Question

A circumferential fin (k= 55 W/m C) of thickness 3 mm and length 3 cm is attached to a pipe of diameter 3 cm. The fin is exposed to a convection environment at 23 C with h=25 W/m2 oC. The fin base temperature is 223 C. Calculate the heat lost (qf) by the fin.

Answer 1

Given: Thickness of the fin, t = 3mm= 0.003 m

Length of the fin, L= 3cm= 0.03 m

Diameter of the pipe, d = 3cm=0.03 m Radius, r₁ = 0.015m

Temperature of the surroundings, = 23 ⁰C

Base temperature of fin, T_b= 223 ⁰C

Convective heat transfer coefficient, h= 25 W/m^{2 0}C

Thermal conductivity of fin material, k=55 W/m⁰C

Assuming that the fin is insulated

For circumferential fin with rectangular profile, the corrected length, L_c of the fin is

L_c = 0.0315 m

Corrected fin radius, r_2c is r_2c=L_c+r₁

r_2c= 0.0315 m+ 0.015 m

r_2c = 0.0465 m

The ratio of corrected radius and actual radius is r_2c/r₁=0.0465/0.015

r_2c/r₁= 3.1

The heat lost by fin, q_f is calculated using the expression

where h is the Convective heat transfer coefficient

A is the Cross sectional Area of fin,

is the Temperature difference,

is the Efficiency of the fin

Efficiency of the fin is obtained from the Figure Efficiencies of Circumferential fins of rectangular profile.

Obtain the value of

where A_m is the Profile Area, A_m=L_ct

The efficiency corresponding to 0.39 and r_2c/r₁= 3.1 is approximately 80%=0.80.

Thus the heat lost by the fin is