ENGR 2213 Thermodynamics

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

2. Exergy (Availability) - Work potential, maximum useful work The term “availability” was made popular in the States by MIT School of Engineering in the 1940s. An equivalent work “exergy” was introduced in Europe in the 1950s. Dead State A system is said to be in the dead state when it is in thermodynamic equilibrium with its surroundings.

3. Exergy At the dead state, 1. A system is at the same temperature and pressure of its surroundings (in thermal and mechanical equilibrium). 2. There are no unbalanced magnetic, electrical and surface tension effects between the system and its surroundings (mechanical equilibrium). 3. It has zero kinetic and potential energy relative to its surroundings (zero velocity and zero elevation above a reference level).

4. Exergy At the dead state, 4. It does not react with the surroundings (chemically inert). Work = f (initial state, process path, final state) To maximize the work output ►the process to be executed is reversible. ►the final state is a dead state.

5. Exergy A system will deliver the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its surroundings, i.e., the dead state. Exergy does not represent the amount of work that a work-producing device will actually deliver upon installation. Rather, it represents the upper limit on the amount of work a device can deliver without violating any thermodynamic laws.

6. Exergy The difference between exergy and the actual work delivered by a device represents the room Engineers have for improvement. Exergy of a system at a specified state depends on the conditions of the surroundings (the dead state) as well as the properties of the system. Exergy is a property of the combined system (the system and its surroundings) and not of the system alone.

7. Vapor Power Cycles Working Fluids High temperature: sodium, potassium, mercury. Low temperature: benzene, freon. Water: low cost, availability, high enthalpy of vaporization. Fuel Types Coal, natural gas, nuclear, geothermal.

8. T 1 2 4 3 S Vapor Power Cycles Carnot Power Cycle 1. Process 2-3 The quality of steam at the turbine exit may be too low. 2. Process 4-1 The pump has to deal with two-phase flows.

9. 1 2 T 3 4 S Vapor Power Cycles Carnot Power Cycle 1. Process 1-2 Isothermal heat transfer at variable pressure.

10. T 3 2 1 4 S Vapor Power Cycles Rankine Cycle Rankine cycle is the ideal cycle for vapor power plants. State 1: saturated liquid State 2: compressed liquid State 3: superheated vapor State 4: saturated mixture

11. 3 T 2 Boiler 3 2 Pump Turbine 1 4 S 1 4 Condenser Ideal Rankine Cycles Process 1-2: isentropic compression in a pump Process 2-3: constant-pressure heat addition in a boiler Process 3-4: isentropic expansion in a turbine Process 4-1: constant-pressure heat rejection in a condenser

12. 3 T 2 1 4 S Ideal Rankine Cycles wp = h2 – h1 = v(p2 – p1) qin = h3 – h2 wt = h3 – h4 qout = h4 – h1

13. Example 1 • Consider a steam power plant operating on the • simple ideal Rankine cycle. The steam enters • the turbine at 3 MPa and 350 ºC and is condensed • in the condenser at a pressure of 75 kPa. • Determine • the thermal efficiency of this cycle. • (b) the back work ratio of this cycle.

14. Example 1 (continued) State 1: saturated liquid at p1 = 75 kPa Table A-5 h1 = hf = 384.39 kJ/kg v1 = vf = 0.001037 m3/kg State 2: compressed liquid at p2 = 3 MPa wp = v(p2 – p1) = (0.001037)(3000-75) = 3.03 kJ/kg h2 = h1 + wp = 384.39 + 3.03 = 387.42 kJ/kg

15. Example 1 (continued) State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K State 4: saturated mixture at p4 = 75 kPa s4 = s3 = sf + x4sfg

16. Example 1 (continued) State 4: saturated mixture at p4 = 75 kPa h4 = hf + x4hfg = 384.39 + 0.886(2278.6) = 2403.2 kJ/kg qin = h3 – h2 = 3115.3 – 387.42 = 2727.88 kJ/kg qout = h4 – h1 = 2403.2 – 384.39 = 2018.81 kJ/kg

17. Example 1 (continued) wt = h3 – h4 = 3115.3 – 2403.2 = 712.1 kJ/kg wnet = wt – wp = 712.1 – 3.03 = 709.07 kJ/kg

18. 3 T 2 1 4 S Real Rankine Cycles Two most common sources of irreversibilities ► Fluid friction ► Undesired heat loss to the surroundings

19. 3 T 3’ 2’ 2 4’ 1 4 S Real Rankine Cycles Process 1-2’ Irreversibilities in the pump Process 2’-3’ Pressure drop due to friction in the boiler Process 3’-4’ Irreversibilities in the turbine Process 4’-1’ Pressure drop due to friction in the condenser

20. T 3 2’ 2 1 4 4’ S Real Rankine Cycles Efficiency of Pump h2’ = (h2 – h1)/ηp + h1 Efficiency of Turbine h4’ = h3 – ηp(h3 – h4)

21. Example 2 Consider a steam power plant operating on the simple ideal Rankine cycle. The steam enters the turbine at 3 MPa and 350 ºC and is condensed in the condenser at a pressure of 75 kPa. Given that ηp = ηt = 0.85, determine the thermal efficiency of this cycle.

