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.