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v Preface This manual contains more or less complete solutions for every problem in the book. Should you find errors in any of the solutions, please bring them to my attention. Over the years, I have tried to enrich my lectures by including historical information on the significant developments in thermodynamics, and biographical sketches of the people involved. The multivolume Dictionary of Scientific Biography, edited by Charles C. Gillispie and published by C. Scribners, New York, has been especially useful for obtaining biographical and, to some extent, historical information. [For example, the entry on Anders Celsius points out that he chose the zero of his temperature scale to be the boiling point of water, and 100 to be the freezing point. Also, the intense rivalry between the English and German scientific communities for credit for developing thermodynamics is discussed in the biographies of J.R. Mayer, J. P. Joule, R. Clausius (who introduced the word entropy) and others.] Other sources of biographical information include various encyclopedias, Asimov’s Biographical Encyclopedia of Science and Technology by I. Asimov, published by Doubleday & Co., (N.Y., 1972), and, to a lesser extent, Nobel Prize Winners in Physics 1901-1951, by N.H. deV. Heathcote, published by H. Schuman, N.Y. Historical information is usually best gotten from reading the original literature. Many of the important papers have been reproduced, with some commentary, in a series of books entitled “Benchmark Papers on Energy” distributed by Halsted Press, a division of John Wiley and Sons, N.Y. Of particular interest are: Volume 1, Energy: Historical Development of the Concept, by R. Bruce Lindsay. Volume 2, Applications of Energy, 19th Century, by R. Bruce Lindsay. Volume 5, The Second Law of Thermodynamics, by J. Kestin and Volume 6, Irreversible Processes, also by J. Kestin. The first volume was published in 1975, the remainder in 1976. vi Other useful sources of historical information are “The Early Development of the Concepts of Temperature and Heat: The Rise and Decline of the Caloric Theory” by D. Roller in Volume 1 of Harvard Case Histories in Experimental Science edited by J.B. Conant and published by Harvard University Press in 1957; articles in Physics Today, such as “A Sketch for a History of Early Thermodynamics” by E. Mendoza (February, 1961, p.32), “Carnot’s Contribution to Thermodynamics” by M.J. Klein (August, 1974, p. 23); articles in Scientific American; and various books on the history of science. Of special interest is the book The Second Law by P.W. Atkins published by Scientific American Books, W.H. Freeman and Company (New York, 1984) which contains a very extensive discussion of the entropy, the second law of thermodynamics, chaos and symmetry. I also use several simple classroom demonstrations in my thermodynamics courses. For example, we have used a simple constant-volume ideal gas thermometer, and an instrumented vapor compression refrigeration cycle (heat pump or air conditioner) that can brought into the classroom. To demonstrate the pressure dependence of the melting point of ice, I do a simple regelation experiment using a cylinder of ice (produced by freezing water in a test tube), and a 0.005 inch diameter wire, both ends of which are tied to the same 500 gram weight. (The wire, when placed across the supported cylinder of ice, will cut through it in about 5 minutes, though by refreezing or regelation, the ice cylinder remains intact.