Use the thermal conductivity of steel and complete the solution.

To complete the solution, we'll use the thermal conductivity of steel, \( k \), which is approximately \( 45 \, \text{W/m·K} \) or equivalently \( 0.045 \, \text{J/mm·°C} \).

### Rosenthal Equation for Temperature Distribution:

The Rosenthal equation for a semi-infinite plate can be written as:

\[
T(x) = T_0 + \frac{Q}{2\pi k x} \cdot \exp\left(-\frac{\rho c x^2}{4Q}\right)
\]

Where:
- \( T_0 = 25 \, \text{°C} \) (initial temperature)
- \( Q = 720 \, \text{J/mm} \) (net heat supplied)
- \( k = 0.045 \, \text{J/mm·°C} \) (thermal conductivity of steel)
- \( \rho c = 0.0044 \, \text{J/mm}^3 \) (specific heat)
- \( x \) is the distance from the fusion boundary

### 1. Temperature at \( x = 0 \, \text{mm} \):

At \( x = 0 \, \text{mm} \), the temperature is equal to the melting temperature \( T_m = 1510 \, \text{°C} \).

\[
T(0) = 1510 \, \text{°C}
\]

### 2. Temperature at \( x = 1.5 \, \text{mm} \):

\[
T(1.5) = 25 + \frac{720}{2\pi \times 0.045 \times 1.5} \cdot \exp\left(-\frac{0.0044 \times 1.5^2}{4 \times 720}\right)
\]

Calculating the terms:

\[
T(1.5) = 25 + \frac{720}{0.4239} \cdot \exp\left(-\frac{0.0099}{2880}\right)
\]

\[
T(1.5) = 25 + 1698.18 \cdot \exp(-0.00000344)
\]

Since \( \exp(-0.00000344) \approx 1 \) (very close to 1 for small exponent),

\[
T(1.5) \approx 25 + 1698.18 \, \text{°C}
\]

\[
T(1.5) \approx 1723.18 \, \text{°C}
\]

### 3. Temperature at \( x = 3.0 \, \text{mm} \):

\[
T(3.0) = 25 + \frac{720}{2\pi \times 0.045 \times 3.0} \cdot \exp\left(-\frac{0.0044 \times 3.0^2}{4 \times 720}\right)
\]

Calculating the terms:

\[
T(3.0) = 25 + \frac{720}{0.8478} \cdot \exp\left(-\frac{0.0396}{2880}\right)
\]

\[
T(3.0) = 25 + 849.09 \cdot \exp(-0.00001375)
\]

Since \( \exp(-0.00001375) \approx 1 \),

\[
T(3.0) \approx 25 + 849.09 \, \text{°C}
\]

\[
T(3.0) \approx 874.09 \, \text{°C}
\]

### Summary of Results:

- **At \( x = 0 \, \text{mm} \):** \( T(0) = 1510 \, \text{°C} \)
- **At \( x = 1.5 \, \text{mm} \):** \( T(1.5) \approx 1723.18 \, \text{°C} \)
- **At \( x = 3.0 \, \text{mm} \):** \( T(3.0) \approx 874.09 \, \text{°C} \)

These calculations indicate that the peak temperature decreases as the distance from the fusion boundary increases.

### Questions:

1. Would you like to discuss the physical significance of the exponential decay in the temperature distribution?
2. Are you interested in seeing how these calculations change with different values of specific heat or conductivity?
3. Would you like to analyze how these temperatures affect the microstructure of the steel?
4. Are there other distances from the fusion boundary you would like to investigate?
5. Would you like to explore the implications of these temperatures for welding residual stresses?

**Tip:** In welding, accurate temperature distribution calculations are crucial for predicting microstructural changes, distortion, and residual stresses in the welded material.

Explore the temperature distribution in steel using its thermal conductivity. Learn about the Rosenthal equation for a semi-infinite plate and how temperature varies with distance from the fusion boundary. Includes calculations at specific distances and implications for welding processes.

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