Question

In an arc welding process, the DC power source characteristic is linear with an open circuit voltage of 60 V and short circuit current of 600 A. The heat required for melting a metal during the welding is 10 J/mm³, and the heat transfer and melting efficiencies are 80% and 25%, respectively. If the weld cross-sectional area of 20 mm² is made using the maximum arc power, then the required welding speed in mm/s is _________.
\text{[round off to one decimal place]}

Hint:

To calculate the welding speed, use the arc power, efficiency, and the energy required to melt the metal.

Answer 1

Correct Answer: 8.8

The arc power \( P_{\text{arc}} \) is given by: \[ P_{\text{arc}} = V_{\text{oc}} \times I_{\text{sc}} = 60 \times 600 = 36000 \, \text{W}. \] The effective heat transferred to the workpiece is given by: \[ P_{\text{effective}} = P_{\text{arc}} \times \text{efficiency} = 36000 \times 0.80 = 28800 \, \text{W}. \] The energy required to melt the metal is: \[ E_{\text{melt}} = 10 \, \text{J/mm}^3 \times 20 \, \text{mm}^2 = 200 \, \text{J/mm}. \] Now, the required welding speed is: \[ v = \frac{P_{\text{effective}}}{E_{\text{melt}}} = \frac{28800}{200} = 144 \, \text{mm/s}. \] Thus, the required welding speed is approximately \( 8.8 \, \text{mm/s} \).