Question

A slab consists of two parallel layers of two different materials of the same thickness having thermal conductivities \[{K_1}\] ​and \[{K_2}\]. Find the equivalent thermal conductivity of the slab.
A. \[\dfrac{{{K_1} + {K_2}}}{{{K_1}{K_2}}}\]
B. \[\dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}\]
C. \[\dfrac{{{K_1} + {K_2}}}{2}\]
D. \[{K_1} + {K_2}\]

Answer 1

Hint:In order to solve this problem we need to understand equivalent thermal conductivity. Equivalent thermal conductivity is defined as the two rods of the same physical dimensions and having the same thermal conductivity k connected to each other in parallel combination is equal to k. Using the concept of this equivalent thermal conductivity we are going to find the solution for this problem.

Formula Used:
To find the heat in the slab the formula is,
\[H = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}\]
Where, A is the cross-sectional area of the slab, L is length of the slab and K is thermal conductivity of the slab.

Complete step by step solution:
Consider a slab that consists of two parallel layers of two different materials of the same thickness that has thermal conductivities \[{K_1}\]​ and \[{K_2}\]. We need to find the equivalent thermal conductivity of the slab.

To find the heat in the slab the formula is,
\[H = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}\]
The two slabs are parallel and of the same dimension then, at steady state,
\[H = {H_1} + {H_2}\]…… (1)
\[\Rightarrow {H_1} = \dfrac{{{K_1}A\left( {{T_2} - {T_1}} \right)}}{L}\]
And, \[{H_2} = \dfrac{{{K_2}A\left( {{T_2} - {T_1}} \right)}}{L}\]

Substitute the value of \[{H_1}\] and \[{H_2}\] in equation (1) becomes,
\[\dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L} = \dfrac{{{K_1}\dfrac{A}{2}\left( {{T_2} - {T_1}} \right)}}{L} + \dfrac{{{K_2}\dfrac{A}{2}\left( {{T_2} - {T_1}} \right)}}{L} \\ \]
\[\Rightarrow \dfrac{{KA}}{L} = \dfrac{{{K_1}\dfrac{A}{2}}}{L} + \dfrac{{{K_2}\dfrac{A}{2}}}{L} \\ \]
\[\therefore K = \dfrac{{{K_1} + {K_2}}}{2}\]
Therefore, the equivalent thermal conductivity of the slab is \[\dfrac{{{K_1} + {K_2}}}{2}\].

Hence, option C is the correct answer.

Note:Here in the given problem it is important to remember that the equation for the heat flow in the rod and using that we are going to calculate the equivalent thermal conductivity as shown above.