—This experiment also provides an opportunity to discuss the movement of glaciers.) Scientific toys, such as “Love Meters” and drinking “Happy Birds”, available at novelty shops, have been used to illustrate how one can make practical use of the temperature dependence of the vapor pressure. I also use some professionally prepared teaching aids, such as the three-dimensional phase diagrams for carbon dioxide and water, that are available from laboratory equipment distributors. Despite these diversions, the courses I teach are quite problem oriented. My objective has been to provide a clear exposition of the principles of thermodynamics, and then to reinforce these fundamentals by requiring the student to consider a great diversity of the applications. My approach to teaching thermodynamics is, perhaps, similar to the view of John Tyndall expressed in the quotation “It is thus that I should like to teach you all things; showing you the way to profitable exertion, but leaving the exertion to you—more anxious to bring out your manliness in the presence of difficulty than to make your way smooth by toning the difficulties down.” Which appeared in The Forms of Water, published by D. Appleton (New York, 1872). 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NJ � - V 7KH�ORDG�RQ�WKH�JDV�FRROHU�LV��IURP�3UREOHP������ � � � � � � 4 1& 7 7 . � u u u �� � � u � S � NJ V J NJ ��� J PRO - PRO�. - V � � � ��� ����� ��� ����� ����� ��� �� 0 5 3.11 (a) This is a Joule-Thomson expansion ⇒ =( ) = = °( ) ≈ = °( ) = $ ? $ . $ . H T H T H T70 bar, 10133 400 1 400 32782 bar, C bar, C kJ kg and T = °447 C , $ .S = 6619 kJ kg K (b) If turbine is adiabatic and reversible &Sgen = 0c h , then $ $ .S Sout in kJ kg K= = 6619 and P = 1013. bar. This suggests that a two-phase mixture is leaving the turbine Let fraction vapor kJ kg K kJ kg K V L x S S = = = $ . $ . 7 3594 13026 Then x x7 3594 1 13026 6619. . .( ) + −( )( ) = kJ kg K or x = 08778. . Therefore the enthalpy of fluid leaving turbine is $ . . . . . $ $ H H H = × + −( ) × = ( ) ( ) 08788 26755 1 08778 417 46 2399 6 V L sat’d, 1 bar sat’d, 1 bar kJ kg Energy balance 0 = + +& $ & $M H M Hin in out out &Q 0 + −&W Ps dV dt 0 but & &M Min out= − ⇒ − = − = & & . . . W M s in kJ kg 3278 2 2399 6 8786 (c) Saturated vapor at 1 bar $ . ; $ . & . . . % . . . & & . . . S H W M S M s = = − = − = ( ) = × = = − = 7 3594 26755 32782 26755 602 7 602 7 100 8786 686% 7 3594 6619 0740 kJ kg K kJ kg kJ kg Efficiency kJ Kh in Actual gen in (d) 0 0 0 1 2 2 1 1 1 2 1 1 2 = + ⇒ = − = − + + − = − + + & & & & & $ $ & & & $ $ & & M M M M M H H W Q P dV dt M S S Q T S sc h c h gen Simplifications to balance equations &Sgen = 0 (for maximum work); P dV dt = 0 (constant volume) & &Q T Q T = 0 where T0 25= °C (all heat transfer at ambient temperature) W Q Water 1 bar 25° C Steam 70 bar 447° C $ .H sat'd liq, 25 C kJ kg °( ) = 10489 ; $ .S T sat'd liq, 25 C kJ kg K = °( ) = 03674 & & $ $Q M T S S= −0 2 1c h ; − = − + − = − − − & & $ $ $ $ $ $ $ $ max W M H H T S S H T S H T Ss 1 2 0 2 1 1 0 1 2 0 2c h c h c h − = − × − − × = + = & & . . . . . . . . . max W M s 3278 2 29815 6619 10489 29815 03674 1304 75 4 65 1309 4 kJ kg 3.12 Take that portion of the methane initially in the tank that is also in the tank finally to be in the system. This system is isentropic S Sf i= . (a) The ideal gas solution S S T T P P N PV RT N PV RT N P V RT N N N f i f i f i R C i i i f f f f i p = ⇒ = F HG I KJ = F H I K = ⇒ = = = = = − = − 300 35 70 1502 1964 6 mol; 1962 17684 8 314 36 . . . . . . K = mol mol∆ (b) Using Figure 2.4-2. 70 bar ≈ 7 MPa, T = 300 K $ . $S Si f= =505 kJ kg K $ . , . . .V m N i i i = = = = × = 00195 07 00195 3590 kg. 35.90 kg 1282 m kg so that m m kg 1000 g kg 28 g mol mol 3 3 3 At 3.5 bar = 0.35 MPa and $ .S Tf = ⇒ ≈505 138 K. kJ kg K Also, $ . , . . . . . V m N f f f = = = = × = 0192 07 0192 3646 kg. 3646 kg 1302 m kg so that m m kg 1000 g kg 28 g mol mol 3 3 3 ∆N N Nf i= − = − = −1302 1282 11518 mol. . 