I am having a heated discussion with a friend about excess solar power being sent into the grid. My friend is an ex navy engineer and says that this can't be done due to the potential of the grid against the generated source (home P.V. system). I see from the posts on the forum, that those with a large enough P.V. systems are being paid for the excess power created. Can somebody please provide me with a technical explanation of this process, so that I can provide it to my engineer friend.
Exporting power into the grid
(22 posts) (8 voices)-
Dave Berry
MemberPosted Sunday 17 Jan 2010 @ 11:55:49 pm from IP # -
rsigmund
MemberNot 'sure' if I can give you a technical explanation but will try anyway...
As I understand things...
The inverter will produce a nominal 240 AC sine wave pushing the power (P=IV) into the grid. Of course 240 Volt AC will not flow if the sine wave is in phase and indentical votage ie. no current will flow..
So the voltage produced by the inverter must be higher than the mains by some amount (not sure how much - presumably small).
The power exported will be used locally i.e it will not be exported 'up' the high voltage distribution lines.
Posted Monday 18 Jan 2010 @ 12:37:30 am from IP # -
Sunshine
MemberMy inderstanding is that electricity flows to the path of lowest resistance, so if the inverter output is say 1000 watts, and the house has a load requirement of 400 watts, then 400 watts out of the 1000 watts will flow to the house, and the surplus 600 watts will flow as an export, i.e. outgoing direction, to the nearest neighbour where there is a load requirement.
My understanding, is that the inverter must have a sine wave in sync with the network, otherwise it will shutdown.
I have observed this once, where the inverter shutdown for several minutes , with an error code mentioning something about "phase mismatch", or similiar (can't quite remember)Posted Monday 18 Jan 2010 @ 1:43:13 am from IP # -
franks
MemberInverter will sample mains voltage, if voltage and freq is ok it will start to output,
it will generate a higher voltage than the mains, grid tie inverter output voltsage and power (using a voltage/power meter) you see the ramp up of voltage and power, typical increase 0.6-1.2 Volt it takes a few secs to ramp up fully.Posted Monday 18 Jan 2010 @ 2:50:32 am from IP # -
rsigmund
MemberYes
Electricity will flow to the path of least resistance but will not flow 'up hill'.
And to add to the discussion - the inverters used for grid con, and stand alone are 'active' systems.
For example
All GC inverters used in australia must use anti-islanding. Which means they will not export during a grid outage. They do this by altering the sine wave output and detecting if it changes the grid sine wave and shuting down...
Posted Monday 18 Jan 2010 @ 2:54:25 am from IP # -
Sunshine
Memberrsigmund,
I know when my inverter and panels were installed and for a few weeks thereafter when the old meter was still in place, the spinning disc was spinning backwards during the day, and forwards at nights, dispelling that theory, that power cannot be exported down the lead-in cable.
The lead-in cable from the street is only a copper cable, so there is no reason for power to not to flow in the opposite direction.
Posted Monday 18 Jan 2010 @ 3:22:51 am from IP # -
rsigmund
MemberSunshine
I am certainly 'not' saying that power cannot flow both ways - it can and does happen - I was just addressing the original question - pointing out that current flow is driven by potential(voltage) differences.
"dispelling that theory, that power cannot be exported down the lead-in cable" I hope you are not refering to my comments? I am sure it can be, just getting paid is the tricky part.
What I meant by the following
"Electricity will flow to the path of least resistance but will not flow 'up hill'"
is that the inverter must produce a higher voltage than that of the grid - you can be sure when your meter was running backwards the output from your inverter was at a higher potential than the grid.
Posted Monday 18 Jan 2010 @ 4:35:40 am from IP # -
Sunshine
Memberrsigmund,
There is no differential in voltage, both the grid in the street and the inverter output is 240v AC @50 Hz, otherwise my inverter will be frying all the appliances in the neighbourhood..
The meter is just wired up to record power flow in both directions, rather than just in one direction.
Current always looks for the path of lowest resistance, not the amount of voltage differential. So as an appliance is switched on in the house, it has less resistance than when it was switched off, so current flows in that direction.
Think of it as a water pipe.. water always tries to find the path of lowest resistance to get out of a pipe, hence sprinkler systems connected to water mains, must have a anti-backflow valve installed, so that potentially contaminated water from the garden can't flow backwards into the water grid.
Posted Monday 18 Jan 2010 @ 5:05:22 am from IP # -
rsigmund
MemberSorry Sunshine
Absolute Hogwash
Basically your analysis is far to simple and in some cases simply incorrect. Let me try to elaborate...
