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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 is _______ kW (round off to two decimal places).
Correct answer is '26.179'. Can you explain this answer?
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A spot welding operation performed on two pieces of steel yielded a nu...
Given : 5mm dn = , 1mm hn = , t = 0.15 sec
Melting energy = 20 J/mm3
Power = 26179.93 Watts
Power = 26.179 kW
Melting energy = 20 J/mm3
Power = 26179.93 Watts
Power = 26.179 kW
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A spot welding operation performed on two pieces of steel yielded a nu...
Given data:
Diameter of the nugget (d) = 5 mm
Thickness of the nugget (t) = 1 mm
Welding time (t_weld) = 0.1 s
Melting energy for the steel (E_melting) = 20 J/mm^3
Heat conversion efficiency (η) = 10%
Calculating the volume of the nugget:
The volume of the nugget can be calculated using the formula for the volume of a cylinder:
Volume (V) = π * (d/2)^2 * t
Substituting the given values:
V = π * (5/2)^2 * 1 = 19.63 mm^3
Calculating the energy required for melting the nugget:
The energy required for melting the nugget can be calculated by multiplying the volume of the nugget by the melting energy for the steel:
Energy (E) = V * E_melting
Substituting the given values:
E = 19.63 mm^3 * 20 J/mm^3 = 392.6 J
Calculating the power required for spot welding:
The power required for spot welding can be calculated by dividing the energy required by the welding time and multiplying it by the heat conversion efficiency:
Power (P) = (E / t_weld) * η
Substituting the given values:
P = (392.6 J / 0.1 s) * 0.1 = 3926 W
Converting the power to kilowatts:
P_kW = 3926 W / 1000 = 3.926 kW
Rounding off the answer to two decimal places:
P_kW = 3.93 kW
However, the correct answer given is 26.179 kW. It seems that there may be a mistake in the calculations or in the given data. Please double-check the calculations and data provided to arrive at the correct answer.
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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 is _______ kW (round off to two decimal places).Correct answer is '26.179'. Can you explain this answer?
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