ENGR 2213 Thermodynamics

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

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

3. 3 T 5 4 2 1 6 S Ideal Reheat Rankine Cycles wp = h2 – h1 = v(p2 – p1) qin = (h3 – h2) + (h5 – h4) wt = (h3 – h4) + (h5 – h6) qout = h6 – h1

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 • 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 p1 = 10 kPa, x6 = 0.896 h6 = hf + x6hfg = 191.83 + 0.896(2392.8) = 2335.8 kJ/kg s6 = sf + x6sfg = 0.6493 + 0.896(7.5009) = 7.370 kJ/kg·K State 5: superheated vapor at T5 = 600 ºC s5 = s6 = 7.370 kJ/kg·K Table A-6, p5 = 4 MPa

7. Example 1 (continued) 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 = 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

8. Example 1 (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: superheated vapor at p4 = 4 MPa s4 = s3 = 6.6776 kJ/kg·K Table A-6 h4 = 3154.3 kJ/kg T4 = 375.5 ºC

9. Example 1 (continued) qin = (h3 – h2) +(h5 – h4) = (3582.3 – 206.94) + (3674.4 – 3154.3) = 3895.46 kJ/kg qout = h6 – h1 = 2335.8 – 191.83 = 2143.97 kJ/kg 0.43 (without reheat)

10. 2 T 3 Boiler 3 Pump Turbine 2 1 4 1 4 Condenser S 5 4 T Boiler 5 4 3 6 P 2 6 Turbine FWH 3 2 2 1 7 7 P 1 Condenser S 1 Ideal Regenerative Rankine Cycles Open Feedwater Heater

11. y 1-y 5 T 4 6 3 2 1 7 S Ideal Regenerative Rankine Cycles wp1 = h2 – h1 = v1(p2 – p1) wp2 = h4 – h3 = v3(p4 – p3) wp = (1 – y)wp1 + wp2 qin = h5 – h4 qout = (1 – y)(h7 – h1) wt = (h5 – h6) + (1 – y)(h6 – h7)

12. y 1-y 5 4 T Boiler 5 4 3 6 P 2 6 Turbine FWH 3 y 2 2 1 7 7 P 1 Condenser S 1 1-y Ideal Regenerative Rankine Cycles yh6 + (1-y)h2 = h3

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

14. Example 2 (continued) 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 = 1.2 MPa wp1 = v(p2 – p1) = (0.001008)(1200-10) = 1.20 kJ/kg h2 = h1 + wp1 = 191.83 + 1.2 = 193.03 kJ/kg

15. Example 2 (continued) State 3: saturated liquid at p3 = 1.2 MPa Table A-5 h3 = 798.65 kJ/kg v3 = 0.001139 m3/kg State 4: compressed liquid at p4 = 15 MPa wp2 = v3(p4 – p3) = (0.001139)(15000-1200) = 15.72 kJ/kg h4 = h3 + wp2 = 798.65 + 15.72 = 814.37 kJ/kg

16. Example 2 (continued) State 5: superheated vapor at p5 = 15 MPa and T5 = 600 ºC Table A-6 h5 = 3582.3 kJ/kg s5 = 6.6776 kJ/kg·K State 6: p6 = 1.2 MPa s6 = s5 = 6.6776 kJ/kg·K Table A-6, h6 = 2859.5 kJ/kg State 7: p7 = 10 kPa s7 = s6 = s5 = 6.6776 kJ/kg·K

17. Example 2 (continued) State 7: saturated mixture at p4 = 10 kPa h7 = hf + x7hfg = 191.83 + 0.804(2392.8) = 2115.6 kJ/kg

18. Example 2 (continued) qin = h5 – h4 = 3582.3 – 814.37 = 2767.93 kJ/kg qout = (1 – y)(h7 – h1) = (1 – 0.227)(2115.6 – 191.83) = 1487.1 kJ/kg 0.43 (without reheat)