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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 12 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 LL 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 2114 3 19407 20275 1 2 150 200 1 2 4 9960 41014 45487 b g b g b g b g With this information, we can now use either the energy of the entropy balance to solve the problem. To compare the results, we will use both (with the linear average Cp in the energy balance and the log mean in the entropy balance. First using the energy balance C H M M M M L f L i L f L i L P vap∆ $ ln . . . exp . . 150 200 4 405 50 20275 010863 010863 089706 −( ) = FHG I KJ = − × = − = −( ) = Now using the entropy balance ln $ ln . . . . ln . . . ln . . . . . . M M C S M M f L i L L f L i L F HG I KJ = + + F H I K = F H I K = F H I K = FH IK = P vap∆ 150 27315 200 27315 4 3793 45487 42315 47315 09628 42315 47315 42315 47315 089805 0 9628 Given the approximations, the two results are in quite good agreement. For what follows, the energy balance result will be used. Therefore, the mass of water finally present (per m3) is M M V M V M V L L L L o 3 3 V 3 V o 3 3 final initial occupying final C m Therefore, the steam occupies 0.8308 m , corresponding to final 0.8308 m C 0.8308 m m kg kg ( ) = × ( ) = = ( ) × = × = ( ) = = = 0897 15506 kg 150 15506 0001091 01692 150 03928 2115 . . $ . . . $ . . b g b g So the fraction of liquid in the tank by mass at the end is 155.06/(155.06+2.12) = 0.9865, though the fraction by volume is 0.1692. Similarly the fraction of the tank volume that is steam is 0.8308, though steam is only 2.12/(155.06+2.12) = 0.0135 of the mass in the tank. (c) Initially there was 6.28 + 172.86 = 179.14 kg of combined steam and water, and finally from the simpler calculation above there is 155.06 + 2.12 = 157.18 kg. Therefore, 87.7% of the total amount of steam + water initially in the tank are there finally, or 12.3% has been withdrawn. This corresponds to 21.96 kg being withdrawn. This is in excellent agreement with the more rigorous finite difference calculations done above. 3.16 (a) dN dt N N N N= = + = −0 1 2 2 1& & ; & & or dU dt N H N H W Q P dV dt W N H N H W N H HS S S= = + + + − = + −0 1 1 2 2 1 1 1 2 1 2 1 & & & & & & & & & or = - dS dt N S N S Q T S S N S S= = − + + = −0 1 1 1 2 0 1 2 1 & & & & & &gen gen CP J mol K = ⋅ 37151. = - = KP 298.15K P & & . W N H H C dT C TS T f f 1 2 1 29815z = ⋅ −c h if the heat capacity is independent of temperature. First consider the reversible case, S S C T dT R dP PT T i f 2 1 1 10 0 gives − = = zz P The solution is 499.14K. Then K J molP & & . . W N CS rev 1 49914 29815 7467= ⋅ −( ) = . The actual work is 25% greater W W C T T act S rev f f = = = ⋅ − = 125 9334 29815 549 39 . . . J mol K The solultion is K P c h (b) Repeat the calculation with a temperature-dependent heat capacity C T T T TP( ) = + ⋅ − ⋅ + ⋅− − −22 243 5977 10 3499 10 7 464 102 5 2 9 3. . . . Assuming reversibility Tf = 479.44K. Repeating the calculations above with the temperature- dependent heat capacity we find Wact = 9191 J, and Tf =520.92K. So there is a significant difference between the results for the constant heat capacity and variable heat capacity cases. 