Power Factor & kWH?
 

wrote in message
...
: chocolatemalt wrote:
:
: ... 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.
:
: No. If your motor used 700 W with a power factor of 1, 700/120
= 5.83 amps
: would flow in your wiring. If the wiring resistance were 0.1
ohms, it would
: dissipate 5.83^2x0.1 = 3.4 watts of heat, and you would pay for
703.4 watts.
:
: With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would
dissipate
: 8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only
penalty
: is the difference, 6.9-3.4 = 3.5 watts. The power company is
less happy
: because they lose more power in their wiring, but they usually
don't
: complain, in the case of houses.
:
: Nick
:
Yeah, I"ve since come across some info that indicates a " very
bad" PF would be in the order of single-digit percentages, as in
a few % or so. I was surprised it could get that low, but I
guess it can.

Besides, it's not jsut heat lost to the wires & system; it's the
fact that the voltage and current never reach peaks at the same
point in time and thus IE never represents true power without
including a PF. Apparently most of the loss is in the energy
required to build the flux fields, interference to it, and then
offset by the collapsing field, which is creating voltage by
trying to increase current, etc etc etc.. Everybody seems to be
missing the phase relationship between the voltage and current
thru a reactive load.

Power Factor & kWH?
 

chocolatemalt writes:

I guess the question is so
simple that the answer is obvious: The energy is consumed within the
motor as non-useful heat.

No.

Power Factor & kWH?
 

chocolatemalt writes:

Perhaps
with superconducting transmission lines someday (assuming it's
achievable), they will no longer care if your PF deviates since their
own energy expenditure will be equal to your real wattage, and reactance
will be irrelevant.

No, there is still an economic penalty to the current capability of the
generation plant being used up for no delivered power. Capital is tied up
without producing power for the customer.

This may be more costly than the line losses today.

It is not much different than the peak demand charges that even small
commerical customers now endure, and which ought to be paid by everyone.
The true cost of electric power is not just fuel and transimission, but the
capital investment for peak capacity.

Power Factor & kWH?
 

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--