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A gun drilling operation is used to drill a 9/64-in diameter hole to a certain depth. It takes 4.5 minutes to perform the drilling operation using high-pressure fluid delivery of coolant to the drill point. The current spindle speed = 4000 rev/min, and feed = 0.0017 in/rev. In order to improve the surface finish in the hole, it has been decided to increase the speed by 20% and decrease the feed by 25%.
How long will it take to perform the operation at the new cutting conditions?

Respuesta :

Answer:

fr = Nf = 4000 rev/min (0.0017 in/rev) = 6.8 in/min Hole depth

d = 4.5 min (6.8 in/min) = 30.6 in New speed v = 4000(1 + 0.20) = 4800 rev/min

New feed f = 0.0017(1- 0.25) = 0.001275 in/min

New feed rate fr = 4800(0.001275) = 6.12 in/min

New drilling time Tm = 30.6/6.12 in/min = 5.0 min

Explanation:

The time taken to perform the operation at the new cutting conditions is 5.0 min.

  • The calculation is as follows:

fr = Nf = 4000 rev/min (0.0017 in/rev) = 6.8 in/min Hole depth

d = 4.5 min (6.8 in/min) = 30.6

In New speed v = 4000(1 + 0.20)

= 4800 rev/min

New feed f = 0.0017(1- 0.25)

= 0.001275 in/min

New feed rate fr = 4800(0.001275)

= 6.12 in/min

New drilling time Tm = 30.6÷6.12 in/min

= 5.0 min

Learn more: brainly.com/question/16911495

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