Question

A copper bar of diameter 200 mm is turned with a feed rate of 0.25 mm/rev with depth of cut of 4 mm. Spindle speed is 160 rpm. The material removal rate (MRR) in mm³/s is :

1. 150
2. 167.55
3. 1500
4. 1675.5

Answer 1

Correct Answer - Option 4 : 1675.5

Concept:

The MRR of a turning process is given by -

MRR = f × d × V

where f = feed (mm/rev), d = depth of cut (mm) and V = cutting velocity (mm/sec).

Calculation:

Given:

Diameter (D) = 200 mm, f = 0.25 mm/rev, d = 4 mm, N = 160 rpm.

Cutting Velocity (V) can be calculated from -

\(V = \frac{\pi{DN}}{60}\)

\(V=\frac{\pi\;{×\;200\;× \;160}}{60}\)

∴ V = 1675.516 mm/sec.

Material removal rate is:

MRR = f × d × V

MRR = 0.25 × 4 × 1675.51

∴ MRR = 1675.516 mm³/sec.