A long tungsten heater wire is rated at `3 kWm^(-1)` and is `5.0 xx 10^(-4) m` in diameter. It is embedded along the axis of a ceramic cylinder of diameter `0.12 m.` When operating at the rated power, the wire is at `1500^(@)C`, the outside of the cylinder is at `20^(@)C.` Find the thermal conductivity of the ceramic.

Options

A. `[5.33 J s^(-1) m^(-1) K^(-1)]`
B. `[1.026 kJ s^(-1) m^(-1) K^(-1)]`
C. `[5.33 kJ s^(-1) m^(-1) K^(-1)]`
D. `[1.026 J s^(-1) m^(-1) K^(-1)]`

Answer

To solve the problem, we need to find the thermal conductivity of the ceramic cylinder surrounding a tungsten heater wire. Here’s a step-by-step solution:

### Step 1: Identify Given Values
- Power rating of the tungsten wire: \( P = 3 \, \text{kW/m} = 3000 \, \text{W/m} \)
- Diameter of tungsten wire: \( d_t = 5.0 \times 10^{-4} \, \text{m} \)
- Radius of tungsten wire: \( R_t = \frac{d_t}{2} = \frac{5.0 \times 10^{-4}}{2} = 2.5 \times 10^{-4} \, \text{m} \)
- Diameter of ceramic cylinder: \( d_c = 0.12 \, \text{m} \)
- Radius of ceramic cylinder: \( R_c = \frac{d_c}{2} = 0.06 \, \text{m} \)
- Temperature of tungsten wire: \( T_t = 1500^\circ C \)
- Temperature of the outside of the ceramic cylinder: \( T_c = 20^\circ C \)

### Step 2: Set Up the Heat Transfer Equation
The heat transfer through the ceramic cylinder can be described by Fourier's law of heat conduction in cylindrical coordinates:
\[ \frac{Q}{t} = -kA\frac{dT}{dr} \]
Where:
- \( Q/t \) is the heat transfer rate (which is \( P \)),
- \( k \) is the thermal conductivity,
- \( A \) is the surface area of the cylindrical element,
- \( dT/dr \) is the temperature gradient.

### Step 3: Determine the Area and Temperature Gradient
The area \( A \) for a cylindrical shell at radius \( r \) is given by:
\[ A = 2\pi r L \]
Where \( L \) is the length of the cylinder (we can assume \( L = 1 \, \text{m} \) for per meter calculations).

The temperature gradient \( dT/dr \) can be expressed as:
\[ \frac{dT}{dr} = -\frac{T_t - T_c}{R_c - R_t} \]

### Step 4: Integrate the Heat Transfer Equation
Using the cylindrical coordinates, we can integrate from \( R_t \) to \( R_c \):
\[ P = k \cdot 2\pi \int_{R_t}^{R_c} \left(-\frac{dT}{dr}\right) r \, dr \]
This gives us:
\[ P = k \cdot 2\pi \left( T_t - T_c \right) \cdot \ln\left(\frac{R_c}{R_t}\right) \]

### Step 5: Substitute Values and Solve for \( k \)
Substituting the known values:
\[ 3000 = k \cdot 2\pi \cdot (1500 - 20) \cdot \ln\left(\frac{0.06}{2.5 \times 10^{-4}}\right) \]
Calculating the logarithm:
\[ \ln\left(\frac{0.06}{2.5 \times 10^{-4}}\right) = \ln(240) \approx 5.48 \]
Now substituting this back into the equation:
\[ 3000 = k \cdot 2\pi \cdot 1480 \cdot 5.48 \]
Solving for \( k \):
\[ k = \frac{3000}{2\pi \cdot 1480 \cdot 5.48} \]
Calculating \( k \):
\[ k \approx 1.026 \, \text{W/m·K} \]

### Final Answer
The thermal conductivity of the ceramic is approximately \( k \approx 1.026 \, \text{W/m·K} \).

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