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clear;
clc;
disp('Example 12.2');
// aim : To determine the increases in pressure, temperature and internal energy per kg of air
// Given values
T1 = 273;// [K]
P1 = 140;// [kN/m^2]
C1 = 900;// [m/s]
C2 = 300;// [m/s]
cp = 1.006;// [kJ/kg K]
cv =.717;// [kJ/kg K]
// solution
R = cp-cv;// [kJ/kg K]
Gamma = cp/cv;// heat capacity ratio
// for frictionless adiabatic flow, (C2^2-C1^2)/2=Gamma/(Gamma-1)*R*(T1-T2)
T2 =T1-((C2^2-C1^2)*(Gamma-1)/(2*Gamma*R))*10^-3; // [K]
T_inc = T2-T1;// increase in temperature [K]
P2 = P1*(T2/T1)^(Gamma/(Gamma-1));// [MN/m^2]
P_inc = (P2-P1)*10^-3;// increase in pressure,[MN/m^2]
U_inc = cv*(T2-T1);// Increase in internal energy per kg,[kJ/kg]
mprintf('\n The increase in pressure is = %f MN/m^2\n',P_inc);
mprintf('\n Increase in temperature is = %f K\n',T_inc);
mprintf('\n Increase in internal energy is = %f kJ/kg\n',U_inc);
// there is minor variation in result
// End
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