Feeding solar power back into municipal grid: Issues and finger-pointing

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On 4/11/2011 6:16 PM, snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

I think you are pushing it....the brushes on a dc motor "guide" the dc to different windings.....it is still dc...
In an ac motor the windings are generally in parallel...all the ac is applied at one time....
I think I got that right....is a long time since I covered motor theory..<grin>...the ac is not "chopped up"....or guided anywhere....
You could say that all electric motors operate the same....as they all depend on magnetism (all generally used motors...there are some operate on static electricity, etc) have fun...sno
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No, it most definitely is *not*. Turn the motor to the next commutator step and you'll see that the current reverses in the winding.

That's why there are no brushes in AC motors? ;-)

...and you need a rotating magnetic field. That is, you need AC. ;-)
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On 4/11/2011 2:42 PM Evan spake thus:

... especially when it's pure BS. But that's our Harry.

Welll, since this is a.h.r, and since you're picking a nit, let me pick yours. I've installed several "higher-end" vent fans (Panasonic), all of which use AC induction motors which are very quiet. Which bath fans use the setup you described? (Besides which, why in the world would you need an "inverter" to run a DC motor from an AC supply? Perhaps you meant "rectifier"?)
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wrote:

Correct. David should refrain from any EE lectures. He hasn't got it in him.
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On Mon, 11 Apr 2011 14:21:20 -0500, The Daring Dufas

To "push" power back into the pipe the generator's phase leads the line's, so in a sense the generator's voltage is greater than the lines (as someone pointed out Kirchhoff didn't lie).
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On 11/04/2011 23:17, harry wrote:

So, you don't increase current by raising the voltage, but you increase current by having a higher potential.
Now, difference in potential is voltage?
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On 4/12/2011 9:47 AM g spake thus:

No, no, no: increasing the current doesn't increase the potential (that's voltage). It increases the *flow* of electricity (= current), at least the maximum possible current. But that's not the same thing as potential difference.
Example: Let's say you run your house off 12 volt batteries (just for illustration). The *potential* of your power circuit is 12 volts (assuming the batteries are fully charged, and they'll actually be closer to 13.something, but let's call it 12).
Now let's say you add some more stuff to your house and find that your lights are going dim because the battery can't provide enough *current* (= amps) to the load. So what you do is add another battery in parallel with the first one. This doubles the available current (= amps), but it does *nothing* to change the voltage; it remains at 12 volts (nominal, as explained above). This is true no matter how many batteries you add *in parallel* with each other. But each battery increases the *available* current (= amps) you can draw from your power source.
Notice that adding more batteries does not "push" more current through the system; it increases the amount of current that can be "pulled" (drawn from) the batteries.
Which is exactly the situation when you connect your photovoltaic system to "the grid". It increases the *available current* to the grid. It does not change the voltage of the grid; there's no need for it to be at a higher voltage than (but it needs to be at about the *same* voltage as) the grid.

Yes. Please refer to any good basic guide to electricity for more details.
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If you do a little research I think you will change your mind.
http://en.wikipedia.org/wiki/Grid_tie_inverter
At the bottom of that page you will find this link
http://www.solarpanelsplus.com/solar-inverters/How-Solar-Inverters-Work-With-Solar-Panels.pdf

More specificly I wrote, "forcing power back into the grid". Power is watts or KW. That's volts times amps. The current will only flow if there is a difference in voltage.

It takes very little voltage difference to flow a lot of current when the "load" is the grid.

