The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivities $K$ and $2 K$ , and thickness $x$ and $4 x$ respectively are $T_{2}$ and $T_{1} (T_{2} > T_{1})$ as shown in figure. The cross-sectional area of the slab is $A$ . The rate of heat transfer through the slab in steady state is $[\frac{A (T_{2} - T_{1}) K}{f x}]$ , where $f$ is equal to

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Solution

Let $T$ be the temperature of the interface.
We know that ,
$\frac{Δ Q}{Δ t} = - K A \frac{Δ T}{Δ x}$

Since the rate of flow of heat across the interface is same,
$K A \frac{(T_{2} - T)}{x} = 2 K A \frac{(T - T_{1})}{4 x}$
$\Rightarrow T = \frac{2 T_{2} + T_{1}}{3}$
Substituting the value of $T$ in $\frac{Δ Q}{Δ t}$ for the slab of thickness $4 x$ we get,(Considering magnitude only)
$\frac{Δ Q}{Δ t} = 2 K \times A (\frac{T - T_{1}}{4 x})$
$= 2 K A \times \frac{(\frac{2 T_{2} + T_{1}}{3} - T_{1})}{4 x}$
$\Rightarrow \frac{Δ Q}{Δ t} = \frac{K A}{3 x} \times (T_{2} - T_{1})$
Comparing this with the given expression $\frac{K A (T_{2} - T_{1})}{f x}$ ,
we get, $f = 3$

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Q. 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

T_{2}

and

T_{1} (T_{2} > T_{1})

. The rate of heat transfer through the slab, in a steady state is

(\frac{A (T_{2} - T_{1}) K}{X}) f,

with

f

equals to

The temperatures of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are $T_{2}$ and $T_{1}$ ( $T_{2} > T_{1}$ ). 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