A hollow tube has a length l, inner radius `R_(1)` and outer radius `R_(2)` . The material has a thermal conductivity K. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperature `T_(1)` and `T_(2)``(T_(2)gtT_(1))` (b) the inside of the tube is maintained at temperature `T_(1)` and the outside is maintained at `T_(2)` .

Explanation

To solve the problem of heat conduction through a hollow tube, we will analyze both scenarios step by step.

### Given:
- Length of the tube: \( l \)
- Inner radius: \( R_1 \)
- Outer radius: \( R_2 \)
- Thermal conductivity: \( K \)
- Temperatures: \( T_1 \) and \( T_2 \) (where \( T_2 > T_1 \))

### Part (a): Flat ends maintained at temperatures \( T_1 \) and \( T_2 \)

1. **Identify the Heat Transfer Equation**:
The heat transfer through conduction can be expressed using Fourier's law:
\[ Q = \frac{K \cdot A \cdot (T_2 - T_1)}{l} \]
where \( A \) is the cross-sectional area through which heat is conducted.

2. **Calculate the Cross-sectional Area**:
The area \( A \) for a hollow tube is given by the difference in the areas of the outer and inner circles:
\[ A = \pi R_2^2 - \pi R_1^2 = \pi (R_2^2 - R_1^2) \]

3. **Substitute the Area into the Heat Transfer Equation**:
Now, substituting the area into the heat transfer equation:
\[ Q = \frac{K \cdot \pi (R_2^2 - R_1^2) \cdot (T_2 - T_1)}{l} \]

4. **Final Expression for Part (a)**:
Thus, the heat flowing through the walls of the tube when the flat ends are maintained at temperatures \( T_1 \) and \( T_2 \) is:
\[ Q = \frac{K \cdot \pi (R_2^2 - R_1^2) \cdot (T_2 - T_1)}{l} \]

### Part (b): Inside of the tube maintained at temperature \( T_1 \) and outside at \( T_2 \)

1. **Identify the Heat Transfer Equation**:
In this case, the heat flows radially from the inner surface to the outer surface. We will use the differential form of Fourier's law:
\[ dQ = -K \cdot A \cdot \frac{dT}{dr} \]

2. **Calculate the Area for Radial Flow**:
The area \( A \) for a cylindrical shell of radius \( r \) and thickness \( dr \) is:
\[ A = 2 \pi r \cdot l \]

3. **Set Up the Differential Equation**:
Substituting the area into the differential equation:
\[ dQ = -K \cdot (2 \pi r \cdot l) \cdot \frac{dT}{dr} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{dQ}{dr} = -\frac{2 \pi K l}{r} \cdot (T_2 - T_1) \]

5. **Integrate**:
Integrate from \( R_1 \) to \( R_2 \) for \( Q \) and from \( T_1 \) to \( T_2 \) for \( T \):
\[ \int_{Q}^{0} dQ = -2 \pi K l \int_{R_1}^{R_2} \frac{(T_2 - T_1)}{r} dr \]

This results in:
\[ Q \cdot \ln\left(\frac{R_2}{R_1}\right) = -2 \pi K l (T_2 - T_1) \]

6. **Final Expression for Part (b)**:
Rearranging gives us:
\[ Q = \frac{2 \pi K l (T_1 - T_2)}{\ln\left(\frac{R_2}{R_1}\right)} \]

### Summary of Results:
- **Part (a)**:
\[ Q = \frac{K \cdot \pi (R_2^2 - R_1^2) \cdot (T_2 - T_1)}{l} \]
- **Part (b)**:
\[ Q = \frac{2 \pi K l (T_1 - T_2)}{\ln\left(\frac{R_2}{R_1}\right)} \]