# Reflection of Buddhism in Contemporary Cinema ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma Ideal Reheat Rankine Cycles 2 Boiler 3 Pump 3 Turbine 1 2 1

4 Condenser 2 Boiler 4 S 3 T 3 5 4 Pump 1 T

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

1 6 qout = h6 h1 w net qout 1 1 (h3 qin qin wp back work ratio w t (h3 5 S h6 h1 h2 ) (h5 h4 )

h2 h1 h4 ) (h5 h6 ) 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. Example 1 Consider a steam power plant operating on the ideal reheat Rankine cycle. The steam enters the turbine at 15 MPa and 600 C and is condensed in the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the lowpressure 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. 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/kgK State 5: superheated vapor at T5 = 600 C s5 = s6 = 7.370 kJ/kgK Table A-6, p5 = 4 MPa 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 Example 1 (continued) State 3: superheated vapor

at p3 = 15 MPa and T3 = 600 C Table A-6 h3 = 3582.3 kJ/kg s3 = 6.6776 kJ/kgK State 4: superheated vapor at p4 = 4 MPa s4 = s3 = 6.6776 kJ/kgK Table A-6 h4 = 3154.3 kJ/kg T4 = 375.5 C 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 qout 2143.97 1 1 0.45

qin 3895.46 0.43 (without reheat) Ideal Regenerative Rankine Cycles 2 Boiler T 3 Pump 3 Turbine 1 2 1

4 Condenser S Open Feedwater Heater 4 Boiler 4 5 T 5 4 P2 3

6 FWH 3 2 Turbine 1 2 P1 1 Condenser 7 6 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) 5 T 4 3 2 1 y 6 1-y 7 S

wt = (h5 h6) + (1 y)(h6 h7) w net (h5 h6 ) (1 y)(h6 h7 ) (1 y)(h2 h1) (h4 h3 ) (h5 h4 ) qin Ideal Regenerative Rankine Cycles w net qout 1 qin qin 5 T 4 (1 y)(h7 h1 ) 1 (h5 h4 )

yh6 + (1-y)h2 = h3 h3 h2 y h 6 h2 3 2 6 y 1-y 1 7 S 4 P2

3 Boiler 5 6 FWH y 2 P1 1 Turbine Condenser 7 1-y

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 C 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. 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 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 Example 2 (continued) State 5: superheated vapor at p5 = 15 MPa and T5 = 600 C Table A-6 h5 = 3582.3 kJ/kg s5 = 6.6776 kJ/kgK State 6: p6 = 1.2 MPa

s6 = s5 = 6.6776 kJ/kgK Table A-6, h6 = 2859.5 kJ/kg State 7: p7 = 10 kPa s7 = s6 = s5 = 6.6776 kJ/kgK Example 2 (continued) State 7: saturated mixture at p4 = 10 kPa s7 sf 6.6776 0.6493 x7 0.804 sfg 7.5009 h7 = hf + x7hfg = 191.83 + 0.804(2392.8) = 2115.6 kJ/kg h3 h2 798.65 193.03 y 0.227 h6 h2 2859.5 193.03 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 qout 1487.1 1 1 0.463 qin 2767.93 0.43 (without reheat)