Question

A spot welding operation performed on two pieces of steel yielded a nugget with a diameter of 5 mm and a thickness of 1 mm. The welding time was 0.1 s. The melting energy for the steel is 20 J/mm³. Assuming the heat conversion efficiency as 10%, the power required for performing the spot welding operation is \(\underline{\hspace{1cm}}\) kW (round off to two decimal places).

Hint:

The power required for a welding operation is determined by the energy to melt the material and the time required for the process, adjusted by the efficiency factor.

Answer 1

Correct Answer: 39 - 40

To calculate the power required for the spot welding operation, follow these steps:

1. **Volume of the Nugget:** The nugget is a cylindrical shape. Use the formula for the volume of a cylinder:
\( V = \pi r^2 h \)
Given, diameter = 5 mm, so radius \( r = 2.5 \) mm, and height \( h = 1 \) mm.
Substituting the values:
\( V = \pi \times (2.5)^2 \times 1 = \pi \times 6.25 \approx 19.63 \) mm³

2. **Total Melting Energy Required:** The energy required to melt the nugget is calculated by multiplying the volume by the melting energy per unit volume:
\( \text{Energy} = V \times \text{Melting energy} = 19.63 \times 20 = 392.6 \) J

3. **Actual Energy Consumption:** The heat conversion efficiency is given as 10%, so:
\( \text{Actual energy consumed} = \frac{392.6}{0.1} = 3926 \) J

4. **Calculate Power Required:** Power is energy divided by time. Given that the welding time is 0.1 seconds:
\( \text{Power} = \frac{3926}{0.1} = 39260 \) W = 39.26 kW

The power required for the spot welding operation is 39.26 kW.