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8 - Prob 6.2 PROBLEM 6.2 PROBLEM STATEMENT Argon(a...

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Prob 6.2 PROBLEM 6.2 PROBLEM STATEMENT: Argon (a monatomic gas with c P =0.125 Btu/lbm-°R) enters a compressor at 10 psia and 100°F and exits at 60 psia and 500°F. If the mass flow rate is 3 lbm/s, determine the rate of entropy increase in the compressor, (Btu/°R-s). m s & DIAGRAM DEFINING SYSTEM AND PROCESS: m = 3 lbm/s & Comp. 1 2 P 2 = 60 psia T 2 = 500°F P 1 = 10 psia T 1 = 100°F GIVEN: Argon, c P = 0.125 Btu/lbm-°R, m = 3 lbm/s & State 1: P 1 = 10 psia , T 1 = 100°F = 560°R State 2: P 2 = 60 psia , T 2 = 500°F = 960°R FIND: Rate of entropy increase, m s & (Btu/°R-s). ASSUMPTION: Ideal gas and constant specific heats. GOVERNING RELATIONS: 2 P 1 1 T s = c ln Rln T P 2 P for IG with constant specific heats QUANTITATIVE SOLUTION: Since all the required quantities are known, we can now compute the entropy increase as follows: 2 2 P 1 1 T P s = c ln Rln T P Btu 960 1.986 Btu/lbmol °R 60 = 0.125 ln ln lbm °R 560 39.95 lbm/lbmol 10 = -0.0217 Btu/lbm °R ± ± ± Thus, the rate of entropy increase in the compressor is ( ) lbm Btu m s = 3 -0.0217 s lbm °R = -0.065 Btu/s °R Negative m s! Is this possible? × & ± & ± DISCUSSION OF RESULTS: The resulting negative entropy change is only possible if there is heat transfer out of the compressor. Since the problem did not state whether there was heat transfer or not, the answer is possible.
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Prob 6.2 PROBLEM 6.3 PROBLEM STATEMENT: What is the change in specific entropy of water (Btu/lbm- ° R) when it is cooled isobarically from 200 psia and 500 ° F to a final specific volume of 2 ft 3 /lbm, and what is the work (Btu/lbm) for the process? DIAGRAM DEFINING SYSTEM AND PROCESS: P 2 = 200 psia v 2 = 2 ft 3 /lbm P 1 = 200 psia T 1 = 500°F GIVEN: H 2 O, isobaric cooling State 1: P 1 = 200 psia , T 1 = 500°F = 960°R State 2: P 2 = 200 psia , v 2 = 2 ft 3 /lbm FIND: and ( ) s Btu/lbm °R ± OUT,1-2 w (Btu/lbm). GOVERNING RELATIONS: Work, w = for isobaric process ( 2 1 v OUT,1-2 2 1 v P dv = P v - v ) QUANTITATIVE SOLUTION: From Table 12e, the specific entropy and volume at State 1 are 1 3 1 s =1.6235 Btu/lbm °R v = 2.7232 ft /lbm ± From Table 11e at O=200 psia it can be seen that State 2 exists as a liquid- vapor mixture since ( ) ( ) ( ) L 2 V v 0.01839 v 2.0 v 2.2883 < < The quality at State 2 can thus be calculated 2 2.0 - 0.01839 x = = 0.873 2.2883 - 0.01839 Thus, the specific entropy at State 2 is ( ) 2 L 2 LV s = s + x s = 0.5440+ 0.873 1.0022 =1.4189 Btu/lbm °R × ± The change in specific entropy is 2 1 s = s - s =1.4189 -1.6235 = -0.2046 Btu/lbm °R ± Work for the process is ( ) ( ) ( ) × × ± OUT,1-2 2 1 3 2 2 2 OUT,1-2 w =P v - v lbf ft Btu 144 in = 200 2.0 - 2.7232 lbm 778.16 ft lbf in ft w = -26.8 Btu/lbm negative value indicates work into the system
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Prob 6.2 DISCUSSION OF RESULTS: Note that the negative change in entropy is possible since there was cooling of the H 2 O during the process.
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