Two identical plates with thermal conductivities K and 3K are joined together to form a single plate of double thickness. The equivalent thermal conductivity of one composite plate so formed for the flow of heat through its thickness is.

Options

A. K
B. 1.5 K
C. 2.5 K
D. 3 K

Answer

To find the equivalent thermal conductivity of the composite plate formed by two identical plates with thermal conductivities \( K \) and \( 3K \), we can follow these steps:

### Step 1: Understand the Configuration
We have two plates:
- Plate 1 with thermal conductivity \( K \) and thickness \( t \).
- Plate 2 with thermal conductivity \( 3K \) and thickness \( t \).

When combined, they form a single plate of double thickness \( 2t \).

### Step 2: Calculate the Thermal Resistance of Each Plate
The thermal resistance \( R \) for a plate is given by the formula:
\[ R = \frac{L}{KA} \]
where \( L \) is the thickness of the plate, \( K \) is the thermal conductivity, and \( A \) is the cross-sectional area.

For Plate 1:
\[ R_1 = \frac{t}{K \cdot A} \]

For Plate 2:
\[ R_2 = \frac{t}{3K \cdot A} \]

### Step 3: Calculate the Total Resistance
Since the two plates are in series, the total thermal resistance \( R_{\text{total}} \) is the sum of the individual resistances:
\[ R_{\text{total}} = R_1 + R_2 = \frac{t}{K \cdot A} + \frac{t}{3K \cdot A} \]

### Step 4: Simplify the Expression
To combine the resistances, we need a common denominator:
\[ R_{\text{total}} = \frac{3t}{3K \cdot A} + \frac{t}{3K \cdot A} = \frac{3t + t}{3K \cdot A} = \frac{4t}{3K \cdot A} \]

### Step 5: Calculate the Equivalent Thermal Conductivity
The equivalent thermal resistance for the composite plate of thickness \( 2t \) can be expressed as:
\[ R_{\text{equivalent}} = \frac{2t}{K_{\text{equivalent}} \cdot A} \]

Setting the total resistance equal to the equivalent resistance:
\[ \frac{4t}{3K \cdot A} = \frac{2t}{K_{\text{equivalent}} \cdot A} \]

### Step 6: Solve for \( K_{\text{equivalent}} \)
Cancelling \( t \) and \( A \) from both sides:
\[ \frac{4}{3K} = \frac{2}{K_{\text{equivalent}}} \]

Cross-multiplying gives:
\[ 4K_{\text{equivalent}} = 6K \]

Thus, we find:
\[ K_{\text{equivalent}} = \frac{6K}{4} = \frac{3K}{2} = 1.5K \]

### Conclusion
The equivalent thermal conductivity of the composite plate is \( 1.5K \).