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+// Example 17_2
+clc;funcprot(0);
+//Given data
+P_1=100;// bar
+T_1=400;// °C
+T_wi=20;// °C
+P_v=71;// cm of Hg
+P_b=76;//cm of Hg
+gradT=10;// Rise in temperature of the cooling water in °C
+rho_w=1080;// kg/m^3
+C_pw=4.6;//kJ/kg.°C
+U=400;// The over all heat transfer coefficient in W/m^2.°C
+P=30;// kW
+
+//Calculation
+P_2=(P_b-P_v)*0.01359;// The pressure in the condenser in bar
+// From h-s chart,
+h_1=3389;// kJ/kg
+h_2=2054;//kJ/kg
+m_s=(P*1000)/(h_1-h_2);// Mass of steam in kg/sec
+x_2=0.782;// dryness fraction from h-s chart
+// Heat lost by steam in condenser = Heat gained by water
+h_f2=159.6;//kJ/kg
+m_w=((h_2-h_f2)*m_s)/(C_pw*gradT);// kg/sec
+CP=m_w;// Capacity of the pump in kg/sec
+Theta_i=(38-20);//°C
+Theta_o=(38-30);//°C
+LMTD=(Theta_i-Theta_o)/log(Theta_i/Theta_o);//°C
+A=((h_2-h_f2)*m_w)/(U*LMTD);// The heat transfer area of the condenser in m^2
+printf('\n(a)The mass of steam supplied to the turbine,m_s=%0.1f kg/sec \n(b)Capacity of the pump=%0.1f kg/sec \n(c) The heat transfer area of the condenser=%0.1f m^2',m_s,m_w,A);
+// The answer vary due to round off error