Negative power factor and PV systems

[mivey]

?????

Voltage and current signs?

If you mean polarity, only voltage has polarity. Current doesn't so you lost me.....

If you don't think so then just look at the current direction indicator on your power quality meter.

A residential power meter makes use of the fact that the current does indeed have a polarity. It reverses the polarity on one of the wires so that the current fluxes will be additive an uses it along with the L-L voltage to get the energy reading (it uses 1/2 windings on the current coils so you don't get twice the kWh).

Also, a load always uses up energy, if it were supplying energy it would not be a load, it would be a source.

I'm afraid it is not as simple as that.

 

[david luchini]

PV inverters make power where the current and voltage are in phase. The load consumes power with some angle between the current and voltage. Whatever is different between the PV inverter and the load -- the PoCo has to supply it. The inverter doesn't do anything to the load, the load is still the same. But it is consuming amps when the PV inverter isn't providing enough of them.

I could be wrong, but I don't think PV inverters "make" power where the current and voltage are in phase. I think they have ratings based on a unity power factor.

Does a 60kW generator with a 0.8 power factor (75kVA) "make" power where the current is 36.9 degrees out of phase with the voltage? If so, what happens when all the load on the generator is resistive?

I would think a 60kW inverter could supply 60kW of power at unity power factor, or 48kW of power at 0.8pf. Either case would supply power at the rated inverter output current.

 

[wiigelec]

I would imagine that if you are supplying all the kW and the utility is supplying all the kVARs the utility would want you to adjust your pf to minimize the amount of kVARs it must supply.

So if you are consuming inductive vars with a lagging-consuming pf angle of 10d (.94 lagging) I would imagine the utility would like you to compensate by running at a leading-supplying pf angle of 170d (-.94 leading).

Conversely if you are consuming capacitive vars with a leading-consuming pf angle of 10d (.94 leading) I would imagine the utility would like you to compensate by running at a lagging-supplying pf angle of -170d (.94 lagging).

...phasor diagrams would sure make this alot easier...

 

[tallgirl]

I could be wrong, but I don't think PV inverters "make" power where the current and voltage are in phase. I think they have ratings based on a unity power factor.

Does a 60kW generator with a 0.8 power factor (75kVA) "make" power where the current is 36.9 degrees out of phase with the voltage? If so, what happens when all the load on the generator is resistive?

I would think a 60kW inverter could supply 60kW of power at unity power factor, or 48kW of power at 0.8pf. Either case would supply power at the rated inverter output current.

There are two different situations, but in the one that is most commonly discussed when it comes to power factor are utility interactive inverters. For those inverters, the voltage and current is in phase, and that's part of their specification -- that they output unity power factor energy back to the grid.

Remember that an inverter is, for the most part, a software device. There aren't the large inductive masses or rotating parts as there are with generators.

 

[K8MHZ]

There are two different situations, but in the one that is most commonly discussed when it comes to power factor are utility interactive inverters. For those inverters, the voltage and current is in phase, and that's part of their specification -- that they output unity power factor energy back to the grid.

Remember that an inverter is, for the most part, a software device. There aren't the large inductive masses or rotating parts as there are with generators.

We have four inverters. Three are interactive and will be grid tied. The other is a stand alone and will not be grid tied.

 

[K8MHZ]

Well, here is what I am looking at:

Nothing in my text mentions anything about having to correct a poor power factor.

My instructor, an electrical engineer with hands on installation experience says he has never heard of having to correct or even look at power factor as a part of the installation.

-----------------

On negative current:

Current is measured in amperes and amperes are a certain amount of electrons passing a fixed point during a period of time. 1 Coulomb per second, IIRC. It doesn't matter if the electrons go left to right or right to left, 1 ampere will equal 1 Coulomb (which is a really big number) of electrons passing the point. Since there is no such thing as a negative Coulomb, there is no such thing as a negative amp.

Current entering a node may be indicated with a + sign, and current leaving a node may be indicated with a -, but that is no indication that current has polarity or can exist as a negative.

Kirchoff says that the current entering a node must equal the current leaving a node so that pretty much says that even though we mark the outflow as negative, it is still a positive number and must equal the current inflow.

 

[K8MHZ]

If you don't think so then just look at the current direction indicator on your power quality meter.

A residential power meter makes use of the fact that the current does indeed have a polarity. It reverses the polarity on one of the wires so that the current fluxes will be additive an uses it along with the L-L voltage to get the energy reading (it uses 1/2 windings on the current coils so you don't get twice the kWh).
I'm afraid it is not as simple as that.

Current can have a direction. I have no problem with that. Direction and polarity aren't the same.

I am intrigued by your description of how a power meter works. How does the meter switch polarity on one of the wires? A meter is one of the few things I haven't torn apart and I am curious as to what is inside. Are you talking about reversing the direction of the windings of one of the coils inside?

 

[K8MHZ]

Current has a direction -- from higher to lower voltage. Just because the voltage rises and falls (we're talkin' AC here ...) doesn't mean the current always goes in the same direction.

