Question

A square cross sectional area beam of area $529m{m}^2$ is protruded from a-furnace which is maintained at $1700{}^{\circ }C$ . The system can be idealized as a fin with base temperature same as that of furnace. At steady state beam length was such that if loses 90 percent of heat which might have lost by any infinite long fin. If the thermal conductivity and convective heat transfer coefficient is given by $80W/mK$ and $210W/m^{2}K$. What was the length of beam? Assuming that it loses heat from end surface also.

• A. $73.21mm$
• B. $68.91mm$
• C. Insufficient data is provided
• D. $63.15mm$

Answer 1

The correct option is D $63.15mm$

Given: $h=210W/m^{2}K,k=80W/mK$

$t^2=529m{m}^2$

$t=23mm$

Heat transfer from infite long fin,

$q_{\infty }=\sqrt{hPkA}\times {\theta }_0$

and heat transfer from actal beam,

$q_{\text{beam}}=\sqrt{hPkA}\times {\theta }_0\times \tan h\left(m{L}_c\right)$

$q_{\text{beam}}=0.9q$

i.e $\tan h\left(m{L}_C\right)=0.9$

$m{L}_C=1.47222$

$m=\sqrt{\frac{hP}{kA}}=\sqrt{\frac{210\times 4\times 22\times {10}^{-3}}{80\times 529\times {10}^{-6}}}$

$=21.366permeter$

From equation (i)

$L_C=0.0689\Rightarrow 68.9mm$

$L_C=L+\frac{A}{P}\Rightarrow 68.9mm$

$L=68.9-\frac{529}{4\times 23}=63.15mm$