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+// Example 22_19
+clc;funcprot(0);
+//Given data
+P=100;// MW
+p_2=80;// bar
+p_3=7;// bar
+p_5=0.05;// bar
+T_4=350;// °C
+
+// Calculation
+// From h-s chart:
+h_2=2990;// kJ/kg
+h_3=2350;// kJ/kg
+h_4=3170;// kJ/kg
+h_5=2180;// kJ/kg
+// From steam tables
+h_f6=138;// kJ/kg
+h_f7=697;// kJ/kg
+function[X]=mass(y)
+    X(1)= (y(1)*(h_3-h_f7))-((1-y(1))*(h_f7-h_f6));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);
+m_p=m*100;// Percentage of bled steam in %
+W=(h_2-h_3)+((1-m)*(h_4-h_5));// kJ/kg
+Q_s=(h_2-h_f7)+((1-m)*(h_4-h_3));// kJ/kg
+n=(W/Q_s)*100;// The efficiency of the power plant in %
+m_b=((P*10^3)/((h_2-h_3)+((1-m)*(h_4-h_5))));// tons/hr
+printf('\nThe percentage of bled steam=%0.1f percentage \nThe thermal efficiency of the cycle=%0.0f percentage \nBoiler generating rate=%0.0f tons/hr',m_p,n,m_b);
+// The answer provided in the textbook is wrong
+