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In an arc welding process the melting efficiency and heat transfer efficiency are 45% and 75% respectively. The area of the weld crossection is 6mm2 and unit energy required to melt the material is 12J/mm3. If welding power is 2.5 kW.
The welding speed is _____mm/s
The welding speed is _____mm/s
(Important - Enter only the numerical value in the answer)
Correct answer is between '11,12'. Can you explain this answer?
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In an arc welding process themelting efficiency and heat transfereffic...
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Given data:
- Melting efficiency (ηm) = 45% = 0.45
- Heat transfer efficiency (ηh) = 75% = 0.75
- Area of the weld cross-section (A) = 6 mm²
- Unit energy required to melt the material (E) = 12 J/mm³
- Welding power (P) = 2.5 kW
Calculating energy input:
The energy input is the product of welding power and time. Since the welding speed is required, we need to find the time taken for welding.
The formula to calculate energy input is:
Energy input (Q) = Power (P) × Time (t)
Given that P = 2.5 kW, we can convert it to watts:
P = 2.5 kW = 2500 W
We can rearrange the formula to solve for time:
t = Q / P
Since we don't have the value of energy input (Q) directly, we can calculate it using the melting efficiency:
Energy input (Q) = ηm × E × A
Substituting the given values, we get:
Q = 0.45 × 12 J/mm³ × 6 mm²
Calculating welding speed:
Now that we have the energy input, we can calculate the time taken for welding:
t = Q / P
Substituting the values of Q and P, we get:
t = (0.45 × 12 J/mm³ × 6 mm²) / 2500 W
Since we want to find the welding speed in mm/s, we can rearrange the formula to solve for speed:
Welding speed = A / t
Substituting the values, we get:
Welding speed = 6 mm² / [(0.45 × 12 J/mm³ × 6 mm²) / 2500 W]
Simplifying the expression, we get:
Welding speed = 2500 W / (0.45 × 12 J/mm³)
Calculating the value, we get:
Welding speed ≈ 11.11 mm/s
Conclusion:
Therefore, the welding speed is approximately 11.11 mm/s.
- Melting efficiency (ηm) = 45% = 0.45
- Heat transfer efficiency (ηh) = 75% = 0.75
- Area of the weld cross-section (A) = 6 mm²
- Unit energy required to melt the material (E) = 12 J/mm³
- Welding power (P) = 2.5 kW
Calculating energy input:
The energy input is the product of welding power and time. Since the welding speed is required, we need to find the time taken for welding.
The formula to calculate energy input is:
Energy input (Q) = Power (P) × Time (t)
Given that P = 2.5 kW, we can convert it to watts:
P = 2.5 kW = 2500 W
We can rearrange the formula to solve for time:
t = Q / P
Since we don't have the value of energy input (Q) directly, we can calculate it using the melting efficiency:
Energy input (Q) = ηm × E × A
Substituting the given values, we get:
Q = 0.45 × 12 J/mm³ × 6 mm²
Calculating welding speed:
Now that we have the energy input, we can calculate the time taken for welding:
t = Q / P
Substituting the values of Q and P, we get:
t = (0.45 × 12 J/mm³ × 6 mm²) / 2500 W
Since we want to find the welding speed in mm/s, we can rearrange the formula to solve for speed:
Welding speed = A / t
Substituting the values, we get:
Welding speed = 6 mm² / [(0.45 × 12 J/mm³ × 6 mm²) / 2500 W]
Simplifying the expression, we get:
Welding speed = 2500 W / (0.45 × 12 J/mm³)
Calculating the value, we get:
Welding speed ≈ 11.11 mm/s
Conclusion:
Therefore, the welding speed is approximately 11.11 mm/s.
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In an arc welding process themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. Can you explain this answer?
Question Description
In an arc welding process themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. 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 themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. 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 themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. Can you explain this answer?.
In an arc welding process themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. 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 themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. 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 themelting efficiency and heat transferefficiency are 45% and 75%respectively. The area of the weldcrossection is 6mm2 and unit energyrequired to melt the material is 12J/mm3. If welding power is 2.5 kW.The welding speed is _____mm/s(Important - Enter only the numerical value in the answer)Correct answer is between '11,12'. Can you explain this answer?.
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