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Thread: Power Factor

  1. #16
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    So would I be correct in thinking that the inverter will be limited by current. So a 3500W inverter/genie with a 1750W load at a .50 PF would be maxed out(though the motor wouldn't be). Its not a highend gennie which "might" be expected to supply 3500W no matter what. The manual says nothing about PF

    RC, look at an AC waveform, https://i.ytimg.com/vi/98FgWHu2eI4/maxresdefault.jpg
    Think of the angels as time stamps.
    The rate of change between 0 and 45 is faster than between 45 and 90.

    Did that help?

  2. #17
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    Power Factor and VAr in Simple Terms

    When electric current (DC or AC) flows through a conductor the current develops a magnetic field about the conductor. The simple experiment of a magnetic compass next to a conductor shows a change of the compass when the current in the conductor is OFF or ON.

    An electric motor has a winding(s) connected to the supply line - be it battery or the "mains." The motor has a rotating shaft to which a load is connected. There is an "air gap" between the two. Energy is transferred across this air gap by a magnetic field - just like the "compass / conductor" experiment.

    The magnetic field is created by current flowing in the conductors (connected to the mains). To create the magnetic field current comes from the mains, and the generators on the network have to supply it.

    In AC systems the current reverses regularly. In a 50 Hertz (cycles per second) the reversal occurs 100 time a second. This means that the magnetic field builds up and collapses regularly. When the magnetic field is building up in the motor - which is really a magnet - it draws current from the mains, but as it collapses - in the next half cycle - current is returnd to the mains.

    As there is no such thing as a free lunch there are some "losses," so not all current (read energy) is returned to the mains. These losses are - the energy taken from the shaft of the motor (Horse Power), the heating losses of the current flowing in the conductors of the motor, the friction losses of the bearings, and the windage losses of the rotating part of the motor, and the cooling fan connected to the motor shaft.

    This "magnetising current" is expressed as VAr (Volt Amps reactive).

    The "Apparent Power" (VA) is the Volts of the mains supply multiplied by the Amps supplied - a simple calculation.

    The Real Power (expressed in Watts) is what is taken from the power system, and you pay for - Shaft mechanical Horse Power plus losses (heating, windage etc.)

    So what is Power Factor? It is a way of expressing this requirement for magnetising current which is needed by AC systems. It is a way of describing the ratios (relationship) between Real Power and Reactive Power - magnetising current.

    If you have a "toaster" there is no magetising current required - well almost none really, because after all a toaster is a conductor carrying current. See the "compass / conductor" experiment. Equally, the conductors supplying your house have a magnetic effect, and the conductors in the street too, and those big transmission lines....

    I have missed a few bits out to make it simpler - after all who in their right mind would want to learn about AC electrical theory!

    I hope this simple explanation makes some sense.
    Regards,
    Alan

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    Around the inner liner/outer skin trim that faces you.
    Nothing in the door.
    Some have heater elements in drains inside frige to ensure the drain hole doesnt freeze over.

  4. #19
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    Quote Originally Posted by BobL View Post
    Think of a moving pendulum - the velocity is greatest when the displacement from its original rest position is zero. And its velocity is zero when it's displacement is greatest.
    I am still not getting how you can have current flow at zero volts. However if the resistance stays the same and the voltage drops, the amps have to increase. So when the volts drop lower and lower, amps should be ramping higher and higher.

    Well that is what Ohm's law says.
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  5. #20
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    Quote Originally Posted by .RC. View Post
    I am still not getting how you can have current flow at zero volts. However if the resistance stays the same and the voltage drops, the amps have to increase. So when the volts drop lower and lower, amps should be ramping higher and higher.

    Well that is what Ohm's law says.
    Ohms law is quite correct for pure DC, but you have to take into account inductive and capacitive currents and voltages which are not in phase. This gives rise to current flow at zero voltage. This is where you will see the word "Impedance" !

    Simply put there is a phase difference which has to be taken into account !

