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A 2-lips twist drill having a diameter of 4 [mm] and point angle 117 rotates at 691 [RPM] and feeds towards the work part at a rate of 0.38 [mm/rev], to drill a through hole of 34 depth. Machining allowance is 4 [mm]. Ignore length of chesil. MRR in [mm3/min] is then given by

Question

A 2-lips twist drill having a diameter of 4 [mm] and point angle 117 rotates at 691 [RPM] and feeds towards the work part at a rate of 0.38 [mm/rev], to drill a through hole of 34 depth. Machining allowance is 4 [mm]. Ignore length of chesil. MRR in [mm3/min] is then given by

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Solution

The Material Removal Rate (MRR) in drilling can be calculated using the formula:

MRR = π * D * f * N

where:

  • D is the diameter of the drill (in mm)
  • f is the feed rate (in mm/rev)
  • N is the rotational speed (in RPM)

Given:

  • D = 4 mm
  • f = 0.38 mm/rev
  • N = 691 RPM

Substituting these values into the formula, we get:

MRR = π * 4 * 0.38 * 691

Calculating this gives:

MRR = 3265.68 mm³/min

So, the Material Removal Rate is 3265.68 mm³/min.

This problem has been solved

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