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8.16 Steam enters the turbine of a simple vapor power plant with a pressure of 12 MPa and a temperature of 600oC and expands adiabatically to condenser pressure, p. Saturated liquid exits the condenser at pressure p. The isentropic efficiency of both the turbine and the pump is 84%. (a) For p = 30 kPa, determine the turbine exit quality and the cycle thermal efficiency. (b) Plot the quantities of part (a) versus p ranging from 6 kPa to 100 kPa. KNOWN: Water is the working fluid in a simple vapor power plant. Data are given at various states in the cycle. The condenser pressure is p. FIND: (a) For p = 30 kPa, determine the turbine exit quality and the cycle thermal efficiency, (b) plot the quantities of part (a) versus p ranging from 6 kPa to 100 kPa. SCHEMATIC AND GIVEN DATA: Q in



p1 = 12 MPa T1 = 600oC 1



Boiler



ht = 84%



T



W t



Turbine



1 2



p2 = 30 kPa (part (a))



4



p4 = p1 = 12 MPa 4 Condenser



hp = 84%



Pump



W p



T1 = 600oC



3



Q out



12 MPa



p



Cooling water



3



2s 2



s



p3 = p2 = 30 kPa (part (a)) x3 = 0 (saturated liquid)



ENGINEERING MODEL: 1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines. 2. Flow through the boiler and condenser occurs at constant pressure. 3. Stray heat transfer in the turbine, condenser, and pump is ignored. 4. Kinetic and potential energy effects are negligible. 5. Condensate exits the condenser as saturated liquid. ANALYSIS: First fix each principal state with p2 = 30 kPa. State 1: p1 = 12 MPa (120 bar), T1 = 600oC → h1 = 3608.3 kJ/kg, s1 = 6.8037 kJ/kg∙K State 2s: p2s = p2 = 30 kPa (0.3 bar), s2s = s1 = 6. 8037 kJ/kg∙K → x2s = 0.8586, h2s = 2295.0 kJ/kg State 2: p2 = 30 kPa (0.3 bar), h2 = 2505.1 kJ/kg (see below)



1



ht 



h1  h2 kJ kJ  h2  h1  ht (h1  h2 s )  3608.3  (0.84)(3608.3  2295.0) = 2505.1 kJ/kg h1  h2 s kg kg



State 3: p3 = 30 kPa (0.3 bar), saturated liquid → h3 = hf3 = 289.23 kJ/kg, v3 = vf3 = 0.0010223 m3/kg, State 4: p4 = p1 = 12 MPa (120 bar), h4 = 303.80 kJ/kg (see below)



hp 



h4  289.23



kJ  kg



v3 ( p4  p3 ) v ( p  p3 )  h4  h3  3 4 h4  h3 hp



(0.0010223



m3 N )(12000  30) kPa 1000 1 kJ kg m2 = 303.80 kJ/kg 0.84 1 kPa 1000 N  m



(a) The turbine exit quality, x2, is x2 



h2  hf 2 hfg 2



Substituting values from Table 3, hf2 = 289.23 kJ/kg and hfg2 = 2336.1 kJ/kg, gives x2 



(2505.1  289.23) kJ/kg = 0.9485 (94.85%) 2336.1 kJ/kg



The thermal efficiency is



h



  Wp / m  (h1  h2 )  (h4  h3 ) Wt / m   Q / m (h  h ) in



1



4



Substituting enthalpy values and solving yield



h



(3608.3  2505.1) kJ/kg  (303.80  289.23) kJ/kg = 0.3294 (32.94%) (3608.3  303.80) kJ/kg



2



The data for the required plots are obtained using IT as follows: IT Code



IT Results for p2 = 30 kPa



p1 = 12000 // kPa T1 = 600 // oC p2 = 30 // kPa eff_t = 0.84 eff_p = 0.84 p3 = p2 x3 = 0 p4 = p1



eta h1 h2 h2s h3 h4 p2s p3 p4 s1 s2s v3 x2 eff_p eff_t p1 p2 T1 x3



h1 = h_PT("Water/Steam", p1, T1) s1 = s_PT("Water/Steam", p1, T1) s2s = s1 p2s = p2 h2s = h_Ps("Water/Steam", p2s, s2s) h2 = h1 - eff_t*(h1 - h2s) x2 = x_hP("Water/Steam", h2, p2)



0.3295 3608 2505 2295 289.9 304.5 30 30 1.2E4 6.803 6.803 0.001022 0.9486 0.84 0.84 1.2E4 30 600 0



h3 = hsat_Px("Water/Steam", p3, x3) v3 = vsat_Px("Water/Steam", p3, x3) h4 = h3 + (v3*(p4 - p3))/eff_p eta = ((h1 - h2) - (h4 - h3))/(h1 - h4)



IT results are consistent with the calculations in part (a). Plots: Condenser Pressure versus Turbine Exit Quality 1



0.4



0.99



0.35



0.98



0.3 Thermal Efficiency



0.97



Quality



Condenser Pressure vs. Thermal Efficiency



0.96 0.95 0.94 0.93



0.25 0.2 0.15 0.1



0.92



0.05



0.91 0.9 0



10



20



30



40



50



60



70



80



90



0



100



0



Condenser Pressure (kPa)



10



20



30



40



50



60



70



80



90



100



Condenser Pressure (kPa)



From the T-s diagram, we see that within the range of pressures considered the working fluid is a liquid-vapor mixture (0 ≤ x ≤ 1) and as p2 increases, points 2s and 2 move to the right. Thus, x2 increases as indicated in the plot above. Further as p2 increases, the average temperature of heat rejection increases, lowering thermal efficiency. Thus, h decreases as indicated in the plot above.



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