ENGR 2213 Thermodynamics

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

2. Increase-in-Entropy Principle Closed systems ΔStotal = ΔSsystem + ΔSsurroundings ≥ 0

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) Table A-4, T = 100 ºC, uf = 418.94 kJ/kg, ug = 2506.5 kJ/kg ΔU = Q – W sf = 1.3069 kJ/kg K, sg = 7.3549 kJ/kg K = - (2506.5 – 418.94) = - 2087.56 kJ/kg Δs = sg – sf = 7.3549 – 1.3069 = 6.048 kJ/kg K

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

6. Example 2 Feedwater Heater: Inlet 1 T1 = 200 ºC, p1 = 700 kPa, Inlet 2 T2 = 40 ºC, p2 = 700 kPa, Exit sat. liquid, p3 = 700 kPa. Find Inlet 2 Inlet 1 Exit

7. Example 2 (continued) Inlet 1: Superheated vapor Table A-6, s1 = 6.8912 kJ/kg K Inlet 2: compressed liquid Table A-4, s2 = 0.6387 kJ/kg K

8. Example 2 (continued) Exit: saturated liquid Table A-5, s3 = 2.0200 kJ/kg K (1 + 4.06)(2.0200) – [(4.06)(0.6387) + (1)(6.8912)] = 0.7369 kJ/kg K

9. Increase-in-Entropy Principle Control volumes (b) Uniform-Flow Processes ΔStotal = ΔSCV + ΔSsurroundings ≥ 0

10. Example 3 A 0.4-m3 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 A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank.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 Determine (a) the amount of heat transfer by the time one-half of the water (in mass) has been withdrawn.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) Discharging Process Uniform-Flow Process Sat. liquid T1 = 200 ºC Q

12. Example 3 (continued) No inlet, one outlet me = (m1 – m2)CV Initial state: Sat. liquid at T1 = 200 ºC Table A-4, v1 = 0.001157 m3/kg, h1 = 850.65 kJ/kg s1 = 2.3309 kJ/kg K Exit: Sat. liquid at T1 = 200 ºC Table A-4, h1 = 850.65 kJ/kg, s1 = 2.3309 kJ/kg K

13. Example 3 (continued) Final state: Sat. mixture at T2 = 200 ºC m2 = ½ m1, v2 = 2v1 v2 = 2v1 = 2(0.001157) = 0.002314 m3/kg u2 = uf + x2ufg = 850.65 + (0.00917)(1744.7) = 866.65 kJ/kg

14. Example 3 (continued) s2 = sf + x2sfg = 2.3309 + (0.00917)(4.1014) = 2.3685 kJ/kg K + (m2u2 – m1u1)CV QCV = mehe + (m2u2 – m1u1)CV = (172.86)(852.45) + (172.86)(866.65) - (345.72)(850.65) = 3077 kJ

15. Example 3 (continued) ΔStotal = (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 p1 = 0.1 MPa, T1 = 15 ºC, and V1 = 0.02 m3, to p2 = 0.42 MPa. Process 2-3: constant-pressue cooling. Process 3-1: isothermal heating to the initial state. Employ the ideal gas model with cp = 1 kJ/kg·K, calculate the change of entropy for each process.

17. p 3 2 1 v Example 4 (continued)

18. Example 4 (continued) State 1: p1 = 0.1 MPa, T1 = 15 ºC, and V1 = 0.02 m3 State 2: p2 = 0.42 MPa, V2 = V1 = 0.02 m3 State 3: p3 = p2 = 0.42 MPa, T3 = T1 = 15 ºC cv = cp – 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 Δstotal = Δs12 + Δs23 + Δs31 = 1.023 – 1.435 + 0.412 = 0