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+// Example 12_8
+clc;funcprot(0);
+// Given data
+T_WB=60.0;// °F
+T_DB=70.0;// °F
+
+// Calculation
+// From Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find
+h_g1=1092.0;// Btu/lbm
+h_fg2=1059.6;// Btu/lbm
+h_f2=28.1;// Btu/lbm
+p_sat=0.2563;// psia
+p_w3=p_sat;// psia
+p_m=14.7;// psia
+w_3=0.622*((p_w3)/(p_m-p_w3));// lbm water per lbm of dry air
+c_pa=0.240;// Btu/lbm.R
+w_1=((c_pa*(T_WB-T_DB))+(w_3*h_fg2))/(h_g1-h_f2);// lbm water per lbm of dry air
+w_1=w_1*7000;// grains of water per lbm of dry air
+printf("\nThe humidity ratio (ω) in the room,w_1=%2.1f grains of water per lbm of dry air",w_1);