question

kenzo avatar image
kenzo asked

AC coupling, factor 1.0 rule and peak load in a 3-phase system

I'm running a Kostal Piko 20 which delivers a max of 18kW in summer; since it's a 3-phase converter, I've started looking at running three MultiPlus II 5000 accordingly. However taking the factor 1.0 rule into account, three would not be enough in theory, since the system may deliver 18kW while those would only cover 15kW. To quote the guide,

(...) At that moment the PV inverter will continue operating at full power until the AC frequency has been increased. Increasing this frequency will take a very short time, but during that time all power will be directed into the batteries as there is no other place for it to go.

According to the data sheet, one MultiPlus II 5000 however can deliver 9000W peak power; am I understanding this right that if

  • the battery isn't fully loaded and
  • it can absorb 18kW of power and
  • the PV inverter "immediately" lowers its power output once the MultiPlus starts increasing the frequency,

I should be fine with such a setup, since each inverter would have to handle "only" 6kW of peak power? If not, what am I missing?
The data sheet unfortunately does not mention the time frame the inverter can handle the peak load of 9kW for; has there any testing been conducted with Kostal or does someone know that this is a setup that works in practice?

Thanks!
J.

AC PV Coupling3 phasepeak
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2 Answers
tnt369 avatar image
tnt369 answered ·

The charging power of the multiplus is max. 70A, so it depends on the battery voltage. With 48V battery the multiplus can charge 3.3kW (47V x 70A). At higher SOC a bit more. To handle 18kW you will need at least 6 units. 2 on each phase.

The continous output power is a little more than 4kW. The peak power is only available for a very short time. I.e. to start a motor or so. Don´t use it for your calculation.

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kenzo avatar image
kenzo answered ·

Thanks for your response!
That means I should be able to do the same setup with 3 x MP 10000 instead (140 A x 47V = 6580, * 3 = 19740). Is there a better/recommended way of doing this instead?

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