Question

A compound slab consists of two parallel plates of different material of same thickness and having thermal conductivities k₁ and k₂. What is the equivalent thermal conductivity of the slab ?

• A. k₁ + k₂
• B. k₁k₂
• C. $\frac{k_{1} + k_{2}}{k_{1}k_{2}}$
• D. $\frac{2k_{1}k_{2}}{k_{1 +}k_{2}}$

Answer 1

Answer: (D)

$\frac{2k_{1}k_{2}}{k_{1 +}k_{2}}$

Answer 2

$\frac{Q}{t} = \frac{KA(\theta_{1} - \theta_{2})}{d}or(\theta_{1} - \theta_{2})\alpha\frac{d}{K}$Now,

$\theta_{1} - \theta_{0}\alpha\frac{d}{K_{1}}$

$\theta_{1} - \theta_{2}\alpha\frac{d}{K_{1}}$

Adding $\theta_{1} - \theta_{2}\alpha d\left\lbrack \frac{1}{k_{1}} + \frac{1}{k_{2}} \right\rbrack$

Also, $\theta_{1} - \theta_{2}\alpha(2d)\left( \frac{1}{k} \right)$

∴ $\frac{2}{k} = \frac{1}{k}_{1} + \frac{1}{k_{2}}ork = \frac{2k_{1}k_{2}}{k_{2} + k_{1}}$

Hence (4) is correct.