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Area and Volume Expansion

Two walls of ...

Question

Two walls of thickness d $_{1}$ and d $_{2}$ ,thermal conductivities K $_{1}$ and K $_{2}$ are in contact.In the steady state if the temperatures at the outer surfaces are T $_{1}$ and T $_{2}$ , the temperature at the common wall will be

A

\frac{K_{1} T_{1} + K_{2} T_{2}}{d_{1} + d_{2}}

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B

\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}

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C

\frac{(K_{1} d_{1} + K_{2} d_{2}) T_{1} T_{2}}{T_{1} + T_{2}}

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D

\frac{(K_{1} d_{1} T_{1} + K_{2} d_{2} T_{2})}{K_{1} d_{1} + K_{2} d_{2}}

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Solution

The correct option is B $\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$
R.E.F.Image.
walls in content so same Area $A_{1} = A_{2} = A$
Suppose the temp at
common surface = T
then heat rate at left wall
will be
$\frac{Q}{t} = \frac{K_{1} A (T - T_{1})}{d_{1}}$
heat rate for second wall $\frac{Q}{t} = \frac{K_{2} A (T_{2} - T)}{d_{2}}$
steady state $\Rightarrow$ heat rate same throughout
the length
so $\frac{K_{1} A (T - T_{1})}{d_{1}} = \frac{K_{2} A (T_{2} - T)}{d_{2}}$
$\Rightarrow K_{1} d_{2} T - K_{1} T_{1} d_{2} = K_{2} T_{2} d_{1} - K_{2} T d_{1}$
$\Rightarrow T (K_{1} d_{2} + K_{2} d_{1}) = K_{2} T_{2} d_{1} + K_{1} T_{1} d_{2}$
$\Rightarrow T = \frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{2}}$

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