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clear;
clc;
disp('Example 5.14');
// aim : To determine the
// (a)heat transfer
// (b)change of internal energy
// (c)mass of gas
// Given values
V1 = .4;// initial volume, [m^3]
P1 = 100;// initial pressure, [kN/m^2]
T1 = 273+20;// temperature, [K]
P2 = 450;// final pressure,[kN/m^2]
cp = 1.0;// [kJ/kg K]
Gamma = 1.4; // heat capacity ratio
// solution
// (a)
// for the isothermal compression,P*V=constant,so
V2 = V1*P1/P2;// [m^3]
W = P1*V1*log(P1/P2);// formula of workdone for isothermal process,[kJ]
// for isothermal process, del_U=0;so
Q = W;
mprintf('\n (a) The heat transferred during compression is Q = %f kJ\n',Q);
// (b)
V3 = V1;
// for adiabatic expansion
// also
P3 = P2*(V2/V3)^Gamma;// [kN/m^2]
W = -(P3*V3-P2*V2)/(Gamma-1);// work done formula for adiabatic process,[kJ]
// also, Q=0,so using Q=del_U+W
del_U = -W;// [kJ]
mprintf('\n (b) The change of the internal energy during the expansion is,del_U = %f kJ\n',del_U);
// (c)
// for ideal gas
// cp-cv=R, and cp/cv=gamma, hence
R = cp*(1-1/Gamma);// [kj/kg K]
// now using ideal gas equation
m = P1*V1/(R*T1);// mass of the gas,[kg]
mprintf('\n (c) The mass of the gas is,m = %f kg\n',m);
// There is calculation mistake in the book
// End
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