LIBRARY 
 
 OF THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
 GIFT OF" 
 
 MRS. MARTHA E. HALLIDIE. 
 Class 
 
WORKS OF PROF. H. W. SPANGLER 
 
 PUBLISHED BY 
 
 JOHN WILEY & SONS. 
 
 Valve-Gears. 
 
 Designed as a Text-book giving those parts of the 
 Theory of Valve-gears necessary to a clear under- 
 standing of the subject. 8vo, xii 4-179 pages, 109 
 figures, cloth, $2.50. 
 
 Notes on Thermodynamics. 
 
 The Derivation of the Fundamental Principles of 
 Thermodynamics and their Application to Numer 
 ical Problems, ismo, vi + 6g pages, 24 figures, 
 cloth, $1.00. 
 
NOTES 
 
 ON 
 
 THERMODYNAMICS. 
 
 BY 
 
 H. W. SPANGLER, 
 
 Whitney Professor of Mechanical Engineering 
 in the University of Pennsylvania. 
 
 PART I. 
 
 SECOND EDITION-. 
 FIRST THOUSAND. 
 
 JOHN WILEY & SONS. 
 
 LONDON: CHAPMAN & HALL, LIMITED. 
 
 1901. 
 
QJ 
 
 *S 
 
 Copyright, 1901, 
 
 BY 
 H. W. SPANGLER. 
 
 HALLIDIE 
 
 ROBERT DRUMMOND, PRINTER, NEW YORK 
 
PREFACE. 
 
 FOR the purpose of covering the theoretical side of 
 thermodynamics more rapidly than could be done with 
 the aid of existing text-books, the author prepared 
 these notes four years ago for use in his classes. 
 
 The results were fairly satisfactory, and as the work 
 is now used by other teachers, a revised edition has 
 been prepared. In this, errors have been corrected, 
 the text has been condensed, and additional problems 
 have been added. 
 
 It is not intended as a reference-book, except for 
 those who have worked it through and have solved the 
 problems. 
 
 There is little that is new in it. All the later 
 writers have been consulted in preparing the work, 
 and whatever has seemed the most satisfactory method 
 of arriving at a result has been made use of. 
 
 The work is not complete in itself, and a good table 
 of the properties of vapors is required to work out 
 many of the problems. The tables prepared by Pro- 
 fessor Peabody are used in the text. 
 
 H. W. SPANGLER. 
 
 UNIVERSITY OF PENNSYLVANIA, 
 June 6, 1901. 
 
 iii 
 
 96050 
 
NOTATION. 
 
 A = Heat equivalent of work = 
 
 c = Specific heat, the subscript indicating the law 
 of the expansion, and is used whether 
 units are foot-pounds or heat-units. 
 H Heat required in heat-units or foot-pounds. 
 J = Mechanical equivalent of heat = 778. 
 K = Constant of equation pv n = K. 
 
 A Total heat required to make I pound of vapor 
 
 from liquid at 32 degrees F. 
 m = Weight.' 
 M= Weight. 
 
 n = Exponent in equation pv n = K. 
 
 p = Pressure in pounds per square foot, absolute. 
 
 q = Heat of liquid. 
 
 r = Total latent heat. 
 
 p Inner latent heat. 
 R Constant for any substance in equation/^=^ T. 
 
 s = Volume of I pound of vapor. 
 
 cr = Volume of I pound of liquid. 
 
 / Temperature Fahrenheit. 
 
 T= Temperature absolute. 
 
 v Difference between the volume of I pound of 
 vapor and I pound of liquid = s cr. 
 
 i) Volume in cubic feet of I pound. 
 
 V = Any volume. 
 W = Work, foot-pounds or heat-units. 
 
NOTES ON THERMODYNAMICS. 
 
 IN Physics a distinction is made between perfect 
 gases and vapors. In this work we will also deal with 
 these two classes of substances, and, for engineering 
 purposes, perfect gases are such as practically obey the 
 laws of Boyle and Charles. Under the head of perfect 
 gases would be classed air, hydrogen, oxygen, super- 
 heated steam, ammonia, carbonic acid, etc., all being 
 sufficiently far from their condensing-point to obey 
 the laws referred to above. 
 
 In the shape of a formula these laws can be best 
 stated as 
 
 This equation is constantly being used in thermody- 
 namics, and the exact meaning of the terms is impor- 
 tant. In all this work English units, pounds, feet, 
 and degrees Fahrenheit will be used. In these units 
 th following definitions may be given to the terms of 
 equation (i): 
 
 p is the absolute pressure in pounds per square foot. 
 
 v is the volume in cubic feet of I pound of the sub- 
 stance dealt with. 
 
NOTES ON THERMODYNAMICS. 
 
 T is the absolute temperature, Fahrenheit degrees. 
 
 R is a constant whose value depends on the sub- 
 stance and the units taken. 
 
 To determine the value of R for any substance, we 
 must have for one given condition of pressure and 
 temperature the corresponding value of the volume of 
 I pound. This we have for many substances. Thus, 
 for air we have, for a pressure of 14.7 pounds per 
 square inch, or / = 14.7 X 144, and a temperature of 
 32 degrees Fahrenheit, or T = 492.7, the volume of 
 I pound of air, or v = 12.39 cubic feet. These are 
 quantities determined by experiment. Putting these 
 values in equation (i), we have for air 
 
 K _ pv 14.7 X H4 X 12.39 . 
 T 492^7 
 
 or, for air, with the units we have taken, we have 
 
 pv = 53,37^- 
 
 This equation is always true for air, and if, at any 
 time or under any conditions, two of the variables in 
 the equation are given, the third can be found. 
 
 Problem 1. 10 pounds of air at 200 degrees F. occupy 120 
 cubic feet; what must be the pressure ? 
 
 Here T = 460.7 + 200 = 660.7 ; v = 12; 
 
 10 
 
 53. vj x 660.7 
 p = - = 2950 pounds per square foot.* 
 
 Probi 2. How many pounds of air does it take to fill 5600 cubic 
 feet at 1 5 pounds pressure per square inch and at 60 
 degrees F. ? 
 
 * The slide rule or three-place logarithms are used in the solu- 
 tion of all problems, and the result is probably correct within 2$. 
 
NOTES ON THERMODYNAMICS. 
 
 Here/ = 15 x 144; T = 460.7 -f .60 = 520.7 ; 
 
 _ 5 2 -7 x 53-37 _ 
 15x144 
 
 of i pound, 5600 cubic feet contain 
 
 5600 
 
 vo i ume 
 
 = 434 
 
 pounds of air. 
 Prob, 3. At what temperature will 10 pounds of air at 1 5 pounds 
 
 pressure per square inch fill 60 cubic feet? 
 Prob. 4. What must be the pressure in a vessel of 4 cubic feet 
 
 if it contains 30 pounds of air at 50 degrees F. ? ^? 
 
 Evidently, if, in equation (i), we are dealing with 
 a substance twice as heavy as air, the value of v in the 
 first member, or the volume of a pound, will be only 
 half as great and, consequently, the value of R would 
 be only half as great. 
 
 
 Substance.* 
 
 Relative 
 Density. 
 
 
 
 R 
 Value. 
 
 Air 
 
 
 14.4 
 
 
 e-7 -17 
 
 CIA 
 
 o 
 
 
 16 
 
 14.4 
 
 X '(I ^7 
 
 48 I 
 
 H 
 
 
 I 
 
 16 
 
 MA 
 
 X %1 ^7 
 
 77O 
 
 N 
 
 
 
 14 
 
 14^4 
 
 X 53-37 
 
 
 rOj 
 
 *X*i 
 
 22 
 
 14 
 
 X 53-37 
 
 
 NH 
 
 JO 
 
 8 5 
 
 22 
 14.4 
 
 X 53-37 
 
 gO 6 
 
 ro 
 
 
 14 
 
 8-5 
 
 X 53-37 
 
 
 
 (steam) .... 
 
 Q 
 9 
 
 14 
 
 X "^ ^7 
 
 54-9 
 85 6 
 
 
 
 
 9 
 
 
 
 * Some of these substances do not act as perfect gases at usual 
 pressures and temperatures, so that care must be exercised in 
 using these constants. 
 
4 NOTES ON THERMODYNAMICS. 
 
 This enables us to apply the formula of equation (i) 
 and its constant, as determined for air, to many other 
 substances. From a table of relative densities one can 
 readily determine the value of R for these substances, 
 as in the table on page 3, and these values are practi- 
 cally correct for engineering calculations. 
 
 Prob, 5. How many pounds of oxygen will a holder contain 
 whose volume is 3 cubic feet, pressure 250 pounds 
 per square inch, and temperature 75 degrees F.? 
 We have for oxygen 
 
 / = 250 x 144 = 36000; 
 
 14.4 x 53.37 
 
 ^^ L x 535-7 
 
 36000 ~ and 
 
 3x36000x16 ^ 
 
 14.4x53.37x535.7 
 
 Prob, 6, What weight of hydrogen will fill a holder of 3.5 
 cubic feet at 200 pounds pressure and / = 80 
 degrees F.? / / 
 
 Prob. 7. What is the temperature at which a cubic foot of CO 2 
 will weigh .2 pound at 100 pounds pressure? / r 
 
 It is convenient to reduce the expression for the 
 weight to a simple formula. If V \s the total volume 
 and v the volume per pound, then 
 
 V pV 
 
 =. weight, or M = ^-^ y 
 
 V 4\. 1 
 
 from which of course, if V = v, the weight is I pound. 
 
NOTES ON THERMODYNAMICS. 5 
 
 Prob. 8. How many pounds of air will fill a vessel of 400 
 cubic feet at 15 pounds pressure if one-half the 
 volume is at 80 degrees F. and the rest at 600 
 degrees F.? 
 The weight of the portion at 80 degrees is 
 
 15 x 144 x 200 
 
 
 
 53.37 x 541 
 
 and the weight of the remainder is 
 
 15 x 144 x 200 
 
 J/a = 2 - = 7-62. 
 
 53.37 x 1061 
 The total weight is 22.65 pounds. 
 
 Prob. 9. What must be the pressure at which 20 pounds of 
 air will fill 270 cubic feet, 180 cubic feet being at 
 500 degrees and 90 cubic feet at 60 degrees F.? 
 
 In defining a perfect gas, there was one peculiarity 
 which was not mentioned and which will now be of 
 use. When a perfect gas is allowed to remain at the 
 same temperature while its volume changes, the 
 amount of heat that must be added to it to change 
 its pressure and volume is that required to do the 
 external work and no more. That is, if a perfect gas 
 is allowed to expand and change its temperature, the 
 quantity of heat which must be added to it is that 
 required to change the temperature, added to that 
 required to do external work. 
 
 As the equation/^ = R T contains three variables, 
 it is not convenient to indicate all the variations of 
 />, v, and T on the same diagram, and for convenience 
 of representation, and because a diagram whose co- 
 ordinates are pressure and volume is a diagram of 
 work, the /, v co-ordinates will be understood unless 
 different co-ordinates are marked on the figure. 
 
6 
 
 NOTES ON THERMODYNAMICS. 
 
 
 b 
 
 
 
 
 a 
 
 
 
 3 
 
 F 
 
 j VOLUMES, rc 
 IG. I. 
 
 Thus, in Fig. I if we call the two axes pressure and 
 volume, and we have a pound of gas in the conditions 
 represented by a, its volume is oy and 
 its pressure is ay, the temperature being 
 fixed from the equation pv = RT. If 
 now the pressure of the gas is increased 
 from ay to by, there being no change 
 in volume, there will be network done , 
 
 A /* 
 
 by the air. As its pressure is increased 
 the temperature is increased in the same proportion, 
 and we must have added enough heat to cause this 
 change in temperature. If, however, instead of in- 
 creasing the pressure, it had been maintained constant 
 and the volume increased from oy to ox, we would 
 have had not only to raise the temperature, but to 
 have done work overcoming a pressure ay through a 
 distance yx. 
 
