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+// Example 11_7
+clc;funcprot(0);
+// Given data
+T_1=100;// °F
+T_2=80;// °F
+P=14.7;// psia
+P_g1=0.9503;// psia
+P_g2=0.5073;// psia
+c_p=0.24;// Btu/lbm-°R
+h_fg2=1048;// Btu/lbm
+h_g1=1105;// Btu/lbm
+h_f2=48.09;// Btu/lbm
+
+// Calculation
+// (a)
+w_2=0.622*(P_g2/(P-P_g2));// lbm H2O/lbm dry air
+w_1=((w_2*h_fg2)+(c_p*(T_2-T_1)))/(h_g1-h_f2);// lbm H2O/lbm dry air
+// (b)
+P_v1=(w_1*P)/(0.622*(1+(w_1)));// psia
+phi=P_v1/P_g1;// The relative humidity in %
+// (c)
+h=(c_p*T_1)+(w_1*h_g1);// Btu/lbm dry air
+printf("\n(a)The humidity ratio,w_1=%0.5f lbm H2O/lbm dry air \n(b)The relative humidity,phi=%0.3f or %2.1f percentage. \n(c)The specific enthalpy of the air,h=%2.1f Btu/lbm dry air",w_1,phi,phi*100,h);