// Theory and Problems of Thermodynamics
// Chapter 8
// Power and Refrigeration Cycles
// Example 14

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))

// Output Results
mprintf('Thermal efficieny = %4.4f ' ,n);