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. SHORT-TITLE CATALOGUE OF THE PUBLICATIONS OF JOHN WILEY & SONS, NEW YORK, LONDON: CHAPMAN & HALL, LIMITED. ARRANGED UNDER SUBJECTS. Descriptive circulars sent on application. 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