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Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately
[ME 2005]
- a)1480 A
- b)3300 A
- c)4060 A
- d)9400 A
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) ...
ρ = 8000 kJ/m3
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µ Ω
I = ?
= 2.356 × 10–4kg.
Heat = 1400 × 8000 × 2.356 × 10–4
= 329.8 Joules
Heat = I2 × 200 × 10–6 × 0.1
I = 4060 Amp.
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µ Ω
I = ?
= 2.356 × 10–4kg.
Heat = 1400 × 8000 × 2.356 × 10–4
= 329.8 Joules
Heat = I2 × 200 × 10–6 × 0.1
I = 4060 Amp.
Most Upvoted Answer
Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) ...
ρ = 8000 kJ/m3
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µ Ω
I = ?
= 2.356 × 10–4kg.
Heat = 1400 × 8000 × 2.356 × 10–4
= 329.8 Joules
Heat = I2 × 200 × 10–6 × 0.1
I = 4060 Amp.
t = 0.1 sec dn = 5 mm
hn = 1.5 mm
LH = 1400 kJ/kg R = 200 µ Ω
I = ?
= 2.356 × 10–4kg.
Heat = 1400 × 8000 × 2.356 × 10–4
= 329.8 Joules
Heat = I2 × 200 × 10–6 × 0.1
I = 4060 Amp.
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Community Answer
Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) ...
The first step is to calculate the mass of the weld nugget. We can use the formula for the volume of a cylinder to find the volume of the weld nugget:
Volume = π * r^2 * h
Where:
r = radius of the weld nugget = 5 mm / 2 = 2.5 mm = 0.0025 m
h = height of the weld nugget = 1.5 mm = 0.0015 m
Volume = π * (0.0025 m)^2 * 0.0015 m
Volume = 1.76714587e-8 m^3
Next, we can calculate the mass of the weld nugget using its volume and the density of steel:
Mass = Volume * Density
Mass = 1.76714587e-8 m^3 * 8000 kg/m^3
Mass = 0.00014137167 kg
Now, we can calculate the energy required to melt this mass of steel using the latent heat of fusion:
Energy = Mass * Latent Heat of Fusion
Energy = 0.00014137167 kg * 1400 kJ/kg
Energy = 0.19792034 kJ
Finally, we can calculate the power used in the welding operation using the energy and the time:
Power = Energy / Time
Power = 0.19792034 kJ / 0.1 s
Power = 1.9792034 kJ/s
Now, we need to calculate the current using the power and the effective resistance:
Power = Current^2 * Resistance
1.9792034 kJ/s = Current^2 * 200 Ω
Solving for Current:
Current^2 = 1.9792034 kJ/s / 200 Ω
Current^2 = 9.896017e-3 A^2
Current = sqrt(9.896017e-3 A^2)
Current = 0.0994806 A
Therefore, the current used in the welding operation is approximately 0.0995 A.
Volume = π * r^2 * h
Where:
r = radius of the weld nugget = 5 mm / 2 = 2.5 mm = 0.0025 m
h = height of the weld nugget = 1.5 mm = 0.0015 m
Volume = π * (0.0025 m)^2 * 0.0015 m
Volume = 1.76714587e-8 m^3
Next, we can calculate the mass of the weld nugget using its volume and the density of steel:
Mass = Volume * Density
Mass = 1.76714587e-8 m^3 * 8000 kg/m^3
Mass = 0.00014137167 kg
Now, we can calculate the energy required to melt this mass of steel using the latent heat of fusion:
Energy = Mass * Latent Heat of Fusion
Energy = 0.00014137167 kg * 1400 kJ/kg
Energy = 0.19792034 kJ
Finally, we can calculate the power used in the welding operation using the energy and the time:
Power = Energy / Time
Power = 0.19792034 kJ / 0.1 s
Power = 1.9792034 kJ/s
Now, we need to calculate the current using the power and the effective resistance:
Power = Current^2 * Resistance
1.9792034 kJ/s = Current^2 * 200 Ω
Solving for Current:
Current^2 = 1.9792034 kJ/s / 200 Ω
Current^2 = 9.896017e-3 A^2
Current = sqrt(9.896017e-3 A^2)
Current = 0.0994806 A
Therefore, the current used in the welding operation is approximately 0.0995 A.
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Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer?
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Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer?.
Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer?.
Solutions for Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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ample number of questions to practice Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) is carried out successfully by passing a certain amount of current for 0.1 second through the electrodes. The resultant weld nugget formed is 5 mm in diameter and 1.5 mm thick. If the latent heat of fusion of steel is 1400 kJ/kg and the effective resistance in the welding operation is 200 μ-ohms, the current passing through the electrodes is approximately[ME 2005]a)1480 Ab)3300 Ac)4060 Ad)9400 ACorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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