1. ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

2. Increase-in-Entropy Principle ΔS total = ΔS system + ΔS surroundings ≥ 0 Closed systems

3. Example 1 Water in a tank, initially a saturated liquid at 100 ºC, undergoes a process to the corresponding saturated vapor state. There is no heat transfer with the surroundings during the process. If the change of states is brought about by the action of a paddle wheel, determine the work required per mass and the total entropy change for the process.

4. Example 1 (continued) = - (2506.5 – 418.94) Table A-4, T = 100 ºC, u f = 418.94 kJ/kg, u g = 2506.5 kJ/kg = - 2087.56 kJ/kg ΔU = Q – W Δs = s g – s f = 7.3549 – 1.3069 = 6.048 kJ/kg K s f = 1.3069 kJ/kg K, s g = 7.3549 kJ/kg K

5. Increase-in-Entropy Principle Control volumes (a) Steady-Flow Processes

6. Example 2 Feedwater Heater: Inlet 1 T 1 = 200 ºC, p 1 = 700 kPa, Inlet 2 T 2 = 40 ºC, p 2 = 700 kPa, Exit sat. liquid, p 3 = 700 kPa. Find Inlet 1 Inlet 2 Exit

7. Example 2 (continued) Inlet 1: Superheated vapor Table A-6,s 1 = 6.8912 kJ/kg K Inlet 2: compressed liquid Table A-4,s 2 = 0.6387 kJ/kg K

8. Example 2 (continued) Exit: saturated liquid Table A-5,s 3 = 2.0200 kJ/kg K (1 + 4.06)(2.0200) = 0.7369 kJ/kg K – [(4.06)(0.6387) + (1)(6.8912)]

9. Increase-in-Entropy Principle ΔS total = ΔS CV + ΔS surroundings ≥ 0 Control volumes (b) Uniform-Flow Processes

10. Example 3 A 0.4-m 3 rigid tank is filled with saturated liquid water 200 ºC. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water from a source at 250 ºC so that the temperature in the tank remains constant. Determine (a) the amount of heat transfer by the time one-half of the water (in mass) has been withdrawn. (b) the total entropy change for this process.

11. Example 3 (continued) Sat. liquid T 1 = 200 ºC Q Discharging Process Uniform-Flow Process

12. Example 3 (continued) No inlet, one outlet m e = (m 1 – m 2 ) CV Initial state: Sat. liquid at T 1 = 200 ºC Table A-4, v 1 = 0.001157 m 3 /kg, h 1 = 850.65 kJ/kg s 1 = 2.3309 kJ/kg K Exit: Sat. liquid at T 1 = 200 ºC Table A-4, h 1 = 850.65 kJ/kg, s 1 = 2.3309 kJ/kg K

13. Example 3 (continued) Final state: Sat. mixture at T 2 = 200 ºC m 2 = ½ m 1, v 2 = 2v 1 v 2 = 2v 1 = 2(0.001157) = 0.002314 m 3 /kg u 2 = u f + x 2 u fg = 850.65 + (0.00917)(1744.7) = 866.65 kJ/kg

14. Example 3 (continued) s 2 = s f + x 2 s fg = 2.3309 + (0.00917)(4.1014) = 2.3685 kJ/kg K + (m 2 u 2 – m 1 u 1 ) CV Q CV = m e h e + (m 2 u 2 – m 1 u 1 ) CV = (172.86)(852.45) + (172.86)(866.65) - (345.72)(850.65) = 3077 kJ

15. Example 3 (continued) ΔS total = (172.86)(2.3309) + (172.86)(2.3685) - (345.72)(2.3309) = 0.616 kJ/K

16. Example 4 A quantity of air undergoes a thermodynamic cycle consisting of three processes in series. Process 1-2: constant-volume heating from p 1 = 0.1 MPa, T 1 = 15 ºC, and V 1 = 0.02 m 3, to p 2 = 0.42 MPa. Process 2-3: constant-pressue cooling. Process 3-1: isothermal heating to the initial state. Employ the ideal gas model with c p = 1 kJ/kg·K, calculate the change of entropy for each process.

17. Example 4 (continued) p v 1 23

18. State 1: p 1 = 0.1 MPa, T 1 = 15 ºC, and V 1 = 0.02 m 3 State 3: p 3 = p 2 = 0.42 MPa, T 3 = T 1 = 15 ºC State 2: p 2 = 0.42 MPa, V 2 = V 1 = 0.02 m 3 c v = c p – R = 1 – 0.287 = 0.713 kJ/kg·K

19. Example 4 (continued) = 1.023 kJ/kg·K = - 1.435 kJ/kg·K

20. Example 4 (continued) = 0.412 kJ/kg·K Δs total = Δs 12 + Δs 23 + Δs 31 = 1.023 – 1.435 + 0.412 = 0