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+// Example 10_9
+clc;funcprot(0);
+// Given data
+m=18.0;// kg/s
+T_b=350.0;// °C
+W=20*10^3;// kW
+// Station 1
+T_1=500.0;// °C
+p_1=3.00;// MPa
+h_1=3456.5;// kJ/kg
+s_1=7.2346;// kJ/kg.K
+// Station 2
+p_2=0.0100;// MPa
+x_2=0.960;// The quality of steam
+h_2f=191.8;// kJ/kg
+h_2fg=2392.8;// kJ/kg
+h_2=h_2f+(x_2*h_2fg);// kJ/kg
+s_2f=0.6491;// kJ/kg.K
+s_2fg=7.5019;// kJ/kg.K
+s_2=s_2f+(x_2*s_2fg);// kJ/kg.K
+// Ground state 
+x_0=0.00;// The quality of steam
+T_0=20.0;// °C
+h_0=83.9;// kJ/kg
+s_0=0.2965;// kJ/kg.K
+
+// Calculation
+a_f1=(h_1-h_0)-((T_0+273.15)*(s_1-s_0));// kJ/kg
+a_f2=(h_2-h_0)-((T_0+273.15)*(s_2-s_0));// kJ/kg
+Q=(W+(m*(a_f2-a_f1)))/(1-((T_0+273.15)/(T_b+273.15)));// kW
+printf("\nThe rate of heat loss from the surface of the turbine,Q=%4.0f kW",Q);
+// The answer vary due to round off error