problem:
No it is not. It is an example with two different voltages.
It is a perfect example of how to supply power to the grid, resulting in higher voltage at the "home-owners end" :)
Wrong. Go back to the example in the link, as I suggested to Homeguy. You have two ideal voltage sources with different voltages connected through two resistors of different values, one 20 ohms, the other 10 ohms. That is a basic model of a battery, where the
10 and 20 ohm resistors represent the different internal resistances of the batteries. So they are batteries with identica voltage, but different internal resistances. Very basic stuff. The Kirchoff equations are right there. All you have to do is change the voltages so that they are equal.Solve the equations for the case where the voltage sources are at identical voltages and you will find that current flows from BOTH sources. One is NOT at a higher potential than the other, which is what Homeguy claims must exist for both to provide current to the load.
I never said current flows from one battery to the other. It's apparently Homeguy and now you who are hung up on that for reasons unknown. If the voltage sources are equal, no current flows between the two, which is the situation that is most desirable when powering a load with two batteries in parallel. They just BOTH supply part of the current to the load resistor.
As I said, go back and solve the equations for the case where the voltages are equal and you will find that they BOTH supply current to the load. Because of the differing resistors which would represent the internal resistance of the batteries, one supplies twice the current of the other. Capiche?
Imperfections do not render a model useless or change the facts. If you want to model the transmission line, then insert some additional resistors after the 10 ohm and 20 ohm resistors to model the line. Make a model that includes capacitance and inductance too, and make it an AC circuit . It changes nothing with regard to the ridiculous requirement that in order to supply current, a source can't just be equal in voltageto another source connected to the same load and that it has to be higher. If you have a model that shows Homeguy's planet, we'd like to see it. Until then, Jim's model is perfectly fine.
I have read your posts and like most other people here have concluded you are wrong, so I don't see why you're calling ME the dumbass.
Uh, huh. I have noted that and if you have any questions about how that works, I'll be happy to answer them. I'm still waiting for your answer about the case with EQUAL voltages. That was the essence of your argument, was it not? That a power source can't provide power in parallel with another unless it raised the voltage? So, leave everything else the same and just make the voltage sources equal. Tell us what current flows through the load resistor and what currents flow through each voltage source. This is EE course circuit theory course 101, about the first week.
BTW, you have an engineering degree?
Yes and since you seem incapable of understanding a simple circuit that represents 2 batteries connected in parallel to a load, no need to add the additional complexities.
I see you do have a question. Simple. With no current flow through battery 1, then the circuit is reduced to an ideal voltage source connected to two resistors in series. One of these represents the internal resistance of battery 2 and is 20 ohms. The other is the load resistor of 40 ohms. So, we have 20 volts across 60 ohms, giving a current of .333Amps. That .333amps produces a voltage of
13.34 across the load resistor. (.333 A X 40ohms.) That means voltage source V1 would have to be at the same potential, 13.34 volts and when it is, no current flows through what represents battery 1.I've answered your question, now answer mine:
What are currents I1, I2, I3 when the voltage sources V1 and V2 are both 20 volts.
I am not sure what you think I am wrong about here. Could you please indicate which of my above statements are wrong?
The topic is " Feeding solar power back into municipal grid". How are the voltage levels when there is a flow into the grid from the PV array?
Maybe I missed a topic change somewhere in the previous posts, which is easy to do since a few posters here does not bother to trim posts.
I did read the example, and yes, I understand Kirchoff's Law.
But you seem to be contradicting yourself, you state above the batteries have different voltages, then you state they are batteries with identical voltages. Maybe I am not understanding what you mean.
I have read all of Homeguy's messages, and I may have missed it, but does he not claim the need to have a higher voltage to push current into the GRID, not the LOAD.
The reason of my "hangup" is the topic: Feeding solar power back into municipal grid.
Your example of batteries with equal voltage might be interesting, but irrelevant to the topic.
If he really means unequal voltage then he's proving my point. He's a dumbass (second point proven).
Where's trader4?
Where did you go, you coward?
You have no response to what I wrote above?
You disappeared from this tangent thread pretty fast, didn't you?
You absolutely loved that link posted above, showing Example 1 - where you thought it was proving me wrong.
Go ahead and substitute 13.33333 volts for Battery 1 in that example and tell me how much current it's supplying to the load.
Wrong.
Motors won't turn faster, but they will take less current. They're doing the same work so will take (roughly) the same power to do it.
Wrong, not that the higher intensity is always useful.
Wrong.
You're batting 1000.
Put the windings in series and it'll run better.
Wrong. You're still batting 1000.
"Co-generation"?
They have to *pay* for that energy, not to mention manage the complexity of the mess and lose money at the same time. Of course they'll opt out, if given the chance. It shouldn't be done, but certainly not for the reasons you suggest.
So what you're saying is this:
Connect 2 batteries of the same voltage together in parallel to the same load and each battey will supply half the current to the load.
