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 $x$ and $4x$ respectively are $T_2$ and $T_1\left(T_2>T_1\right)$ . 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 $f$ equal to:-

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

Answer 1

Answer: (D)

For slabs in series, we have :
$R_{eq}=R_1+R_2$
i.e., $\frac{5x}{K_{eq.}A}=\frac{4x}{2KA}+\frac{x}{KA}$
$K_{eq.}=\frac{5K}{3}$
Now, in steady state, rate of heat transfer through the slab
$=\frac{K_{eq.}A\left(T_2-T_1\right)}{5x}=\left(\frac{A\left(T_2-T_1\right)K}{x}\right)f$
Putting the value of $K_{eq}$ . we get
$f=\frac{1}{3}$