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clear;
clc;
disp('Example 5.22');
// aim : To determine the
// (a) new pressure of the air in the receiver
// (b) specific enthalpy of air at 15 C
// Given values
V1 = .85;// [m^3]
T1 = 15+273;// [K]
P1 = 275;// pressure,[kN/m^2]
m = 1.7;// [kg]
cp = 1.005;// [kJ/kg*K]
cv = .715;// [kJ/kg*K]
// solution
// (a)
R = cp-cv;// [kJ/kg*K]
// assuming m1 is original mass of the air, using P*V=m*R*T
m1 = P1*V1/(R*T1);// [kg]
m2 = m1+m;// [kg]
// again using P*V=m*R*T
// P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so
P2 = P1*m2/m1;// [kN/m^2]
mprintf('\n (a) The new pressure of the air in the receiver is = %f kN/m^2\n',P2);
// (b)
// for 1 kg of air, h2-h1=cp*(T1-T0)
// and if 0 is chosen as the zero enthalpy, then
h = cp*(T1-273);// [kJ/kg]
mprintf('\n (b) The specific enthalpy of the air at 15 C is = %f kJ/kg\n',h);
// End
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