how do you tell where current flows?

SparkyTris

A solar PV salesman came round the other day, and gave his pitch on a rooftop grid tie system. I am amazed that consumers are "going for it" at 59 degrees North Latitude in temperate (read cloudy) climate, but apparently they are. (!)

The salesman explained that an advantage of a small generator (ie <=4kW) is that it (the grid tie inverter output) could be fed into a spare way in my CU, allowing me to get "first dibs" on the energy collected from the PV panels. spare energy from the panels is exported, via an export meter.

forgetting all the IEE regulations/FIT/supply/demand politico-economics for the purposes of this argument,

why does it matter where the grid tie inverter is connected? if it is connected to the 100A cutout position, I think I will still use my solar energy first, due to Kirchoff's law(?). As I understand it, the Grid Tie Inverter keeps its volts out just higher than line volts in order to get its current to flow out of it? (how it does this I have no idea)
Again due to kirchoffs law, if (due to the high loading/low resistance in the final circuits in my house), this high Inverter output voltage will cause the current to flow "preferentially" around my final circuits. the current will tend to resist the higher apparent impedance on the outside of the house, because there is high volts already on that side (due to power stations).

If I disconnect all my finals, the house will be high (almost infinite) impedance, and then my inverter volts will push current to (I presume) the nearest neighbour's load?

I could tie my inverter output into a point on the underground 240VAC cable in the back garden which is equi-ohmic between me and the neighbour, then we both get an equal take of the inverter output?

taking the same argument to wind power, there has been a high uptake of FIT windpower, with 6kw eoltec machines. farmers put them up in a field, and they go through a grid tie inverter onto the 240VAC line. the owners only export electricity in this scheme (they do not believe that they get "first dibs" on the wind energy collected in the field). However, I contend that if the inverter is situated very close to the high electrical demand presented by a modern farm, then they WILL be using their wind energy first.

I know this is a bit esoteric.

but am I wrong in any of it? please could greater minds than my own explain where I am wrong?. Thanks in advance.

I'd like to get funded for a European coulomb migration survey. how does one tag a coulomb? that would clear it all up.

As long as its connected to the right side of the meter, it doesnt matter, and he is talking toss and nonsense. But then again, he is a sales man!

Daniel

the current will tend to resist the higher apparent impedance on the outside of the house, because there is high volts already on that side (due to power stations).
Please don't confuse voltage and impedance.

Your grid connection has a low impedence probablly well under an ohm so a small change in voltage results in a large change in current flow but it also has a voltage close to what your inverter is generating.

Your loads have a relatively high impedance so a small change in voltage results in a fairly small change in current flow.

Your inverter will adjust it's output voltage to maintain the desired current flow.
The current flow in your service cable will be the current you are generating minus the current you are using. That flow may be in either direction*

If I disconnect all my finals, the house will be high (almost infinite) impedance, and then my inverter volts will push current to (I presume) the nearest neighbour's load?

I could tie my inverter output into a point on the underground 240VAC cable in the back garden which is equi-ohmic between me and the neighbour, then we both get an equal take of the inverter output?
The current will flow based on demand in the network. If you introduce at a point upstream of both you and your neighbour then either you will send current to both you+neighbour and the grid or your current will mix with current from the grid to supply you+neighbour.

If there are loops in the network things get a lot more complex with various voltages and impedances pushing fractions of the current down different paths.

Once currents have co-mingled into the same conductor there is no meaningful way to seperate them.

taking the same argument to wind power, there has been a high uptake of FIT windpower, with 6kw eoltec machines. farmers put them up in a field, and they go through a grid tie inverter onto the 240VAC line. the owners only export electricity in this scheme (they do not believe that they get "first dibs" on the wind energy collected in the field). However, I contend that if the inverter is situated very close to the high electrical demand presented by a modern farm, then they WILL be using their wind energy first.

There are really three seperate issues

1: The pysical flows of current and associated energy
2: The "logical flows" created by buying and sellinc electricty
3: The disconnection policy in the event of a fault.

1 is dictated by the voltages and impedences of the system. Mostly people don't care too much about it unless they are worried it might cause overvoltage/undervoltage issues.

2 is dictated by the metering arrangements and the contracts, markets, regulations and subsidies surrounding those meters.

