation becoming, in present 
 symbols, 
 
 t\ — t%— (<2 — ^4) 
 
 ^ 
 
 \og,- 
 
 f2 — 
 
 Prob. 256. A condenser handles 17 400 lb. of steam per hr. 
 at 0.696 lb. pressure, 0.85 dry. The condensed steam is dis- 
 charged at 80°. The circulating water enters at 60° and leaves 
 at 74°. Find the weight of water. (Ans., 1 113 000 lb. per hr.) 
 
 Prob. 257. The condenser in Prob. 256 has 600 tubes in 
 each "pass" — see Fig. 42. These each have an internal cross- 
 sectional area of 0.219 sq. in. If the weight of water is 64 lb. 
 per cu. ft., find the water velocity through the tubes. (Ans., 
 5.3 ft. per sec.) 
 
 Prob. 258. In the foregoing, find the values of p, tm and A. 
 (Ans., p = 922; t^ = 14.9, A = 1140.) 
 
 Prob. 259. In Probs. 256-258, each lineal ft. of tube gives 
 0.1635 sq. ft. of siu-face. There are two passes. What is the 
 length of tubes? (Ans., 5.8 ft.) 
 
 In FEED WATER HEATERS, p is somctimes increased by the use 
 of retarders or by employing corrugated tubes. Sometimes a 
 double corrugated tube is used, the water passing through the 
 narrow space between the inner and outer walls. The object 
 of increasing p is to decrease surface and thereby to save space 
 and cost. Many special forms of heater increase p without 
 decreasing total bulk or cost, because the surface is of such 
 shape as to be of large bulk per unit of transmitting area. Con- 
 densers generally have plain straight tubes. 
 
 130. Evaporator. Essentially, an evaporator is a shell con- 
 taining salt water in which there is immersed a coil containing 
 steam. Pure steam is driven off from the sea water and later 
 condensed, thus furnishing a supply for boiler feed and domestic 
 
166 
 
 Thermodynamics, Abridged 
 
 purposes. Fig. 45 shows considerable elaboration on this. The 
 salt feed is first used as cooling water in a small condenser or 
 "distiller." The bulk of it is then wasted, but a part feeds the 
 evaporator shell. Its temperature having been somewhat raised 
 
 To main engine 
 condenser or hot welK, 
 
 Separator-.^ 
 From boilerihrough presscire X^L^h^f ^ 
 reducing valve or from receiver " 
 of compoun d engine — ^ 
 
 Fvaporafor— 
 
 Wc 
 
 SalfiVater 
 /'Supply 
 
 I {W2+W4-I-Ws+Wg) 
 
 (rDisfi/ier 
 
 ■Coils 
 
 ■Drinking 
 Wafer 
 
 Drip'' \wg V'BlowDown 
 
 W6 
 
 ^ he 
 
 \^2 
 
 Fio. 45. Evaporator. 
 
 at the distiller, less heat will now be required to evaporate than 
 would otherwise be needed. In the evaporator the salt water is 
 boiled: and, if some of the coil is unsubmerged, slightly super- 
 heated. As the sea water is vaporized, the salt left behind con- 
 tinually strengthens the brine in the evaporator shell. Hence the 
 contents must be frequently or steadily blown down to keep the 
 salinity within reasonable limits. 
 
 The vapor leaving the evaporator passes through a separator 
 where any moisture is baffled off. It may then go to the distiller 
 
Heat Transfee Appliances 167 
 
 and be condensed for use as drinking water, etc. If it is desired 
 to use the condensed pure water for boiler make-up, it can be 
 carried directly to the main engine condenser. The whole 
 evaporator will then carry a vacuum: instead of "blowing" 
 down, a pump must be used to withdraw concentrated brine: 
 and the heat in the vapor (though not the vapor itself) will be 
 discharged outboard (and therefore wasted) with the condenser 
 circulating water. A better plan is to carry the vapor to the main 
 hot well for condensation. Its heat as well as its substance is 
 then transferred to the boiler feed. 
 
 Assume the distiller to be in use. Then with symbols as in 
 Fig. 45, 
 Qi = heat supplied by live steam at evaporator 
 
 = ^8(^8 + XgLs — hi), 
 
 Qi = heat absorbed at evaporator 
 
 = (W6 + Wi){h + XsLi — h) + WiQh — h), 
 Qi = 0,2 -\- radiation at evaporator, 
 
 Qz = heat discharged at separator = wji^, 
 
 I — Xi 
 
 Wi = Wi , 
 
 Qi — heat rejected at distiller = w^iHs — h^), 
 Qs = heat absorbed at distiller = {w^ + Wi+ Wi+ W6){h2 — h{), 
 Q4 = Qs + radiation at distiller. 
 
 A test on a small machine gave the following values: ps = 80, 
 xs = 0.99, ti = 292.7, t«8 = 2800, ps = 25, X3 = 0.99, ti = 126.15, 
 W5 = 2340, Wi = 200, fs = 237.8, h = 209.55, h = 80. The coil 
 transmitting surface in the evaporator was 70 sq. ft. Weights 
 are in lb. per hr. The following are tabular properties: 
 
 h = 282, U = 900.3, hi = 262.1, A3 = 208.4, L3 = 952, 
 hi = 94, /jj = 206.1, h = 177.5, hi = 48.03. 
 Then 
 
 Wi = 2340 X 9V = 23.6, (wj + Wi) = 2363.6, 
 Qi = 2800 (282 + 0.9 9 X 900.3 - 262.1) = 2 552 352 B.t.u., 
 Qi = 2363.6 (208.4 + 0.99 X 952 - 94) + 200(206.1 - 94) 
 
 = 2 522 420 B.t.u. 
 
168 Thermodynamics, Abridged 
 
 Then the radiation loss at the evaporator is Qi— Q2 = 29 932 
 
 B.t.u., 
 
 Qi = 2340 (208.4 + 952 - 177.5) = 2 300 000 B.t.u., 
 
 Qs = (^2 + 2363.6 + 200) (94 - 48.03) 
 
 = 45.97W2 + 118 000 B.t.u. 
 
 Disregarding radiation at the distiller (it should be very small), 
 put Qi = Qi. Then wa = 47 400 and the weight of circulating 
 water pumped to the distiller is 47 400 + 2563.6 = 49963.6 lb., 
 the greater part of which is wasted. 
 
 The heat transmitted at the evaporator was 2 522 420 B.t.u. 
 per hr. The mean temperature difference (counter flow) was 
 tm = 114. The value of p realized was then 2 522 420 -J- (70 
 X 114) = 316. Assuming the same value to have been realized 
 at the distiller, with indeterminate flow and tm = 145, the neces- 
 sary surface there is 2 300 000 ^ (145 X 316) = 50 sq. ft. 
 
 Higher values of p may be realized for a short period, but the 
 accumulation of scale is rapid. 
 
 131. Multiple Effect Evaporator. In Art. 130, 2340 lb. of dis- 
 tilled water were produced per 2800 lb. of live steam condensed, 
 or 0.833 lb. of water per lb. of steam. A large waste of heat 
 occurred at the point Wi, h^, Fig. 45. If the vapor leaving the 
 separator had been conducted to the coil of a second evaporator, 
 in which the shell pressure was well below 25 lb., additional 
 evaporation might have been produced there. Such an arrange- 
 ment would constitute a multiple effect machine. 
 
 Thus, in Art. 130, the vapor leaving the separator is at 25 lb. 
 pressure and dry. Conduct to a second evaporator in which 
 the shell pressure is 10 lb., the supply temperature 90° and the 
 emerging vapor dry. Assume the coil drip to be at 230.6°. 
 The weight of vapor delivered to this coil is w^ = 2340 lb. The 
 heat it delivers is 2340 {His - ^230-6) = 2340 (1160.4 - 198.8) 
 = 2 250 000 B.t.u. The heat required to vaporize 1 lb. of 
 liquid fed to the shell is Hw — ho = 1143.1 - 58 = 1085.1 
 B.t.u. The weight vaporized is then (ignoring radiation and 
 blow down losses) 2 250 000 H- 1085.1 = 2070 lb. In the two 
 
Heat Transfer Appliances 169 
 
 effects combined, the distilled water produced per lb. of original 
 steam condensed is (2070 + 2340) -f- 2800 = 1.57. By using 
 THREE effects, the economy might have been still further im- 
 proved. Actual rates range up to 3.5 or 4 lb. of vapor per lb. 
 of steam with six effects. The number of stages is limited by 
 the total range of steam pressures. There must be a consider- 
 able temperature difference (and hence pressure difference) 
 between coil and shell. The use of exhaust steam for the first or 
 "primary" coil would make the complication and increased cost 
 of multiple effect apparatus unnecessary: but with exhaust 
 steam the mean temperature difference is low, the surface large 
 and the apparatus expensive. Recent capital ships provide for 
 the use of exhaust, but the amount of exhaust available is in- 
 adequate for continuous running. Weight limitations also pre- 
 clude the exclusive use of exhaust steam. 
 
 When several stages are used, the latter ones are employed not 
 for actual vaporization, but for heating the incoming salt water, 
 just as the distiller heats it in Fig. 45. Matters are so adjusted 
 that there is no surplus of water to waste between effects. As 
 with heat transfer in general, parallel flow economizes surface 
 while counter flow leads to a higher final temperature at the 
 distiller discharge and hence permits of more stages (and more 
 economy) with a given pressure range between primary steam 
 supply and condenser. In parallel flow, the coldest salt water 
 enters the first (highest pressure) stage : in counter flow, it enters 
 the last (lowest pressure) stage. 
 
 Prob. 260. In Art. 130, check the values of C at evaporator 
 and distiller. 
 
 Prob. 261. In Fig. 45, Wg = 2000, ps = 75, Xs = 1.0, U = 
 302°, radiation at evaporator = 20 000B.t.u.perhr.,<2 = 153.01, 
 P3 = 30, xz = 1.0, We = 0. Find weight of distilled water pro- 
 duced per lb. of steam supplied. (Ans., 0.86 lb.) 
 
 Prob. 262. In Prob. 261, h = 90, ts = 205.87. Radiation at 
 distiller = 14 000 B.t.u. per hr. Find weight of salt water 
 pumped through the distiller, and weight of this water that is 
 wasted. (Ans., 26 900 lb. and 25 180 lb.) 
 12 
 
170 Thermodynamics, Abridged 
 
 Prob. 263. With p = 300, find surface necessary at evapor- 
 ator and at distiller in Probs. 261 and 262. (Ans., 62 sq. ft. 
 and 53 sq. ft.) 
 
 Prob. 264. In the foregoing, instead of using a distiller, the 
 vapor from the evaporator is carried to a coil in a second evapor- 
 ator. The pressure in the shell is here 20 lb. The coil drip is 
 at 240.1°, the salt water supply at 141.52° and the vapor pro- 
 duced is dry. If radiation here is 18 000 B.t.u. per hr., and 
 p = 300, find the necessary surface in this second effect. (Ans., 
 104 sq. ft.) How much distilled water is produced per lb. of 
 original steam by the whole machine? (Ans., 1.635 lb.) 
 
 132. Radiation Loss from Steam Pipes and Vessels. For 
 bare pipes or steam vessels with little or no air circulation and 
 with only such steam circulation as is due to condensation, 
 p = 1.8, as for steam radiators in the table of Art. 125. The 
 value may rise to 5 or more if the steam temperature is very high 
 and if there is a flow of steam. Commercial coverings lead to 
 values of p below 0.5. 
 
 Prob. 265. A 10-in. pipe (lOf in. outside diameter, 81.55 
 sq. in. internal cross-sectional area) carries steam initially dry 
 at 200 lb. pressure, a distance of 300 ft. The steam velocity 
 entering the pipe is 8000 ft. per min. The surrounding air is at 
 50° F. Take p = 3, the pipe being bare. How much heat is 
 lost per hoiu-? What weight of steam flows per hr.? What 
 proportion of its original heat (above 32°) does the steam lose? 
 Disregarding drop of pressure, what is the dryness of the steam 
 at the far end of the pipe? (Ans., 841 000 B.t.u. lost per hr.: 
 118 400 lb. flow per hr.: loss = 0.59 per cent.: dryness = 0.99.) 
 Note: the loss of heat goes on whether the pipe is delivering 
 steam at full capacity or not: hence the percentage of loss is 
 great when the delivery of steam is small. 
 
CHAPTER VIII 
 
 FLUID FLOW AND THE STEAM TURBINE 
 
 Flow of Steam 
 133. New Assumption. Thus far, Equation (3) has been our 
 basis. There is one effect of heat not recognized in that Equa- 
 tion as heretofore applied, i.e., the production of kinetic energy. 
 Where the addition of heat produces appreciable fluid velocities, 
 the general equation must be written 
 
 H=T+I+W+V, (73) 
 
 where V stands for that part of the heat H which is converted 
 into kinetic energy or velocity energy. 
 
 If 1 lb. of steam works from condition i to condition 2 without 
 reception or emission of heat, the total energy being constant, 
 and U representing velocity, 
 
 {T+ I+Wh + ^= (,T+ I+Wh + ^. 
 
 Now (T + / + W) may be taken as the total heat of steam 
 above 32°, whether the steam be wet, dry or superheated. 
 Designating such total heat by Q, 
 
 778Q.+ ^=778Q.+ ^. 
 
 In most cases, Z7i^ is negligibly small* as compared with J72^. 
 Hence, nearly, 
 
 U2' = 778 X 2ff(Qi - Q2), 
 
 Cr2 = 224VQi-Q2, (74) 
 
 where U2 is the final velocity in ft. per sec. Values of Qi and Q2 
 are to be taken as for adiabatic constant-entropy expansion 
 
 * In a steam turbine, Ui is the velocity of steam at the entrance to the 
 nozzle and Ui is the velocity at the nozzle outlet. The ratio of Ui to Ui 
 may be about 20: hence that of Ui" to t/i^ is about 400. 
 
 171 
 
172 Thermodynamics, Abridged 
 
 from the total heat-entropy diagram, Fig. 26. The corresponding 
 value of U2 is the velocity attained by expansion in an ideal 
 frictionless nozzle. 
 
 Prob. 266. Dry steam at 200 lb. pressure expands adiabat- 
 ically to 1 lb. pressure. What is the ideal velocity attained? 
 (Ans., 4100 ft. per sec.) 
 
 134. Effect of Friction. In all actual nozzles, there is a fric- 
 tion effect due to the rubbing of steam against metal. If there 
 is sufficient friction, as when the orifice is very small, the velocity 
 may be entirely destroyed. What then becomes of the heat 
 developed by friction? It can only go back into the steam. 
 The law of such process (when the velocity is wholly destroyed) 
 is, Qi = Qi'. the total heat is constant. The expansion valve 
 operation in vapor refrigeration, path def, Fig. 36, illustrates 
 this. No heat has been absorbed or emitted from without: 
 but the entropy has changed. Heat has been converted into 
 velocity, and then reconverted, by means of friction, into heat. 
 • In considering the injector (Prob. 240), nothing was said of 
 velocity. The heat lost by the steam is first converted into 
 velocity. There then follows an exchange of energies between 
 water and steam, the kinetic energy is greatly reduced and the 
 reduction is applied to increase the heat existing as such. In 
 Prob. 240, it was assumed that the total heat was the same after 
 mixture as before. This is not strictly true. A small part has 
 become kinetic energy, represented by the moving body of warm 
 water on its way to the boiler. At a boiler pressure of even 200 
 lb., each lb. of water must have velocity (kinetic energy) merely 
 sufiicient to overcome a head of about 460 ft., or the necessary 
 kinetic energy is 460 ft. lb.— little more than half of 1 B.t.u. 
 The error in Prob. 240 is therefore exceedingly small. 
 
 135. Throttling Calorimeter. This device consists essentially 
 of a very small orifice through which steam is expanded. The 
 velocity is wholly reconverted to heat, so that the law of the 
 process is Qi = Qa- Inspection of Fig. 26 will show that if the 
 total heat remains constant while the pressure decreases, the 
 
Fluid Flow and the Steam Turbine 173 
 
 steam becomes drier and is finally superheated. The process 
 is represented on Fig. 26 by a horizontal line drawn from left to 
 right. The calorimeter is used to determine the quantity of 
 moisture in steam. The pressure is measured on the inlet side, 
 and pressure and temperature are noted on the outlet side. 
 Then if the amount of expansion is sufficient to produce super- 
 heat at the outlet, 
 
 {h+xL)i = {H+k{t' -t)}2. (75) 
 
 Values of hi, Li, ti and H^ are taken from the table, for the pres- 
 sures noted. The observed value of W then indicates the 
 ORIGINAL dryness of the steam. The method fails when t' ^ t, 
 i.e., when no superheating occurs. This may result either from 
 insufficient expansion or from the use of steam too wet to begin 
 with. If a measured quantity of heat Q is added to each lb. of 
 steam on the high-pressure side, the range of the instrument can 
 be increased: 
 
 q+{h+ xLh = {H + h{t' - t)h. (76) 
 
 Problems may readily be solved by the use of Fig. 26, noting 
 that a constant heat process implies a horizontal path. 
 
 Prob. 267. Steam at 100 lb. pressure is expanded to 15 lb. 
 pressure in a throttling calorimeter. The temperature is then 
 found to bs 216°. What was the original dryness? (Ans. 
 0.961.) 
 
 Prob. 268. What is the wettest steam at 100 lb. pressure 
 that can be tested in an ordinary throttling calorimeter using 
 pressures as in Prob. 267? (Ans., 0.959 dry.) 
 
 Prob. 269. Steam at 100 lb. pressure has added to it 100 
 B.t.u. per lb. When expanded to 15 lb. its temperature is 
 found to be 263°. What was its dryness? (Ans., 0.874.) 
 
 136. Velocity and Flow in Actual Nozzles. In Fig. 46, let a 
 be the initial condition of the steam, determined by its pressure 
 and quality. Let h indicate its condition at the same entropy 
 but at some lower pressure, reached after expansion in a nozzle. 
 Suppose friction to be such that 10 per cent, of the velocity 
 
174 
 
 Thermodynamics, Abridged 
 
 energy is re-converted to heat energy. Then lay off hk = 0.10 
 Xah= 0.10(Qa - Qh). The heat content after expansion is 
 Qk, but the pressure is ph: hence the condition of the steam is 
 indicated by the point j, found by drawing kj horizontally. 
 
 Fig. 46. Expansion in the Turbine. 
 
 Thus, if Pa = 200, superheat at a = 200°, and ph = 180: Fig. 
 26 gives Qa = 1308, Qh = 1297. Then hk = 0.10 X 11 = 1.1, 
 Q^. = Q- = 1297 + 1.1 = 1298.1. Fig. 26 now shows the steam 
 to be 186° superheated at the point j. The velocity due to 
 expansion from 200 lb. to 180 lb. is 224Vl308 - 1298^1 = 705 
 ft. per sec. Note that the heat drop is Qa — Qj instead of 
 Qa — Qh, in an actual nozzle with friction. 
 
 Fig. 46a plots resulting velocities (C/2) for these conditions, 
 with various pressures attained during expansion. The specific 
 
Fluid Flow and the Steam Turbine 
 
 175 
 
 volumes, v, of steam are also plotted on this diagram. Thus 
 for an expanded pressure of 180 lb., with 186° of superheat, the 
 specific volume is (Art. 83) 3.3 cu. ft. 
 
