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+// Example 8_8
+clc;funcprot(0);
+// Given data
+P_1=0.01;// MPa
+P_3=0.2;// MPa
+P_4=4;// MPa
+P_5=30;// MPa
+T_6=600;// °C
+T_8=T_6;// °C
+T_10=350;// °C
+mdot=1;// kg/s
+
+// Calculation
+// The enthalpies are found frrom the steam tables to be
+h_1=192;// kJ/kg
+h_2=h_1;// kJ/kg
+h_4=1087;// kJ/kg
+h_5=h_4;// kJ/kg
+h_8=3674;// kJ/kg
+h_3=505;// kJ/kg
+h_6=3444;// kJ/kg
+h_10=3174;// kJ/kg
+s_6=6.2339;// kJ/kg.K
+s_7=s_6;// kJ/kg.K
+h_7=(((6.2239-6.0709)/(6.3622-6.0709))*(2961-2801))+2801;// kJ/kg
+s_8=7.3696;// kJ/kg.K
+s_9=s_8;// kJ/kg.K
+h_9=(((6.2239-6.0709)/(6.3622-6.0709))*(2961-2801))+2801;// kJ/kg
+s_10=8.0636;// kJ/kg.K
+s_11=s_10;// kJ/kg.K
+s_f11=0.6491;// kJ/kg.K
+s_fg11=7.5019;// kJ/kg.K
+x_11=(s_11-s_f11)/s_fg11;// The quality of steam at state 11
+h_f11=192;// kJ/kg
+h_fg11=2393;// kJ/kg
+h_11=h_f11+(x_11*h_fg11);// kJ/kg
+mdot7=(h_5-h_3)/(h_7-h_3);// kg/s
+mdot9=(((1-mdot7)*h_3)-h_2+(mdot7*h_2))/(h_9-h_2);// kg/s
+W_T=((mdot)*(h_6-h_7))+((1-mdot7)*(h_8-h_9))+((1-mdot7-mdot9)*(h_10-h_11));// The power from the turbine in kW
+Q_B=((mdot)*(h_6-h_5))+((1-mdot7)*(h_8-h_7))+((1-mdot7-mdot9)*(h_10-h_9));// The boiler energy input in kW
+n=W_T/Q_B;// The cycle efficiency 
+printf("\nThe maximum possible cycle efficiency,n=%0.3f or %2.1f percentage.",n,n*100);