Q. 225.0( 2 Votes )

A composite slab is prepared by pasting two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Answer :

Here the slabs are placed in such a way that their conductivities are in series as shown in the fig below


Thicknesses of the slabs as L1 and L2

Thermal conductivities as K1 and K2.


Hence, their equivalent conductivity is similar to 2 resistors connected in series.

From above fig.

Given thermal conductivities of the slabs as K1 andK2

The of rate thermal conduction through first slab is given by -

Q1 = (1)


= thermal conductivity of the first slab

= area of first slab

= length of first slab

= temperature of the first slab and

T= junction temperature

Similarly the rate thermal conduction through second slab is given by –

Q2 = (2)

Since they are connected end to end

Qeqv = Q1 = Q2

Where Qeqv is given by

Qeqv =

Also, their area of cross section are equal ie,

A1 = A2 (3)

From (1) ,(2) and (3)


Solving for T

T = (4)

Substituting (4) in (1) and (2) –

Keqv =

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