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

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

diy

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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