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// Given:-
T1 = 300.00 // in kelvin
AV = 5.00 // volumetric flow rate in m^3/s
p1 = 100.00 // in kpa
pr = 10.00 // compressor pressure ratio
T3 = 1400.00 // turbine inlet temperature in kelvin
Wt_ms = 706.9 // kJ/kg
Wc_m = 279.7
// Analysis
// At state 1, the temperature is 300 K. From Table A-22,
h1 = 300.19 // in kj/kg
pr1 = 1.386
// Interpolating in Table A-22,
h2 = 579.9 // in kj/kg
// From Table A-22
h3 = 1515.4 // in kj/kg
pr3 = 450.5
// Interpolating in Table A-22, we get
h4 = 808.5 // in kj/kg
// calculations
Wtbym = 0.8*Wt_ms
Wcbym = Wc_m/0.8
h2 = 300.19 + Wcbym
//pr2 = pr*pr1
//pr4 = pr3*1/pr
// Part(a)
//eta = ((h3-h4)-(h2-h1))/(h3-h2) // thermal efficiency
Qinbym = h3 - h2
n = (Wtbym-Wcbym)/Qinbym
// Result
printf( '\n The thermal efficiency is : %.3f ',n)
// Part(b)
//bwr = (h2-h1)/(h3-h4) // back work ratio
bwr = Wcbym/Wtbym
// Result
printf( '\n The back work ratio is : %.3f',bwr)
// Part(c)
R = 8.314 // universal gas constant, in SI units
M = 28.97 // molar mass of air in grams
// Calculations
//mdot = AV*p1/((R/M)*T1) // mass flow rate in kg/s
Wcycledot = 5.807*(Wcbym-Wtbym) // The net power developed
// Result
printf( '\n The net power developed, is : %.f kW .',-Wcycledot)
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