So if I extrapolate that situation:
If my service voltage is currently sitting at 120.0 volts, and if I adjust my PV invertor output voltage to 120.0 volts, and then connect my PV output to my service connection, then my PV system will somehow magically supply half the current to to the load (the load being my house and all other houses sharing the same service line).
Wow. That sounds like a really good bargain. Just by matching the power companies voltage at my service input, my PV system will supply half the current - always! And it doesn't matter how many PV panels I have!
Wow. Who knew it would be that simple and effective?
That's good for a first order approximation. The fact is that water doesn't run down hill and electricity does *not* flow from low potential to high. The inverter's instantaneous voltage *must* be higher than the grid for current to flow into the grid.
Think about turning a bicycle crank, one without a clutch makes the point better. You're not doing any work unless you're applying pressure to the pedals. If you do nothing it drives you.
No, you can't get on the "freeway" unless you're going faster than 60. If you're going slower, they're actively pushing you off.
It's OK for a first order, but not for discussing the details people are trying to get into here.
Well, in a simplified scenario, (Grid, load, PV array) there will be "load sharing" between the grid and the PV array. With the voltage pretty much equal. I say pretty much equal, because there are line resistance losses to take into account.
I think there is some misunderstanding about the concepts here.
trader4 talks about 2 batteries supplying a load. The example is good for calculating load sharing between 2 voltage sources with load resistor and line resistance.
The way I see it is that the LOAD with this battery analogy is the house load. One battery models the PV array, the other the grid.
One thing should be clear: To get power into the grid at all, the house load must be lower than what the PV array can supply. If you remove the house load, then all the available PV array current will flow into the grid, with the inverter at a higher voltage.
To get back to that Kirchoff's Law example, if we remove the 40 ohms resistor (the load), there are basically 2 voltage sources opposing each other. When these voltages are equal, no current flows.
To allow current to flow into the grid, the PV array voltage has to be higher, whether there is a house load or not.
That's exactly what I've been saying - that you "turn it up" (the inverter's voltage output) to maximize the PV's current (I) supply into the grid.
But everyone else (or most everyone else) is saying no - that simply matching the grid voltage (as measured at your service connection) is all that happens (and is all that needs to happen) for the entire PV current (I) capacity of the PV system to be "injected" into the grid.
So now that we agree that PV systems need to raise the grid voltage if they're going to "force" their maximal available supply capacity into the grid, it's a moot or academic question as to what exactly their supply situtation would be (how much current they'd supply into the grid) if the invertors simply matched the grid voltage.
You can only guess at what the inverter is forcing and sensing without looking at the schematic and uP code or a technical explanation of it. The Wiki type explanations are oversimplified for general readers, engineers have better sources.
I was hired to decipher and troubleshoot several lead-acid and lithium battery charging circuits after the designers quit, and found a couple of different approaches in use. Generally they compared voltage and current measurements to a model and used the result to pulse-width- modulate the output control.
Power factor control is similar to the design issues of a grid-tie inverter, with a large enough market to support custom ICs:
ICs:
Motors aren't the only things that need PF correction. Capacitors aren't the only, or often the, way of doing it.
An incorrect assumption. Those are for more complex loads like switching power supplies with rectifier inputs.
jsw
Ok, subsitute the words "gets wasted and performs no useful work". That is what you are saying.
That sure must be news to all the power companies that are paying people for having PV arrays generate power to the grid.
Actually, they do turn very slightly faster when supplied by a slightly higher voltage. Heres' a good reference that covers HVAC compressors:
Agreed. At higher voltage they do burn a tiny bit brighter. And that would appear to be an example of what you could consider wasted energy. Unless you want to factor in that in winter at least, in some cases, it adds to the available heat.
Because 240V is out of the range of operation for a 120V AC motor. Stick to the case at hand. We're talking about running an AC motor at 120V or 121V. I say running it at 121V means the current will be slightly less, resulting in the motor operating at the same HP output, but at slightly higher voltage and slightly lower current. And/or part of the voltage increase will result in more power being delivered by the motor to the AC comptressor. You say what? The motor justs takes that extra volt and turns it into pure heat? How does it know to do that?
Well they do consume power in relation to their voltage and current. Take a look at these formulas:
Look at the one "To find horsepower." Clearly I can get the same HP output by raising the voltage slightly while the current gets reduced. Also take a look at the previous graph, which clearly shows that full load current DECREASES if you increase voltage slightly.
Again, you'd think that if most or all of that net energy that is put onto the grid by PV arrays is being wasted, we'd have heard about it from someone long before this.
You guys need to get over your basic understanding of electricity.
When you connect to a grid the voltage is always **EXACTLY*** the same as the connect point. They are in parallel.
If your impedance source is lower than the grid's at that point you will supply the majority of the current. If your impedance source is higher (to the load) then you will supply less than the grid.
