Question

Two $8mm$ thick steel plates are placed $5mm$ apart and welded by a butt joint. Welding is carried out at $20V$ and speed of $5mm/s$. Heat transfer efficiency is $0.80$. If the heat required to melt steel is $10J/m{m}^3$ and melting efficiency is $0.625$. The weld current (in amperes) will be

• A. $100$
• B. $300$
• C. $400$
• D. $200$

Answer 1

The correct option is D $200$

$\text{Vol of weld puddle}=8\times 5\times x=40xm{m}^3$

$x\text{= a gap}$

$\text{Heat required}=10\times 40\times x=400xJ\text{... (i)}$

$\text{Heat generated}={\eta }_1{\eta }_2\times \frac{VIx}{v}$

$=\frac{0.625\times 20\times I\times x\times 0.8}{5}$

$400x=\frac{0.625\times 20\times I\times x\times 0.8}{5}$

$\text{[Given}{\eta }_1=0.8,{\eta }_2=0.62\text{]}$

$I=200A$