"There is no differential in voltage, both the grid in the street and the inverter output is 240v AC @50 Hz, otherwise my inverter will be frying all the appliances in the neighbourhood.."
You should say the grid is not often exactly at 240v AC @50 Hz but often somewhere close to that value! Also 242v AC will not fry your appliances! the grid fructuates quite a bit in reality...And yet another point: 240v AC @50 Hz (produced by your inverter) will be attenuated by the time it gets to your neighbour and become less than 240v AC @50 Hz...
"Current always looks for the path of lowest resistance, not the amount of voltage differential."
Mmmmm... Hard to know where to start with this one. Yes the first part about resistance is true (the last part is NOT), however NO current will flow if there is no electrical potential and current flow can be reversed by reversing the polarity.
No current flow through either resister
12v 1 Ohm Resister
100 Ohm Resister 12vConventional current flows left to right (more through 1 Ohm resister)
12v 1 Ohm Resister
100 Ohm Resister 0vConventional current flows right to left (more through 1 Ohm resister)
0v 1 Ohm Resister
100 Ohm Resister 12vYes think of a water pipe - connect a pump to the sprinkler and pump at a pressure higher than the water mains and you will make the water flow backwards
Dave Berry
This is starting to feel ironic
Posted Monday 18 Jan 2010 @ 6:40:47 am from IP # -
Sunshine
Memberrsigmund,
What?
You are saying the only way current can move in a particular direction is to vary the voltage?
The voltage in the street is supposed to be 240v AC @50 Hz, same as the inverter, so you are saying no power will flow into the house until your voltage in the house is lower than the street!
No way.
The voltage in the grid does vary during the course of the day, but this does not control if an inverter is able to export power or not.
Absolutely correct, "Yes think of a water pipe - connect a pump to the sprinkler and pump at a pressure higher than the water mains and you will make the water flow backwards", so if the pressure (electrical resistance) in the house is greater than the grid, current will flow to the grid and you will export power!
Posted Monday 18 Jan 2010 @ 7:02:20 am from IP # -
Sunshine
Memberrsigmund,
So to make it simple.. if a grid connected inverter has all the fuses pulled going to the house, therefore there is high resistance going to house, as cable is open circuit, then there is no way power can flow to house.. Agree? Ok, then current will flow to grid as grid will have lower resistance than all the fuses pulled in the house. Agree?
Ok, then that proves current will flow to grid regardless if the grid is 239v or 241v.
If you then insert a fuse in the meterbox and switch on a clock radio, there will be a small amount of current flow back into the house, as the clock radio will only draw a small amount of current, the rest will be exported back into the grid.I used to use power line monitors in my line of work years ago, and the grid voltage varies considerably, roughly 218v to 252v I have observed.
This is due usually to the power load of the consumers during the course of the day/night.. In peak power times, voltage will drop, in middle of night, voltage will go up.
So technically, if an appliance uses 2400w, and at 240v, will use 10 amps.
However if the voltage drops to 220v, then appliance will draw 10.9 amps, and as the power meter in the meter box measures the current flow in amps, then really you will be paying more for power when the voltage is lower, as the meter will register more power (current) being registered, and paying less than when grid voltage is say at 250v.Posted Monday 18 Jan 2010 @ 7:50:55 am from IP # -
ghostgum
MemberSunshine, as an electrical engineer, I prefer rsigmund's explanations.
If you want to use the water analogy, then Voltage is pressure (or height difference). Current is the water flow rate. Resistance is the restriction to flow caused by the size and length of the pipe. Fat short pipe has less resistance than a long narrow pipe.
To get current to flow from the inverter to the grid, the inverter voltage must be slightly higher than grid voltage. Since the resistance of the wire from the street to your switchboard is small, and the current isn't huge, the voltage difference will also be small. The voltage at the street wire, at your meter box, at your switchboard, at the inverter output, and at your appliance will all be different if there is current flowing. We are talking about fractions of a volt.
Resistivity of copper is 1.7 x 10^-8 ohm metre. As a quick calculation, the 16mm2 copper wire 8m long from my meter box to my switchboard has a resistance of 1.7x10^-8 * 8 / 16x10^-6 = 0.0085 ohm. With 10A flowing, the voltage difference across this wire is V = IR = 10 * 0.0085 = 0.085V or 85 millivolts. Wherever current is flowing, there are voltage drops, except for superconductors and you won't find any of those at room temperatures in the home.
