Question

consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2K. The equivalent conductivity of the slab is
options are
(a) 2K/3 (b) (2^1/2)K (c) 3K (d) 4K/3

Answer 1

Answer on Question #42638-Physics-Molecular Physics-Thermodynamics

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2K. The equivalent conductivity of the slab is

Options are

(a) $2 \mathrm{~K} / 3$ (b) $(2^{\wedge} 1 / 2) \mathrm{K}$ (c) $3 \mathrm{~K}$ (d) $4 \mathrm{~K} / 3$

Solution

Let length is $L$ , area is $A$ , $T_{1}$ be the temperature of the sink, $T_{2}$ be the temperature of the source, $T$ be the temperature of the junction.

For the first conductor, the thermal rate is given by

$$\frac {H _ {1}}{t} = \frac {K A (T - T _ {1})}{L}.$$

For the second conductor, the thermal rate is given by

$$\frac {H _ {2}}{t} = \frac {2 K A (T _ {2} - T)}{L}.$$

The two slabs are connected in series so the heat current flowing through will be same. So,

$$\frac {K A (T - T _ {1})}{L} = \frac {2 K A (T _ {2} - T)}{L} \rightarrow (T - T _ {1}) = 2 (T _ {2} - T) \rightarrow \mathrm {T} = \frac {1}{3} (2 \mathrm {T} _ {2} + \mathrm {T} _ {1}).$$

Let the equivalent coefficient of thermal conductivity be $K^{\prime}$

Adding thermal rate for the first and second conductors

$$\frac {(H _ {1} + H _ {2})}{t} = \frac {K A (T - T _ {1})}{L} + \frac {2 K A (T _ {2} - T)}{L} \rightarrow \frac {H}{t} = \frac {\left[ \mathrm {K A} \left\{\frac {1}{3} (2 \mathrm {T} _ {2} + \mathrm {T} _ {1}) - \mathrm {T} _ {1} \right\} + 2 \mathrm {K A} \left\{\mathrm {T} _ {2} - \frac {1}{3} (2 \mathrm {T} _ {2} + \mathrm {T} _ {1}) \right\}\right]}{\mathrm {L}}$$ $$\frac {H}{t} = \frac {K A \frac {4}{3} (T _ {2} - T _ {1})}{L} = \frac {K ^ {\prime} A (T _ {2} - T _ {1})}{L}.$$

So, equivalent $K^{\prime} = \frac{4}{3} K$

Answer: (d) 4K/3.

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