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Lecture Companion site to accompany thermodynamics: An engineering approach (7/e): Chapter 7.2 - Yunus Çengel, Michael A. Boles

Chapter 7.2 - Entropy: A measure of disorder. The objectives of this chapter are to: Derive the reversible steady-flow work relations; develop the isentropic, or adiabatic, efficiencies of various steady-flow engineering devices and apply the definitions to turbines, compressors, and nozzles; introduce and apply the entropy balance to various systems. | Chapter 7 Continued Entropy: A Measure of Disorder Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 7th edition by Yunus A. Çengel and Michael A. Boles Example 7-9 Air, initially at 17oC, 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 For air, k = 1.4, and a pressure ratio of 8:1 means that P2/P1 = 8 b.Variable specific heat method Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = 1.2311. Interpolating in the air table at this value of Pr2, gives T2 = 522.4 K = 249.4oC c. A second variable specific heat method. Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = 1.66802 kJ/kg K. b. Variable specific heat method For the isentropic process At this value of soT2, the air table gives T2 = 522.4 K= 249.4oC. This | Chapter 7 Continued Entropy: A Measure of Disorder Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 7th edition by Yunus A. Çengel and Michael A. Boles Example 7-9 Air, initially at 17oC, 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 For air, k = 1.4, and a pressure ratio of 8:1 means that P2/P1 = 8 b.Variable specific heat method Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = 1.2311. Interpolating in the air table at this value of Pr2, gives T2 = 522.4 K = 249.4oC c. A second variable specific heat method. Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = 1.66802 kJ/kg K. b. Variable specific heat method For the isentropic process At this value of soT2, the air table gives T2 = 522.4 K= 249.4oC. This technique is based on the same information as the method shown in part b. d.Using the EES software with T in oC and P in kPa and assuming P1 = 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 = 249.6oC Example 7-10 Air initially at 0.1 MPa, 27oC, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is 0.5 MPa, 227oC. (b) Find the entropy change when the final state is 0.5 MPa, 180oC. (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. b. c. c b a 1 s T P1 P2 2 The T-s plot is Give an explanation for the difference in the signs for the entropy changes. 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 .