 Again, if neither the pressure nor the volume re- 
 mains constant, we have in Fig. 2 the condition a for 
 the initial condition and d for the 
 final condition, and the amount of 
 heat which must have been added 
 from a to d must have been enough 
 to change the temperature from 
 that at a to that at d, and also to 
 do an amount of work equal to the 
 area ayxd. 
 
 To express the relation between these quantities 
 we must have units in which to measure them. 
 
 The unit in which the quantity of heat is measured 
 is the amount of heat which must be added to I 
 
 - 
 
 y x 
 FIG. 2. 
 
NO TES ON T HER MOD YNA MICS. 7 
 
 pound of water at 62 degrees F. to raise its tempera- 
 ture to 63 degrees, and is called the British Thermal 
 Unit, or simply B.T.U. 
 
 The unit of work is the foot-pound, arid, experi- 
 mentally, it has been determined that one B.T.U. is 
 equivalent to 778 foot-pounds. 
 
 The number of heat-units which must be added to 
 I pound of any substance to raise it I degree in tem- 
 perature is called the specific heat. 
 
 Referring now to Fig. i, if c v is the amount of 
 heat which must be added per degree to raise the 
 temperature from a to b, then c v is the specific heat 
 for constant volume, and the total heat required is 
 c v (T b T a ) when T b and T a are the temperatures cor- 
 responding to the conditions b and a respectively. 
 The value c v for air is .169 heat-unit, or 132 foot- 
 pounds. 
 
 Prob. 10, If 5 cubic feet of air at 30 pounds pressure per 
 square inch and 60 degrees F. has 20 heat-units 
 added to it at constant volume, and if the heat 
 required to raise the temperature of i pound t 
 degree at constant volume is .169 heat-unit, what, 
 is the resulting temperature ? 
 
 The weight of air is ^ = .774 pound. The 
 
 53-37X52I 
 
 heat required to raise this i degree is . 774 x. 169 
 = .131 heat-unit. The rise in temperature is 
 
 therefore - =153 degrees. 
 
 Prob. 11. If 15 cubic feet of air at 100 pounds pressure per 
 square incii is raised from 60 degrees to 100 de- 
 grees F. at constant volume, how much heat is 
 inquired? 5" Z <J 
 
8 NOTES O.V THERMODYNAMICS. 
 
 Similarly, if c p is the total amount of heat pet- 
 degree which must be added to the pound of gas at 
 a, Fig. I, to cause the gas to expand from a to c, 
 then c p is the specific heat for constant pressure, 
 and the total heat required is c p (T c T a ). In this 
 case, however, the heat has been used partly in rais- 
 ing the temperature, the remainder being required to 
 do external work. We can therefore write for the 
 quantity required to change the temperature only, 
 c v (T c T a ), and for the quantity required to do the 
 work, ay(ox oy) or f a (v c v a )> an d, as all the heat 
 must be accounted for, we can write 
 
 For air c p is .238 heat-unit* or 185 foot-pounds. 
 
 Prob. 12. If i pound of air is changed from 20 degrees to 30 
 . degrees F. at a constant pressure of 100 pounds 
 per square inch, how much heat must be added if 
 to raise the temperature alone required that the 
 equivalent of 132 foot-pounds of work be added 
 for each degree ? 
 
 The heat to change the temperature only is the 
 equivalent of (30 20) x 132 = 1320 foot-pounds. 
 The amount of work to be done is to overcome 
 the pressure of 100x144 pounds per square foot 
 through the difference in volume. The initial 
 
 volume is ^^ = 1.79 and the final vol- 
 
 100 x 144 
 
 ume = - ^ = 1.8^. The work is then 100 
 100 x 144 
 
 x 144(1.83 1.79) = 576.0 foot-pounds. The total 
 heat required is therefore the equivalent of 1320 
 + 576 = 1896 foot-pounds. 
 
NOTES ON THERMODYNAMICS. 9 
 
 Prob. 13. If c v = 132 foot-pounds, prove, by using equation (2), 
 that cp= 185 foot-pounds. j t $ 
 
 Taking now the third case, if we call c n the total 
 amount of heat per degree which must be supplied 
 from a to d, Fig. 2, then c n is the specific heat for the 
 law represented in the figure. This is used up partly 
 in changing the temperature, which will account for 
 the amount c v (T d T a ), and the balance in doing the 
 work represented by the area ayxd. We can there- 
 fore write 
 
 c n (T d -T a } = c v (T d ~T a } + ay*d. . . (3) 
 
 The two equations above can be written in the gen- 
 eral form, 
 
 Total Heat = 
 
 Heat required to raise temperature -J- work done, 
 
 or, in the differential form, 
 
 dH c v dt +pdv, .... (4) 
 
 the latter term being the calculus method of indicating 
 the elementary area ay'x'd'. 
 
 This equation is the fundamental one of the thermo- 
 dynamics of gases. 
 
 Equation (2) can be written as below from the fact 
 that p c v c -RT C , and p a v a = R T a : 
 
 / ( T T \ /- ( T T\ _L E>( T T\ 
 
 Cp\l c 1 a ) C v \ 1 c 1 a ) -{- /v^ 1 c 1 a ), 
 
 or 
 
1 NO TES ON T HER MOD YNAMICS. 
 
 This equation represents the relation between the 
 quantities which it is important to remember. 
 
 Experimentally, it has been shown that, for perfect 
 gases, 
 
 ^ =1.41, 
 
 c v 
 
 and we can write 
 
 c p \c v \ R :: 1.41 : i : .41, 
 or 
 
 Heat added at constant pressure : Heat required to 
 raise the temperature : the work done : : 1.41 : I : .41. 
 
 Prob. 14. If 5 pounds of air at 170 degrees F. has 16 heat-units 
 added to it at constant pressure, how much work 
 is done ? What is the final temperature ? 
 To find the work done we have 5 x c .(ti 170) 
 
 = i6h. u. 
 
 Work = 5 x R(t* 170) = x 16 x 778 ft.-lbs. 
 
 = 3620 ft.-lbs. 
 
 Work 3620 
 
 -The rise in temperature = - = - = 13 .6. 
 
 5^ 5 x 53-37 
 
 The final temperature is 170 + 13.6 = 183.6. 
 
 Prob. .15. A given weight of air expanding at constant pressure 
 does 1000 foot-pounds of work. What heat must 
 have been added to the air ? How much heat was 
 used to raise the temperature? 
 
 Prob. 16. 15 cubic feet of air expands to 40 cubic feet under a 
 constant pressure of 30 pounds per square inch. 
 How much heat was required ? 
 
 Now/z> RT, and if all are variables, we can write 
 pdv + vdp = Rdt, 
 
NOTES ON THERMODYNAMICS. II 
 
 and substituting this value of pdv in (4), we have 
 
 dH = c v dt + Rdt vdp, 
 or, from (5), 
 
 dH = c p dt - vdp (6) 
 
 The two equations (4) and (6) are often spoken of as 
 the two fundamental equations of the thermodynamics 
 of perfect gases. 
 
 The quantity of heat required to cause I pound of 
 air to expand doing work can then be written as 
 follows : 
 
 \ pdv, . . . 
 J *i 
 
 H=c,(T,- r,)+ I pdv, ... (7) 
 
 fc/P, 
 
 in which T 2 is the final temperature, v 2 the final vol- 
 ume, and 7\ and z/ t the corresponding initial condi- 
 tions. 
 
 Prob. 17. If the initial condition is such that 5 pounds of air 
 occupy 50 cubic feet at 30 degrees F., and the final 
 condition such that it occupies 120 cubic feet at 40 
 degrees F., and the expansion takes place along a 
 straight line, how much work is done and how 
 much heat added ? 
 It is first necessary to find the pressure. From pv=RT 
 
 we have for the initial state, p = ' - = 2630 
 
 pounds per square foot. For the final condition, 
 p _ 53-37 _ _ II2Q p 0unc i s p er S q U are foot. 
 
 ~5~ 
 
 The work done is therefore, from a diagram, 
 2630+ ii2o i2o _ Q _ jooot-ies h. u. 
 
 ' 
 
 OF 
 
 :? 
 
12 
 
 NOTES ON THERMODYNAMICS. 
 
 The heat required to raise the temperature is 
 
 and the total heat required is 
 
 1 68 4- 8 = 176. 
 
 To determine the value of area abed or the 
 
 I pdv, 
 
 we must know the law connecting 
 3pvT 2 the pressure and the volume of 
 the path ab. If we call this pv n 
 = K, we have, knowing p v v l and 
 71, and/ 2 , v 2 and 7" 2 , 
 
 d c 
 
 FIG. 3. 
 
 or 
 
 n = 
 
 log A- lQ g A 
 
 The value of K is obtained from either of the above 
 equations. 
 
 Prob. 18^ What is the value of n that the expansion curve 
 passing through the same initial and final points 
 as in problem (17) should be/z/ M = K1 
 
 Area of 
 
 T kdv K 
 
 dv K r i i*. 
 
 *j* = ^ = [- 
 
 n 
 
NOTES ON THERMODYNAMICS. 13 
 
 Putting in the value of K for the equation above, we 
 have the work 
 
 w= 
 
 I n 
 
 Prob. 19. Having given, in problem (17), that the law of the 
 expansion is /z/- 975 = K, how much work is done 
 if the final condition is / = 40 ? 
 
 Work = 5 x ^ 3 ' 37 (40 30) = 107000 ft.-lbs. 
 The total quantity of heat required is therefore 
 
 c v (T 2 -T l ) + Tlt (T 2 -T l \ . . . (10) 
 c v -nc.+R ,_ c p- nc 
 
 l _ n 
 
 when C H is the specific heat according to the law 
 pv" = K. 
 
 Equation (10) is worth committing to memory as it 
 is here given. 
 
 Prob, 20, How much heat would be required in problem (19)? 
 
 From problem (17) the heat required to change the 
 
 temperature is 8 heat-units. From (19) the work 
 
 done = I07 8 00 = 137 heat-units. The total heat 
 required is 137 4- 8 = 145 heat-units. 
 
14 NOTES ON THERMODYNAMICS. 
 
 The special cases already treated of and some others 
 may readily be derived from equations (9) and (10). 
 
 In n = o, pv n K becomes/ constant, the work 
 done, frpm (9), is, evidently, R(T 2 7^), and the heat 
 required, from (10), is (c v + R)(T 2 TJ= c P (T 2 TJ 
 as before. 
 
 If = o, we have v = constant, and the work done, 
 n 
 
 from (9), is evidently o, as 
 
 The heat required, from (10), is c v ( T 2 T^. 
 
 If the heat is constant, we have - --(T 2 7' 1 )=o, 
 
 and one solution of this is = n. This expansion, 
 
 where no heat is added nor taken away but work is 
 done, is called adiabatic expansion, and its equation is 
 
 pv c v= K y or, as for air -- = 1.41, we have 
 
 C v 
 
 pv lAl = K. (n) 
 
 The work done is 
 
 R (T T\ (T T^ 
 
 *-n ( 2 ~" 1 >-7^~1 ( l " 2) * 
 
NOTES ON THERMODYNAMICS. I 5 
 
 Evidently, as c p ~ R + c v , and c p = 1.41^, we have 
 R = -4it\,, and the work done is, for adiabatic expan- 
 sion, c v (T 2 7\), and the heat given up is c v (T 2 7^) 
 to do this work. 
 
