Question

In an arc welding process, the voltage and current are $25V$ and $300A$ respectively. The arc heat transfer efficiency is $0.85$ and welding speed is $8mm/s$. The net heat input (in $J/mm)$ is

• A. $64$
• B. $79700$
• C. $1103$
• D. $797$

Answer 1

The correct option is D $797$

$\text{Given data:}$

$\text{Voltage:}V=25volt$

$\text{Current:}I=300amp$

$\text{Efficiency:}\eta =0.85$

$\text{Weld speed:}v=8mm/s$

$\text{Power generated due to}arc=VI$

$\text{Power utilized as heat}=\eta VI$

$=0.85\times 25\times 300=6375J/s$

$\text{The net heat input}=\frac{6375}{v}J/mm=\frac{6375}{8}J/mm$

$=769.8J/mm\approx 797J/mm$