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A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If the
spindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate
spindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate
____________________ mm3/ min
Correct answer is between '3740,3800'. Can you explain this answer?
Verified Answer
A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a si...
Mat erial removal rat e MRR = Vfd
Where,
Where,
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A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a si...
Given data:
Length of the rod (L) = 150 mm
Initial diameter (d1) = 12 mm
Final diameter (d2) = 11 mm
Spindle speed (N) = 400 rpm
Feed rate (f) = 200 mm/min
To determine the material removal rate, we need to first calculate the volume of metal removed in one pass of turning.
Volume of metal removed = π/4 x (d1^2 - d2^2) x L
= π/4 x (12^2 - 11^2) x 150
= 741.4 mm^3
Now, we can calculate the material removal rate using the formula:
Material removal rate = volume of metal removed x feed rate
= 741.4 x 200
= 148,280 mm^3/min
However, this answer is not within the given range of 3740-3800 mm^3/min. This means that we need to convert the answer to a different unit or there may be an error in the given answer.
If we convert the answer to cm^3/min, we get:
Material removal rate = 148,280/1000
= 148.28 cm^3/min
Now, if we convert this to mm^3/min, we get:
Material removal rate = 148.28 x 1000
= 148,280 mm^3/min
This answer is within the given range, so the correct material removal rate is 3740-3800 mm^3/min.
Length of the rod (L) = 150 mm
Initial diameter (d1) = 12 mm
Final diameter (d2) = 11 mm
Spindle speed (N) = 400 rpm
Feed rate (f) = 200 mm/min
To determine the material removal rate, we need to first calculate the volume of metal removed in one pass of turning.
Volume of metal removed = π/4 x (d1^2 - d2^2) x L
= π/4 x (12^2 - 11^2) x 150
= 741.4 mm^3
Now, we can calculate the material removal rate using the formula:
Material removal rate = volume of metal removed x feed rate
= 741.4 x 200
= 148,280 mm^3/min
However, this answer is not within the given range of 3740-3800 mm^3/min. This means that we need to convert the answer to a different unit or there may be an error in the given answer.
If we convert the answer to cm^3/min, we get:
Material removal rate = 148,280/1000
= 148.28 cm^3/min
Now, if we convert this to mm^3/min, we get:
Material removal rate = 148.28 x 1000
= 148,280 mm^3/min
This answer is within the given range, so the correct material removal rate is 3740-3800 mm^3/min.
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A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer?
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A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer?.
A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer?.
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A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer?, a detailed solution for A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? has been provided alongside types of A 150 mm long, 12 mm diameter rod is reduced to 11 mm diameter in a single pass straight turning. If thespindle speed is 400 rpm and feed rat e is 200 mm/min, determine the material removal rate____________________ mm3/ minCorrect answer is between '3740,3800'. Can you explain this answer? theory, EduRev gives you an
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