Question

An iron fin$(k=210\frac{kJ}{m{h}^{\circ }C})$ has length 50 mm, width 100 mm and thickness 5 mm. Assume, $h=42kJ/m^2{h}^{\circ }C$ and the temperature at the base of the fin as ${80}^{\circ }C$. If the atmosphere temperature is ${20}^{\circ }C$, the temperature at the tip of the fin will be

• A. ${74.21}^{\circ }C$
• B. ${34.21}^{\circ }C$
• C. ${24.3}^{\circ }C$
• D. ${60}^{\circ }C$

Answer 1

The correct option is A ${74.21}^{\circ }C$


$\text{Perimeter ,}p=2(W+t)=2(0.1+0.005)=0.21m$

$\text{Area},A_c=0.1\times 0.005=5\times {10}^{-4}{m}^2$

$m=\sqrt{\frac{hp}{k{A}_c}}=\sqrt{\frac{42\times 0.21}{210\times 5\times {10}^{-4}}}=9.1651$

$mL=9.1651\times 0.05=0.458,mL<2.67$

Fin is heated as finite with insulated tip

$\frac{T-T_{\infty }}{T_o-T_{\infty }}=\frac{\cos hm(L-x)}{\cos hmL}$

At tip , x = L, then
$\frac{T_L-T_{\infty }}{T_o-T_{\infty }}=\frac{1}{\cos h\left(mL\right)}$

$\frac{T_L-20}{80-20}=\frac{1}{\cos h\left(0.458\right)}$

$\Rightarrow T_L={74.21}^{\circ }C$