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In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is
[ME 2006]
- a)64
- b)797
- c)1103
- d)79700
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In an arc welding process, the voltage and current are 25 V and 300 A ...
V = 25 Volt, I = 300 A, η = 0.85
Weld speed = 8 mm/s.
Power Generated at arc = V.I
= 25 × 300 = 7500 J/S
Heat transferred = 0.85 × 7500
= 6375 J/S
So net Heat Input = 6375/8 = 797 J/mm.
Weld speed = 8 mm/s.
Power Generated at arc = V.I
= 25 × 300 = 7500 J/S
Heat transferred = 0.85 × 7500
= 6375 J/S
So net Heat Input = 6375/8 = 797 J/mm.
Most Upvoted Answer
In an arc welding process, the voltage and current are 25 V and 300 A ...
The given data:
- Voltage (V) = 25 V
- Current (I) = 300 A
- Arc heat transfer efficiency (η) = 0.85
- Welding speed (Vw) = 8 mm/s
Calculating the net heat input:
The net heat input (Q) can be calculated using the formula:
Q = V × I × η ÷ Vw
Substituting the given values:
Q = 25 V × 300 A × 0.85 ÷ 8 mm/s
Simplifying the expression:
Q = 6375 J/s ÷ 8 mm/s
Converting units:
Since the welding speed is given in mm/s, we need to convert the heat input to J/mm. To do this, we divide the heat input by the welding speed:
Q = 6375 J/s ÷ 8 mm/s = 796.875 J/mm
Rounding off the value to the nearest whole number, we get:
Q ≈ 797 J/mm
Therefore, the net heat input is approximately 797 J/mm.
Conclusion:
The net heat input in the given arc welding process is approximately 797 J/mm.
- Voltage (V) = 25 V
- Current (I) = 300 A
- Arc heat transfer efficiency (η) = 0.85
- Welding speed (Vw) = 8 mm/s
Calculating the net heat input:
The net heat input (Q) can be calculated using the formula:
Q = V × I × η ÷ Vw
Substituting the given values:
Q = 25 V × 300 A × 0.85 ÷ 8 mm/s
Simplifying the expression:
Q = 6375 J/s ÷ 8 mm/s
Converting units:
Since the welding speed is given in mm/s, we need to convert the heat input to J/mm. To do this, we divide the heat input by the welding speed:
Q = 6375 J/s ÷ 8 mm/s = 796.875 J/mm
Rounding off the value to the nearest whole number, we get:
Q ≈ 797 J/mm
Therefore, the net heat input is approximately 797 J/mm.
Conclusion:
The net heat input in the given arc welding process is approximately 797 J/mm.
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In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. Can you explain this answer?
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In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. 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 In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. 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 In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. Can you explain this answer?.
In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. 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 In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. 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 In an arc welding process, the voltage and current are 25 V and 300 A respectively. The arc heat transfer efficiency is 0.85 and welding speed is 8 mm/s. The net heat input (in J/mm) is[ME 2006]a)64b)797c)1103d)79700Correct answer is option 'B'. Can you explain this answer?.
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