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To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.
Correct answer is '0.942'. Can you explain this answer?
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To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm...
Given data:
- Hole diameter (d) = 20 mm
- Rotational speed (N) = 450 rpm
- Feed (f) = 0.2 mm
- Specific power (P) = 0.03 kW-min/cm3
- Motor efficiency (η) = 90%
Calculating material removal rate (MRR):
MRR = π * (d/2)^2 * f * N
= π * (20/2)^2 * 0.2 * 450
= π * 10^2 * 0.2 * 450
= 3141.59 mm^3/min
Calculating power required:
Power required (Preq) = MRR * P
= 3141.59 * 0.03
= 94.248 kW-min
Converting power to kW:
Preq = 94.248 kW-min * (1/60) hr/min
= 1.5708 kW-hr
Calculating power of electric motor:
Pmotor = Preq / η
= 1.5708 / 0.9
= 1.7453 kW
Conclusion:
The power of the electric motor required for the drilling machine is approximately 1.7453 kW, which rounds off to 0.942 kW.
- Hole diameter (d) = 20 mm
- Rotational speed (N) = 450 rpm
- Feed (f) = 0.2 mm
- Specific power (P) = 0.03 kW-min/cm3
- Motor efficiency (η) = 90%
Calculating material removal rate (MRR):
MRR = π * (d/2)^2 * f * N
= π * (20/2)^2 * 0.2 * 450
= π * 10^2 * 0.2 * 450
= 3141.59 mm^3/min
Calculating power required:
Power required (Preq) = MRR * P
= 3141.59 * 0.03
= 94.248 kW-min
Converting power to kW:
Preq = 94.248 kW-min * (1/60) hr/min
= 1.5708 kW-hr
Calculating power of electric motor:
Pmotor = Preq / η
= 1.5708 / 0.9
= 1.7453 kW
Conclusion:
The power of the electric motor required for the drilling machine is approximately 1.7453 kW, which rounds off to 0.942 kW.
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Community Answer
To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm...
Volume of metal removed during drilling operation = V
∴ V= π/4 X D2 X f X N = π/4 X (2)2 X 0.02 x 450
= 28.274 cm3 /min
Power required at the drill = V x S
= 28.274 x 0.03
= 0.848 kW
Power of electric motor = 0.848/η = 0.848/0.9 = 0.942 kW
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To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. Can you explain this answer?
Question Description
To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. 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 To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. 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 To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. Can you explain this answer?.
To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. 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 To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. 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 To drill a 20 mm diameter hole in CI work piece at 450 rpm and 0.2 mm feed. The specific power is 0.03 kW - min/ cm3 and motor efficiency is 90%. Then the power of electric motor (in kW) for a drilling machine is___________.Correct answer is '0.942'. Can you explain this answer?.
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