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A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)
Correct answer is between '6.6,7.4'. Can you explain this answer?
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A 250 mm thick slab of a nickel alloy is subjected to cold rolling usi...
Initial thickness (hi) = 250 mm.
Roller diameter (D) = 450 mm
Draft or Reduction in thickness ∆h = D (1 - cosα)
Where α is angle of bite.
∆h = 450 (1 – cos 10°) = 6.83 mm
Most Upvoted Answer
A 250 mm thick slab of a nickel alloy is subjected to cold rolling usi...
°, determine the final thickness of the slab.
We can use the formula for the reduction in thickness during cold rolling:
Reduction in thickness = (1 - length of arc of contact / length of initial slab) x initial thickness
The length of arc of contact can be calculated using the formula:
Length of arc of contact = angle of bite x roll diameter
Substituting the given values, we get:
Length of arc of contact = 10° x 450 mm = 4,500 mm
The length of the initial slab can be assumed to be the same as the length of the arc of contact, since the slab is rolled in a continuous process. Therefore:
Length of initial slab = 4,500 mm
Substituting these values, we get:
Reduction in thickness = (1 - 4,500 mm / 4,750 mm) x 250 mm = 5.26 mm
Therefore, the final thickness of the slab after cold rolling is:
Final thickness = 250 mm - 5.26 mm = 244.74 mm (rounded to two decimal places)
We can use the formula for the reduction in thickness during cold rolling:
Reduction in thickness = (1 - length of arc of contact / length of initial slab) x initial thickness
The length of arc of contact can be calculated using the formula:
Length of arc of contact = angle of bite x roll diameter
Substituting the given values, we get:
Length of arc of contact = 10° x 450 mm = 4,500 mm
The length of the initial slab can be assumed to be the same as the length of the arc of contact, since the slab is rolled in a continuous process. Therefore:
Length of initial slab = 4,500 mm
Substituting these values, we get:
Reduction in thickness = (1 - 4,500 mm / 4,750 mm) x 250 mm = 5.26 mm
Therefore, the final thickness of the slab after cold rolling is:
Final thickness = 250 mm - 5.26 mm = 244.74 mm (rounded to two decimal places)
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A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer?
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A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. 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 A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. 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 A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer?.
A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. 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 A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. 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 A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer?.
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A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer?, a detailed solution for A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer? has been provided alongside types of A 250 mm thick slab of a nickel alloy is subjected to cold rolling using a roll of diameter 450 mm. If the angle of bite during rolling is 10°, the maximum reduction (in mm) during rolling is ___ (answer upto two decimal places)Correct answer is between '6.6,7.4'. Can you explain this answer? theory, EduRev gives you an
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