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+clc;
+clear;
+//Example 3.42
+Ts=373 //[K]
+rho_l=957.9 //rho*l[kg/m^3]
+Cpl=4217 //[J/kg.K]
+mu_l=27.9*10^-5 //[kg/(m.s)]
+rho_v=0.5955 //[kg/m^3]
+Csf=0.013
+sigma=5.89*10^-2 //[N/m]
+Nprl=1.76
+lambda=2257 //[kJ/kg]
+lambda=lambda*1000 //in [J/kg]
+n=1 //for water
+m_dot=30 //Mass flow rate [kg/h]
+m_dot=m_dot/3600 //[kg/s]
+D=30 //Diameter of pan [cm]
+D=D/100 //[m]
+g=9.81 //[m/s^2]
+A=%pi*(D/2)^2 //Area in [sq m]
+Q_by_A=m_dot*lambda/A //[W/sq m]
+//For nucleate boiling point we have:
+dT=(lambda/Cpl)*Csf*(((Q_by_A)/(mu_l*lambda))*sqrt(sigma/(g*(rho_l-rho_v))))^(1.0/3.0)*(Nprl^n) //[K]
+Tw=Ts+dT //[K]
+printf("\n Temperature of the bottom surface of the pan is %f W/(sq m)",Tw);