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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/see. The net heat input (in J/mm) is
- 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 ...
Effective heat transfer per second (P) = η VI =0.85x25x300 J/s
Speed (c) = 8 mm/s
Net heat input 
Most Upvoted Answer
In an arc welding process, the voltage and current are 25 V and 300 A ...
Given:
Voltage (V) = 25 V
Current (I) = 300 A
Arc heat transfer efficiency (η) = 0.85
Welding speed (v) = 8 mm/sec
To find:
Net heat input (Q) in J/mm
Solution:
The net heat input (Q) can be calculated using the formula:
Q = V * I * η / v
Calculating the Net Heat Input:
Q = 25 V * 300 A * 0.85 / 8 mm/sec
= 6375 J/sec
Since we need to find the net heat input in J/mm, we need to convert the welding speed from mm/sec to mm.
Calculating the Net Heat Input in J/mm:
Welding speed (v) = 8 mm/sec
To convert it to mm, we can multiply it by 1000 (as there are 1000 mm in 1 meter):
v = 8 mm/sec * 1000 mm/m
= 8000 mm/min
Now we can calculate the net heat input (Q) in J/mm:
Q = 6375 J/sec / 8000 mm/min
= 0.796875 J/mm
Rounding off to the nearest whole number, we get:
Q = 0.797 J/mm
Hence, the correct answer is option 'B' - 797 J/mm.
Voltage (V) = 25 V
Current (I) = 300 A
Arc heat transfer efficiency (η) = 0.85
Welding speed (v) = 8 mm/sec
To find:
Net heat input (Q) in J/mm
Solution:
The net heat input (Q) can be calculated using the formula:
Q = V * I * η / v
Calculating the Net Heat Input:
Q = 25 V * 300 A * 0.85 / 8 mm/sec
= 6375 J/sec
Since we need to find the net heat input in J/mm, we need to convert the welding speed from mm/sec to mm.
Calculating the Net Heat Input in J/mm:
Welding speed (v) = 8 mm/sec
To convert it to mm, we can multiply it by 1000 (as there are 1000 mm in 1 meter):
v = 8 mm/sec * 1000 mm/m
= 8000 mm/min
Now we can calculate the net heat input (Q) in J/mm:
Q = 6375 J/sec / 8000 mm/min
= 0.796875 J/mm
Rounding off to the nearest whole number, we get:
Q = 0.797 J/mm
Hence, the correct answer is option 'B' - 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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)79700Correct answer is option 'B'. Can you explain this answer?
Question Description
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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)79700Correct answer is option 'B'. Can you explain this answer?, a detailed solution 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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)79700Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of 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/see. The net heat input (in J/mm) isa)64 b)797 c)1103 d)79700Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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