Engineering, 23.04.2020 21:02 wjdelrieovie5s

In an external cylindrical grinding operation on a hardened steel tube whose outside radius = 42.5 mm, the grinding wheel diameter = 125 mm and wheel width = 20 mm. The work piece rotates at a surface speed of 25 m/min, the wheel rotates at 1800 rev/min, infeed (depth of cut) = 0.05 mm, and traverse feed = 0.50 mm/rev. There are 50 active grits/cm2 of wheel surface, and the operation is performed dry.

Determine the (a) volume rate of metal removed, (b) number of chips formed per unit time, and (c) average volume per chip. (d) If the tangential cutting force on the work = 45 N, compute the specific energy in this operation.

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Answer from: derrickgatson48

The volume rate of metal removed

V_R= 625mm^2/mm

The number of chips formed per unit time

N_C = 176,715 chips/min

The average volume per chip

V= 0.003537 mm^3 / chip

The the specific energy in this operation

$Q= 50.8938 N \cdot m / mm^3$

Explanation:

From the question we are told that

The outside radius is R = 42.5mm

The grinding wheel diameter is $D = 125 \ mm$

The grinding wheel width is w = 20 mm

The surface speed of the work piece rotation of $v = 25 m/min =25 * \frac{1000mm}{m} = 25 *10^3 mm/min$

The speed of rotation of the wheel is $N = 1800 \ rev/min$

The depth of cut is D = 0.05mm

The transverse feed is T_r = 0.50 mm/ rev

The number of grit /cm^2 of wheel surface is $C = 50\ grits/cm^2 = 50 * [\frac{10^{-2} cm }{mm^2} ] = 50 *10^{-2} grits/mm^2$

The cutting force is F = 45 N

The volume rate of the metal removed is mathematically represented as

V_R = v * D* T

Substitution value

V_R = 25*10^3 * 0.05 *0.5

V_R= 625mm^2/mm

The speed of the wheel is mathematically represented

$v = N \pi D$

$= 1800 * \pi * 125$

= 706,858mm/min

The number of chips formed per unit time is mathematically represented as

$N_C = 706858 * 0.5 * 50 *10^{-2}$

= 176,715 grits/min

N_C = 176,715 chips/min

The average volume per is mathematically represented as

$V = \frac{V_R}{N_C}$

$= \frac{625 }{176,715}$

V= 0.003537 mm^3 / chip

The specific energy is the operation mathematically

$Q = \frac{F v }{V_R}$

$= \frac{(45)[ (706,858.3) * \frac{1m}{1000mmm} ]}{625}$

$Q= 50.8938 N \cdot m / mm^3$

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