Experiment related to on grid solar - Page 1
Forum section: electronics / general

author: evlo
author_rank: Contributor
timestamp: August 02, 2022, 08:08:00 pm
content: I don't currently have equipment to do experiment related to solar power, but I think it is quite interesting - can you put a load on two power supplies at the same time, but one power supply will have slightly higher voltage than the other one and see the current flowing from each supply?At least in my country, they do on-grid solar like this: grid > breaker box < inverter and say that equipement connected via breaker box will use solar power because the inverter is supplying slightly higher voltage.I was unable to verify this with my shitty multimeter, on DC, two batteries and a piece of led strip as a load.

author: Someone
author_rank: Super Contributor
timestamp: August 02, 2022, 11:39:07 pm
content: Quote from: evlo on August 02, 2022, 08:08:00 pmThis is a conceptual description, it isnt the whole story. The grid is AC rather than DC so you need to use complex impedance's and AC analysis, they use a phase shift in the "two" voltages to direct current across the (low loss) link.Applying the same to batteries isnt possible, you'd need to put a resistive (lossy) link between them.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 03, 2022, 02:42:54 am
content: You can't do the same thing with batteries but you can with DC power supplies. The equivalent of a grid tied inverter is not another battery (voltage source) but more like a current supply. The current source has a voltage comparator that shuts off if the voltage from the grid goes out of range.

author: evlo
author_rank: Contributor
timestamp: August 03, 2022, 09:08:06 am
content: That is why i'm asking someone (David?) could do this experiment.I mean probably it works, somehow, as people do not complain about it, thought it would be interesting to me to do experiment and measure it, however currently I do not have access to equipment needed nor solar installation I could play with.I also think it is more phase shift then the voltage thing ...But how come phase shift does not clause issues in the rest of the distribution network

author: Berni
author_rank: Super Contributor
timestamp: August 03, 2022, 09:53:21 am
content: AC distributions grids are indeed not just simply like batteries.The cables that feed grid power to your house simply look like a very low impedance AC voltage source. Nothing you can do can affect the grid voltage or frequency (apart from resistive losses on cable going to your house) because the grid has gigawatts of power running trough it.So what the solar inverter does is actually send current into the grid that has the same sine shape and phase as the voltage on the grid. It could actually send any shape and magnitude of current into the grid and the current would just flow into it, but doing so would not deliver power into the grid efficiently, so your electricity meter wouldn't count all of it, also likely breaking the contract you made with the electrical company that requires you to send power into the network with a sufficiently high power factor (a measure of how in phase sine shaped your power is), so they can sue you or disconnect your service. This is because sending power in this inefficient current shape causes extra losses in the electric companies power network.Now when you plug in a resistive load (like a heating element) it will also take a sine wave shaped current from the grid, but this one is flowing in the opposite direction, so the electricity meter counts that as taking power from the grid. If you also have an inverter running, this means it cancels out the current produced by the inverter and lowers it by that amount.So if both currents are say 10A then they cancel out completely leaving 0A to be sent out to the grid. Hence the electricity meter has 0A flowing trough it hence it is counting 0W and 0kWh. If the inverter is producing 15A and the load is still using 10A then that leaves 5A of current that needs somewhere to go, because there are not enough loads to use it up the only direction it can flow is out to the grid via the meter so it will measure those -5A or -1200W, counting that as power sent to the grid.The result your power does in the grid is those 5A flow trough the wires back towards the distribution transfer, canceling out more loads along the way until it is used up. This means the distribution transformer doesn't have to supply the 5A you gave to it, so it will also draw those 1.2kW less on the high voltage line feeding it. This means the power plant feeding that high voltage line will also see 1.2kW less load, so the generators in that power plant will take 1.2kW less power to spin, so the control system there will have to close the valves slightly to let slightly less steam into the turbines. If it didn't do that the turbines would keep trying to spin the generator at the same power but nothing is taking the power, so the leftover power stays as kinetic energy going into spinning the generator faster and faster. This would raise the mains frequency from 50Hz to 51Hz then 52Hz etc.. This doesn't actually happen because the grid is carefully controlled to make sure this does not happen and also it takes a LOT of power to make all generators in Europe to spin faster by any measurable amount.

