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+// Example 7_3
+clc;funcprot(0);
+// Given data
+m_CO2=0.1;// lbm of CO_2
+m_N2=0.1;// lbm of N_2
+T_0=77+460;// °R
+P=30;// psia
+P_0=14.7;// psia
+T=440;// °F
+R=1.986;// Btu/lbmol-°R
+
+// Calculation
+// Use table F-4E,for CO_2
+h=7597.6;// Btu/lbmol
+h_0=4030.2;// Btu/lbmol
+phi=56.070;// Btu/lbmol-°R
+phi_0=51.032;// Btu/lbmol-°R
+X_CO2=(m_CO2/44)*[(h-h_0)-(T_0*((phi-phi_0)-(R*log(P/P_0))))];// The availability of CO_2 in Btu
+printf("\nThe availability of CO_2,X=%1.2f Btu",X_CO2);
+// Use table F-4E,for N_2
+h=6268.1;// Btu/lbmol
+h_0=3279.5;// Btu/lbmol
+phi=49.352;// Btu/lbmol-°R
+phi_0=45.743;// Btu/lbmol-°R
+X_N2=(m_N2/28)*[(h-h_0)-(T_0*((phi-phi_0)-(R*log(P/P_0))))];// The availability of N_2 in Btu
+printf("\nThe availability of N_2,X=%1.2f Btu",X_N2);