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 _____ kW (round off to two decimal places).

Answer 1

Concept:

Melting efficiency is given by:

\({η _m} = \;\frac{{{H_m}}}{{{H_i ×η_h}}}\)

where, Hm = Heat required to melt the metal, Hi = Heat input required, η_m = melting efficiency, η_h = heat conversion efficiency.

Calculation: 

Given:

Heat required to melt (Hm) = 20 J/mm3

Heat transfer efficiency, (ηth) = 10 % = 0.1

Diameter d = 5 mm and thickness h = 1 mm, time =  0.1 s, ηm = 100 % = 1, η_h = 10 % = 0.1

Volume of nugget = \(V = \frac{{\pi}}{{{{4}}{{}}}} ×\ d^2 × h\)

\(V = \frac{{\pi}}{{{{ 4\;}}{{}}}} ×\ 5^2 × \ (1)\)

V = 19.63 mm3

Hm = 20 J/mm3

H_m = 20 × 19.63 = 392.7 J.

Power required to melt \(= \frac{{392.7}}{Time}= \frac{{392.7}}{{{{ 0.1\;}}{{}}}} =3927\;W\)

\({η _m} = \;\frac{{{H_m}}}{{{H_i ×η_h}}}\)

\({1} = \;\frac{{{3927}}}{{{H_i×0.1}}}\)

Hi = 39270 Watt = 39.27 kW