Re: 7-1/2HP 3-Ph. MTR on Single Phase
- From: John <amdinc@xxxxxxxxxxxxxxx>
- Date: Fri, 15 Jun 2007 21:33:34 -0400
Robert Swinney wrote:
Eric wrote:
"> Greetings Robert,
I'm a little confused about power factor correction and electrical
code. Users of big induction motors sometimes used another induction
motor to correct power factor. They may still do. These were sometimes
called "Rotary Capacitors". I thought that a VFD would do the same
thing. Which is one of the advantages of VFDs. So how does the poor
power factor of a lightly motor get reflected back to the utility? And
why is running a 7.5 HP motor with a 3 HP VFD a code violation? If the
VFD is overloaded it shuts off."
Eric,
Power factor cannot be "corrected", per se, with a vfd. Power factor in any
system is the ratio of kVAr (kilovolt-amps, reactive) to kVA (kilo-volt
amps). That means pf is the ratio of wattless power to real power; "Real
power", that is, power consumed resistively, is kilo-watts, (kW).
(Visualize a rt. triangle with kVAr as one side, kW as another side and kVA
as hypotenuse). In the triangle, say, the angle between kVA (hypotenuse)
and the kW side is 41 degrees. The triangle as described, represents an
electrical load with a phase angle of kW / kVA = 41 degrees. It is commonly
said that, "pf = the cosine of the phase angle." This means simply the
phase angle between current and voltage in any load is the cosine of the
angle between them.
It can be said the power factor of the load described is cos 41 deg.= 0.75.
Any load placed on an electrical system has power delivered to it from the
system. That is true whether the source is a simple DC battery or the
"grid". In essense, all loads are seen by the source. Saying pf is
reflected back to the source may be a bit misleading. As an AC load comes
on line, the line has to supply current to the resistive element of the load
(kW) as well as supply current to the wattless element (kVAr) of the load.
Remember the triangle. Total power consumed is the hypotenuse kVA. Power
always equals volts x amps. In an AC system: Power (VA) = the square root
of the sum of (W) and (VAr). Pythagorean theorem, pure and simple.
The "wattless" element, though not seen by a power meter, (watt-hour meter)
is still a a part of the load that is demanding its share of current from
the source (grid). That current is combined with the torque-causing current
in a motor resulting in extra current flowing to the motor. Magnetizing
current in a motor accounts for most of the (kVAr) extra current drain over
and above that current which results in torque. Magnetizing current in a
motor is largely responsible for the kVAr (wattless) current drain whether
the motor is delivering torque or not. This is why it is said that an
unloaded (idling) induction motor has a "low" power factor - the magnetizing
current is flowing even though torque is low. Even though the pf is low
compared to the pf of the motor under load, total current drain is still
quite low in an idling induction motor. Some texts say a properly designed
induction motor has magnetizing current that is aprox. 1/10 of full-load
current.
Bob Swinney
If you know about SWR in rf transmission lines... power factor is the
same thing. The transmission line from the source to the load is not
balanced, it usually has more inductance so you add caps to balance it.
The heavier the load the more the phase shift.
John
.
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