I could see you could have two different arranagement, an arrangement where the customers power (both demand and generation) was all mereged together then metered as one flow or a system where the demand and generation are metered seperately.

3 is usually dictated by safety and cost considerations. It is generally a bad idea to try and feed a dead grid unless you have the power to connect and disconnect parts of that grid (e.g. you are a large power station working in conjunction with national grid). It would be possibel to deisgn

You could design a system to switch to "off-grid mode" in the event of a failure of the public grid however such a system would be complex and expensive (you would essentially have to have a load prioritisation system to direct the limited available energy to the most important loads and/or switch in a battery or generator system to take up the slack.

In the farmers case financially there are most likely two seperate transactions going on, the wind farm selling power and the farm buying power. Furthermore in the event that the grid dies there is almost certainly no mechanism to move to a disconnected operating mode and the wind farm will shut down.

So although physically current may well be flowing from the wind farm to the farmer he is not in not in any meaningful sense getting "first dibs" on the electricity

*Yes I know it is AC but one can still consider it as having a direction as long as one considers the current waveform relatively to the voltage waveform.

Once currents have co-mingled into the same conductor there is no meaningful way to seperate them.
What about filters? Isn't PV electricity green?

Once currents have co-mingled into the same conductor there is no meaningful way to seperate them.
What about filters? Isn't PV electricity green?

Its probably very pale green since if you take the cost of producing the panels, the raw materials, transportaion, and then fixing it on a rooftop, the cost of other associated materials used in installation, further more cost of producing hamburgers, energy to cook thyem, eaten by electricians! lol

You forgot the methane the electricians will make after eating them.

Please don't confuse voltage and impedance.
quite, this was slack of me. but high(er) open-circuit volt implies high(er) z, no?

Your inverter will adjust it's output voltage to maintain the desired current

is this how the inverters work out what output Voltage to make? How DO they do that? current transformers on the output lines?

The current will flow based on demand in the network. If you introduce at a point upstream of both you and your neighbour then either you will send current to both you+neighbour and the grid or your current will mix with current from the grid to supply you+neighbour.

... so if both me and the neighbour go all finals disconnected (open circuit), any available current from our microgeneration inverter will be sucked up through the pole transformer, onto the 11kV lines, possibly even onto the 33kV lines and so on, then back down to wherever the (nearest?) demand is?

Regarding the windfarms; I see your 3 separate issues, but I'm only considering the physics at this stage! (by the way, all the grid tie inverters Ive seen have a disconnect function in the event of undervolt on the line, specifically to protect linesmen - I think this might be a stipulation of the FIT installs)

So although physically current may well be flowing from the wind farm to the farmer he is not in not in any meaningful sense getting "first dibs" on the electricity

Quite so, he is exporting ALL of his wind energy on one meter and importing ALL his energy consumed on a separate meter. so even though hes got a bran new 6kW windmill thats always turning, he sees no difference in his electric bills.
However, my contention is that, because the windmill export meter tails are spliced into the same service cable that feeds his import meter, the actual coulombs coming out of the windmill current source will go right into the current sink presented by the farm, and preferentially so..... but I don't know why. Am I right? if so could you explain why?
I was about to say because the farm is lower impedance than the alternative (which is to go up onto the pole trafo, thence onto 11kV line) but this is clearly wrong, going by your last post.

Surely, a live grid trafo would see the farm load as low resistance
and as 50hz is close to dc, dc resistance =ish= impedance at 50hz

to get max power transfer, impedance of source & load should be equal. (?)

Plugwash, you said the current will flow based on demand in the network. Which is fairly intuitive. But, following that:

I know that Aberdeen city permanently has a much higher demand than the farm Im thinking of. does this mean that all of the coulombs generated by the 6kW windmill in Orkney (and conditioned by its associated grid tie inverter) naturally flow to the massive current sink of Aberdeen?

I think not.... (??)

all of this leads on to my next q. Big political issue here is that the grid infrastructure in Orkney & Northern SCotland does not have the capacity to export (renewable) energy south to the demand centres of Edinburgh, Glasgow, points south. this is apparently a limiting factor in the development of windfarms, wave, tidal energy.