    Ever had a tingle from a 1.5 volt battery and a coil ? Well that is caused by the out of phase difference between the DC current from the battery and the inductance of the coil resisting the current flow and building up a magnetic field in the coil. The collapse of this magnetic field induces a a high voltage across the coil, this generating the tingle that you felt.
    Best Regards:
    Baron J.

  6. #21
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    [QUOTE=.RC.;1975471]I am still not getting how you can have current flow at zero volts.

    An inductor is a coil of wire. it can be wrapped around nothing (air), or a core made of a magnetic material. If it was being used at Mains frequency it would be made of iron - usually with a fair bit of silicon in it. The iron composition is not important here.

    When you apply a voltage to the inductor (coil) a current will flow. Initially it will be small, but will grow with time. The growth in current will not be linear because as the current flows it creates a magnetic field around the conductor. This magnetic field will resist a "change" in the magnetic field!

    If it was a DC supply the current would reach a steady state. The magnetic field would be constant (not changing).

    If it was an AC supply the voltage of supply would start at zero, reach an instantaneous peak voltage, then begin to fall back to zero before reversing polarity, and doing the same again but in the opposite polarity.

    When the AC supply reaches an instantaneous peak, and commences to fall the magnetic field created by the current would resist the fall in supply voltage. The magnetic field would begin to collapse (creating a current to oppose the falling voltage) in an effort to maintain the voltage. In essence the magnetic field is returning energy stored in the magnetic field to the power system (the supply).

    So how can you get current flow when the (supply) voltage is zero?

    The collapsing magnetic field is acting in respose to the falling supply voltage (instantaneous voltage) in an effort to maintain the voltage.

    Because it is acting in response to the falling instantaneous voltage the magnetic field will continue to supply it's stored energy to the system until it is exhausted (no magnetic field left) even if the instantaneous (supply) voltage falls to zero, and even reverses! A reversed voltage is still a falling voltage as far as the inductor is concerned. It is the collapsing magnetic field that is supplying the current, not the "Mains"supply voltage. Hence you can get current flow at zero (supply) voltage! Also, zero volts is an instantaneous transient thing. It doesn't last long!

    An analogy or two may help.

    If you had two 12 volt car batteries in parallel. There terminal voltages will be very close, but as one battery's terminal volts fall, the other battery will supply current in an effort to maintain the terminal volts - until they are both flat. This is similar to the supply / inductor (coil) situation above, but doesn't help explain current flow at zero volts. Battery voltage cannot fall below zero in this analogy.

    The effect a flywheel has on a rotating shaft is analagous to a magnetic field. If the motor supplying the flywheel is switched off, the flywheel whould expend the energy "stored" in it to maintain the rotational speed of the shaft.

    Bits of AC theory have been deliberately left out, as it is the concept of storing energy to (and recovering it from) a magnetic field that is important.

  7. #22
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    Hi Alan, Guys,

    AC theory has some quite difficult concepts to put together in order to understand it !

    You did a good job there !
    Best Regards:
    Baron J.

  8. #23
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    Thanks BaronJ.

    I always wonder if what I am writing makes sense, and I am glad that what I wrote helps Forum members to understand a complicated concept!

    I have great admiration for those who came before us and worked out the theory, while we that follow all stuggle to understand what they said!

    Best Regards,
    Alan.

  9. #24
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    So the question remaining for me is.

    What happens to the unused energy, or the reactive power?

    One information source says the unused energy is returned to the source. This is why bad power factor is not helpful, it requires bigger electrical wires then would ordinarily be needed, as more current is flowing then needs to be.

    For generators does it lead to increased heating in the windings causing overloading?
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  10. #25
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    Quote Originally Posted by .RC. View Post
    So the question remaining for me is.

    What happens to the unused energy, or the reactive power?

    One information source says the unused energy is returned to the source. This is why bad power factor is not helpful, it requires bigger electrical wires then would ordinarily be needed, as more current is flowing then needs to be.

