Question:
An arc welding operation is performed at 25 V and 200 A at welding speed of 2 mm/s. The heat used for melting is 80% of the total heat generated. The unit melting energy of the metal to be joined is 10 J/mm$^3$. The volume of the weld metal produced per unit time, in mm$^3$/s, is ................. (in integer).
An arc welding operation is performed at 25 V and 200 A at welding speed of 2 mm/s. The heat used for melting is 80% of the total heat generated. The unit melting energy of the metal to be joined is 10 J/mm$^3$. The volume of the weld metal produced per unit time, in mm$^3$/s, is ................. (in integer).
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The volume of weld metal produced is directly proportional to the heat used for melting and inversely proportional to the unit melting energy of the metal.
Updated On: Sep 4, 2025
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Solution and Explanation
Given:
Voltage, \( V = 25 \, \text{V} \)
Current, \( I = 200 \, \text{A} \)
Welding speed, \( v = 2 \, \text{mm/s} \)
Heat used for melting = 80% of total heat
Unit melting energy of the metal = 10 J/mm$^3$
The total heat generated per second is: \[ Q_{\text{total}} = V \times I = 25 \times 200 = 5000 \, \text{W} = 5000 \, \text{J/s} \] The heat used for melting is 80% of the total heat generated: \[ Q_{\text{melt}} = 0.8 \times 5000 = 4000 \, \text{J/s} \] The volume of the weld metal produced per second is given by: \[ \text{Volume per second} = \frac{Q_{\text{melt}}}{\text{Unit melting energy}} = \frac{4000}{10} = 400 \, \text{mm}^3/\text{s} \] Thus, the volume of the weld metal produced per unit time is 400 mm$^3$/s.
Voltage, \( V = 25 \, \text{V} \)
Current, \( I = 200 \, \text{A} \)
Welding speed, \( v = 2 \, \text{mm/s} \)
Heat used for melting = 80% of total heat
Unit melting energy of the metal = 10 J/mm$^3$
The total heat generated per second is: \[ Q_{\text{total}} = V \times I = 25 \times 200 = 5000 \, \text{W} = 5000 \, \text{J/s} \] The heat used for melting is 80% of the total heat generated: \[ Q_{\text{melt}} = 0.8 \times 5000 = 4000 \, \text{J/s} \] The volume of the weld metal produced per second is given by: \[ \text{Volume per second} = \frac{Q_{\text{melt}}}{\text{Unit melting energy}} = \frac{4000}{10} = 400 \, \text{mm}^3/\text{s} \] Thus, the volume of the weld metal produced per unit time is 400 mm$^3$/s.
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