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+// Example 8_9

+clc;funcprot(0);

+// Given data

+P_1=10;// kPa

+P_2=2;// MPa

+T_3=600;// °C

+n_T=80/100;// The efficiency of the turbine

+

+// Calculation

+// From the steam tables we find

+h_2=192;// kJ/kg

+h_1=h_2;// kJ/kg

+h_3=3690;// kJ/kg

+s_3=7.7032;// kJ/kg.K

+s_4a=s_3;// kJ/kg.K

+s_f4a=0.6491;// kJ/kg.K

+s_fg4a=7.5019;// kJ/kg.K

+x_4a=(s_4a-s_f4a)/s_fg4a;// The quality of steam at state 4'

+h_f4a=192;// kJ/kg

+h_fg4a=2393;// kJ/kg

+h_4a=h_f4a+(x_4a*h_fg4a);// kJ/kg

+w_a=n_T*(h_3-h_4a);// kJ/kg

+q_B=h_3-h_2;// kJ/kg

+n=w_a/q_B;// The cycle efficiency

+h_4=h_3-w_a;// kJ/kg

+// The temperature is interpolated to be

+T_4=(((2692-2688)/(2783-2688))*(150-100))+100;// °C

+printf("\nThe cycle efficiency,n=%0.3f or %2.1f percentage. \nThe temperature of steam at tthe turbine outlet,T_4=%3.0f°C",n,n*100,T_4);