Wheel diameter = 150 mm, and infeed = 0.06 mm in a surface grinding operation. Wheel speed = 1600 m/min, work speed = 0.30 m/s, and crossfeed = 5 mm. The number of active grits per area of wheel surface = 50 grits/cm2. Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work.

Question

Keaton Higgins

Engineering

15 January, 04:49

Wheel diameter = 150 mm, and infeed = 0.06 mm in a surface grinding operation. Wheel speed = 1600 m/min, work speed = 0.30 m/s, and crossfeed = 5 mm. The number of active grits per area of wheel surface = 50 grits/cm2. Determine (a) average length per chip, (b) metal removal rate, and (c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work.

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Answers (1)

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Todd Kennedy · Answer 1

Answer: a) 3mm

b) 5400mm^3/min

c) 4000000chips/min

Explanation:

Wheel diameter (D) = 150mm

Infeed (W) = 0.06mm

Wheel speed (V) = 1600m/min

Work speed (Vw) = 0.3m/s

Cross feed (d) = 5mm

Number of active grits per area of wheel surface = 50grits/cm^2

Average length per chip (Lc) = ?

Metal removal rate (Rmr) = ?

Number of chips formed per unit time (nc) = ?

a) Lc = (Dd) ^0.5

Lc = (150*0.06) ^0.5

Lc=3mm

b) Rmr=VwWd

Rmr = (0.3m/s) * (10^3mm/m) * (5mm) * (0.06mm)

Rmr=5400mm^3/min

c) nc=VWc

nc = (1600m/min) (10^3) (5mm) (50grits/cm^2) (10^-2)

nc=4,000,000chips/min.