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+// Example 8_2
+clc;funcprot(0);
+// Given data
+P_1=10;// kPa
+P_2=4;// MPa
+T_3=400;// °C
+h_2=192;// kJ/kg
+h_3=3214;// kJ/kg
+s_3=6.7698;// kJ/kg.K
+
+// Calculation
+s_f4=0.6491;// kJ/kg.K
+s_fg4=7.5019;// kJ/kg.K
+s_4=s_3;// kJ/kg.K
+x_4=(s_4-s_f4)/s_fg4;// The quality of steam at state 4
+q_B=h_3-h_2;// The heat input in kJ/kg
+h_f4=192;// kJ/kg
+h_fg4=2393;// kJ/kg
+h_4=h_f4+(x_4*h_fg4);// kJ/kg
+w_T=h_3-h_4;// kJ/kg
+n_2=w_T/q_B;// The cycle efficiency
+// From example 8.1
+n_1=0.3232;// The power cycle efficiency at P_2=2 MPa
+Pi=((n_2-n_1)/n_1)*100;// The percentage increase in efficiency
+printf(" The percentage increase in efficiency is %1.2f percentage.",Pi);