22. Example 2 (continued) State 1: saturated liquid at p1 = 75 kPa Table A-5 h1 = hf = 384.39 kJ/kg v1 = vf = 0.001037 m3/kg State 2: compressed liquid at p2 = 3 MPa ws = v(p2 – p1) = (0.001037)(3000-75) = 3.03 kJ/kg h2’ = h1 + wp/ηp = 384.39 + 3.03/0.85 = 387.95 kJ/kg

23. Example 2 (continued) State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K State 4: saturated mixture at p4 = 75 kPa ws = 712.1 kJ/kg h4’ = h3 – ηtws = 3115.3 – 0.85(712.1) = 2510 kJ/kg

24. Example 2 (continued) qin = h3 – h2’ = 3115.3 – 387.95 = 2727.35 kJ/kg qout = h4’ – h1 = 2510 – 384.39 = 2125.61 kJ/kg

25. T S Increase the Efficiency of a Rankine Cycle 1. Lowering the condenser pressure 2. Superheating the steam to a higher temperature 3. Increasing the boiler pressure

26. Example 3 Consider a steam power plant operating on the simple ideal Rankine cycle. The steam enters the turbine at 3 MPa and 350 ºC and is condensed in the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this cycle, (a) if the condenser pressure is lowered to 10 kPa (b) in addition to the change in (a) if the steam is superheated to 600 ºC (c) in addition to the change in (b) if the boiler pressure is raised to 15 MPa

27. Example 3 (continued) (a) State 1: saturated liquid at p1 = 10 kPa Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg State 2: compressed liquid at p2 = 3 MPa wp = v(p2 – p1) = (0.001008)(3000-10) = 3.01 kJ/kg h2 = h1 + wp = 191.83 + 3.01 = 194.84 kJ/kg

28. Example 3 (continued) State 3: superheated vapor at p3 = 3 MPa and T3 = 350 ºC Table A-6 h3 = 3115.3 kJ/kg s3 = 6.7428 kJ/kg·K State 4: saturated mixture at p4 = 10 kPa s4 = s3 = sf + x4sfg

29. Example 3 (continued) (a) State 4: saturated mixture at p4 = 10 kPa h4 = hf + x4hfg = 191.83 + 0.812(2392.8) = 2134.8 kJ/kg qin = h3 – h2 = 3115.3 – 194.84 = 2920.46 kJ/kg qout = h4 – h1 = 2134.8 – 191.83 = 1942.97 kJ/kg

30. Example 3 (continued) (b) State 1 and State 2 remain the same State 3: superheated vapor at p3 = 3 MPa and T3 = 600 ºC Table A-6 h3 = 3682.3 kJ/kg s3 = 7.5085 kJ/kg·K State 4: saturated mixture at p4 = 10 kPa s4 = s3 = sf + x4sfg

31. Example 3 (continued) (b) State 4: saturated mixture at p4 = 10 kPa h4 = hf + x4hfg = 191.83 + 0.914(2392.8) = 2378.8 kJ/kg qin = h3 – h2 = 3682.3 – 194.84 = 3487.46 kJ/kg qout = h4 – h1 = 2378.8 – 191.83 = 2186.97 kJ/kg

32. Example 3 (continued) (c) State 1 remains the same State 2: compressed liquid at p2 = 15 MPa wp = v(p2 – p1) = (0.001008)(15000-10) = 15.11 kJ/kg h2 = h1 + wp = 191.83 + 15.11 = 206.94 kJ/kg State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC

33. Example 3 (continued) State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC Table A-6 h3 = 3582.3 kJ/kg s3 = 6.6776 kJ/kg·K State 4: saturated mixture at p4 = 10 kPa s4 = s3 = sf + x4sfg

34. Example 3 (continued) (c) State 4: saturated mixture at p4 = 10 kPa h4 = hf + x4hfg = 191.83 + 0.804(2392.8) = 2115.7 kJ/kg qin = h3 – h2 = 3582.3 – 206.94 = 3375.36 kJ/kg qout = h4 – h1 = 2115.7 – 191.83 = 1923.87 kJ/kg