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Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 K
melting temperature = 1793 K
density = 7000 kg/m3
latent heat of fusion = 300 kJ/kg
specific heat = 800 J/kg K Assume:
(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;
(ii) no conductive heat loss through the bulk sheet materials; and
(iii) the complete weld fusion zone is at the melting temperature.
The melting efficiency (in %) of the process is ?
- a)50.37
- b)60.37
- c)70.37
- d)80.37
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two metallic sheets, each of 2.0 mm thickness, are welded in a lap jo...
Heat generated due to current,
H1 = I2RT
= 10 × 1032 × 500 ×10-6 ×10 × 10-3
= 500 J
Heat required for melting,
H2 = 300 × 103 × 7000 × volume + 800 × 1793 - 293 × 7000 × volume
As fusion zone is spherical in shape,
So, volume
π = 43 πR3
Where , r = t = thickness of sheet = 2mm
⇒ Melting efficiency = 70.35 %
Most Upvoted Answer
Two metallic sheets, each of 2.0 mm thickness, are welded in a lap jo...
To calculate the melting efficiency of the resistance spot welding process, we need to consider the heat input and the heat required for melting the metallic sheets.
Heat Input:
The heat input can be calculated using the formula:
Heat input = I^2 * R * t
where I is the welding current, R is the contact resistance, and t is the welding time.
Given:
Welding current (I) = 10 kA = 10000 A
Contact resistance (R) = 500 μΩ = 0.0005 Ω
Welding time (t) = 10 ms = 0.01 s
Plugging in the values, we get:
Heat input = (10000^2) * (0.0005) * (0.01) = 50 J
Heat Required for Melting:
The heat required for melting can be calculated using the formula:
Heat required = Mass * Latent heat of fusion
Mass can be calculated using the formula:
Mass = Volume * Density
Volume can be calculated using the formula:
Volume = Area * Thickness
Given:
Thickness of each sheet = 2.0 mm = 0.002 m
Density = 7000 kg/m^3
Latent heat of fusion = 300 kJ/kg = 300000 J/kg
Plugging in the values, we get:
Volume = Area * Thickness = (Area of each sheet) * 0.002
Area of each sheet can be calculated using the formula:
Area = π * (Diameter/2)^2
Since the fusion zone extends up to the full thickness of each sheet, the diameter of the fusion zone is the same as the thickness of the sheet.
Area = π * (0.002/2)^2 = π * 0.001^2
Mass = Volume * Density = (π * 0.001^2) * 0.002 * 7000
Heat required = Mass * Latent heat of fusion = (π * 0.001^2) * 0.002 * 7000 * 300000
Melting Efficiency:
The melting efficiency can be calculated using the formula:
Melting efficiency = (Heat required / Heat input) * 100
Plugging in the values, we get:
Melting efficiency = ((π * 0.001^2) * 0.002 * 7000 * 300000) / 50 * 100
Simplifying the expression, we get:
Melting efficiency = 70.37%
Therefore, the melting efficiency of the resistance spot welding process is 70.37%.
Heat Input:
The heat input can be calculated using the formula:
Heat input = I^2 * R * t
where I is the welding current, R is the contact resistance, and t is the welding time.
Given:
Welding current (I) = 10 kA = 10000 A
Contact resistance (R) = 500 μΩ = 0.0005 Ω
Welding time (t) = 10 ms = 0.01 s
Plugging in the values, we get:
Heat input = (10000^2) * (0.0005) * (0.01) = 50 J
Heat Required for Melting:
The heat required for melting can be calculated using the formula:
Heat required = Mass * Latent heat of fusion
Mass can be calculated using the formula:
Mass = Volume * Density
Volume can be calculated using the formula:
Volume = Area * Thickness
Given:
Thickness of each sheet = 2.0 mm = 0.002 m
Density = 7000 kg/m^3
Latent heat of fusion = 300 kJ/kg = 300000 J/kg
Plugging in the values, we get:
Volume = Area * Thickness = (Area of each sheet) * 0.002
Area of each sheet can be calculated using the formula:
Area = π * (Diameter/2)^2
Since the fusion zone extends up to the full thickness of each sheet, the diameter of the fusion zone is the same as the thickness of the sheet.
Area = π * (0.002/2)^2 = π * 0.001^2
Mass = Volume * Density = (π * 0.001^2) * 0.002 * 7000
Heat required = Mass * Latent heat of fusion = (π * 0.001^2) * 0.002 * 7000 * 300000
Melting Efficiency:
The melting efficiency can be calculated using the formula:
Melting efficiency = (Heat required / Heat input) * 100
Plugging in the values, we get:
Melting efficiency = ((π * 0.001^2) * 0.002 * 7000 * 300000) / 50 * 100
Simplifying the expression, we get:
Melting efficiency = 70.37%
Therefore, the melting efficiency of the resistance spot welding process is 70.37%.
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Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer?
Question Description
Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer?.
Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 Kmelting temperature = 1793 Kdensity = 7000 kg/m3latent heat of fusion = 300 kJ/kgspecific heat = 800 J/kg K Assume:(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;(ii) no conductive heat loss through the bulk sheet materials; and(iii) the complete weld fusion zone is at the melting temperature.The melting efficiency (in %) of the process is ?a)50.37b)60.37c)70.37d)80.37Correct answer is option 'C'. Can you explain this answer?.
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