 If the temperature is kept constant we have T l ~T 2J 
 pv = RT = K, and n = I. The amount of heat re- 
 quired is then, from equation (9), 
 
 <^ (T r)^*, 
 
 I I V 2 O' 
 
 which is indeterminate. We can, however, determine 
 the quantity of work done and of heat added by going 
 back to the original equation, 
 
 - 73.+ 
 
 Here T z = T v and pv K. Consequently 
 
 crW 
 
 e ^ (12) 
 V \ 
 
 Evidently, from the equation pv = RT, we can put 
 ^ for either/^ or/ 2 2/ 2 , and, from /z/ = 7T, we can 
 
 put for the value . In solving problems, that form 
 
 v \ A 
 
 of equation should be used which covers the greatest 
 
 amount of given data. 
 
 Prob, 21. If i pound of air has 40 heat-units added to it and 
 25 heat-units are the equivalent of the external 
 work, what is the value of n in the equation 
 
1 6 NO7^ES ON THERMODYNAMICS. 
 
 As the external work = 25 h. u. = (7^ 3 Ti), the 
 
 I ~* ft 
 
 remainder, 15 h. u., =c v (Ti 7^), or 
 
 11 = Cv ^ 1 ~ n ) l ~ H 
 25 ~ R .41 
 
 = .754. 
 
 Prob. 22. If 10 heat-units are added to i pound of air at con- 
 stant pressure, what work is done and what is the 
 rise in temperature ? We have 
 
 cp\c v \ 7v' : : Heat added : Heat to raise tempera- 
 ture : work 
 
 10 10 x.4i 
 : : 1.41 : r : .41 : : 10 : : 
 
 1.41 1.41 
 
 10 x .41 x 778 
 
 Work = - -f-L. = 2270 ft.-lbs. 
 
 1.41 
 
 As we are dealing with i pound, the rise in temper- 
 ature 
 
 Work 
 
 ., - = 42.4 degrees. 
 
 Prob. 23. If 40 heat-units are added to 5 pounds of air having 
 a pressure of 25 pounds per square inch and a 
 volume of 30 cubic feet, what is : (i) final v, /, / ; 
 (2) the work done if (A) it is added at constant 
 pressure, (B) at constant volume, (C) at constant 
 temperature, (D) according to the law/z/* = A"? 
 
 The value of ;/ when no heat is added could have 
 been determined directly from the fundamental equa- 
 tions as follows : When no heat is added we can 
 write 
 
 dH = c^t -f- pdv o, or c v dt = pdv, 
 and 
 
 dH Cpdt vdp = o, or c p dt vdp, 
 
NOTES ON THERMODYNAMICS. 
 
 and dividing one by the other we have 
 
 c v pdv c,, dp dv 
 
 CP 
 
 vdp' 
 
 or . 
 
 or integrating between limits we have 
 
 or, dropping the logarithms, 
 
 or 
 
 or 
 
 To determine whether the temperature will rise or 
 fall during expansion, whether work must be done by 
 the air or on the air, and whether heat must be added 
 or taken away, Fig. 4 will be of service. Through 
 
 
 
 n=o 
 
 FIG. 4. 
 
 the initial point A, Fig. 4, we have drawn a series 
 of curves for different values of n. n o is at 
 
 constant pressure, = o is at constant volume, n = I 
 
 is at constant temperature, and n = 1.41 is an adia- 
 batic. 
 
 Evidently all expansion curves having n positive will 
 
1 8 NOTES ON THERMODYNAMICS. 
 
 fall between a and d, all having n negative will fall 
 between a and //. All compression curves having n 
 positive will fall between e and h, and negative values 
 will fall between d and e. 
 
 Starting at A, if the path of the air is to the right, 
 work is done by the air, or is positive; if to the left, 
 work is done on the air, or is negative. The following 
 table should be mastered by the student. From A, 
 then, calling rise in temperature, heat added, or work 
 done by the air positive, we have, if curve falls between 
 the limits, 
 
 n. Temp. Heat. Work. 
 
 a to b o < n < I + + -j- 
 
 b to c I < n < 1. 4 1 -j- -|- 
 
 c to d 1.41 < n + 
 
 d\.o e n < o 
 
 e to/ o < n < I 
 
 ftog I < n < 1.41 + 
 
 gto h 1.41 < n + + 
 
 hto a n < o + + + 
 
 We are now ready to take up the question of the 
 amount of heat expended and the amount of work 
 done when the gas under consideration goes through 
 a series or cycle of changes, being 
 at the end in the same condition as 
 at the beginning. In Fig. 5, sup- 
 pose we have a pound of the gas 
 starting at the condition p\T l and 
 ~ expanding according to the law 
 
 pv m == K until it reaches a point 
 Suppose now it expands along the line 
 
NOTES ON THERMODYNAMICS. 1$ 
 
 pv" = K^ to the condition p s v s T 3 . It is then com- 
 pressed along the \me pv m = K^ to p^v^T^ , which is 
 such a point that, if the compression is continued 
 along the line/z/ 1 = K% , it will again reach its initial 
 condition. 
 
 There are certain algebraic relations between the 
 quantities in this diagram which should first be de- 
 duced. They are : 
 
 A_A. ^_^3. j^2_ ^3 
 
 A~A ' ^i~V T,~'' TV 
 From the given data we have 
 
 and multiplying tfe^ese equations together we have 
 
 V - Z> 3 / N 
 
 ?- <'3) 
 
 Again, 
 
 A 
 A = 
 
 or 
 
 rr = f^ = x > from Os); 
 
 A = A ..... (I4) 
 
 Pi A 
 
 Multiplying (13) by (14) we have 
 
 ^ = g = .^ = ^ _ _ > (I5) 
 
 These relations should be kept in mind, as they 
 often lead to an easy solution of problems. Equations 
 
2O NOTES ON THERMODYNAMICS. 
 
 (13) and (14) are true if the figure is bounded by any 
 two similar (algebraic) sets of curves, and equation (15) 
 is only true for substances having/^ =: RT for their 
 equations. 
 
 The work done in a cycle similar to that in the 
 figure is evidently the area of the diagram, or it is the 
 heat added from 4 to 2, less that taken away from 2 
 to 4. 
 
 From 4 to i, 
 
 T t ); Work=(r 1 - T t ). 
 
 From I to 2, 
 
 From 2 to 3, 
 
 Heat = (<:+ ^) ( 7-3- 7y ; Work^y-^ T s - 
 From 3 to 4, 
 
 The net work done is therefore the area of the 
 diagram, or 
 
 - T t - T t ) + -^-(- T- T s + 
 
 i w v i m 
 
 The total quantity of heat which must be added is 
 
NOTES ON THERMODYNAMICS. 21 
 
 that required to raise the body from T t to T 2 through 
 T or 
 
 = total heat added ; or, calling c m and r w the specific 
 heats according to the laws I, 2 and 4, I, we have 
 
 Total heat = c n (T, - T t ) + c m (T t - T,). 
 
 The efficiency, which is the ratio of the work done 
 to the heat expended, is then 
 
 If either set of curves is adiabatic we have, say for 
 n = 1.41, for the efficiency 
 
 As R = .41^ , we have 
 
 o 
 
 ^ -T t 
 
 T, - r; 
 
 ~ 
 
 T T 
 Putting T 3 = -^ , we have for the efficiency 
 
 06) 
 
22 NOTES ON THERMODYNAMICS. 
 
 That is, in any such cycle, the efficiency is the drop 
 in temperature along either adiabatic divided by the 
 highest temperature on that adiabatic. The ammint 
 of work done in such a cycle can be determined by mul- 
 tiplying the heat added by this efficiency. 
 
 Prob. 24. A cycle is made up of two adiabatics and two 
 curves pw* = K. If 10 heat-units are added to 
 i pound of air, /i = 3000 pounds per square 
 foot, Vi = 10 cubic feet, how much work will be 
 done, the lowest temperature in the cycle being 
 o degrees F., and what is the highest tempera- 
 ture in the cycle ? 
 In Fig. 6 we have the data given as shown. To 
 
 3000 x 10 
 determine T \ , we have 7\ = - = 561. 
 
 53-37 
 The work done is 
 
 561 461 
 10 x - x 778 = 1390 ft.-lbs. 
 
 To determine 7\, we know that 10 heat-units are 
 
 added from T\ to 7" 2 ac- 
 cording to the law pv^ 
 = K, or 
 
 IG = 
 
 7781.41 
 
 7^-7^ = 32.8, T, = 593-8. 
 
 Prob. 25. A cycle is made up of two isothermals and two con- 
 stant-volume lines. The extreme volumes are 40 
 and 10 cubic feet, and the extreme pressures are 
 15 and 100 pounds per square inch. How much 
 work is done and how much heat is required ? 
 
 Prob. 26. A cycle is made up of two constant-pressure and 
 two isothermal lines. The extreme pressures are 
 
NOTES ON THERMODYNAMICS. 2$ 
 
 15 and 10 pounds per square inch, and the ex- 
 treme volumes are 10 and 70 cubic feet. How 
 much work is done and how much heat is re- 
 quired ? 
 
 Prob, 27. Having given 2 pounds of air at /i 3000 pounds, 
 z/i= 15 cubic feet, 7^=460, 7" 3 = 420, and/z/- 7 =A", 
 how much work is done, the other curves being 
 adiabatics ? 
 
 In a cycle such as we have just been considering it 
 can be shown that the work done 
 may be expressed in a number of 
 ways. In Fig. 7 the heat added 
 from T, to T 2 = c H (T 2 - T,) = Q r 
 
 The heat taken away from T 9 to T t 
 
 FIG 7 
 
 = c.(T t - T t )=Q 2 . The first of 
 
 these divided by 7^ is equal to the second divided 
 
 by 7' 4 . For we have the relation 
 
 T T 
 
 IJ _3 
 
 T ~" T ' 
 
 f 1 ^4 
 
 and, therefore, 
 
 f .(T t - 
 
 T T 
 
 2 I 2 4 
 
 In the same way the heat along the top line di- 
 videcl by T 2 is equal to the heat along the bottom 
 line divided by T y The work in such a cycle can 
 therefore be stated as the heat added along either 
 line divided by the temperature at either end of the 
 line taken and multiplied by the range oi tempera- 
 
24 NOTES OK THERMODYNAMICS. 
 
 ture along the adiabatic passing through the point at 
 which the temperature was taken, or the work is 
 
 ||(7\ -T t ) = Q (T* - TV) = %(T, - TV) 
 
 /I A 7 4 
 
 = ? (TV -TV). (. 7 ) 
 
 3 
 
 This relationship should be entirely understood. 
 
 Having shown that the work done in any cycle 
 having adiabatic curves for two of the bounding 
 curves is equal to the heat added times the range in 
 temperature along one adiabatic divided by the max- 
 imum temperature along that adiabatic, it can be 
 shown that, if the heat is added at constant tempera- 
 ture, the maximum range in the 
 cycle being the same, the amount of 
 work done or the efficiency will be 
 the greatest. In Fig. 8 let 1,2 and 
 FlG g 3, 4 be isothermals, and 2, 3 and 
 
 I, 4 be adiabatics. Then I and 2 will 
 be at the highest temperature, and 3 and 4 at the 
 
 'p j" 
 
 lowest. The efficiency is then -^-= 4 . Now, sup- 
 
 f\ 
 
 pose the heat, instead of being added along an iso- 
 thermal, is added according to any law as 1,5. The 
 temperature of 5 is evidently below I or 2, as we have 
 assumed that 7i is the highest obtainable temperature. 
 As 7" 3 is the lowest temperature, it is evident that a 
 curve similar to I, 5 passing through 3 will cut I, 4 at 
 a higher temperature than 7" 4 . The efficiency is then 
 
NOTES ON THERMODYNAMICS. 
 
