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+
+clc;
+clear;
+//Example 3.51
+rho=975 //[kg/m^3]
+k=0.871 //[W/m.K]
+dT=10   //[K]
+mu=380.5*10^-6  //[N.s/m^2]
+lambda=2300 //[kJ/kg]
+lambda=lambda*1000  // Latent heat of condensation [J/kg]
+Do=100  //Outer diameter [mm]
+Do=Do/1000  //[m]
+g=9.81  //[m/s^2]
+//for horizontal tube
+h1=0.725*((rho^2*lambda*g*k^3)/(mu*Do*dT))^(1/4)    //Average heat transfer coefficient
+//for vertical tube
+//h2=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)    //Average heat transfer coefficient
+h2=h1   //For vertical tube
+//implies that
+L=(0.943*((rho^2*lambda*g*k^3)^(1/4))/(h1*((mu*dT)^(1/4))))^4   //[m]
+L=0.29      //Approximate in book
+h=0.943*((rho^2*lambda*g*k^3)/(mu*L*dT))^(1/4)  //[W/(sq m.K)]
+A=%pi*Do*L  //Area in [m^2]
+Q=h*A*dT    //Heat transfer rate [W]
+mc_dot=Q/lambda  //[Rate of condensation]in [kg/s]
+mc_dot=mc_dot*3600  //[kg/h]
+printf("\n Tube length is %f m\n",L);
+printf("\n Rate of condemsation per hour is %f kg/h",mc_dot);