Question

If a Metal Inert Gas welding, two stainless steel workpieces are joined with electrode diameter of 2.4 mm, wire feed rate of 3.2 m/min and total area of weld bead of 40 mm${}^2$, then the welding speed used in the welding process will be _________ mm/sec.

• A. 3.04
• B. 7.88
• C. 2.50
• D. 6.03

Answer 1

The correct option is D 6.03

$\text{Melting rate of electrode}$

$\frac{\pi }{4}\times {2.4}^2\times 3200=14476.45m{m}^3/min$

$\text{Melting rate of electrode = filling rate of weld bead}$

$\text{14476.45 = area of weld bead}\times \text{velocity of welding}$

$\text{14476.45 = 40}\times \text{V}$

$\Rightarrow \text{=V = 14476.45/40 = 361.9 mm/min}$

$\text{Welding speed (mm/sec) =}\frac{361.9}{60}\text{= 6.03 mm/sec}$