 IS 
 
 y _ y y y 
 
 ^-~, 6 , which is evidently less than -^--TR 4 , as 7" 6 i 
 1 1 *i 
 
 greater than 7^. As the efficiency is less, the work 
 done by the same quantity of heat is less. Therefore 
 the greatest efficiency is obtained when heat is added 
 at constant temperature, which also implies that heat 
 must be taken away at constant temperature. 
 
 The diagrams we have drawn heretofore have shown 
 the amount of work done, but have given us no 
 graphical idea of the quantity of heat which enters the 
 cycle. This quantity of heat, as well as the quantity 
 of work, can be shown by a definite area on this 
 diagram. By a definite area is meant one that can be 
 measured by a planimeter. 
 
 Suppose 1,2, Fig. 9, to be the path representing 
 the changes in pressure and volume. We have the 
 
 10 11 
 
 FIG. 9. 
 
 work done A -f- B -f- C, the letters referring to the 
 spaces in which they occur. 
 
 AL i the total energy in the gas can be represented 
 by drawing the adiabatic I, 12, 6 and continuing it 
 indefinitely to the right. The area under this curve, 
 or C -\- F -{- G, is the equivalent of the energy in the 
 substance at i. because it is the amount of work 
 
26 NOTES ON THERMODYNAMICS, 
 
 which would be done if it was allowed to expand at 
 the expense of its own heat until it reached the abso- 
 lute zero. At 2 the energy remaining in the gas can 
 be represented by the total area under the adiabatic 2, 3 
 drawn through 2. This is equal to D-\-E-\-F-\-H-\-G. 
 We have then that the amount of heat added is equal 
 to the energy remaining at 2 plus the work done from 
 I to 2 and minus the energy at i, or 
 
 Heat added 
 
 or the area between the path i, 2 and two adiabatics 
 drawn through the extremities of the path and indefi- 
 nitely extended. 
 
 We have already seen that the work done by a 
 pound of air expanding adiabatically can be repre- 
 sented by 
 
 where T 2 is the final and 7i is the initial temperature. 
 The energy in a pound of gas at I can be determined 
 
 TT) rri 
 
 " by making T 2 in the above equation o, and - - or 
 is the energy. Similarly at 2 the energy in a pound 
 
 of the gas is - . If 2, 4 is an isothermal through 
 .41 
 
 2, the energy in the gas at 2 is the same as at 4, or 
 
NOTES ON THERMODYNAMICS. 2f 
 
 RT 
 
 , and if I, 3, 5 is- an isothermal through I, the 
 
 .41 
 
 energy in the gas at I, 3, or 5 is 
 
 .41 .41 
 
 Evidently, if, after expansion takes place from I to 
 2, we allow it to continue adiabatically to 3, the air 
 has as much energy at 3 as it had at I, and whatever 
 heat we have added has all gone to do work. The 
 total work done is (A + B + C + D + E + F), and 
 this is equal to the heat added from I to 2. 
 
 The area D + E + F is equal to the area K -\- L, 
 
 7? T* 
 
 for at 2 the energy in the gas is -- 2 , and at 4 it is the 
 
 .41 
 
 r> y 
 
 same. At 3 the energy is -- -, and at 5 it is the same. 
 
 .41 
 
 Passing from 2 to 3 the energy converted into work is 
 
 r> 
 
 (T z 7\), and from 4 to 5 it is the same. But the 
 
 work done is in one case D -\-E-\- F, and in the other 
 K-\-L\ and as they are the equivalent of the same 
 amount of energy, they are equal to each other. 
 
 Prob. 28. How much energy is there in I pound of air after it 
 has expanded adiabatically to 20 cubic feet, if its 
 initial conditions were/ = 2000 pounds, v = 16 
 cubic feet ? 
 
 Prob. 29. What is the energy in 10 cubic feet of oxygen at 
 IGU pounds pressure per square inch and 100 
 degrees F. ? 
 
28 
 
 NOTES ON THERMODYNAMICS. 
 
 There is another method of illustrating graphically 
 the heat added under any conditions. If we attempt 
 to draw a diagram having absolute temperature T for 
 ordinatesand Q, the heat, for the area under any curve 
 
 - 
 
 to the other axis of co-ordinates, the abscissa is 
 
 because Q = 
 
 The quantity / - 
 
 called _enr< 
 
 Evidently on such a diagram an adiabatic is repre- 
 sented by a line parallel to the T axis, because no 
 heat is .added along an adiabatic. The diagrams 
 shown in Figs. 10 and n represent a-/, v diagram 
 
 and a 
 
 3000- 
 
 10 go 
 
 VOLUME IN CUBIC FEET. 
 
 FIG. 10. 
 
 The data assumed in drawing these diagrams are 
 PA 3000> T A = 561, V A = 10, V B = 20; for AB, 
 n = o; AC, n i ; AD, n = 1.41 ; CE y n 1.41; 
 and for DE, n = i. 
 
NOTES ON THERMODYNAMICS. 2$ 
 
 In locating points in Fig. n, the point A is taken 
 at any point on the T = 561 line. To determine the 
 distance to C, we have, as this is a constant-tempera- 
 ture line, dQ = pdv, and 
 
 f -/?-*/* =5=3;., 
 
 To locate the point B, we have dQ c p dt and 
 
 These diagrams are drawn to such a scale that the 
 area represents foot-pounds in either diagram. In the 
 first diagram, Fig. 10, the area under AB is the work 
 done at constant pressure, and in the second diagram, 
 
 1.41 
 Fig. u, it is the heat added and is - as great. In 
 
 _ , - -- _ i ._ _ , - - __ '~ ..... -r4> t - . |T< 1M . rr| . mfm _ ___^p 
 
 the first the area under AC is the work done at con- 
 stant temperature, and in the second it is the heat 
 added and is exactly equal to it. In the first the 
 area under AD is the work done adiabatically, and in 
 the second it is zero, as it should be. 
 
 If we draw through D an isothermal as shown by 
 the line DE, the point E completes a cycle, and for 
 
 the second figure evidently ^ ^r^> as proved 
 
 1 AC ^ DE 
 
 above, and the areas AC ED in the two figures are 
 equal. 
 
 Prob. 30. Draw diagrams, similar to Figs. 10 and n, to 
 scale representing the expansion of i pound of air 
 
3O NOTES ON THERMODYNAMICS. 
 
 at 60 pounds pressure and 100 degrees F. (A) 
 adiabatically, (B) along the isothermal, (C) at con- 
 stant pressure, until the volume is doubled, and in 
 each case, if possible, represent by a definite area 
 the amount of work done and energy expended. 
 
 GENERAL EQUATIONS. 
 
 In taking up the portions of thermodynamics treat- 
 ing of substances generally, certain matters which we 
 have already deduced apply, while certain others do 
 not. Thus, Fig. 12, if AB\s the path of the substance 
 under discussion (any substance), the 
 external work done is here, as before, 
 *. the area ABDC. The total amount 
 
 B- 
 
 of head added to cause the substance 
 to pass from A to B is again repre- 
 sented by the area between AB and 
 PT^ D TO F two adiabatics at the extremities A 
 
 I 1G. 12. 
 
 and B indefinitely extended to the 
 right. Here, however, the adiabatics are not neces- 
 sarily curves whose equation is pv lAl = K, as this rela- 
 tion only applies to perfect gases. They are curves, 
 however, so drawn that from B to E, for instance, the 
 area BEFD, which is the external work done, is the 
 exact equivalent of the heat-energy which has disap- 
 peared as such between B and E. 
 
 We have called certain lines isothermals, and made 
 certain statements about these lines. That is, in Fig. 
 13, if AB is an isothermal fora perfect gas, it is a rect- 
 angular hyperbola, the heat added from A to B is 
 the area L'BAL and is exactly equal to the area 
 
NOTES ON THERMODYNAMICS. 
 
 c 
 FIG. 13. 
 
 A BCD representing the external work. Hereafter 
 AB, if it is an isothermal, is 
 only a line of constant tem- 
 perature; it need not be and 
 often is not a rectangular hy- 
 perbola. The heat added is 
 equal to L'BAL but is not 
 necessarily equal to ABCD. 
 The work done is equal to ABCD and may or may 
 not be equal to L'BAL. 
 
 The attempt will be made hereafter to use the terms 
 adiabatic and isothermal in the general sense spoken 
 of above. 
 
 Fig. 14 shows the work done, and Fig. 15 the heat 
 added isothermally to any substance. In Fig. 14 the 
 
 VOLUME 
 
 FIG. 14. 
 
 ENTROPY 
 
 FIG. 15. 
 
 isothermal may be a rectangular hyperbola if we are 
 dealing with air, a constant-pressure line if we are deal- 
 ing with a mixture of liquid and vapor, or it is the line 
 which represents the relation between / and v at con- 
 
32 NOTES ON THERMODYNAMICS. 
 
 stant temperature. In Fig. 15 it must be a line perpen- 
 dicular to the T axis. ALand B' areadiabatics; in Fig. 
 14 they are curves, and in Fig. 1 5 they must be straight 
 lines parallel to the Taxis. The heat //added from A to 
 B in both diagrams is the area ABL'L. Draw any other 
 isothermal A' B' in both diagrams so that its tempera- 
 ture is dt degrees below AB. Evidently, from Fig. 1 5, 
 
 JT 
 
 the area ABB' A' is equal to -^dt. From Fig. 14, the 
 
 equal area. ABB' A' is / dp dv, and these two quantities 
 are equal to each other, or 
 
 H i 
 
 dt I dpdv, . . . . (18) 
 
 where dp is the vertical distance between AB and 
 A'B 1 ' , or dv is the horizontal distance between these 
 lines, but not. both at the same time. We can write 
 the equation in either of the following forms: 
 
 H C V B C P B 
 
 Y dt = / (dp) dv \ (dv)dp, 
 
 the quantity in the parenthesis meaning that the value 
 of (dp] is fixed by the isothermals and that dv is the 
 other independent variable, or in the last member the 
 reverse is the case. 
 
 As it is the quantity ab in Fig. 14 that we must 
 
NOTES ON THERMODYNAMICS. 33 
 
 insert in the equation for (dp), we can determine its 
 value from the equation of the substance by deter- 
 
 lAp \* 
 mining ljj~J* which gives us the rate of charge of / 
 
 with T, and multiplying this by dt, or ab = f \ dt. 
 
 Similarly (dv) cd = \-^\ dt, or writing these values 
 
 \ zit ]p 
 
 in the original equations, we have 
 
 // r 
 T dt : " 'J 9 
 
 or differentiating, 
 
 We see, then, that, if heat is added to any substance 
 along an isothermal, the quantity of this heat can be 
 represented by either of the two quantities in equa- 
 tion (19). 
 
 *This form is chosen to clearly indicate that we wish to obtain 
 a number (or an expression) giving the ratio of the simultaneous 
 changes of/ and 7' at constant volume, and this in no way de- 
 pends on the value of dt. 
 
34 NOTES ON THERMODYNAMICS. 
 
 Prob. 31. Prove from equation (19) that if heat is added to air 
 at constant temperature, the heat required is 
 
 For air PV = AT and [4f\ = (*} .from this equa- 
 
 \4t/ v \dt) v 
 
 tion, gives 
 
 vdp = Rdt, 
 
 dp\ R 
 
 From equation (19), 
 
 As the temperature is to be constant, we have 
 
 Prob, 32. How much heat must be added at constant tem 
 perature to a substance whose equation is 
 
 6 i - 273 
 = I0 6 ' 1 ~T 
 
 to change its volume at constant temperature 
 from z/i to Vi ? 
 Prob. 33. Having given 
 
 , = __ _ 
 P ~~ v ' TV* 
 
 as the relation between the pressure, volume, and 
 temperature of a substance, how much heat must 
 be added at constant temperature to change its 
 volume from Vi to 7/9, having given the values of 
 A, B,pi, and T? . 
 