Right now I'm sending 868 watts back to the main service panel from the inverters (where it is promptly consumed by the A/C compressor ...) When current comes FROM that panel it has a "positive" sign, when it goes TO that panel it has a "negative" sign. Just like with DC power ...

(And yes, AC current has a sign -- don't let your Fluke be your teacher.)

I still think that direction and polarity are being used interchangeably when they shouldn't be.

Polarity in DC is a function of magnetism. It has nothing to do with current direction. Polarity is independent of current and exists without it's presence. IOW, polarity has nothing to do with current.

 

[K8MHZ]

PV inverters make power where the current and voltage are in phase. The load consumes power with some angle between the current and voltage. Whatever is different between the PV inverter and the load -- the PoCo has to supply it. The inverter doesn't do anything to the load, the load is still the same. But it is consuming amps when the PV inverter isn't providing enough of them.

I hate to ask this question because it's akin to asking "do you know how electricity works", but do you understand that there are two different components to AC power -- voltage and current?

For a non-reactive ("linear") load, current is a function of voltage only. If it's a 120 watt non-reactive load, on a 120 volt AC system, it draws 1 amp at 120 volts. 1 Amp x 120 volts = 120 volt-amps, and conveniently 120 watts. When the voltage is 60 volts, it draws 0.5 amps (I = E / R, R is constant). For a reactive load it may draw 0.5 amps at 120 volts and 1 amp at 60 volts when the voltage is either rising or falling (depending on whether it is inductive or capacitive reactance).

The situation with PV inverters is that they produce power so that the current and the voltage rise and fall together. There is no difference. In the above example, the difference between the 1 amp at 120 volts being produced, and the 0.5 amps at 120 volts being consumed (and so on for the 60 volt case), is the responsibility of the PoCo.

Of course I understand the components of power.

So the PV is producing one amp, a half an amp more than being consumed, and the poco also has to kick in a half an amp?

Tell me this: We have a load and it is reactive and we connect a grid tied PV system to it and it supplies enough energy that the meter stops. Now, if the poco is still having to make amps to fill the reactive void made by the inverter, are they measurable?

What happens with the meter stopped when you remove the meter and take the poco out of the picture? Does this reactive load now fail because the inverter can only supply power that is not reactive?

Perhaps you and mivey have stumbled across something not discussed in our text. I will gladly share the title and author with you. Something of that importance needs to be part of a college text on PV installation, don't you think?

Let me know how that works out for you. If the author concedes and the text is updated, I'll buy into your allegations. Until then, I fail to see how a PV system can only supply watts and the poco gets dinged for the VArs.

 

[mivey]

Well, here is what I am looking at:

Nothing in my text mentions anything about having to correct a poor power factor.

My instructor, an electrical engineer with hands on installation experience says he has never heard of having to correct or even look at power factor as a part of the installation.

I don't think voltage control has been required for solar inverters, even though manufacturers make voltage/var control inverters. Having volt/var control will allow us to connect more of these to the grid while keeping the voltage stable.

That does not mean you won't get billed for excess var load if you don't control the vars.

 

[K8MHZ]

I don't think voltage control has been required for solar inverters, even though manufacturers make voltage/var control inverters. Having volt/var control will allow us to connect more of these to the grid while keeping the voltage stable.

That does not mean you won't get billed for excess var load if you don't control the vars.

Show me where someone or some company has been billed for excess var loads as a result of a PV installation.

I know that industrial customers are penalized for poor power factors. What I am waiting for is any kind of proof that PV systems are detrimental to the power factor in question.

 

[SAC]

The situation with PV inverters is that they produce power so that the current and the voltage rise and fall together. There is no difference. In the above example, the difference between the 1 amp at 120 volts being produced, and the 0.5 amps at 120 volts being consumed (and so on for the 60 volt case), is the responsibility of the PoCo.

Can you provide a circuit diagram of such an inverter - I'd like to see how it accomplishes that behavior. For example, if one were to connect such an inverter to an inductor (without another source in parallel), what happens with the voltage and current waveforms? Will they be in phase, or will the physics of an inductor still hold?

 

[K8MHZ]

Can you provide a circuit diagram of such an inverter - I'd like to see how it accomplishes that behavior. For example, if one were to connect such an inverter to an inductor (without another source in parallel), what happens with the voltage and current waveforms? Will they be in phase, or will the physics of an inductor still hold?

Not only am I curious, I may be able to do some lab measurements.

Because of this discussion, I feel that it would be worthy of merit to take some actual measurements rather than to make assumptions. Since the class I am in is the first of it's kind at this college and the instructor is a hands on kind of guy, I think we should be flexible enough for such a test.

Now, if others have already done such tests, please post the results for comparison.

 

[wiigelec]

Can you provide a circuit diagram of such an inverter...

It would have a capacitive reactance such that Xc=Xl in series with the inductor...

 

[K8MHZ]

It would have a capacitive reactance such that Xc=Xl in series with the inductor...

Where is the capacitance of such an inverter listed? It's not on the label. I don't think it's in the manual either, but I'll check on Thursday.

