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Conduction Law

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 $\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$
$\begin{matrix} U n d e r s t e a d y s t a t e h e a t f l u x p e r u n i t a r e a k (\frac{d T}{d x}) i s s a m e a c r o s s t w o w a l l s . h e n c e, w e h a v e K_{1} \frac{T_{1} - T_{c}}{d_{1}} = K_{2} \frac{T_{c} - T_{2}}{d_{2}} w h e r e T_{c} i s c o m m o n w a l l t e m p e r a t u r e . s o l v i n g f o r T_{c} w e w i l l g e t T_{c} = \frac{T_{1} + α T_{2}}{α + 1} W h e r e α = \frac{d_{1}}{d_{2}} \frac{k_{2}}{k_{2}} \end{matrix}$
$\begin{matrix} O n p u t t i n g t h e v a l u e o f α = \frac{d_{1}}{d_{2}} . \frac{k_{1}}{k_{2}} T h e n, T_{c} = \frac{k_{1} T_{1} d_{2} + k_{2} T_{2} d_{1}}{k_{1} d_{2} + k_{2} d_{1}} \end{matrix}$
Hence,Option $B$ is the correct answer.

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