A hole of radius `r_(1)` is made centrally in a uniform circular disc of thickness d and radius `r_(2)` . The inner surface (a cylinder of length d and radius `r_(1)`) is maintained at a temperature `theta_(1)` and the outer surface (a cylinder of length d and radius `r_(2)`) is maintained at a temperature `theta_(2)(theta_(1)gttheta_(2))` .The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Explanation

To solve the problem of calculating the heat flowing per unit time through a circular disc with a hole, we can follow these steps:

### Step 1: Understand the Setup
We have a circular disc with:
- Inner radius \( r_1 \)
- Outer radius \( r_2 \)
- Thickness \( d \)
- Inner surface temperature \( \theta_1 \)
- Outer surface temperature \( \theta_2 \) (where \( \theta_1 > \theta_2 \))
- Thermal conductivity \( K \)

### Step 2: Apply Fourier's Law of Heat Conduction
According to Fourier's law, the heat transfer \( Q \) through a material is given by:
\[ Q = -K A \frac{dT}{dx} \]
For a cylindrical shell at radius \( r \) with thickness \( dr \), the area \( A \) is given by the lateral surface area of the cylinder:
\[ A = 2 \pi r \cdot d \]
Thus, we can express the heat transfer through the ring as:
\[ dQ = -K (2 \pi r d) \frac{dT}{dr} \]

### Step 3: Rearranging the Equation
Rearranging gives us:
\[ dQ = -2 \pi K d \cdot r \frac{dT}{dr} \]

### Step 4: Integrate to Find Total Heat Transfer
To find the total heat transfer \( Q \) from the inner radius \( r_1 \) to the outer radius \( r_2 \), we integrate:
\[ Q = \int_{T_1}^{T_2} dQ = -2 \pi K d \int_{r_1}^{r_2} r \frac{dT}{dr} dr \]

### Step 5: Change of Variables
We can express \( dT \) in terms of \( dr \):
\[ \frac{dT}{dr} = \frac{\theta_2 - \theta_1}{r_2 - r_1} \]
Substituting this into the integral gives:
\[ Q = -2 \pi K d \left( \frac{\theta_2 - \theta_1}{r_2 - r_1} \right) \int_{r_1}^{r_2} r dr \]

### Step 6: Evaluate the Integral
The integral \( \int r dr \) evaluates to:
\[ \int_{r_1}^{r_2} r dr = \left[ \frac{r^2}{2} \right]_{r_1}^{r_2} = \frac{r_2^2}{2} - \frac{r_1^2}{2} = \frac{1}{2}(r_2^2 - r_1^2) \]

### Step 7: Substitute Back into the Equation
Substituting the evaluated integral back into the equation for \( Q \):
\[ Q = -2 \pi K d \cdot \frac{(\theta_2 - \theta_1)}{(r_2 - r_1)} \cdot \frac{1}{2}(r_2^2 - r_1^2) \]

### Step 8: Final Expression for Heat Transfer
The negative sign indicates the direction of heat flow (from higher to lower temperature). Thus, the total heat transfer per unit time through the disc is:
\[ Q = \frac{2 \pi K d (\theta_1 - \theta_2)}{\ln\left(\frac{r_2}{r_1}\right)} \]