Question

Two rods, one of length $L$ and the other of length $2L$ are made of the same material and have the same diameter. The two ends of the longer rod are maintained at ${100}^{\circ }C.$ One end of the shorter rod is maintained at ${100}^{\circ }C$ while the other end is insulated. Both the rods are exposed to the same environment at ${40}^{\circ }C.$ The temp at the insulated end of the shorter rod is measured to be ${55}^{\circ }C.$ The temp at the midpoint of the longer rod would be.

• A.
  ${40}^{\circ }C.$
• B.
  ${50}^{\circ }C$
• C.
  ${55}^{\circ }C$
• D.
  ${100}^{\circ }C$

Answer 1

The correct option is C

${55}^{\circ }C$


Temperature distribution along the rod is given by

$\theta =T-T_{\infty }$
$\theta =\frac{{\theta }_{1}sinhm(l-x)+{\theta }_{2}sinhmx}{sinhml}$

where $x=L:l=2L{\theta }_1=100-40={60}^{\circ }C{\theta }_2=100-40={60}^{\circ }C{\theta }_1={\theta }_2$

$\theta =\frac{{\theta }_1[sinhml+sinhml]}{sinh2ml}$

$\theta ={\theta }_1\frac{2sinhml}{sinh2ml}......\left(i\right)$

For rod of length $l$ with insulated tip temperature distribution

$\theta ={\theta }_0\frac{cosh\left[m\right(l-x\left)\right]}{coshml}$
for $x=L$
${\theta }_L={\theta }_0\frac{1}{coshml}$
$coshml=\frac{{\theta }_0}{{\theta }_L}=\frac{100-40}{50-40}=\frac{60}{15}=4ml=\left(cos{h}^{-1}4\right)=2.063$
For same material fins
By Eqs.(i) and (ii), we get

${\theta }_{l/2}=60\times \frac{2sinh2.063}{sinh4.126}$
$T_{l/2}-T_{\infty }=15T_{l/2}={55}^{\circ }C$