In practice the inverter may actually be measuring the voltage and controlling its output current, rather than the other way around. That is, it may be acting as a current source rather than a voltage source. It monitors the resistance of the grid, and if this exceeds certain limits then it will shut down to both protect itself and to avoid generating power when there is a grid fault.
Regarding your 2400W example. If the voltage drops to 220V and the appliance is unchanged, then the current will drop from 10A to 9.2A. The power meter in your meter box does not measure current alone, it measures real power, which depends on the voltage, the current, and the phase relationship between them. It gets complicated if we have non-sinusoidal waveforms.
Posted Monday 18 Jan 2010 @ 10:20:21 am from IP # -
Sunshine
Memberghostgum,
Interesting, so what you are saying is that the inverter has to continually adjust it's output voltage to make it able to import or export power? So if the inverter voltage is lower than the grid, and there is no load in the house where does the power end up?
So if you were able to modify the inverter to it's maximum voltage output, then it will really be configured for gross feed-in rather than net feed-in.
That sounds like a simple mod to me.
I was advised by the solar company that it needs to be rewired to go from net to gross feed-in.Posted Monday 18 Jan 2010 @ 11:43:36 am from IP # -
ghostgum
MemberThe inverter always exports power, never imports. Unlike transformers which can transfer power in either direction, the circuitry in an inverter is likely to only supply power, not consume. Some of the circuitry (diodes, transistors) only allows power to flow one way.
Yes, the inverter voltage is always changing, but in an effort to export the maximum possible power from the solar panels. I think it more likely that the inverter is behaving more like a current source, and is continually adjusting its current rather than its voltage.
If the inverter voltage is lower than the grid, then no current flows from the inverter to the grid, and so the inverter doesn't generate power. You can put a solar panel in the sun and not connect anything to it. If you do this, then your 170W panel doesn't generate any power because there is nowhere for the current to flow.
If the house doesn't use the power, then it goes out onto the grid and is used by others. The current flows along the easiest path, which is likely to be to nearby load, in other words your neighbours. Using the water analogy, you and your neighbours have pipes feeding from a dam. The inverter pumps water back into the pipe, which is not going to end up in the dam, it will be used nearby.
Net feed-in and gross feed-in are a separate issue, and nearly always requires a wiring change to go from one to the other.
For net feed-in, the inverter and the house connect to the same terminal on the meter. The meter measures house consumption minus inverter output, which can be positive (net consumption of power) or negative (net generation of power).
For gross feed-in, the inverter connects to one terminal on the meter, and the house connects to another. Effectively the meter is actually two meters. One measuring the inverter power, and the other measuring the house power.
For some digrams showing net and gross metering, see
http://www.ue.com.au/customers/download/Co-generationElectricityMeteringdiagrams.pdf
These are from my distributor in Victoria.
We will use diagram 4 for net metering. NSW would use something like diagram 1 for gross metering.Posted Monday 18 Jan 2010 @ 12:04:16 pm from IP # -
LanceTurner
Key MasterSeems to be a lot of confusion here, you are all thinking of the grid as if it's a DC source. Remember, it is an AC sinewave, so when an inverter is feeding into the grid it has to match the grid's voltage and frequency, but it has to lead the phase angle slightly. By doing this, at any given point the inverter's voltage is slightly higher than the grids on the rising part of the wave, and lower on the falling part (but sinewaves go both positive asnd negative with respect to earth, so it can be hard to conceptualise), hence the flow of power to the grid.
Inverters control their power output to match the incoming power available, they constantly adjust their output voltage against the mains grid hundreds or thousands of times per second, and just shift the voltage so that the appropriate current flow is achieved.
It helps if you consider the energy flow in discrete steps, as instantaneous points in time. It can be a little hard to get your head around, but if you picture a sinewave (grid voltage) and then overlay a second one that is slightly leading it, you will see the difference in voltage between the two waves, hence the flow between the inverter and the grid.
Also remember that it's not just voltage that changes value in a sinewave, the current also changes direction 100 times a second, ie it flows backwards and forwards in the same circuit constantly. It's totally different to DC flows...
Posted Monday 18 Jan 2010 @ 11:59:51 pm from IP # -
LanceTurner
Key MasterUnfortunately with AC systems you can't really keep it simple, if you explain it as if it's DC then people end up with a very incorrect understanding of how it works. Unfortunately there's a great deal of misinformation out there about all aspects of electrical and electronic systems.
Posted Tuesday 19 Jan 2010 @ 5:17:42 am from IP #
