Question

Inner surface of a cylindrical shell of length $l$ and of material of thermal conductivity $k$ is kept at constant temperature $T_1$ and outer surface of the cylinder is kept at constant temperature $T_2$ such that $(T_1>T_2)$ as shown in figure. Heat flows from inner surface to outer surface radially outward. Inner and outer radii of the shell are $R$ and $2R$ respectively. Due to lack of space this cylinder has to be replaced by a smaller cylinder of length $\frac{l}{2}$ inner and outer radii $\frac{R}{4}$ and $R$ respectively and thermal conductivity of material $nk$. If rate of radially outward heat flow remains same for same temperatures of inner and outer surface i.e., $T_1$ and $T_2$, then find the value of $n$.

Answer 1

Rate of heat flow through the two cylinders is the same.

i.e $i_{t{h}_1}=i_{t{h}_2}$ (thermal currents equal)

Take a thin cylindrical element of thickness $dr$ at radius $r$

$i_{th}=-k\left(2\pi xl\right)\frac{dT}{dx}$
$\frac{dx}{x}{i}_{th}=-k\left(2\pi l\right)dT$

Integrating both sides,
${\int }_{R_1}^{R_2}\frac{dx}{x}{i}_{th}=-k\left(2\pi l\right){\int }_{T_1}^{T_2}dT$

$ln\frac{R_2}{R_1}{i}_{th}=-k.2\pi l(T_2-T_1)$
$i_{th}=\frac{-k.2\pi l(T_2-T_1)}{ln\frac{R_2}{R_1}}$

For cylinder 1,
$i_{t{h}_1}=\frac{-k.2\pi l(T_2-T_1)}{ln2}$

For cylinder 2,
$i_{t{h}_1}=\frac{-nk.2\pi \frac{l}{2}(T_2-T_1)}{ln4}$

Now, $i_{t{h}_1}=i_{t{h}_2}$
$\Rightarrow n=4$