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Diffstat (limited to '3685/CH12/EX12.8/Ex12_8.sce')
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diff --git a/3685/CH12/EX12.8/Ex12_8.sce b/3685/CH12/EX12.8/Ex12_8.sce new file mode 100644 index 000000000..60add7144 --- /dev/null +++ b/3685/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,83 @@ +clc +// Part (a) +h1 = 2758 // Enthalpy at state 1 in kJ/kg +h2 = 1817 // Enthalpy at state 2 in kJ/kg +h3 = 192 // Enthalpy at state 3 in kJ/kg +h4 = 200// Enthalpy at state 4 in kJ/kg +Wt = h1-h2 // turbine work +Wp = h4-h3 // Pump work +Q1 = h1-h4 // Heat addition +Wnet = Wt-Wp // Net work doen +n1 = Wnet/Q1 // First law efficiency +WR = Wnet/Wt // Work ratio +Q1_ = 100 // Heat addition rate in MW +PO = n1*Q1_ // power output +cpg = 1000 // Specific heat capacity in J/kg +wg = (Q1_/(833-450)) // mass flow rate of gas +EIR = wg*cpg*((833-300)-300*(log(833/300)))/1000 // Exergy input +n2 = PO/EIR // Second law efficiency + +printf("\n Example 12.8\n") +printf("\n Part (a)") +printf("\n The first law efficiency n1 is %f",n1*100) +printf("\n The second law efficiency n2 is %f",n2*100) +printf("\n The work ratio is %f",WR) +// Part (b) +h1b = 3398 // Enthalpy at state 1 in kJ/kg +h2b = 2130 // Enthalpy at state 2 in kJ/kg +h3b = 192 // Enthalpy at state 3 in kJ/kg +h4b = 200// Enthalpy at state 4 in kJ/kg +Wtb = 1268 // turbine work in kJ/kg +Wpb = 8 // Pump work in kJ/kg +Q1b = 3198// Heat addition rate in kW +n1b = (Wtb-Wpb)/Q1b //first law efficiency +WRb = (Wtb-Wpb)/Wtb // WOrk ratio +EIRb = 59.3 // Exergy input rate in MW +Wnetb = Q1_*n1b // net work done + +n2b = Wnetb/EIRb // Second law efficiency +printf("\n Part (b)") +printf("\n The first law efficiency n1 is %f",n1b*100) +printf("\n The second law efficiency n2 is %f",n2b*100) +printf("\n The work ration is %f",WRb) + +// Part (c) +h1c = 3398 // Enthalpy at state 1 in kJ/kg +h2c = 2761 // Enthalpy at state 2 in kJ/kg +h3c = 3482 // Enthalpy at state 3 in kJ/kg +h4c = 2522 // Enthalpy at state 4 in kJ/kg +h5c = 192 // Enthalpy at state 5 in kJ/kg +h6c = 200// Enthalpy at state 6 in kJ/kg +Wt1 = 637 // Turbine work in kJ/kg +Wt2 = 960 // Turbine work in kJ/kg +Wtc = Wt1+Wt2 // Net turbine work in kJ/kg +Wp = 8 // Pump work in kJ/kg +Wnetc = Wtc-Wp // net work done +Q1c = 3198+721 // Heat addition +n1c = Wnetc/Q1c// First law efficiency +WRc = Wnetc/Wtc// Work ratio +POc = Q1_*n1c// Power output +EIRc = 59.3// Exergy input in MW +n2c = POc/EIRc // Second law efficiency +printf("\n Part (c)") +printf("\n The first law efficiency n1 is %f",n1c*100) +printf("\n The second law efficiency n2 is %f",n2c*100) +printf("\n The work ration is %f",WRc) + +// Part (d) +T3 = 45.8 // saturation temperature at 0.1 bar in degree celsius +T1 = 295 // saturation temperature at 80 bar in degree celsius +n1d = 1-((T3+273)/(T1+273)) // First law efficiency +Q1d = 2758-1316 // Heat addition +Wnet = Q1d*n1d // Net work output +Wpd = 8 // Pump work in kJ/kg +Wtd = 641// Turbine work in kJ/kg +WRd = (Wt-Wp)/Wt // Work ratio +POd = Q1_*0.439// Power output +EIRd = (Q1_/(833-593))*cpg*((833-300)-300*(log(833/300)))/1000 //Exergy Input rate in MW +n2d = POd/EIRd // Second law efficiency +printf("\n Part (d)") +printf("\n The first law efficiency n1 is %f",n1d*100) +printf("\n The second law efficiency n2 is %f",n2d*100) +printf("\n The work ration is %f",WRd) +//The answers vary due to round off error |