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Two walls of thickness $d_1$ and $d_2$ and thermal conductivities $K_1$ and $K_2$ are in contact. In the steady state if the temperatures at the inner surface are T₁ and the temperatures at the outer surface $T_2$ , the temperature at the common wall is
Option: 1
$\frac{K_1T_1d_2 + K_2T_2d_1}{K_1d_2 + K_2d_1}$

Option: 2
$\frac{K_1T_1 + K_2T_2}{d_2 + d_1}$

Option: 3
$\frac{K_1d_2 + K_2d_1}{T_1 + T_2}$

Option: 4
$\frac{K_1T_1d_1 + K_2T_2d_2}{K_1d_1 + K_2d_2}$

Answers (1)

best_answer

Temperature of Interface (Junction Temperature) -

$\theta=\frac{\frac{K_{1}}{l_{1}}\theta_{1}+\frac{K_{2}}{l_{2}}\theta_{2}}{\frac{K_{1}}{l_{1}}+\frac{K_{2}}{l_{2}}}$

Let the temperature at the junction be $T$

$\Rightarrow \frac{K_1(T_1 - T)A}{d_1} = \frac{K_2(T - T_2)A}{d_2}$

$\Rightarrow K_1T_1d_2 - K_1Td_2 = K_2d_1T - K_2d_1T_2$

$\Rightarrow T = \frac{K_1T_1d_2 + K_2d_1T_2}{K_1d_2 + d_1K_2}$

Posted by

Rishabh

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