A gas tungsten arc welding operation is performed using a current of 250 A and an arc voltage of 20 V at a welding speed of 5 mm/s. Assuming that the arc efficiency is 70%, the net heat input per unit length of the weld will be______kJ/mm (round off to one place).

The correct answer is 0.7.

Step-by-Step Explanation:

To find the net heat input per unit length of the weld, we need to calculate the total power generated by the welding arc, apply the arc efficiency to find the net heat entering the workpiece, and then divide this net heat by the welding speed.

1. Identify the given parameters:
Welding current, I = 250 A
Arc voltage, V = 20 V
Welding speed, v = 5 mm/s
Arc heat transfer efficiency, η = 70% = 0.7

2. Calculate the total power generated by the arc:
The total electric power (P) generated by the arc is given by:
$P=V\times I$
Substituting the given values:
$P=20\text{V}\times 250\text{A}=5000\text{W}=5000\text{J/s}$

3. Calculate the net heat input rate:
The net heat input rate (power actually transferred to the weld) accounts for the arc efficiency (η):
$P_{\text{net}}=\eta \times P$
Substituting the efficiency value:
$P_{\text{net}}=0.7\times 5000\text{J/s}=3500\text{J/s}$

4. Calculate the net heat input per unit length:
The net heat input per unit length (H_s) is the ratio of the net power to the welding speed:
$H_s=\frac{P_{\text{net}}}{v}$
Substituting the values:
$H_s=\frac{3500\text{J/s}}{5\text{mm/s}}=700\text{J/mm}$

5. Convert the result to kJ/mm:
Since 1 kJ = 1000 J:
$H_s=\frac{700}{1000}=0.7\text{kJ/mm}$

Thus, the net heat input per unit length of the weld is 0.7 kJ/mm.