One very simple rule. Two supplies in parallel output the same voltage. When you measure each of them you will measure across the same two points for both measurements.
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Well, in a simplified scenario, (Grid, load, PV array) there will be "load sharing" between the grid and the PV array. With the voltage pretty much equal. I say pretty much equal, because there are line resistance losses to take into account.
I think there is some misunderstanding about the concepts here.
trader4 talks about 2 batteries supplying a load. The example is good for calculating load sharing between 2 voltage sources with load resistor and line resistance.
The way I see it is that the LOAD with this battery analogy is the house load. One battery models the PV array, the other the grid.
One thing should be clear: To get power into the grid at all, the house load must be lower than what the PV array can supply. If you remove the house load, then all the available PV array current will flow into the grid, with the inverter at a higher voltage.
To get back to that Kirchoff's Law example, if we remove the 40 ohms resistor (the load), there are basically 2 voltage sources opposing each other. When these voltages are equal, no current flows.
To allow current to flow into the grid, the PV array voltage has to be higher, whether there is a house load or not.
That was where I was coming from too. However, having gone back and revisited that basic circuit diagram of two voltage source driving a load, I have come around to where I see Homeguy's point that if a new source wants to deliver power onto the grid, it can raise the voltage at the load. Let's say I have a solar array that is covered up and its sunny outside. It's connected via distribution lines to a load that is a block away. Another power source is located a similar distance away from the load on the other side. In other words, a case like the circuit example Jim Wilkens provided.
When I uncover the PV array, for it's additional X KW of power to make it to the load down the block, at least one of 3 things needs to happen:
1 - The PV raises the voltage on it's end of the distribution system slightly. 2 - the load increases3 - the power source on the other end of the distribution system lowers it's voltage.
Here's the circuit example again that Jim provided:
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.
I think this is what Homeguy has been saying all along. So, I've come full circle here and now agree with him on that issue. I still disagree that the slight increase in voltage in a distribution system means that the power is wasted. The issue there is how the loads respond to being given
121V instead of 120V. HG claims that except for resistance heaters, that energy goes to waste. And I say there he is wrong, but that topic is being covered in another part of this thread.All that "load increases" is a bunch of baloney!
A fixed load is just that....***FIXED***
Sources share the load between themselves according to the impedance between and including the source and the load. If ht e grid were a perfect conductor and had zero impedance no co-gen source could work.
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That was where I was coming from too. However, having gone back and revisited that basic circuit diagram of two voltage source driving a load, I have come around to where I see Homeguy's point that if a new source wants to deliver power onto the grid, it can raise the voltage at the load. Let's say I have a solar array that is covered up and its sunny outside. It's connected via distribution lines to a load that is a block away. Another power source is located a similar distance away from the load on the other side. In other words, a case like the circuit example Jim Wilkens provided.
When I uncover the PV array, for it's additional X KW of power to make it to the load down the block, at least one of 3 things needs to happen:
1 - The PV raises the voltage on it's end of the distribution system slightly. 2 - the load increases3 - the power source on the other end of the distribution system lowers it's voltage.
Here's the circuit example again that Jim provided:
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) = .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=13.2V, I1=0, the voltage on the load is 13.2 volts, and I2= .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.
I think this is what Homeguy has been saying all along. So, I've come full circle here and now agree with him on that issue. I still disagree that the slight increase in voltage in a distribution system means that the power is wasted. The issue there is how the loads respond to being given
121V instead of 120V. HG claims that except for resistance heaters, that energy goes to waste. And I say there he is wrong, but that topic is being covered in another part of this thread.
Define a fixed load.
As a teaser, say I buy a 2 kW heater to heat my office in my home on those cold winter nights when I am reading posts in alt.energy.homepower. Is that 2 kW heater a fixed load?
I disagree. But I have been known to be wrong :)
Could you state why it does not work?
Is it possible to hook up any power source to such a grid? If the answer is yes, why is co-gen not possible?
there is a VERY slight increase in local voltage.
If you want to push 5 kW back into the gird, the local voltage rises by the amount of voltage drop in the wires leading to the grid with 5 kW flowing through them. Its the same amount as it drops when 5 kW flows out.
For example, if the grid is 120.0 and your house is pulling 5 kW, then the local voltage at your house may drop to 119.9.
If your house pushes 5 kW into the grid the local voltage at your house may rise to 120.1.
The 5 kw is not wasted, the rest of the grid reduces its generation by that 5 kW to keep the grid at 120.0.
Another analogy is tandem bikes. If the back person pushes harder, the front person has to push less to go at the same speed. For synchronous AC motors and generators this is really a good analogy, they are all running at exactly the same speed and the PHASE slips ahead or behind slightly depending on which way the power flows. You can think of it as a bit of stretch in the bike chain one way or the other.
A lot of the engineering of power systems goes into how the load is shared among multiple sources.
But in any case, a 5 kW load or source is very small compared to the overall power flow in the grid.
Mark
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