Thermodynamics: An Engineering Approach 6th Edition, Yunus A. Cengel, Michael A.

Question

Thermodynamics: An Engineering Approach 6th Edition, Yunus A. Cengel, Michael A. Boles, problem 10-78

Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and (450 Celsius) and expands to a pressure of 0.4MPa. At this pressure, 60 percent of the steam is extracted from the turbine, and the remainder expands to 10kPa. Part of the extracted steam is used to heat the feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. Assuming the turbines and the pumps to be isentropic, and determine the mass flow rate of steam through the boiler for a net power output of 15 MW.
Answser: 17.7 kg/s

Explanation / Answer

h1=hf@10Kpa=191.81KJ/Kg   v1=vf@10kpa=0.00101m3/kg

pump-1,W,in=v1(P2-P1)=0.00101(400-10)=0.39Kj/Kg

h2=h1+W,pump1=192.20Kj/kg

h3=h4=h9=hf@0.4Mpa=604.6Kj/Kg

v4=vf@0.4Mpa=0.001084m3/kg

pump-2,W,in=v4(P5-P4)=0.001084(6000-400)=6.07Kj/Kg

h5=h4+W,pump2=610.7Kj/kg

p6=6Mpa   T6=450C   =>h6=3302.9KJ/Kg   s6=6.7129KJ.KG.K

P7=0.4Mpa ,S7=S6   =>x7=(s7-sf)/s,fg=(6.7129-1.7765)/5.1191=0.9661

h7=hf+x7.Xgf=604.66+(0.9661)2133.4=2665.7Kj/Kg

P8=10Kpa ,S6=S8 =>x8=(s8-sf)/s,fg=(6.7129-0.6492)/7.4996=0.8097

h8=hf+x8.Xgf=191.81+(0.8097)2392.1=2128.7Kj/Kg

then per Kg of steam flowing through the boiler:

wt,out=(h6-h7)+0.4(h7-h8)=852Kj/Kg

wp,in=0.4W,p1,in+W,p2,in=6.23Kj/Kg

w,net=Wt,out-Wp,in=845.8Kj/Kg

m'=W',net/w,net=15000/845.8=17.7 Kg/s

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