NOTES OAT THERMODYNAMICS. 35 
 
 If, however, the heat, instead of being added along 
 an isothermal, is added along any other line, the follow- 
 ing method will determine the 
 quantity of heat. Let AB (Fig. 
 
 1 6) be the line of the expansion, 
 the co-ordinates being p and v. 
 Let A and C be points dt degrees 
 apart. The heat added between 
 the points A and C is represented FIG. 16. N 
 
 by the area A^C, and this area = dH. Through C 
 draw the isothermal CD until it cuts the line of con- 
 stant pressure through A. The heat added from A to 
 C is equal to that added from A to D, minus that 
 from D to C. Or, it is more nearly true to say that 
 the latter quantity becomes more and more nearly 
 equal to the heat added from A to C t as the tempera- 
 ture difference between A and C becomes smaller. 
 Calling the difference in temperature dt, then AD is 
 
 +7-1 dt, and CE is dp, as the point C is fixed by the 
 
 atjt 
 
 intersection of the isothermal dt degrees above A and 
 the given curve of expansion AB. The area $ADi is 
 the heat added from A to D, or is by definition c p dt. 
 The heat from D to C is the area 21 DC, or, from 
 
 (/Ji'\ 
 -r- \ dp, and we have taken this form 
 
 because AD or dv is the quantity fixed by the two 
 isothermals dt degrees apart. The heat from A to C is 
 
 =c p dt-T(-~\ dp = dH, (20) 
 
 ' 
 
3<> NOTES OAT THERMODYNAMICS. 
 
 which is one form of the general thermodynamic 
 equation. 
 
 Another form of this equation 
 is obtained as follows: Draw AF 
 (Fig. 17) at constant volume until 
 it cuts the isothermal through C. 
 2 Then the area iAC2 differs from 
 iAF 4 + 4FC2 by the area AFC, 
 which disappears as dt is made 
 
 smaller. Then 
 
 Ag=dv, AF=$\dt, 
 
 \AIJ V 
 
 and we have the areas 
 
 AF 4 i = cjt, 4 FC2 = T (^]dv, 
 and 
 
 )dv,. . . . (21) 
 
 which is a second form of the fundamental equation. 
 In these two equations the terms r- and j depend 
 only on the equation of the substance, and could 
 
 have been written -3 and 7-, while the other terms 
 at at 
 
 depend on the law of the expansion. That is, in the first 
 one we have made dt and dp depend on the law which 
 we have chosen to assume for the expansion, but the 
 
 Av 
 value - depends only on the substance which is to 
 
NOTES ON THERMODYNAMICS. 37 
 
 expand. In using these formulae it must be remem- 
 bered that the units must be the same for all the 
 terms. That is, if the area dpdv is in foot-pounds, it 
 represents a certain part of the diagram, and the c p or 
 c v must be in foot-pounds also ; or if c p and c v are in 
 heat-units, the value of dpdv must be in heat-units 
 also. 
 
 Prob. 34. Suppose that there is a substance which, in the 
 state we propose using it, is a gas, and that the 
 relation between its pressure, temperature, and 
 volume, as determined by experiment, can be 
 
 T) 
 
 expressed by the equation pv = A T -- =. What 
 
 will it do under various methods of expanding it ? 
 First calling p constant, we have 
 
 = Adt 
 or 
 
 tdv\ _A B 
 
 \dt) p -p + rp> 
 
 and calling v constant, 
 
 -4_J B 
 
 r v + rv 
 
 and the two forms of the fundamental equation 
 are therefore 
 
 (A) 
 (B) 
 
 If now the substance is to expand at constant vol- 
 ume, we have, from (B), H = c v dt. If at constant 
 
33 NOTES ON THERMODYNAMICS. 
 
 pressure, from (A), // = c p dt. If at constant tem 
 perature, from (A), 
 
 AT+ ,or f from (B), 
 
 If it is to expand adiabatically, we have dH ' o for 
 
 __dp 
 
 both equations and = - V which is in the 
 c v dv 
 
 / ^ 
 
 same form as the equation for the adiabatic ex- 
 
 pansion of air. 
 
 In using the fundamental formula we must remem- 
 ber that the formula gives us the heat added from A 
 to B in the figures, and that when we speak of the 
 heat in a substance we are measuring for some datum. 
 Ordinarily this is taken at 32 degrees F., and, as this 
 is the temperature at which a change of state in water 
 takes place, we must define more particularly, so that 
 if we are dealing with water or its vapor it is cus- 
 tomary to measure the heat from that in water at 32 
 degrees. 
 
 Heat in Water and Steam. The application of the 
 general formula to the heat in a liquid and its vapor is 
 as follows : When heat is added to a liquid (water, for 
 instance) at 32 degrees, its temperature rises and its 
 volume changes slightly. This continues until the 
 temperature reaches such a point that vapor begins to 
 
NOTES ON THERMODYNAMICS. 39 
 
 form. This is always a definite point for a given 
 pressure. For water, 15 pounds pressure and 213 de- 
 grees correspond, 100 pounds pressure and 327 degrees ; 
 for ammonia, 37.8 pounds pressure and 10 degrees, 
 1 80 pounds pressure and 90 degrees, etc. The addi- 
 tion of any further quantity of heat to the liquid which 
 is ready to boil does not increase the temperature, but 
 vapor begins to form, part of the heat being used up 
 in increasing the volume, and part in some sort of 
 internal work required to change the liquid water into 
 vapor. This condition of affairs continues until suffi- 
 cient heat has been added to convert all the liquid 
 into vapor. Any further addition of heat again raises 
 its temperature and continues to increase the volume. 
 The addition of heat, therefore, at constant pres- 
 sure takes place in three successive stages : first, 
 while it is entirely a liquid ; second, while part is 
 liquid and part vapor; and third, after it is entirely a 
 vapor. We have generally 
 
 While it is a liquid v is practically constant and 
 dH = c v dt, H c v (T l T 32 ) and is called q, or the 
 heat of the liquid. 
 
 In reality there is a certain amount of work done 
 and dv is not strictly zero, but the ordinary value of 
 the specific heat of liquids includes the very small 
 amount of heat necessary to do the external work. 
 
 c v is not necessarily constant and / c v dt is not neces- 
 sarily equal to c r (Tj T^). If we know the relation 
 
40 NOTES ON THERMODYNAMICS. 
 
 between c v and T, it should be inserted before inte- 
 grating and the exact value found. It is customary 
 to say that for water c v = I, while in reality c 
 I -j- .00004^ -j- .0000009/ 2 , t being in the centigrade 
 scale, and we have 
 
 H ' q i cdt / (i + .00004/ -f- .oooooo9/ 2 X/ 
 
 t.y o / o 
 
 = / + 00002 / 2 -f- . 0000003 / 3 , 
 
 which is the true value of the heat of the liquid in 
 French units. To get the corresponding quantity in 
 English units, enter this equation with the centigrade 
 temperature, and $ the value of the quantity obtained 
 is the value in B.T.U. for the corresponding Fahren- 
 heit temperature. 
 
 Prob. 35. The specific heat of liquid anhydrous ammonia is 
 given by the equation (French units) 
 
 c = 
 
 How much heat must be added to i kilogram to 
 
 raise its temperature from 20 to 40 degrees C. ? 
 Prob. 36, What is the specific heat of liquid ether at 30 de- 
 grees C. if the equation for q (French units) is 
 
 q = 
 
 Prob. 37. How much heat is required to raise i pound of 
 water from 60 to 160 degrees F., using the specific 
 heat of water ? 
 
 Prob. 38. What will be the temperature of i pound of water 
 at 60 degrees if 10 heat-units are added to it ? 
 
 Prob. 39. Using the data of problem (34), how much heat must 
 be added to i pound of liquid ether to raise its 
 temperature from 40 to 50 degrees F.? 
 
NOTES ON THERMODYNAMICS. 41 
 
 It is interesting to note just what proportion of 
 this value of q is actually used for heating and what 
 proportion goes to do outside work, because the part 
 that does work may or may not be available if, for 
 any reason, we have to make use of the heat in the 
 water. 
 
 One pound of water at 50 degrees occupies .016 
 cubic foot. 
 
 One po.und of water at 140 degrees occupies .01627 
 cubic foot. 
 
 The amount of work done if the water is under, 
 say, 100 pounds pressure per square inch is .00027 
 X 100 X 144 3.89 foot-pounds, or .005 heat-units. 
 The total heat required to raise I pound of water 
 from 50 degrees F. to 140 degrees F. is 90.1 heat- 
 units, or a practically negligible amount is used for 
 doing work and we can say that all the heat added 
 while it is still a liquid remains in it. 
 
 When the water reaches the boiling-point the tem- 
 perature no longer rises, and we must again apply our 
 general formula, as the conditions under which it was 
 originally applied no longer hold. We have 
 
 Now 
 
 dt-o and H= 
 
 the total latent heat, as it is called. As T is con- 
 stant, we could have written 
 
4 2 NOTES ON THERMODYNAMICS. 
 
 To apply this formula it is necessary to know the 
 relation between p and / for the vapor to determine 
 
 the value of -jr) , and it is also necessary to know 
 
 the limiting values of v. Experimentally, the rela- 
 tion between/ and t can be easily obtained. The 
 value of v when the liquid is all vapor is difficult to 
 determine experimentally, and as rcan be determined 
 readily by experiment, this formula is of more value 
 in determining the limiting value of v than in deter- 
 mining the value of r. In applying the formula 
 
 dp 
 
 either way, we know that -y- does not depend on dv, 
 
 as for each pressure there is a definite temperature 
 and the equation might have been written 
 
 - T (%f*- T (I) <.-.' 
 
 where v z is the volume of I pound of vapor, and v l the 
 volume of I pound of liquid. 
 
 Prob. 40. What is the volume of I pound of saturated steam 
 at 100 pounds pressure per square inch if 
 r = 1113.9 .695/, and 
 
 p 99 = 99X144, T 99 =326.86+460.7, 
 
 /ioo = 100X144, Tioo = 327.58 + 460.7, 
 
 p lol = ioi x 144, Tin 328.30 + 460.7, 
 
 Ap = /101 p 99 = 2X144, 
 
 AT = 7\ui T a = 1.44, 
 
 rioa _= 1113.9 .695 x 327. 58 = 884. 
 
NOTES ON THERMODYNAMICS. 43 
 
 Aft 
 From the formula ?= T~^(vi r u^) we have 
 
 884 x 778=788.28 x 2 X I44 (z/ 8 vi), 
 1.44 
 
 884 x 778 x 1.44 
 or *'-*"= 2 x. Jx 788.38 = 4 '3 6 
 
 and z/ a = 4. 36 + .016 = 4.38. 
 
 Prob. 41. What is the volume of i kilogram of saturated vapor 
 
 of ether at 50 C., using the first five columns of 
 
 Table IV, Peabody?* 
 Prob. 42. What is the value of r in English units for carbon 
 
 bisulphide at 50 F., using only columns i, 2, 3, 9, 
 
 10, ii of Table VII, Peabody? 
 
