chocolatemalt wrote:


I understand the theory of PF, but the question is much simpler: If the
power company is feeding you 1000W on a straight V*A basis, and your
motor is seeing just 700W of useful work on a PF*V*A basis, there are
300W of energy that have "disappeared". I guess the question is so
simple that the answer is obvious: The energy is consumed within the
motor as non-useful heat.


In a motor. the current creates a magnetic field which varies in
magnitude and direction during the cycle. Some energy is stored in the
magnetic field for part of the cycle and sent back toward the utility on
another part of the cycle. The 300 watts of reactive power is from
energy being stored and returned 120 times per second and is not power
that disappears. Because of the properties an inductor the current is
shifted and the current flow sine wave is later than the voltage
sinewave and the reactive power is 90 degrees out of phase with the real
power.

A lightly loaded motor will have a lower PF than fully loaded motor
because the real power of the fully loaded motor will be higher and will
be a higher percentage of the total power. My memory of dim teachings is
that the reactive power doesn't change much with loading.

The higher current caused by the magnetic field interchange does cause
real I squared R losses in wire as detailed in other posts.

In a capacitor energy is stored in an electric field and is also stored
and released 120 times per cycle. But in a capacitor the current leads
the voltage. Power factor correction capacitors cancel the effects of
inductance.

Over correction - too much capacitance to balance the inductance can
cause resonance (if I remember right), which can cause the voltage to
rise as in the post by Beachcomber. This can be made worse by harmonics
which can be generated by VFDs, power supplies, dimmers.

bud--