Re: [CAD_CAM_EDM_DRO] 3 Phase/VFD question

In a message dated 6/1/2004 1:31:00 PM Central Standard Time,
ccq@... writes:
motor that was rated, if I remember correctly, at 1792 RPM. A VFD I've
looked at has a frequency rating of .5 to 360 Hz. If combined, what would be the
output range of RPMs
from the motor? Do I divide 1792 by 60, and then multiply that by 360, or
what?

Bill: I may be wrong, but I THINK this strange number, 1792, is the result
of taking the 1800 RPM "sync. speed" and deducting some (theoretical quantity?)
value "speed derating". You absolutely sure you have the number exactly
correct? Because 1782 = 99% of 1800 RPM. If you run a standard "1750 RPM" motor
on 60 Hz. with no load, or VERY little load, it will run at exactly 1800 RPM.
Same for a "3450 RPM" motor: 3600 RPM. ANYway, you would get the correct
result with your algorithm, but the CORRECT proceedure would be to divide the
VFD's output-f by 60 (or 50, if in almost anywhere in the rest of the world),
and then multiply the 1792 or 1750 or 3450 or whatever "name-plate RPM" by the
result. (I understand that 360 Hz. is generally the highest f you can get "in
the U.S." (60 Hz. mains), but in Europe with 50 Hz, you can get up to 400 Hz.
Has something to do with the "explosion speed" of motors or some-such
old-wives' tale spec. I dealt with a rotary f-converter I installed in England,
once, and I PROMISE, though "all went well, and worked, and I was paid", I don't
EVER wanna hafta mess with such again! Whew! TOO Much iron and copper for ME!
Jan Rowland


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