First slide

Heat transfer

Question

A composite slab consists of two slabs A and B of different materials but of the same thickness placed in contact as shown in figure. The thermal conductivity of A and B are $k_{1}$ and $k_{2}$ respectively. A steady temperature difference of $12^{\circ} C$ is maintained across the composite slab. If $k_{1} = \frac{k_{2}}{2}$ , the temperature difference across slab A will be:

Moderate

A

$4^{\circ} C$
B

$8^{\circ} C$
C

$12^{\circ} C$
D

$16^{\circ} C$

Solution

Let temperature of junction be ${T_{o}}^{\circ} C$
Temperature difference across slab $A = {(T_{1} - T_{o})}^{\circ} C$
Temperature difference across slab $B = {(T_{o} - T_{2})}^{\circ} C$
Since both slabs are connected in series, heat current through both will be equal
$\begin{array}{l} \frac{k_{1} A (T_{1} - T_{0})}{\frac{l}{2}} = \frac{k_{2} A (T_{o} - T_{2})}{\frac{l}{2}} \\ As k_{1} = \frac{k_{2}}{2} \\ \Rightarrow \frac{k_{2}}{2} (T_{1} - T_{o}) = (T_{o} - T_{2}) k_{2} \\ \Rightarrow T_{1} - T_{o} = 2 T_{o} - 2 T_{2} \end{array}$
$\begin{array}{l} \Rightarrow T_{1} + 2 T_{2} = 3 T_{0} \dots . (1) \\ As T_{1} - T_{2} = 12^{\circ} C \\ \Rightarrow T_{2} = T_{1} - 12 \dots . . (2) \\ From (1) and (2), we get \\ \Rightarrow T_{1} + 2 (T_{1} - 12) = 3 T_{o} \\ \Rightarrow 3 T_{1} - 24 = 3 T_{0} \\ \Rightarrow T_{1} - T_{0} = 8^{\circ} C \end{array}$

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