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

[krw]

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.

[krw]

Oh, but according to Homeguy, that doesn't work because batteries are VOLTAGE sources.

Not on Homeguy's and Vaughn's planet. One of the batteries will be charging the other.

[David Nebenzahl]

On 4/14/2011 5:48 AM [e-mail] 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 ...

[Home Guy]

There's your problem right there.

It's usually not a good idea to connect batteries together in parallel, unless they are exactly of the same type, age, condition, etc. If you get a weak cell in one of the batteries it will turn into a load.

And what happens when they are not exactly at the same voltage before being connected together? How do you insure that you always get current flowing out of both of them?

The IEEE paper I posted earlier today shows exactly that - that PV systems raise local grid voltage and the utility company must compensate by reducing primary supply voltage to down-regulate the secondary voltage coming from the distribution transformer.

Take a 12 V car battery and wire it up in parallel with 8 AAA batteries connected in series. Then connect a load and tell me how much current the AAA batteries will supply vs their what their potential current supply could be if they were connected to their own isolated load.

If they are unequal in capacity, then yes that will eventually happen.

[krw]

What a dumbass!

They will share in proportion to their capacity. Electricity, water, nor shit flow uphill. ...though you have been pumping enough of the latter here.

Wrong!

[Home Guy]

" [e-mail]" unnecessarily full-quoted:

I'm not saying that it dissapears.

I'm saying that if your local grid is sitting at 120V and your panels come on and raise it to 121V, and if the utility company doesn't down-regulate their side to bring the local grid back to 120V, then the current that your panels are injecting is wasted. It's wasted because all the linear loads on the grid that are designed for 120V will not operate any better at 121 volts. Motors won't turn faster, lights won't really burn brighter. They will just give off a little more heat thanks to the extra current the panels are supplying to the grid.

But sure - electric heaters will get hotter. They're the only devices on the grid that are intended to convert electrical energy into heat.

So why not run a 120V motor with 240 volts then?

AC Motors are not simple loads like a resistor, but they will still "consume" power (V x I) as a function of their supply voltage.

All the power companies in the world are in the business of generating electricity in the thousands of volts and sending it out over high-tension wires. That's what they'd rather do if they weren't being hamstrung by crazy ideas and new rules / laws made by politicians about small-scale co-generation.

Look at the microFIT program in Ontario. When the rules were changed to allow local utilities to veto hookups based on "network capacity" or "substation insufficiency", they were only too happy to start swinging their veto left and right. They don't want to see this small-scale shit coming on-line if they have a choice.

[Jim Wilkins]

How to properly analyze the problem:

Look at Example 1.

jsw

[trader4]

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.

[trader4]

This is what he said:

"If they are unequal in capacity, then yes that (one will be charging the other) will eventually happen. "

Which of course is total nonsense and I hope you agree.

[trader4]

Thank you Jim. I was sitting here wishing I could draw a simple circuit into the newsgroup that models what we are talking about and shows what really happens. You've gone one better and found a perfect example from an independent and credible source. That circuit is EXACTLY the model for the dual battery example I brought up. Each battery is modeled as an ideal voltage source in series with a resistor. And each supplies part of the current flowing through the load. One never charges the other, nor does one need to have a higher voltage to "push" current.

I suspect we'll be hearing soon from Homeguy about how this isn't a valid way two batteries connected in parallel to a load can be modeled.

[Home Guy]

You're the dumbass!

Read the following:

============= SMUD wanted to investigate what effect reverse power flow from exporting PV systems would have on service and substation voltage regulation. Figure 1 shows the Home-to-Substation voltage difference (top) and solar irradiance (bottom) on a clear, cool day, Saturday, March 7, 2009, representative of a day with relatively low load and high local PV penetration. Penetration is defined as the amount of PV output divided by the load at a particular point in time.

At night the substation voltage ranged between 0.4 V to 0.7 V higher than the home voltage. This is representative of a typical circuit with voltage drops through the line and transformer impedances. During daylight hours, this reversed and the home voltage rose as high as 0.7 V, or 0.6%, greater than the substation voltage. ============

Did you read the last sentence dumbass?

The differential between substation and home voltage went from -.7 to

+.7 volts - a difference of almost 1.5 volts due to the effect of the PV panels injecting current into the grid.

If, according to you, there was no such phenomena of an increase in local grid voltage caused by PV panels, then there should be no basis for the point of this IEEE research paper I posted yesterday.

But engineers know that there will be a voltage increase because of these panels, and the excercise now is to figure out how much PV power can come on-line before the substation becomes unable to properly regulate it's output voltage levels.

============== Since the PV penetration levels were relatively low, there were no adverse effects on voltage regulation. It was possible to see the effects of the PV systems on the voltage at the individual homes and the distribution transformers. ==============

There will be more PV-equipped homes coming on-line in that project and it's not yet known if in total their operation will cause poorly controlled or unstable grid voltage.

Capacity is the "quantity of electrons" - which by itself tells you nothing of the potential (voltage).