Yes, I do. Now let's see if you can understand this.
Take two 12 volt car batteries with one discharged to 11 volts. Connect them in parallel and check the voltage. It will be some value between what they measured seperately. While they are connected like this current is flowing from the charged battery to the discharged battery. Power is being forced into it raising its state of charge. If the two batteries had been at exactly the same voltage there would have been no current flow. Now take 8 AAA batteries connected in series to give 12 volts. Disconnect the charged battery and connect the AAAs to the 11 volt battery. Measure the voltage. It will be for all practical purposes unchanged from 11 volts. The AAA cells are charging the bigger battery but they are so small compared to it that they seem insignificant. That is how your PV system looks to the grid.
The voltage on the grid can vary by up to + or - 10%. It is usually kept withing + or - 5%. Take a volt meter and check the voltage at your wall outlet. Check it several times during the day and you will find that it varies. Don't try to claim that it doesn't, check it and you will find that it does. As loads are put on the grid it drags the voltage down. The power company responds by generating more power to bring the voltage back up. As loads are taken off the voltage will climb, and it is brought back down by producing less power.
The point here is that there is no such thing as the grid not being able to accept the power you have produced. As long as you are connected you can always force your KW in.
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Bruce Richmond wrote:

I don't think the issue is whether or not you can force current into the grid via your 120/208 VAC service connection.
The question is:
a) does your power source need to overcome the instantaneous line voltage in order to achieve a flow of current (answer: yes, and to the extent that your power source has the capacity to do so, you raise the output voltage as high as you can, because if you don't - then you have excess capacity that is not going to make it out to the grid and hence you won't gain revenue for the entire potential of your generating system)
b) by raising the voltage on your local 120/208 grid, can your local stepdown transformer adjust it's own operation by sensing that higher voltage and reduce it's own output voltage in an attempt to regulate the system back down to the desired setpoint? (answer: I don't know - probably not. The neighborhood stepdown transformers probably weren't designed to compete with sources of current being connected to their distribution outputs).
c) So if the voltage on your local 120/208 grid is being raised slightly because of your PV system and it's desire to push as much current back into the grid as it can generate, then will this actually reduce the amount of current that the regional sub-station is sending to your local step-down transformer? (answer: the substation probably doesn't have a direct line to your local stepdown transformer, and any alterations it can make to it's output voltage is probably seen by many step-down transformers including yours that are all wired to the same circuit. So in reality it's doubtful that the regional substation would even sense that your PV system has raised the local grid voltage).
d) So your PV system is raising the local grid voltage, and you're probably pushing out 40 amps at 120 VAC or 20 amps at 240 VAC on a sunny summer day. So what is that extra juice doing? Well, it's flowing through the compressor motors of 10 to 20 of your neighbor's AC units - whether they need it or not. Because you've raised the local grid voltage slightly, that translates into a few extra watts (maybe 250 watts for each house that's fed from the same stepdown transformer). So all the fridge compressors and AC compressor motors, lights - all linear loads are going to blow away that extra line voltage as heat - instead of useful work.
Nuf said?
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Then you should look back and see that is precisely the issue being discussed here. You wrote, "But still - you can't push more electricity onto a network than the load is asking for (given that your invertors are functioning correctly I guess)."
David wrote, "That second statement is correct: you can't "push" electrons into the grid. But it doesn't matter *how* your inverters are working; it's a basic law of physics."
I have explained in simple terms (without getting into power factors, phase shifting, pulse width modulation, etc) the physics behind how you can push your power onto the grid whether it is asking for it or not.

You do not raise it up as high as you can, you raise it just enough to do the job. The ignition coil in your car raises the 12 volts of its battery to many thousands of volts to force a spark across the gap of its sparkplugs. You don't need thousands of volts to feed power into the grid.

Go back to the example I provided using two batteries. As I said, when they are connected in parallel the 12 volt battery charges the 11 volt battery and the voltage across them will measure somewhere between 11 and 12 volts. Connect a light to the batteries. You will measure a slight drop in the voltage but it will still be over 11 volts. That means the 11 volt battery is still being charged and all the power to light the light and charge the 11 volt battery is coming from the 12 volt battery. Connect more lights (load) to the batteries and you can drag the voltage down so that it is just over 11 volts. So long as the voltage across the two batteries is higher than the stand alone voltage of the 11 volt battery all the current going through the lights will be coming from the 12 volt battery. And it doesn't matter that the 12 volt battery has been dragged down to within a small fraction of a volt over the 11 volt battery, the lights see 11+ volts. Can you see now how the inverter can pump its power into the system? By having its voltage just a bit higher than the transformer, but well within the normal range of the line voltage, it can take over feeding the local water heaters, cooking stoves, air conditioners, lights, etc. No additional controlls are needed to reduce the current coming from the transformer. The voltage difference takes care of it.

