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

Options

A. `2/3 K`
B. `sqrt(2) K`
C. `3 K`
D. `4/3 K`

Explanation

To find the equivalent thermal conductivity of a compound slab consisting of two different materials with equal thickness and thermal conductivities \( K \) and \( 2K \), we can follow these steps:

### Step 1: Understand the Configuration
We have two slabs of equal thickness \( L \):
- The first slab has thermal conductivity \( K \).
- The second slab has thermal conductivity \( 2K \).

### Step 2: Calculate the Thermal Resistance
The thermal resistance \( R \) for each slab can be calculated using the formula:
\[ R = \frac{L}{K \cdot A} \]
where \( L \) is the thickness of the slab, \( K \) is the thermal conductivity, and \( A \) is the cross-sectional area.

For the first slab (thermal conductivity \( K \)):
\[ R_1 = \frac{L}{K \cdot A} \]

For the second slab (thermal conductivity \( 2K \)):
\[ R_2 = \frac{L}{2K \cdot A} \]

### Step 3: Combine the Thermal Resistances
Since the two slabs are in series, the total thermal resistance \( R_{total} \) is the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 \]
Substituting the values we calculated:
\[ R_{total} = \frac{L}{K \cdot A} + \frac{L}{2K \cdot A} \]

### Step 4: Simplify the Expression
To simplify \( R_{total} \):
\[ R_{total} = \frac{L}{K \cdot A} + \frac{L}{2K \cdot A} = \frac{2L}{2K \cdot A} + \frac{L}{2K \cdot A} = \frac{3L}{2K \cdot A} \]

### Step 5: Calculate the Equivalent Thermal Conductivity
The equivalent thermal conductivity \( K_{eq} \) can be found using the relationship:
\[ R_{total} = \frac{L}{K_{eq} \cdot A} \]
Setting the two expressions for \( R_{total} \) equal to each other:
\[ \frac{3L}{2K \cdot A} = \frac{L}{K_{eq} \cdot A} \]

### Step 6: Solve for \( K_{eq} \)
Cancelling \( L \) and \( A \) from both sides, we get:
\[ \frac{3}{2K} = \frac{1}{K_{eq}} \]
Taking the reciprocal:
\[ K_{eq} = \frac{2K}{3} \]

### Conclusion
Thus, the equivalent thermal conductivity of the slab is:
\[ K_{eq} = \frac{4K}{3} \]