On 4/15/2011 8:53 PM snipped-for-privacy@att.bizzzzzzzzzzzz spake thus:
Whoa. Where do you get that (using Smitty's analogy)? Sure, you can't be going slower; that's self-evident. You say you have to be going faster. But you say nothing about going (approximately) *the same speed*, which is what Smitty's example was saying. (And is what ever article I've ever read about inverters, grid interties, etc., has said. (*None* of them say "the voltage of the contributing system has to be slightly higher than the grid in order to feed current into it". None of 'em.)
(Which, by the way, is eggs-ackley the same thing I've been saying here ...)
I was on your side of this issue before I went back and looked at the circuit model for two power sources driving a load that Jim Wilkins provided. I made a post today in reply to Mho where I went through the math. Sadly, Mho didn't even bother to go through the detailed circuit analsysis I provided. If you look for the post, I encourage you to review the analysis and the math and see what you conclude. The conclusion I came to is HomeGuy and KRW are right. To get power onto the grid, the additional source coming online has to be slightly higher that the voltage on the grid.
Look at it this way. Suppose the utility pole at my house ia at
120.000V. I hook up a power source, be it a generator, PV, whatever on the end of the line inside my house. I place exactly 120.000V on my end of that line. The wire from my power source to the pole has some small resistance, R. With 120.000V on one end of a resistor and 120.000V on the other end, by ohm's law, how much current will flow? Zero.
How do I get current to flow? At least one of three things must happen:
1 - I raise the voltage on my end of the wire slightly.
2 - The load increases on the rest of the system which in turn lowers the voltage at the utility pole outside my house
3 - Whatever else is driving the load reduced it's voltage slightly, which in turn lowers the voltage at the utility pole by house.
Those are the only choices to get a potential difference across the line connecting the source in my house to the grid and only then can current flow and power from my house make it onto the grid. And with option 1, my raising the voltage slightly on the house end, in turn must raise the voltage at the utility pole slightly as well. In other words, I've raised the voltage of the grid at the utility pole at my house. It's a very small amount, but it's real.
You're wasting your time with Mho. He obviously isn't interested in figuring out anything. I provided a detailed circuit analysis that shows bringing an additonal power source online will result in a slightly higher voltage at the grid and the load, provided everything else stays the same. The example was a FIXED load. I only pointed out that one alternative to raising the grid voltage is that the load could instead increase
On 4/17/2011 2:20 PM snipped-for-privacy@optonline.net spake thus:
[cut to the chase, i.e., trader4's example:]
All I can say is, good example, but wrong conclusion.
Let's change the example just a little. We'll have a large AC source (call it "the grid"), and a smaller AC source (the solar system's inverter), connected *in parallel*, and then some resistance (the total load of "the grid") after all that, completing the circuit.
In this example, "the grid" and the PV inverter are operating at
*exactly* the same voltage.
Let me argue this negatively, and see what you say to it: If you're saying that this will not work (i.e., that the PV inverter cannot contribute any power to the circuit because it isn't at a higher voltage), then *no* circuit where you have two power sources in parallel could ever work under the same circumstances.
No electric vehicle would ever work, because the battery cells in them are (pretty close to) exactly the same voltage, so how would each tiny cell (tiny in comparison to the total power of its siblings) ever be able to "push" electricity into the circuit?
That elementary electronics tutorial (Kirchoff's Law, etc.) explains everything you need to know here. And it doesn't require any higher voltage.
But what exactly is wrong with this example, which closely conforms to what we have been discussing? We have one end of a power wire connected to the PV array in my house. The other end is connected
100 feet away to the grid, ie the utility pole by my house. The sun is behind an eclipse, the array isn't generating any power. The voltage at the utility pole is 120.000V. The sun comes out. To put power on the grid, current must flow from my house, through that wire, to the utility pole. The wire has some small resistance, R. According to physics in my world, the only way to get current flowing in that wire is to have the end in my house at a HIGHER voltage than it is at the pole. It doesn't have to be a lot higher and it actually will be only slightly higher. But without that difference, tell me how could current ever flow?
The instant it does flow, I'm now pushing current out onto the grid. Assuming the rest of the grid stays the same, ie the load is fixed, the other power sources don't change, that means that the voltage at the pole now increase slightly as well. Net result is the voltage in my house is now say 120.1V, the voltage at the pole is now 120.05V and I have in fact raised the voltage of the grid.
