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+// Example 22_6
+clc;funcprot(0);
+//Given data
+p_1=215;// bar
+T_1=500;// °C
+p_2=40;// bar
+T_2=280;// °C
+p_3=p_2-1;// bar
+p_4=8;// bar
+T_4=270;//°C
+p_5=p_4-0.5;// bar
+p_6=0.07;// bar
+m=10;// The flow of steam in kg/sec
+
+//Calculation
+// From h-s diagram
+h_1=3234;// kJ/kg
+h_2a=2822;// kJ/kg
+h_2=2910;// kJ/kg
+h_3=3435;// kJ/kg
+h_4a=2977;// kJ/kg
+h_4=2998;// kJ/kg
+h_5=3473;// kJ/kg
+h_6a=2444;// kJ/kg
+h_6=2578;// kJ/kg
+//From steam tables,
+h_f7=162;// kJ/kg
+W=(h_1-h_2)+(h_3-h_4)+(h_5-h_6);// Work done per kg of steam kJ/kg
+Q=(h_1-h_f7)+(h_3-h_2)+(h_5-h_4);// Heat supplied per kg of steam kJ/kg
+n_th=(W/Q)*100;//The thermal efficiency of the cycle in %
+P=(W*m);// Power developed by the plant in kW
+n_i1=((h_1-h_2)/(h_1-h_2a))*100;//Isentropic efficiency of the first stage in %
+n_i2=((h_3-h_4)/(h_3-h_4a))*100;//Isentropic efficiency of the second stage in %
+n_i3=((h_5-h_6)/(h_5-h_6a))*100;//Isentropic efficiency of the third stage in %
+printf('\n(a)The thermal efficiency of the cycle=%0.1f percentage \n Power developed by the plant=%0.0f kW \n(b)Isentropic efficiency of the first stage=%0.1f percentage \n Isentropic efficiency of the second stage=%0.1f percentage \n Isentropic efficiency of the third stage=%0.0f percentage',n_th,P,n_i1,n_i2,n_i3);