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+// Example 8_5
+clc;funcprot(0);
+// Given data
+P_3=600;// psia
+T_3=1000;// °F
+P_4=40;// psia
+T_4=600;// °F
+P_5=2;// psia
+
+// Calculation
+// From Table C-2E
+h_2=94;// Btu/lbm
+h_1=h_2;// Btu/lbm
+// From Table C-3E
+h_3=1518;// Btu/lbm
+s_3=1.716;// Btu/lbm-°R
+s_4=s_3;// Btu/lbm-°R
+h_4=(((1.716-1.712)/(1.737-1.712))*(1217-1197))+1197;// Btu/lbm
+// At 40 psuia and 600°F
+h_5=1333;// Btu/lbm
+s_5=1.862;// Btu/lbm-°R
+s_6=s_5;// Btu/lbm-°R
+s_f6=0.175;// Btu/lbm-°R
+s_fg6=1.745;// Btu/lbm-°R
+x_6=(s_6-s_f6)/s_fg6;// The quality of steam at state 6
+h_f6=94;// Btu/lbm
+h_fg6=1022;// Btu/lbm
+h_6=h_f6+(x_6*h_fg6);// Btu/lbm
+q_B=(h_5-h_4)+(h_3-h_2);// The energy input in Btu/lbm
+w_T=(h_5-h_6)+(h_3-h_4);// The energy output in Btu/lbm
+n=w_T/q_B;// The thermal efficiency
+printf("\nThe thermal efficiency,n=%0.3f or %2.1f percentage.",n,n*100);