Question

A slab is made up of two layers of different materials of the same thickness and having terms conductivities K1 and K2. The equivalent thermal conductivity of the composite slab is

Answer 1

Answer:

The equivalent thermal conductivity of the composite slab is

$k_{eq} = \frac{2K_1K_2}{K_1 + K_2}$

Explanation:

As we know that rate of heat flow through the slab of given thickness is

$\frac{dQ}{dt} = (kA)\frac{\Delta T}{x}$

Here we know that

k = thermal conductivity of rod

A = area

x = thickness

Now if two such slabs are placed over each other then the rate of heat flow is given as

$\frac{dQ}{dt} = k_1A\frac{\Delta T_1}{x} = k_2A\frac{\Delta T_2}{x}$

now the equivalent slab will have same area and double thickness

so we will have

$\frac{dQ}{dt} = k_{eq}A\frac{\Delta T}{2x}$

now we have

$\frac{dQ}{dt} = k_{eq}A\frac{\Delta T}{2x} = k_1A\frac{\Delta T_1}{x} = k_2A\frac{\Delta T_2}{x}$

Now we know that total length is sum of length of two slabs

$\Delta T_1 + \Delta T_2 = \Delta T$

$\frac{x\frac{dQ}{dt}}{k_{1}A} + \frac{\frac{dQ}{dt}}{k_{2}A} = \frac{2x\frac{dQ}{dt}}{k_{eq}A}$

so we have

$k_{eq} = \frac{2K_1K_2}{K_1 + K_2}$

#Learn

Topic : Thermal conduction

https://brainly.in/question/7322025