I can understand that. A cable can only carry a finite qty of Amps.

however, there has never been any political whinging that the infrastructure cannot support the increasing electrical demand in Orkney. (demand is definitely increasing year on year).

can the same cable that connects Orkney to the UK Mainland be used to export power as import it?

I dont see why not.
OFGEM tell us that it doesnt have the capacity for (much) more export to be connected. Now, does this mean that Orkney is already a net exporter of current, and that every new generator further increases the current flowing south in "our" cable?

I may be wrong, but I dont believe that there is that much energy in Orkney to export. BUT, there definitely is energy being produced here. Now, surely the effect of all that is to actually reduce the current being drawn northbound through our subsea interconnector??
I realise that wind, tide, wave energy is "bursty" by nature. but I dont think that really applies in Orkney. there is zero wave energy at present, minimal (sub 500kW tidal). the windmills of which there are many, are virtually constantly turning, because it is almost always windy.

unless I am wrong about the whole thing and Orkney is no longer drawing current north, Why are OFGEM telling us the capacity is not there?
maybe the electricity transmission system is designed to be one way only - but how? it's all just wires and transformers, innit?

[Once currents have co-mingled into the same conductor there is no meaningful way to seperate them.
I think the word "meaningful" probably needs be be dropped from that statement - electrons don't come with labels

Kind Regards, John.

SparkyTris, you are right about the way the farm produced energy is connected to the grid where by all of it is exported, it will export all of this energy only if its potential is higher than the grid, be it just a few mili volts above, but because the so called the impediance of the grid is extremely small, quite simply it is almost a dead short circuit! as there are thousands up on thousands of parralell loads connected on the grid! so your generator will have no problem exporting all of its energy provided its voltage is slightly higher than the grid voltage.

Take this analogy, where a small stream is flowing into a large dam, if the dam level goes higher, the stream will not be able to flow out but rather flow back in! so the dam could stsrt to flood the stream! The dam being the grid, and the stream being the farm generator, so it has to be at a higher level to fall into the larger pool of water in the dam, and even if the stream (farm generator) is a few volts above the dam (grid volts) it will spalash into the big dam and disperse so thinly that its effect will be hardly noticable overall, someone on the other side of the dam might not even see any ripples!
nOnly perhaphs someone close to the stream may hear the stream fall into the dam and see any ripples although these will be minute comparably as the stream is far far too small.

As for metering, yes, if you were generating electricity not just for your own benefit, but to export the surplous that you are not using, makes sense why not!, so you would connect the farm generator whereby it is exporting 100% of it and then whatever you are using is being metered as imported, so the net result is you pay for the difference in the exported and imported meter readings, apart from any further incentives you get for exporting, like you might get higher terrif for exporting than what you have to pay for importing.

so the point is the inverter must produce a higher voltage than the grid, the inverter cannot adjust its voltage that it is producing but rather the grid will clamp it down to its level, as the grid is like a huge load on it, so as long as the inverter is designed to produce say around 5 to 10 volts above the grid voltage, the inverter's voltage will automatically get clamped down to the grid's level by its extremely low impedience.

(Inverters can detect when the grid voltage drops out, this is because if the grid looses its power, then effectively your farm generator simply cannot produce enough current to keep the voltage to its nominal levels and will soon get overloaded to almost zero volts! this overloading could be detected and the farm generator isolated from the grid and won't trip back in unless the grid vvoltage/power is restored again)

quite, this was slack of me. but high(er) open-circuit volt implies high(er) z, no?
You can have a high open circuit voltage with a low impedance (e.g. the grid) or you can have zero open circuit voltage with a high impedence (e.g. your houses loads).

Transformers change both voltage and impedance. A step down transformer will reduce both the voltage and impedance of the source. Similarlly a step down transformer will increase both the voltage and impedance of the source.

is this how the inverters work out what output Voltage to make? How DO they do that? current transformers on the output lines?
I dunno exactly how grid tie inverters work but my guess is they have a number of sensors. Probablly at the very least input and output voltage and either input or output current (you don't really need to measure both input and output current as they will be strongly related to each other but I bet many inverters do anyway if only for fault detection) . Which are then fed into a control program that drives the switching elements of the inverter.The program must decide how much voltage to apply, where in the cycle to apply it and ultimately whether it is safe to continue operating.

