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Thread: Power Factor

  1. #31
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    Correct -industrial customers. Depends on the customer's consumption. Big customers usually have a connection at a higher voltage - 22kV and higher is common, Special metering at the supply point records PF, as well as KW used.

    Generally, domestic customers are not big enough for a supply authority to bother with. The special metering required is somewhat expensive. Having said that, the newer domestic meters are capable. It all depends on money! Is it worth the effort if the return is small? And the domestic customers complain to their Member of Parliament, and the whole thing gets into the press and the electricity supplier's reputation is tarnished! The situation may change in time - who knows.

  2. #32
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    So where did we come down on a 3.4kW inverter supplying a 1.7kW load with a .50PF. As far inverter is concerned it's a max output?

    Come to think of it, even if the inverter isn't at max output, the wiring almost is. I'm sure running 28A in a 15A cable would be frowned upon.

  3. #33
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    Hi Stuart,

    Energy in equals energy out ! If there is no load then there are only losses to take into account. If the load is 1.5 Kw, then the energy in is only 1.5 Kw plus losses.
    Best Regards:
    Baron J.

  4. #34
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    Hi Baron,

    I get that the engine of the generator wouldn't have trouble supplying a 1.7kW load with a 0.50PF. As it only cares about the 1.7kW part of that.

    But the inverter while supplying the same 1.7kW load with a 0.50PF will also have a 1.7kW apparent power load. Right?

    So even if the inverter could supply 3.4kW with a bad PF the 15amp wiring is at its limit.
    So a bigger generator isn't going to fix the problem.
    Either fix the PF or get a bigger generator AND increase the wiring amp rating. correct?

  5. #35
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    I think you are saying .... The 3.4kW inverter is running at maximum output. It is supplying a 1.7kW 0.5PF load.

    If my assumption is correct, then 1.7kW is going into the load connected at the inverter (within the facility where the solar panels and inverter are located) - and the rest is losses plus an export to the "mains," and that total being 1.7kW. ie inverter output is being divided between the load (1.7kW), losses, and the "other" load - the Mains.

    As BaronJ said energy (out of the inverter) will equal the "load" plus the losses. Consider the "mains" to be a load, and it should make sense.

    As for the current rating of the conductors, 3.7kW at 240 volts, is a current of 14.1666 amps.

    As for the Power Factor, most domestic solar panel installations don't supply anything but kW, so the current component of the 0.5PF load will come from the "mains."

    Putting it mathematically, kW (load) = V x I(load) x PF (load),
    Therefore, I (load) = kW (load) divided by V x PF(load),
    so using your data, with kW = 1.7kW, PF = 0.5, and assuming V = 240 volts,
    I = 1700 watts divided by (240 x 0.5),
    ie. I = 1700 divided by 120.
    Therefore I = 14.1666 amps.

    Yes, the current relating to the Power Factor must be supplied - by some source.

    Now for the inverter:
    kW (inverter) = V x I x PF (inverter).
    Using your data, with kW = 3.4kW, V = 240 volts, and ASSUMING PF = 1.0 - totally resistive - because the inverter cannot supply anything but a resistive load.
    I = 3400 watts divided by (240 x 1.0),
    ie. I = 3400 divided by 240.
    Therefore I = 14.1666 amps.

    If you are struggling with this - don't feel alone! I know lots of electrical engineering type people who find difficulty grasping the concept.

    As our Treasurer (Paul Keating) once said when dealing with Government and a matter relating to the economy:
    ".... they all needed a Bex and a good lie down." Bex is hard to come by now!

    Think of it this way, then have a "good lie down":

    There is a load with two sources of supply - the inverter and the mains. One supplies the kW component, the other the Power Factor related component.

  6. #36
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    I thought power factor of an electric motor increases with the load put on it. Really bad under no load, less bad under load.
    Gold, the colour of choice for the discerning person.

  7. #37
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    "As for the current rating of the conductors, 3.7kW at 240 volts, is a current of 14.1666 amps."
    Just to be 100% sure, this is a typo? 7 instead of a 4, easy typo to make, if it's not my head will explode, you have been warned!

