# 8-3536 - 8-35&36EquationsSet 24 Problem 8.35Number...

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Unformatted text preview: 8-35&36EquationsSet 24: Problem 8.35Number states as theEESdiagramwindow.W\$ =‘Steam’(1)T1= 480 [C](2)P1= 8000 [kPa](3)P3= 8 [kPa](4)\$Ifnot ParametricTableP2= 700 [kPa]feed water heater pressure(5)\$endifPower= 100 [MW]·1000kWMW(6)ηpump= 1(7)ηturbine= 1(8)Turbine 1s1= s (W\$,T=T1,P=P1)(9)h1= h(W\$,T=T1,P=P1)(10)s1=sid,2(11)hid,2= h(W\$,P=P2,s=sid,2)(12)Wt1= (h1-hid,2)·ηturbine(13)Wt1=h1-h2(14)T2= T(W\$,P=P2,h=h2)(15)x2= x(W\$,P=P2,h=h2)(16)s2= s (W\$,P=P2,h=h2)(17)Turbine 2sid,3=s2(18)hid,3= h(W\$,P=P3,s=sid,3)(19)wt2= (h2-hid,3)·ηturbine(20)wt2=h2-h3(21)T3= T(W\$,P=P3,h=h3)(22)x3= x(W\$,P=P3,h=h3)(23)s3= s (W\$,P=P3,h=h3)(24)Condenserx4= 0(25)P4=P3(26)h4= h(W\$,P=P4,x=x4)(27)T4= T(W\$,P=P4,x=x4)(28)s4= s (W\$,P=P4,x=x4)(29)Qc= (h3-h4)·(1-y)(30)v4= v (W\$,P=P4,x=x4)(31)Tw,in= 15 [C]temperature of entering cooling water(32)Tw,out= 35 [C]temperature of exiting cooling water(33)Patm= 101.3 [kPa]pressure of cooling water(34)˙Qcond= ˙mwater·(h(W\$,T=Tw,out,P=Patm)-h(W\$,T=Tw,in,P=Patm))this equation fixes mdot,w(35)Pump 1P5=P2(36)wp1=v4·P5-P4ηpump(37)h5=h4+wp1(38)T5= T(W\$,h=h5,P=P5)(39)s5= s (W\$,h=h5,P=P5)(40)Feedwater Heatery·h2+ (1-y)·h5-h6= 0energy balance on feedwater heater - steady-state(41)h6= h(W\$,P=P6,x= 0)saturated liquid at exit(42)s6= s (...
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## This note was uploaded on 07/29/2008 for the course ME 361 taught by Professor Martin during the Spring '07 term at University of Wisconsin.

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8-3536 - 8-35&36EquationsSet 24 Problem 8.35Number...

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