Question

Q.Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is,
(A) √2K (B) 3K
(C) (4/3)K (D) (2/3)K​

Answer 1

Explanation:

Correct option is

C

3

4

K

The thermal resistances can be added as they are in series. But, note that the width of the combined slab is twice the width of individual slab.

So, we have:

⇒

K

eq

A

2L

=

KA

L

+

2KA

L

⇒K

eq

=

3

4K

Answer 2

Answer:

option (C) (4/3)k is correct

Explanation:-

The thermal resistances can be added as they are in series. Here the width of the combined slab is twice the width of individual slab

Hence, we have:

2L/KeqA = (L/KA) + {L/(2K)A}

⇒ (2K + k)/ 2K2 = 2/Keq

⇒ Keq = (4/3)K

Therefore,

The correct option is (C)