 B C 
 
 Rankine gives log p = A -= for the rela- 
 tion between the pressure and the temperature, and 
 the above equation can be written 
 
 r = p(v<> 
 
 Regnault's experiments give the following for the 
 relation between the latent heat and the temperature: 
 
 r = 1113-9 ~ .695/1 
 
 and Peabody has deduced constants for Regnault's 
 formula in the form of log/ = a ba n -4- cfi n for the 
 relation between pressure and temperature which can 
 be used for determining the value of z/ 2 v^ 
 
 The value of r above given consists of two parts, 
 one of which does external work and the other internal 
 
 * Peabody's Tables of the Properties of Saturated Steam and 
 other'* Vapors, 
 
44 NOTES ON THERMODYNAMICS. 
 
 work. Calling u the difference in volume 1*% v i , p the 
 pressure, and A the heat equivalent of work, the ex- 
 ternal work is Apu, and the internal is r Apu = p. 
 The relations between p and Apu are very different 
 from the corresponding quantities while in a liquid 
 state, as the Apu is about -^p. 
 
 It is to be remembered that the external work has 
 been done, and while the heat to~do it has been ex- 
 pended, this heat no longer exists in the steam formed. 
 It may have been expended in pumping water, and 
 may exist as potential energy stored in water in some 
 distant reservoir. That is, r has been expended and p 
 remains in the vapor, and the Apu is not in the steam 
 and is not available for any future work. When the 
 pound of water at 32 F. is heated and entirely evapo- 
 rated under constant pressure, we have added to it 
 q -{- r = A. heat-units, and this is called the total heat. 
 It is often written as total heat " in the steam." This 
 expression is incorrect, as it is the total heat required 
 to form the steam. The amount of heat "in the 
 steam" is only q -f- p. 
 
 The steam being entirely formed, the addition of 
 more heat at constant pressure superheats it, and it 
 has been found that the specific heat of superheated 
 steam at constant pressure is .48. That is, if the 
 steam is raised / degrees above its point of saturation 
 the heat added is .48* ^(^ S up. ^sat.)- 
 
 Of this heat added, only a portion remains in the 
 steam. A certain amount of external work must be 
 done, and while we have expended .48(7" sup ^ sa t.) 
 heat-units, a quantity of work has been done equal to 
 
NOTES ON THERMODYNAMICS. 45 
 
 /(^sup. z'sat.)- The heat remaining in the steam is 
 therefore 
 
 To determine the value of this quantity we must 
 have the relation between the pressure, volume, and 
 temperature of superheated steam. 
 
 This relation determined experimentally can be ex- 
 pressed by the following equation (Peabody) : 
 
 /^ = 93 . 5 r- 97 1/*, 
 
 from which either T or v can be readily found if the 
 remaining two quantities are given. In tabular form 
 we then have, starting with water at 32 degrees and 
 ending at the state given below : 
 
 ALL LIQUID. 
 
 Heat added q\ 
 Heat remaining = q. 
 
 MIXTURE OF LIQUID AND VAPOR. 
 (x = parts vapor.) 
 
 Heat added = q + xr\ 
 Heat remaining = q + xp ; 
 Work done xApu. 
 
 ALL VAPOR. 
 
 Heat added = q + r + .4*(T mp . - T sat ) ; 
 Heat remainin 
 
 Work done Apu-\-p(v^ ~ ^at. 
 
46 NOTES ON THERMODYNAMICS. 
 
 Prob. 43. How much external work is done in converting I 
 pound of water at 60 degrees into a mixture hav- 
 ing x = .6 at 150 pounds pressure? 
 How much heat is expended ? 
 
 Prob. 44. How much heat is in I pound of superheated steam 
 at 150 pounds pressure and 400 degrees F., count- 
 ing from 32 degrees, and how much work has been 
 done ? 
 
 Prob. 45. If 80,000 foot-pouncls of external work is done in 
 converting i pound of water into steam at 150 
 pounds pressure, what must be the condition of 
 the steam ? 
 
 The distinction between the heat added and the 
 heat remaining in a substance can be perhaps better 
 understood by the following example : Suppose B 
 (Fig. 1 8) is the initial state of I 
 pound of water at 15 pounds pres- 
 A sure and 213 degrees F., and A is 
 
 Iits final condition at 100 pounds 
 pressure and 327.58 degrees F. 
 c At B the water has in it q 1 8 1. 8 
 
 ; heat-units. At A it will have in 
 
 FIG. 18. 
 
 it as steam 
 
 qJfp=i 297.9 + 802.8 = 1 100.7. 
 
 To pass from B to A we must do a certain amount of 
 work. The difference in volume between B and A \ 
 4.387 cubic feet. 
 
 Suppose that the volume A is first filled at 1 5 pounds 
 pressure, and that afterwards heat is added and the 
 
NOTES ON THERMODYNAMICS. tf 
 
 pressure is raised to 100 pounds from C to A : the 
 amount of work done is equal to 
 
 15 x 144 x 4-387 
 
 - = 12.2 heat-units. 
 
 The total heat that must be expended is therefore 
 1 100.7 + I2 - 2 1 8 1. 8 = 931.1 heat-units. 
 
 Suppose again that the pressure is first raised to D 
 and the volume is then increased to A. The work 
 done in this case is equal to 
 
 IPO X 144 X 4-387 , 
 
 3 = 81.2 heat-units, 
 
 775 
 
 and the heat required is 
 
 1 100.7 + 81.2 1 8 1. 8 1000.1. 
 
 It is therefore to be noted that the amount of heat 
 which must be expended depends upon the way in 
 which it is expended, but that the portion of the heat 
 added which remains in the substance is, in the exam- 
 ple above given, always 
 
 
 1100.7 181.8 = 918.9 heat-units, 
 or 
 
 0100 + ^100 - ?15' 
 
48 NOTES ON THERMODYNAMICS. 
 
 Prob, 46. Four pounds of a mixture of steam and water at 60 
 pounds pressure per square inch fill a vessel A of 
 10 cubic feet capacity, and 6 pounds of mixture fill 
 another vessel, B, of 10 cubic feet at 100 pounds 
 pressure. If the contents of the two vessels are 
 intimately mixed, the volume not changing, wrnt 
 will be the final pressure, assuming no radiation ? 
 First determine the heat in vessel A. We have 
 
 4_*r x 7.096 + 4(1 x) .016 = 10, x .35. 
 
 - Heat = 4(261.9 + .35 x 830.7) =2212. 
 To determine the heat in vessel B : 
 
 6x x 4.403 + 6(1 x) .016 =10, x = .376. 
 Heat = 6(297.9 + .376 x 802.8) = 3600, 
 The heat per pound of the mixture is then 
 
 22,2 + 3 600 = 
 10 
 
 and the volume occupied per pound is f & = 2 cubic 
 feet. We have then two equations to satisfy : 
 
 x x s + (i ;r).oi6 = 2, 
 q + *P = 5 8l - 2 > 
 
 and these can best be solved by trial. 
 
 Prob, 47. What heat must be added at constant volume to 
 raise the pressure of one pound of a mixture of 
 steam and water occupying 3.8 cubic feet from 
 100 to 150 pounds pressure per square inch ? 
 
 Prob. 48. A vessel of 10 cubic feet capacity has in it 4 pounds 
 of a mixture of steam and water at 100 pounds 
 pressure ; 25 pounds of water at 60 degrees F. are 
 pumped into the vessel. What is the resulting 
 temperature, assuming no radiation? 
 
 Prob. 49. If 10 cubic feet of dry saturated steam at 100 pounds 
 pressure per square inch is allowed to pass from a 
 
NOTES ON THEKMODYNAMrCS. 
 
 49 
 
 boiler into an open vessel having in it 25 pounds 
 of water at 60 degrees F., what is the resulting 
 temperature ? 
 
 Adiabatics. We have already proved that if a sub- 
 stance expands at constant tenv" 
 perature between two adiabatics, 
 the heat added divided by the tern,- jjj 
 perature is constant. To repeat in 2 
 as lightly different form, let the dia- % 
 gram, Fig. 19, be a heat diagram, 
 in which AB and EF are constant- 
 temperature lines, and A C and BD 
 are two adiabatics. Then the area ABDC divided by 
 
 TT TT 
 
 rr* /-> 7-k /~* i i t i T> J ^ AB " EF J 
 A = area EFDC divided by T E , or -=- ^r- dl ~ 
 
 ENTROPY 
 
 FlG * I9 ' 
 
 rectly from the figure. We can also write 
 
 /A f*A f*E 
 
 dff_ I djf_ I dH_ 
 ~T~J F ~T~J B ~r" 
 
 as each of these quantities is the horizontal distance 
 between the lines AC and BD. That is, it makes no 
 difference how much heat is added between E and B, 
 
 for instance, nor how it is added, the quantity / =- 
 
 H PT 
 
 is constant and, if we please, is equal to 
 
 or is equal to 
 
 f B j(ff EGB ) 
 
 Jz ' T 
 
 or 
 
 the T in the latter case 
 
 being a variable, and is the temperature at which 
 i s added. 
 
5O NOTES ON THERMODYNAMICS. 
 
 Along theadiabatic, as dH = o, we have / .=0. 
 
 /dH 
 -^- is constant between two 
 
 adiabatics for any substance gives us 
 another method of obtaining the equa- 
 
 <F~ c tion to the adiabatic for air. In Fig. 
 
 20 suppose a to be a point on one 
 
 FIG 20. adiabatic, and b and ^points on another. 
 
 CdH . 
 As / is constant from a to b, or to <:, suppose ab 
 
 to be a constant-volume line and ac a constant-pres- 
 sure line. 
 
 We have for ab 
 
 CdH 
 J ~Y"' 
 
 for ac 
 
 rdH = f r 'c j # = lop T, 
 
 and 
 
 T t T c 
 
NOTES ON THERMODYNAMICS. 
 
 or, as/z' = RT, 
 
 \v> A 1 - 41 ^ 1 - 41 ' 
 
 or 
 
 which we have before deduced in an entirely different 
 way. 
 
 When we come to apply this method to liquids and 
 vapors the problem is rather more complicated. In 
 Fig. 21 suppose a to represent the 
 pressure and volume of I pound of 
 water, and suppose the temperature 
 to be 7\. Let be be an adiabatic 
 curve such that at b we have x b 
 pounds of steam and I ^pounds 
 of water, and suppose that at c we 
 have x e pounds of steam and I x c pounds of water. 
 We know that 
 
 FIG. 21. 
 
 r*H r 
 
 / ~r" '' / 
 
 tS a * t/ a 
 
 dH 
 
 from what has just been proved. 
 
 On the path from a to b suppose first the tempera- 
 ture is raised to d, and then that x b pounds of steam 
 are made. From a to d, dH = cdt, because in the 
 general formula, 
 
 dH = c.4t 4- 
 
 we have dv = o, and hence / 
 
 H 
 T 
 
52 NOTES ON THERMODYNAMICS. 
 
 From d to b the heat dH added is rdx, and we can 
 write 
 
 d 
 and 
 
 /r r b 
 off _ I cat 
 T " ' IT T 
 */ a * 
 
 From a to r we can write 
 
 ^- 
 
 T 'Jr. 
 
 4_ 
 
 ' ' 
 
 T. T T f ' 
 and as these are equal, we can write 
 
 / V /+/J / ->* >* f . ' * /*/T / -V <>* 
 
 / ff , -3y*_ / fff /,,-N 
 
 l'r~T~T~~ 7 F''^~T~l~~T r ' ' ( 2 3) 
 
 *y y 'b y y a-* * 
 
 In this equation ^r is the specific heat of water, or is 
 
 ^ 
 
 -77, and if we know one value of x, we can deter- 
 
 at 
 
 mine any other. Ordinarily the value of / can 
 
 be calculated with sufficient accuracy by calling <: = I, 
 
 //// T 
 
 is then c log e ?; but Peabody's tables give 
 
 the value of this quantity using the exact value of c, 
 so that it need not be calculated. 
 