3.17 T T Pi f i = 300 K, = 800 K, and = 1.0 bar C T T T TP -2 -5 2 -9 3( ) = 29.088 - 0.192 10 + 0.4 10 - 0.870 10 J mol K × × × ⋅ C T T dT P dP P P W C T dT T T P P f rev T T i f i f i f P K K 6 P K K Calculated final pressure = 3.092 10 Pa. J mol ( ) ( ) . = = = = = z z z = × = = × 300 800 1 300 800 41458 10 3.18 Stage 1 is as in the previous problem. Stage 2 Following the same calculation as above. Stage 2 allowed pressure = 9.563 10 Pa = 1.458 10 J mol = Stage 2 work , 7 rev -4 P W f 2 × × Stage 3 Following the same calculation method = 2.957 10 Pa = Stage 3 allowed pressure. = 1.458 10 J mol = Stage 3 work , -9 rev 4 P W f 3 × × Question for the student: Why is the calculated work the same for each stage? 3.19 The mass, energy and entropy balances are dM dt M M M M dU dt M H M H Q W M H H W W M H H dS dt M S M S Q T S M S S S S M S S = + = = − = = + + + − + = = + − = = + + + = − + = = − & & , & & & $ & $ & & ; & $ $ & ; & & $ $ & $ & $ & & & $ $ & & & $ $ 1 2 2 1 1 1 2 2 1 1 2 1 2 1 1 1 2 2 1 1 2 1 2 1 0 0 0 0 0 s s s gen gen gen c h c h c h c h 300 05° =C, 5 bar MPa. $ .H1 3064 2= kJ kg $ .S1 7 4599= kJ kg K 100 01° =C, 1 bar MPa. $ .H2 26762= kJ kg $ .S2 7 3614= kJ kg K & & . . W M s kJ kg 1 2676 2 3064 2 388 = − = satisfied the energy balance. & & $ $ . . . S M S Sgen kJ kg K 1 2 1 7 3614 7 4599 00985= − = − = − can not be. Therefore the process is impossible. 3.20 Steam 20 bar 2= ° MPa and 300 C $ .H = 30235 kJ kg $ .S = 67664 kJ kg (from Steam Tables) $ .U = 2772 6 kJ kg Final pressure = 1 bar. For reference saturation conditions are P T= =01 99 63. . MPa, $ .U L = 417 36 $ .H L = 417 46 $ .S L = 13026 $ .U V = 25061 $ .H V = 26755 $ .S V = 7 3594 (a) Adiabatic expansion valve & &W Q= =0 and 0 M.B.: dM dt M M= + =& &1 2 0 ; & &M M2 1= − ; E.B.: dU dt M H M H= + =& $ & $1 1 2 2 0 ; $ $H H2 1= From Steam Tables ⇒ H2 30235= . T = °250 C $ .H = 29743 kJ kg $ .S = 80333 kJ kg K P = 01. MPa T = °300 C $ .H = 3074 3 kJ/kg $ .S = 8 2158 kJ/kg K By interpolation T H= ° ⇒275 C gives = 3023.5 kJ / kg$ all vapor $ . & $ & $ & & & $ $ . S dS dt MS M S S S M S S = = + + = = − − = 81245 0 1359 1 2 2 2 1 kJ kg K = 8.1254 6.7664 kJ kg Kgen gen gen (b) Well designed, adiabatic turbine E.B.: & $ & $ &M H M H W1 1 2 2 0+ + = ; & $ $W H H= −2 1c h S.B.: & $ & $M S M S1 1 2 2 0+ = ; $ $S S2 1= ; $ .S2 67664= kJ kg K ⇒ Two-phase mixture. Solve for fraction of liquid using entropy balance. x x x ⋅ ( ) + −( ) ⋅ = = ( ) 7 3594 1 13026 6 7664 0 902 . . . . not good for turbine! $ . . . . . & & . . . & & . H W M W M 2 0902 26755 0098 417 46 2454 2 2454 2 30235 569 3 569 3 = × + × = = −( ) = − − = kJ kg kJ kg kJ kg (c) Isorthermal turbine ⇒ superheated vapor T P = ° = O QP 300 01 C MPa final state . $ . $ . H S = = 3074 3 82158 kJ kg kJ kg K E.B.: & $ & $ & &M H M H Q W1 1 2 2 0+ + + =s S.B.: & $ & $ & M S M S Q T1 1 2 2 + + +Sgen 0 = 0 & & $ & $ & $ $ & & $ $ . . . . & & & & $ $ . . . . 