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Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is -
A
$\dfrac{2}{6}K$
B
$\sqrt{2}K$
C
3K
D
$\dfrac{4}{3}K$
Detailed Solution
Heat flows through the two layers one after the other, so they are in series. Thermal resistance of a slab: $R = \dfrac{l}{KA}$.
For series connection: $R_{eq} = R_1 + R_2$
$\dfrac{2l}{K_{eq}A} = \dfrac{l}{KA} + \dfrac{l}{2KA}$
$\dfrac{2}{K_{eq}} = \dfrac{1}{K} + \dfrac{1}{2K} = \dfrac{3}{2K}$
$\Rightarrow K_{eq} = \dfrac{4}{3}K$
For series connection: $R_{eq} = R_1 + R_2$
$\dfrac{2l}{K_{eq}A} = \dfrac{l}{KA} + \dfrac{l}{2KA}$
$\dfrac{2}{K_{eq}} = \dfrac{1}{K} + \dfrac{1}{2K} = \dfrac{3}{2K}$
$\Rightarrow K_{eq} = \dfrac{4}{3}K$