Question

A fin has $5mm$ diameter and $100mm$ length. The thermal conductivity of fin material is $400W{m}^{-1}{K}^{-1}.$ One end of the fin is maintained at ${130}^{\circ }C$ and its remaining surface is exposed to ambient air at ${30}^{\circ }C$. If the convective heat transfer coefficient is $40W{m}^{-2}{K}^{-1},$ the heat loss (in W) from the fin is

• A.
  $0.08$
• B.
  $5.0$
• C.
  $7.0$
• D.
  $7.8$

Answer 1

The correct option is B

$5.0$
Given data:
diameter of fin,
$d=5mm=0.005m$
$\therefore$ cross-sectional area,
$A=\frac{\pi }{4}{d}^2=\frac{3.14}{4}\times (0.005{)}^2$
$=1.96\times {10}^{-5}{m}^2$
Length of fin,
$L=100mm=0.1m$
Thermal conductivity
$K=400W/mK$
$T_{\circ }={130}^{\circ }C$
$T_{\infty }={30}^{\circ }C$
$h=40W/m^{2}K$

Perimeter: $P=\pi d=3.14\times 0.005=0.0157m$
$m=\sqrt{\frac{hP}{kA}}=\sqrt{\frac{40\times 0.0157}{400\times 1.95\times {10}^{-5}}}$
$=8.97m^{-1}$
$mL=8.97\times 0.1=0.897$
For long fin, $mL>2.64$
So, the fin is insulated tip,

$Q_{fin}=\sqrt{hPkA}(T_{\circ }-T_{\infty })tanhmL$
$=\sqrt{40\times 0.0157\times 400\times 1.95\times {10}^{-5}}$
$(130-30)tanh(8.97\times 0.1)$
$=0.0699\times 100\times tanh0.897$
$=0.0699\times 100\times 0.7148=4.99W\simeq 5W$