0.0732A=AD1/4d3/4V
v
ground. A depth of 0.003 in. is to be removed
from a cylindrical section 8 in. long and with a
3-in. diameter. If each part is to be ground in
not more than one minute, what is the approxi-
mate power requirement for the grinder? What
if the material is changed to a hard titanium
alloy?
or 0.226 in3. Therefore, the minimum metal re-
9.3), the power requirement is
P= (10 hp-min/in3)(0.226 in3/min) = 2.26 hp
9.73 A grinding operation is taking place with a 10-
in. grinding wheel at a spindle rotational speed
of 4000 rpm. The workpiece feed rate is 50
ft/min, and the depth of cut is 0.002 in. Con-
tact thermometers record an approximate max-
imum temperature of 1800◦F. If the workpiece
is steel, what is the temperature if the spindle
speed is increased to 5000 rpm? What if it is
increased to 10,000 rpm?
or
∆T=AD1/4d3/4V
5000 rpm, or a surface speed of 2080 ft/min,
the temperature rise will be
∆T=AD1/4d3/4V
v1/2
= (18,500)(10)1/4(0.002)3/42080
is
v1/2
ing point of steel (see Table 3.3 on p. 106).
Clearly, the temperature cannot increase above
the melting point of the workpiece material.
This indicates that the 10,000 rpm speed, com-
bined with the other process parameters, would
not be a realistic process parameter.
9.74 The regulating wheel of a centerless grinder is
rotating at a surface speed of 25 ft/min and is
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