Re: Frequency and Voltage

Toller (Toller@xxxxxxxxx) said...
>
>
><trader4@xxxxxxxxxxxxx> wrote in message
>news:1129903959.258882.160720@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> Just to further clarify, regarding frequency, the OP is correct.
>> Frequency of an AC load does not affect power
>>
>Are you sure about that? Frequency affects induction, and if induction
>affects affects power, then frequency affects power.

Frequency does not effect power, as a purely inductive or capacitive
load does not draw any power (not counting any losses due to the tiny
resistance of the conductors themselves, which means that you can never
really have a purely inductive or capacitive load!).

Since the impedance will be affected by frequency, the total current
draw will be changed as a result of a frequency change.

The cosine of the angle between current and voltage is known as the
"power factor". In a purely resistive load, it is 1. In a purely inductive
or capacitive load, it is zero. Typical loads lie inbetween, but it is
best to keep it as close to 1 as possible.

Industrial customers of electric utilities must maintain as high a power
factor as possible. They usually have banks of capacitors that can be
automatically switched on and off as needed as they will have heavy
inductive loads from motors. If they maintain too low a power factor,
they will be charged for kVA-hours instead of kW-hours.

For instance, if you maintained a 0.5 power factor, then a 100 A load
at 120 V would be 100 x 120 x 0.5 = 6000 watts. An hour of this would
be 6 kWh, but it is 12 kVAh.

Why be charged for more energy than you actually used? Simply because you
are a burden on the system. Even though your load was only 6000 watts, you
drew double the current than was really necessary for that amount of
power. Therefore, the infrastructure needed to deliver that power had to
have twice the capacity than was really necessary.

Low power factor loads tend to be inductive, so a bank of capacitors
can cancel it out. Inductors cause current to lag behind the voltage, while
capacitors cause current to lead voltage. The two cancel each other out.
In the example above, since capacitors store and release current, they
supply the "extra" current needed for the inductive load, so the only
current draw on the supply is for the current actually needed to provide
power. If perfectly matched, the 6000 watt load would draw 50 amps from
the supply while the other 50 amps would be current between the inductive
load and the capacitors.

As an interesting side note: your home probably has a leading power factor
most of the time. The wiring in your home actually acts as a capacitor.
When I was a student, I worked on weekends as a watchman at a factory.
There was a power factor meter where the bank of capacitors was. When the
factory was shut down and the only load was lighting, the power factor
was usually 0.7-0.8 leading. As machines were started up and the inductance
of the load increased, the PF would rise to 1 then start dropping on the
lagging side. I believe a bank of capacitors would be switched in when it
dropped to 0.7. In addition to the kWh meter, there was a kVAh meter.

--
Calvin Henry-Cotnam
"Never ascribe to malice what can equally be explained by incompetence."
- Napoleon
-------------------------------------------------------------------------
NOTE: if replying by email, remove "remove." and ".invalid"

.

Relevant Pages

  • Re: Flyback transformer design confusion!!
    ... When applying a constant DC voltage across a resistive ... load, just *how* can I increase the power into that load, other than by ... Iout is to change the resistor. ...
    (sci.electronics.design)
  • Re: Flyback transformer design confusion!!
    ... would have to calculate the average current flowing through the load. ... and unless something is horribly wrong with your smps, the two should be roughly equal - losses mean the "inductor power" dE*F is somewhat larger than the load power. ... The power given by the expression above is that of the inductance ... its a power supply, so the voltage, by definition, is not supposed to change. ...
    (sci.electronics.design)
  • Re: Flyback transformer design confusion!!
    ... When applying a constant DC voltage across a resistive load, just *how* can I increase the power into that load, other than by changing the resistor? ... only way to change load power if the output voltage remains constant. ...
    (sci.electronics.design)
  • Re: RL in Class A SE, an alternative view was Re: Class-A RL determination in SE amps.
    ... "What determines output power of a tube?" ... try an amp built to the other outlook. ... The load value tolerance for SET isn't too bad in fact, ...
    (rec.audio.tubes)
  • Re: Flyback transformer design confusion!!
    ... When applying a constant DC voltage across a resistive ... load, just *how* can I increase the power into that load, other than by ... Iout is to change the resistor. ...
    (sci.electronics.design)