Question

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivity K and 2K,respectively.The equivalent thermal is conductivity of the slab is
​

Answer 1

Answer:

K=2K1K2K1+K2=2.K.2KK+2K=43K.

Explanation:

Answer 2

Q]____?

=> 4/3 K

Explanation:

The quantity of heat flowing across a slab in

time t,

Q = KA∆θ / l

Where,

K = thermal conductivity

∆Q = Change in temperature

A = area of slab

l = thickness

For same heat flow through each slab and ( composite slab ), we have

K1A ( ∆θ1 ) / l = K2A ( ∆θ2 ) / l = [ K' A ( ∆θ1 + ∆θ2)] / 2l

or K1∆θ1 = K2 ∆θ2 = K' /2 ( ∆θ1 + ∆θ2 ) = C

__[ say ]

So, ∆θ1 = C/K1, ∆θ2 = C/k2

=> ( ∆θ1 + ∆θ2 ) = 2C / K'

=> C / K1 + C/K2 = 2C/ K'

=> C [ K1 + K2 / K1 K2 ] = 2C / K'

=> K' = 2K1 K2 / K1 + K2

Given, K1 = K.K2 = 2K

So, K' = 2K × 2K / ( K + 2K ) = 4/3 K

The Equivalent thermal is conductivity of the slab is 4/3 K !