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P P dV dt dU dt W Q P dV dtu S u = + = - - Þ = + -0 0 0a f or U U W Q P V V S S Q T S Q T S TS T S u2 1 0 2 1 2 1 0 0 2 0 1 0 - = + - - - = + Þ = - - a f gen gen and U U W T S T S T S P V V W U PV T S U PV T S T S T S W U PV T S u u u 2 1 0 2 0 1 0 0 2 1 2 0 2 0 2 1 0 1 0 1 0 0 2 1 0 0 0 - = + - - - - = + - - + - + ³ = - = + - gen gen gensince where a f a f a f max ,A A A (b) 0 0 1 2 1 2 0 = - + + = - + + & & & & & & & M H M H Q W M S M S Q T S S gen Here & &W WS u= Þ = - - = - + - + Þ = - - - + ³ = - = - & & & & & & & & & & & & &max W M H M H Q M H M H MT S MT S T S W M H T S M H T S T S T S W M H T S u u u 2 1 2 1 0 1 0 2 0 2 0 2 1 0 1 0 0 2 1 0 0 gen gen genSince where a f a f b gB B B (c)Using the Steam Tables we find i) at 30 bar =3 MPa and 600°C $ .U = 32850 kJkg, $ .S =75085 kJkg K, $ .V = 013243 m kg3 $ $ $ $ . . . . . . A1 0 0 232850 1013 013243 10 29815 75085 105976 = + - = + + ´ × - ´ = U P V T S bar m kg kJbarm kJkg 3 3 ii) at 5 bar =05. MPa and 300°C $ .U = 28029 kJkg, $ .S =74599 kJkg K, $ .V =05226 m kg3 $ . . . . . .A2 228029 10130522610 298157459963167= + ´ ´ - ´ = kJkg $ $ $ ( . . ) .Wu = - = - = -A A2 1 63167 105976 42809 kJkg kJkg This is the maximum useful work that can be obtained in the transformation with the environment at 25°C and 1.013 bar. It is now a problem of clever engineering design to develop a device which will extract this work from the steam in a nonflow process. (d) Since the inlet and exit streams are at 25°C and P =1013. bar, any component which passes through the power plant unchanged (i.e., the organic matter, nitrogen and excess oxygen in the air, etc.) does not contribute to the change in availability, or produce any useful work. Therefore, for each kilogram of coal the net change is: 07 5833 5833 . . . kg of carbon mol of C mol of O to produce 58.33 mol CO 2 2 = + also 015 833. . kg of water mol of HO undergoes a phase change from liquid to vapor 2= Therefore & $ & . . . . . . & $ & . . . . . . . max M N M N W W i i i i i i u u B B B B in in carbon oxygen liquid water out out carbon dioxide water vapor actual kJkg coal kJkg coal kJkg coal kJkg coal kW-hrkg coal kJkg coal = = ´ + ´ + ´ - + ´( ) = - = = ´ -( )+ ´ - + ´( ) = - = - - ( ) = = - = å å ( ) ( ) ( ) ( ) ( ) b g b g 5833 0 5833 0 833 6831729815 0039 1976 5833 94052 833 578 29815 00106 24858 248581976 22882 22 7920 Efficiency in % = ´ = 7920100 22882 346%. Thus a coal-fired electrical power generation plants converts slightly more than 1/3 of the useful work obtainable from the coal it consumes. This suggests that it would be useful to look for another method of generating electrical power from coal . . . for example, using an electro- chemical fuel cell. Considering the amount of coal consumed each year in power generation, and the consequences (strip mining, acid rain, greenhouse effect, etc.) the potential economic savings and environmental impact of using only 1/3 as much coal is enormous. ���� $VVXPSWLRQV� �� 7XUELQH�DQG�SXPS�RSHUDWH�UHYHUVLEO\ DQG�DGLDEDWLFDOO\ �� 1R�SUHVVXUH�GURS�DFURVV�FRQGHQVHU�DQG ERLOHU �� 2QO\�KHDW�WUDQVIHU�RFFXUV�DW�FRQGHQVHU DQG�ERLOHU /RFDWLRQ 3DWK 6WDWH 7 q� �& 3��EDU� �+� N- NJ �6� N- NJ. ���P � NJ��9 � VXSHUKHDWHG YDSRU ��� �� ������ ������ �� ’6 � � VXSHUKHDWHG YDSRU �� ��� ������ � ����� �� 3 FRQVWDQW � VDW©G OLT ���� ��� ����� ������ ������� �� ’6 � � FRPS�OLT ���� �� ����� ������ �� 3 FRQVWDQW � ��� �� ������ ������ >$W�HDFK�VWDWH�WKH�SURSHUWLHV�LQ�ER[HV�ZHUH�NQRZQ�DQG�XVHG�ZLWK�WKH�VWHDP�VWDEOHV�WR�ILQG�WKH�UHPDLQLQJ SURSHUWLHV�@��1RZ�UHDG\�WR�DQVZHU�TXHVWLRQV� �D� 