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authorpriyanka2015-06-24 15:03:17 +0530
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+//Variable declaration:
+Nf = 125 //Array of fins per meter
+w = 1 //Length of fin (m)
+//From example 17.15:
+t = 4/10**3 //Thickness of fin (m)
+Do = 50/10**3 //Outside diameter of tube (m)
+Af = 7.157*10**-3 //Fin surface area (m^2)
+h = 40 //Heat transfer coefficient (W/m^2.K)
+DTb = 180 //Excess temperature at the base of the fin (K)
+Qf = 50 //Fin heat transfer rate (W)
+
+//Calculation:
+ro = Do/2 //Radius of tube (m)
+wb = w-Nf*t //Unfinned exposed base length (m)
+Ab = 2*%pi*ro*wb //Area of the base of the fin (m^2)
+At = Ab+Nf*Af //Total heat transfer surface area (m^2)
+Qw = h*(2*%pi*ro*w)*DTb //Heat rate without fin (W)
+Qb = h*Ab*DTb //Heat rate from the base (W)
+Qft = Nf*Qf //Heat rate from the fin (W)
+Qt = Qb+Qft //Total heat rate (W)
+Qm = h*At*DTb //Maximum heat transfer rate (W)
+n = Qt/Qm //Overall fin efficiency
+E = Qt/Qw //Overall fin effectiveness
+Rb = 1/(h*Ab) //Thermal resistance of base ( C/W)
+Rf = 1/(h*Nf*Af*n) //Thermal resistance of fin ( C/W)
+
+//Result:
+printf("The rate of heat transfer per unit length of tube is : %.1f W .",Qt)
+printf("Or, the rate of heat transfer per unit length of tube is : %.2f kW .",Qt/10**3)
+printf("The overall fin efficiency is : %.1f %%",n*100)
+printf("The overall fin effectiveness is : %.2f .",E)