Answer:
a
The volume rate of metal removed
[tex]V_R= 625mm^2/mm[/tex]
b
The number of chips formed per unit time
[tex]N_C = 176,715 chips/min[/tex]
c
The average volume per chip
[tex]V= 0.003537 mm^3 / chip[/tex]
d
The the specific energy in this operation
[tex]Q= 50.8938 N \cdot m / mm^3[/tex]
Explanation:
From the question we are told that
The outside radius is [tex]R = 42.5mm[/tex]
The grinding wheel diameter is [tex]D = 125 \ mm[/tex]
The grinding wheel width is [tex]w = 20 mm[/tex]
The surface speed of the work piece rotation of [tex]v = 25 m/min =25 * \frac{1000mm}{m} = 25 *10^3 mm/min[/tex]
The speed of rotation of the wheel is [tex]N = 1800 \ rev/min[/tex]
The depth of cut is [tex]D = 0.05mm[/tex]
The transverse feed is [tex]T_r = 0.50 mm/ rev[/tex]
The number of [tex]grit /cm^2[/tex] of wheel surface is [tex]C = 50\ grits/cm^2 = 50 * [\frac{10^{-2} cm }{mm^2} ] = 50 *10^{-2} grits/mm^2[/tex]
The cutting force is F = 45 N
The volume rate of the metal removed is mathematically represented as
[tex]V_R = v * D* T[/tex]
Substitution value
[tex]V_R = 25*10^3 * 0.05 *0.5[/tex]
[tex]V_R= 625mm^2/mm[/tex]
The speed of the wheel is mathematically represented
[tex]v = N \pi D[/tex]
[tex]= 1800 * \pi * 125[/tex]
[tex]= 706,858mm/min[/tex]
The number of chips formed per unit time is mathematically represented as
[tex]N_C = 706858 * 0.5 * 50 *10^{-2}[/tex]
[tex]= 176,715 grits/min[/tex]
[tex]N_C = 176,715 chips/min[/tex]
The average volume per is mathematically represented as
[tex]V = \frac{V_R}{N_C}[/tex]
[tex]= \frac{625 }{176,715}[/tex]
[tex]V= 0.003537 mm^3 / chip[/tex]
The specific energy is the operation mathematically
[tex]Q = \frac{F v }{V_R}[/tex]
[tex]= \frac{(45)[ (706,858.3) * \frac{1m}{1000mmm} ]}{625}[/tex]
[tex]Q= 50.8938 N \cdot m / mm^3[/tex]