 0} 
 
 o. 
 
 ^2000-14 
 
 loool-z 
 
 -flo2zle5itesFSc0zh 
 
 0.002 1 
 
 PressuTEip 
 
 200 180 160 140 120 100 80 
 Flow in Nozzle 
 
 Fig. 46a. Flow in Nozzle. 
 
 If Wo = weight of steam discharged, lb. per sec, the cross- 
 sectional area of the nozzle, at a point where the pressure is p, 
 is «Wo -^ C^2 sq. ft. Values of vlXJ^ are plotted in Fig. 46a: 
 which is based on an initial pressure of 200 lb., 200° initial super- 
 heat and 10 per cent, reconversion of kinetic energy to heat. 
 
 Suppose the " nozzle " is merely a hole in a plate. In this 
 as in any other case, the area of cross-section is mathematically 
 infinite on the high-pressm-e side, for here JJi. = 0. In practice, 
 this side is simply well chamfered. If the pressure of discharge 
 is 180 lb., the area for 1 lb. discharge per sec. is (from Fig. 46a) 
 0.0047 sq. ft. If the discharge pressure is 140 lb., the area per 
 lb. discharged per sec. is 0.0031 sq. ft. Suppose the actual area 
 to be 1 sq. ft. Then the weights discharged per sec. are 1 -^ 
 0.0047 = 212 lb. if the discharge pressure is 180 lb., and 1 -h 
 0.0031 = 323 lb. if it is 140 lb. If the discharge pressure is 116 
 lb., the weight is a maximum, and is 1 -^ 0.00291 = 344 lb. per 
 sec. The diagram shows clearly that the area necessary to dis- 
 charge a given weight is a minimum when the discharge pressure 
 
176 Thermodynamics, Abridged 
 
 is 116 lb. Conversely, the weight discharged through an orifice 
 of given area is a maximum when the discharge pressure is 116 lb. 
 A further decrease in discharge pressure does not increase the 
 weight flowing. The discharge pressure at which maximum 
 weight of flow is realized is called the critical pressure. For 
 steam, the critical pressure is always 58 per cent, of the initial 
 pressure (it is 116 lb., in this case). For air, it is 53 per cent. 
 
 Now consider an actual nozzle designed for 200 lb. initial 
 pressure and 200° initial superheat, with 10 per cent, recon- 
 version. It will have a well rounded entrance. If it is a con- 
 verging nozzle, it will decrease in cross-sectional area from 
 inlet to outlet. Suppose, in such case, the outlet area to be 1 
 sq. ft. The weight of flow will depend on the outlet pressure and 
 will be (as above) 212 lb. if the discharge pressure is 180 lb., 
 323 lb. if the pressure is 140 lb., and 344 lb. if it is the critical 
 pressure of 116 lb. The corresponding outlet velocities, also 
 from Fig. 46a, are 705, 1290 and 1590 ft. per sec. If the dis- 
 charge pressure is the critical pressure (116 lb.) then the pressure 
 will be 140 lb. at a point in the nozzle where the cross-sectional 
 area is 0.0031/0.00291 = 1.065 sq. ft., and the velocity at that 
 point will be 1290 ft. per sec. 
 
 If the discharge pressure is less than the critical pressure, the 
 nozzle should first contract to a throat and then diverge 
 (diverging nozzle). Thus in Fig. 46a, the pressure in the throat 
 (narrowest section) is 116 lb. The ordinate of the nozzle area 
 curve is there 0.00291. At a point beyond the throat, where the 
 cross-sectional area is 1.047 sq. ft., the ordinate corresponding 
 is 1.047 X 0.00291 = 0.00305, so that the pressure at that 
 point is 80 lb. and the velocity is 2030 ft. per sec. Note that 
 the divergent portion of the nozzle does not affect the 
 
 WEIGHT of flow, BUT ONLY THE FINAL (OUTLET) PRESSURE 
 
 AND VELOCITY. The weight of flow is determined by the throat 
 area, and the throat pressure is the critical pressure; if the final 
 pressure is less than the critical pressure. 
 
 The reason for these relations is suggested by Fig. 46a. As 
 expansion proceeds, pressure falls and volume and velocity both 
 
Fluid Flow and the Steam Turbine 177 
 
 increase. The volume at first increases more slowly than the 
 velocity, and afterward more rapidly. The stream of steam 
 reaches its minimum cross-section where the quotient of volume 
 by velocity is a minimum. 
 
 The long equation of Art. 83 need be used for specific volumes 
 ONLY WHEN THE STEAM IS SUPERHEATED. After the expansion 
 passes the saturation curve (Fig. 46), the dryness may be read 
 from Fig. 26 and the corresponding wet steam volume computed 
 directly. 
 
 Approximate formulae for flow in lb. per hour are 
 
 „.?0^'fe„.„,,ed steam 
 
 "" = H-0.00065«„-0 °' ^"P^^^^^t^'i ^*^^°^ 
 
 (77) 
 
 where At = throat area in sq. in., (<„ — t) = superheat. These 
 apply only when the discharge pressure is less than 0.58 the 
 initial, and are based on a reconversion of about 7 per cent, 
 instead of 10 per cent. For air, when the discharge pressure 
 is equal to or less than 0.53 the initial. 
 
 The velocity in the throat is then that of sound. 
 
 Prob. 270. (a) Steam at 200 lb. pressure, 200° superheated, 
 is discharged through a converging nozzle with 10 per cent, 
 reconversion to a chamber where the pressure is 40 lb. The 
 outlet area is 0.1 sq. ft. What is the pressure in the outlet? 
 (Ans., 116 lb.) What weight of steam will be discharged? 
 (Ans., 34.4 lb. per sec.) What is the outlet velocity? (Ans., 
 1590 ft. per sec.) (6) The discharge pressure is changed to 140 
 lb. State outlet pressure, velocity and weight discharged. 
 (Ans., 140 lb., 1290 ft. per sec, 32.3 lb.) (c) A diverging 
 nozzle having the same reconversion is used for the same initial 
 steam conditions. It discharges 2000 lb. per hr. with a discharge 
 pressure of 1 lb. Find pressure, velocity, specific volume and 
 
178 Thermodynamics, Abridged 
 
 area at the throat and at the outlet. (Ans., at throat, 116 lb., 
 1590 ft. per sec., 4.6 cu. ft., 0.00161 sq. ft. At outlet, 1. lb., 
 4130 ft. per sec, 290 cu. ft., 0.0389 sq. ft.) (d) Under case (c), 
 state values of total volume per hr. -t- velocity in ft. per 
 SEC. at throat (Ans., 5.79), outlet (Ans., 140), and at a point in 
 the expansion where the heat content is 1274.7 and the specific 
 volume is 4 (Ans., 6.2). 
 
 Prob. 271. Check Prob. 270, Case (c), as to weight flowing, 
 by Equation (77). (Ans., Equation gives 2100 lb. per hr.) 
 What weight of dry steam at 200 lb. pressure would flow through 
 the same nozzle? (Ans., 2400 lb.) What weight of air at this 
 pressure and at 40° F.? (Ans., 3900 lb.) What would be the 
 throat pressure with air? (Ans., 106 lb.) What is the throat 
 velocity with air? (Ans., 1180 ft. per sec.) 
 
 Prob. 272. Determine throat and outlet areas of a nozzle to 
 discharge 500 lb. of dry steam at 200 lb. pressure, per hr., if 
 the discharge pressure is 2 lb. and the reconversion is 10 per cent. 
 (Ans., 0.0528 and 0.768 sq. in.) 
 
 137. Nozzle Efficiency. The percentage of reconversion is 
 determined by the design of the nozzle. If there were no fric- 
 tion, there would be no reconversion and the nozzle efficiency 
 would be 100 per cent. Good efficiency requires smooth interior 
 finish, a well rounded entrance and a proper angle (15° or so) 
 between the diverging sides. Most important, the outlet area 
 and throat area should be in suitable ratio for the pressures 
 used. If either pressure (inlet or outlet) differs from that con- 
 templated in design, eddies will be produced and a loss of effi- 
 ciency will result. If e = nozzle efficiency, 1 — e = proportion 
 of reconversion and U2 = 224Ve(Qo — Qb), Fig. 46. The value 
 of U2 varies, thus, not with e but with Ve. Values of e up to 
 0.94 may be expected. Note that e = ZJz^ -^ 50 056 {Qa 
 -Qb). 
 
 Prob. 273. The nozzle efficiency is 0.94. How much is 
 velocity reduced by friction? (Ans., 3 per cent.) 
 
Fluid Flow and the Steam Turbine 
 
 179 
 
 Simple Impulse Turbines 
 138. Simple Velocity Diagram. The small sketch in Fig. 466 
 indicates the operation of a single-stage impulse turbine, like 
 the De Laval. The wheel W carries curved blades or buckets, 
 B and revolves with its shaft S. The nozzle N directs the jet 
 at an acute angle (about 20°) with a plane of rotation. 
 
 B-' 
 
 
 Fig. 466. Impulse Turbine; Single Expansion. 
 
 The larger diagram shows the velocities involved. This 
 diagram lies in an imaginary plane which, in the upper of the 
 sketch views, passes through the axis of the nozzle and is per- 
 pendicular to the plane of the paper. 
 
 Draw gh representing the line of motion of the bucket. Draw 
 ab representing the direction of motion of the steam jet, the two 
 lines intersecting at the nozzle angle a. Let the lengths ah 
 and he represent the magnitudes of steam and (peripheral) 
 wheel velocities. Now the relative* steam velocity (velocity 
 
 * Absoltjte and Relative Velocities. (1) A ship sails west at 10 knots 
 speed. The wind blows west at 14.17 knots (absolute velocity of wind). 
 The RELATIVE velocity of the wind with kespect to the ship is 4.17 knots. 
 
180 
 
 Thermodynamics, Abridged 
 
 as it would seem to be to an observer moving with the wheel) 
 is the resultant ac. This is deflected by the bucket to the direc- 
 tion cd. Friction and eddy losses make the ultimate relative 
 
 ^Bucket 
 
 \^ 
 
 sJ~^ 
 
 ^" 
 
 Siafionary 
 Bucket 
 
 ^^' 
 
 in^^-^y^i 
 
 f 1 , 
 
 \ 
 
 n1 
 
 ^'Moving Bucket 
 
 1 
 
 ' 
 
 gr^p 
 
 Fig. 47. Impulse Turbine Triple Expansion. 
 
 exit velocity less than the initial relative velocity, the relation 
 of the two being as in Fig. 48. Let ch denote the magnitude of 
 the relative exit velocity. Combine this with the wheel velocity 
 {hj = be) to obtain the absolute or true velocity, cj, of steam 
 leaving the buckets. 
 
 Force is change of momentum. For 1 lb. of steam, the initial 
 momentum is abjg. The momentum in the direction of 
 MOVEMENT is bg/g. The final momentum in the same direction 
 is — ckjg. The negative sign indicates that this is a rearward 
 momentum. The corresponding force is a reactive impulse, or 
 
 westward. (2) The wind blows bast at 14.17 knots. The absolute wind 
 velocity is 14.17 knots, the relative velocity is 24.17 knots, east. (3) The 
 wind blows sotith at 10 knots absolute velocity. The relative velocity 
 will be 14.17 knots, southeast. (4) The wind blows southwest at 14.17 
 knots absolute velocity. The relative velocity is 10 knots, south. In each 
 case, the relative velocity is directed aft as compared with the absolute velocity. 
 
Fluid Flow and the Steam Turbine 
 
 181 
 
 reaction, while the force corresponding with the momentum 
 bg/g is an impulse. The change of momentum, or force im- 
 pelling the bucket, is {bg + ck)/g. The speed of the bucket is be. 
 
 400 800 2001600 20002400 
 Equivalent Ultimate Velocity=ch ormj 
 
 Fig. 48. Bucket Friction. 
 
 Work is force times speed. Hence the work obtained per lb. 
 of steam is 
 
 W=—{bg+ ck) ft. lb. 
 
 bc(bg + ck) bc{bg + ck) 
 
 B.t.u. (79) 
 
 32.17 X 778 25028 
 
 Prob. 274. Steam at 150 lb. pressure, and dry, expands to 
 1 lb. pressure with a nozzle efficiency of 0.93. Find the resulting 
 velocity. (Ans., 3850 ft. per sec.) 
 
 Prob. 275. With steam velocity = 3850, bucket velocity 
 = 600, a = 20°, construct to some convenient scale a velocity 
 
182 Thermodynamics, Abridged 
 
 diagram like that of Fig. 466, the relative angles acg and dcg being 
 equal. Measure the necessary lines and state the work done per 
 lb. of steam. (Ans., 125 B.t.u.) 
 
 139. Kinetic Efficiency. The efficiency of conversion of kinetic 
 energy into work at the buckets is what we may call the 
 
 KINETIC EFFICIENCY, 
 
 e,= W-- ^"' ^ 50056^ 
 
 ^* 2gX 778 UJ ' ^^^^ 
 
 It is easily shown that the maximum value with equal relative 
 angles is attained when be = ^bg, and that that value is 
 
 + F 
 
 _ faby 1_ 
 
 2 ' 
 
 where F = bucket friction coefficient = ch/ca = eh/cd, to be 
 taken from Fig. 48. If there is no bucket friction, F = 1 and 
 ^Amax= {iglohy. If a = 0, then the point k would coincide 
 with the point /and e^^^^ would be 100 per cent, when be = \ba. 
 Note that a is a necessary evil and that the ideal performance is 
 realized when cj is vertical. 
 Maximum kinetic efficiency occurs when the peripheral 
 
 SPEED OF the buckets IS HALF THE ROTATIVE COMPONENT OF 
 
 THE STEAM SPEED. High peripheral speeds are essential to 
 efficiency. 
 
 Prob. 276. Compute e^ in Prob. 275. (Ans., 0.42.) 
 
 Prob. 277. With a = 0°, Fig. 47, gcd = 0°, F = 1.0, what 
 would have been the best value of be, W and Ck for Prob. 275? 
 (Ans., be = 1925 ft. per sec, W = 297 B.t.u., e^ = 1.0.) 
 
 Prob. 278. In Prob. 275, for a = 20°, F = l.Q, state the 
 best value of be and the resulting values of W and e*, the relative 
 angles being equal. (Ans., be = 1810 ft. per sec, W = 261 
 B.t.u., ek = 0.88.) 
 
 Prob. 279. With the bucket friction and value of a assumed 
 in Prob. 275, and with equal relative angles, state the best values 
 of be, W and e^. (Ans., be = 1810, e^ = 0.76, W = 226 B.t.u.) 
 
Fluid Flow and the Steam Turbine 
 
 183 
 
 Fig. 49 shows the variation in W with buclcet speed, be, for 
 the conditions of Prob. 275, except that the friction factor F is 
 taken constantly at 0.75. The variation of i^ is such that the 
 work is actually a little greater than is indicated by the curve, 
 at high speeds, and a little lower at low speeds. Note that the 
 curve of Fig. 49 represents e^ (to another scale). 
 
 
 j^l20 
 
 ^ 600 800 1000 1200 1400 1600 1800 2000 
 
 Peripheral Speed of Buckets.f t per sec. 
 
 Fig. 49. Effect of Bucket Speed on Work at Buckets and on Kinetic 
 
 Efficiency. 
 
 140. Primary Difficulties in Turbine Design. Let u (= be, 
 Fig. 466) denote peripheral speed in ft. per sec, N = r.p.m., 
 d = wheel diameter (pitch diameter) in ft. Then wdN = 60w or 
 dN = 19.1m. For high values of u, either d or N must be high. 
 The smaller the capacity, the smaller preferably, is d: hence N 
 is high in small machines. Flexible shafts and reduction gears 
 may be essential in turbines of this type. The largest values 
 of u in actual use are around 1300; of N, around 24 000. 
 
 The capacity of an impulse turbine is determined by the nozzle 
 
184 Thermodynamics, Abridged 
 
 area and steam conditions rather than by any dimension of the 
 wheel. The height of the buckets should somewhat exceed the 
 nozzle outlet diameter, to prevent lateral spreading of steam. 
 The wheel runs in a chamber containing steam at condenser 
 pressure, and (in this type) need not make a close fit with its 
 casing. No high pressure steam is carried beyond the nozzle. 
 The pressure is constant across the face of the buckets. 
 
 Prob. 280. In Prob. 275, what is the pitch diameter of the 
 bucket wheel if it makes 10 000 r.p.m.? (Ans., 1.146 ft.) 
 
 141. Turbine Efficiency. The turbine must be regarded as 
 operating in the complete expansion (Rankine) cycle: Art. 80. 
 Its ideal efficiency is therefore to be computed as in Equations 
 (45), (49). We have seen that the conversion of heat into 
 kinetic energy is accompanied with a nozzle friction loss, and 
 that the conversion from kinetic energy into work at the buckets 
 is accompanied with a further loss. Let Bj = efficiency of ideal 
 cycle, e^ = nozzle efficiency, e^ = kinetic efficiency. Then the 
 efficiency from steam to work at the buckets is BjX e^X e*. 
 Further loss now occurs. 
 
 The revolving disk and the rows of buckets both undergo 
 friction against the steam. For wheels under 36 in. diameter 
 revolving in a chamber containing dry steam, empirically, 
 
 Ld = cPu"-^ -f- 4400?) B.t.u. per hr., 
 
 Lb = dh^-^'w"-^ -=- 990d B.t.u. per hr., 
 
 where v = specific volume of dry steam at nozzle outlet pressure, 
 h = radial height of buckets, in.. La = friction loss at disk. 
 Lb = friction loss for one row of buckets. For m rows of buckets 
 on one disk, the total rotation loss per disk is Ld + mLt. 
 This is to be multiplied by the factors of Fig. 50 if the steam is 
 (as usually) not dry in the wheel chamber. In Fig. 46, the steam 
 enters the chamber as it leaves the nozzle, at the state d. It 
 leaves the chamber at a state h, defined by Q^ — Qk = W, the 
 pressure at h being the same as at d. The mean of the drynesses 
 at d and h determines the abscissa of Fig. 50. 
 
 (80) 
 
Fluid Flow and the Steam Turbine 
 
 185 
 
 1.0 
 
 ] 
 
 — V n 
 
 t 
 
 3 1.7 
 
 oo 
 
 r 
 
 in 
 
 t 
 
 O 1.0 
 
 1 
 
 iTiR 
 
 I 
 
 LU 1.9 
 
 1 
 
 «^ 1 X 
 
 J 
 
 _3 1.T 
 
 1 
 
 o 
 
 1;;;n 
 
 r 
 
 L 
 
 (0 1.0 
 O 
 
 f 
 
 
 J 
 1 
 
 t°i.Z 
 
 t 
 
 ^ il 
 
 / 
 
 i2 '•' 
 
 05 
 
 
 > 7 
 
 
 o i.u ^ 
 
 
 O /' 
 
 
 "'...r^ 
 
 
 y^ 
 
 
 100 50 
 SuperheatF Mo 
 
 k — ■ — >< — 
 
 4 8 12 
 isture, per cent 
 
 Fig. 50. Correction Factors for Rotation Loss. 
 
