A single pass full penetration weld pass is made on steel plates having specific heat, ρc (0.0044 J/mm3), thickness, t (5 mm), initial temperature, to (25 ⁰C), melting temperature, tm (1510 ⁰C), Net heat supplied, Q (720 J/mm). Calculate the peak temperature at 3.0 mm, 1.5 mm and 0 mm distance from the fusion boundary

To calculate the peak temperature at various distances from the fusion boundary for a single-pass full penetration weld, we use the Rosenthal equation for a semi-infinite plate. The formula for the peak temperature \( T(x) \) at a distance \( x \) from the fusion boundary is:

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

Where:
- \( T_0 \) is the initial temperature (25 °C)
- \( Q \) is the net heat supplied per unit length (720 J/mm)
- \( k \) is the thermal conductivity (not provided, but we'll assume the term \( \frac{Q}{k} \) is combined within the formula)
- \( x \) is the distance from the fusion boundary (3.0 mm, 1.5 mm, and 0 mm)
- \( \rho c \) is the specific heat (0.0044 J/mm³)

### Calculation at Different Distances

1. **At \( x = 0 \) mm (Fusion Boundary):**

   \[
   T(0) = T_0 + \frac{Q}{2\pi k \cdot 0} \cdot \exp\left(-\frac{\rho c \cdot 0^2}{4Q}\right)
   \]

   Since \( x = 0 \), the distance term becomes undefined in this form, but the peak temperature will actually be the melting temperature \( T_m \) (1510 °C) due to full penetration. 

   Therefore, \( T(0) = 1510 \, \text{°C} \).

2. **At \( x = 1.5 \) mm:**

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

3. **At \( x = 3.0 \) mm:**

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

### Note:
To get exact numeric values for \( T(1.5) \) and \( T(3.0) \), we need the thermal conductivity \( k \) or we could proceed under the assumption that \( \frac{Q}{k} \) is a known quantity. If you can provide this value, or if we assume it cancels in a simplified expression, we can compute it. Alternatively, we can proceed with further simplifications or assume a normalized form.

Let me know how you’d like to proceed or if you have the missing thermal conductivity value.

### Questions:

1. Would you like to simplify the calculation with assumptions, or do you have the thermal conductivity \( k \)?
2. Do you need more detailed explanations on how the Rosenthal equation is derived or applied?
3. Are you interested in understanding how the distance \( x \) affects the cooling rate after welding?
4. Would you like to explore how variations in specific heat \( \rho c \) influence the temperature distribution?
5. Are there any specific applications where you would apply this calculation?

**Tip:** When calculating temperature distribution in welding, understanding the material's thermal properties, like conductivity, is crucial as they significantly affect the heat flow and cooling rates.

Learn how to calculate peak temperature at different distances from the fusion boundary in welding using the Rosenthal equation. This problem involves specific heat, thickness, initial and melting temperatures, and net heat supplied. Explore detailed calculations and understand the impact of distance on temperature distribution.

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