Question

Consider a compound slab consisting of two pieces of same length and different materials having equal thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal comductivity of the slab is?

Ans (4/3)K

How?

Answer 1

Let length = L

Area = A

T₁ be the temperature of the sink

T₂ be the temperature of the source

T be the temperature of the junction

For the first conductor, the thermal rate is given by

H₁/t = KA (T – T₁)/L ---1

For the first conductor, the thermal rate is given by

H₂/t = 2KA (T₂ – T)/L ---2

The two slabs are connected in series so the heat current flowing through will be same,

So,

KA (T – T₁)/L = 2KA (T₂ – T)/L

=> (T – T₁) = 2(T₂ – T)

=> T = 1/3(2T₂ +T₁) ---3.

Let the equivalent coefficient of thermal conductivity be K’

Adding 1 and 2

(H₁+H₂)/t = KA (T – T₁)/L + 2KA (T₂ – T)/L

=> H/t =[ KA{1/3(2T₂ +T₁)– T₁} + 2KA{ T₂ - 1/3(2T₂ +T₁)}]/L

=> H/t = KA 4/3(T₂ – T₁)/L

=> H/t = (4/3K) (T₂ – T₁)/L

=> H/t = K’ (T₂ – T₁)/L

So, equivalent K’ = 4/3 K