Question

A compound slab is made up of two slabs. If thermal conductivity of their material are Kâ and Kâ respectively and their cross-sectional areas are the same, the equivalent thermal conductivity of the slab is ............ (Consider series connection)
(A) Kâ + Kâ
(B) (Kâ - Kâ) / 2
(C) (Kâ + Kâ) / KâKâ
(D) 2KâKâ / (Kâ + Kâ)

Answer 1

Hey dear,

◆ Answer- (D)

k = 2ka.kb/(ka+kb)

● Explaination-

Let two slabs of same thickness t and area of cross section A be joined.

Equivalent resistance is given By -

R = R1 + R2

(t+t)/kA = t/kaA + t/kbA

2/k = 1/ka + 1/kb

2/k = (ka+kb)/(ka.kb)

k/2 = ka.kb / (ka+kb)

k = 2ka.kb / (ka+kb)

Thermal conductivity of the combination is 2ka.kb/(ka+kb).

Hope this helps you.

Thanks for asking.