Question 710: Thermodynamics & KTG

Question

A composite slab consists of two materials having coefficients of thermal conductivity K and $2 K$, thickness $x$ and $4 x$ respectively. The temperatures of two outer surfaces of a composite slab are $\mathrm{T}_2$ and $\mathrm{T}_1$ respectively $\left(\mathrm{T}_2>\mathrm{T}_1\right)$. The rate of heat transfer through the slab in a steady state is $\left[\frac{A\left(T_2-T_1\right) K}{x}\right] f$, where $f$ is equal to

Accepted Answer

Let the two slabs have:
Slab 1:
Thermal conductivity $K$, thickness $x$
Slab 2:
Thermal conductivity $2K$, thickness $4x$
Let the temperature at the interface be $T$.
Step 1: Write heat transfer equations for each slab
For steady-state, the
rate of heat transfer ($Q$) through each layer is same
:
For
Slab 1
:
$$ Q = \frac{K\,A\,(T_2 - T)}{x} $$
For
Slab 2
:
$$ Q = \frac{2K\,A\,(T - T_1)}{4x} $$
Step 2: Equate the rates and solve for $T$
Set the two $Q$ values equal:
$$ \frac{K\,A\,(T_2 - T)}{x} = \frac{2K\,A\,(T - T_1)}{4x} $$
Simplify:
$K$ and $A$ cancel out.
$2/(4x) = 1/(2x)$:
$$ \frac{T_2 - T}{x} = \frac{T - T_1}{2x} $$
Cross-multiplied:
$$ 2(T_2 - T) = T - T_1 $$
$$ 2T_2 - 2T = T - T_1 $$
$$ 2T_2 + T_1 = 3T $$
$$ T = \frac{2T_2 + T_1}{3} $$
Step 3: Substitute $T$ back to find $Q$ in terms of $T_2$ and $T_1$
Recall:
$$ Q = \frac{K\,A\,(T_2 - T)}{x} $$
Use $T = \dfrac{2T_2 + T_1}{3}$:
$$ T_2 - T = T_2 - \frac{2T_2 + T_1}{3} = \frac{3T_2 - 2T_2 - T_1}{3} = \frac{T_2 - T_1}{3} $$
So,
$$ Q = \frac{K\,A}{x} \cdot \frac{T_2 - T_1}{3} $$
Or,
$$ Q = \left[\frac{A (T_2 - T_1) K}{x} \right] \cdot \frac{1}{3} $$
Step 4: Compare with the question's expression
Here,
$$ f = \frac{1}{3} $$
Final Answer:
Option D: $f = \dfrac{1}{3}$

Answer

1

Answer

$\frac{2}{3}$

Answer

$\frac{1}{2}$

Answer

$\frac{1}{3}$

Thermodynamics & KTG

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