Question

Consider a compound slab consisting of two
different materials having equal thicknesses and
thermal conductivities K and 2K, respectively.
The equivalent thermal conductivity of the slab
is
(a) 4 K (b) 2 K
3 3
(c) √3 K (d) 3 K

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively.

The equivalent thermal conductivity of the slab is

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Thermal conductivities are K and 2K

$\large\underline{\underline{\sf To\:Find:}}$

• Equivalent thermal conductivity

Thermal Resistance (R)

❏$\large{\boxed{\bf \blue{R=\dfrac{l}{KA}} }}$

Lenght is same for both material

So ,

$\implies{\sf \dfrac{l}{K_{eq}}=\dfrac{l}{K_1}+\dfrac{l}{K_2}}$

$\implies{\sf \dfrac{2l}{K_{eq}}= \dfrac{l}{K}+\dfrac{l}{2K}}$

$\implies{\sf \dfrac{2}{K_{eq}}=\dfrac{1}{K}+\dfrac{1}{2K} }$

$\implies{\sf \dfrac{2}{K_{eq}}=\dfrac{2K+K}{2K^2}}$

$\implies{\sf \dfrac{2}{K_{eq}}= \dfrac{3K}{2K^2} }$

$\implies{\sf \dfrac{2}{K_{eq}}=\dfrac{3}{2K}}$

$\implies{\sf \dfrac{K_{eq}}{2}=\dfrac{2K}{3}}$

$\implies{\sf K_{eq}=\dfrac{2×2K}{3} }$

$\implies{\bf \red{K_{eq}=\dfrac{4K}{3}}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

The equivalent thermal conductivity of the slab

is ${\bf \red{\dfrac{4K}{3}}}$.