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+// Example 4_18
+clc;funcprot(0);
+// Given data
+P_2=4000;// kPa
+T_2=600;// °C
+P_1=20;// kPa
+v=0.001;// m^3/kg
+m=1;// kg
+h_1=251.4;// kJ/kg
+
+// Calculation
+w_P=(P_2-P_1)*v;// kJ/kg
+h_2=w_P+h_1;// kJ/kg
+// From steam tables
+h_3=3674;// kJ/kg
+h_4=2610;// kJ/kg
+q_B=h_3-h_2;// kJ/kg
+w_T=h_3-h_4;// The work output in kJ/kg
+n=(w_T-w_P)/q_B;// The thermal efficiency
+printf("\nThe thermal efficiency,n=%0.3f or %2.1f percentage",n,n*100);