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+// Example 12_13
+clc;funcprot(0);
+// Given data
+V_a1=2000;// ft^3/min
+T_DB1=50.0+459.67;// R
+phi_1=80.0/100;// The relative humidity
+V_a2=1000;// ft^3/min
+T_DB2=100.0+459.67;// R
+phi_2=40.0/100;// The relative humidity
+R_a=53.34;// ft.lbf/(lbm.R)
+p_m=14.7// lbf/in^2
+
+// Calculation
+p_sat1=0.178;// psia
+p_w1=phi_1*p_sat1;// psia
+p_a1=p_m-p_w1;// lbf/in^2
+v_a1=(R_a*T_DB1)/(p_a1*144);// ft^3/(lbm dry air)
+p_sat2=0.9503;// psia
+p_w2=phi_2*p_sat2;// psia
+p_a2=p_m-p_w2;// lbf/in^2
+v_a2=(R_a*T_DB2)/(p_a2*144);// ft^3/(lbm dry air)
+m_a1=V_a1/v_a1;// lbmdry air/min
+m_a2=V_a2/v_a2;// lbmdry air/min
+m_a3=m_a1+m_a2;// lbmdry air/min
+// Then, from the psychrometric chart (Chart D.5), we find
+w_1=44/7000;// lbm water vapor/(lbm dry air)
+w_2=115/7000;// lbm water vapor/(lbm dry air)
+h_1=19;// Btu/(lbm dry air)
+h_2=42;// Btu/(lbm dry air)
+w_3=((m_a1/m_a3)*w_1)+((m_a2/m_a3)*w_2);// grains of water vapor/(lbm dry air)
+h_3=((m_a1/m_a3)*h_1)+((m_a2/m_a3)*h_2);// Btu/(lbm dry air)
+// From the point where the lines ω = 65.8 grains/(lbm dry air) = constant and h = 26 Btu/(lbm dry air) = constant intersect on the psychrometric chart, we can read from this chart that
+T_DB=63;// °F
+T_WB=59;// °F
+phi=75;// %
+T_DP=56;// °F
+printf("\nThe dry bulb temperature of the outlet mixture,T_DB=%2.0f°F \nThe relative humidity of the outlet mixture,phi=%2.0f percentage",T_DB,phi);