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// Theory and Problems of Thermodynamics
// Chapter 8
// Power and Refrigeration Cycles
// Example 3
clear ;clc;
//Given data
P1 = 2.5 // entering pressure of superheated steam in MPa
T1 = 523.15 // entering temperature of superheated steam in K
P2 = 10 // pressure of dry saturated steam in kPa
// Steam at 2.5 MPa and 523.15 K
h4 = 2880.1 // in kJ/kg
s4 = 6.4085 // in kJ/kg K
// Steam at 10 kPa
vf = 0.001010 // in m^3/kg
hf = 191.83 // in kJ/kg
sf = 0.6493 // in kJ/kg K
hfg = 2392.8 // in kJ/kg
sfg = 7.5009 // in kJ/kg K
h1 = 191.83 // in kJ/kg
//s4=s5 => s4 = sf+X5*sfg
X5 = (s4-sf)/sfg
h5 = h1 + X5*hfg // in kJ/kg
W_p = vf*(P2*1e3-P1) // work done by turbine
h2 = h1+W_p
n = ((h4-h5)-(h2-h1))/(h4-h2)
// Output Results
mprintf('Thermal efficiency of power plant = %4.3f' ,n);
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