Chapter_7_2

Chapter_7_2 - Student Study Guide for 5th edition of...

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Student Study Guide for 5 th edition of Thermodynamics by Y. A. Ç engel & M. A. Boles 7-41 Chapter 7 - 41 Example 7-9 Air, initially at 17 o C, is compressed in an isentropic process through a pressure ratio of 8:1. Find the final temperature assuming constant specific heats and variable specific heats, and using EES. a. Constant specific heats, isentropic process T T P P s const kk 2 1 2 1 1 F H G I K J = F H G I K J = . () / For air, k = 1.4, and a pressure ratio of 8:1 means that P 2 / P 1 = 8. TT P P K KC o 21 2 1 1 14 1 14 17 273 8 5253 252 3 = F H G I K J =+ == / (. ) /. .. bg b. Variable specific heat method P P PP P P s const ref ref s const r r 2 1 2 1 2 1 F H G I K J = F H G I K J = = = . . / / Using the air data from Table A-17 for T 1 = (17+273) K = 290 K, P r 1 = 1.2311. P P rr 2 1 12311 9 8488 = .( 8 ) . Interpolating in the air table at this value of P r 2 , gives T 2 = 522.4 K = 249.4 o C
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Student Study Guide for 5 th edition of Thermodynamics by Y. A. Ç engel & M. A. Boles 7-42 Chapter 7 - 42 c. A second variable specific heat method. Using the air table, Table A-17, for T 1 = (17+273) K = 290 K, s o T1 = 1.66802 kJ/kg K. For the isentropic process ssR P P kJ kg K kJ kg K kJ kg K oo 21 2 1 166802 0 287 8 2 26482 =+ = + = ln .. l n . bg At this value of s o T 2 , the air table gives T 2 = 522.4 K= 249.4 o C. This technique is based on the same information as the method shown in part b. d. Using the EES software with T in o C and P in kPa and assuming P 1 = 100 kPa. s_1 = ENTROPY(Air, T=17, P=100) s_2 = s_1 T_2 = TEMPERATURE(Air, P=800, s=s_2) The solution is: s_1 = 5.668 kJ/kg K s_2 = 5.668 kJ/kg K T_2 = 249.6 o C
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Student Study Guide for 5 th edition of Thermodynamics by Y. A. Ç engel & M. A. Boles 7-43 Chapter 7 - 43 Example 7-10 Air initially at 0.1 MPa, 27 o C, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is 0.5 MPa, 227 o C. (b) Find the entropy change when the final state is 0.5 MPa, 180 o C. (c) Find the temperature at 0.5 MPa that makes the entropy change zero. Assume air is an ideal gas with constant specific heats. Show the two processes on a T-s diagram. a. ssC T T R P P kJ kg K K K kJ kg K MPa MPa kJ kg K pav 21 2 1 2 1 1005 227 273 27 273 0 287 05 01 0 0507 −= = + + F H G I K J F H G I K J =+ , ln ln .l n () n . . . b. T T R P P kJ kg K K K kJ kg K MPa MPa kJ kg K 2 1 2 1 180 273 27 273 0 287 0 0484 = + + F H G I K J F H G I K J =− , ln ln n n . . .
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Student Study Guide for 5 th edition of Thermodynamics by Y. A. Ç engel & M. A. Boles 7-44 Chapter 7 - 44 c. TT P P K MPa MPa KC kk o 21 2 1 1 14 1 14 27 273 05 01 4754 202 4 = F H G I K J =+ F H G I K J == () / (. ) /. . . .. The T-s plot is Give an explanation for the difference in the signs for the entropy changes. c b a 1 s T P 1 P 2 2
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Student Study Guide for 5 th edition of Thermodynamics by Y. A. Ç engel & M. A. Boles 7-45 Chapter 7 - 45 Example 7-11 Nitrogen expands isentropically in a piston cylinder device from a temperature of 500 K while its volume doubles. What is the final temperature of the nitrogen, and how much work did the nitrogen do against the piston, in kJ/kg?
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Chapter_7_2 - Student Study Guide for 5th edition of...

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