1HW�ZRUN�RXWSXW�SHU�NLORJUDP � � � � � �� � � �� � � � � � � � � � � � + + + + � � � � ���� � ����� ���� ���� ���� � 2 7 2 7 � N- NJ �E� +HDW�GLVFDUGHG�E\�FRQGHQVHU � � �� � � � �+ + � � ���� ����� ������ N- NJ �F� )UDFWLRQ�RI�ZRUN�XVHG�E\�SXPS���� �� � u � �� � ���� ���� ��� ���� � ����� ����� � � � � � �G� +HDW�DEVRUEHG�LQ�ERLOHU� �� � ���� � ���� ���� ��� � � N- NJ �H� 7KHUPDO�HIILFLHQF\���� � u u 1HW�ZRUN�RXW +HDW�LQ ��� ���� � ��� ���� � ����� � � � &RQGHQVHU : %RLOHU 3XPS 7XUELQH � � � � Carnot efficiency (1100°C and 15°C) = - = +( ) - +( ) +( ) ´ =T T T 1 2 1 110027315 15 27315 110027315 100 7902% . . . . For comparison, Carnot efficiency (700°C and 60.1°C) which are the temperature levels of working fluid (steam) in the closed-loop power cycle = +( ) - +( ) + ´ = 70027315 601 27315 70027315 100 6576% . . . . . which is almost twice as high as the actual efficiency. 3.39 Three subsystems: unknowns T f1 , P f 1 , T f 2 , P f 2 , T f 3 , P f 3 (6 unknowns) After process P P Pf f f1 2 3= = (2 equations) Subsystem 1 has undergone a reversible adiabatic expansion Þ =S Sf i1 1, or T T P P f i f i R C 1 1 1 1 = F HG I KJ P (1 equation)(#1) Subsystem 3 has undergone a reversible adiabatic compression Þ =S Sf i3 3, or T T P P T P P f i f i R C i f i R C 3 3 3 3 3 3 = F HG I KJ = F HG I KJ P P (1 equation)(#2) Mass balance subsystems 1 + 2 N N N N P V T P V T P V T P V T f f i i f f f f i i i i i1 2 1 2 1 1 1 2 2 2 1 1 1 2 2 2 + = + Þ + = + or P T V T f f f f 05 10 05 29315 1 025 29315 0017909 1 2 2 . . . . . .+ F HG I KJ = ´ + ´ = (1 equation)(#3) Energy balance on subsystems 1 + 2 + 3 N U N U N U N U N U N U P V RT C T P V RT C T P V RT C T P V RT C T P V RT C T P V RT C T P V V V P V P V P V f f f f f f i i i i i i f f f f f f f f f i i i i i i i i i f f f i i i 1 1 2 2 3 3 1 1 2 2 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 2 3 1 1 2 2 3 3 + + = + + + + = + + + + = + + V V V V V V c h but V V V V V Vf f i i1 2 3 1 2 3 05 025 025 1+ + = + + = + +( ) =. . . m 3 P f = ´ + ´ + ´ = 10 05 1 025 1 025 1 55 . . . . bar using this resultin eqn. (1) ® =T f1 25245. K in eqn. (2) ® =T f3 44893. K V V P P T T V f i i f f i f 3 3 3 3 3 3 2 025 1 55 44893 29315 006961 025 2 00696104304 = × = ´ ´ = = ´ - = . . . . . . . . m m 3 3 Now using Eqn. (#3) P T V T T Tf f f f f f05 55 05 25245 04304 0017909 33741 1 2 2 2 2 . . . . . . .+ F HG I KJ = + F HG I KJ = Þ = K Thus the state of the system is as follows Initial Final T1 293.15 K 252.45 K P1 10 bar 5.5 bar T2 293.15 K 337.41 K P2 1 bar 5.5 bar V2 025. m3 04304. m3 T3 293.15 K 448.93 K P3 1 5.5 bar V3 025. m3 00696 m. 3 Work done on subsystem 3 Energy balance N U N U W PdV P V RT C T P V RT C T P V C R P V C R W W C R R P V P V f f i i f f f f i i i i f f i i f f i i 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 355 006961 025 03984 3984 - = = - - = - = = - - = ´ - ´( ) = × = z V V V V P 3 barm kJ c h . . . . . 3.40 For the mass and energy balances, consider the composite system of can + tire as thesystem. Also gas is ideal for this system Q = 0 and W =0 mass balance: N N N N P V T P V T P V T P V T f f i i f f f f i i i i1 2 1 2 1 1 1 2 2 2 1 1 1 2 2 2 + = + Þ + = + (1) energy balance N U N U N U N U P V P V P V P Vf f f f i i i f f i i1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2+ = + Þ + = + (2) (see derivation of eqn. (c) of Illustration 2.5-5) AlsoP Pf f1 2 26 bar= = ¬( ). 