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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 2

+

+clear ;clc;

+

+//Given data

+r = 1.4;         // r = CP/CV

+P1 = 0.5;        // pressure at state 1 in MPa

+T1 = 300;        // temperature at state 1 in K

+V_21 = 2;        // ratio of V2 and V1

+T2 = T1 * V_21;  // temperature at state 2 in K

+T3 = T1;       // temperature at state 3 in K

+R = 8.314;       // gas constant in J/K

+

+//For reversible adiabatic path

+T_23 = T2 / T3                // T_23 ratio of T2 and T3

+V_32 = T_23 ^ (1/(r-1));      // ratio of V3 and V2

+

+/// Q = Q12 = Q23 = Cp*T1

+Q_A = R * r * T1 / (r -1)     // heat interactions for adiabatic path

+Q_A = Q_A / 1000;             // units conversion J to kJ

+

+// W = W12 + W23 = R*T1 + Cv*T1 = Cp*T1

+W_A = R * r * T1 / (r -1)     // work done for adiabatic path

+W_A = W_A / 1000;             // units conversion J to kJ

+

+// for reversible isothermal process

+// W_I = integrate('P','V',3,1) = R*T1*log(V1/V3) = R*T1*ln(V1/V2)*(V2/V3)

+W_I = R * T1 * log(1 / (V_21 * V_32) );     // work for reversible isothermal process

+W_I = W_I / 1000;             // units conversion J to kJ

+

+// Results

+mprintf('heat interaction for reversible adiabatic path = %4.2f kJ', Q_A)

+mprintf('\n work done for adiabatic path = %4.2f kJ', W_A)

+mprintf('\n work done for reversible isothermal process = %5.3f kJ', W_I)

+