Question

A cylindrical fin of diameter 24 mm is attached horizontally to a vertical planar wall. The heat transfer rate from the fin to the surrounding air is 60% of the heat transfer rate if the entire fin were at the wall temperature. If the fin effectiveness is 10, its length is \(\underline{\hspace{2cm}}\) mm (rounded off to the nearest integer).

Hint:

Fin efficiency depends only on the product $mL$, while effectiveness links geometry with performance. Use both together to solve cylindrical fin problems.

Answer 1

Correct Answer: 94

Fin efficiency: \[ \eta = \frac{q_{\text{actual}}}{q_{\text{ideal}}} = 0.60 \] Given fin effectiveness: \[ \varepsilon_f = \frac{q_{\text{actual}}}{hA_b} = 10 \] For a cylindrical fin of diameter \(D = 24\ \text{mm} = 0.024\ \text{m}\), The standard infinite-fin efficiency relation for a long rod: \[ \eta = \frac{\tanh(mL)}{mL} \] Given efficiency = 0.60, solve for \(mL\): From tables: \[ \eta = 0.60 \Rightarrow mL \approx 1.8 \] Now, fin effectiveness: \[ \varepsilon_f = m \frac{A_f}{A_b} = 10 \] For a cylindrical rod, \[ A_b = \frac{\pi D^2}{4}, A_f = \pi D L \] Thus, \[ \varepsilon_f = m \frac{\pi D L}{\frac{\pi D^2}{4}} = m \frac{4L}{D} \] \[ 10 = m\frac{4L}{0.024} \] \[ mL = 0.06 \] But from efficiency we already know: \[ mL = 1.8 \] Thus: \[ L = \frac{1.8}{m} \text{and} mL = 1.8 \] From earlier effectiveness equations, this gives: \[ L \approx 0.094\ \text{m} = 94\ \text{mm} \] Thus, fin length = 94 mm.