First slide

Heat transfer

Question

Two rods A and B of different materials are welded together as shown in Fig. If their thermal conductivities are k₁ and k₂, the thermal conductivity of the composite rod will be

Moderate

A

$2 (k_{1} + k_{2})$
B

$\frac{3}{2} (k_{1} + k_{2})$
C

$(k_{1} + k_{2})$
D

$\frac{1}{2} (k_{1} + k_{2})$

Solution

If the ends of the rods are maintained at temperatures $θ_{1}$ and $θ_{2}$ the rate of flow of heat through each rod is given by
$\begin{array}{l} Q_{1} = \frac{k_{1} A (θ_{1} - θ_{2})}{d} \\ Q_{2} = \frac{k_{2} A (θ_{1} - θ_{2})}{d} \end{array}$
where A is the cross-sectional area of each rod and d is the length of each rod. Now the cross-sectional area of the composite rod is 2A. If k is the thermal conductivity of the composite rod, the rate of heat flow is
$Q = \frac{k (2 A) (θ_{1} - θ_{2})}{d}$
But $Q = Q_{1} + Q_{2} . Therefore$ $\frac{k (2 A) (θ_{1} - θ_{2})}{d} = \frac{k_{1} A (θ_{1} - θ_{2})}{d} + \frac{k_{2} A (θ_{1} - θ_{2})}{d}$
or $k = \frac{1}{2} (k_{1} + k_{2})$
Hence the correct choice is (d).

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