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+// Example 22_28
+clc;funcprot(0);
+//Given data
+p_1=80;// bar
+T_1=470;// °C
+p_2=7;// bar
+T_1=350;// °C
+p_3=0.035;// bar
+m_s=100;// kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3350;// kJ/kg
+h_2=2770;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2220;// kJ/kg
+// From steam tables
+h_f5=112;// kJ/kg
+h_f6=697;// kJ/kg
+function[X]=mass(y)
+    X(1)= (y(1)*(h_2-h_f6))-((1-y(1))*(h_f6-h_f5));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+m_b=m*100;// The fraction of steam bled for reheating in %
+Q_s=(h_1-h_f6)+((1-m)*(h_3-h_2));// Heat supplied in the boiler and reheater in kJ/kg
+W=(h_1-h_2)+((1-m)*(h_3-h_4));// Power output in kJ/kg
+P=(W*100)/1000;// Capacity of the plant in MW
+n=(W/Q_s)*100;// The efficiency of the plant in %
+printf('\n(a)Fraction of steam bled for feed heating=%0.0f percentage \n(b)Heat supplied per kg of steam in boiler and turbine=%0.0f kJ/kg \n(c)Power output of the plant=%0.0f MW \n(d)Thermal efficiency of the plant=%0.1f percentage',m_b,Q_s,P,n);
+// The answer vary due to round off error