Fundamentally the only way to push power into a grid connection is to produce a voltage higher than the open circuit voltage of said grid connection.

... so if both me and the neighbour go all finals disconnected (open circuit), any available current from our microgeneration inverter will be sucked up through the pole transformer, onto the 11kV lines, possibly even onto the 33kV lines and so on,
Right.

then back down to wherever the (nearest?) demand is?
Assuming the network was a tree your powert would flow back up the tree (with some of it being taken off to supply any loads along the way and some of it being lost along the way) until it reached a point where the backflow from your inverter merges from the outflow of the grid to supply a customer.

Of course the real network is not a tree so the splitting and merging is going to be a bit messier than that.

by the way, all the grid tie inverters Ive seen have a disconnect function in the event of undervolt on the line, specifically to protect linesmen - I think this might be a stipulation of the FIT installs
Indeed, grid tie inverters are designed to shut down completely if anything goes out of line that may indicate that the grid connection has been lost (undervoltage, overvoltage, underfrequency, overfrequency).

Surely, a live grid trafo would see the farm load as low resistance
Low is relative. The impedance of the farm load is probablly between 1 ohm and 100 ohms depending on what is turned on. The impedance of the grid connection is likely a small fraction of an ohm.

However, my contention is that, because the windmill export meter tails are spliced into the same service cable that feeds his import meter, the actual coulombs coming out of the windmill current source will go right into the current sink presented by the farm, and preferentially so..... but I don't know why. Am I right? if so could you explain why? .
Let us suppose the farm has a resistance of 10 ohms and zero open circuit voltage while the grid has a resistance of 0.1 ohms and an open circuit voltage of 240V. Both the farm and the grid are connected to our inverter.

Now suppose the voltage at the point where grid, inverter and farm load join is 239V. By ohms law the current to the farm will be 23.9A while the current from the grid will be 10A. So by kirchoff's current law (KCL) the current from the inverter must be 13.9A

Now suppose the voltage at the point where grid, inverter and farm load join is 240V. By ohms law the current to the farm will be 24A while the current from the grid will be 0A. So by kirchoff's current law (KCL) the current from the inverter must be 24A

Finally suppose the voltage at the point where grid, inverter and farm load join is 241V. By ohms law the current to the farm will be 24.1A while the current from the grid will be 10A. So by kirchoff's current law (KCL) the current from the inverter must be 34.1A

Clearly we can see that small changes in the inverter output voltage dramatically change the current flow to/from the grid but only have a small impact on current flow to the farm.

If the desired output current of the inverter is higher than the farms demand then the inverters output voltage will settle just above the grid's open circuit voltage and current will flow from inverter to both farm and grid. If the desired output current of the inverter is lower than the farms demand then the inverters output voltage will settle just below the grid's open circuit voltage and current will flow from both grid and inverter to farm.

to get max power transfer, impedance of source & load should be equal. (?)
Assuming a source behaves linearly you will get the maximum power out of it by having a load with equal impedence to the source. Of course at this point the voltage will be reduced to half the open circuit voltage and the efficiency of the source is likely to be terrible. Most practical power sources* will usually trip out long before they reach this point to prevent them going up in smoke.

If efficiency (rather than maximum power transfer) is your goal then you want the source to have a much lower impedance than the load.

*Radio gear is rather different, they do often use matched impedances there for other reasons and they are prepared to put up with low efficiency.

MikefromLondon, thanks for the easy-to-understand analogy. indeed, the trickles into a reservoir were very much the picture I was forming, some being just that, trickles, with a few (hinkley point, sizewell etc) being substantial streams). and the consumers being drains, more or less, on the face of the dam.
Of course this analogy implies the possibility of on-grid storage which is sadly not a reality yet!