 It should be noted that the amount of rotation loss depends on 
 the dimensions of the wheel and not on the weight of steam 
 flowing. Hence it is relatively large for machines of small output 
 or with large diameter wheels. If two rows of buckets are 
 placed on one disk, Ld is not increased. Hence the total rota- 
 13 
 
186 Thermodynamics, Abridged 
 
 tion loss is not doubled. Rotation loss is reduced when the 
 steam is initially superheated. It is very great where buckets 
 revolve in very wet steam. For wheels larger than 36 in., Ld may 
 be 5 per cent, greater than is indicated by Equation (80) . 
 
 Prob. 281. A wheel of 1.146 ft. dia. revolves at 10 000 
 r.p.m. in dry steam at 1 lb. pressure. Buckets are 1.3 in. high. 
 Find Ld and h. (Ans., 103 and 560 B.t.u. per hr.) 
 
 Prob. 282. The wheel in Prob. 281 revolves in a chamber in 
 which the pressure is 1 lb. The nozzle inlet pressure was 150 lb. 
 (dry steam) and the nozzle efBciency 0.93. What is the dryness 
 at the nozzle outlet? (Ans., 0.80.) If the work at the buckets 
 is 125 B.t.u., find Qh- (Ans., 1069.) What then is cc/.? (Ans., 
 0.966.) What is the mean dryness in the wheel chamber? 
 (Ans., 0.883.) What is the rotation loss correction? (Ans., 
 1.765.) What is the corrected rotation loss? (Ans., 1170 B.t.u. 
 per hr.) 
 
 142. Efficiency and Steam Rate. The bearing friction and 
 shaft packing losses of the turbine may be kept down to 1 per 
 cent, of the shaft output. Radiation losses are small. Losses 
 incurred in speed reduction are perhaps properly chargeable to 
 the turbine in comparing its performance with that of a re- 
 ciprocating engine. Efficiencies of reduction gears range up 
 to 0.98: hydraulic transmissions and electric drive are tech- 
 nically less efficient. The work at the buckets corresponds 
 with the Ihp. of an engine, but cannot be measured directly. 
 Turbine capacities and economics are therefore expressed on the 
 basis of brake hp. or shaft hp. The ratio of shaft hp. to bucket 
 hp. may be called the mechanical efficiency, although the 
 meaning of this term is then not quite the same as with steam 
 engines. This usage lumps the rotation loss with the mechanical 
 friction loss. 
 
 Let W = work at buckets per lb. of steam, in B.t.u., w = lb. 
 of steam per hr. Then bucket hp. = Ww s- 2545. Let L = 
 total rotation loss, B.t.u. per hr. Then shaft input = Ww — L 
 B.t.u. per hr. Let 1-1= proportion of shaft input lost in 
 
Fluid Flow and the Steam Turbine 
 
 187 
 
 mechanical friction (say 1 per cent.). Then l{Ww — i) = shaft 
 output, B.t.u. per hr., or l{Ww — i)/2545 = shaft hp. and the 
 mechanical efficiency, as above defined, is 
 
 7/TTr T\ TT7 2545 X shaft hp. 
 Ww -L)-^Ww = ^^ = e^. 
 
 The efficiency from steam to work at the shaft is e = e^ X Sjv 
 X Cj. X Bj^. What corresponds with the " relative efficiency " 
 of the steam engine (Art. 93) 
 is the product 6^ X Cjy. This 
 may be higher than in steam 
 engines, and is chiefly deter- 
 mined by the value of e^. If 
 (Fig. 46) the steam and feed 
 water conditions are Qa and 
 h, the weight of steam per 
 brake hp.-hr. is as in Art. 
 93, w/hp. = 2545 -^ e{Qa - 
 hb) lb. Fig. 51 gives ideal 
 steam rates for the complete 
 expansion cycle (Arts. 80, 
 84). Results from actual 
 trials are given below. Art. 
 148. 
 
 ZOO m 140 
 Absolute Initial Steam Pressure lb.persq.in. 
 
 Fig. 51. Steam Rates for Ideal Tur- 
 bine Cycles. 
 
 Prob. 283. The turbine gives 125 B.t.u. work at buckets per 
 lb. of steam, and the rotation loss is 1170 B.t.u. per hr. Me- 
 chanical friction is 1 per cent. The shaft hp. is 100. Find 
 weight of steam per hr. (Ans., 2053 lb.) 
 
 Prob. 284. In Prob. 283, what is the "mechanical efficiency"? 
 (Ans., 0.985.) If the initial steam was at 150 lb. pressure, and 
 dry, and the back pressure was 1 lb., with e^^ = 0.93, find ej, 
 Bk, and e. (Ans., 0.28, 0.42, 0.108.) What is the "relative 
 efficiency"? (Ans., 0.39.) What weight of steam is used per 
 brake hp.-hr.? (Ans., 20.53 lb.) Check the last result from 
 Fig. 51. 
 
188 
 
 Thermodynamics, Abridged 
 
 Eniering 
 Sfeam 
 
 Multi-stage Ttirbines 
 143. Velocity Compounding. In Fig. 466, there is a consider- 
 able exit velocity, cj, which carries away with it kinetic energy. 
 This velocity should be kept small, or else means should be 
 
 provided for the further util- 
 ization of the wasted kinetic 
 energy. The method is sug- 
 gested in Fig. 52. The steam 
 leaving the buckets A is re- 
 versed by stationary buckets 
 
 B, projecting inward from 
 the casing, and used again 
 in a second set of buckets 
 
 C. To decrease rotation loss 
 (Art. 141), these additional 
 buckets are mounted on the 
 original disk. 
 
 In Fig. 47, the triangles 
 ahc, chj duplicate those of 
 Fig. 466. Now lay off jm 
 parallel with ba, implying equality of angles a for both rows 
 of buckets. The length of jm is that of cj, reduced by bucket 
 friction as in Fig. 48. Lay off mn, parallel and equal to 6c. 
 (Both rows of buckets move in the same direction at the same 
 speed.) Draw jn and draw np = jn so that the relative an- 
 gles jnt and pnt are equal. (This equality does not always 
 exist.) Lay off nq, less than np, the ratio of the two being given 
 by Fig. 48. Lay off qr parallel and equal to mn. Then the 
 additional work gained by the second row of moving buckets, 
 AT NO ADDITIONAL COST for heat, is 
 
 Fig. 52. Arrangement of Buckets in 
 a Velocity-Compounded Turbine. 
 
 w. 
 
 mn{mt + ns) 
 25 028 
 
 B.t.u. per lb. of steam. 
 
 (81) 
 
 If the point r lies to the left of a vertical line through n, the 
 component ns is negative. 
 
Fluid Flow and the Steam Turbine 
 
 189 
 
 Such velocity compounding increases work and efficiency. 
 In the present illustration, only two stages are used. This is, in 
 fact, all that it would be profitable to use under the velocity 
 relations shown by Fig. 47 to be existing. With lower values 
 of u, the desirable number of stages increases. By velocity 
 compounding, we may increase the kinetic eflBciency at a given 
 value of u, or may maintain some desired efficiency without 
 using high values of u. Bucket friction establishes a limit to 
 this procedure. 
 
 Prob. 285. In Prob. 275, a second stage is added as described 
 in Art. 143. What is the additional work obtained per lb. of 
 steam? (Ans., 35 B.t.u., or 28 per cent.) 
 
 144. Pressure-Compounded Impulse Turbine (Curtis Type). 
 It is the RATIO of bucket speed to steam speed which determines 
 
 steam Admission] jlSvS I* rIS^ 
 
 1 Velocity 
 
 -! — Pressure 
 
 iri 
 
 ^■£31 
 
 Fig. 53. Arrangement of Nozzles and Buckets in a Two-Stage Curtis Turbine. 
 
190 
 
 Thekmodynamics, Abridged 
 
 kinetic efficiency, ratherthan the bucket speed itself. Velocity- 
 compounding, as just described, is equivalent to increase of 
 bucket speed but is limited by bucket friction. Peessuke 
 COMPOUNDING attacks the problem from another angle, by 
 decreasing the steam speed. This is done by expanding in 
 stages. One set of nozzles expands the steam part way down 
 to condenser pressure. If half the " heat drop " ab, Fig. 46, 
 
 // 
 
 
 Fig. 54. Pressure-Compounded Turbine. 
 
 is realized in this stage (or set of nozzles), the outlet velocity 
 resulting will be that in a single-expansion nozzle divided by ^2 
 (Equation (74)). This velocity is then used in a wheel which 
 may be either simple or velocity-compounded. Leaving this 
 wheel, the steam expands down to condenser pressure through 
 the second set of nozzles and then strikes a second wheel. Fig. 53 
 suggests the arrangement with two pressure stages (sets of 
 nozzles) and two-stage velocity compounding in each pressure 
 
Fluid Flow and the Steam Turbine 
 
 191 
 
 stage. The number of pressure stages should increase as the 
 peripheral speed, u, decreases, in order to maintain an advan- 
 tageous ratio between u and the steam velocity. 
 
 Assume steam at 150 lb. pressure, dry, with a condenser 
 pressure of 1 lb. and a nozzle efficiency of 0.93. Locate the 
 initial state, o, on the total heat-entropy diagram, as in Fig. 54. 
 Draw ab vertically, locating b. Read off Qa = 1194, Qb = 877. 
 
 a 
 
 "^ 
 
 r "7 
 
 Fig. 55. Two-Stage Velocity Diagram, Pressure-Compounded Turbine. 
 
 Assume two pressure stages: therefore bisect ab, determining 
 Pc = 17 lb., Qc = 1035.5. The first stage nozzles will expand 
 from 150 to 17 lb. pressure and the velocity resulting is 224 
 V0.93(1194 - 1035.5) = 2720. Assume u = 400, k = 20° and 
 equal relative angles. Construct the velocity diagram for two 
 velocity stages as in Fig. 55, following Fig. 47. The work per 
 lb. of steam in the first pressure stage is 
 
 ^^{bg+ck+mt+ns) 
 
 = 2!^ (2550 -f- 1230 + 1095 -f 215) = 81 B.t.u. 
 
 The heat in the steam entering the second pressure stage is 
 (Fig. 54) Qe = Qa — 81 = 1113. This determines the point e on 
 
192 
 
 Thermodynamics, Abeidged 
 
 the line of 17 lb. pressure. Adiabatic expansion in the second 
 pressure stage is represented by the vertical line ef, determining 
 Qf = 941 B.t.u. Then the heat to be converted into velocity in 
 this second pressure stage is Qe — Q/ = 1113 — 941 = 172 B.t.u. 
 In the first stage it was 1194 - 1035.5 = 158.5 B.t.u. The 
 second stage will therefore have a higher initial velocity and 
 will give more work per lb. of steam. 
 
 It is usually desirable that the stage conversions be equal: 
 i.e., that Qa — Qo = Qe — Q/. This leads to similar blading in 
 the two stages. The conversions are unequal in this case 
 because of reheat, i.e., the reconversion of velocity not utilized, 
 into heat. To allow for the influence of reheat, we may arbi- 
 trarily increase the drop in the first stage. This implies lowering 
 the discharge pressure of the nozzles in that stage. An approxi- 
 mation to the desired condition is obtained by starting with a 
 heat drop which is a mean between Qa — Qc and Qe — Qt- 
 
 Prob. 286. What is the mean value of the two heat drops in 
 Art. 144? (Ans., 165.3.) If this drop is used for the first stage, 
 what is the value of QJ (Ans., 1028.7.) What is then the 
 
 = 400 
 |300 
 
 fzoo 
 
 ^100 
 
 3 6 9 12 15 18 
 No.ofS 
 
 21 24 27 30 
 age 
 
 Fig. 56. Steam Velocity, Reaction Turbine. 
 
 value of 'Pol (15.7.) Of the outlet velocity? (Ans., 2780.) 
 Assume that the use of this velocity in the first stage velocity 
 diagram. Fig. 55, gives work equal to 84 B.t.u. per lb. of steam. 
 What then is the value of Qe? (Ans., 1110 B.t.u.) Of p,,? 
 (Ans., 15.7.) Of Qjt (Ans., 944.7 B.t.u.) Of the heat drop 
 in the second stage? (Ans., 165.3.) Of the work in the second 
 stage? (Ans., 84 B.t.u.) What is the total work at the buckets 
 
Fluid Flow and the Steam Turbine 
 
 193 
 
 of the whole turbine? (Ans., 168 B.t.u. per lb. of steam.) What 
 is the gain over the machine of Prob. 275? (Ans., 43 B.t.u. or 
 34 per cent.) Note that this gain is secured even though u has 
 been decreased. 
 
 145. Reaction Turbine. The impact due to the velocity cj, 
 Fig. 466, is a reactive impact. The steam is flowing away 
 from the bucket. This reaction is less than the impulse produced 
 by the inlet velocity ab. Under ordinary conditions, it must 
 be less. If, however, the buckets are so shaped that some 
 expansion (lowering of pressure) occurs in them, there may be 
 an increase of velocity instead of a decrease, from ac to ch, in 
 spite of bucket friction. The reaction work may then equal or 
 even exceed the impulse work. 
 
 Fig. 57. Velocity Diagram, Reaction Turbine. 
 
 In Fig. 57, assume that the relative exit angle gcd is equal 
 to the absolute inlet angle, a. Assume also that there is 
 expansion in the buckets, so great that the relative exit velocity 
 without friction would be cd > ab. Assume further that the 
 amount of expansion is such that this velocity, reduced by friction 
 as in Fig. 48, is ch = ab. Transpose the triangle chj to the posi- 
 tion ah'f. Here it will be symmetrically opposed to the tri- 
 angle abc. 
 
194 Thermodynamics, Abridged 
 
 The work per lb. of steam, following Equation (79), is 
 
 ^^ /I. 1 N ^g X ch' 
 25028^^^+^^) = "25028" ^•*-"- 
 
 About gf as a center describe a circular arc bl. Draw cf vertically. 
 Then, from geometry, be X ch' = c/^ and 
 
 ^ = 2W8=(il:5y^-*- ^«2) 
 
 This modified system of velocities is approximated in turbines 
 of the REACTION (Parsons) type. There is a further character- 
 istic of reaction turbines. Since expansion is to occur in the 
 blades, no nozzles are necessary. The steam entering the 
 first row of blades is simply at steam pipe velocity, around 100 
 to 200 ft. per sec. Hence peripheral speeds (which should be 
 about half the steam speed) may be very low without sacrifice 
 of efficiency. Each row of blades may be regarded as a set of 
 nozzles designed for a very small heat drop. The reaction 
 turbine is essentially a low speed turbine. It contains rows of 
 moving buckets alternated with rows of stationary reversing 
 buckets (like those of Fig. 47), projecting inward from the casing. 
 
 Prob. 287. In Fig. 57, ab = 224, be = 100, tan a = 0.5. 
 Find W. (Ans., 1.2 B.t.u.) 
 
 146. Steam Velocities in Reaction Turbines. With low steam 
 velocities, the work done in a row of buckets is necessarily very 
 small. This is suggested by Prob. 287. Hence the number of 
 rows is very large. If the average work per row is 5 B.t.u., and 
 the total bucket work of the turbine 150 B.t.u., 30 stages or 
 rows of buckets are required. 
 
 Fig. 57 shows that the absolute exit velocity cj is less than 
 the absolute entrance velocity ab. In passing to the second row 
 of moving buckets, however, further expansion occurs in the 
 fixed blades: so that the absolute entrance velocity of the 
 second row is made somewhat greater than ab. Hence (if 
 the value of u is the same for both rows) a greater amount of 
 work will be done in the second row than in the first. 
 
Fluid Flow and the Steam Turbine 
 
 195 
 
 A reaction turbine, unlike one of the impulse type, carries its 
 wheel chamber constantly full of steam. Its capacity is there- 
 fore determined by the area of the annulus comprising the 
 buckets, as well as by the steam conditions. Tip clearances 
 must be low. This is not necessary in impulse turbines. 
 
 yelodfy 
 
 Fig. 58. Parsons Reaction Type Steam Turbine. 
 
 Fig. 58 shows a marine type of reaction turbine. Steam 
 enters at jE and moves to the left. The buckets are mounted on 
 three wheels or drums, of increasing diameter toward the exhaust 
 end. This increase is made in order to avoid the use of very 
 long buckets (long radially) toward the low pressure end. The 
 valve V is employed to admit high-pressure steam to the lower 
 pressure stages. This is done when overloads are encountered 
 
196 Thermodynamics, Abridged 
 
 but involves some loss of eflSciency, there being an impaired 
 relation of velocities. The three short drums to the right of the 
 steam inlet E may be provided with reversed buckets, then 
 constituting an astern turbine. Economy is not necessary 
 when moving astern: hence this system contains only a few 
 stages. The reversing action is not considered in the velocity 
 and pressure curves below the illustration. Means must be 
 provided for directing the steam either to the right or to the 
 left from E. Values of u for the various drums have ratios equal 
 to those of drum diameters. A common successive ratio is ^2^ 
 The work done by the successive drums increases toward the 
 exhaust end. With three drums, the relative amounts of work 
 are often in the ratio 1 : 1 : 1|. 
 
 Suppose the adiabatic drop to be as in Art. 144, S17 B.t.u. 
 Bucket friction should be low because of the low steam speeds 
 used, but high because of the large number of buckets. There 
 is a leakage loss past the blade tips, which reduces efficiency. 
 Allowing for leakage and rotation loss (Art. 141), the work at 
 the buckets may be expected to be around 75 per cent, of the 
 adiabatic heat drop in carefully designed machines. 
 
 Assume the probable work at the buckets to be in our case 
 230 B.t.u., and that 60 B.t.u. are to be derived from a first drum. 
 How many stages (rows of buckets) shall that drum have? 
 
 Assume u = 100, a = 26°, initial steam velocity 257 ft. per 
 sec. By the method of Fig. 57, the work done in the first stage 
 is 1.45 B.t.u. It will be greater in later stages. The number of 
 stages will therefore be less than 60 -J- 1.45, or less than 41. 
 
 Assume a probable variation of steam velocities from beginning 
 to end of drum as in Fig. 56. In general, this velocity increases 
 more and more rapidly toward the low-pressure end, as here 
 indicated. The curve may be represented by an equation in the 
 form 
 
 U= aX e'", ^ (83) 
 
 where U = steam velocity, ft. per sec, 
 s = the number of the stage, 
 
 e = base of Napierian system of logarithms, and a and b 
 are constants. 
 
Fluid Flow and the Steam Turbine 197 
 
 The curve is established when the steam velocities at two 
 stages are known. We have assumed U = 257 when s = 1. 
 Assume also that U = 364 when s = 30. Then 
 
 ^" = 5-'=^^^ loge||=296, 6 = 0.012. 
 
 Ui = 257 = aeOoiS log. 257 = log. a + 0.012, a = 254. 
 Then the equation of the whole curve is U = 254 X e""^^'. 
 
 Referring to Fig. 57, the work done at any stage is (Equation 
 (82)), 
 
 25 028 25 028 25 028 
 
 _ (U cos a)^ — {TJ cos a— uY _ 2uU cos a — u^ 
 ^ 25 028 ~ 25 028 
 
 li U = velocity at any stage, Ui = velocity at first stage, 
 
 c = Ui/u, then 
 
 U 
 
 Ui e' 
 
 = e^-^'-" and U = cuX e^'-'-^K 
 
 Then 
 
 ^= 25028 = 25028 (2'''=°^«><^'"-l) 
 
 The work of the whole drum is 
 
 For the conditions assumed, 
 
 1 n noo f / gooi^ca-i) — 1 \ 
 
 ^^ = 25028 P X 2-57 X 0.8988 ( -^;0l2— )- ^ ) • 
 To find the value of e^^^^^'-", put this quantity equal to x. Then 
 logs X = 0.012(5 - 1), 
 
 SPF= 0.4(4.62^'-.)= 60, 
 
 ^•^Q^;~^^ = 150 + 5= 385(0:- 1), 
 log (535 + s)- 0.005225 = 2.581. 
 
