First slide

Conduction

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity k and 2 K and thickness x and 4 x respectively are T₂ and T₁ (T₂ > T₁). The rate of heat transfer through the slab, in a steady-state is $(\frac{A (T_{2} - T_{1}) K}{x})$ f with f equal to:

Difficult

A

1
B

$\frac{1}{2}$
C

$\frac{2}{3}$
D

$\frac{1}{3}$

Solution

For slabs in series, we have :
$R_{eq} = R_{1} + R_{2}$
$i.e., \frac{5 x}{K_{cq.} A} = \frac{4 x}{2 KA} + \frac{x}{KA}$
$K_{eq} = \frac{5 K}{3}$
Now, in steady state, rate of heat transfer through the slab
$= \frac{K_{eq \cdot} A (T_{2} - T_{1})}{5 x} = (\frac{A (T_{2} - T_{1}) K}{x}) f$
Putting the value of K_eq. we get $f = \frac{1}{3}$

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