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 slab.

Question

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 slab. (A) Ke q=(L1+L2/(L1)K1+(L2)K2 (B) Ke q=(L1+L2/K1 K2) (C) Ke q=(L1/K1+K2)+(L2/K1+K2) (D) Ke q=(L1 L2/L1+L2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/86aa87c9f806dc003784e1c57239f8d2-.png" /> Re q =R1+R2 ⇒ (L1+L2/Ke q A) =(L1/K1 A)+(L2/K2 A) ⇒ Ke q =(L1+L2/(L1)K1+(L2)K2

Q. A composite slab is prepared by pasting two plates of thicknesses $L_{1}$ and $L_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ . The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.

$K_{e q} = \frac{L _{1} + L _{2}}{\frac{L _{1}}{K _{1}} + \frac{L _{2}}{K _{2}}}$

$K_{e q} = \frac{L _{1} + L _{2}}{K _{1} K _{2}}$

$K_{e q} = \frac{L _{1}}{K _{1} + K _{2}} + \frac{L _{2}}{K _{1} + K _{2}}$

$K_{e q} = \frac{L _{1} L _{2}}{L _{1} + L _{2}}$

$R_{e q} = R_{1} + R_{2}$
$\Rightarrow \frac{L _{1} + L _{2}}{K _{e q} A} = \frac{L _{1}}{K _{1} A} + \frac{L _{2}}{K _{2} A}$
$\Rightarrow K_{e q} = \frac{L _{1} + L _{2}}{\frac{L _{1}}{K _{1}} + \frac{L _{2}}{K _{2}}}$

Q. 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 slab.

Keq​=K1​L1​​+K2​L2​​L1​+L2​​

Keq​=K1​K2​L1​+L2​​

Keq​=K1​+K2​L1​​+K1​+K2​L2​​

Keq​=L1​+L2​L1​L2​​