    Sorry gentlemen, I believe I may have inadvertently caused some confusion.
    I am talking about loads placed on a 3.5kW* inverter/generator(supposedly pure sine wave. *I thought it was 3.4kW).
    While I do have solar/inverter, it is connected to the grid even when the house is on the generator.
    I have 15A wiring from the gennie to the house. So I am limited to 15A of apparent power be that 15A = 3.6kW / (240 V x 1.0PF) or 15A = 1.8kW / (240 x 0.5PF)?
    Although the motor on the gennie only needs to produce the real power.


    To ask the same questions and a few more with the new information.
    Even if the inverter could supply 3.5kW* with a bad PF the 15amp wiring is at its limit?(pretty sure we have covered that)
    Because of the above a bigger generator isn't going to fix the problem.
    Either fix the PF and leave the wiring alone or get a bigger generator and increase the wiring amp rating?
    Would improving PF make life a little easier for the inverter as the current would go down?(though not necessarily viable)

    Further I assume the inverters output would be limited by max current, if for no other reason than to put the "biggest number" on the box for a given price you would use 1.0PF.

    And perhaps I should add, I don't currently have an issue, my place runs on the gennie just fine(with a few limits). I am just making sure I have my head around it, so that if my geenie does start shutting down while only outputting 2kW I would know why. I do have a friend who's gennie wont run his house, which made no sense to me given the loads he is telling me he has, until I thought about PF.


    Quote Originally Posted by .RC. View Post
    I thought power factor of an electric motor increases with the load put on it. Really bad under no load, less bad under load.
    That's what I thought and isn't that what AJ said in posts 4 and 29?

  8. #38
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    Good Morning Gentlemen,

    Power factor is a function of the load not the source !

    The generator is rated to supply that amount of energy, the load only takes the amount of energy that that it needs to function.

    In a resistive load the voltage and current is in phase and the power factor is 1.

    An inductive load has the voltage and current out of phase resulting in a power factor of less than 1.

    In large factory installations where there are lots of inductive loads (motors), there is often a large bank of capacitors that are used to compensate for the inductance by cancelling out the phase difference.

    The effect of this so call wattless current is heating ! There are other losses as well but this is the main one.
    Best Regards:
    Baron J.

  9. #39
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    Evening Baron,

    Quote Originally Posted by BaronJ View Post
    Power factor is a function of the load not the source !
    Agreed
    Quote Originally Posted by BaronJ View Post
    The generator is rated to supply that amount of energy,
    Agreed(at least for the motor. If the inverter is rated to supply that amount of energy regardless of PF aren't currents going to get out of hand?)
    Quote Originally Posted by BaronJ View Post
    the load only takes the amount of energy that that it needs to function.

    In a resistive load the voltage and current is in phase and the power factor is 1.

    An inductive load has the voltage and current out of phase resulting in a power factor of less than 1

    In large factory installations where there are lots of inductive loads (motors), there is often a large bank of capacitors that are used to compensate for the inductance by cancelling out the phase difference.

    The effect of this so call wattless current is heating ! There are other losses as well but this is the main one.
    Agreed.

    But if that wattless current results in a current flow that is double the rated current of the wiring, that would be bad no?

    Am I missing something?

  10. #40
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    Hi Stuart, Guys,

    Quote Originally Posted by Stustoys View Post
    Evening Baron,

    At least for the motor. If the inverter is rated to supply that amount of energy regardless of PF aren't currents going to get out of hand?)
    Generators, inverters etc, are designed to supply a given amount of power/energy. As long as the load doesn’t exceed this value there isn't a problem. However the generator/inverter cannot absorb the so called "Wattless" energy, so it gets reflected back towards the load. In doing so it gets used up in heating the conductors/wires adding to the losses already present in the system.


    But if that wattless current results in a current flow that is double the rated current of the wiring, that would be bad no?

    Am I missing something?
    Bad, yes ! But if you have reached this point the load is heading towards a short circuit.

    The amount of current in an inductive or capacitive circuit is a function of the frequency and the values of either or any combination of them. It is this fact that makes understanding AC theory so difficult. Most loads are inductive, some are just resistive, very few are capacitive.
    Best Regards:
    Baron J.

  11. #41
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    It looks like I did not properly understand Stuart's original post at the beginning of this thread.

    I missed the bit about "genie" meaning a rotating machine with a prime mover.

    Am I now correct - there is a generator (a rotating machine with a prime mover) plus solar panels with an inverter, and a load (part of which are motors). All this is connected to the "Mains." Have I got that right now?

