Question

A cylindrical aluminum fin has a 1cm diameter and is 30 cm in length. It is attached to a surface at 80 and is exposed to the air at 220. The thermal conductivity is 237 W/m*k and the heat transfer coefficient is 18 W/m*k. Assume the fin to be very long. What is the rate of heat transfer from the fin effectiveness?

Answer 1

Answer:

Q = 42.877

e = 76.87 %

Explanation:

Given:

- The diameter of the Aluminium rod d = 1 cm

- The Length of the Aluminium rod L = 30 cm

- The base Temperature of rod Tb = 80°C

- The temperature of ambient air T∞ = 220°C

- The thermal conductivity of rod k = 237 W/mK

- heat transfer coefficient is h = 18 W/m*k

Find:

What is the rate of heat transfer from the fin

What is the fin effectiveness?

Solution:

- The perimeter P of the fin:

                  P = pi*d

                  P = pi*0.01 = 0.03141 m

- The area of rod cross section A:

                  A = pi*d^2 / 4 = pi*0.01^2 / 4

                  A = 0.00007 m^2

- The heat transfer rate of fin:

                  $\\Q = \sqrt{h*P*k*A}*(T_i_n_f - Tb)$                  

Then:

                  $Q = \sqrt{18*237*0.0007*0.03141}*(220 - 80)\\\\Q =42.877 W$

- The effectiveness e of an infinitely long pin is as follows:

                  $e = \sqrt{\frac{k*P}{h*A} } \\\\e = \sqrt{\frac{237*0.03141}{18*0.00007} } \\\\e = 76.87%$