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

Page 7 of 9  
On 15/04/2011 05:47, snipped-for-privacy@optonline.net wrote:

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" :)
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
" snipped-for-privacy@att.bizzzzzzzzzzzz" wrote:

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
> Every

Take a look at the dual voltage source circuit diagram that Jim Wilkins supplied a couple posts back. It's example #1.
http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
" snipped-for-privacy@optonline.net" wrote:

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?
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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. http://www.allaboutcircuits.com/vol_2/index.html http://www.circuit-magic.com/laws.htm
jsw
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

This is a good example: http://www.allaboutcircuits.com/vol_2/chpt_5/4.html "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. http://www.microwaves101.com/encyclopedia/Smithchart.cfm
jsw
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

If he really means unequal voltage then he's proving my point. He's a dumbass (second point proven).
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On 15/04/2011 07:41, snipped-for-privacy@optonline.net wrote:

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On 15/04/2011 10:46, snipped-for-privacy@optonline.net wrote:

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
" snipped-for-privacy@optonline.net" wrote:

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?
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On 15/04/2011 16:56, Home Guy wrote:

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.
--------------------------
"g" wrote in message
On 15/04/2011 16:56, Home Guy wrote:

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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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 increases
3 - the power source on the other end of the distribution system lowers it's voltage.
Here's the circuit example again that Jim provided:
http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
It's example one.
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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.
--------------------------------
wrote in message wrote:

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 increases
3 - the power source on the other end of the distribution system lowers it's voltage.
Here's the circuit example again that Jim provided:
http://www.electronics-tutorials.ws/dccircuits/dcp_4.html
It's example one.
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. V1.2V, 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.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On 17/04/2011 10:09, Mho wrote:

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?
> If ht e grid were a

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?
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

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
His silly short response makes no sense at all.
Add pictures here
βœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Related Threads

    HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.