    Regards,
    Alan.

  12. #42
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    Sorry if you guys feel you are hitting your head against a wall.

    Hi Baron,

    Quote Originally Posted by BaronJ View Post
    Bad, yes ! But if you have reached this point the load is heading towards a short circuit.
    But doesn't my maths show that the wiring is maxed out(15A) by a 1.8kW load with a 0.50PF?(If is matters, I'm using 0.50 just because its easier, 2.16kW would max the wiring out if you use the 0.60PF as per first post)


    Hi Alan,

    Quote Originally Posted by A J in WA View Post
    It looks like I did not properly understand Stuart's original post at the beginning of this thread.

    I missed the bit about "genie" meaning a rotating machine with a prime mover.
    Sorry about that.
    Then a generator with an inverter output added to the confusion.


    Quote Originally Posted by A J in WA View Post
    Am I now correct - there is a generator (a rotating machine with a prime mover) plus solar panels with an inverter, and a load (part of which are motors). All this is connected to the "Mains." Have I got that right now?
    Not exactly.

    There are all of the above but.
    The solar panels with their inverter are always connected to the mains.
    The load(part of which are motors) is the house.
    The house can be connected to the mains OR the generator(a motor/alternator with an inverter)

  13. #43
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    Quote Originally Posted by Stustoys View Post

    But if that wattless current results in a current flow that is double the rated current of the wiring, that would be bad no?

    Am I missing something?

    When you apply an a.c. voltage to a load that comprises Inductors and capacitors you get a somewhat curious result. The current in the inductance and capacitance will not be in phase with the voltage applied. When you do your calculations of the power at various instants through the cycle there will be times when the result shows that the watts value is negative this is because the voltage is going in a positive direction when the current is going negative or vice versa.

    So, we can have a load where at some instant the power dissipated or consumed is positive and at some instant it is negative. If we add these positive and negative values up over a full cycle you can calculate a wattage figure that approximates zero even though an apparently large amount of current has been flowing during the cycle.

    The explanation for how this happens is that inductors, motors, transformers and the like, store energy in their magnetic field and at some point release it again. Capacitors are similar but they store energy in an electric field between the plates.
    Best Regards:
    Baron J.

  14. #44
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    Quote Originally Posted by BaronJ View Post
    When you apply an a.c. voltage to a load that comprises Inductors and capacitors you get a somewhat curious result. The current in the inductance and capacitance will not be in phase with the voltage applied. When you do your calculations of the power at various instants through the cycle there will be times when the result shows that the watts value is negative this is because the voltage is going in a positive direction when the current is going negative or vice versa.

    So, we can have a load where at some instant the power dissipated or consumed is positive and at some instant it is negative. If we add these positive and negative values up over a full cycle you can calculate a wattage figure that approximates zero even though an apparently large amount of current has been flowing during the cycle.

    The explanation for how this happens is that inductors, motors, transformers and the like, store energy in their magnetic field and at some point release it again. Capacitors are similar but they store energy in an electric field between the plates.
    Yes I think I get that, at least on a superficial level, at least I think I do
    But my question is(just to be sure I have the maths correct), what current does the wiring see if a 1.8kW load with a 0.50PF is connected to it?

  15. #45
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    VA is the "apparent" power in an AC circuit. If you put a voltmeter across a load, and an amp meter in series with the load you would get a voltage and a current reading. Multiplying these together is the "VA" or "apparent power."
    VA (apparent power) = Volts x Amps.
    See post#3 for the relationship between these.
    Look at the second part: cos q = adjacent divided by hypotenuse.

    PF = 0.5 = cos q = kW divided by VA.

    Now for the question: With a load of 1.8kW (1800 watts), PF = 0.5 What is the current?
    Substituting the known elements in the equation:

    0.5 = 1800 divided by VA
    Cross multiplying, we get:
    VA = 1800 divided by 0.5
    VA = 3600

    Now we know the “apparent power” (VA), and we know the voltage (240 volts).
    Substituting these two known quantities into the equation:
    VA = Volts x Amps

    Cross multiplying, we get:
    Amps = VA divided by Volts
    Amps = 3600 divided by 240
    Amps = 15

    Current on a 1.8kW load at 240 volts and PF = 0.5 has a current of 15 amps.

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