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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 in 200 Ω , the current passing through the electrodes is approximately    
  • a)
    1480A  
  • b)
    3300 A  
  • c)
    4060 A  
  • d)
    9400 A 
Correct 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) ...
Heat required for melting = (mL) =
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Spot welding of two 1 mm thick sheets of steel (density = 8000 kg/m3) ...
Given:
Thickness of each sheet (t) = 1 mm
Density of steel (ρ) = 8000 kg/m3
Time taken for welding (t) = 0.1 s
Diameter of weld nugget (d) = 5 mm
Thickness of weld nugget (h) = 1.5 mm
Latent heat of fusion of steel (L) = 1400 kJ/kg
Effective resistance (R) = 200 Ω

We need to find the amount of current passed through the electrodes.

First, we can calculate the volume of the weld nugget using the formula for the volume of a cylinder:

V = πr2h

where r is the radius of the weld nugget (d/2) and h is the thickness of the weld nugget. Substituting the given values, we get:

V = π(2.5 mm)2(1.5 mm) = 29.32 mm3

Next, we can calculate the mass of the weld nugget using the formula:

m = ρV

Substituting the given values, we get:

m = 8000 kg/m3 × 29.32 mm3 × (10-6 m3/mm3) = 0.2346 kg

The amount of heat required to melt the weld nugget can be calculated using the formula:

Q = mL

where L is the latent heat of fusion of steel. Substituting the given values, we get:

Q = 0.2346 kg × 1400 kJ/kg = 328.44 kJ

The power required to melt the weld nugget can be calculated using the formula:

P = Q/t

where t is the time taken for welding. Substituting the given values, we get:

P = 328.44 kJ / 0.1 s = 3284.4 kW

The current required for welding can be calculated using the formula:

I = √(P/R)

Substituting the given values, we get:

I = √(3284.4 kW / 200 Ω) = 81.3 kA

Therefore, the amount of current passed through the electrodes is 81.3 kA.
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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 in 200 Ω , the current passing through the electrodes is approximately a)1480A b)3300 A c)4060 A d)9400 ACorrect answer is option 'C'. Can you explain this answer?
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