Two walls of thicknesses d1 and d2 and thermal conductivities K1 and K2 are in contact….

Q: Two walls of thicknesses d1 and d2 and thermal conductivities K1 and K2 are in contact. In the steady state, if the temperature at the outer surfaces are T1 and T2 , the temperature at the common wall is

(a) $\displaystyle \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1} $

(b) $\displaystyle \frac{K_1 T_1 + K_2 T_2 }{d_1 + d_2} $

(c) $\displaystyle (\frac{K_1 d_1 + K_2 d_2 }{T_1 + T_2})T_1 T_2 $

(d) $\displaystyle \frac{K_1 d_1 T_1 + K_2 d_2 T_2}{K_1 d_1 + K_2 d_2} $

Click to See Answer :
Ans: (a)
Sol: Let T be the temperature of common wall .

Since Heat current $\displaystyle i = \frac{\Delta T}{R}$ is same through both the walls .

$\displaystyle \frac{T_1 -T}{R_1} = \frac{T -T_2}{R_2}$

$\displaystyle \frac{T_1 -T}{d_1/K_1 A} = \frac{T -T_2}{d_2/K_2 A}$

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

K1 d2 T1 – K1 d2 T = K2 d1 T – K2 d1 T2

$\displaystyle T = \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1} $