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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)
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