Question

Three rods AB, BC and BD having thermal conductivities in the ratio $1:2:3$ and lengths in the ratio $2:1:1$ are joined as shown in Fig. The ends A, C and D are at temperature ${T_1}$, ${T_2}$ and ${T_3}$ respectively. Find the temperature of junction B. Assume steady state.

Answer 1

Hint: In this question, we are given the ratios of thermal conductivity and the lengths of the rods AB, BC, and CD. And we have to find the temperature of the junction B. First step is to let the thermal conductivity and length be $k$ and $l$. Now, apply Kirchhoff’s junction rule as the heat current is entering from AB and leaving from BC and BD. Solve further.

Formula used:
Kirchhoff’s Junction rule –
The sum of all currents entering a junction must equal the sum of all currents leaving the junction.
Heat current –
$H = \dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{kA\Delta T}}{d}$
$\Delta Q = $small amount of heat transferred in time $\Delta t$
$A = $ Cross- sectional area
$\Delta T = $Temperature difference
$k = $Thermal conductivity
$l = $ Length

Complete Step by Step Solution:
Let, the thermal conductivity and the length of the rods be $k$, $l$ respectively.
Given that, the thermal conductivity of the rods AB, BC and BD are in the ratio $1:2:3$ and the lengths in the ration $2:1:1$.

Therefore, the thermal conductivity of the rods AB, BC and BD will be $k$, $2k$ and $3k$
The lengths of the rods AB, BC and BD are $2l$, $l$ and $l$.
Let, T be the temperature at junction B.

Assuming, ${T_1} > T > {T_2}$ and $T > {T_3}$
Now, using Kirchhoff’s Junction rule
${\left. {\dfrac{{\Delta Q}}{{\Delta t}}} \right]_{AB}} = {\left. {\dfrac{{\Delta Q}}{{\Delta t}}} \right]_{BC}} + {\left. {\dfrac{{\Delta Q}}{{\Delta t}}} \right]_{BD}}$
$\dfrac{{kA\left( {{T_1} - T} \right)}}{{2l}} = \dfrac{{2kA\left( {T - {T_2}} \right)}}{l} + \dfrac{{3kA\left( {T - {T_3}} \right)}}{l}$
$\dfrac{{{T_1} - T}}{2} = 2\left( {T - {T_2}} \right) + 3\left( {T - {T_3}} \right)$
${T_1} - T = 10T - 4{T_2} - 6{T_3}$
$T = \dfrac{1}{{11}}\left( {{T_1} + 4{T_2} + 6{T_3}} \right)$

Hence, at steady state the temperature in the junction B is $\dfrac{1}{{11}}\left( {{T_1} + 4{T_2} + 6{T_3}} \right)$.

Note: The key concept involved in solving this problem is the good knowledge of Kirchhoff’s law. Students must remember that there are two laws. First is the junction rule and the second one is the loop rule. First one we have stated in the above solution and the loop rule states that the sum of all the potential differences (electric) around the loop is equal to zero.