Cylindrical Rods Heat Conduction Analysis

The problem requires finding the interface temperature ($T_{int}$) between two cylindrical rods (A and B) made of different materials when they are joined end-to-end and subjected to different temperatures at their free ends. This is a problem of steady-state heat conduction.

Physics Principle: Steady State Heat Flow

Under steady-state conditions, the rate of heat transfer ($H$) is constant throughout the composite system. The heat flow rate through a cylindrical rod is given by the formula:

$H = \frac{KA \Delta T}{L}$

Where:

• $K$ represents the thermal conductivity.
• $A$ is the cross-sectional area ($A = \pi r^2$).
• $\Delta T$ is the temperature difference across the rod.
• $L$ is the length of the rod.

Applying Heat Flow Equality

For the joined rods A and B, the rate of heat flow must be the same at the interface:

$H_A = H_B$

Using the heat flow formula:

$\frac{K_A A_A (T_A - T_{int})}{L_A} = \frac{K_B A_B (T_{int} - T_B)}{L_B}$

Since the cross-sectional area $A = \pi r^2$, we can substitute this in:

$\frac{K_A (\pi r_A^2) (T_A - T_{int})}{L_A} = \frac{K_B (\pi r_B^2) (T_{int} - T_B)}{L_B}$

Canceling $\pi$ from both sides yields:

$\frac{K_A r_A^2 (T_A - T_{int})}{L_A} = \frac{K_B r_B^2 (T_{int} - T_B)}{L_B}$

Substituting Given Ratios and Values

The problem provides the following information:

• Length ratio: $\frac{L_A}{L_B} = 1$, which implies $\frac{L_B}{L_A} = 1$.
• Radius ratio: $\frac{r_A}{r_B} = 2$, which implies $\frac{r_A^2}{r_B^2} = (\frac{r_A}{r_B})^2 = 2^2 = 4$.
• Thermal conductivity ratio: $\frac{K_A}{K_B} = \frac{1}{2}$.
• Temperatures: $T_A = 400$ K and $T_B = 200$ K.

Rearranging the equation to group the ratios:

$\frac{K_A}{K_B} \times \frac{r_A^2}{r_B^2} \times \frac{L_B}{L_A} \times (T_A - T_{int}) = (T_{int} - T_B)$

Calculation of Interface Temperature

Substitute the known ratios and temperatures into the equation:

$(\frac{1}{2}) \times (4) \times (1) \times (400 \text{ K} - T_{int}) = (T_{int} - 200 \text{ K})$

Simplify the left side:

$2 \times (400 - T_{int}) = (T_{int} - 200)$

Expand the terms:

$800 - 2 T_{int} = T_{int} - 200$

Rearrange to solve for $T_{int}$:

$800 + 200 = T_{int} + 2 T_{int}$

Combine terms:

$1000 = 3 T_{int}$

Calculate the final value:

$T_{int} = \frac{1000}{3}$ K

$T_{int} \approx 333.33$ K

The calculated interface temperature is $\frac{1000}{3}$ K.