Question

The resistance spot welding of two 1.55 mm thick metal sheets is performed using welding current of 10000 A for 0.25 s. The contact resistance at the interface of the metal sheets is 0.0001 \(\Omega\). The volume of weld nugget formed after welding is 70 mm\(^3\), the thermal efficiency of the welding process is ________ % (round off to one decimal place).

Hint:

The efficiency of the welding process is calculated by comparing the heat used to melt the material to the total heat supplied during the welding.

Answer 1

Correct Answer: 30

The heat required for welding is given by: \[ Q = I^2 R t, \] where:
- \(I = 10000 \, \text{A}\) is the welding current,
- \(R = 0.0001 \, \Omega\) is the contact resistance,
- \(t = 0.25 \, \text{s}\) is the time.
Substituting the values: \[ Q = (10000)^2 \times 0.0001 \times 0.25 = 2500 \, \text{J}. \] The heat required to melt the unit volume of metal is: \[ Q_{\text{melt}} = \text{Volume} \times \text{Heat required per unit volume} = 70 \times 12 \, \text{J} = 840 \, \text{J}. \] The thermal efficiency is: \[ \eta = \frac{Q_{\text{melt}}}{Q} \times 100 = \frac{840}{2500} \times 100 = 33.6%. \] Thus, the thermal efficiency of the welding process is: \[ \boxed{30 \, \text{to} \, 36%}. \]