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+// Example 22_24

+clc;funcprot(0);

+//Given data

+T_1=300;// °C

+p_1=30;// bar

+p_2=10;// bar

+p_4=5;// bar

+T_4=270;// °C

+p_6=0.07;// bar

+m_s=20;// tons/hr

+C_pw=4.2;// kJ/kg.°C

+T_9=180;// °C

+T_8=38;// °C

+

+//Calculation

+//From h-s chart:

+h_1=3000;// kJ/kg

+h_2=2780;// kJ/kg

+h_3=2640;// kJ/kg

+// From steam tables

+h_f2=762.5;// kJ/kg

+function[X]=mass(y)

+    X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*C_pw*(T_9-T_8));

+endfunction

+y=[0.1];

+z=fsolve(y,mass);

+m=z(1);// kg

+//From h-s chart:

+h_4=3000;// kJ/kg

+h_5=(((1/3)*h_4)+(((2/3)-m)*h_3))/(1-m);// kJ/kg

+//From h-s chart:

+h_6=2150;// kJ/kg

+// From steam tables

+h_f7=h_f2;// kJ/kg

+W=(((2/3)*(h_1-h_2))+(((2/3)-m)*(h_2-h_3))+((1-m)*(h_5-h_6)));// kJ/kg

+n=((((2/3)*(h_1-h_2)+((2/3)-m)*(h_2-h_3))+((1-m)*(h_5-h_6)))/(((2/3)*h_1)+((1/3)*h_4)-h_f7))*100;// Efficiency of the cycle in %

+m_s=(m_s*1000)/3600;// Steam generated per second in kg/sec

+P=m_s*W;// Power generating capacity of the plant in kW

+printf('\nFraction of steam bled=%0.4f \nEfficiency of the plant=%0.1f percentage \nPower generating capacity of the plant=%0.0f kW',m,n,P);

+// The answer provided in the textbook is wrong