3; using eqn. (3) in eqn. (2) yields P P V P V P V V i f f i 1 1 1 2 2 2 2 1 2 2 4 26 406 10 1 4 10 6 10 10927= + - = ´ ´ - ´ ´ ´ = - - - . . . c h bar To use eqn. (1) to get final temperatures, need another independent equation relating T f1 and T f 2 . Could do an energy balance around tank 1, as in derivation of eqn. (f) of Illustration 2.5-5. A more direct way is to do an entropy balance around a small fluid element, as in Illustration 3.5-2 and immediately obtain Eqn. (e) of that illustration T T P P f i C R f i 1 1 1 1 F HG I KJ = F HG I KJ P Thus T T P P f i f i R C 1 1 1 1 831430 295 26 10927 10469= F HG I KJ = FH IK = P K . . . . (very cold!) Using this result in eqn. (1) gives T f2 30326 K= . . 3.41 (a)System: Gas in the tank — system boundary is just before exit to the tank. System is open, adiabatic, and of constant volume. M.B.: S.B.: constant since dN dt N d N S dt N S S dN dt N dS dt N S dN dt S N dS dt S N = ( ) = Þ + = = Þ = = ¹( ) & & & 0 or 0 Note: Gas just leaving system has the same thermodynamic properties as gas in the system by the “well- mixed” assumption. For the ideal gas this implies T T P P R C 2 1 2 1 = F HG I KJ P (b)T2 831430 22 27315 1 13352= +( )FH IK =. . . bar 17.5 bar K (c)System: Gas in the tank + engine (open, constant-volume, adiabatic) M.B.: N N Nf i- = D DN =amount of mass (moles) that left the system E.B.: N U N U NH Wf f i i S- = +D out Note: Hout= constant, since gas leaving engine is of constant properties. Thus + = - - = - - + = - - + = = - + W N U N U N H N U N U N H N H N C N C T N C T N C T T T N C T C T N RT S f f i i f f i i f i f i i f i i f f i i i D out out out V V P out P out out V P but ; B = - = - - UVW reference temperature see Eqn. (2.4-8)P V H C T T U C T T RT 0 0 0 a f a f Now PV NRT N PV RT = Þ = N N W i f s = ´ ´ ´ × = = ´ ´ ´ = = -( ) ´ - ´ + ´ ´ = - - 175 29515 831410 03566 kmol 1 13352 831410 004504 00450430 8314 13352 30 29515 03566831429515 6066 kJ 2 2 . . . . . . . . . . . . . . . bar0.5 m K barm kmol K bar0.5 m K kmol 3 3 3 Since WS >0, work must be put into the engine if the outlet temperature is to be maintained at 22°C. (Alternatively, heat could be added and work extracted.) We should check to see if the process considered above is indeed possible. Can do this by using the entropy balance and ascertaining whether Sgen³ 0. Entropy balance N S N S N N S S N C T T N R P P N C T T N R P P S f f i i f i f f f f i i i i - = - + Þ - - + = c h out gen P out out P out out genln ln ln ln But T Tiout K= = 29515. , and P Pfout bar= =1 so N C T T N R P P Sf f i i i f P genln ln+ = or Sgen kJK= ´ + ´ =00450530 13352 29515 035668314 175 1 7414. ln . . . . ln . . Thus, Sgen> 0, and the process is possible! (d)Similar process, but now isothermal: system = gas in tank and engine. M.B.: N N Nf i- = D E.B.: N U N U N H Q W N N H Q Wf f i i s f i s- = + + = - + +D out outc h S.B.: N S N S N S Q T S N N S Q T Sf f i i f i- = + + = - + +D out gen out genc h Set Sgen= 0, since we want maximum work (see Sec. 3.2). Thus Q T N S N S N N T S TN S S TN S S N TR P P N RT P P f f i i f i f