Question
Download Solution PDFIn an arc welding process, the voltage and current are 25 Volts and 300 amperes respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s, the net heat input (in J/mm) will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The net heat input in arc welding is calculated using:
\( Q = \frac{V \times I \times \eta}{\text{welding speed}} \)
Given:
- Voltage, \( V = 25 \, \text{Volts} \)
- Current, \( I = 300 \, \text{Amps} \)
- Efficiency, \( \eta = 0.85 \)
- Welding Speed, \( 8 \, \text{mm/s} \)
Calculation:
\( Q = \frac{25 \times 300 \times 0.85}{8} = \frac{6375}{8} = 796.875 \approx \boxed{796.8 \, \text{J/mm}} \)
Last updated on Jul 15, 2025
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