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clear;
clc;
disp('Example 5.19')
// aim : To determine the
// (a) Gamma,
// (b) del_U
// Given Values
P1 = 1400;// [kN/m^2]
P2 = 100;// [kN/m^2]
P3 = 220;// [kN/m^2]
T1 = 273+360;// [K]
m = .23;// [kg]
cp = 1.005;// [kJ/kg*K]
// Solution
T3 = T1;// since process 1-3 is isothermal
// (a)
// for process 1-3, P1*V1=P3*V3,so
V3_by_V1 = P1/P3;
// also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence
// and process process 2-3 is iso-choric so,V3=V2 and
V2_by_V1 = V3_by_V1;
// hence,
Gamma = log(P1/P2)/log(P1/P3); // heat capacity ratio
mprintf('\n (a) The value of adiabatic index Gamma is = %f\n',Gamma);
// (b)
cv = cp/Gamma;// [kJ/kg K]
// for process 2-3,P3/T3=P2/T2,so
T2 = P2*T3/P3;// [K]
// now
del_U = m*cv*(T2-T1);// [kJ]
mprintf('\n (b) The change in internal energy during the adiabatic expansion is U2-U1 = %f kJ (This is loss of internal energy)\n',del_U);
// End
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