Only if you assume the connection between the two is zero ohms, ie a perfect conductor. That would be akin to declaring the line connecting my house to the pole to be a perfect conductor and changing the model from what it is in the real world. The model that corresponds to what we have and also to your new proposed example is essentially the circuit example that Wilkins provided. The two resistors R1 and R2 represent the internal impedance of the two power sources. If you want to model the grid connecting them, then you could add two resistors, one after R1, the other after R2 to model the resistance of the grid between each of the power sources and the load. It doesn't change the analysis.
You can model a two cell battery with that circuit diagram as well. Let's leave V2 at 20V, making it a 20V cell with an internal resistance of
20 ohms modeled by R2. It's companion cell is V1 with internal resistance of 10 ohms. Do the math and you'll find that with V1 at 13.2 volts, no current flows through the V1 half of the circuit. All the load current comes from V2. Start increasing V1 and only then does current flow through V1 and through the load. The consequences of that are then that the voltage at the load increases, which in turn decrease the current flowing from V2.
In your real battery with cells connected in parallel, R1 and R2, the internal resistances, would be very close or equal. And V1 would be close in value to V2, both would be supplying about half the power. But it doesn't change the application of Kirchoff's Law or the conclusions.
Did you look at the detailed analysis I provided in an earlier post where I went through the analysis of that circuit example? Here it is repeated. Take a look at that circuit and go through it step by step. It is a model of two power sources driving a load.
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Voltage source V1 and R1 represent a simple battery, with R1 being the internal resistance of the battery. Same with V2 and R2. For our purposes a suitable model for a PV array and another power source on the other end of the distribution lines.
Let's leave V2 as is at 20V, supplying all the current to the load, with no current coming from V1. You then have a simple series circuit consisting of resistors R2 and R3 connected to voltage source V2. A current of 20/(40+20) =3D .33A is flowing, which is I2 in the drawing. The only way for no power to be flowing through the other half of the circuit encompassing V1 is if V1 is at the exact same potential as the load resistor. With .33A flowing through the load, R3, you have R3 at 13.2 V. That means V1 must be 13.2 volts. With V1 at 13.2 volts the voltage across R1 is zero and no current flows. So, everything is in balance. V1=3D13.2V, I1=3D0, the voltage on the load is 13.2 volts, and I2=3D .33A is flowing from V2 through R2, R3.
Now, if we want V1 to start supplying part of the power, what has to happen? V1 has to increase ABOVE 13.2 volts. And when it does, the voltage across the load resistor R3 will also increase. As that happens, current will start to flow now from V1 through the load resistor and at the same time the current from V2 will decrease slightly, as the potential drop across R2 is decreased slightly.
The net result of this is that the voltage across the load has increased. Current I1 is now flowing from V1, I2 from V2 is now slightly less and the combined currents of I1 and I2 which together are I3 has increased slightly.
The other ways to get V1 to supply power would be for either the load resistor R3 to decrease in value or for voltage source V2 to decrease.
If you want to more closely model the situation, we could add two more resistors to model the distribution line resistance. A resistor could be added after R1 and after R2. But if you go through the analysis, it doesn't change the basics of the above analysis. For source V1 to supply power, the voltage on it's portion of the distribution system and across the load must increase.
Again, physics doesn't care what you like and don't like. It is.
Well, I guess you could say that "currents" push against each other, but it requires a difference in voltage to have a current. Think of the intersection of two rivers.
You assume wire has zero resistance. Bad assumption.
That is "said" has little bearing on physics.
The biggest flaw is that resistance is not zero and you take what people "say" too literally. Analogies are always flawed. That's why they're called "analogies". ;-)
It makes more sense if you think of the inverter as forcing a constant CURRENT and let the voltages be whatever the source (wire etc) resistance makes them at that current.
The grid may or may not act like an infinite sink. The continual load variations will probably swamp out any voltage measurement you might make, so it's reasonable to consider it an infinite sink unless you have a very large inverter. The GTI wants to dump all the current from the array onto the line and will adapt itself to the line voltage, whatever it may be.
If you connect a PV panel to a 12V battery the panel will source as much current as the sunlight produces, at the voltage of the battery even if the panel's open circuit voltage is above 20V. The battery voltage will rise a little because of the IR drop in its internal resistance.
Picture two water tanks connected with a loop of pipe underneath. Put a ball in the connecting pipe. The two tanks are connected in parallel. The ball in the pipe doesn't go anywhere because it has equal pressure on both sides. Now pour some water into one of the tanks to raise its level (voltage). There will now be more pressure applied to the ball by that tank and the water will flow into the other tank, pushing the ball in that direction.
