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+// Example 22_26
+clc;funcprot(0);
+//Given data
+p_1=80;// bar
+T_1=470;// °C
+p_2=7;// bar
+T_1=350;// °C
+p_3=0.35;// bar
+m_s=50;// kg/sec
+
+//Calculation
+//From h-s chart:
+h_1=3310;// kJ/kg
+h_2=2780;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2220;// kJ/kg
+// From steam tables
+h_f2=697;// kJ/kg
+h_6=h_f2;// kJ/kg
+h_5=111.85;// kJ/kg
+h_f4=h_5;// kJ/kg
+function[X]=mass(y)
+    X(1)= (y(1)*(h_2-h_f2))-((1-y(1))*(h_f2-h_f4));
+endfunction
+y=[0.1];
+z=fsolve(y,mass);
+m=z(1);// kg
+m_b=m*100;// Amount of steam bled off in %
+m_l=(100-m_b);// Amount of steam supplied to L.P turbine in %
+Q_b=h_1-h_6;// kJ/kg
+Q_r=(1-m)*(h_3-h_2);// kJ
+Q_s=Q_b+Q_r;// Total amount of heat supplied by the boiler and reheater in kJ/kg
+W=(h_1-h_2)+((1-m)*(h_3-h_4));// kJ/kg
+n=(W/Q_s)*100;
+P=(m_s*W)/1000;// Power developed by the steam in MW
+printf('\n(a)Amount of steam bled off for feed heating=%0.0f percentage \n(b)Amount of steam in LP turbine=%0.0f percentage \n(c)Heat supplied in the boiler and reheater=%0.1f kJ/kg \n(d)Cycle efficiency=%0.1f percentage \n(e)Power developed by the steam=%0.1f MW',m_b,m_l,Q_s,n,P);
+// The answer provided in the textbook is wrong