I am proposing to install a Multiplus II 48/5000va GX in my home.
Some of my loads have poor power factor eg freezer PF=0.5
If I add capacitors to the compressor will this reduce the current drawn from the battery when running off grid.
Thanks
Bob
The starting torque is always the decisive factor for refrigerators and freezers. After that, everything is no problem. I wouldnât worry about this with a 5000 Multi. The power is completely sufficient.
Nothing can be improved here. Capacitors or extension cables as resistors wonât help here.
So no worries with a 5000 
I think you missunderstood the question.
BoB is refering to a bad Powerfactor of some consumers. If you imagine a consumer, requiring 230W in a 230V system, a PowerFactor of 1 would result in 1A on the AC Side.
If said consumer has a powerfactor of 0.5, the MP would need to provide 2A to deliver 230W of power.
In Grid-Situations, private people are not charged for reactive power - however in Off-Grid situations, the power has to come from âsomewhereâ.
So, the question is, if a bad powerfactor would result in a higher DC-Discharge, and if improving the power factor would therefore help to increase overall battery duration.
I understand this question, because I thought about this as well, but answering it is beyond my knowledge in that area.
@dognose Spot On.
As you say the VA has got to come from somewhere but I canât quite get my head round whether the battery current depends on PF of the load.
If I had the hardware here I could connect it up and make measurements but Iâve not pressed the trigger on the order yet.
Bob
Reactive currents are cyclic, and theoretically wont normally add to the DC current the inverter consumes. Except that the poor pf causes an increase in resistive losses - in the AC cables, compensation capacitors and inductive windings. These losses DO add to the dc current - however in a domestic situation, they are normally quite small. One off grid site I worked on had a particularly bad power factor - about 0.68, and adding a power factor correction unit did lead to an increase in generator efficiency. The thing to do is measure the dc current while switching the correction capacitor into and out of circuit.
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I am posting this to check if my understanding is correct. We live off-grid and I am interested in really understanding battery consumption.
The graphs are from Home Assistant.
Peak #1 was heating water on the electric kitchen hob. Load is purely resistive, therefore apparent load = load.
Peak #2 was house water pump jumping on. Load is inductive, therefore apparent load has reactive component.
@dognose So my batteries delivered the shown 972 W, correct?
@MikeD I am aware that this is not the measurement that you proposed, because the peak is just a lucky observation and not a proper integral to determine the energy consumed (in kWh) required to calculate the power factor.
The Victron MQTT broker has sensor.victron_mqtt_c06xxxx3d76f_vebus_276_vebus_energy_inverter_to_ac_out
Wouldnât it be cool to have the same for reactive power (and apparent power)?
I guess thatâs what solidstore (sorry, new users can only mention 2 users, LOL) meant in VRM only show half AC output - #10 by solidstore
Excited to read your comments!
Was trying to tag @solidstore before. 
Thatâs the key: bad pf wonât hurt too much at the lower end of the inverters power capabilities but will bite you in the b#tt when running them at full throttle. Capacitor banks can help but Iâd be surprised if that would be more economical then sizing up the inverter itself. Not that you seem to need that, if those graphs are representative you still have a lot of headroom.
TLDR: No, in your situation improved pf will not significantly reduce battery current / improve DC to AC efficiency. It will help keep things cooler a bit. 10% heat loss versus 20% heat loss is âonlyâ 10% difference in current but double the heat.
@ricardocello pointed me to correct MQTT topic for power in/out Multiplus in ac coupled ESS setup - VictronEnergy - thanks!
Looks like we have apparent power already.
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Electric motors only give a really poor power factor when running on light / no load. When running under load, the power factor tends to be pretty good -like for a water pump. Power factor needs a proper meter to measure, as it relates to the phase between the voltage and current wave forms. Electric motors also have a strong start up surge, that most meters donât really capture. If you want a power meter that can measure power factor and load peaks, have a look at the EKM range of meters.