Question

Two metallic plate of equal thickness and coefficient thermal conductivity K1 and K2 put together face to face. A common plate is constructed. What is equivalent K?
A) K1k2/k1+k2
B) 2k1k2/k1+k2
C) K1+k2/k1k2
D) None

Answer 1

The correct answer is option B) 2k1 k2/k1 + k2

Solution:

We know that

Equivalent k = L1 + l2/ (L1/K1) + (L2/K2)

where  L1 and L2 are the thickness of the plate 1 and plate 2.

K1 and K2 are coefficients of thermal conductivity.

In this question, L1 = l2 = L

So, K = L + L / ( L/K1 + L/K2) = 2K1 K2/ K1 + k2.

Hence proved.