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

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

3. Ideal Regenerative Rankine Cycles Closed Feedwater Heater 1 2 3 7 4 5 6 8 Trap TurbineBoiler Condenser P FWH y 1-y T S 1 2 3 4 5 7 68 y 1-y

4. Ideal Regenerative Rankine Cycles q in = h 4 – h 3 q out = (1 – y)(h 6 – h 1 ) + y(h 8 – h 1 ) w t = (1 – y)(h 5 – h 6 ) + (h 4 – h 5 ) w p = h 2 – h 1 = v 1 (p 2 – p 1 ) T S 1 2 3 4 5 7 68 y 1-y y(h 5 – h 7 ) = h 3 – h 2 = yh 8 + (1 – y)h 6 – h 1

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 (a) the fraction of steam extracted from the turbine, (b) the thermal efficiency of this cycle.

6. Example 1 (continued) State 2: compressed liquid at p 2 = 12 MPa w p1 = v(p 2 – p 1 ) = (0.001006)(12000-6) = 12.07 kJ/kg h 2 = h 1 + w p1 = 151.53 + 12.07 = 163.6 kJ/kg Table A-5 h 1 = h f = 151.53 kJ/kg v 1 = v f = 0.001006 m 3 /kg State 1: saturated liquid at p 1 = 6 kPa

7. Example 1 (continued) State 3: compressed liquid at p 3 = 12 MPa and T 3 = 170 ºC Table A-7 h 3 = 725.86 kJ/kg State 4: superheated vapor at p 4 = 12 MPa and T 4 = 520 ºC Table A-6 h 4 = 3401.8 kJ/kg s 4 = 6.5555 kJ/kg·K State 5: p 5 = 1 MPa and s 5 = s 4 = 6.5555 kJ/kg·K Table A-5 s f = 2.1387 kJ/kg·K s g = 6.5865 kJ/kg·K

8. Example 1 (continued) State 5: saturated mixture at p 5 = 1 MPa h 5 = h f + x 5 h fg = 762.81 + 0.993(2015.3) = 2764.2 kJ/kg State 6: saturated mixture at p 6 = 6 kPa, s 6 = s 4 h 6 = h f + x 6 h fg = 151.53 + 0.773(2415.9) = 2018.3 kJ/kg

9. Example 2 (continued) State 7: saturated liquid at p 7 = 1 MPa State 8: p 8 = 6 kPa and h 8 = h 7 Table A-5 h 5 = 762.81 kJ/kg (b) q in = h 4 – h 3 = 3401.8 – 725.86 = 2675.9 kJ/kg (a) y(h 5 – h 7 ) = h 3 – h 2

10. Example 1 (continued) w t = (h 4 – h 5 ) + (1 – y)(h 5 – h 6 ) = (3401.8 – 2764.2) + (1 – 0.281)(2764.2 – 2018.3) = 1174.0 kJ/kg q out = yh 8 + (1 – y)h 6 – h 1 = 0.281(762.81) + (1 – 0.281)2018.3 – 151.53 = 1514.1 kJ/kg

11. Example 1 (continued) w net = w t – w p = 1174.0 – 12.07 = 1161.9 kJ/kg

12. Ideal Regenerative Rankine Cycles Open Feedwater Heater Closed Feedwater Heaters 1. Simple Complex 2. Inexpensive to build Expensive 3. Good heat transfer Less effective in heat characteristics transfer 4. Require a separate Do not require a separate pump for each heater pump for each heater

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

14. Ideal Regenerative Rankine Cycles Combined Feedwater Heaters 7 1 2 5 3 4 8 6 y 1-y 9 FWH P2P1 TurbineBoiler FWH Condenser 5 T S 1 2 3 4 7 6 8 y 1-y 9 w p1 = h 2 – h 1 = v 1 (p 2 – p 1 ) w p2 = h 4 – h 3 = v 3 (p 4 – p 3 ) h 4 = h 3 + v 3 (p 4 – p 3 ) h 9 = h 3 + v 3 (p 4 – p 3 ) p 4 = p 5 = p 9, T 9 = T 3 h 4 = h 5 = h 9 y(h 7 – h 3 ) = (1-y)(h 9 – h 2 )