author: evlo
author_rank: Contributor
timestamp: August 03, 2022, 10:23:15 am
content: Now when you plug in a resistive load (like a heating element) it will also take a sine wave shaped current from the grid, but this one is flowing in the opposite direction, so the electricity meter counts that as taking power from the grid. If you also have an inverter running, this means it cancels out the current produced by the inverter and lowers it by that amount.From what I have read it does not work like that in Czechia, but people lie a lot about that stuff here.What you describe is exactly what would MAKE SENSE to me.Sending power to the grid and if it equals out power that is being drawn it is just not counted.So basically thishttps://std.iec.ch/terms/terms.nsf/3385f156e728849bc1256e8c00278ad2/57729bc6f1abad96c125787f00246517which I do not understand, but makes sense, not anything about voltage and phase shift that would make device use this or that power.So that would mean that inverter generates 25hz shifted phase power?!Still interesting experiment with two different voltage power sources and one consumer.

author: Berni
author_rank: Super Contributor
timestamp: August 03, 2022, 11:28:07 am
content: Quote from: evlo on August 03, 2022, 10:23:15 amThis is an explanation of reactive/aparant/real power.You don't need to worry about this because household electricity meters use the power of math to only count the real power (W), not apparent power(VA) or reactive power(VAr). The electricity company will also only generally allow you to use inverters with a power factor of around >0.95 (max 5% reactive) and this means that means the inverter pretty much only outputs real power (W)Quote from: evlo on August 03, 2022, 10:23:15 amNo, the inverter must follow the 50Hz mains in both voltage and phase.You CAN'T have sources with different voltages connected to the grid, because the grid is the one dictating the voltage. You can only choose how much current you want to push/pull from the grid at any given moment, this current is what electricity meters measure and multiply with grid voltage to get power in Watts.The only way you can have sources on the grid at different voltages is for the sources to be able to supply so much current that the voltage difference is created in the cables connecting them. The mains network tries to use as low resistance wires as possible so this takes a lot of current. Something like 0.1 Ohm source impedance is not unusual on the mains feeders where your grid comes in. So in order to overpower your local mains network you need to force the 300V down to -300V (in places where your 25Hz source goes opposite phase of the 50Hz grid), so you would need to push 6000A into the grid, so you need an inverter rated for about 1 MW. So you better have a really REALLY big solar array handy, or perhaps a diesel generator the size of a locomotive.But lets say you did have such a big inverter what would happen? Well your main fuses would pop instantly and disconnect you from the grid as they should. If you still ware not happy and replaced the fuses with solid copper bars then you would manage to force your new 25Hz over the whole street, soon after the fuses at the distribution transformer would blow and your 1MW inverter would become the new and only power plant for the street using your new 25Hz power. After some time the cable running to your house would likely get very hot, melting the insulation and letting the wires contact shorting out, drawing much more current and overloading the inverter making it shut down, leaving the whole street without power as its last power source went offline. After some time the power companies control center would attempt reset the fuses, they would blow again because the cables to your house are still melted and shorted. Some time after a technician would arrive to figure out what happened, ending up with the power company very upset about your 25Hz power experiment.

author: evlo
author_rank: Contributor
timestamp: August 03, 2022, 11:37:09 am
content: 50hz, but 25hz shifted, at least that is how i undertand 4qudrand 3hase electric metering

author: Berni
author_rank: Super Contributor
timestamp: August 03, 2022, 11:59:46 am
content: Quote from: evlo on August 03, 2022, 11:37:09 amPhase is measured in degrees or radians (or perhaps gradians if you are into obscure units). Hertz is a measure of frequency.The relationship between the two is that Hertz defines how many full 360° phase rotations are done per second. So if if your inverter is shifted by 25Hz it means it is producing either 25Hz or 75Hz