NOTES ON THERMODYNAMICS. 53 
 
 Prob, 50, If i pound of a mixture of steam and water occupying 
 3.8 cubic feet at a pressure of 100 pounds absolute 
 expands adiabatically to 15 pounds pressure, what 
 is its volume ? 
 
 We have from the steam-table 
 7\ o = 327.58 + 460.7; Vol. i Ib. steamioo = 4403 cu. ft.; 
 
 Q Q A 
 
 Calling 7 = 32 + 460.7, we have 
 -y- = 2.3026 (loj 788.28 log 49 2 .7) = 470 
 
 approximately, or .4733 from the tables. 
 r 1B = 213.03 4- 460.7; Vol. i Ib. steam is = 26.15 cu - ft-J 
 fcdl 
 
 K lf = 965.^ J =.3143. 
 
 To determine x\> we have, as .016 is the volume of 
 a pound of water, 
 
 (i x b } .016 +x b 4403 = 3' 8 ; 
 
 We can then write 
 
 ' T cdt_ .861 x 884 _ P T "cdt ^965.1 
 
 r~" 788.28 ~~J Tn '^r + 673^3 ; 
 
 4733 + .964= .3143 + 1431^; 
 
 x c = .782 ; 
 
 Vol. = .782 x 26.15 4- (i .782) .016 =20.5 cu. ft. 
 Prob, 51 1 If i pound of a mixture containing 40 per cent of 
 water is compressed adiabatically from 20 to 60 
 pounds pressure, what is the percentage of mois- 
 ture at the higher pressure? 
 
 Prob. 52. A pound of a mixture is expanded adiabatically, so 
 that it has the same percentage of water at 60 and 
 15 pounds. What must have been the percentage 
 at 60 pounds pressure? 
 
54 NOTES ON THERMODYNAMICS. 
 
 Whenever a body expands adiabatically, or at the 
 expense of its own heat, the amount of external work 
 done must be the difference in the quantity of heat in 
 it at the beginning and at the end of the expansion. 
 
 If we have a mixture of steam and water at the 
 beginning of the expansion so that the portion of 
 steam is x lt the heat present is q l -\- x^p l . At the end 
 of the expansion the heat is q^ + ;r 2 p 2 , and the amount 
 of work done is therefore 
 
 Prob. 53. In problem (50) how much work is done in the 
 
 expansion ? 
 From the tables 
 
 ?i = 297.9, pi =802.8, 
 ^2 = 181.8, pi = 892. 6, and 
 
 the work = 297.9 +.861 x 802.8 181.8 .782 x 892.6 = 107 h. u., 
 or 107 x 778 = 83200 ft.-lbs. 
 
 Prob. 54. What work is done if 20 cubic feet of a water mix- 
 ture weighing 6 pounds expands adiabatically from 
 80 pounds to 20 pounds pressure ? 
 
 Prob. 55. i pound of steam at 100 pounds pressure expands 
 adiabatically to 15 pounds. How much work is 
 done? 
 
 Prob. 56. i pound of water at 327 degrees F. expands adia- 
 batically to 15 pounds pressure. How much work 
 is done ? 
 
 If we are dealing with superheated steam instead of 
 3. mixture, we have for the value of / =- three parts: 
 
NOTES ON THERMODYNAMICS. 55 
 
 cdt 
 
 one 
 
 ct 
 while it is still a liquid or / ^-, one while it is 
 
 becoming steam at constant temperature, or (as it 
 is all converted into steam), and a third portion, 
 
 f, 
 
 7sat. 2 
 
 and we can write 
 
 CdH Ccdt 
 
 When superheated steam expands adiabatically, we 
 have, for the amount of work done, the difference in 
 the quantity of heat at the beginning and end of 
 expansion. 
 
 The heat at the beginning is 
 
 , 
 
 4l +P 1 
 
 The heat at the end of expansion is 
 
 ,r cr T \ A(^.UP. -- F-QI 
 
 P 2 + ^(^sup. -- 7*t.) ~ , 
 
 on the assumption that it remains superheated until 
 the end of the expansion. 
 
 Prob. 57, i pound of steam at 150 pounds pressure occupies a 
 volume of 3.3 cubic feet. What is its condition 
 after it expands adiabatically to 15 pounds pres- 
 sure, and what work is done'? ^^^R A R>^ 
 
 UN1VEK rx/ 
 
56 NOTES ON THERMODYNAMICS. 
 
 As i pound of saturated steam at 150 pounds pressure 
 occupies 3.011 cubic feet, the steam in the problem 
 must be superheated, and from the equation of 
 superheated steam we have 
 
 93-5 ' 
 
 or 
 
 = I5ox 144x3.3 + 97i xdsox 144)* go 
 93-5 
 
 Saturated steam at 150 pounds pressure has a tempera- 
 ture of 358.26 degrees F. = 818.96, or the steam is 
 superheated 71 degrees. We have then 
 
 dH cdt r 80 
 
 861.2 890 
 
 - + -48 x 2.3026 log 
 
 = 1.6055. 
 At 15 pounds pressure 
 
 >cdt r 
 
 As the sum of these two = 1.7473 is greater than 
 1.6055, tne steam is evidently not superheated at 
 the lower pressure and we have 
 
 1.6055 = -S^ + ^XMSS. 
 x = .901. 
 
 To determine the amount of work done during this 
 expansion, we have the heat at the initial condition 
 
 = ?I60+/,50 + I ^(8 9 0-8l9)- ^g 44 (3.3-3.0II) I 
 
 [Heat added less work done] 
 = 330 + 778.1+26.07 = 1134.17 
 
NOTES ON THERMODYNAMICS. S7 
 
 At the final condition the heat in the steam is 
 ? 18 + .90i/o IB = i8i.8 + .90i x 892.6 = 985.03. 
 The work done is 
 
 1134.17 985.03 = 149.14 h. u. = 116000 ft.-lbs. 
 
 Prob. 58, If in the above problem the volume had been 4 cubic 
 feet at 150 pounds pressure, what would have been 
 the condition and how much work would have 
 been done if it had expanded to 15 pounds pres- 
 sure ? 
 
 Prob. 59. If i pound of steam at 15 pounds pressure super- 
 heated 60 degrees is adiabatically compressed to 
 100 pounds pressure, what is its temperature and 
 volume? 
 
 Curve of Constant Steam Weight. If I pound of 
 saturated steam expands in such a manner that we 
 have always I pound of saturated steam whatever its 
 pressure, the expansion curve is called a curve of con- 
 stant steam weight. Or if a mixture of steam and 
 water having a given proportion of steam expands in 
 such a way that, whatever its pressure, there is always 
 the same proportion of steam present, the curve of 
 expansion is called a curve of constant steam weight. 
 Prob. 60, If i pound of a mixture of steam and water at 120 
 pounds pressure expands so that 30 per cent is 
 always steam, what are the volumes at 120, 90, 60, 
 and 30 pounds pressure ? 
 
 At 120 pounds we have for the volume of the steam 
 .30x3.711, and for the water .7ox.oi6, and the total 
 volume is 1.1133 + . 0112 = 1-1245 cubic feet. 
 Prob, 61, A mixture of 60 per cent steam and 40 percent water 
 expands from 90 to 15 pounds pressure, so that 
 there is always 60 per cent steam. What is the 
 volume at every 15 pounds pressure, if the total 
 weight is 5 pounds? 
 
5 NOTES ON THERMODYNAMICS. 
 
 To Determine the Work Done. The amount of work 
 done by such an expansion can only be approximately 
 determined by calculation. The most convenient way 
 of doing it is to assume that the expansion curve is in 
 the form pv n =K' , and find the most probable value of 
 n y and from the equation of the curve determine the 
 area. 
 
 To determine the most probable value of n, it is 
 not correct to determine several values of n and 
 average them. The following, from the method of 
 least squares, gives the most probable value of n and 
 is not at all difficult to follow out. Determine as 
 many values of / and v as desired, and write these 
 values in the logarithmic equation as below : 
 
 > K " J 
 
 + n lo g ^2 = K" \ 
 + n log 7' 3 = K", etc. 
 
 Add these equations together and we have 
 
 (A) 
 
 Now multiply each of the original equations by the 
 coefficient of n in that equation and we have 
 
 log A log z/j + n (log i\ ) z = K " lo g v i 5 
 
 log/2 lo g V* + n ( 10 S ^2 ) 2 K" log V*\ 
 
 log/3 log v z + n (log v z ) 2 K" log v s . 
 Adding these equations together we have 
 
 2 log / log v + n2 (log vf = ZK" log 7-. . (B) 
 
NOTES ON THERMODYNAMICS. 59 
 
 Solving (A) and (B) will give the most probable value 
 of n. Ordinarily three-place logarithms are not accu- 
 rate enough for this work. The amount of work 
 is then 
 
 I n 
 
 To determine the quantity of heat that will be re- 
 quired to produce this expansion, we know that the 
 heat at the end of the expansion added to the work 
 done must be equal to the heat in the steam at the 
 beginning of the expansion added to the heat sup- 
 plied. We have already shown how to determine 
 three of these quantities so that the heat supplied can 
 be determined. 
 
 Prob. 62. i pound of steam at 60 pounds pressure expands to 
 40 pounds along a curve of constant steam 
 weight. How much work is done and how much 
 heat must be supplied ? We have the following 
 for the pressures and volumes : 
 
 At 60 Ibs. V = 7.096 cu. ft.; 
 50 Ibs. F= 8.414 cu. ft.; 
 40 Ibs. V = 10.37 cu. ft. 
 
 To determine the law of expansion write : 
 
 log p + n log v = K" 
 
 1.778+ .851;* = K" 1.513 + .724^= .851^" 
 
 1*099+ .925;* = K" 1.572 + .856;;= .925/4"' 
 
 1.602 + 1.016;; = K" 1.628 + 1.032;; = i.oi6A^" 
 
 5.079 + 2.792;; = $K" (A) 4.713 + 2.6i2 = 2.792A"' (B) 
 
 n = 1.07. 
 
 Work = 60x144x7.09^-40x144x10.37 = 2 ft lbs> 
 
 1.07 i 
 
60 NOTES ON THERMODYNAMICS. 
 
 21500 
 Heat required = q** + p*o + -^-~o -- ? 60 ~~ P M> 
 
 236.4 + 850.3 + 27.6 261.9 8 3-7 2I -7 n - u - 
 
 Rectangular Hyperbola. In many cases a rectan- 
 gular hyperbola practically represents the expansion 
 taking place in a mixture of steam and water under 
 actual conditions. This is in no sense a theoretical 
 expansion line for a steam expansion, but it practi- 
 cally represents what actually takes place in many 
 steam-engine cylinders. The law of the expansion 
 here is/z/ = K y and the amount of work done is 
 
 The amount of heat required is 
 
 e Ml q\ *iPi, 
 \ 
 
 the subscript 2 referring to the final condition, and I 
 to the initial condition. 
 
 Prob. 63. i pound of a water mixture containing 30 per cent 
 of moisture expands from 100 pounds to 20 
 pounds, so that 30 per cent of moisture is always 
 present. How much work is done, and must 
 heat be added or taken away, and how much ? 
 
 Prob. 64. i pound of a mixture containing 30 per cent of 
 moisture expands from 100 pounds to 30 pounds 
 along a rectangular hyperbola. How much 
 work is done, what is the condition at the end of 
 the expansion, and how much heat must be 
 added or taken away ? 
 
