Question

A $2m$ long wire of resistance $4ohm$ and diameter $0.64mm$ is coated with plastic insulation of thickness $0.06mm$. When a current of $5A$ flows through the wire, find the temperature difference across the insulation in steady state.
If $K=0.16\times {10}^{-2}cal/c{m}^{\circ }Csec$.

Answer 1

Consider a concentric cylindrical shell of radius $r$ and thickness $dr$ as shown in figure. The radial rate of flow of heat through this shell in steady state will be
$H=\frac{dQ}{dt}=-KA\frac{d\theta }{dt}$ Negative sign is used as with increase in $r,\theta$ decreases
Now as for cylindrical shell $A=2\pi rL$
$H=-2\pi rLK\frac{d\theta }{dt}$
or, ${\int }_a^n\frac{dt}{r}=-\frac{-2\pi LK}{H}{\int }_{{\theta }_1}^{{\theta }_2}d\theta$
which on integration and simplification gives
$H=\frac{dQ}{dt}=-\frac{-2\pi LK(\theta -{\theta }_2)}{ln\left(\frac{b}{a}\right)}.....\left(1\right)$
Here, $H=\frac{I^{2}R}{4.2}=\frac{(5{)}^2\times 4}{4.2}=24\frac{cal}{sec};L=2m=200cm$
$r_1=\left(0.64/2\right)mm=0.032cm$ and
$r_2=r_1+d=0.032+0.006=0.038cm$
$({\theta }_1-{\theta }_2)=\frac{24\times ln\left(\frac{38}{32}\right)}{2\times 3.14\times 200\times 0.16\times {10}^{-2}}$

$\Rightarrow ({\theta }_1-{\theta }_2=\frac{55\times [1.57-1.50]}{2}=2^{\circ }C$.