198 
 
 Thermodynamics, Abridged 
 
 By trial and error, s is found to be between 33 and 34. A whole 
 number must be chosen. Select 34. Then in Equation (84), 
 
 gO.396 2 
 
 zw 
 
 (gO.396 _ 1 \ 
 
 ^•62 -0:012--^^) 
 
 = 0.4(186.3 - 34) = 60.92 B.t.u., 
 
 instead of 60, as originally desired. The steam velocity at the 
 34th stage may now be computed. It will be the initial steam 
 velocity of the next succeeding drum. 
 
 Fig. 59. Westinghouse-Parsons Double Flow Turbine. 
 
 The whole rotor of a reaction turbine is subject to an end 
 thrust due to the static pressure of the steam toward the low- 
 pressure end. This is often counterbalanced by " dummy 
 pistons " on which the areas are equivalent to the exposed 
 drum areas while steam pressures are in the opposite direction. 
 End thrust can also be eliminated, as in Fig. 59, by supplying 
 steam near the middle of the length and allowing it to flow both 
 ways. This illustration shows another common feature, the 
 use of an impulse wheel (one set of nozzles, two velocity stages) 
 preceding the reaction elements. A large proportion of the 
 heat is converted into work at the impulse wheel, at a point 
 where the steam is fairly dry and the rotation loss (Art. 141) 
 consequently small. Hence the reaction part of the turbine is 
 reduced in size and contains fewer stages. Furthermore, tip 
 
Fluid Flow and the Steam Turbine 199 
 
 leakage (which is most objectionable at the high-pressure end 
 of the reaction turbine) is limited to the low-pressure reaction 
 stages. 
 
 Prob. 288. Check the following values of Art. 146: 1.45, 
 0.012, 254. What would be the work of the drum if 33 stages 
 were used? (Ans., 58.9 B.t.u.) What is the st^am velocity at 
 the 34th stage? (Ans., 382 ft. per sec.) 
 
 Turbine Economy 
 
 147. Low Pressure Turbines. The turbine works with com- 
 plete expansion while the engine does not. The former can do a 
 great deal with steam rejected by an engine. Thus, in Fig. 26, 
 suppose steam at 5 lb. pressure, 0.80 dry, to be available. Its 
 heat content is 933 B.t.u. Expanded adiabatically to 1 lb. 
 pressure, its content becomes 849 B.t.u. The ideal work is then 
 84 B.t.u. per lb. of steam, and the actual work might easily be 
 60 per cent, of this. 
 
 Prob. 289. A steam engine receives dry steam at 270 lb. 
 pressure and expands it to 5 lb. pressure. A turbine then 
 expands it down to f lb. pressure. Ideally, what proportion of 
 the total power is derived from the turbine? What is the ideal 
 steam rate for the combination? (Ans., the turbine gives 42 per 
 cent, of the power of the engine. Steam rate, 6.6 lb. per Ihp.-hr.) 
 What would have been the ideal rate of a turbine alone, using 
 dry steam at 270 lb. pressure and expanding to J lb. back pres- 
 sure? (Ans., 6.6 lb.) 
 
 148. Tests of Steam Turbines. Following are ranges of steam 
 rates from many trials : 
 
 Lbs. Steam per Brake hp.-hr. Vacuum, in. 
 
 Saturated steam: Simple impulse type ... 14 to 23 25 to 28 
 
 Pressure compounded . . 16 to 20 27 to 29 
 
 Reaction type 14 to 15 27 to 28 
 
 Superheated steam: Simple impulse type. . .12i to 15| 27 to 28 
 
 Pressure compounded . . 12i to 16| 28 to 29| 
 
 Reaction type llj to 14^ 27 to 29 
 
 Many recent tests show improvement on these figures, which in no case 
 involved very high superheat. 
 
200 
 
 Thermodynamics, Abridged 
 
 Normal atmospheric pressure is 14.696 lb. per sq. in. or 29.92 
 in. of mercury. Vacuum is measured downward from atmos- 
 pheric pressure. If the vacuum gauge reading is po in. of mercury 
 the absolute pressure is (14.696/29.92) (29.92 - po) = 0.491 (29.92 
 — Po) lb. per sq. in. 
 
 .-^20 
 
 rkw- 
 rveE 
 
 18 
 
 <U 3 
 
 
 Cl-O 
 
 
 l--~^ 
 
 
 o 
 
 16 
 
 e-o" 
 
 
 -^rrf 
 
 
 ^-f 
 
 14 
 
 <o iR 
 
 
 iZ 03 
 
 
 •5' 
 
 
 m 
 
 \l 
 
 E 
 
 
 (0 
 
 
 
 10 
 
 ^ 
 
 
 0.2 
 
 0.6 0.8 1.0 1.2 1.4 L6 
 Proportion of Rated Load 
 
 Fig. 60. Turbine Efficiency as Affected by Load. 
 
 Fig. 60 shows the effect of variable load on turbine economy, 
 the curves being similar to those for steam engines (Fig. 29). 
 Particulars follow: 
 
 A and B, 1050 hp. Westinghouse-Parsons machine, 165 lb. 
 pressure, 28 in. vacuum. A, saturated steam; B, steam 
 superheated 100°. 
 C and D, 400 hp. Westinghouse turbine, '28 in. vacuum: 
 
 C, saturated steam; D, steam superheated 90°. 
 E, 9000 kw. 5-stage Curtis turbine, 215 lb. pressure, 125° 
 
 superheat, 29 in. vacuum. 
 Fig. 61 illustrates the effect of varying superheat. The steam 
 rate tends to bear a straight line relation with the amount of 
 superheat. The lines are. 
 
Fluid Flow and the Steam Turbine 
 
 201 
 
 A, 2000 hp. Curtis turbine. 
 
 B and C, average results: B for 26 in. vacuum, C for 28 in. 
 
 vacuum. 
 D and E, 400 hp. Westinghouse machine, 28 in. vacuum. D, 
 
 1.3 full load; E, full load. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^~ 
 
 
 ■--. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 — ■ 
 
 
 
 
 
 
 
 
 
 
 — ■ 
 
 / 
 
 — , 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 — ■ 
 
 — 
 
 . . 
 
 
 
 
 
 8 
 
 
 , 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ■ 
 
 
 
 C 
 
 
 
 
 
 *-^. 
 
 
 — - 
 
 
 
 
 
 
 
 
 
 
 
 
 
 t 16 
 
 
 
 
 
 
 
 
 
 
 ' 
 
 ■— ■ 
 
 —- 
 
 — 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ""■ 
 
 A 
 
 
 f 14 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^^ 
 
 
 — 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 F 
 
 S. 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 :::; 
 
 — 
 
 — . 
 
 , 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^"* 
 
 ' — 
 
 —- 
 
 ^ 
 
 zz 
 
 
 to 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 IP 
 
 ( 
 
 ) 
 
 Fi 
 
 2 
 
 3. 6 
 
 
 
 1. 
 
 4 
 
 Tu] 
 
 
 
 rbin 
 
 6 
 Su 
 
 eE 
 
 
 
 IP( 
 
 fici 
 
 8 
 
 encj 
 
 
 he 
 
 T as 
 
 IC 
 
 AfE 
 
 
 
 ecte 
 
 12 
 ah 
 
 db 
 
 
 r. 
 
 ySi 
 
 M 
 
 ipei 
 
 K) 
 
 hea 
 
 t. 
 
 )0 
 
 1 
 
 Fig. 62 shows a straight line relation between vacuum and 
 economy. ^ is a 2000 hp. Curtis machine ; B, a 4750 kw. turbine ; 
 C and Z) are 1500 kw.: P is at | load, C at full load. 
 
 Prob. 290. A recent turbine is guaranteed to use not over 
 13 lb. of steam per kw.-hr. with steam at 300 lb. pressure, super- 
 heated 300° and at 28.9 in. vacuum. Are there any rates shown 
 in Art. 148 or Figs. 60-62 indicating that this is possible? What 
 is the ideal steam rate per hp.-hr. under these steam conditions? 
 (Ans., 5.44 lb.) Per kw.-hr.? (Ans., 7.3 lb.) If the efficiency 
 from shaft to switchboard is 0.93, the guarantee implies what 
 steam rate per brake hp.-hr.? (Ans., 9 lb.) If the mechanical 
 efficiency is 0.95, what is the relative efficiency? (Ans., 0.64.) 
 14 
 
202 
 
 Thermodynamics, Abridged 
 
 Prob. 291. Find the vacuum in in. of mercury at normal 
 atmospheric pressure when the back pressure is 1 lb. absolute. 
 (Ans., 27.88 in.) 
 
 Prob. 292. The barometer reads 29.4 in. The vacuum gauge 
 reads 27 in. What is the absolute back pressure? (Ans., 1.18 lb.) 
 
 Prob. 293. In Fig. 60, what percentage reduction in steam 
 rate results at normal load by substituting for saturated steam, 
 steam superheated 90° or 100°? (Ans., 8.6 and 9.5 per cent.) 
 
 Prob. 294. What is the weight of steam used per hour at 100 
 per cent, overload by turbine B, Fig. 60? (Ans., 29,000 lb.) 
 
 2Ir 
 ■|20 
 ■£l9 
 ll8 
 
 E 
 m 
 
 17 
 
 16 
 15 
 
 289 2839 27.89 27.37 26.87 26.37 25;85 
 
 Vacuum In. 
 0.5 0.75 1 1.25 1.5 i75 2 
 Exhaust Pressure-Lb;perSq.ln. 
 
 Fig. 62. Txirbine Efficiency as Affected by Vacuum. 
 
 Prob. 29S. In Fig. 60, which is the most economical turbine 
 at normal load? At loads varying from 60 to 140 per cent, of 
 rated capacity? 
 
 Prob. 296. In Fig. 61, state for all cases the percentage 
 saving in weight of steam due to superheating 100°. (Range 
 is from 7 to 1 1 per cent.) 
 
 Prob. 297. In Fig. 61, with dry steam, what is the percentage 
 reduction in steam rate by increasing the vacuum from 26 to 
 28 in.? What is it when the steam is superheated 100°? 160°? 
 (Ans., 8, 7, 61 per cent.) 
 
 Prob. 298. State in words the significance of the crossing of 
 curves C and D, Fig. 62. 
 
CHAPTER IX 
 
 INTERNAL COMBUSTION ENGINES 
 
 Four-Cycle Engines with Explosive Charge 
 149. The Otto Cycle. Fig. 63 is an indicator diagram from a 
 gas engine. Note that it represents four successive strokes. 
 At the point c, the piston is at the left-hand end of its stroke, 
 the exhaust valve is closed and the inlet valve is open. The 
 pivSton moves one full stroke to the right, drawing in a charge of 
 
 P 
 
 Afmospherk 
 Pressure- 
 
 Fig. 63. Gas Engine Indicator Diagram. 
 
 COMBUSTIBLE MIXTURE. This Operation is represented by the 
 line ci, along which the pressure is slightly below that of the 
 atmosphere. 
 
 At d, the inlet valve closes. The piston now makes a stroke 
 to the left, COMPRESSING the combustible mixture along tZe • • • . 
 At or near the end of this stroke, the charge is ignited by appro- 
 priate means, and there is a rapid combustion accompanied by a 
 sudden rise of pressure. The piston then moves to the right 
 under the expansive action (falling pressure, increasing volume) 
 of the heated gas. At or near the end of this stroke, the exhaust 
 valve opens and the burnt mixture is discharged (line he) at a 
 pressure slightly above that of the atmosphere. Fig. 65 shows 
 
 the engine. 
 
 203 
 
204 
 
 Thermodynamics, Abeidged 
 
 150. Ideal Cycle. This operation is idealized in Fig. 64, 
 diagram abed. Here the counter-clockwise (power-coNSUMiNG) 
 area due to intake and exhaust is disregarded, the combustion 
 is assum'ed to be perfectly instantaneous and compression and 
 expansion are taken to be adiabatic. The cycle is therefore 
 bounded by two adiabatics and two lines of constant volume. 
 
 800 
 700 
 600 
 500 
 400 
 300 
 200 
 100 
 
 -Atmospheric Pressure 
 
 Fig. 64. Ideal Otto Cycle. 
 
 Remembering that efficiency is (heat absorbed — heat re- 
 jected) -T- HEAT ABSORBED, and that there is neither absorption 
 nor rejection of heat along an adiabatic, 
 
 Hbc — Hda 
 
 c 
 
 5/34° 
 
 
 
 
 
 
 
 -Constant Spec if/'cHeatMf7 
 
 
 \ 
 
 -Variable 5peci tic Heats 
 0=^0.24 3'0. 16. b'0.00003 
 
 
 
 
 
 
 
 2380?' 
 
 A 
 
 
 
 
 
 
 1 \ \ 
 1 ^ \ 
 
 
 
 
 
 
 i \ \ 
 1 \ V 
 
 ..-pyy- 
 
 const 
 
 
 
 
 1 ^ \ 
 
 . 
 
 
 
 
 
 ! \ 
 1 \ 
 
 \ 
 
 
 
 
 
 
 \ \ 
 
 
 
 
 
 
 \ \ 
 \ \ 
 
 
 
 
 
 
 
 \ 
 
 
 
 b 
 
 UO/4' 
 
 
 'K 
 
 
 o 
 
 \ 
 
 1 
 
 K 
 
 
 ^^ 
 
 LVVU 
 
 
 
 ^--*, 
 
 2 
 
 -^^^ 
 
 S00'~ 
 
 ^ ei 
 
 
 
 - 
 
 71 -i 
 
 
 ^ 
 
 
 e = 
 
 Hh 
 
 1 _ :^« = 1 _ ^KTd- Ta) _ _ Ti- Ta 
 Hbc wl{Tc- Tb) Tc- Tb' 
 
Internal Combustion Engines 205 
 
 the specific heat at constant volume being denoted by I and the 
 weight of mixture by w. Now the paths cd and ba are isodiabatics 
 (Art. 31). Hence Ta/Ta = TjTt, 
 
 rp rp n^ n^ n^ 
 
 _ _ 1 — i-5 _ 1 i d— i a _ J_a 
 
 r^ rp -*■» rp rp rp s 
 
 a lb ic — J-b lb 
 
 Ta Tb- 
 
 ' = ^-Yb=—fr'=^-\c) ^^-v^) '(^^) 
 
 if the mixture is regarded as a perfect gas. In these expressions, 
 
 C = COMPRESSION RATIO = "Pblfa', C = CLEARANCE = Vbl{Va 
 
 — Vb). One of the expressions looks like that previously given 
 (Art. 34) for the efficiency of the Carnot cycle. It has a different 
 meaning: Tb and Ta are not the extreme temperatures of the 
 cycle. The efficiency of the Otto cycle is less than that of the 
 Carnot cycle. It is much greater than that of the Rankine 
 cycle, used for the steam engine and turbine: mainly because of 
 the very great temperature range employed in gas engines. 
 Clearance in gas engines is made to suit the compression desired. 
 It is not to be viewed as a " necessary evil," as it is in steam 
 engines. 
 
 Prob. 299. Given C = 10, find efficiency and clearance, using 
 y = 1.4. (Ans., e = 0.48, c = 0.24.) 
 
 151. Mean Effective Pressure. Still considering the sub- 
 stance to be a perfect gas, 
 
 _ 2^ _ PcVc - PdVd - VbVb + VaVa , , , „„, 
 
 P" - 144(Fa - Vb) " (y- l){Va - Vb) ^'^^ ^*- "^^^ 
 
 _ Vcipc — Pb) + Vaipa - Tpd) ' 
 
 {y- l)(Fa - Vb) 
 But VcKVa -Vb) = c: while 
 
 Fi. /I \^/" 
 
 iVa-Vb)lVa=i-^^=i-[^) =i-c-^'iy\ 
 
 Then, remembering that pa = Pb/C and pd = pJC, 
 
206 Thermodynamics, Abridged 
 
 cjpc — Pb) I P b — Vc 
 
 Pm = 
 
 2/-1 
 
 A more useful expression is 
 
 778Sff 5.403wZ(n - Tb)e 
 
 Pm = 
 
 144(Fa - Vb) 
 
 Va- n 
 
 This has a special meaning. The quantity wl{Tc — Tb) repre- 
 sents the heat available in the cylinder for raising the tempera- 
 ture of w lb. of mixture, at constant volume, from Tb to Tc. 
 
 Fig. 65. Otto Cycle Gas Engine. 
 
 It is the HEAT VALUE of the mixture present. All of that heat, 
 under ideal conditions, is used to increase the temperature 
 along be. Now consider the hitherto ignored intake stroke, ea, 
 Fig. 64. At the beginning of this stroke, the volume of exhaust 
 gas in the cylinder was Ve- At the end of the intake stroke, the 
 volume of exhaust gas and fresh mixture was Va- Hence the 
 volume of fresh mixture admitted, and present during com- 
 bustion, was Va— Ve = Va— Vb- Then the heat value 
 
Internal Combustion Engines 207 
 
 OF 1 CU. FT. OF MIXTURE SUPPLIED WaS wl{Tc — Tt) ^ {Va — Vb) ■ 
 
 Call this B. Then 
 
 Pm = 5.403Se, lb. per sq. in. (87) 
 
 Here e == efficiency, from Equation (85). 
 
 Prob. 300. In Prob. 299, find p„ from Equation (86) if 
 Pa = 14.696, po = 346.96. (Ans., 58 lb.) What heat value 
 per CU. ft. of mixture does this imply? (Ans., 22 B.t.u.) 
 
 152. Variable Specific Heat. At the high temperatures pre- 
 vailing in the cylinder of an internal combustion engine, the 
 specific heats are highly variable, although the gases are not 
 so imperfect as to depart from Equation (12). Some refinement 
 is therefore necessary over the foregoing. 
 
 Suppose I = a+ bT, where a and b are constants. Then 
 
 Htc= ridT= radT+ CbTdT 
 
 Jb Jb Jb 
 
 Similarly, Had = a{Ta- Ta) + {b/2)iTJ' - TJ), and 
 
 _ Hbc — Had _ ^ _ Had 
 Hbc Hbc 
 
 a{Td-Ta) + l{Td'-Ta') 
 
 = 1 T . (88) 
 
 a(n - n) + 2 (Tc' - Tb') 
 
 Under these conditions, adiabatics are not isodiabatics, for the 
 equation of the adiabatic is not pV^ = const. On the contrary, 
 if we may write k= o+bT, R/778 =R'=k-l=o-a, 
 the equation of the adiabatic (with variable specific heats) is 
 thus derived: 
 
 In an adiabatic process, the external work done equals the 
 
 loss of internal energy: ot T = — W (Art. 28). Applying this 
 
 to a very brief process, during which the pressure is substantially 
 
 .constant, IdT = - Pdi>l778 for 1 lb. weight. Writing 1= a 
 
 -\- bT, P = RT/v, this may be reduced to 
 
208 
 
 Thermodynamics, Abridged 
 
 T V 
 
 Now differentiate the perfect gas equation, Pv = RT. This 
 gives Pdv + vdP = RdT. Dividing by Pv, 
 
 dv dP_dT 
 
 v+P~T- 
 Substituting this value in the first term of the preceding equation, 
 we obtain 
 
 Integrating, (a + R') Iog« v + a loge P + bT = const. 
 