NOTES ON THERMODYNAMICS. 6 1 
 
 CYCLES PASSED THROUGH BY VAPORS. 
 
 When a vapor is used in a cylinder the amount of 
 work done and the amount of heat required can be 
 determined as follows: Suppose that at #, Fig. 22, we 
 have I pound of a mixture of vapor and liquid, x a parts 
 being vapor, and suppose that, at 
 b, x b parts are vapor, the pressure re- 
 maining constant. 
 
 From b to c let the expansion be 
 according to any law, and at c let x c 
 
 be the proportion of the vapor. Let a ' _?' b ' c ' 
 
 FIG. 22. 
 heat be taken away first at constant 
 
 pressure, and then according to the same law as the 
 expansion curve be, so that we have at the end of the 
 cycle the same condition of affairs as at the beginning. 
 The amount of work done is the area of the figure 
 abed. It can be most easily calculated by finding the 
 separate areas and combining them so that 
 
 W = abb' a' + bcc'b' - cdd'c' - add' a'. 
 
 1^*7**** 
 
 The area abb' a' = (x b x^)Ap a u a . 
 
 ~^\ > - ^ Z. ^** 
 
 The area cdd'c' = (x c x^)Ap c u c . 
 
 areas under be and ad depend upon the law of 
 the expansion and can be determined as shown before. 
 The amount of heat required to do this work is the 
 heat required from a to c and is equal to 
 
 (& + *cpc) - (3 a + *a?a) + 
 
62 NOTES ON THERMODYNAMICS. 
 
 \ The amount of heat which must be taken away is 
 
 (q a + *apa] + adcc'a' + (q c + x c p c ). 
 The relation between the various values of x depends 
 on the law of the expansion. 
 
 If the expansion is adiabatic, the value of x c and x d 
 can be determined if x a and x b are given. We have 
 
 s. 
 
 cdt x b r b x c r c 
 
 T t ' T c 
 all tl 2 terms of which are known except x c ; and 
 
 x d r d ^ 
 
 J d T T a ' T d 
 
 from which x d can be determined. 
 
 As Y b and T b are equal to Y a and T a , and similarly 
 for c and d, we can write from the last two equations 
 
 r a Y C 
 
 \%6 % a) y v*< *</ 
 
 a -L c 
 
 The work done can then be written 
 
 W= Ap b u b (x b x a ) + \q b + x b p b q c x c p^\ 
 
 - Ap c ii c (x c x^) \q a + x a p a q d x d p d ~\ 
 
 = x b Y b x a Y b x c r c + x d Y c 
 (x b x a }r b (x c x d )r c 
 
 (x b - x*)r a 
 
NOTES ON THERMODYNAMICS. 63 
 
 In the last equation (x # a ) r a is the heat added 
 
 T a T c 
 from a to b. The efficiency is therefore a ' c , which 
 
 a 
 
 is Carnot's efficiency, as might have been expected as 
 this is a Carnot cycle. When this condition of affairs 
 exists in a cylinder, the cylinder fulfils the functions 
 of boiler, engine, and condenser, as we have assumed 
 that the given weight of the substance is in the 
 cylinder at all the points of the cycle. 
 
 Prob, 65. How much work is done in the cycle of Fig. 3$, if 5 
 pounds of a mixture of steam and water expands 
 having p a = 100 pounds per square inch, p a =15 
 pounds per square inch,;r a = .1, x b .9 ? 
 From the tables 
 
 ; a = 884.0 h. u. ; T a 788.3 ; T d = 673.7. 
 The heat added from a to b is 
 
 M(x b - x a }r a = 5 x (.9 - . i) x 884.0 = 3536 h. u. 
 The work done is 
 
 = 40200 t '- 
 
 Prob, 66. i pound of NH 3 expands through a cycle, as in 
 Fig. 22. If / a = 60 degrees F., id = 10 degrees F., 
 x a .1, x\ i, how much work is done and how 
 much heat is required ? 
 
 Prob. 67. If in a cycle, like Fig. 22, v e = 10 cubic feet, v a = i 
 cubic foot,/ a = 1 50 pounds per square inch, p c - 15, 
 how much work is done and how much heat is 
 required if i pound of steam is used? 
 
 In an actual engine the conditions are different 
 from those in the last figure, as from a to b there is 
 not the same weight in the cylinder, and from c to d 
 
64 NOTES ON THERMODYNAMICS. 
 
 the weight also varies. And in addition there is con- 
 stant interchange of heat between the cylinder walls 
 and the steam. 
 
 First, neglecting the action of the cylinder-walls, 
 suppose Fig. 23 represents what takes 
 place in the cylinder. At a the 
 clearance volume is filled with steam 
 whose steam proportion is x a . The 
 steam from the boiler is admitted 
 and fills the cylinder to c. Expan- 
 
 rlG. 23. 
 
 sion takes place to d, and exhausts 
 to a again. 
 
 Let m pounds be in the cylinder at a, and M pounds 
 be added from the boiler. Let x' be the value for 
 the steam coming from the boiler. If we know the 
 
 v 
 
 volume at c, we have -^> volume of I pound 
 
 m -f- M 
 
 and 
 
 from which x e can be found. 
 
 To find the work from b to c we have that the heat 
 at c added to the work done is equal to the heat at a 
 added to the heat received from the boiler, or 
 
 Work be = m(q a + x a p a - q c - x c p c ) + M(x'r e - *>,). 
 We might have written 
 
NOTES ON THERMODYNAMICS. 65 
 
 but it has been written in the form first given to call 
 attention to the fact that the last term in the first 
 equation contains r c and not p c . The reason is that 
 the heat brought into the cylinder from the boilers 
 includes not only q c and x' ' p c , but also the external 
 work which must be done in forcing this steam out of 
 the boiler, or x 'Ap c u c . 
 
 The work under cd is determined as before shown, 
 and the work under da is the area of the rectangle 
 under ad, or 
 
 Prob. 68. In an engine having Fig. 23 for a card, let V a = .4 
 cubic feet, V* = 8 cubic feet, p b = 100 x 144, 
 p a = 1 5 x 144, x a = .9, Xd = -8, cd being an adia- 
 batic. How much work is done ? 
 First find x c . 
 
 / 1 
 
 cdt^ JCcTmo _ .8r 16> 
 T T^ioo T\* 
 
 To find the volume at c we must know the weight 
 along cd and we have 
 
 i6 = 8. 
 From the tables s* 26.15 I 
 
 ^ = .8X26.15 + . 2X.OI6 = -^2 = ' 3Sl lbS ' ; 
 
 V c = .381 (.88x4.403 + .12 x.oi6) = 1.47 cu. ft. 
 
66 NO TES ON THE R MOD YNA MICS. 
 
 The work ^ = (1.47 4)100x144 =15400 
 
 The work cd =. . 381^100 + . 88p 100 q .8p 15 ]x778 = 32000 
 
 474o 
 
 The work da = (8 .4)15 x 144 = 16400 
 
 Work in cycle = 31000 
 
 Prob. 69. In the above problem, how much steam must the 
 
 boiler have furnished if x' = i ? 
 
 Prob. 70. How much steam was in the cylinder at b, and what 
 was the value of Xb if there was no loss of heat 
 through or to the cylinder-walls ? 
 
 Prob. 71. In problem 68, how much heat must have been 
 taken up by the cylinder-walls if x' = i and 
 x c = .88 ? 
 
 If, instead of the exhaust continuing to a, it had 
 stopped at e of Fig. 24, the above formula will apply 
 by putting in the corresponding 
 values of pressures and temperatures, 
 etc., for the new point a, and the 
 amount of work will be reduced by 
 the area aef, which must be deter- 
 
 mined as already shown. 
 In all engines using vapors, the quantity of heat re- 
 jected along the line da of Fig. 23, or de of Fig. 24, 
 is a large proportion of the total heat supplied to an 
 engine. To use the same working substance over and 
 over again in an engine, it must be liquefied, pumped 
 into a boiler, and evaporated again. All the heat re- 
 jected from the engine less the amount which remains 
 in the working substance as a liquid cannot be again 
 utilized for doing work in the same engine. The 
 quantity of heat which must be supplied to the work- 
 ing substance for each cycle is therefore the amount 
 
NOTES ON THERMODYNAMICS. 6? 
 
 which must be added to it as a liquid at the tempera- 
 ture of its discharge from the engine. 
 
 Prob. 72. In Fig. 24, using steam, if x c = .7, m c = i, x e = i t 
 m e =.i t p b = ioo x 144, ^=15x144, / a = 3ox 144, 
 the curves cd and ea are rectangular hyperbolas, 
 how much work is done per cycle and how much 
 heat is expended ? 
 
 Prob. 73. If, having given the data of problem 72, the sub- 
 stance is anhydrous ammonia, what work is done 
 and how much heat is expended per cycle ? 
 
 Prob. 74. If, having given the data of problem 72, the sub- 
 stance is SO 2 ,what is the work done and what the 
 heat expended per cycle ? 
 
 When the action of the cylinder-walls is taken into 
 account, the following analysis might be made after 
 the method of Hirn. Assume that at c, Fig. 23, we 
 have steam with a given proportion of moisture and 
 that the expansion is a rectangular hyperbola, and 
 assume further that saturated steam without moisture 
 has been supplied, which is nearly true, and that the 
 steam discharged is steam without moisture, which 
 may or may not be true. 
 
 From c to d the cylinder-walls must give up heat per 
 pound equal to 
 
 all the terms of which are known except x d . This 
 can be calculated from 
 
68 NOTES ON THERMODYNAMICS. 
 
 From d to a the cylinder-walls must give up heat to 
 the amount 
 
 r a ) - a 
 
 This is, of course, on the assumption that no heat 
 is radiated. The amount radiated can be accounted 
 for and the formula made exactly true. 
 
 P rob. 75. Suppose we have, Fig. 23, volume </ = 7.2 cubic 
 feet, volume a = .14 cubic feet, volume c = 1.08 
 cubic feet; weight steam used = .35 pound ; pres- 
 sure c = loo, pressure a = 15 ; x e = -64, x a = .9. 
 What should theoretically be the condition of the 
 exhaust steam if the boiler supplies steam having 
 x f = i, and the expansion curve is a rectangular 
 hyperbola, assuming no radiation from the cylinder. 
 The heat received from the boiler less that rejected to 
 the condenser or air is the work done, as we have 
 assumed no radiation. 
 
 The heat received is M(q c + r c ). 
 The work done is 
 
 100 x 144 x .88 + loo x 144 x i. 08 log e -Q 7.06 x 144 x 15 
 
 I Oo 
 
 = 27000 ft.-lbs. = 34.7 h. u. 
 The heat rejected is M(q a -\-x a r a ) and 
 
 M(q a + x a r a ) = M(q c + r c } - 34.7 ; 
 
 r e ga) 34-7 
 
 Mr a 
 
 _ '35( 2 97-9 + 884 181.8) 34.7 _ 
 35x965-1 
 
 showing tbat under these conditions the exhaust 
 steam will have 6.9 per cent moisture in it. 
 
NOTES ON THERMODYNAMICS. 69 
 
 Prob, 76. A condensing engine working between 150 and 4 
 pounds pressure requires 15 pounds of dry satu- 
 rated steam per indicated horse-power per hour. 
 If no heat is radiated from the cylinder, what 
 must be the average condition of the exhaust ? 
 
 Prob. 77. Draw a diagram showing the quantity of dry satu- 
 rated steam that must be used per horse-power per 
 hour in order that the exhaust at 4 pounds pres- 
 sure may be dry saturated steam, if the steam- 
 pressure is 80, 100, 120, 140, and 160 pounds per 
 square inch, there being no radiation. 
 

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