ENGR 2213 Thermodynamics

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

2. 4 5 Boiler 3 6 P 2 Turbine FWH 5 2 T 7 Condenser P 1 4 1 6 3 2 1 7 S Ideal Regenerative Rankine Cycles Open Feedwater Heater

3. T 3 4 4 Boiler Turbine y 7 1-y 5 3 5 • 6 2 y 1-y Condenser 1 8 6 2 FWH 8 1 P 7 S Trap Ideal Regenerative Rankine Cycles Closed Feedwater Heater

4. T 4 7 5 3 • 2 y 1-y 1 8 6 S Ideal Regenerative Rankine Cycles wp = h2 – h1 = v1(p2 – p1) qin = h4 – h3 qout = (1 – y)(h6 – h1) + y(h8 – h1) = yh8 + (1 – y)h6 – h1 wt = (1 – y)(h5 – h6) + (h4 – h5) y(h5 – h7) = h3 – h2

5. Example 1 • Consider a steam power plant operating on the ideal • regenerative Rankine cycle. The steam enters the • turbine at 12 MPa and 520 ºC and is condensed in the • condenser at a pressure of 6 kPa. Some steam leaves • the turbine at a pressure of 1 MPa and enters the • closed feedwater heater. While condensate exits the • feedwater heater as saturated liquid at 1 MPa, the • feedwater exits the heater at a temperature of 170 ºC. • Determine • the fraction of steam extracted from the turbine, • the thermal efficiency of this cycle.

6. Example 1 (continued) State 1: saturated liquid at p1 = 6 kPa Table A-5 h1 = hf = 151.53 kJ/kg v1 = vf = 0.001006 m3/kg State 2: compressed liquid at p2 = 12 MPa wp1 = v(p2 – p1) = (0.001006)(12000-6) = 12.07 kJ/kg h2 = h1 + wp1 = 151.53 + 12.07 = 163.6 kJ/kg

7. Example 1 (continued) State 3: compressed liquid at p3 = 12 MPa and T3 = 170 ºC Table A-7 h3 = 725.86 kJ/kg State 4: superheated vapor at p4 = 12 MPa and T4 = 520 ºC Table A-6 h4 = 3401.8 kJ/kg s4 = 6.5555 kJ/kg·K State 5: p5 = 1 MPa and s5 = s4 = 6.5555 kJ/kg·K Table A-5 sf = 2.1387 kJ/kg·K sg = 6.5865 kJ/kg·K

8. Example 1 (continued) State 5: saturated mixture at p5 = 1 MPa h5 = hf + x5hfg = 762.81 + 0.993(2015.3) = 2764.2 kJ/kg State 6: saturated mixture at p6 = 6 kPa, s6 = s4 h6 = hf + x6hfg = 151.53 + 0.773(2415.9) = 2018.3 kJ/kg

9. Example 2 (continued) State 7: saturated liquid at p7 = 1 MPa Table A-5 h5 = 762.81 kJ/kg State 8: p8 = 6 kPa and h8 = h7 (a) y(h5 – h7) = h3 – h2 (b) qin = h4 – h3 = 3401.8 – 725.86 = 2675.9 kJ/kg

10. Example 1 (continued) qout = yh8 + (1 – y)h6 – h1 = 0.281(762.81) + (1 – 0.281)2018.3 – 151.53 = 1514.1 kJ/kg wt = (h4 – h5) + (1 – y)(h5 – h6) = (3401.8 – 2764.2) + (1 – 0.281)(2764.2 – 2018.3) = 1174.0 kJ/kg

11. Example 1 (continued) wnet = wt– wp = 1174.0 – 12.07 = 1161.9 kJ/kg

12. Ideal Regenerative Rankine Cycles Open Feedwater Heater Closed Feedwater Heaters 1. SimpleComplex 2. Inexpensive to buildExpensive 3. Good heat transferLess effective in heat characteristicstransfer 4. Require a separateDo not require a separate pump for each heaterpump for each heater

13. 6 5 6 T Boiler Turbine 1-y y 4 7 8 5 • FWH 7 • 9 y 3 2 1-y Condenser 2 1 8 4 9 FWH 1 3 P 2 P 1 S Ideal Regenerative Rankine Cycles Combined Feedwater Heaters

14. Ideal Regenerative Rankine Cycles 6 T Combined Feedwater Heaters 4 wp1 = h2 – h1 = v1(p2 – p1) 5 • 7 • 9 y 3 2 1-y wp2 = h4 – h3 = v3(p4 – p3) 8 1 p4 = p5 = p9, T9 = T3 S h4 = h3 + v3(p4 – p3) h4 = h5 = h9 h9 = h3 + v3(p4 – p3) 5 6 Boiler Turbine y(h7 – h3) = (1-y)(h9 – h2) 1-y y FWH 8 7 2 Condenser 4 9 FWH 1 P2 P1 3