 -loge V + loge P + -T= const., 
 ffl "> 
 
 PyOiagiTia ^ const., (89) 
 
 where e is the base of the Napierian system of logarithms. 
 
 5 6 7 8 9 10 11 
 Compression Ratio C=Pb/Pa 
 
 Fig. 66. Otto Cycle Efficiencies, Constant and Variable Specific Heats. 
 
Internal Combustion Engines 209 
 
 In the ideal case, the mixture is considered to be pure air. 
 It is, in fact, mostly air. Values of a, b and o are not exactly 
 known: but approximately, a = 0.1801, b = 0.0000283, o = 
 0.2511. Fig. 64 shows by dotted lines the resulting cycle, to 
 scale, though not for these particular values of the constants. 
 In both diagrams, pa = 14.696,, Ta = 500, C = 10, and the 
 heat evolved during combustion is 700 B.t.u. per lb. of mixture. 
 The chief difPerence is with respect to the rise of temperature 
 along be. With I = 0.17, this is 700 -i- 0.17 = 4120°. 
 
 Fig. 66 compares efficiencies of the two cycles, the broad band 
 of results from Equation (88) covering the whole range of pro- 
 posed values of a, b and o. The heavy dotted line within this 
 band represents an efficiency which is 81.8 per cent, of that by 
 Equation (85) . This line may be regarded as representing results 
 of Equation (88) as closely as they can be known at present. 
 In other words, the ideal efficiency is 81.8 per cent, of that 
 given by the simple (constant specific heat) expression of Equa- 
 tion (85). 
 
 Prob. 301. In Art. 152, show that the heat absorbed along 
 be' is approximately 700 B.t.u. Take values of constants from 
 Fig. 64. 
 
 Prob. 302. From Equation (89), show that 
 
 0, Tb b — a Pi 
 
 -iog„jr+-(n-r.) = ^ioge-- 
 
 Prob. 303. Does Charles' law apply when the specific heats 
 are variable as in Art. 152? 
 
 Prob. 304. What is the probable true ideal efficiency when 
 C = 10? (Ans., 0.397.) 
 
 153. Variations of Ideal Efficiency with Load: Governing. 
 When the load on an engine decreases, the supply of heat should 
 be decreased. This is accomplished in a constant speed engine 
 one of four ways in : 
 
210 Thermodynamics, Abridged 
 
 (1) Vary the time of ignition. This simply throws away heat 
 
 not needed. 
 
 (2) Omit drawing in a charge for one or more strokes (" hit 
 
 or miss ")• This leads to irregular speed and is applic- 
 able only to small engines. 
 
 (3) Dilute the charge with an excess of air. 
 
 (4) Decrease the weight of charge drawn in. 
 
 Methods (3) and (4) are important. We have seen that 
 efficiency depends on compression (Art. 150, Equations (85), 
 Fig. 66) : strictly, on the compression ratio, C. Compression 
 is limited by the temperature attained at b, Fig. 64. We must 
 
 Typical Ignition Temperatures, °F.* 
 
 Ether, 750 Acetylene, 760-820 Benzene, 780 Alcohol, 950 
 
 Liquid Fuels, 950-1020 Equal Parts H and O, 957 Benzol, 970 
 
 Ethylene, 1005-1015 Hydrogen, 1075-1100 
 
 Illuminating Gas, 1100 Methane, 1200-1240 
 
 not reach the ignition temperature by compression, else pre- 
 ignition (self-ignition) will occur. For each possible mixture, 
 there is a limiting permissible compression and hence a limiting 
 efficiency. We should work as close to this compression as 
 we can. In fact, without compression it would be diflficult or 
 impossible to ignite some fuels at all. The accompanying table 
 lists the more common mixtures in the approximate order of 
 allowable compressions, and therefore in the approximate order 
 of efficiencies. (The values of the last column are discussed 
 later.) 
 
 Now if we dilute the charge with excess of air, we raise its 
 ignition temperature (and also make ignition occur more slowly) . 
 Hence we are operating at an efficiency below the best possible 
 efficiency. 
 
 If we decrease the weight of charge, we also dilute it. This 
 would occur, if for no other reason, because the charge is mixed 
 with an approximately constant quantity of exhaust gas left 
 
 * Except as noted, in air at normal pressure. 
 
Internal Combustion Engines 
 
 211 
 
 in the clearance space. Here also we are operating, when at 
 reduced load, at less than maximum possible efficiency. 
 
 Usual Compression and Mean Effective Pressures for Various Fuels* 
 
 Fuel and Engine 
 
 Compression 
 Pressures 
 
 Usual Range Average 
 
 Kerosene in small hot bulb engines .... 
 
 Gasoline, automobile 
 
 Kerosene, carburettor engine 
 
 Fuel oil, hot bulb engine 
 
 Gasoline, stationary and marine engines 
 
 Heavy fuel oils 
 
 Rich gas mixtures 
 
 Lean gas mixtures 
 
 Natural gas, large engines 
 
 Coal gas, small engines 
 
 Carburetted water gas, small engines . . 
 
 Producer gas, general 
 
 Coke oven gas, large engines 
 
 Blast furnace gas, very large engines . . . 
 Alcohol, carburettor engine 
 
 45- 90 
 60-110 
 60-100 
 
 75-120 
 
 75-135 
 
 70- 90 
 
 90-135 
 
 90-175 
 
 90-135 
 
 90-120 
 
 115-175 
 
 120-150 
 
 135-205 
 
 135-225 
 
 75 
 80 
 80 
 60 
 85 
 
 135 
 115 
 105 
 145 
 135 
 170 
 165 
 
 50- 80 
 70- 90 
 50- 75 
 50- 85 
 80- 95 
 50- 70 
 70- 85 
 65- 80 
 80-110 
 60-100 
 60-100 
 55- 90 
 80-110 
 80-120 
 130 
 
 In either case, the efficiency at light loads is less than it should 
 be, and (actually) much less than at full load. Methods (3) 
 and (4) for governing are generally (and unavoidably) used 
 simultaneously. When the mixture is diluted by throttling the 
 gas, the intake suction is increased and the pressure at a, Fig. 64, 
 is reduced. The weight is therefore reduced. When the weight 
 of charge is decreased, the increased intake suction modifies the 
 proportions of the charge. 
 
 An important defect in gas engine governing arises from the 
 fact that the governing action (by methods 2, 3 or 4) is exerted 
 one full revolution before its effect is felt. 
 
 Governing by dilution of charge naturally decreases the 
 B.t.u. per cu. ft. of mixture and hence the value of pm (Art. 151). 
 Governing by decreasing weight of charge reduces the B.t.u. 
 PER cu. FT. OF DISPLACEMENT and hcuce the value of pm- 
 
 Prob. 305. What is the best ideal efficiency to be expected 
 with producer gas, if the pressure at a, Fig. 64, is 14.6 lb.? 
 (Ans., 0.417.) 
 
 * Mostly from Lucke. 
 
212 Thermodynamics, Abridged 
 
 Prob. 306. State an average value of C for alcohol, if pa 
 = 14.35. (Ans., Hi) 
 
 Prob. 307. In an engine governed by varying the weight of 
 charge, how would the efSciency vary with the load if the clear- 
 ance were perfectly adjustable? 
 
 154. The Actual Engine. In the actual engine, the intake 
 pressure is somewhat below atmospheric and -pa = 8§ to 14|. 
 The compression is not precisely adiabatic. It may be repre- 
 sented by pj)" = const., with n between 1.25 and 1.35. Ignition 
 is not instantaneous and the combustion line has a perceptible 
 slope to the right as it rises. This is accentuated at high piston 
 speeds or with slow ignition (improper mixture) conditions. 
 The speed of flame propagation is of the order of 7 to 20 ft. per 
 sec. at the pressures existing and with correct mixtures. The 
 maximum temperature and pressure are much less than those 
 computed as for Fig. 64 by either method. This is partly because 
 the volume increases during combustion and partly because of 
 the cooling which occurs. The maximum temperature is rarely 
 much over 3000° F. The expansion curve is a polytropic with 
 n between 1.3 and 1.5. The temperature at the point when the 
 exhaust valve opens may be anywhere from 1200° to 2000°. 
 The exhaust valve opens considerably before the end of the 
 stroke. The average exhaust pressure is 16 or 17 lb. The 
 temperature in the exhaust pipe is 700° to 1000°. 
 
 155. Diagram Factor and Volumetric Efficiency. The actual 
 indicated thermal efficiency is usually from 40 (occasionally down 
 to 30) to 55 per cent, of that given by Equation (85), or 49 to 67 
 per cent, of that by Equation (88), for the compression used. 
 The " relative efficiency " (Art. 93) is therefore about the same 
 as for steam engines: but its variations have not been as thor- 
 oughly studied. The lower values of relative efficiency are 
 obtained with engines employing carburettors to vaporize the 
 fuel. This percentage which the actual indicated thermal 
 efficiency bears to the ideal efficiency is also called the diagram 
 
 FACTOR. 
 
Internal Combustion Engines 213 
 
 The VOLUMETRIC EFFICIENCY is the ratio of the volume of 
 gas drawn in (measured at normal barometer and 62°) to the 
 corresponding piston displacement. In discussing the ideal 
 cycle, Art. 151, it was considered to be 1.0. In practice, it is 
 decreased by the excess of the exhaust pressure over 14.696, by 
 the suction below that amount, and by any intake temperature 
 above 62°. Values of the volumetric efficiency are e» = 0.50 
 to 0.87. It is lowest in high speed engines and with automatic 
 valves. It is particularly low in airplane engines when flying 
 at high altitudes. Supercompression involves filling the 
 cylinder by means of a special external pump: by which method 
 e« may be greatly increased. 
 
 The actual mean effective pressure is then to be obtained 
 from a modification of Equation (87) : 
 
 Vm' = 5.403/£ee«, (90) 
 
 where B = B.t.u. per cu. ft. of mixture, / = diagram factor, 
 e = ideal efficiency. Values of pm are given in the second table 
 of Art. 153. The ratio of actual to ideal mean effective pressure 
 iPm'/pm) is about equal to the ratio of actual and ideal pressure 
 rises, pc — Pb, Fig. 64. 
 
 Prob. 308. What maximum mean effective pressure may be 
 expected with blast furnace gas (60 B.t.u. per cu. ft. of mixture) 
 in a large slow-running engine with mechanically operated valves? 
 (Ans., 87 lb.) 
 
 Prob. 309. An automobile engine gives 65 lb. mean effective 
 pressure. It has average compression. Assume 100 B.t.u. per 
 cu. ft. of mixture. Make an estimate of its volumetric efficiency. 
 (Ans., 0.82.) 
 
 Prob. 310. Compute B for blast furnace gas, from the follow- 
 ing table. (Ans., 57.) 
 
 Prob. 311. Assume gasoline to be represented by CeHu, and 
 to contain 19 600 B.t.u. per lb. What is its heat value per cu. ft. 
 OF VAPOR at standard conditions (32°, 14.696 lb.)? Compute B. 
 (Ans., 4710 and 96 B.t.u.) 
 
214 
 
 Thermodynamics, Abridged 
 Properties of Gas Fuels (Lucke) 
 
 
 Composition by Volume 
 
 B.t.u. 
 
 
 CHi 
 
 CtHs 
 
 CO 
 
 02 
 
 Hi 
 
 COa 
 
 Ns 
 
 per 
 Cu. Ft.* 
 
 Natural gas 
 
 (Kansas) 
 
 Coke oven gas 
 
 Retort coal gas 
 
 Oil gas 
 
 0.982 
 
 0.343 
 
 0.3135 
 
 0.525 
 
 0.002 
 
 0.044 
 
 0.002 
 
 0.001 
 0.040 
 0.022 
 0.235 
 
 0.001 
 
 0.0025 
 0.0600 
 0.0860 
 0.0100 
 0.2861 
 0.4485 
 
 0.2610 
 
 0.0025 
 0.0110 
 0.0035 
 0.0050 
 
 0.0050 
 
 0.0020 
 
 0.420 
 0.525 
 0.185 
 0.027 
 0.456 
 
 0.150 
 
 0.025 
 0.015 
 0.005 
 0.114 
 0.045 
 
 0.053 
 
 0.101 
 0.035 
 
 0.571 
 0.001 
 
 0.532 
 
 950.1 
 
 579.8 
 
 542.6 
 
 1196.5 
 
 Blast furnace gas . . 
 Water gas 
 
 107.5 
 329.9 
 
 Producer gas (from 
 hard coal) 
 
 134.7 
 
 156. Mechanical Efficiency : Overload Capacity. If we follow 
 steam engine practice, the mechanical efficiency is ej^ = bhp. 
 -^ ihp., where ihp. refers to the net diagram of Fig. 63; i.e., to 
 
 * Low value, i.e., heat evolved when the products of combustion are not 
 condensed in the calorimeter: taken at 32° and normal barometer. Note: 
 the values given in the last column are not those of B, Equation (90). That 
 symbol stands, not for the heat value of the fuel, but for the heat value of 
 1 cu. ft. of combustible mixture, formed by adding the proper amount of air 
 to the fuel. The weight of chemically ideal mixture is computed by utilizing 
 the law that combining proportions by volume are those op the number 
 of molecules entering into the reaction. Thus CO and H2 require 
 half their volume of oxygen, C2H1 three times its volume of oxygen, CH4 
 twice its volume and CeHe seven and one half times. From the total oxygen 
 requirement thus computed deduct the oxygen in the fuel and multiply the 
 remainder by 4.75 to obtain the volume of air which must be added to 1 cu. ft. 
 of fuel to produce a perfect mixture. Thus, for natural gas: 
 
 0.982 X 2 = 1.964 
 (CeHs) 0.001 X 7| = 0.0075 
 
 0.0025 X ^ = 0.0013 
 
 1.9728 
 0.0025 
 
 1.9703 X 4.75 = 9.37 
 1.0 
 
 At 62° and normal barometer, 
 B =92X 
 
 950.1 -r- 10.37 = 92 B.t.u. per cu. ft. mixture 
 at 32° F. 
 
 460 + 32 
 460 + 62 
 
 86* B.t.u. 
 
 The mixture obtained in practice is never exactly right. Hence B is less 
 than this. If the mixture is far from correct, combustion is slow. If very 
 weak or very strong, the mixture may refuse to ignite at all. 
 
Internal Combustion Engines 215 
 
 the clockwise cycle minus the negative loop. It is logical, how- 
 ever, to regard the ihp. as the larger value due to considering the 
 clockwise cycle alone, without deduction for the loop, because 
 that loop is due to friction. The friction is of course fluid friction 
 rather than mechanical friction. The loop consumes 5 to 10 
 per cent, of the power of the engine. If we regard the clockwise 
 diagram as representing the ihp., the mechanical efficiency ranges 
 from 0.75 to 0.85, increasing for large engines and being higher 
 for double-acting than for single-acting engines (Art. 7). Let 
 A = gross Lhp., C = loop ihp., D = brake hp.: then as now 
 defined e^^ = D/A whereas as defined for a steam engine 
 e^ = D/{A - C). 
 Equation (90) shows that pm' varies with/, B, e and e„. Natur- 
 ally, / and e„ will be kept as high as possible at the rated load 
 condition. The value of e depends on the compression alone, 
 and that will be as high as the nature of the fuel will permit. 
 The effort will be made to have the mixture proportions right. 
 Hence all factors have maximum values at rated load. In a 
 constant speed engine, there is consequently no way of increasing 
 power beyond that for which the engine was built. Efficiency 
 and output vary in the same direction, not at all as in a steam 
 engine. Maximum efficiency occurs at maximum power. An 
 Otto cycle engine has, strictly speaking, no overload capacity. 
 A margin of power for emergencies can be provided only by 
 rating the engine at a power less than that obtainable at maxi- 
 mum efficiency (Art. 165). 
 
 Prob. 312. The gross ihp. is 100, loop hp. = 8, brake hp. 
 = 80. Find two values of ej^. (Ans., 0.87 and 0.80.) Which 
 is adopted in the text? (Ans., 0.80.) 
 
 Prob. 313. Sketch a curve of thermal efficiency against brake 
 hp. Compare Fig. 71. 
 
 Other Otto Cycle Engines 
 157. Two-Cycle Engine. The action of the two-cycle engine 
 is suggested (and somewhat exaggerated) in Fig. 67. The 
 
216 
 
 Thermodynamics, Abridged 
 
 exhaust opens at d, quite early in the expansion stroke. When 
 the pressure has fallen to pe, the inlet opens to admit a charge 
 of mixture under pressure. This flows in along efg. The 
 exhaust does not close until point a is reached. Compression 
 then begins. A separate pump or its equivalent is provided for 
 
 Fig. 67. Two-Cycle Engine. 
 
 delivering the mixture to the cylinder. Its indicator diagram, 
 shown dotted and with the V axis reversed, corresponds with 
 the negative loop of Fig. 63: but the loop now consumes 7 to 12 
 per cent, of the engine power and the mechanical efficiency as 
 defined in Art. 156 is reduced. It is from 0.65 to 0.70, being 
 high for large engines. 
 
 Along efg, inlet and exhaust valves are both open. Hence 
 some fuel will be lost. This is mitigated by scavenging, i.e., 
 
Internal Combustion Engines 217 
 
 by having a blast of pure air precede the injection of the charge : 
 but it is never entirely eliminated. The two-cycle gas engine is 
 about 20 per cent, less efficient than the four-cycle. If it were 
 not for its larger fluid friction loss, it would give twice the power, 
 because only two strokes, instead of four, are necessary to com- 
 plete a cycle. (Exhaust and intake are crowded into a small 
 portion of the expansion and compression strokes.) Actually, 
 the gain in power is about 70 or 80 per cent. 
 
 The necessary outside compression of mixture may be pro- 
 vided by (o) using the idle end of an otherwise single-acting 
 cylinder : (b) utilizing the crank case as a compression chamber : 
 (c) separate pumps. Methods (a) and (b) make the parts of the 
 engine inaccessible and complicate lubrication and cooling. A 
 two-cycle engine may be valveless, the passages being slots in 
 the cylinder wall, opened and closed by the piston. 
 
 In the smaller and cheaper two-cycle engines, e» is particularly 
 low. The ideal efficiency is that given by Equations (85) or (88) . 
 Diagram factors and mean effective pressures are about 0.8 those 
 of four cycle engines. 
 
 Prob. 314. State the probable actual indicated thermal and 
 brake thermal efficiency, and mean effective pressure, of a large 
 slow-running two-cycle engine using blast-furnace gas. See Art. 
 155, Prob. 308, Prob. 310, Art. 156. (Ans., 0.25, 0.165, 63 lb.) 
 