3.13 dS C dT T R dV V = + eqn. (3.4-1) ∆S a R bT cT dT e T dT T R dV V = −( ) + + + +LNM O QP + zz 2 3 2 so that S T V S T V a R T T b T T c T T d T T e T T R V V 2 2 1 1 2 1 2 1 2 2 1 2 2 3 1 3 2 2 1 2 2 1 2 3 2 , , ln ln a f a f a f c h c h c h − = −( ) + − + − + − − − +− − Now using PV RT V V T T P P S T P S T P a T T b T T c T T d T T e T T R P P = ⇒ = ⋅ ⇒ − = + − + − + − − − −− − 2 1 2 1 1 2 2 2 1 1 2 1 2 1 2 2 1 2 2 3 1 3 2 2 1 2 2 1 2 3 2 , , ln ln a f a f a f c h c h c h Finally, eliminating T2 using T T P V PV2 1 2 2 1 1= yields S P V S P V a P V PV b R P V PV c R P V PV d R P V PV eR P V PV R P P 2 2 1 1 2 2 1 1 2 2 1 1 2 2 2 2 1 1 2 3 2 2 3 1 1 3 2 2 2 2 1 1 2 2 1 2 3 2 , , ln ln a f a f a f a f a f a f a f d i d i − = F HG I KJ + − + − + − − − − − − 3.14 System: contents of valve (steady-state, adiabatic, constant volume system) Mass balance 0 1 2= +& &N N Energy balance 0 1 1 2 2= + +& &N H N H &Q 0 +Ws 0 −P dV dt 0 ⇒ =H H1 2 Entropy balance 0 1 1 2 2= + + +& & &N S N S Sgen &Q T 0 ⇒ = − =∆S S S S N2 1 & & gen (a) Using the Mollier Diagram for steam (Fig. 2.4-1a) or the Steam Tables T P P H T S 1 1 2 2 2 2 600 K 35 7 30453 293 7 277 = = = = ⇒ ≈ ° = bar bar J g C J g K$ . $ . $ $ .H H1 2 3045 3= = J g . Thus $ .S1 65598 = J g K ; Texit C= °293 ∆ $ $ $ .S S S= − =2 1 0717 J g K (b) For the ideal gas, H H T T1 2 1 2 600 K= ⇒ = = ∆ ∆ S S T P S T P C T T R P P R P P S = − = − = − = ⇒ = 2 2 1 1 2 1 2 1 2 1 1338 0743 , , ln ln ln . $ . a f a f p J mol K J mol K 3.15 From the Steam Tables At 200oC, P V V U U H H S S H S L V L V L L L V = = = = = = = = ⋅ = ⋅ = = ⋅ 15538 MPa 0001157 012736 m 85065 25953 852 45 27932 2 3309 64323 19407 41014 . $ . / $ . / $ . / $ . / $ . / $ . / $ . / $. / $ . $ . m kg kg kJ kg kJ kg kJ kg kJ kg kJ kg K kJ kg K kJ / kg kJ / kg K 3 3 vap vap∆ ∆ (a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties of the steam and water will be At 150oC, P V V U U H H S S H S L V L V L V L V = = = = = = = = ⋅ = ⋅ = = ⋅ 04578 MPa 0001091 03928 m 63168 kJ 25595 632 20 kJ 27465 18418 kJ 68379 2114 3 4 9960 kJ . $ . / $ . / $ . / $ . / $ . / $ . / $ . / $ . / $ . / $ . / m kg kg kg kJ kg kg kJ kg kg K kJ kg K kJ kg kg K 3 3 vap vap∆ ∆ (b) For simplicity of calculations, assume 1 m3 volume of tank. Then Mass steam initially = 0.8 m 0.12736 m kg kg Mass water initially = 0.2 m 0.001157 m kg Weight fraction of steam initially = 6.2814 179.14 Weight fraction of water initially = 6.2814 179.14 3 3 3 3 / . / . . . = = = = 62814 17286 kg 003506 096494 The mass, energy and entropy balances on the liquid in the tank (open system) at any time yields dM dt M dM U dt M H dM S dt M S M dU dt U dM dt M H H dM dt M dU dt dM dt H U L L L L L V L L L V L L L L L V V L L L L V L = = = + = = = − & ; $ & $ $ & $ $ $ & $ $ $ $ $ ; and or c h Also, in a similar fashion, from the entropy balance be obtain M dS dt dM dt S S dM dt SL L L V L L$ $ $ $ = − =c h ∆ vap There are now several ways to proceed. The most correct is to use the steam tables, and to use either the energy balance or the entropy balance and do the integrals numerically (since the internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature. This is the method we will use first. Then a simpler method will be considered. Using the energy balance, we have dM M dU H U M M M U U H U M M U U H U L L L V L i L i L i L i L i L i V i L i L i L i L i L i V i L = − − = − − = + − − F HG I KJ + + + + $ $ $ , $ $ $ $ $ $ $ $ or replacing the derivatives by finite differences or finally 1 1 1 11 So we can start with the known initial mass of water, then using the Steam Tables and the data at every 5oC do a finite difference calculation to obtain the results below. i T (oC) $UiL (kJ/kg K) $HiV (kJ/kg K) MiL (kg) 1 200 850.65 2793.2 172.86 2 195 828.37 2790.0 170.88 3 190 806.19 2786.4 168.95 4 185 784.10 2782.4 167.06 5 180 762.09 2778.2 165.22 6 175 740.17 2773.6 163.42 7 170 718.33 2768.7 161.67 8 165 696.56 2763.5 159.95 9 160 674.87 2758.1 158.27 10 155 653.24 2752.4 156.63 11 150 631.68 2746.5 155.02 So the final total mass of water is 155.02 kg; using the specific volume of liquid water at 