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clear;
clc;
disp('Example 5.13');
// aim : To determine the
// final volume, work done and the change in internal energy
// Given values
P1 = 700;// initial pressure,[kN/m^2]
V1 = .015;// initial volume, [m^3]
P2 = 140;// final pressure, [kN/m^2]
cp = 1.046;// [kJ/kg K]
cv = .752; // [kJ/kg K]
// solution
Gamma = cp/cv;
// for adiabatic expansion, P*V^gamma=constant, so
V2 = V1*(P1/P2)^(1/Gamma);// final volume, [m^3]
mprintf('\n The final volume of the gas is V2 = %f m^3\n',V2);
// work done
W = (P1*V1-P2*V2)/(Gamma-1);// [kJ]
mprintf('\n The work done by the gas is = %f kJ\n',W);
// for adiabatic process
del_U = -W;// [kJ]
mprintf('\n The change of internal energy is = %f kJ',del_U);
if(del_U>0)
disp('since del_U>0, so the the gain in internal energy of the gas ')
else
disp('since del_U<0, so this is a loss of internal energy from the gas')
end
// End
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