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

2. Ideal Reheat Rankine Cycles T S 1 2 3 4 Turbine Boiler Condenser Pump 1 2 3 4 Boiler Condenser Pump 1 2 3 6 4 5 T S 1 2 3 4 5 6

3. Ideal Reheat Rankine Cycles q in = (h 3 – h 2 ) + (h 5 – h 4 ) q out = h 6 – h 1 T S 1 2 3 4 5 6 w t = (h 3 – h 4 ) + (h 5 – h 6 ) w p = h 2 – h 1 = v(p 2 – p 1 )

4. Ideal Reheat Rankine Cycles The reheat process in general does not significantly change the cycle efficiency. The sole purpose of the reheat cycle is to reduce the moisture content of the steam at the final stages of the expansion process.

5. Example 1 Consider a steam power plant operating on the ideal reheat Rankine cycle. The steam enters the turbine at 15 MPa and 600 ºC and is condensed in the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low- pressure turbine is not to exceed 10.4%, determine (a) the pressure at which the steam should be reheated. (b) the thermal efficiency of this cycle.

6. Example 1 (continued) State 6: saturated mixture at p 1 = 10 kPa, x 6 = 0.896 State 5: superheated vapor at T 5 = 600 ºC s 5 = s 6 = 7.370 kJ/kg·K h 6 = h f + x 6 h fg = 191.83 + 0.896(2392.8) = 2335.8 kJ/kg s 6 = s f + x 6 s fg = 0.6493 + 0.896(7.5009) = 7.370 kJ/kg·K Table A-6, p 5 = 4 MPa

7. Example 1 (continued) State 2: compressed liquid at p 2 = 15 MPa w p = v(p 2 – p 1 ) = (0.001008)(15000-10) = 15.11 kJ/kg h 2 = h 1 + w p = 191.83 + 15.11 = 206.94 kJ/kg Table A-5 h 1 = h f = 191.83 kJ/kg v 1 = v f = 0.001008 m 3 /kg State 1: saturated liquid at p 1 = 10 kPa

8. Example 1 (continued) State 3: superheated vapor at p 3 = 15 MPa and T 3 = 600 ºC Table A-6 h 3 = 3582.3 kJ/kg s 3 = 6.6776 kJ/kg·K State 4: superheated vapor at p 4 = 4 MPa s 4 = s 3 = 6.6776 kJ/kg·K Table A-6 h 4 = 3154.3 kJ/kg T 4 = 375.5 ºC

9. Example 1 (continued) q in = (h 3 – h 2 ) +(h 5 – h 4 ) = (3582.3 – 206.94) + (3674.4 – 3154.3) = 3895.46 kJ/kg q out = h 6 – h 1 = 2335.8 – 191.83 = 2143.97 kJ/kg 0.43 (without reheat)

10. 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 Turbine Condenser Pump 1 2 3 4 Boiler T S 1 2 3 4 Open Feedwater Heater

11. Ideal Regenerative Rankine Cycles q in = h 5 – h 4 q out = (1 – y)(h 7 – h 1 ) w t = (h 5 – h 6 ) + (1 – y)(h 6 – h 7 ) w p1 = h 2 – h 1 = v 1 (p 2 – p 1 ) T S 1 2 3 4 5 7 6 w p2 = h 4 – h 3 = v 3 (p 4 – p 3 ) 1-y y w p = (1 – y)w p1 + w p2

12. Ideal Regenerative Rankine Cycles T S 1 2 3 4 5 7 6 1-y y 1 2 3 7 4 5 6 P 1 Turbine Boiler Condenser P 2 FWH y 1-y yh 6 + (1-y)h 2 = h 3

13. Example 2 Consider a steam power plant operating on the ideal regenerative Rankine cycle using open feedwater heater. The steam enters the turbine at 15 MPa and 600 ºC and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the feedwater heater. Determine (a) the fraction of steam extracted from the turbine. (b) the thermal efficiency of this cycle.

14. Example 2 (continued) State 2: compressed liquid at p 2 = 1.2 MPa w p1 = v(p 2 – p 1 ) = (0.001008)(1200-10) = 1.20 kJ/kg h 2 = h 1 + w p1 = 191.83 + 1.2 = 193.03 kJ/kg Table A-5 h 1 = h f = 191.83 kJ/kg v 1 = v f = 0.001008 m 3 /kg State 1: saturated liquid at p 1 = 10 kPa

15. Example 2 (continued) State 3: saturated liquid at p 3 = 1.2 MPa Table A-5 h 3 = 798.65 kJ/kg v 3 = 0.001139 m 3 /kg State 4: compressed liquid at p 4 = 15 MPa w p2 = v 3 (p 4 – p 3 ) = (0.001139)(15000-1200) = 15.72 kJ/kg h 4 = h 3 + w p2 = 798.65 + 15.72 = 814.37 kJ/kg

16. Example 2 (continued) State 5: superheated vapor at p 5 = 15 MPa and T 5 = 600 ºC State 6: p 6 = 1.2 MPa s 6 = s 5 = 6.6776 kJ/kg·K Table A-6, h 6 = 2859.5 kJ/kg Table A-6 h 5 = 3582.3 kJ/kg s 5 = 6.6776 kJ/kg·K State 7: p 7 = 10 kPa s 7 = s 6 = s 5 = 6.6776 kJ/kg·K

17. Example 2 (continued) State 7: saturated mixture at p 4 = 10 kPa h 7 = h f + x 7 h fg = 191.83 + 0.804(2392.8) = 2115.6 kJ/kg

18. Example 2 (continued) q in = h 5 – h 4 = 3582.3 – 814.37 = 2767.93 kJ/kg q out = (1 – y)(h 7 – h 1 ) = (1 – 0.227)(2115.6 – 191.83) = 1487.1 kJ/kg 0.43 (without reheat)