 158. Size and Power of Engine. Following Art 7, for a four- 
 cycle single-acting cylinder (which yields power at one-fourth 
 the rate of a steam cylinder) 
 
 ihp. - jg ct p„ X 33 QQQ 264 000 
 
 _ pJAS _ pJLAN 
 
 132 000 66 000 ' 
 
 For other types, use the following multipliers for the above: 
 
 2 for four-cycle double-acting, 
 2 for two-cycle single-acting, 
 4 for two-cycle double-acting. 
 15 
 
218 Thermodynamics, Abridged 
 
 Most engines have a plurality of cylinders, all of which develop 
 equal power. Double-acting engines are difficult to lubricate 
 and cool. 
 
 Prob. 315. In Prob. 314, find the diameter and stroke of an 
 engine to develop 500 brake hp. at 600 ft. piston speed and 
 110 r.p.m., if two-cycle, two-cylinder, single-acting. (Ans., 
 23.6 by 32.7 in.) 
 
 Prob. 316. The commercial formula for brake hp. of an 
 automobile engine is n(P/2.5, where n = no. of cylinders, d = dia. 
 in inches. If this is based on 1000 ft. piston speed and gj^ = 0.75, 
 for single-acting four-cycle engines, what mean effective pressure 
 is assumed in the cylinders? (Ans., 90 lb.) 
 
 159. Oil Engines. Engines employing carburettors to vapor- 
 ize a volatile fuel are not to be classed as oil engines, although 
 kerosene, and some few heavier oils, may be used in such engines 
 if the carburettor is heated. An oil engine, properly so called, 
 draws in pure air on its suction stroke and the fuel is injected 
 separately. A two-cycle oil engine may therefore avoid the 
 loss of fuel to the exhaust (Art. 157) which is the largest factor 
 in producing low efficiencies of two cycle engines. 
 
 The less volatile fuels are employed for oil engines. They can- 
 not be ignited by electric spark. Hence such engines are pro- 
 vided with an uncooled " hot bulb," " hot tube " or plate, 
 which communicates with the clearance space. This is kept red 
 hot. It is heated by a torch for starting. It ignites the com- 
 bustible mixture only after the temperature of the air in such 
 mixture has been somewhat increased by compression. 
 
 Some oil engines inject the fuel during the suction stroke or 
 the early part of the compression stroke. The whole charge 
 is then heated by the compression before it actually ignites. 
 The time of ignition is determined by the proportioning of the 
 uncooled surface, the compression, etc., and is apt to be irregular 
 at variable loads. In other engines, there is timed injection 
 of the fuel, close to the end of the compression stroke. Ignition 
 cannot then be premature. Further, the compression may 
 
Internal Combustion Engines 219 
 
 be carried very high and high efficiency secured. On the other 
 hand, the fuel must be injected against a high resisting pressure. 
 Heavy oils must be finely subdivided in order that ignition may 
 occur. Hence the best method of delivering them to the cylinder 
 is by a compressed air blast. The necessary air pressure for 
 maximum subdivision of the oil particles is about twice the 
 pressure in the cylinder (Art. 136). Hence with high com- 
 pressions, very high air pressures and multi-stage air com- 
 pression are necessary for feeding the fuel. 
 
 Oil engines of this type are easily governed by varying the 
 quantity of fuel injected. The diagram factor decreases some- 
 what at light loads. Overloads lead to over-rich mixtures, 
 overheating and a smoky exhaust. Compression pressures 
 (j)6. Fig. 64) as high as 300 lb. are being used, but these lead to 
 extremely high maximum pressures. 
 
 For liquid fuels in general, the specific gravity is 
 
 _ 140 
 *~130+6^ 
 
 where b = reading of Baum6 hydrometer. Water weighs 8.25 
 lb. per gal. at 62°. The weight of a gallon of oil at this tempera- 
 ture is 8.25s lb. and its heat value is very nearly 
 
 18 650 -t- 40(6 - 10) B.t.u. per lb. 
 
 Oils are sold by bulk. They expand 0.00055 of the original 
 volume per degree rise in temperature. Fuel oil contains C, 
 0.80 to 0.87 and H2, 0.10 to 0.14, the balance being O2 and N2 
 with (frequently) very small traces of sulphur. 
 
 Prob. 317. How much heat is contained in 50 gal. of fuel oil 
 of 30° Baum6 density at 62° F.? (Ans., 7 000 000 B.t.u.) 
 What will be the bulk of this oil at 92° F.? (Ans., 50.83 gal.) 
 
 The Diesel Cycle 
 160. Ideal Cycle. The ideal Diesel cycle draws in a charge of 
 pure au- at atmospheric pressure (eo, Fig. 68) and compresses it 
 adiabatically to about 500 lb. pressure. The temperature thus 
 
220 
 
 Thekmodynamics, Abridged 
 
 attained is so high that upon injection of the fuel it immediately 
 ignites. The delivery of fuel is so retarded that combustion 
 occurs at constant pressure, be, hence there is no explosive 
 action. The supply of fuel is cut off and combustion ceases at c. 
 
 Fig. 
 
 Ideal Diesel Cycle. 
 
 Adiabatic expansion follows and the exhaust valve opens at d. 
 The pressure drops instantaneously to that of the atmosphere, 
 at which pressure the burnt gases are expelled. 
 Under these ideal conditions, the eflBciency is. 
 
 Hi 
 
 e = 
 
 ; — Hda - Hda 
 
 Hbc Hbc 
 
 ^ ^_ wl{Ta- Tg) _ ^ 
 
 Td — Ta 
 
 wk{Tc - n) y{Tc- Ti) 
 
 The ideal mean effective pressure is (following Art. 151) 
 
 HboB 
 
 Vrr. 
 
 5.403 
 
 F„- Vi 
 
 = 5.403Be. 
 
 (92) 
 
 (93) 
 
 where B = heat supplied per cu. ft. of piston displacement. 
 For average fuel oil containing C 0.85, H2 0.12, O2 0.01, the vol- 
 ume of air at standard conditions required for ideal mixture is 
 about 200 cu. ft. per lb. of fuel. Hence in the ideal cycle with 
 e„ = 1.0 (Art. 155), B = 19 500 -^ 200 = 97|, very nearly, 
 and fm = 536e. Actual values are less than half this (Art. 162) . 
 
Internal Combustion Engines 
 
 221 
 
 Prob. 318. Given ta = 40° F., C = pt/p, = 34, T, = 2Ti, 
 y = 1.4, find e and p„. (Ans., 0.574, 308 lb.) 
 
 161. Variation of Ideal Efficiency. The Diesel cycle is more 
 efficient than the Otto. It is rather unique among heat engine 
 cycles, in that its efficiency 
 
 INCREASES AT LIGHT LOADS. 
 
 Fig. 69 plots values of e 
 against the ratio of expan- 
 sion, r= Fd/F. = ValVc, 
 Fig. 68. This is based on 
 U = 40° F., C = 34, as in 
 Prob. 318, with y = 1.4. 
 High ratios of expansion im- 
 ply low mean effective pres- 
 sures. Compare the diagrams 
 abed and abc'd', Fig. 68. Since 
 the ideal value of pm varies 
 directly with e (Equation 
 
 0.63 
 a) 
 
 "0.6Z 
 
 S' 
 
 .|0.61 
 
 jPO.60 
 
 (D 
 
 ^0,58 
 0.57 
 
 N 
 
 
 
 
 
 
 
 
 v^- 
 
 
 
 
 
 
 
 v^ 
 
 
 
 
 
 
 
 % 
 
 
 
 y 
 
 ^ 
 
 
 
 % 
 
 / 
 
 / 
 
 
 
 
 
 y 
 
 / 
 
 
 
 
 
 <& 
 
 K\ 
 
 \ 
 
 
 
 
 ,# 
 
 
 \ 
 
 
 
 y' 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 \ 
 
 120 
 
 110 = 
 
 ioo| 
 
 90^: 
 
 80 i 
 
 70^ 
 
 60i 
 
 ti 
 
 50 £ 
 
 40== 
 
 30 
 
 5 6 7 8 9 10 II. 12 
 
 Ratio of Expansion r=Va/Vc 
 
 (93)) the underload condition ^^ gO. Diesel Cycle at Various Loads, 
 must be one of (a) poor mix- 
 ture, (6) low volumetric efficiency or (c) low relative efficiency. 
 
 162. The Diesel Engine in Practice. In actual engines, the 
 efficiency occasionally, though not frequently, shows an increase 
 with decrease of load until the load has been reduced to about 
 half its normal value. The heat losses then begin to offset the 
 cyclic gain, and below this point the efficiency decreases. In 
 most cases, the efficiency falls off with decrease of load over the 
 whole range. Most Diesel engines are four-cycle, single-acting. 
 The two-cycle arrangement is sometimes used, requiring that 
 the air charge be delivered at 20 to 30 lb. pressure. This con- 
 sumes about 4 per cent, of the engine power. The two-cycle 
 Diesel engine gives 70 to 80 per cent, more output than the 
 four cycle, but its relative and actual efficiencies are about 10 
 per cent. less. Mechanical efficiencies of Diesel engines are 
 from 0.70 to 0.75. The high compression used requires that the 
 
222 Thermodynamics, Abridged 
 
 fuel be supplied by a compressed air blast (Art. 159) at about 
 1000 lb. pressure. From 16 to 34 cu. ft. of free air (Art. 42) 
 are required per brake hp.-hr., and the compressor absorbs 4 to 7 
 per cent, of the power of the engine. 
 
 Diagram factors in well-designed engines are around / = 0.50 
 to 0.62. The corresponding value of ^J, the actual mean 
 effective pressure, is 5.403 e„/e5', where B' = B.t.u. per cu. ft. 
 of combustible mixture at standard conditions. The value 
 of 'pm may also be computed from Fig. 68, 
 
 , _ 2?i(l + c) / w(l + c-c r) _ C- r"\ 
 
 ^'" ~ jx - 1 V Kl + c) Gr^ )' 
 
 in which 
 
 n = exponent of polytropic curves ah and ci, 
 
 c = clearance = F^ ^ {Va - Vb) = I ^ (C^'" - 1), 
 
 r = ratio of expansion = Vd/Vc, 
 
 C = compression ratio = pb/pa- 
 
 For the usual conditions pb = 500, C = 34, if we take n = y 
 = 1.4 and compute c as 0.088, this becomes 1360(1. 4/r - l/r^-*) 
 — 114. The curve is plotted in Fig. 69 for n = 1.3 (which 
 alters the value of c) and with which 
 
 (94) 
 
 A r r^V 
 
 p„'= 1785(^— -^3J-101. (95) 
 
 This last expression may be used with good results in actual 
 engines, and there is a satisfactory reason why. It is based 
 simply on the shape of the pV diagram, and is subject only to 
 such error as may arise from minor irregularities or abnormally 
 shaped curves. The cut-off at rated load is usually made to 
 occur at about 1/10 stroke, i.e., {Vc - W) = 0.10(Fa - Ft), 
 Fig. 68. With the usual compression, this corresponds with a 
 value of r around 6. Fig. 69 shows' the corresponding value of 
 Pm' to be 109. In ordinary practice the value of pj is from 100 
 to 120 lb. at rated load. 
 
 Prob. 319. Find c and r in Prob. 318. (Ans., c = 0.088, 
 r = 6.22.) 
 
Internal Combustion Engines 223 
 
 Prob. 320. In Prob. 318, if / = 0.50 and e^ = 0.75, what 
 weight of oil (19 500 B.t.u. per lb.) will be used per brake hp.-hr.? 
 (Ans., 0.61 lb.) 
 
 Prob. 321. In Prob. 318, if B' = 97^ and / = 0.50, what is 
 the volumetric efficiency when pm' = 100? (Ans., e„ = 0.66.) 
 
 Prob. 322. If C = 34, w = 1.3, find c and r when cut-off is 
 at 10 per cent, of stroke. (Ans., c = 0.071, r = 6.26.) 
 
 Prob. 323. In Prob. 322, find fj from Equation (95). (Ans., 
 105 lb.) 
 
 Prob. 324. With pm as in Prob. 323, find the dimensions of a 
 six-cylinder four-cycle single-acting engine to develop 2400 hp. 
 at 1000 ft. piston speed and 200 r.p.m. (Ans., diameter, 25.3 in.; 
 stroke, 30 in.) 
 
 Tests of Internal Combustion Engines 
 163. Methods of Stating Efficiency. If Q = low heat value 
 of fuel (Art. 155), B.t.u. per lb., W = weight of fuel used per 
 Ihp.-hr., the actual indicated thermal efficiency is 
 
 2545 
 
 ^''~ wq 
 
 If e = efficiency of ideal cycle by Equation (85), the relative 
 efficiency is e^ = ea -f- e. If e^ = mechanical efficiency, the 
 BRAKE thermal EFFICIENCY is BaBu and the fuel rate per brake 
 hp.-hr. is TF -r ejf lb. The indicated heat rate is 42.42 -f- Ca. 
 B.t.u. per Ihp.-min. The brake heat rate is 42.42 -^ eaBu, 
 Indicator practice is less satisfactory than with steam engine, 
 hence results are often referred to brake hp. Engines using 
 producer gas are often reported on the basis of the fuel rate. 
 If ep = efficiency of producer and Qp = heat value of producer 
 fuel, B.t.u. per lb., the efficiency of the plant from fuel to cylinder 
 is epBa'. from fuel to brake, it is BpCaej^. The fuel rate is 2545 
 -j- QpCpBa lb. per Ihp.-hr. The value of ep is around 0.75 in good 
 operation. 
 
 Prob. 32S. In Probs. 318 and 320, state the actual indicated 
 thermal, relative and brake thermal efficiencies, the brake fuel 
 
224 
 
 Thermodynamics, Abridged 
 
 rate and the indicated and brake heat rates. (Ans., 0.287: 
 0.50: 0.215: 0.61 lb.: 148 B.t.u.: 197 B.t.u.) 
 
 Prob. 326. An engine using producer gas, generated at an 
 efficiency of Cp = 0.72, has an actual indicated thermal efficiency 
 of 0.30 and a mechanical efficiency of 0.85. The coal used con- 
 tains 12 725 B.t.u. per lb. What weight of coal is consumed 
 per brake hp.-hr.? (Ans., 1.09 lb.) 
 
 164. Economy Determined by Compression. As indicated by 
 theory (Art. 150), the efficiency at full load is chiefly a matter 
 of compression. Good four-cycle gasoline and kerosene engines, 
 with the low compressions prevailing for those fuels, give values 
 of Ba from 0.15 to 0.24, the first figure being realized even in 
 
 ^0.9 
 
 fcO.6 
 105 
 !S04 
 
 \ 
 
 
 
 
 
 
 
 
 
 \ 
 
 s, 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 ■\ 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 "~- 
 
 ~-~ 
 
 
 
 
 
 
 
 
 
 
 
 
 50 100 150 ZOO 250 300 350 400 450 500 
 Approximate CompressionLb.per5qJn.Ab5olute 
 (p^,,Figs,64and68) 
 
 Fig. 70. Economy vs. Compression in Oil Engines. 
 
 very poor installations. With higher compressions, on producer 
 gas, natural gas, blast furnace gas and some illuminating gases, 
 full load values of Ca are from 0.25 to 0.32, occasionally rising to 
 nearly 0.40. Fig. 70 illustrates guaranteed performances of 
 a number of oil engines, including Diesel engines. A law is clearly 
 suggested. Diesel engines attain 0.40 actual efficiency on fuel 
 oil or kerosene. Some of the very high-pressure Otto cycle oil 
 engines almost reach this figure. 
 
 165. Economy in Relation to Load. Fig. 71 shows a uniform 
 type of curve, widely different from that obtained with steam 
 
Internal Combustion Engines 
 
 225 
 
 engines or steam turbines (Figs. 29 and 60). In all cases the 
 efficiency is a maximum at maximum load. This applies 
 whether governing is as in Otto cycle gas engines (Art. 153), 
 by the Diesel variable cut-ofP method, or by varying speed as in 
 the case of the Daimler engine. None of the curves for Diesel 
 engines show the high efficiency at light load referred to in Art. 
 162. 
 
 Prob. 327. In Fig. 70, what is the probable brake fuel rate 
 of an engine using 7 atmospheres compression? (Ans., 0.834 lb.) 
 If the pressure at the beginning of compression is one atmosphere, 
 what is the corresponding ideal efficiency by the dotted curve 
 
 34 
 
 30 
 
 §26 
 
 uj 22 
 
 CD 
 
 18 
 
 14 
 
 10 
 
 20 
 
 00 110 
 
 40 50 60 70 80 90 
 Percent of Rated (Brake)h.p. 
 
 Fig. 71. Variation of Economy with Load, Internal Combustion Engines. 
 
 of Fig. 66? At a mechanical efficiency of 0.80, with fuel con- 
 taining 19 500 B.t.u. per lb., what relative efficiency as com- 
 pared with Equation (88) is thus implied? (Ans., 0.56.) 
 
 Prob. 328. What weight of 30° fuel oil should be used per 
 brake hp.-hr. by the best Diesel engine of Fig. 71, at full load? 
 At half load? (Ans., 0.42 lb. : 0.51 lb.) 
 
226 Theemodynamics, Abridged 
 
 Prob. 329. If the Daimler engine of Fig. 71 used 68° gasoline, 
 how many gallons would it require per minute at 70 per cent, of 
 its full load? (Ans., 0.0213.) 
 
 166. Heat Balance. The steam engine discharges the largest 
 proportion of the heat it receives to the condenser. The gas 
 engine divides its heat supply into three (very roughly equal) 
 parts: one-third becomes indicated work, one-third is discharged 
 at the exhaust, one-third is carried off by the cooling water 
 discharged from the jackets. More accurately, indicated work 
 accounts for 15 to 46 per cent.: jackets, for 32 to 41 per cent.: 
 and exhaust for 24 to 33 per cent. : with a small unaccounted-for 
 balance. The proportion to the jackets is low in very large 
 engines. 
 
 Jacket water enters at 60^ to 80° and emerges at 120° to 170°. 
 The former exit temperature must be maintained if the fuel 
 contains much hydrogen, in order that preignition maybe avoided. 
 Water at these discharge temperatures has very little value 
 for heating purposes. The heat is there, but the availability 
 is not. If , Ca = efficiency of engine, j = proportion of heat 
 supplied which is carried off by the jacket water, the heat 
 removed by the water is 2545j -f- Ca, B.t.u. per hp.-hr. This is 
 equal to Wyi{U — *i) where w„ = weight of jacket water per 
 hp.-hr. and h and U are its inlet and outlet temperatures. The 
 water pressure must be sufficient to ensure through circulation. 
 
 The exhaust heat (the amount of which may be computed in a 
 similar way) is at high temperature (Art. 154) and is hence 
 much more available. The transfer of this heat to water (pre- 
 ferably the jacket water itself) is the usually attempted method 
 of utilizing it. This is done in a device like an economizer 
 (Art. 124). The high gas velocity causes much difficulty in 
 handling it, and it is a troublesome matter to avoid producing 
 serious back pressure at the engine. Some 30 per cent, of the 
 heat in the exhaust gases may be recovered under good condi- 
 tions. If this is recovered as hot water, the best use to make of 
 this water is for heating buildings. If the heat is recovered as 
 steam, to be used in an engine, the very low efficiency of the 
 steam engine enters into the matter. 
 
Internal Combustion Engines 227 
 
 Prob. 330. How much water per hour is required in the 
 jackets of a 1000 hp. engine having an actual efficiency of 0.25 
 and a jacket loss of 0.35, if the water temperature rises from 70° 
 to 140°? (Ans., 50 900 lb.) 
 
