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+// Example 17_17
+clc;funcprot(0);
+//Given data
+m_s=250;// tons/hr
+T_s=40;// °C
+T_wi=30;//°C
+T_wo=36;//°C
+U_o=2.5;//kW/m^2°C
+P_t=0.078;// bar
+v=1.8;// m/s
+d_i=23;// mm
+d_o=25;// mm
+rho_w=1000;// kg/m^3
+moisture=12;// Percentage
+x_2=(100-12)/100;// Dryness fraction
+p_t=0.078;// bar
+C_pw=4.2;// kJ/kg.°C
+R=287;// J/kg°C
+v=1.8;// m/s
+
+//Calculation
+//From steam tables,at 40°C\
+p_sat=0.074;//bar
+h_fg2=2407;// kJ/kg
+v_g2=19.54;// m^3/kg
+//gradh=H_2-h_3
+gradh=x_2*h_fg2;// kJ/kg
+m_s=(250*1000)/3600;// kg/sec
+m_w=(m_s*gradh)/(C_pw*(T_wo-T_wi));// kg/sec
+p_air=p_t-p_sat;// bar
+v_s2=x_2*v_g2;// m^3/kg
+m_a=(m_s*v_s2*p_air*10^5)/(R*(T_s+273));// kg/sec
+Theta_i=(T_s-T_wi);// °C
+Theta_o=(T_s-T_wo);// °C
+LMTD=(Theta_i-Theta_o)/(log(Theta_i/Theta_o));//Logrithemic mean temperature difference in °C
+A_s=(m_s*gradh)/(U_o*LMTD);// m^2
+n=(m_w)/((%pi/4)*(d_i/1000)^2*rho_w*v);// Number of tubes 
+L=A_s/(%pi*(d_o/1000)*n);// Length in m
+printf('\nQuantity of water circulation=%0.0f kg/sec \nAir leakage in the condenser=%0.2f kg/sec \nThe length of each tube,L=%0.1f m \nNumber of condenser tubes,n=%0.0f',m_w,m_a,L,n);
+// The answer vary due to round off error