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// Theory and Problems of Thermodynamics
// Chapter 8
// Power and Refrigeration Cycles
// Example 15
clear ;clc;
//Given data
T1 = 300 // intial temperature of air in K
P1 = 0.1 // initial pressure of air in MPa
P2 = 5.0 // final pressure of air in MPa
q1 = 20 // energy injected per mole of air in kJ
R = 8.314 // gas constant
Cp = 3.5*R // specific heat ratio at constant pressure in kJ/kg K
gam = 1.4 // heat capacity ratio
// calculations
r0 =(P2/P1)^(1/gam) // Otto cycles
T2 = T1*(r0)^(gam-1)
//q1 = Cp * (T3-T2)
deff('y=temp(T3)', 'y = q1*1e3 - Cp*(T3-T2)')
T3 = fsolve(10,temp) // maximum cycle temperature
rc = T3/T2 // cut off ratio
// Thermal efficiency of cycle
n = 1-(1/(gam*(r0^(gam-1))))*(((rc^gam)-1)/(rc-1))
W = n*q1 // work done per mole of air in kJ/mol
v1 = R*T1/P1
//v1-v2 = v1*(1/v2/v1) = v1*(1-1/r0)
v1_v2 = v1*(1-1/r0) // v1_v2 = v1 - v2
Pm = W*1e3/v1_v2 // Mean effective pressure in MPa
// Output Results
mprintf('\Work done per kg of air = %4.3f kJ/mol' ,W);
mprintf('\n Mean effective pressure = %4.4f MPa' ,Pm);
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