Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$, thickness $x$ and $4x$ respectively, are $T_2$ and $T_1(T_2>T_1)$. Area of cross-section $A$ is the same. The rate of heat transfer through the slab, in steady state, is $F\left[\frac{A(T_2-T_1)K}{x}\right]$. The value of $F$ is
(answer up to two decimal places)

Answer 1

Using Fourier's law of conduction,
$\frac{dQ}{dt}=\left[\frac{A(T_2-T_1)K}{x}\right]=\frac{T_2-T_1}{R}$........(1)
where $R=\frac{x}{KA}$ is the thermal resistance.

Here, the two slabs are in series.
$\Rightarrow R_{eq}=R_1+R_2=\frac{x}{KA}+\frac{4x}{2KA}=\frac{3x}{KA}$

$\therefore$ Rate of heat transfer through the composite slab
$i=\frac{dQ}{dt}=\frac{T_2-T_1}{R_{eq}}$
$=\frac{(T_2-T_1)}{3x/KA}=\frac{KA(T_2-T_1)}{3x}$
$=\frac{1}{3}\left[\frac{A(T_2-T_1)K}{x}\right]$..........(2)

Comparing (2) with the given rate of heat transfer,
$\Rightarrow F=\frac{1}{3}=0.33$