    For generators does it lead to increased heating in the windings causing overloading?
    try this; https://electrical-engineering-porta...system-healthy

  11. #26
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    Simple answer is yes. The current is "travelling" out to the inductor and returns every half of the cycle (more or less), so the conductors and switchgear and generator needs to be bigger for poor Power Factor loads. And to make it worse the customer with a poor Power Factor does not get a bill!

    This problem (not paying) has been dealt with by legislation - in W.A. A customer wanting to connect to the system must present a Power Factor (to the system) that is acceptable - PF = 0.8, but in the Goldfields of W.A. customers are required to present a PF = 0.9 because of electrical constraints in that part of the network. Penalties apply for non compliance.

    Heating of the conductors, switchgear and generators is the biggest problem. There are some other electrical issues - like generator under excitation, and what happens in faults on the system too but I won't mention them.
    Last edited by A J in WA; 6th Oct 2020 at 02:07 PM. Reason: Correction to sentence 2, and last paragraph.

  12. #27
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    Quote Originally Posted by A J in WA View Post
    This problem (not paying) has been dealt with by legislation - in W.A. A customer wanting to connect to the system must present a Power Factor (to the system) that is acceptable - PF = 0.8, but in the Goldfields of W.A. customers are required to present a PF = 0.9 because of electrical constraints in that part of the network. Penalties apply for non compliance.
    So would they need active PF correction? or would they add PF correction caps to fridges etc?

    My other fridge has a PF of 0.55 its about 3 years old, so the manufactures don't seem to have fixed things.

  13. #28
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    Quote Originally Posted by .RC. View Post
    So the question remaining for me is.

    What happens to the unused energy, or the reactive power?

    One information source says the unused energy is returned to the source. This is why bad power factor is not helpful, it requires bigger electrical wires then would ordinarily be needed, as more current is flowing then needs to be.

    For generators does it lead to increased heating in the windings causing overloading?
    I think he has got it ! Paraphrasing from Eliza DoLittle.

    In actuality there isn't any unused energy ! It all gets used somewhere and heating is the main result.
    Best Regards:
    Baron J.

  14. #29
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    [QUOTE=Stustoys;1975561]So would they need active PF correction? or would they add PF correction caps to fridges etc?

    Power Factor correction on appliances such as fridges is taken care of by the appliance manufacturer. The requirement for a PF = 0.8 or better is when it it working hardest. Hottest ambient temperature, compressor working at the highest head pressure etc. So when it is anything less, the PF will be worse. Unfortuneately their is no perfect solution. Besides this on a cool day when your fridge is running, your neighbour's fridge is off. It kind of balances out.

    PF correction is usually fixed at the substation that supplies your area. This global approach is better in most respects.

    Active PF correction is usually taken care on a bigger scale again (not just your home or area). There is electrical equipment to automatically correct PF - installed at substations. They are called Static VAr Compensators (SVC). They act in a manner similar to the way a voltage regulator - on older cars with DC generators works, but they regulate the PF, not voltage.

  15. #30
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    Quote Originally Posted by A J in WA View Post
    Power Factor correction on appliances such as fridges is taken care of by the appliance manufacturer. The requirement for a PF = 0.8 or better is when it it working hardest. Hottest ambient temperature, compressor working at the highest head pressure etc. So when it is anything less, the PF will be worse. Unfortuneately their is no perfect solution. Besides this on a cool day when your fridge is running, your neighbour's fridge is off. It kind of balances out.
    So then how are customers going to meet this spec? "A customer wanting to connect to the system must present a Power Factor (to the system) that is acceptable - PF = 0.8, but in the Goldfields of W.A. customers are required to present a PF = 0.9 ". Is it perhaps averaged PF? MY place isnt even close to .8 let alone .9. Or are you talking about industrial customers? They have always had to keep an eye on PF in VIC from what I understand.

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