Question

A cylindrical pipe of length $L$ has two layers of material of conductivity $k_1$ and $k_2$ (see figure). If the innermost surface of the cylinder is maintained at $T_1$ and the outermost surface is at $T_2(<T_1)$, calculate the radial rate of heat flow.

• A. $\frac{2\pi k_{1}L(T_1-T_2)}{\text{ln}\left(\frac{l_2}{l_1}\right)}$
• B. $\frac{2\pi k_{2}L(T_1-T_2)}{\ln \left(\frac{l_2}{l_1}\right)}$
• C. $\frac{(T_1-T_2)}{\frac{1}{2\pi k_{1}L}\ln \left(\frac{r_2}{r_1}\right)+\frac{1}{2\pi k_{2}L}\ln \left(\frac{r_3}{r_2}\right)}$
• D. $\frac{(T_1-T_2)2\pi \left(k_1{k}_2\right)}{(k_1+k_2)\ln \left(\frac{r_2}{r_1}\right)}$

Answer 1

The correct option is C $\frac{(T_1-T_2)}{\frac{1}{2\pi k_{1}L}\ln \left(\frac{r_2}{r_1}\right)+\frac{1}{2\pi k_{2}L}\ln \left(\frac{r_3}{r_2}\right)}$

Given, both concentric cylinders are made up of different materials are connected in series combination.
In this case, two thermal resistances are in series.
Thus we can write that,
$R=R_1+R_2$
Thermal resistance offered by a cylindrical shell of inner radius $a$ and outer radius $b$ and thermal conductivity $k$ is given by
$R=\frac{\ln \left(\frac{b}{a}\right)}{2\pi kL}$

$\therefore$ From the given diagram by using $\left(1\right)$, we get
$R=\frac{1}{2\pi k_{1}L}\ln \left(\frac{r_2}{r_1}\right)+\frac{1}{2\pi k_{2}L}\ln \left(\frac{r_3}{r_2}\right)$

i.e Heat current $I=\frac{T_1-T_2}{\frac{1}{2\pi k_{1}L}\ln \left(\frac{r_2}{r_1}\right)+\frac{1}{2\pi k_{2}L}\ln \left(\frac{r_3}{r_2}\right)}$
Thus, option (c) is the correct answer.