Question

A steel rod $(k=32W/m^{\circ }C)$ 12 mm in diameter and 60 mm long having insulated end is to be used as spine. It is exposed to surrounding with a temperature of ${60}^{\circ }C$ and heat transfer coefficient of $55W/m^{2\circ }C$. If the temperature at the base of fin is ${95}^{\circ }C$, the fin efficiency is determined as

• A. 78%
• B. 62.18 %
• C. 69.63 %
• D. 32 %

Answer 1

The correct option is B 62.18 %

Given data,

D = 0.012 m, L= 60 mm,

$T_{\infty }={60}^{\circ }C$ $h=55W/m^{2\circ }C$

$T_o={95}^{\circ }C$ $k=32W/m^{\circ }C$

$\text{Perimeter(p)}=\pi D=\pi \times 0.012=0.0377m$

$\text{Area}{A}_c=\frac{\pi }{4}{D}^2=\frac{\pi }{4}\times {0.012}^2=1.1305\times {10}^{-4}{m}^2$

$m=\sqrt{\frac{4h}{kD}}=\sqrt{\frac{4\times 55}{32\times 0.012}}=23.93$

$mL=23.93\times 60\times {10}^{-3}=1.436$

$\tan h\left(mL\right)=\tan h\left(1.436\right)=0.8929$

$\text{The fin efficiency,}\eta =\frac{\tan h\left(mL\right)}{mL}$

$=\frac{0.8929}{1.436}=62.18\%$