Chapter 7 (part 1) - Entropy: A measure of disorder. In this chapter, the learning objectives are: Examine a special class of idealized processes, called isentropic processes, and develop the property relations for these processes; derive the reversible steady-flow work relations; develop the isentropic efficiencies for various steady-flow devices; 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, 8th edition by Yunus A. Çengel and Michael A. Boles 53 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 = , and a pressure ratio of 8:1 means that P2/P1 = 8 specific heat method 54 Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = . Interpolating in the air table at this value of Pr2, gives T2 = K = c. A second variable specific heat method. Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = kJ/kg K. b. Variable specific heat method For the isentropic process 55 At this value of soT2, the air table gives T2 = K= . This . | Chapter 7 Continued Entropy: A Measure of Disorder Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 8th edition by Yunus A. Çengel and Michael A. Boles 53 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 = , and a pressure ratio of 8:1 means that P2/P1 = 8 specific heat method 54 Using the air data from Table A-17 for T1 = (17+273) K = 290 K, Pr1 = . Interpolating in the air table at this value of Pr2, gives T2 = K = c. A second variable specific heat method. Using the air table, Table A-17, for T1 = (17+273) K = 290 K, soT1 = kJ/kg K. b. Variable specific heat method For the isentropic process 55 At this value of soT2, the air table gives T2 = K= . This technique is based on the same information as the method shown in part b. 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 = kJ/kg K s_2 = kJ/kg K T_2 = Example 7-10 Air initially at MPa, 27oC, is compressed reversibly to a final state. (a) Find the entropy change of the air when the final state is MPa, 227oC. (b) Find the entropy change when the final state is MPa, 180oC. (c) Find the temperature at MPa that makes the entropy change zero. Assume air is an ideal gas with constant specific heats. 56 Show the two processes on a T-s diagram. a. b. 57 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. 58 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
