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 $\left( \frac{A\left( T_{2} - T_{1} \right)K}{x} \right)f$, with $f$ equals to

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 $T_{2}$ and $T_{1}$ $(T_{2} > T_{1})$. The rate of heat transfer through the slab, in a steady state is $\left( \frac{A\left( T_{2} - T_{1} \right)K}{x} \right)f$, with $f$ equals to

Dr. Abubacker · Accepted Answer

(d) Let the temperature of common interface be $T℃$. Rate of heat flow
$H = \frac{\mathcal{Q}}{t} = \ \frac{KA\mathrm{\Delta}T}{l}$
$	herefore$ $H_{1} = \left( \frac{\mathcal{Q}}{t} ight)_{1}$=$\frac{2KA(T - T_{1)}}{4x}$
And $H_{2} = \left( \frac{\mathcal{Q}}{t} ight)_{2}$=$\frac{KA(T_{2} - T)}{x}$
In steady state, the rate of heat flow should be same in whole system $ie,$
$H_{1} = \ H_{2}$
$\Rightarrow$ $\frac{2KA(T - T_{1})}{4x}$ = $\frac{KA(T_{2} - T)}{x}$
$\Rightarrow$ $\frac{T - T_{1}}{2} = \ T_{2} - T$
$\Rightarrow$ $T - T_{1} = \ {2T}_{2} - 2T$
$\Rightarrow$ $T = \frac{2T_{2} + T_{1}}{3}$ …(i)
Hence, heat flow from composite slab is
$H = \ \frac{KA(T_{2} - T)}{x}$
= $\frac{KA}{x}\ \left( T_{2} - \frac{2T_{2} + T_{1}}{3} ight) = \ \frac{KA}{3x}\ (T_{2} - T_{1})$ …(ii)
[from Eq. (i)]
Accordingly, $H = \left\lbrack \frac{A\left( T_{2} - T_{1} ight)K}{x}\  ightbrack f$ … (iii)
By comparing Eqs. (ii) and (iii), we get
$\Rightarrow$ $f = \ \frac{1}{3}$

Explanation

Reviewed by Dr. Abubacker

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