This question asks us to find the formula for the thermal resistance of a hollow cylinder when heat is conducted radially.

Understanding Thermal Resistance

Thermal resistance is a property that describes how well a material or object resists the flow of heat. A higher value indicates greater resistance to heat transfer. It's analogous to electrical resistance in an electrical circuit.

The general formula for thermal resistance ($R_{th}$) is the ratio of the temperature difference ($\Delta T$) across the object to the rate of heat flow ($Q$) through it:

$$R_{th} = \frac{\Delta T}{Q}$$

Radial Conduction in a Hollow Cylinder

For a hollow cylinder with an inner radius $r_1$, outer radius $r_2$, length $L$, and thermal conductivity $K$, heat transfer occurs radially. This means heat flows outwards from the inner surface to the outer surface (or vice versa).

We use Fourier's Law of Heat Conduction to describe this process. For radial heat flow, Fourier's Law is expressed as:

$$Q = -K \cdot A \cdot \frac{dT}{dr}$$

Where:

• $Q$ is the rate of heat transfer.
• $K$ is the thermal conductivity of the cylinder's material.
• $A$ is the surface area perpendicular to the direction of heat flow. For a cylinder, this area changes with radius $r$, and is given by $A = 2\pi r L$.
• $\frac{dT}{dr}$ is the temperature gradient in the radial direction.

Substituting the area $A$ into the equation:

$$Q = -K (2\pi r L) \frac{dT}{dr}$$

Deriving the Thermal Resistance Formula

To find the thermal resistance, we need to integrate this equation over the radius of the cylinder, from the inner radius $r_1$ to the outer radius $r_2$. We rearrange the equation to separate the variables $T$ and $r$:

$$dT = -\frac{Q}{2\pi KL} \frac{dr}{r}$$

Now, we integrate both sides. We assume the temperature at the inner radius $r_1$ is $T_1$ and at the outer radius $r_2$ is $T_2$. Typically, $r_1 < r_2$, and heat flows from the higher temperature to the lower temperature.

$$\int_{T_1}^{T_2} dT = \int_{r_1}^{r_2} -\frac{Q}{2\pi KL} \frac{dr}{r}$$

Performing the integration:

$$T_2 - T_1 = -\frac{Q}{2\pi KL} \int_{r_1}^{r_2} \frac{1}{r} dr$$

The integral of $\frac{1}{r}$ is the natural logarithm, $\ln(r)$:

$$T_2 - T_1 = -\frac{Q}{2\pi KL} [\ln(r)]_{r_1}^{r_2}$$

$$T_2 - T_1 = -\frac{Q}{2\pi KL} (\ln(r_2) - \ln(r_1))$$

Using the logarithm property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

$$T_2 - T_1 = -\frac{Q}{2\pi KL} \ln\left(\frac{r_2}{r_1}\right)$$

The temperature difference is $\Delta T = T_1 - T_2$. Multiplying the equation by -1:

$$T_1 - T_2 = \frac{Q}{2\pi KL} \ln\left(\frac{r_2}{r_1}\right)$$

Now, we can rearrange this to find the thermal resistance $R_{th} = \frac{T_1 - T_2}{Q}$:

$$R_{th} = \frac{T_1 - T_2}{Q} = \frac{1}{Q} \left( \frac{Q}{2\pi KL} \ln\left(\frac{r_2}{r_1}\right) \right)$$

This simplifies to:

$$R_{th} = \frac{\ln(r_2 / r_1)}{2\pi KL}$$

Comparing with Options

We compare our derived formula with the given options:

• Option 1: $\ln (r_1/r_2)/2\pi KL$
• Option 2: $\ln (r_2 /r_1 )/2\pi KL$
• Option 3: $2\pi KL/\ln (r 1 /r 2 )$
• Option 4: $2\pi KL/\ln (r_2 /r_1 )$

The derived formula, $\frac{\ln(r_2 / r_1)}{2\pi KL}$, matches Option 2 exactly.