Question

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

Answer 1

The equivalent conductivity of the composite slab is $\mathrm{K}_{\mathrm{s}}=\left(L_{1}+L_{2}\right) K_{1} K_{2} /\left(L_{1} K_{2}+L_{2} K_{1}\right)$

Explanation:

Two slabs are in series with each other so the total resistance can be calculated as series combination.

Total resistance $R_{s}=R_{1}+R_{2}$

Resistance in terms of thermal conductivities is given as  

              $\mathrm{R}=\mathrm{I} / \mathrm{KA}$

Given,

Thickness of plates = L1 and L2

Thermal conductivities of plates = K1 and K2

Step 1:-  

             $\frac{\left(L_{1}+L_{2}\right)}{A K_{s}}=\frac{L_{1}}{A K_{1}}+\frac{L_{2}}{A K_{2}}$

Both the slabs have equal area of cross -section so area cancels out from both sides

Step 2:-

              $\frac{\left(L_{1}+L_{2}\right)}{K_{s}}=\frac{L_{1}}{K_{1}}+\frac{L_{2}}{K_{2}}$

Step 3:-

              $\frac{\left(L_{1}+L_{2}\right)}{K_{s}}=\left(L_{1} K_{2}+L_{2} K_{1}\right) / K_{1} K_{2}$

By solving the above equation we get, the equivalent conductivity of composite slab

Step 4:-

              $\mathrm{K}_{\mathrm{s}}=\left(L_{1}+L_{2}\right) K_{1} K_{2} /\left(L_{1} K_{2}+L_{2} K_{1}\right)$