First slide

Conduction

Question

Two slabs A and B of different materials but of the same thickness are joined as shown in fig. (2). The thermal conductivities of A and B are K₁ and K₂ respectively. The thermal conductivity of the composite slab is

Moderate

A

$(K_{1} + K_{2}) / 2$
B

$\sqrt{(K_{1} K_{2})}$
C

$(K_{1} + K_{2})$
D

$2 K_{1} K_{2} / (K_{1} + K_{2})$

Solution

The thermal resistance of a slab of length I, area A and thermal conductivity K is given by $R = l / (KA)$
For slab A, $R = (l / 2) / (K_{1} A)$
For slab B, $R_{2} = (l / 2) / (K_{2} A)$
The two rods are joined in series. Hence, their thermal resistance $R = R_{1} + R_{2}$
or $\frac{l}{KA} = \frac{(l / 2)}{K_{1} A} + \frac{(l / 2)}{K_{2} A}$ or $\frac{1}{K} = \frac{1}{2} [\frac{1}{K_{1}} + \frac{1}{K_{2}}]$
or $K = \frac{2 K_{1} K_{2}}{(K_{1} + K_{2})}$

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