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18 January, 00:02

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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  1. 18 January, 01:52
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    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.
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