 Prob. 331. The loss to the exhaust in the above engine is 
 0.30, of which amount 28 per cent, is recovered. How many 
 B.t.u. are recovered per hr.? (Ans., 853 000.) If this is all 
 transferred to the jacket water, how much will that water be 
 heated above 140°? (Ans., 16.8°, or to 156.8°.) If, on the 
 contrary, it is used to generate steam at 100 lb. pressure from 
 some of that water, how many B.t.u. must be added to each 
 lb. of water? (Ans., 1078.4.) How many lb. of steam will be 
 generated per hr.? (Ans., 790.) If this steam is used in a 
 turbine which consumes 15.8 lb. of steam per Ihp.-hr., what hp. 
 will the turbine develop? (Ans., 50.) What is the per cent, 
 saving or gain by this complication? (Ans., 5 per cent.) 
 
Properties of Saturated Steam 
 
 (Condensed from Steam Tables and Diagrams, by Marks and Davis, with 
 
 the permission of the publishers, Messrs. Longmans, Green, & Co.) 
 
 p 
 
 t 
 
 V 
 
 h 
 
 L 
 
 H 
 
 r 
 
 nw \ n. 
 
 M, 
 
 0.089 
 
 32 
 
 3294 
 
 
 
 1073.4 
 
 1073.4 
 
 1019.3 
 
 
 
 2.1832 
 
 2.1832 
 
 0.S63 
 
 70 
 
 871 
 
 38.06 
 
 1052.3 
 
 1090.3 
 
 994.0 
 
 0.0745 
 
 1.9868 
 
 2.0613 
 
 0.505 
 
 80 
 
 636.8 
 
 48.0 
 
 1046.7 
 
 1094.8 
 
 987.4 
 
 0.0932 
 
 1.9398 
 
 2.0330 
 
 0.696 
 
 90 
 
 469.3 
 
 68.0 
 
 1041.2 
 
 1099.2 
 
 980.8 
 
 0.1114 
 
 1.8944 
 
 2.0068 
 
 1 
 
 101.83 
 
 333.0 
 
 69.8 
 
 1034.6 
 
 1104.4 
 
 972.9 
 
 0.1327 
 
 1.8427 
 
 1.9754 
 
 2 
 
 126.15 
 
 173.5 
 
 94.0 
 
 1021.0 
 
 1115.0 
 
 956.7 
 
 0.1749 
 
 1.7431 
 
 1.9180 
 
 3 
 
 141.52 
 
 118.5 
 
 109.4 
 
 1012.3 
 
 1121.6 
 
 946.4 
 
 0.2008 
 
 1.6840 
 
 1.8848 
 
 4 
 
 153.01 
 
 90.5 
 
 120.9 
 
 1005.7 
 
 1126.5 
 
 938.6 
 
 0.2198 
 
 1.6416 
 
 1.8614 
 
 5 
 
 162.28 
 
 73.33 
 
 130.1 
 
 1000.3 
 
 1130.5 
 
 932.4 
 
 0.2348 
 
 1.6084 
 
 1.8432 
 
 6 
 
 170.06 
 
 61.89 
 
 137.9 
 
 995.8 
 
 1133.7 
 
 927.0 
 
 0.2471 
 
 1.5814 
 
 1.8285 
 
 7 
 
 176.85 
 
 53.56 
 
 144.7 
 
 991.8 
 
 1136.5 
 
 922.4 
 
 0.2579 
 
 1.5582 
 
 1.8161 
 
 8 
 
 182.86 
 
 47.27 
 
 150.8 
 
 988.2 
 
 1139.0 
 
 918.2 
 
 0.2673 
 
 1.5380 
 
 1.8053 
 
 9 
 
 188.27 
 
 42.36 
 
 156.2 
 
 985.0 
 
 1141.1 
 
 914.4 
 
 0.2756 
 
 1.5202 
 
 1.7968 
 
 10 
 
 193.22 
 
 38.38 
 
 161.1 
 
 9S2.0 
 
 1143.1 
 
 910.9 
 
 0.2832 
 
 1.5042 
 
 1.7874 
 
 11 
 
 197.75 
 
 35.10 
 
 165.7 
 
 979.2 
 
 1144.9 
 
 907.8 
 
 0.2902 
 
 1.4895 
 
 1.7797 
 
 12 
 
 201.96 
 
 32.36 
 
 169.9 
 
 976.6 
 
 1146.5 
 
 904.8 
 
 0.2967 
 
 1.4760 
 
 1.7727 
 
 13 
 
 205.87 
 
 30.03 
 
 173.8 
 
 974.2 
 
 1148.0 
 
 902.0 
 
 0.3025 
 
 1.4639 
 
 1.7664 
 
 14 
 
 209.55 
 
 28.02 
 
 177.5 
 
 971.9 
 
 1149.4 
 
 899.3 
 
 0.3081 
 
 1.4523 
 
 1.7604 
 
 14.696 
 
 212 
 
 26.79 
 
 180.0 
 
 970.4 
 
 1150.4 
 
 897.6 
 
 0.3118 
 
 1.4447 
 
 1.7565 
 
 IS 
 
 213.0 
 
 26.27 
 
 181.0 
 
 969.7 
 
 1150.7 
 
 896.8 
 
 0.3133 
 
 1.4416 
 
 1.7549 
 
 16 
 
 216.3 
 
 24.79 
 
 184.4 
 
 967.6 
 
 1152.0 
 
 894.4 
 
 0.3183 
 
 1.4311 
 
 1.7494 
 
 17 
 
 219.4 
 
 23.38 
 
 187.5 
 
 965.6 
 
 1153.1 
 
 892.1 
 
 0.3229 
 
 1.4215 
 
 1.7444 
 
 18 
 
 222.4 
 
 22.16 
 
 190.5 
 
 963.7 
 
 1154.2 
 
 889.9 
 
 0.3273 
 
 1.4127 
 
 1.7400 
 
 19 
 
 225.2 
 
 21.07 
 
 193.4 
 
 961.8 
 
 1155.2 
 
 887.8 
 
 0.3316 
 
 1.4045 
 
 1.7360 
 
 20 
 
 228.0 
 
 20.08 
 
 196.1 
 
 960.0 
 
 1156.2 
 
 885.8 
 
 0.3356 
 
 1.3965 
 
 1.7320 
 
 21 
 
 230.6 
 
 19.18 
 
 198.8 
 
 958.3 
 
 1157.1 
 
 883.9 
 
 0.3393 
 
 1.3887 
 
 1.7280 
 
 22 
 
 233.1 
 
 18.37 
 
 201.3 
 
 956.7 
 
 1158.0 
 
 882.0 
 
 0.3430 
 
 1.3811 
 
 1.7241 
 
 23 
 
 235.5 
 
 17.62 
 
 203.8 
 
 955.1 
 
 1158.8 
 
 880.2 
 
 0.3465 
 
 1.3739 
 
 1.7204 
 
 24 
 
 237.8 
 
 16.93 
 
 206.1 
 
 953.5 
 
 1159.6 
 
 878.5 
 
 0.3499 
 
 1.3670 
 
 1.7169 
 
 25 
 
 240.1 
 
 16.30 
 
 208.4 
 
 952.0 
 
 1160.4 
 
 876.8 
 
 0.3532 
 
 1.3604 
 
 1.7136 
 
 26 
 
 242.2 
 
 15.72 
 
 210.6 
 
 950.6 
 
 1161.2 
 
 875.1 
 
 0.3564 
 
 1.3542 
 
 1.7106 
 
 27 
 
 244.4 
 
 15.18 
 
 212.7 
 
 949.2 
 
 1161.9 
 
 873.5 
 
 0.3694 
 
 1.3483 
 
 1.7077 
 
 28 
 
 246.4 
 
 14.67 
 
 214.8 
 
 947.8 
 
 1162.6 
 
 872.0 
 
 0.3623 
 
 1.3425 
 
 1.7048 
 
 29 
 
 248.4 
 
 14.19 
 
 216.8 
 
 946.4 
 
 1163.2 
 
 870.5 
 
 0.3662 
 
 1.3367 
 
 1.7019 
 
 30 
 
 250.3 
 
 13.74 
 
 218.8 
 
 945.1 
 
 1163.9 
 
 869.0 
 
 0.3680 
 
 1.3311 
 
 1.6991 
 
 31 
 
 252.2 
 
 13.32 
 
 220.7 
 
 943.8 
 
 1164.5 
 
 867.6 
 
 0.3707 
 
 1.3257 
 
 1.6964 
 
 32 
 
 254.1 
 
 12.93 
 
 222.6 
 
 942.5 
 
 1165.1 
 
 866.2 
 
 0.3733 
 
 1.3205 
 
 1.6938 
 
 33 
 
 255.8 
 
 12.57 
 
 224.4 
 
 941.3 
 
 1165.7 
 
 864.8 
 
 0.3759 
 
 1.3155 
 
 1.6914 
 
 34 
 
 257.6 
 
 12.22 
 
 226.2 
 
 940.1 
 
 1166.3 
 
 863.4 
 
 0.3784 
 
 1.3107 
 
 1.6891 
 
 35 
 
 259.3 
 
 11.89 
 
 227.9 
 
 938.9 
 
 1166.8 
 
 862.1 
 
 0.3808 
 
 1.3060 
 
 1.6868 
 
 36 
 
 261.0 
 
 11.58 
 
 229.6 
 
 937.7 
 
 1167.3 
 
 860.8 
 
 0.3832 
 
 1.3014 
 
 1.6846 
 
 37 
 
 262.6 
 
 11.29 
 
 231.3 
 
 936.6 
 
 1167.8 
 
 859.5 
 
 0.3855 
 
 1.2969 
 
 1.6824 
 
 38 
 
 264.2 
 
 11.01 
 
 232.9 
 
 935.5 
 
 1168.4 
 
 858.3 
 
 0.3877 
 
 1.2926 
 
 1.6802 
 
 39 
 
 265.8 
 
 10.74 
 
 234.5 
 
 934.4 
 
 1168.9 
 
 857.1 
 
 0.3899 
 
 1.2882 
 
 1.6781 
 
 40 
 
 267.3 
 
 10.49 
 
 236.1 
 
 933.3 
 
 1169.4 
 
 855.9 
 
 0.3920 
 
 1.2841 
 
 1.6761 
 
 41 
 
 268.7 
 
 10.25 
 
 237.6 
 
 932.2 
 
 1169.8 
 
 854.7 
 
 0.3941 
 
 1.2800 
 
 1.6741 
 
 42 
 
 270.2 
 
 10.02 
 
 239.1 
 
 931.2 
 
 1170.3 
 
 853.6 
 
 0.3962 
 
 1.2769 
 
 1.6721 
 
 43 
 
 271.7 
 
 9.80 
 
 240.5 
 
 930.2 
 
 1170.7 
 
 852.4 
 
 0.3982 
 
 1.2720 
 
 1.6702 
 
 44 
 
 273.1 
 
 9.59 
 
 242.0 
 
 029.2 
 
 1171.2 
 
 851.3 
 
 0.4002 
 
 1.2681 
 
 1.6683 
 
 45 
 
 274.5 
 
 9.39 
 
 243.4 
 
 928.2 
 
 1171.6 
 
 850.3 
 
 0.4021 
 
 1.2644 
 
 1.6665 
 
 46 
 
 275.8 
 
 9.20 
 
 244.8 
 
 927.2 
 
 1172.0 
 
 849.2 
 
 0.4040 
 
 1.2607 
 
 1.6647 
 
 47 
 
 277.2 
 
 9.02 
 
 246.1 
 
 926.3 
 
 1172.4 
 
 848.1 
 
 0.4059 
 
 1.2571 
 
 1.6630 
 
 48 
 
 278.5 
 
 8.84 
 
 247.5 
 
 925.3 
 
 1172.8 
 
 847.1 
 
 0.4077 
 
 1.2536 
 
 1.6613 
 
 49 
 
 279.8 
 
 8.67 
 
 248.8 
 
 924.4 
 
 1173.2 
 
 846.1 
 
 0.4095 
 
 1.2502 
 
 1.6597 
 
 50 
 
 281.0 
 
 8.51 
 
 250.1 
 
 923.5 
 
 1173.6 
 
 845.0 
 
 0.4113 
 
 1.2468 
 
 1.6581 
 
Steam and Other Vapoes 
 
 229 
 
 Properties of Saturated Steam — Continued 
 
 (Condensed from Steam Tables and Diagrams, by Marks and Davis, with 
 the permission of the publishers, Messrs. Longmans, Green, & Co.) 
 
 p 
 
 t 
 
 V 
 
 A 
 
 L 
 
 H 
 
 r 
 
 nw 
 
 It, 
 
 n. 
 