f i i f f i i = - - - = - - - = - + d i c h d i a f out out out out out ln ln But P Pf = =out bar1 and Q W N U N U N H N H Q N RT N RT Q N N RT Q s f f i i f i f i i f = ´ ´ FH IK = = - - + - = - + - = - - 03566831429515 175 1 25046 kJ. . . ln . . out out c h but N PV RTf = = ´ ´ ´ = ´- -1 05 29515 831410 2038102 2. . . . kmol and Ws = -( ) ´ ´ - = -0356600204831429515 25046 16796 kJ. . . . . . . In this case we obtain work! 3.42 a) For each stage of the compressor, assuming steady-state operation and reversible adiabatic operation we have from the mass, energy and entropy balances, respectively 0 0 0 = + = - = - = + + = - = + = & & & & & & $ & $ & & & ($ $ ) & $ & $ $ $ M M M M M M H M H W W M H H M S M S S S in out out in in in out out out in in in out out out in or or and or So through each compressor (but not intercooler) stage, one follows a line on constant entropy in Fig. 2.4-2 Therefore, for first compressor stage we have $ ( ) $ ( ) . $ ($ . ) &( H T P S T P H S P T W in in out out = = = = = = = = = = = - = 200 K,1 767 200 K,1 65 65 5 963 295 963767 bar kJ/kg and bar kJ/kg K kJ/kg K, bar kJ/kg and K Therefore the first stage work per kg. of methane flowing through the compressor is first stage) kJ/kg 196 kJ/kg After cooling, the temperature of the methane stream is 200 K, so that for the second compressor stage we have $ ( ) $ ( ) . $ ($ . ) &(sec H T P S T P H S P T W in in out out = = = = = = = = = = = - = 200 K,5 760 kJ 200 K,5 565 565 25 960 kJ 300 K 960760 kJ bar /kg and bar kJ/kg K kJ/kg K, bar /kg and Therefore the second stage work per kg. of methane flowing through the compressor is ond stage) /kg 200 kJ/kg Similarly, after intercooling, the third stage compressor work is found from $ ( ) $ ( ) . $ ($ . ) &( H T P S T P H S P T W in in out out = = = = = = = = = = = - = 200 K,25 718 kJ 200 K,25 465 465 100 bar855 288 K 855718 kJ bar /kg and bar kJ/kg K kJ/kg K, kJ/kg and Therefore the third stage work per kg. of methane flowing through the compressor is third stage) /kg 137 kJ/kg Consequently the total compressor work through all three stages is &W = + + =196200137 533 kJ/kg b) The liquefaction process is a Joule-Thomson expansion, and therefore occurs at constant enthalpy. The enthalpy of the methane leaving the cooler at 100 bar and 200 K is 423 kJ/kg. At 1 bar the enthalpy of the saturate vapor is 582 kJ/kg, and that of the liquid is 71 kJ/kg. Therefore from the energy balance on the throttling valve and flash drum we have $ $ $( ) ( )$( $( ( ) H H H x H xH x x in out= = - = - × + × or sat'd. vapor, 1 bar) + sat'd. liquid, 1 bar) 423 kJ kg kJ kg kJ kg 200 K,100 bar1 1 71 582 where x = 0.689 is the fraction of vapor leaving the flash drum, and (1-x) = 0.311 is the fraction of the methane that has been liquefied. Therefore, for each kilogram of methane that enters the simple liquefaction unit, 689 grams of methane are lost as vapor, and only 311 grams of LNG are produced. Further, since 533 kJ of work are required in the compressor to produce 311 grams of LNG, approximately 1713 kJ of compressor work are required for each kg. of LNG produced. c) As in the illustration, we choose the system for writing balance equations to be
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