And height is exactly equivalent to voltage potential.

So PV panels can't push current into the grid unless the invertors raise their voltage higher than the grid voltage. Just matching the grid voltage gets you to the point where your current flow is ZERO. Every millivolt you adjust your output voltage higher than the grid voltage means some small increment in current outflow from your panels. Since you can't store the energy coming from the panels via battery bank, then it's in your best interest to always maximize the amount of current you're injecting into the grid up to the full potential of the panel's output capacity. That means raising the output voltage as high as you need to so that every milliamp is squeezed out of them and onto the grid.

[Home Guy]

You dumbass.

Look more closely at the current flow in battery 1. It's NEGATIVE.

=================== The negative sign for I1 means that the direction of current flow initially chosen was wrong, but never the less still valid. In fact, the

20v battery is charging the 10v battery. ==================

[trader4]

Total nonsense and such a basic failure at elementary circuit basics that it discredits just about anything else you have to say. The voltage of the array IS the voltage of the grid at the point it's connected. How can it be anything else, unless you want to include the resistance of the wire used to make the connection, which is immaterial for the discussion.

Take a look at the dual voltage source circuit diagram that Jim Wilkins supplied a couple posts back. It's example #1.

It's a simple diagram of two ideal voltage sources with series resistors connected to a load. That serves as a basic model for two batteries or two generators or two PV arrays, etc connected to a load. The example gives the full equations for what would be two batteries of differing voltages and internal resistance connected in parallel to a load. Change the voltages so that they are equal, make them 20V. Solve those equations and you'll find that BOTH sources are supplying current to the load. The voltage on the "grid", ie across the load resistor is just one value. One source is not at a higher value to "push" current.

And I'll bet if you do the equations with the voltage sources at the same value, you'll find that twice as much current flows from the voltage source with the 10 ohm resistor as the one with the 20 ohm resistor.

[trader4]

Maybe it wasn't clear what I meant when I said:

"Just take two 12V batteries and connect them in parallel to a 12ohm resistor. Under the laws of physics the rest of us use the voltage would remain at

12 volts and BOTH batteries would be supplying part of the 1 AMP flowing through the resistor."

I didn't mean two batteries labled nominally as 12V that are actually at different voltages because one is fully charged, the other only partially charged. We were trying to discuss a simple a comparison as possible of using two power sources on a circuit.

So, do you agree that under the condition of two identical fully charged batteries at exactly 12V, connected in parallel to a load, the current will flow from both batteries through the load? I hope you do. As for Homeguy he apparently believes one has to be at a higher voltage to "push" current. I have yet to hear him explain how the batteries then decide which one it will be and how they will change their voltage to obtain the allegedly necessary "push" to get the current flowing.

[Home Guy]

I did, you dumbass #2.

Go and read my last post.

And note what happens when the voltage sources have unequal voltages.

And note that we are not talking about batteries here in the case of a municipal power grid and a PV system.

If that diagram shows reverse current flow because Battery 1 has a lower voltage than Battery 2 (and current I1 is negative), then at what point does current I1 become zero? What would the voltage of battery 1 have to be for current I1 to be zero?

[bud--]

Use of "per-unit" values for voltage, current, impedance, ... is common in the electric power field. It makes calculations easier, particularly as the system gets more extensive. One of the "per-unit" values is "% impedance".

A utility transformer is likely to be rated in "% impedance".

Daestrom has written about this. Looks like mII (or whoever) is familiar with it. That leaves you. If you knew as much as you think you know you would be familiar with % impedance.

You really need to go for some instruction. These things can't be worked out by lying on your bed and thinking about it.

[bud--]

(Which is why I didn't try.)

Duh...

A motor running at a constant RPM creates a fixed amount of mechanical power for a given load. RPM of induction motors is not very sensitive to voltage. The electrical power used is tied to the mechanical power consumed. Raising the voltage a little lowers the current a little.

[Jim Wilkins]

Considering the level of this discussion when the solutions are simple algebra, you aren't ready to handle the analytical geometry and differential equations of AC circuit analysis.

jsw

[g]

Your example, while it is correct, have little bearing on real life problems. The grid can in no circumstances be looked at as identical to a PV array voltage converted to AC voltage.

Two identical batteries have identical inner resistance, that is why your example works.

He is correct. Why? Because he wants to "push" current into the grid. With your 2 batteries at the exact same voltage, how do you get current flowing from the "home battery" to the "grid battery"?

He does not have to explain that, it is self explanatory when one understand that there are no perfect conductors with zero resistance in a power distribution system.

To get power into the grid from a local generated power the voltage has to be higher. HOW much higher depends on the impedance in the systems.

[Jim Wilkins]

This is a good example:

"Now that we've seen how series and parallel AC circuit analysis is not fundamentally different than DC circuit analysis,..."

Electrical engineers use j instead of i for the square root of -1 because i has long been the standard for Current (intensity). In this instance the imaginary number is an excellent tool to analyze real- world physical phenomena.

jsw