 51 
 
 282.3 
 
 8.35 
 
 261.4 
 
 922.6 
 
 1174.0 
 
 844.0 
 
 0.4130 
 
 1.2436 
 
 1.6666 
 
 52 
 
 283.5 
 
 8.20 
 
 262.6 
 
 921.7 
 
 1174.3 
 
 843.1 
 
 0.4147 
 
 1.2402 
 
 1.6649 
 
 53 
 
 284.7 
 
 8.05 
 
 253.9 
 
 920.8 
 
 1174.7 
 
 842.1 
 
 0.4164 
 
 1.2370 
 
 1.6534 
 
 54 
 
 285.9 
 
 7.91 
 
 256.1 
 
 919.9 
 
 1175.0 
 
 841.1 
 
 0.4180 
 
 1.2339 
 
 1.6519 
 
 55 
 
 287.1 
 
 7.78 
 
 266.3 
 
 919.0 
 
 1176.4 
 
 840.2 
 
 0.4196 
 
 1.2309 
 
 1.6605 
 
 56 
 
 288.2 
 
 7.65 
 
 257.5 
 
 918.2 
 
 1176.7 
 
 839.3 
 
 0.4212 
 
 1.2278 
 
 1.6490 
 
 57 
 
 289.4 
 
 7.52 
 
 258.7 
 
 917.4 
 
 1176.0 
 
 838.3 
 
 0.4227 
 
 1.2248 
 
 1.6475 
 
 58 
 
 290.5 
 
 7.40 
 
 259.8 
 
 916.5 
 
 1176.4 
 
 837.4 
 
 0.4242 
 
 1.2218 
 
 1.6460 
 
 59 
 
 291.6 
 
 7.28 
 
 261.0 
 
 915.7 
 
 1176.7 
 
 836.5 
 
 0.4267 
 
 1.2189 
 
 1.6446 
 
 60 
 
 292.7 
 
 7.17 
 
 262.1 
 
 914.9 
 
 1177.0 
 
 836.6 
 
 0.4272 
 
 1.2160 
 
 1.6432 
 
 61 
 
 293.8 
 
 7.06 
 
 263.2 
 
 914.1 
 
 1177.3 
 
 834.8 
 
 0.4287 
 
 1.2132 
 
 1.6419 
 
 62 
 
 294.9 
 
 6.95 
 
 264.3 
 
 913.3 
 
 1177.6 
 
 833.9 
 
 0.4302 
 
 1.2104 
 
 1.6406 
 
 63 
 
 295.9 
 
 6.85 
 
 266.4 
 
 912.5 
 
 1177.9 
 
 833.1 
 
 0.4316 
 
 1.2077 
 
 1.6393 
 
 64 
 
 297.0 
 
 6.75 
 
 266.4 
 
 911.8 
 
 1178.2 
 
 832.2 
 
 0.4330 
 
 1.2050 
 
 1.6380 
 
 65 
 
 298.0 
 
 6.65 
 
 267.6 
 
 911.0 
 
 1178.6 
 
 831.4 
 
 0.4344 
 
 1.2034 
 
 1.6368 
 
 66 
 
 299.0 
 
 6.56 
 
 268.6 
 
 910.2 
 
 1178.8 
 
 830.6 
 
 0.4368 
 
 1.2007 
 
 1.6356 
 
 67 
 
 300.0 
 
 6.47 
 
 269.6 
 
 909.5 
 
 1179.0 
 
 829.7 
 
 0.4371 
 
 1.1972 
 
 1.6343 
 
 68 
 
 301.0 
 
 6.38 
 
 270.6 
 
 908.7 
 
 1179.3 
 
 828.9 
 
 0.4386 
 
 1.1946 
 
 1.6331 
 
 69 
 
 302.0 
 
 6.29 
 
 271.6 
 
 908.0 
 
 1179.6 
 
 828.1 
 
 0.4398 
 
 1.1921 
 
 1.6319 
 
 70 
 
 302.9 
 
 6.20 
 
 272.6 
 
 907.2 
 
 1179.8 
 
 827.3 
 
 0.4411 
 
 1.1896 
 
 1.6307 
 
 71 
 
 303.9 
 
 6.12 
 
 273.6 
 
 906.5 
 
 1180.1 
 
 826.5 
 
 0.4424 
 
 1.1872 
 
 1.6296 
 
 72 
 
 304.8 
 
 6.04 
 
 274.6 
 
 905.8 
 
 1180.4 
 
 826.8 
 
 0.4437 
 
 1.1848 
 
 1.6286 
 
 73 
 
 305.8 
 
 6.96 
 
 275.5 
 
 905.1 
 
 1180.6 
 
 825.0 
 
 0.4449 
 
 1.1825 
 
 1.6274 
 
 74 
 
 306.7 
 
 6.89 
 
 276.5 
 
 904.4 
 
 1180.9 
 
 824.2 
 
 0.4462 
 
 1.1801 
 
 1.6263 
 
 75 
 
 307.6 
 
 6.81 
 
 277.4 
 
 903.7 
 
 1181.1 
 
 823.5 
 
 0.4474 
 
 1.1778 
 
 1.6252 
 
 80 
 
 312.0 
 
 6.47 
 
 282.0 
 
 900.3 
 
 1182.3 
 
 819.8 
 
 0.4636 
 
 1.1666 
 
 1.6200 
 
 85 
 
 316.3 
 
 6.16 
 
 286.3 
 
 897.1 
 
 1183.4 
 
 816.3 
 
 0.4690 
 
 1.1661 
 
 1.6151 
 
 90 
 
 320.3 
 
 4.89 
 
 290.6 
 
 893.9 
 
 1184.4 
 
 813.0 
 
 0.4644 
 
 1.1461 
 
 1.6105 
 
 95 
 
 324.1 
 
 4.65 
 
 294.6 
 
 890.9 
 
 1185.4 
 
 809.7 
 
 0.4694 
 
 1.1367 
 
 1.6061 
 
 100 
 
 327.8 
 
 4.429 
 
 298.3 
 
 888.0 
 
 1186.3 
 
 806.6 
 
 0.4743 
 
 1.1277 
 
 1.6020 
 
 105 
 
 331.4 
 
 4.230 
 
 302.0 
 
 885.2 
 
 1187.2 
 
 803.6 
 
 0.4789 
 
 1.1191 
 
 1.5980 
 
 110 
 
 334.8 
 
 4.047 
 
 305.6 
 
 882.6 
 
 1188.0 
 
 800.7 
 
 0.4834 
 
 1.1108 
 
 1.5942 
 
 115 
 
 338.1 
 
 3.880 
 
 309.0 
 
 879.8 
 
 1188.8 
 
 797.9 
 
 0.4877 
 
 1.1030 
 
 1.5907 
 
 120 
 
 341.3 
 
 3.726 
 
 312.3 
 
 877.2 
 
 1189.6 
 
 796.2 
 
 0.4919 
 
 1.0964 
 
 1.5873 
 
 125 
 
 344.4 
 
 3.583 
 
 315.5 
 
 874.7 
 
 1190.3 
 
 792.6 
 
 0.4959 
 
 1.0880 
 
 1.6839 
 
 130 
 
 347.4 
 
 3.462 
 
 318.6 
 
 872.3 
 
 1191.0 
 
 790.0 
 
 0.4998 
 
 1.0809 
 
 1.5807 
 
 140 
 
 353.1 
 
 3.219 
 
 324.6 
 
 867.6 
 
 1192.2 
 
 785.0 
 
 0.6072 
 
 1.0676 
 
 1.5747 
 
 150 
 
 358.5 
 
 3.012 
 
 330.2 
 
 863.2 
 
 1193.4 
 
 780.4 
 
 0.5142 
 
 1.0650 
 
 1.5692 
 
 160 
 
 363.6 
 
 2.834 
 
 335.6 
 
 868.8 
 
 1194.6 
 
 775.8 
 
 0.5208 
 
 1.0431 
 
 1.6639 
 
 170 
 
 368.5 
 
 2.676 
 
 340.7 
 
 864.7 
 
 1196.4 
 
 771.5 
 
 0.5269 
 
 1.0321 
 
 1.5590 
 
 180 
 
 373.1 
 
 2.533 
 
 346.6 
 
 860.8 
 
 1196.4 
 
 767.4 
 
 0.5328 
 
 1.0215 
 
 1.5643 
 
 190 
 
 377.6 
 
 2.406 
 
 360.4 
 
 846.9 
 
 1197.3 
 
 763.4 
 
 0.6384 
 
 1.0114 
 
 1.6498 
 
 200 
 
 381.9 
 
 2.290 
 
 354.9 
 
 843.2 
 
 1198.1 
 
 759.6 
 
 0.6437 
 
 1.0019 
 
 1.6456 
 
 210 
 
 386.0 
 
 2.187 
 
 369.2 
 
 839.6 
 
 1198.8 
 
 766.8 
 
 0.6488 
 
 0.9928 
 
 1.5416 
 
 220 
 
 389.9 
 
 2.091 
 
 363.4 
 
 836.2 
 
 1199.6 
 
 752.3 
 
 0.6638 
 
 0.9841 
 
 1.5379 
 
 230 
 
 393.8 
 
 2.004 
 
 367.5 
 
 832.8 
 
 1200.2 
 
 748.8 
 
 0.5586 
 
 0.9768 
 
 1.5344 
 
 240 
 
 397.4 
 
 1.924 
 
 371.4 
 
 829.6 
 
 1200.9 
 
 746.4 
 
 0.5633 
 
 0.9676 
 
 1.5309 
 
 250 
 
 401.1 
 
 1.850 
 
 375.2 
 
 826.3 
 
 1201.5 
 
 742.0 
 
 0.5676 
 
 0.9600 
 
 1.5276 
 
 2947 
 
 689.0 
 
 0.05 
 
 ? 
 
 
 
 ? 
 
 
 
 ? 
 
 
 
 ? 
 
INDEX 
 
 Reference is to Article numbers unless otherwise specified. 
 
 absolute temperature, 11 
 
 absolute zero, 11 
 
 action of engine, 7 
 
 adiabatic, 28, 29, 72, 79 
 
 air, 18 
 
 air compressor, 40-49 
 
 air consumption, air engine, 52, 53 
 
 air cooler, 125 
 
 air engine, 38, 39, 50-55 
 
 air machinery, 38-62 
 
 air refrigeration, 57-62 
 
 air-steam mixture, 92 
 
 alcohol, 153 
 
 ammonia, 112, 113, page 184 
 
 ammonia condenser, 125 
 
 automobile engine, 157 
 
 avaUabihty of heat, 36 
 
 back pressure, 88 
 desirable, 102 
 Baum6 scale, 159 
 binary vapor engine, 103 
 blast furnace gas, 163, 155 
 boiler, 125, 126 
 boiling points of gases, 18 
 boiUng points of liquids, 63 
 Boyle's law, 9, 20 
 brine, 115 
 brine cooler, 125 
 brine system, 115 
 British thermal unit, 3 
 bucket friction, 138, 143 
 
 calcium chloride brine, 115 
 calorie, 3 
 
 calorimeter, throtthng, 135 
 carbon dioxide, 65, 112 
 carbon dioxide refrigerating machine, 
 119-121 
 
 carbon monoxide, 18 
 Carnot cycle, 34-37, 88-89 
 Centigrade temperature, 2 
 Charles' laws, 10 
 clearance, 7, 23, 95, 104 
 
 gas engine, 150 
 coal gas, 153, 165 
 
 coeflBcient of heat transmission, 125 
 coke oven gas, 153, 155 
 combination turbine, 146 
 complete expansion cycle, 80 
 compound steam engine, 94, 98, 100, 
 
 102, 106-109 
 compound turbine, 143-146 
 compressed air, 38-56 
 compressed air blast, oil engine, 159 
 compressed air plant, 55, 56 
 compressed air transmission, 56 
 compression, 95, 97, 98 
 compression, air engine, 50 
 
 Du-esel engine, 160, 162 
 
 internal combustion engine, 164 
 compression ratio, 38 
 
 Otto cycle, 160, 153, 169 
 compressor, air, 40-49 
 compressor, refrigeration. 111 
 condensation, 78 
 
 cyhnder, 95-100 
 condenser, 49, 59, 92, 101, 129 
 
 refrigeration, 111 
 condensing engine, 88, 94 
 constant dryness curve, 72 
 
 entropy, 79 
 
 heat content, 76 
 
 volume, 75 
 convergent nozzle, 136 
 cooling in air compressor, 41, 43, 47 
 
 air engine, 60, 54 
 counter flow, 124 
 
 230 
 
Index. 
 
 231 
 
 critical pressure, 136 
 critical temperature, 65 
 Curtis turbine, 144 
 cycle, 21, 22, 34 
 
 Diesel, 160, 161 
 
 Ericsson, 62 
 
 ideal vapor, 80 
 
 incomplete expansion, 95 
 
 Otto, 149-151 
 
 Rankine, 80, 95 
 
 for refrigeration, 111 
 
 regenerative, 61, 62 
 
 Stirling, 62 
 
 superheated steam, 84-87 
 cylinder condensation, 95-100 
 cylinder ratio, 106, 108 
 
 Dalton's law, 13, 90 
 Davis formula, 66 
 De Laval turbine, 138-142 
 dense air ice machine, 60 
 diagram, indicator, 95 
 diagram factor, 104 
 
 compound engine, 107, 108 
 
 gas engine, 155 
 Diesel engine, 160-162 
 difference of specific heats, 15, 16 
 disgregation work, 2, 18, 63 
 direct expansion, 115 
 displacement, 7, 23, 42, 115 
 
 specific, 116 
 distiUer, 130 
 divergent nozzle, 136 
 double-acting engine, 7 
 double-flow turbine, 146 
 double-valve engine, 94 
 drop, compound engine, 108 
 dry compression, 117 
 dryness, effect on effi[ciency, 88 
 dry vapor, 64 
 dummy piston, 146 
 
 economizer, 125, 128 
 economy, gas engine, 165 
 effects of heat, 2 
 efficiency, 5, 6 
 air engine, 64 
 
 efficiency, air refrigeration, 57, 59 
 
 Carnot cycle, 34 
 
 compressed air plant, 55, 56 
 
 cycle, 22 
 
 dense air ice machine, 60 
 
 Diesel engine, 160, 161 
 
 evaporation, 131 
 
 heat engine, 34, 36 
 
 ideal steam engine, 88 
 
 incomplete expansion cycle, 81 
 
 internal combustion engine, 163- 
 165 
 
 Joule cycle, 38 
 
 mechanical (gas engine), 156, 157 
 
 nozzle, 137, 141 
 
 Otto cycle, 150 
 
 variable specific heat, 152 
 Rankine cycle, 80, 84-87 
 for refrigeration, 114, 115, 118 
 
 regenerative cycle, 62 
 
 relative, 93, 94, 97, 155 
 
 steam cycle, 82 
 
 steam engine, 93-103 
 
 Stirling cycle, 62 
 
 turbine, 141, 142, 147, 148 
 
 two-cycle gas engine, 157 
 
 volumetric, 42, 46, 155, 157 
 end thrust, 146 
 engine action, 7 
 entropy, 67-73 
 
 of hqmd, 71 
 
 superheated steam, 83 
 entropy of vaporization, 71 
 equivalent evaporation, 73 
 equivalent heat and mechanical units, 
 
 73 
 equivalent mean effective pressure, 
 
 107 
 ether, 112 
 ethyl chloride, 112 
 evaporator, 125, 130, 131 
 exhaust loss, gas engine, 166 
 exhaust steam, quality of, 101 
 expander, 60 
 expansion in air engine, 50 
 
 in nozzle, 136 
 
 ratio of, 96, 104 
 
232 
 
 Index. 
 
 expansion valve, 111, 113 
 exponentials, valves of, page 12 
 external work, 2 
 
 any path, 29 
 
 graphical representation, 19 
 
 poljdiropic, 26 
 
 superheating, 83 
 
 vapor adiabatic, 79 
 
 vaporization, 65, 74, 77 
 
 factor, diagram, 104 
 
 factor of evaporation, 73, 83 
 
 Fahrenheit temperature, 2 
 
 feed water heater, 125, 129 
 
 feed water temperature, 73 
 
 first law of thermodynamics, 4 
 
 flow of air, 136 
 
 flow of steam, 133-137 
 
 fluid friction, 156 
 
 fluids for refrigeration, 112 
 
 four-cycle engine, 149-156 
 
 four-valve engine, 94 
 
 free air, 42, 52 
 
 free expansion, 136 
 
 friction, nozzle, 134-137 
 
 turbine, 141, 142 
 fuel ofl, 153, 159 
 furnace efficiency, 126 
 
 gases, composition, 155 
 
 heat value, 155 
 gasoline, 153 
 gauge pressure, 9 
 governing gas engines, 153 
 gun, gases in, 32 
 
 heat absorbed, any path, 29 
 graphical representation, 33 
 polytropic, 27 
 
 heat balance, gas engine, 166 
 
 heat chart, 86, 87 
 
 heat expenditure, forming vapor, 73 
 
 heat of liquid, 63 
 
 heat rate, 82, 93, 163 
 
 heat transfer appliances, 123-132 
 
 heat unit, 3 
 
 heat value, gases, 155 
 
 liquids, 159 
 heating of liquid, 63, 68 
 helium, 18 
 high heat value, 155 
 hit-or-miss governing, 153 
 horse power, air engine, 51 
 
 air refrigeration, 58, 59 
 
 compressing air, 42, 45 
 
 dense air ice machine, 60 
 
 engine, 7 
 
 impulse turbine, 140 
 
 reaction turbine, 146 
 
 steam boiler, 126 
 hot air engine, 61, 62 
 humidity, 91 
 hydrogen, 18 
 
 ice melting effect, 59 
 ideal air engine, 53 
 
 air refrigerating machine, 58-62 
 
 steam rate, 82 
 ignition, 153, 159 
 imperfect gas, 32 
 impulse turbine, 138-142 
 incomplete expansion cycle, 81, 95 
 indicated horse power, 7 
 indicator, 7 
 
 indicator diagram, 7, 95, 149, 154 
 indirect refrigeration, 115 
 initial pressure, 88, 147 
 injector, 134 
 
 intercooler, 44, 47, 48, 125 
 interoooler pressure, 45, 48 
 internal combustion engine, 149-166 
 internal energy, 63, 65, 74 
 isodiabatic, 31 
 isothermal, 20, 29 
 
 jacket, air compressor, 43, 47 
 
 gas engine, 166 
 
 steam, 97, 98, 104 
 Joule cycle, 38, 39, 57 
 Joule's law, 18 
 
 kerosene, 153 
 
 kinetic efficiency, 139, 143 
 
Index. 
 
 233 
 
 latent heat of fusion, 59 
 of vaporization, 64, 65 
 law of perfect gas, 12 
 load vs. efficiency, turbine, 148 
 low heat value, 155 
 low-pressure turbine, 147 
 
 marine reaction turbine, 146 
 
 steam engine, 99 
 ■ mass-flow theory, 125 
 maxirmim efficiency, heat engine, 
 34 
 
 flow, nozzle, 136 
 mean B.t.u., 3 
 
 mean effective pressure, 7, 23 
 air engine, 53 
 Carnot cycle, 37 
 dense air ice machine, 60 
 Diesel engine, 160, 162 
 equivalent simple engine, 107' 
 Joule cycle, 39 
 Otto cycle, 151, 153, 155 
 Rankine cycle, 80, 84, 87 
 refrigeration, 114, 115, 118 
 steam engine, 80, 81, 104, 105 
 Stirling cycle, 62 
 superheated steam, 105 
 two-stage air compressor, 45 
 mean temperature difierence, 124 
 mechanical eflBciency, 6 
 Diesel engine, 162 
 gas engine, 156, 157 
 refrigeration, 115 
 turbine, 142 
 mechanical equivalent of heat, 4 
 methane, 18 
 mixtures, 123 
 air and steam, 90-92 
 gases, 13 
 moist air, 90-92 
 molecular weights, 18 
 Mollier diagram, 86, 87 
 multiple effect evaporator, 131 
 multiple expansion engine, 100, 110 
 multi-stage air compressor, 48 
 refrigeration, 117 
 turbine, 143-146 
 16 
 
 natural gas, 153, 155 
 negative heat of liquid, 63 
 negative loop, 3 
 negative specific heat, 14 
 nitrogen, 18 
 
 non-condensing engine, 94 
 nozzle, 136, 137 
 nozzle angle, 138 
 
 oil engine, 159 
 
 oil gas, 155 
 
 Otto cycle, 149-151 
 
 outside compression, 157 
 
 overload capacity, 7, 156 
 
 oxygen, 18 
 
 parallel flow, 124 
 
 Parsons turbine, 145, 146 
 
 parts of engine, 7 
 
 perfect gas, 8-37 
 
 peripheral speed, 139, 140, 144, 146 
 
 permanent gas, 8-37 
 
 pipe covering, 132 
 
 piston pressures, 106 
 
 piston speed, 7, 104, 115 
 
 polytropic, 24-27, 29 
 
 power-speed curve, marine engine, 99 
 
 preheat, 54, 55 
 
 preheater, 54 
 
 pressure, 8 
 
 pressure-compounded turbine, 144 
 
 pressure-temperature curve, 63 
 
 producer gas, 153, 155 
 
 properties of gases, 18 
 
 of steam, 66, 74 
 propulsion, power for, 6 
 
 quadruple-expansion engine, 100,103 
 
 110 
 quaUty governing, 153 ^ 
 
 quality of steam, 135 
 quantity governing, 153 
 
 radiant heat, 126 
 
 radiation from steam pipes, 132 
 
 radiator, 125 
 
 Rankine cycle, 80, 84-87, 95 
 
 for refrigeration, 111, 114, 118 
 
234 
 
 Index. 
 
 ratio of expansion, 88, 96, 104, 161, 
 
 162 
 ratio of specific heats, 17 
 reaction turbine, 145, 146 
 receiver, 106, 125 
 reduction gear, 142 
 refrigeration, 111-122 
 refrigeration, air, 57-62 
 Carnot cycle, 35 
 coils, 125 
 regenerative engine, 61, 62, 103 
 reheater, 100 
 relative efficiency, 53, 55, 93, 94, 97, 
 
 142, 155 
 relative humidity, 91 
 relative velocity, 138 
 releasing valve gear, 94 
 rotation loss, 141 
 
 saturated air, 90 
 saturated vapor, 64 
 scavenging, 157 
 
 second law of thermodynamics, 36 
 simple steam engine, 104, 105 
 simple impulse turbine, 138-142 
 single-acting engine, 7 
 single-stage turbine, 138-142 
 single-valve engine, 94 
 specific displacement, 116 
 specific gravity, liquid fuels, 159 
 specific heat, 3 
 
 any path, 29 
 
 gases, 14, 15 
 
 polytropic, 27 
 
 variable, 152 
 speed, piston, 104 
 speed-power curve, 99 
 speed, steam engine, 99, 104 
 
 turbine, 140 
 steam, 63-92 
 
 steam consumption, engine, 94, 96 
 steam engine, action, 95 
 
 compound, 94, 98, 100, 102 
 
 condensing, 94 
 
 efficiency, 93-103 
 
 heat rate, 93 
 
 mechanical efficiency, 100 
 
 steam engine, non-condensing, 94 
 
 piston speed, 104 
 
 power, 93-100 
 
 quadruple, 100, 103 
 
 regenerative, 103 
 
 triple, 94, 100 
 
 Uniflow, 103 
 
 vacuum, 102 
 
 valves, 94 
 steam jacket, 97, 98, 104 
 steam power plant, 80 
 steam rate, 82, 142 
 superheated, 94, 98 
 table, 66 
 turbine, 133-148 
 straight line law, boiler, 126 
 sulphur dioxide, 112 
 
 refrigerating machine, 122 
 supercompression, 155 
 superheat, 83, 88, 109, 110 
 in refrigeration, 117, 118 
 superheater, 125, 127 
 superheated steam, efficiency, 94, 98, 
 148 
 
 mean effective pressure, 105 
 superheated vapor, 64, 83-87 
 superheating, 72 
 supersaturated air, 90 
 surface condenser, 125, 129 
 surface efficiency, 126 
 
 temperature, 2, 3 
 
 temperature-entropy diagram, 67-73 
 temperature ranges, compoimd en- 
 gine, 106 
 terminal drop, 108 
 thermal efficiency, 6 
 thermodynamics, definition, 1 
 thermodynamic surface, 12 
 throttling calorimeter, 135 
 timed injection of oil, 159 
 tonnage, 69, 115, 116 
 total heat of vapor, 65 
 total heat-entropy diagram, 86, 87 
 total heat-pressure diagram, 87 
 transfer of heat, 123-132 
 triple-expansion engine, 94, 100, 110 
 
Index. 
 
 235 
 
 turbine, 133-148 
 
 two-cycle engine, 157, 158, 159, 162 
 
 two-stage air compressor, 44H18 
 
 Uniflow engine, 103 
 
 vacuum, 92, 148 
 
 best for engine, 102 
 
 pump, 49 
 valve, steam engine, 94 
 valveless gas engine, 157 
 Van der Waals equation, 32 
 vapor, 8, 63-92 
 
 for heat engine, 89 
 
 refrigeration, 111-122 
 vaporization, 64, 65, 72, 74 
 variable specific heat, 152 
 velocity, adiabatic flow, 133 
 velocity compounding, 143, 144 
 
 velocity diagram, 138, 144, 145 
 energy, 133 
 nozzles, 136, 137 
 steam, in reaction turbine, 146 
 volume of gas, 8, 12 
 liquid, 65, 77 
 superheated steam, 83 
 vapor, 64, 74, 75 
 volumetric efficiency, 42, 46, 115, 155 
 157 
 
 water gas, 153, 155 
 water required air compressor, 43 
 Westinghouse-Parsons turbine, 146 
 wet compression, 117 
 wet vapor, 64, 72, 74-76 
 wire-drawing, 134^-137 
 work of turbine buckets, 138, 142 
 143, 145, 146 
 
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