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During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.
- a)39.58
- b)45.25
- c)49.38
- d)54.98
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
During a steady gas metal arc welding with direct current electrode p...
Melting efficiency is the ratio of the heat used for melting the wire electrode to the total heat input. To determine the melting efficiency, we need to calculate the heat input and the heat required for melting the wire electrode.
1. Heat input:
The heat input can be calculated using the formula:
Heat input = welding current × welding voltage × welding speed × welding efficiency
Given:
Welding current = 150 A
Welding voltage = 30 V
Welding speed = 6 m/min
To calculate the welding efficiency, we need to know the arc voltage. Assuming an average arc voltage of 20 V, we can calculate the welding efficiency using the formula:
Welding efficiency = arc voltage / (arc voltage + welding voltage)
Arc voltage = welding voltage × welding efficiency
Arc voltage = 30 V × welding efficiency
Substituting the values, we get:
Arc voltage = 30 V × welding efficiency
Now, we can calculate the heat input:
Heat input = 150 A × 30 V × 6 m/min × welding efficiency
2. Heat required for melting the wire electrode:
The heat required for melting the wire electrode can be calculated using the formula:
Heat required = (wire mass flow rate × specific heat × (melting temperature - ambient temperature)) / melting efficiency
Given:
Wire diameter = 1.2 mm
Wire feed rate = 12 m/min
Density of wire electrode = 7000 kg/m3
Specific heat of wire electrode = 500 J/kg°C
Melting temperature of wire electrode = 1530°C
Ambient temperature = 30°C
Melting efficiency = 2/3
To calculate the wire mass flow rate, we need to determine the wire cross-sectional area:
Wire cross-sectional area = (π/4) × (wire diameter)2
Substituting the values, we get:
Wire cross-sectional area = (π/4) × (1.2 mm)2
Now, we can calculate the wire mass flow rate:
Wire mass flow rate = wire cross-sectional area × wire feed rate × density
Substituting the values, we get:
Wire mass flow rate = (π/4) × (1.2 mm)2 × 12 m/min × 7000 kg/m3
Finally, we can calculate the heat required for melting the wire electrode:
Heat required = (wire mass flow rate × specific heat × (melting temperature - ambient temperature)) / melting efficiency
Substituting the values, we get:
Heat required = (wire mass flow rate × specific heat × (1530°C - 30°C)) / (2/3)
Now, we can calculate the melting efficiency:
Melting efficiency = (heat required / heat input) × 100
Substituting the calculated values, we get the melting efficiency as 39.58%.
Therefore, the correct answer is option A) 39.58.
1. Heat input:
The heat input can be calculated using the formula:
Heat input = welding current × welding voltage × welding speed × welding efficiency
Given:
Welding current = 150 A
Welding voltage = 30 V
Welding speed = 6 m/min
To calculate the welding efficiency, we need to know the arc voltage. Assuming an average arc voltage of 20 V, we can calculate the welding efficiency using the formula:
Welding efficiency = arc voltage / (arc voltage + welding voltage)
Arc voltage = welding voltage × welding efficiency
Arc voltage = 30 V × welding efficiency
Substituting the values, we get:
Arc voltage = 30 V × welding efficiency
Now, we can calculate the heat input:
Heat input = 150 A × 30 V × 6 m/min × welding efficiency
2. Heat required for melting the wire electrode:
The heat required for melting the wire electrode can be calculated using the formula:
Heat required = (wire mass flow rate × specific heat × (melting temperature - ambient temperature)) / melting efficiency
Given:
Wire diameter = 1.2 mm
Wire feed rate = 12 m/min
Density of wire electrode = 7000 kg/m3
Specific heat of wire electrode = 500 J/kg°C
Melting temperature of wire electrode = 1530°C
Ambient temperature = 30°C
Melting efficiency = 2/3
To calculate the wire mass flow rate, we need to determine the wire cross-sectional area:
Wire cross-sectional area = (π/4) × (wire diameter)2
Substituting the values, we get:
Wire cross-sectional area = (π/4) × (1.2 mm)2
Now, we can calculate the wire mass flow rate:
Wire mass flow rate = wire cross-sectional area × wire feed rate × density
Substituting the values, we get:
Wire mass flow rate = (π/4) × (1.2 mm)2 × 12 m/min × 7000 kg/m3
Finally, we can calculate the heat required for melting the wire electrode:
Heat required = (wire mass flow rate × specific heat × (melting temperature - ambient temperature)) / melting efficiency
Substituting the values, we get:
Heat required = (wire mass flow rate × specific heat × (1530°C - 30°C)) / (2/3)
Now, we can calculate the melting efficiency:
Melting efficiency = (heat required / heat input) × 100
Substituting the calculated values, we get the melting efficiency as 39.58%.
Therefore, the correct answer is option A) 39.58.
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Community Answer
During a steady gas metal arc welding with direct current electrode p...
Heat generated = VI
= 150*30
= 4500 J/s
Heat available for melting = ⅔ * heat generated
= ⅔ * 4500
= 3000 J/s
Heat required for welding = ?/4 d2 * R * pCp ΔT
= ?/4 * 1.22 * 10-6 * 12/60 * 7000 * 500 (1530-30)
= 1187.52J/s
Melting efficiency = 1187.52 /3000 *100
= 39.58%
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During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer?
Question Description
During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. 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 During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. 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 During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer?.
During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. 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 During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. 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 During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer?.
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During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer?, a detailed solution for During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice During a steady gas metal arc welding with direct current electrode positive polarity; the welding current, voltage and weld speed are 150 A, 30 V and 6 m/min respectively. A metallic wire electrode of diameter 1.2 mm is being fed at a constant rate of 12 m/min. The density, specific heat and melting temperature of the wire electrode are 7000 kg/m3, 500J/kg and 1530°C respectively. The melting efficiency (in %) of the wire electrode is Assume the ambient temperature to be 30°C and neglect the latent heat of melting. Further consider that 2/3rd of the total electrical power is available for melting of the wire electrode.a)39.58b)45.25c)49.38d)54.98Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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