As you saw above the current flow through the transformer will be reduced. The substation sees that as a reduced load and will behave the same way it would any other time the load goes away. No additional controls are needed.

Yes, you have said enough to make it clear how little you know.
When an electric motor is running it produces a back EMF that counters the flow of the current.
http://en.wikipedia.org/wiki/Back_EMF
That AC compressor requires the same power it did before. Keeping it simple that power is volts times amps equals watts. Divide the power required by the higher volts now provided and you will find fewer amps are required. The wasted heat is proportional to the square of the current. So raising the voltage means there will be less waste heat from the motor.
With resistance heaters the higher voltage flows more current, so you get more heat, which is what you wanted anyway.
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wrote:

But watts is *not* volts times amps, in an AC circuit. There is a power factor in there to worry about. In the capacitor example, watts dissipated is zero (or close to it) but VA might be rather high.

Correct. Ohms Law.
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On 4/12/2011 9:06 PM snipped-for-privacy@att.bizzzzzzzzzzzz spake thus:

That is *not* Ohm's Law. Where do you get that? Sheesh--you're trying to lecture *me* on this stuff???
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wrote:

E=IR, certainly *IS* Ohm's law. I and E are proportional. You can't increase I without increasing E. Get it? I suppose not.
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Wrong. You CAN increase I without increasing E. You have 3 variables in that formula, not just 2.
>Get it? I suppose not.
Apparently not.
Vaughn

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wrote:

Dumbass, it's a fixed circuit.

You've only proved that you're just as stupid as David.
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Names? Didn't your mother tell you how babyish that is?

You never defined the circuit, except perhaps in your own mind. I was responding to your statement about E=IR.

And you have proven yourself as a troll.
Bye Vaughn
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wrote:

She taught me to tell the truth. I did.

You're illiterate, then. Not surprising either.

No, you insist on demonstrating just how stupid you really are, though.
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On Apr 13, 9:07pm, " snipped-for-privacy@att.bizzzzzzzzzzzz"

I think he gets it directly from Ohms Law. V=IR.
Or, I = V/R
If V, the voltage is zero, then I, the current must be zero. Or, in other words, current will only flow if there is a difference in voltage.

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On 4/14/2011 5:48 AM snipped-for-privacy@optonline.net spake thus:

But that's not Ohm's Law (the statement "current will flow only if there is a difference in voltage"). Actually, that is a *tautology* (look it up). In other words, that's the very definition of current, which requires a potential difference (voltage > 0) to flow. Ohm's law didn't establish that, because it was already established by the time he came along.
You've correctly stated Ohm's Law, but that's not what it says. Strictly speaking, what Ohm determined was that the current flowing in a circuit is proportional to the voltage and inversely proportional to the resistance--but only for certain resistors. Specifically, his carefully calibrated metal resistances, at a certain temperature. So "Ohm's law"--what he determined experimentally and published--is only this:
I = E / R
and that only at fixed temperature. Turns out "Ohm's law" does *not* hold for a lot of things that look like resistances in the real world (for example, any humble tungsten filament fails to observe it).
But that's going waaaaay deeper into it than we need to here ...
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I never said Ohm established it. Only that from Ohm's law for the circuit under discussion you can directly verify that with zero potential you get zero current.

And if E is 0, what does this say I will be? Zero. Yes, Ohm isn;t the first guy to discover that current only flows from a potential difference. But his law clearly reflects it and shows it to be true.

Total nonsense. Just because a filament changes resistance with temperature does not mean Ohm's Law doesn't apply. Ohms law applies at every discrete temperature/resistance point the filament has. If we followed your logic almost nothing would behave according to Ohm's Law. Even the simplest resistor changes resistance slightly as current flows through it and it's temperature rises slightly. That means the resistance has changed, not that Ohm's Law no longer applies.

At least one step deeper than you should have gone.
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