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+// Example 22_23
+clc;funcprot(0);
+//Given data
+P=120;// MW
+p_1=86;// bar
+p_2=7;// bar
+p_3=0.35;// bar
+T_1=350;// °C
+
+//Calculation
+//From h-s chart:
+h_1=2980;// kJ/kg
+h_2=2520;// kJ/kg
+h_3=3170;// kJ/kg
+h_4=2550;// kJ/kg
+// From steam tables
+h_f1=304.3;// kJ/kg(liquid heat at 0.35 bar)
+T_s1=72.7;// °C
+h_f2=697;// kJ/kg(liquid heat at 7 bar)
+T_s2=165;// °C
+h_f4=h_f1;// 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);// tons/hr
+S=(1/m);// The ratio of steam bled to steam generated
+m_s=((P*10^3)/((h_1-h_2)+((1-m)*(h_3-h_4))))*(3600/1000);// kg/sec
+n_th=(((h_1-h_2)+((1-m)*(h_3-h_4)))/((h_1-h_f1)+((1-m)*(h_3-h_2))))*100;
+printf('\n(a)The ratio of steam bled to steam generated=%0.2f \n(b)The boiler generating capacity=%0.1f tons/hr \n(c)The thermal efficiency of the cycle=%0.1f percentage',S,m_s,n_th);
+// The answer vary due to round off error