Question

the plane surfaces of two sheets of different metals are kept in contact with each other . the thicknesses of sheets are 2.5 cm and 3.0 cm respectively and the ratio of their thermal conductivities in the same order is 5:6 . If the outer surfaces of sheets are at constant temperature of 100 C and 10 C respectively, calculate the temperature of interface????​

Answer 1

$T_i=55^{\circ}C$ is the interface temperature.

Explanation:

Given:

• thickness of sheet 1, $x_1=0.025\ m$

• thickness of sheet 2, $x_2=0.03\ m$

• ratio of their respective thermal conductivity, $\frac{k_1}{k_2} =\frac{5}{6}$

• temperature of the outer surface of the sheet 1, $T_1=100^{\circ}C$

• temperature of the outer surface of the sheet 2, $T_2=10^{\circ}C$

Using Fourier's law:

$\dot Q=k.A.\frac{dT}{dx}$

$\frac{\dot Q}{A} =k\times \frac{dT}{x}$

let the thermal conductivity of the sheet 1 & 2 be 5z and 6z respectively.

Then using electrical analogy:

$\frac{\dot Q}{A} =\frac{dT}{\frac{x}{k} }$

$\frac{\dot Q}{A} =\frac{dT}{(\frac{x_1}{k_1}+\frac{x_2}{k_2}) }$

$\frac{\dot Q}{A} =\frac{100-10}{(\frac{0.025}{5z}+\frac{0.03}{6z}) }$

$\frac{\dot Q}{A} =9000\ z$

Now for the interface temperature we apply the Fourier's law on sheet 1:

$\frac{\dot Q}{A} =\frac{dT}{\frac{x_1}{k_1} }$

Heat flux being constant throughout.

$9000 z =\frac{100-T_i}{\frac{0.025}{5z} }$

$T_i=55^{\circ}C$

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TOPIC: thermal conductivity, Fourier's Law

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