Plugwash,
Finally suppose the voltage at the point where grid, inverter and farm load join is 241V. By ohms law the current to the farm will be 24.1A while the current from the grid will be 10A. So by kirchoff's current law (KCL) the current from the inverter must be 34.1A

Did you mean that the 10A would flow TO the grid in this case? (ie, inverter exporting)???

if so, then your excellent illustration of how inverter V affects current flow, seems to vindicate my assertion that - provided the inverter volts are slightly higher than (or equal to) grid open circuit voltage - the farm will indeed be preferentially using its own wind energy.
so that's why - voltage level of the energy source

you keep mentioning the desired output current of the inverter. but I cant see how that would work. eg - I desire my little inverter to supply 100A, please. That won't work, at least not all of the time.

rather, I desire my little inverter to "make the best" of the dc from the windmill. If it's got 6kW available due to max wind, then I could (ideally) expect 25A to be provided at nominal 240V. following on your example, I could then expect the inverter to make 241V on its terminals. 24.89A flows to the farm (cos its closest)
no
no
the grid is apparentyl virtually a short circuit, so it will immediately suck up anything the little inverter produces.
but its sitting at 240V already. farm's at 0V, presenting 10Ohms. inverter's outputting at 241. so its going to send 241/10 =24.1A into the farm. the remainder (6kW available at 241V) 0.79A goes to the grid.
why did I choose it to make 241V I wonder?

am I right?

radio users are not prepared to put up with low efficiency, and certainly not in a power transmission line! impedances must match along feedlines precisely to improve efficiency. it is electromagnetic energy being transmitted in a RF feedline as well as a domestic power feedline, it's just the wavelengths that are different.

there must be a link here but Im blowed if I can see it.

efficiency in a power transmission line IS maximum power transfer.

but I do see your point.

perhaps it is that domestic Z's are actually about as low as the trafo's facing them. must be. its all basically similar: low impedance source at a power station. transformed up to high kV's, high-Z for transmission, then back down again. Any mismatch in Z results in poor efficiency (cf running single-phase 300kV on lines designed for 11kV. the massive E-field in close proximity to the other conductor would cause heating I think?? of course catastrophic arcing would ruin it entirely in another manner, but I feel that you probably could "get away" with 300kV separated by 2.5m in dry air. rather, its the inherent inefficiency due to the field strength that makes it a bad idea. but I'm out of my depth there.

[/b]

radio users are not prepared to put up with low efficiency, and certainly not in a power transmission line! impedances must match along feedlines precisely to improve efficiency. it is electromagnetic energy being transmitted in a RF feedline as well as a domestic power feedline, it's just the wavelengths that are different. there must be a link here but Im blowed if I can see it.
efficiency in a power transmission line IS maximum power transfer.
This might be totally off-track, but .... is not one of the issues/differences perhaps that, in the case of RF power transmission, one can employ impedance-matching techniques (e.g transformers) at the ends of the transmission line, because one does not care (within reason) what the voltages and currents in the line are - a luxury which a domestic power feed does not have?

Kind Regards, John.

plugwash said:
Finally suppose the voltage at the point where grid, inverter and farm load join is 241V. By ohms law the current to the farm will be 24.1A while the current from the grid will be 10A. So by kirchoff's current law (KCL) the current from the inverter must be 34.1A
Did you mean that the 10A would flow TO the grid in this case? (ie, inverter exporting)???
Yes sorry.

Those numbers were mostly to illustrate how a small change of voltage results in a MASSIVE change in current flow to/from the grid but only a small change in current flow to/from the local load.

you keep mentioning the desired output current of the inverter. but I cant see how that would work. eg - I desire my little inverter to supply 100A, please. That won't work, at least not all of the time.
Indeed, the inverter must adjust based on what is going on with both the grid and the source so that it finds the operating point that gets the most out of the source and then delivers that power succesfully to the grid.

Thinking about it the main thing* that the inverter actually needs to be monitor is the input voltage. If the input voltage is too low then the inverter is overloading the source so it should reduce the output voltage. if the input voltage is too high then the inverter is underloading the source so it should increase the output voltage.

but its sitting at 240V already. farm's at 0V, presenting 10Ohms. inverter's outputting at 241. so its going to send 241/10 =24.1A into the farm. the remainder (6kW available at 241V) 0.79A goes to the grid.
why did I choose it to make 241V I wonder?
If the inverter can only supply 25A and the grid has an open circuit voltage of 240V and an impedance of 0.1 ohm impedance then it wouldn't reach as high as 241V. It would settle at approximately 240.1V

*Obviously the inverter must also monitor the position in the output cycle so it delivers the power at the right time in the cycle and it must monitor a lot of other things to for safety reasons.

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