Question

Find the equivalent thermal conductivity of the given rod.

• A. $\frac{2K}{3}$
• B. $3K$
• C. $\frac{3K}{2}$
• D. $\frac{4K}{3}$

Answer 1

The correct option is D

$\frac{4K}{3}$

Step 1: Given Data

Let, $A=$area, $T_1$ is the temperature of the sink, $T_2$ is the temperature of the source, $T$ is the temperature of the junction, $L$ be the thickness of the junction

Let the equivalent coefficient of thermal conductivity be $K$.

The given thermal conductivities are $K$ and $2K$ each having length $L$.

Let $H_1$ be the amount of heat taken from the source and $H_2$ be the amount of heat given to the sink that is transferred in time $t$.

Step 2: Express the Thermal Rates of the Rod

Thermal conductivity is given as,

$\frac{△Q}{△t}=-K\frac{△T}{△x}$

where $△Q$ is the change in heat and $△x$ is the change in length.

For the first conductor, the thermal rate is given by

$\frac{H_1}{t}=\frac{KA\left(T-T_1\right)}{L}$ $\left(1\right)$

For the second conductor, the thermal rate is given by

$\frac{H_2}{t}=\frac{2KA\left(T_2-T\right)}{L}$ $\left(2\right)$

Step 3: Evaluate Temperature of the Junction

The two slabs are connected in series so the heat rate flowing through them will be the same

So, $\frac{H_1}{t}=\frac{H_2}{t}$

$\Rightarrow \frac{KA\left(T-T_1\right)}{L}=\frac{2KA\left(T_2-T\right)}{L}$

$\Rightarrow$ $T-T_1=2\left(T_2-T\right)$

$\Rightarrow$ $3T=2T_2+T_1$

$\Rightarrow$ $T=\frac{2T_2+T_1}{3}$

Step 4: Find the Equivalent Thermal Conductivity

Adding equations $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{H_1+H_2}{t}=\frac{KA\left(T-T_1\right)+2KA\left(T_2-T\right)}{L}$

$=\frac{KA}{L}\left[\left\{\frac{1}{3}\left(2T_2+T_1\right)-T_1\right\}+2\left\{T_2-\frac{1}{3}\left(2T_2+T_1\right)\right\}\right]$

$=\frac{KA}{L}\left(\frac{2T_2+T_1-3T_1+6T_2-4T_2-2T_1}{3}\right)$

$=\frac{4}{3}K.\frac{A}{L}\left(T_2-T_1\right)$

Hence, the equivalent thermal conductivity is $\frac{4K}{3}$.

$\therefore$Option $\left(d\right)$ is the correct answer.