Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $\text{X}$ and $4\text{X}$, respectively are $T_2$ and $T_1(T_2>T_1)$. The rate of heat transfer through the slab, in a steady-state, is $\left[\frac{A(T_2-T_1)K}{\text{X}}\right]f$, with $f$ equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{3}$

Given,
Length of slabs are $\text{X}$ and $4\text{X}$ having conductivity $K$ and $2K$ respectively.

In the steady state, temperature will be constant (with time) at all points on the slab.

Heat transfer will be the same in both the slabs to maintain the constant temperature. So, both slabs are in series.

Effective thermal resistance in series combination,

$R_{eff}=R_1+R_2$

As, $R=\frac{l}{kA}$

$R_{eff}=\frac{\text{X}}{KA}+\frac{4\text{X}}{2KA}$

Where, $A$ is the cross-sectional area of the slab.

$\Rightarrow R_{eff}=\frac{3\text{X}}{KA}$

The rate of heat transfer is given as,

$H=\frac{dQ}{dt}=\frac{\Delta T}{R_{eff}}$

Two slabs can be replaced by the slab having resistance equal to $R_{eff}$.

$\Rightarrow H=\frac{T_2-T_1}{\left(3\text{X}/KA\right)}$

$\Rightarrow H=\left(\frac{A(T_2-T_1)K}{\text{X}}\right)\times \frac{1}{3}$

Comparing it with $\left(\frac{A(T_2-T_1)K}{\text{X}}\right)f$

$\Rightarrow f=\frac{1}{3}$

Hence, option $\left(C\right)$ is correct.

Why this question? Concept: In the series combination, effective thermal resistance is the sum of the individual thermal resistance.