150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3 leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found to be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, the final volume fraction of water is 16.91%, and the fraction of the initial steam + water that has been withdrawn is (172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steam has withdrawn, and 87.7% of the original mass of steam and water remain in the tank. For comparison, using the entropy balance, we have dM M dS S S M M M S S S M M S S S L L L V L i L i L i L i L i L i i L i L i L i L i = − − = − = + −F HG I KJ + + + + $ $ $ , $ $ $ $ $ $ or replacing the derivatives by finite differences of finally vap vap 1 1 1 11 ∆ ∆ So again we can start with the known initial mass of water, then using the Steam Tables and the data at every 5oC do a finite difference calculation to obtain the results below. i T (oC) $SiL (kJ/kg K) $SiL (kJ/kg K) MiL (kg) 1 200 2.3309 6.4323 172.86 2 195 2.2835 6.4698 170.86 3 190 2.2359 6.5079 168.92 4 185 2.1879 6.5465 167.02 5 180 2.1396 6.5857 165.17 6 175 2.0909 6.6256 163.36 7 170 2.0419 6.6663 161.60 8 165 1.9925 6.7078 159.87 9 160 1.9427 6.7502 158.18 10 155 1.8925 6.7935 156.53 11 150 1.8418 6.8379 154.91 So the final total mass of water is 154.91 kg; using the specific volume of liquid water at 150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3 leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found to be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, the final volume fraction of water is 16.90%, and the fraction of the initial steam + water that has been withdrawn is (172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steam has withdrawn, and 87.7% of the original mass of steam and water remain in the tank. These results are similar to that from the energy balance. The differences are the result of round off errors in the simple finite difference calculation scheme used here (i.e., more complicated predictor-corrector methods would yield more accurate results.). A simpler method of doing the calculation, avoiding numerical integration, is to assume that the heat capacity and change on vaporization of liquid water are independent of temperature. Since liquid water is a condensed phase and the pressure change is small, we can make the following assumptions $ $ $ $ $ $ $ ; $ U H H H H dU dt dH dt C dT dt dS dt C T dT dt L L V L L L L L L L L ≈ − = ≈ ≈ ≈ and and vap P P ∆ With these substitutions and approximations, we obtain from the energy balance M dU dt dM dt H U M dH dt dM dt H M C dT dt dM dt H C H C H dT dt M dM dt C H M M L L L V L L L L L L L L L L L L f L i L $ $ $ $ $ $ $ $ $ ln = − → = = = −( ) = FHG I KJ c h Now using an average value of and over the temperature range we obtain or vap P vap P vap P vap P vap ∆ ∆ ∆ ∆ ∆ 1 150 200 and from the entropy balance M dS dt dM dt S M C T dT dt dM dt S C S C T S dT dt M dM dt C S M M L L L L L L L L L L L f L i L $ $ $ $ $ $ ln . . ln = → = = + + F H I K = F HG I KJ ∆ ∆ ∆ ∆ ∆ vap P vap P vap P vap P vap Now using an average value of and over the temperature range we obtain or 1 150 27315 200 27315 From the Steam Table data listed above, we obtain the following estimates: C U T U T C T T S T S T C S T S T L L L P o o o o P P o o C C C - C kJ kg K or using the ln mean value (more appropriate for the entropy calculation) based on C C kJ kg K = = − = = − = ⋅ F HG I KJ = − = = − = + + FH IK = − FH IK = ⋅ $ ( ) $ ( ) . . . ln $ $ $( ) $( ) ln . . . . ln . . . 200 150 200 150 852 45 632 20 50 4 405 200 150 200 27315 150 27315 2 3309 18418 47315 42315 4 3793 2 1 2 1a f a f Also, obtaining average values of the property changes on vaporization, yields ∆ ∆ ∆ ∆ ∆ ∆ $ $ $ . . . $ $ $ . . . H H T H T S S T S T vap vap o vap o vap vap o vap o C C kJ kg C C kJ kg K = × = + = = × + = = × = + = = × + = ⋅ 1 2 150 200 1 2
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