IF two metallic plates of equal thickness and thermal conductivities `K_(1) and K_(2)` are put together face to face anda common plate is constructed, then the equivalent thermal conductivity of this plate will be

Options

A. `(K_(1)K_(2))/(K_(1) + K_(2))`
B. `(2K_(1)K_(2))/(K_(1) + K_(2))`
C. `((K_(1)^(2) _ K_(2)^(2))^(3//2))/(K_(1)K_(2))`
D. `((K_(1)^(2) + K_(2)^(2))^(3//2))/(2K_(1)K_(2))`

Answer

To find the equivalent thermal conductivity \( K_{eq} \) of two metallic plates with equal thickness and thermal conductivities \( K_1 \) and \( K_2 \) placed face to face, we can follow these steps:

### Step 1: Understand the Setup
We have two plates:
- Plate 1 with thermal conductivity \( K_1 \)
- Plate 2 with thermal conductivity \( K_2 \)
Both plates have the same thickness \( L \) and cross-sectional area \( A \).

### Step 2: Apply Fourier's Law of Heat Conduction
According to Fourier's law, the heat transfer \( Q \) through a material is given by:
\[ Q = K \cdot A \cdot \frac{\Delta T}{L} \]
where \( \Delta T \) is the temperature difference across the material.

### Step 3: Define the Temperature Difference
Let:
- The temperature at one end of Plate 1 be \( T_1 \)
- The temperature at the junction (interface) of the two plates be \( T \)
- The temperature at the end of Plate 2 be \( T_2 \)

The temperature differences for the two plates are:
- For Plate 1: \( \Delta T_1 = T_1 - T \)
- For Plate 2: \( \Delta T_2 = T - T_2 \)

### Step 4: Write the Heat Transfer Equations
For Plate 1:
\[ Q = K_1 \cdot A \cdot \frac{T_1 - T}{L} \]
For Plate 2:
\[ Q = K_2 \cdot A \cdot \frac{T - T_2}{L} \]

### Step 5: Set the Heat Transfer Equal
Since the heat transfer \( Q \) is the same through both plates (steady state), we can set the equations equal to each other:
\[ K_1 \cdot A \cdot \frac{T_1 - T}{L} = K_2 \cdot A \cdot \frac{T - T_2}{L} \]

### Step 6: Simplify the Equation
Cancel out \( A \) and \( L \) from both sides:
\[ K_1 (T_1 - T) = K_2 (T - T_2) \]

### Step 7: Rearrange the Equation
Rearranging gives:
\[ K_1 T_1 - K_1 T = K_2 T - K_2 T_2 \]
\[ K_1 T_1 + K_2 T_2 = (K_1 + K_2) T \]

### Step 8: Solve for the Junction Temperature \( T \)
From the above equation, we can express \( T \):
\[ T = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2} \]

### Step 9: Find the Equivalent Thermal Conductivity
The equivalent thermal conductivity \( K_{eq} \) can be defined as:
\[ Q = K_{eq} \cdot A \cdot \frac{T_1 - T_2}{L} \]
Substituting \( T \) into this equation and simplifying will lead us to:
\[ \frac{T_1 - T_2}{L} = \frac{(T_1 - T) + (T - T_2)}{L} \]
Substituting the expressions for \( T_1 - T \) and \( T - T_2 \) gives:
\[ \frac{T_1 - T_2}{L} = \frac{Q}{K_{eq} \cdot A} \]

### Step 10: Final Expression for \( K_{eq} \)
After manipulating the equations, we find:
\[ \frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} \]
Thus, the equivalent thermal conductivity is given by:
\[ K_{eq} = \frac{2 K_1 K_2}{K_1 + K_2} \]

### Conclusion
The equivalent thermal conductivity of the combined plate is:
\[ K_{eq} = \frac{2 K_1 K_2}{K_1 + K_2} \]