 

[gar]

110524-1945 EDT

K8MHZ:

If I build a device that produces a nearly instantaneous current of a value proportional to a reference voltage with no phase shift of the current relative to the voltage, then that device produces a real power component to the system and no reactive component.

Now change the design of the device so that the current lags the voltage by 90 deg, then it only supplies the reactive component of current.

Because PV energy is so expensive one wants all components of the PV system to work as efficiently as possible. Thus, the inverter needs to be designed for unity power factor output. So long as what a PV inverter is connected is a moderately stiff voltage system, then it is possible for the inverter to operate as a current source and not cause a major distortion to the big system voltage.

The result is that the PV supplies some or all of the real power to the local load, and the grid (big system) supplies any required reactive power.

Relative to a comment in an earlier post I want to clarify the definition of power factor. PF = real power (watts i.e., work expended) divided by (Vrms * Irms). In a special case this also = Vrms * Irms * cos theta where both current and voltage are sine waves and theta is the phase angle between current and voltage.

Current does have a direction. Consider electrons in a vacuum tube. In electron terms the current flows from the emitting cathode to the anode (plate). These electrons really do move from one place to another. The defined direction is opposite when using the positive current flow convention. The positive convention is usually used in circuit analysis. Within a wire the current flow is more a flow of energy with any one electron moving only a short distance. But it is still useful to talk about a current flow, and it has direction.

I once had to solve a class problem where an electron beam with a certain velocity was injected into a box with an electric and magnetic field present. The problem was to adjust the two fields so as to have the electron beam pass thru a hole on the opposite side of the box. This hole was so located that it was not opposite the entry point. Note: an electron beam moving thru a uniform magnetic field perpendicular to the field follows a circular path. An electric field causes the electron to move in a linear path. a correct combination of the two fields will cause a spiral path.

When talking about a power meter or a watt-hour meter there is a direction of energy flow. This is determined by taking the instantaneous product of voltage and current. The KWH meter on my house would spin backwards if I connected a generator on the house side and powered the generator with a sufficiently large engine to produce more power than the power consumed in the house. This meter will run backwards. The direction is automatic because the direction of rotation is a function of the phase relationship of the current to voltage.

In M. B. Stout's book on Basic Electrical Measurements there is a discussion on wattmeters. This book might be in the Western Michigan University library, possibly at Sate, and definitely at Michigan. Another book that discusses wattmeters and watt-hour meters is "Electrical Circuits and Machinery", by Hehre and Harness. This is certainly at the U of M Engineering library. Most libraries in the state have borrowing capability.

.

 

[K8MHZ]

Because PV energy is so expensive one wants all components of the PV system to work as efficiently as possible. Thus, the inverter needs to be designed for unity power factor output. So long as what a PV inverter is connected is a moderately stiff voltage system, then it is possible for the inverter to operate as a current source and not cause a major distortion to the big system voltage.

The result is that the PV supplies some or all of the real power to the local load, and the grid (big system) supplies any required reactive power.

What I don't get is how the inverter can run inductive loads when the grid is not present if the grid is needed to supply reactive volt amps.

I totally respect your, miveys and tallgirls assertions. The only way to put this to rest is to perform actual testing.

If those tests confirm that a PV system adversely affects power factor, that needs to be stated in the text. If those tests do not confirm any adverse power factor effect, the text is accurate with proof to support it. If it needs to be stated in the text, rest assured I will be dropping the publisher a line.

 

[gar]

110524-2222 EDT

SAC:

If you connect an ideal current source to any non-zero impedance the result will be infinite voltage across the current source. We have no such real world device. Something will limit the voltage.

But we can connect a real world current source (at least an approximation) to a voltage source and the voltage will be defined by the voltage source and the current by the current source. For example a storage battery and a suitable current source. Use a small storage battery and a small regulated power supply with current limiting control for the experiment. The power supply has to be capable of a voltage greater than the battery. Should be an adjustable voltage supply. It will be more interesting if the power supply current limit function is also adjustable. You want a power supply without a fold-back current limiter. It should limit at a fixed current. You probably need to use a battery and power supply and not two power supplies because it is not usually possible to backfeed a regulated power supply.


K8MHZ:

It will be easy to run your experiment. Get two sets of a suitable wattmeter, RMS voltmeter, and RMS current meter. Connect one set to the output of the PV inverter. The other set to the input from the grid ( the wall outlet ). Use a micro-inverter because its size will be smaller. Put a single load on this system. Choose a small induction motor with no mechanical load, poor power factor. You will need to figure the sizing to make your experiment most informative. I will guess at a 1/3 HP motor.

Run experiments with and without the inverter connected. One result will be an indication of how effective the inverter is in producing a PF = 1. Additionally you will be able to show the change in grid PF with and without the inverter source.

.

 

[gar]

110524-2253 EDT

K8MHZ:

If the inverter is not designed to operate in a generated voltage mode, then there is a problem if it is operated without the grid voltage source. The inverter will shut down.

.

 

[wiigelec]

Where is the capacitance of such an inverter listed?

Same place as the switching KVAR banks that come with large motor installations...