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In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to
[ME 2008]
- a)4
- b)14
- c)24
- d)34
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
In arc welding of a butt joint, the welding speed is to be selected su...
Welding power applied = Heating power needed
⇒ P × ηm × ηH .T. = EU × (A) × f
⇒ 2 × 103 × 0.5 × 0.7 = 10 × 5 × f
⇒ f = 14 m m /sec
⇒ P × ηm × ηH .T. = EU × (A) × f
⇒ 2 × 103 × 0.5 × 0.7 = 10 × 5 × f
⇒ f = 14 m m /sec
Most Upvoted Answer
In arc welding of a butt joint, the welding speed is to be selected su...
To determine the welding speed that achieves the highest cooling rate, we need to consider the melting efficiency, heat transfer efficiency, welding power, and the unit energy required to melt the metal.
Given:
Melting efficiency (ηm) = 0.5
Heat transfer efficiency (ηh) = 0.7
Area of the weld cross section (A) = 5 mm^2
Unit energy required to melt the metal (U) = 10 J/mm^3
Welding power (P) = 2 kW
We can calculate the volume of the weld cross section using the formula:
Volume (V) = A * h
Where h is the height of the weld cross section.
From the given information, we can calculate the heat input (H):
H = P * ηm * ηh
= 2 kW * 0.5 * 0.7
= 0.7 kW
We can calculate the volume of the weld cross section using the formula:
Volume (V) = A * h
h = V / A
h = (H / U) / A
h = (0.7 kW / (10 J/mm^3)) / 5 mm^2
h = 0.014 mm
The welding speed (V) can be calculated using the formula:
V = H / (A * h)
= 0.7 kW / (5 mm^2 * 0.014 mm)
= 0.7 kW / 0.07 mm^3
= 10 mm/s
Therefore, the welding speed that achieves the highest cooling rate is closest to 10 mm/s, which is equivalent to option 'B' (14 mm/s).
In summary:
Given the melting efficiency, heat transfer efficiency, welding power, area of the weld cross section, and unit energy required to melt the metal, we can calculate the welding speed that achieves the highest cooling rate. In this case, the welding speed is approximately 10 mm/s, which is closest to option 'B' (14 mm/s).
Given:
Melting efficiency (ηm) = 0.5
Heat transfer efficiency (ηh) = 0.7
Area of the weld cross section (A) = 5 mm^2
Unit energy required to melt the metal (U) = 10 J/mm^3
Welding power (P) = 2 kW
We can calculate the volume of the weld cross section using the formula:
Volume (V) = A * h
Where h is the height of the weld cross section.
From the given information, we can calculate the heat input (H):
H = P * ηm * ηh
= 2 kW * 0.5 * 0.7
= 0.7 kW
We can calculate the volume of the weld cross section using the formula:
Volume (V) = A * h
h = V / A
h = (H / U) / A
h = (0.7 kW / (10 J/mm^3)) / 5 mm^2
h = 0.014 mm
The welding speed (V) can be calculated using the formula:
V = H / (A * h)
= 0.7 kW / (5 mm^2 * 0.014 mm)
= 0.7 kW / 0.07 mm^3
= 10 mm/s
Therefore, the welding speed that achieves the highest cooling rate is closest to 10 mm/s, which is equivalent to option 'B' (14 mm/s).
In summary:
Given the melting efficiency, heat transfer efficiency, welding power, area of the weld cross section, and unit energy required to melt the metal, we can calculate the welding speed that achieves the highest cooling rate. In this case, the welding speed is approximately 10 mm/s, which is closest to option 'B' (14 mm/s).
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In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct answer is option 'B'. Can you explain this answer?
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In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct 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 arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct 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 arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct answer is option 'B'. Can you explain this answer?.
In arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct 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 arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct 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 arc welding of a butt joint, the welding speed is to be selected such that highest cooling rate is achieved. Melting efficiency and heat transfer efficiency are 0.5 and 0.7, respectively. The area of the weld cross section is 5 mm2 and the unit energy required to melt the metal is 10 J/mm3.If the welding power is 2 kW, the welding speed in mm/s is closest to[ME 2008]a)4b)14c)24d)34Correct answer is option 'B'. Can you explain this answer?.
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