Question

Q. 18 Two rods A & B are connected in series as shown in fig. the conductivity of A
Is Ki = 100 w/m -°C and conductivity of Bis K = 50 W/m - °C. The free ends
of the rods A & B has temp. 30° & 60° respectively the temp. of common meeting
point will be.​

Answer 1

$T_i=40^{\circ}C$ is the interface temperature at the junction of the rods.

Explanation:

Given:

• thermal conductivity of rod A, $K_A=100\ W.m^{-1}.^{\circ}C^{-1}$

• thermal conductivity of rod B, $K_B=50\ W.m^{-1}.^{\circ}C^{-1}$

• free-end temperature of rod A, $T_A=30^{\circ}C$

• free-end temperature of rod B, $T_B=60^{\circ}C$

• length of each rod be, $L\ meters$

Now using the Fourier's law of thermal conduction:

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

$\Rightarrow \frac{\dot Q}{A} =k.\frac{dT}{dx}$ .............................(1)

where:

A = area of the medium normal to the direction of the temperature gradient

dT = temperature difference across the surfaces

dx = distance between the surfaces

k = thermal conductivity of the medium

Now putting the respective values in eq. (1) for the heat flux throughout the joint rod:

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

using electrical analogy of resistances in series

$\frac{\dot Q}{A} =(60-30)\div (\frac{L}{100} +\frac{L}{50} )$

$\frac{\dot Q}{A} =30\times \frac{100}{3L}$

$\frac{\dot Q}{A} =\frac{1000}{L}$

Now this heat flux is constant throughout the cross-section.

So, apply Fourier's Law in the rod A:

$\frac{1000}{L} =(T_i-30)\div \frac{L}{100}$

$T_i=40^{\circ}C$

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TOPIC: Fourier's Law of conduction

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