A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm^-1K^-1.
A fin has 5 mm diameter and 100 mm length. The thermal conductivity of fin material is 400 Wm^-1K^-1. One end of the fin is maintained at 130°C and its remaining surface is exposed to ambient air at 30°C. If the convective heat transfer coefficicent is 40 Wm^-2K^-1, what is the heat loss from the fin?
Diameter of fin is 5mm and length is 100 mm, m=\sqrt{\frac{hP}{KA}},P = \pi d,A = \frac{\pi}{4}d^{2}
\Rightarrow \frac{P}{A}=\frac{4}{d}
\Rightarrow m = \sqrt{\frac{4h}{Kd}}=\sqrt{\frac{4\times40}{400\times5\times10^{-3}}}=8.94
mL = 8.94\times0.1=0.894
mL<2.64
Q_{f_{in}} = mAK\left ( T_{0}-T_{\infty } \right )tan h\left ( mL \right )
\Rightarrow Q_{f_{in}} = 8.94\times\frac{\pi}{4}\left ( 5\times10^{-3} \right )^{2}\times400\times100\times tanh\left ( 8.94\times0.1 \right )=5.008\,W
