mattwrightzen wrote: Hi, a friend has a 5kW and 8.2kW fronius primo inverter with 17kW of panels.
When he loses the grid due to blackout can he tie those inverters to the mpp pip4048
So presumably he also has a separate PIP system with batteries that he can switch the loads to.
The fronius supports a ramp down of its output as grid frequency rises from 51-52.7hz
As I indicate below, I don't believe that this will come into play with the PIP.
Does anyone know how the mpp 4048 will behave if there is grid feed in on the load side
I assume you mean what happens if he connects his grid tie inverter(s) to the PIP outputs, which have loads connected, during a blackout only. I don't think that there is a practical and safe way to connect the PIP output to the mains/utility.
My wild guess, expanded below, is
6 something else
Firstly, although I've worked on a research microgrid project, I was not there for the design, and took virtually no part in the design. I did have to think about power flows and AV vector diagrams a bit, so I have a little knowledge in this area.
I'm confident enough that I think we can take as facts:
* The PIP inverter makes no attempt to synchronise with the AC output terminals. In other words, it's not a grid interactive inverter.
* The PIP inverter always seems to be aiming for 230 VAC at the output. I'm assuming "battery mode", where the inverter hardware is configured as an actual inverter (as opposed to being configured as an AC charger, when in "line mode").
* When there is AC present at the AC input terminal, and if and when it passes some quality checks, the inverter synchronises to the AC input terminals. My guess is that this is to make easier the switching of the loads to the AC input (either by going to line mode, using its internal relay, or using an external relay or contactor).
I'm fairly confident of this point:
* When there is no AC present at the AC input terminals, the PIP inverter operates at 50.0 Hz (or 60 Hz if so configured), and makes no attempt to synchronise with anything else.
So now to my guess as to what will happen if a grid interactive inverter is connected to the PIP's output, when powering a load. For the sake of example, let's say the load is 2.0 kW, all supplied by the PIP from a battery. Let's say the grid interactive inverter has 3.0 kW of power to provide, but we'll imagine that it ramps up the power slowly (say 200 W per second), starting at zero power.
While there is less than 2.0 kW from the grid interactive inverter, it seems to me that the phase of the power to the loads will shift slightly (and no-one will notice that unless making very careful measurements), the load AC voltage will remain very close to 230 V, and the PIP will gradually supply less and less power to the loads. It won't find anything unexpected (ignoring the very real possibility of higher frequency spikes mucking up its load voltage and current measurements), it will just look like the load is gradually reducing.
At the point where the grid interactive inverter is providing 2.0 kW, all the power to the load will come from the grid interactive inverter, and the PIP will feel as though it has no load.
Of course, the interesting part is what happens next, as the grid interactive inverter attempts to provide more power to its output terminals.
The PIP is targetting 230 VAC at its output. So it looks like a low impedance 230 VAC source with an inductor (very likely the one we're talking about in other posts recently) between that AC source and its output terminals. With say 2.1 kW coming from the grid interactive inverter, and only 2.0 kW absorbed by the loads, the extra 0.1 kW will have to go somewhere. It seems to me that the load voltage will increase only slightly (due to voltage drop across the inductor), but it will only be a volt or two at most, much less while there is only 0.1 kW of excess power. But the PIP in targeting 230 VAC will experience a rise in the voltage across its output full bridge. In other words, the excess power will tend to charge the capacitor feeding the output full bridge. In other words, the power supply for the output MOSFETs or IGBTs that make up the full bridge making the 50 Hz 230 VAC. I'll refer to this capacitor (there may be more than one, but they would be equivalent to one) as the "output capacitor" for brevity. This is not a capacitor doing output filtering; its filtering the power supply for the full bridge.
The PIP has an unusual architecture, in that there is a battery side full bridge (50 V to 400 V) feeding a buck converter feeding the output full bridge. There is a "bus voltage" as reported by the QPIGS (general status query) command of the PIP likely refers to the voltage at the input to the buck converter. It always seems to be quite close to 8 times the battery voltage, so that means the battery side converter has a transformer with a 1:8 turns ratio.
Here is the important thing: the buck converter is unidirectional. The output full bridge, and the battery side full bridge are both bidirectional; they can push or pull power in either direction (to/from the battery, and in theory, to/from the load terminals). So I think what will happen is that the 100 W of excess power from the grid interactive inverter will very quickly charge up the capacitors at the output of the buck converter, and there will be nowhere else for that energy (over a few cycles) to go. The PIP monitors "bus voltage", but it seems to me that this could be on the input side of the buck converter. So maybe it won't notice, the output capacitor will go to 600 V, destroying the output bridge and the output capacitor. Or maybe it will notice, and will shut down the inverter. If it just stops switching the output full bridge, this won't help, as the free-wheeling diodes of those devices will still pump up the output capacitor. [ Edit: actually, but not very much. ] If it disconnects the output relay, the grid interactive inverter will cause the load AC voltage to rise. Probably the load voltage will quickly exceed a limit and they will trip off. Perhaps the PIP will switch to line mode, which will at least protect the inverter hardware, but in a blackout situation, there will be nothing to power the loads other than the grid interactive inverter. In all these scenarios, the loads get too much voltage or no power at all.
So unfortunately, I can't see a way that this will work, unless there is a "zero export" system in place, so the grid interactive inverter never provides too much power for the load. Those things are always a bit tricky, and tend to under or overshoot with sudden load changes, so the zero export system would have to be very good not to cause overvoltage of the PIP's output capacitor.
Finally, the anti-islanding provisions of the grid interactive inverters would have to regard the PIP's output as being the presence of a grid. I hear rules of thumb like "the grid interactive inverter has to be no higher in rated power than that of the stand alone inverter", but I'm dubious about this.
When trying to find a diagram to illustrate the topology, I realise I've misremembered how the buck converter works. It seems that it's only active in battery charge mode (line mode). In battery (inverter) mode, it's effectively just an inductor. So that affects my answer; see my next post.
[ Edit: minor rewording, energy -> power, battery side inverter -> battery side converter, etc. ]
[ Edit: integral diode -> free-wheeling diode ]
[Moderator note: Coulomb's follow-up post with topology diagrams
has been moved to the PIP repairs and hardware modifications topic.]
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