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2 years ago

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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 cros

sfeed 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
c. number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work.

1 answer:

gogolik [260]2 years ago

4 0

Answer:

a) the average length per chip is 3 mm

b) the metal removal rate MR is 90 mm³/sec

c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

Explanation:

Given that;

wheel diameter = 150 mm

infeed = 0.06 mm

wheel speed = 1600 m/min = 16000000 mm/s

work speed = 0.30 m/s = 300 mm/s

cross feed = 5 mm

active grits per area = 50 grits/cm²

a)

Average length per chip

Average chip length is given by

Lc = √fd

f is infeed and d is diameter of the wheel

so we substitute

Lc = √( 0.06 × 150

Lc = √ 9

Lc = 3 mm

Therefore the average length per chip is 3 mm

b)

metal removal rate MR is expressed as;

MR = fvc

v is work speed, c is cross feed and f is infeed

so we substitute

MR = 0.06 × 300 × 5

MR = 90 mm³/sec

Therefore the metal removal rate MR is 90 mm³/sec

c)

number of chips formed per unit time is expressed as

No = Vw × c × G

Vw is wheel speed and G is active grits per area

so we substitute

No = 1600000 × 5 × 50/10²

= 4,000,000 chips/min

Therefore number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

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