Attach a single drain pipe to the middle of the connector pipe above. If the tanks start at the same level and the resistance to flow in the pipes is the same then water will flow out of both pipes at the same rate. If one tank is higher than the other it is pushing harder and more water will flow out of it than the lower tank.
The power company has many stations and wants them to all contribute their share. We OTOH have spent big $$$ on our PV system and if it produces 5KW we want that whole 5KW to go to some load so that we get paid for it. We are not interested in playing nice and sharing a load if it means we don't get to contribute the full 5KW. So we raise the voltage just enough to flow the current we need to. That will result in less current flowing from the grid because the load is only going to accept so much flow and the PV system is taking more than its share.
Yes there will be flow at 119, 120 or 121, but from what source? Like I said, we aren't interested in playing nice. We are entitled to pump the power we produced into the line and get paid for it. If the power company has to reduce their production a bit to keep the voltage from climbing too high so be it.
Let me say it again, perhaps you'll catch on. Physics doesn't care what you like. It is what it is.
Because, like electricity, water always flows "down hill" - high to low.
OK.
OK.
Voltage is dropped across a resistance. Not all points in the grid go up because your solar cells output more voltage because there is non-zero resistance between all points. If you're generating electricity, your house will be at a *higher* voltage than the pole. If the resistance of the wire were zero this couldn't happen.
But... but... but... EVERY car insurance company is cheaper than its competitors and every alkaline cell is the longest lasting... LOL
Seriously, the power stations are not higher than each other, but every one is higher than the grid, imagine three power stations, or 3 generators in the same station all putting out exactly 48KV, the substation/transformer they are connected to is getting slightly less, say 47,990 volts. If they then need to bring a fourth generator online they use a syncroscope to adjust the prime mover to sync up the frequency and phase, adjust the field to match the voltage IE 47,990, close the breakers, and slowly increase the field and prime mover so that the load is shared and now all 4 are at 48KV.
It *sounds*--and I'm sure you'll correct me if I'm wrong--as if you're agreeing with me, and with Smitty, and others when we say that it is
*not* required that the photovoltaic inverter supply a higher voltage in order to transfer current to the grid. (I take this from the last sentence in the next-to-last paragraph, where you say " ... will adapt itself to the line voltage, whatever it may be".)
The arguments against this, with all the pseudo-science being thrown around (most of it by the ones who are also slinging insults) are getting quite tiresome here.
No, the voltage will STILL be higher if you're supplying current to the grid. Wires have resistance. Current sources don't go against physics and really are the same thing as voltage sources (Norton/Thevenin duality). Physics doesn't lie.
You're throwing around the pseudo-science. If you don't like the treatment, you can easily leave.
Pigs-arse you were "teasing". You were sucked in by the GymmyBob troll.. another of a few hundred of his 'victims'. Deservedly so on this occasion.. you prove a clueless dolt.
Get yourself educated on "how to" in reading news, rube. netscape.public.mozilla.general
Then go check out your 'victor' :-/ "Gymy Bob" /
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"John P. Bengi" /
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"Solar Flare" /
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"Pizzza Girl" /
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"Janice" /
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"Joesepi" /
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.. mind how you go. Don't want you bleeding all over the place:-D
You can't have two voltage sources EXACTLY in parallel, its like dividing by 0. There is always some small resistance between voltage sources or the sources themselves have resistance.
The sources have to be higher then the grid, not higher then eaach other. Lets say the grid is some point in the middle of a square and it is at 120.0. Lets say there are 4 sources feeding that center point from the 4 corners. Each feed has a small resistance between it and the "grid point". Each one can be at 120.1 for example and power will flow from each to the grid point. If feed point one is a big nuclear plant it might be at 120.3 and your small solar plant at point
2 might be at 120.01. But each will feed power TO the grid point load. Each plant adjusts itself to the right slight increse in voltage to feed the amount of power it has avaialbe.
(This is an oversimplification of what really happens. What really happens also has a lot to do with frequency and phase.)
If the load is at the grid point and is at 120.0, then any source under 120.0 will pull power and not source it., but the load as a reistor will pull current no matter what voltage is on it. A resistor can never geneate power. But a motor connected to a grid can pull power if you load the shaft or it can push power if you hook an engine to the shaft and drive it.
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