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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