Question

A weld is made using $MIG$ welding process with the following welding parameters:

$\text{Current:}200A\text{, Voltage :}25V$

$\text{Welding speed:}18cm/min$

$\text{Wire diameter:}1.2mm\text{,}$

$\text{Wire feed rate:}4m/min\text{.}$

$\text{Thermal efficiency of the processes}65\text{\%}$

$\text{The area of cross-section of weld bead in}m{m}^2\text{is}$

• A. $38.6$
• B. $16.3$
• C. $30.3$
• D. $251$

Answer 1

The correct option is D $251$

$\text{Volume of weld}=\frac{\pi }{4}{d}^2\times l$

$d=\text{wire diameter,}l=\text{wire feed rate}$

$V=\frac{\pi }{4}\times {1.2}^2\times 4000$

$=4521.6m{m}^2/min$

$\text{Area of weld}=\frac{4521.6}{\text{speed}}=\frac{4521.6}{18\times 10}=25.12m{m}^2$