Hello,
In the Victron Quattro 48/10000/140-100/100 datasheet, it states that the inverter’s power factor range is ±0.8. Does this mean the Quattro cannot supply a resistive load with a power factor greater than 0.8?
Additionally, it mentions that the continuous output power at 25 °C is 10,000 VA and 8,000 W. Is this power specified as DC or AC?
If the load’s power factor exceeds 0.8, can the inverter supply a load requiring active power greater than 8,000 W?
thank you
Hi the powerfactor is related to the multiplus, not the load
Resistive loads have a power factor of 1
Only inductive and or capactive loads have a leed efficient power factor
Output power is AC
Your third question has nothing to do with each other, multiplus can deliver 8000W continue at 25 degrees, depending on your setup and use case (on/off grid) it can use grid and power assist from battery
Thank you for your response, but I still don’t understand what is meant by the power factor range of ±0.8. Could you please clarify?
do you think below statment is correct ?
Victron specifies 10 000 VA and 8 000 W for their inverters:
10 000 VA: Apparent power limit (total capacity the inverter can handle, including reactive and real power).
8 000 W*: Real power limit (active power delivered to the load at PF = 0.8).
This means:
At PF = 1 (resistive load), the inverter can provide up to 10 000 W.
At PF < 0.8 (inductive load), the real power drops to 8 000 W to protect the components from stress caused by reactive power.
No the multiplus itself has an power factor of 0.8
In short:
So apperant power is 10000w
Active power is 8000w
And 2000w is blindcurrent
If you connect a resistive device 1.0pf to the multi that draws 1000w, its 1000w real power
If you connect a inductive device whit 0.9pf to the multi that needs 1000w, its real power will be 1100w (about)
Theoretically if a connected device to the multi is 0.9pf and draws a total of 9000w (1000w more than 8000w multi) the other 1000w comes from grid
The multi can however supply loads with more than 8000w for a short moment (its in the specs)
Honestly i would not worry to much about this all…
Your multi can supply 8000w at 25 degrees for a long time, so keep your loads under that (if you dont have grid support)
In off-grid mode, the power factor is a characteristic of the load.
In on-grid (ESS) mode, the power factor is a characteristic of the inverter injecting power into the grid. The grid operator might have special requirements regarding the power factor when injecting power into the grid.
Imagine a diesel generator with a diesel engine that provides 8 kW mechanical power and an alternator that can pass 44 amps through its winding. This generator would be capable of providing up to 8000W and up to 10000 VA at 230V.
The active power (W) is limited by the mechanical engine power.
The apparent power (VA) is limited by the alternator winding.
In off-grid mode, the Quattro can supply loads with any power factor, as long as the active power does not exceed 8 kW and the apparent power does not exceed 10 kVA.
Exactly that!
There is one more possible situation.
The Quattro could be in AC charger mode. It draws AC power from the grid to charge the battery with DC current. In this case, the Quattro is a load for the grid and, as with any AC load, it also has a power factor, depending on its internal electronics.
PS: power factor is NOT efficiency!
You are right about its not efficiency, i was trying to explain 
Thank you for your response. I acknowledge that I am having some difficulty fully understanding the implications of a power factor of 0.8, especially since the inverter is theoretically capable of powering any type of load, regardless of its power factor.
Let’s take the example of a Quattro inverter with the following characteristics: 48/10000/140-100/100.
To better visualize the load power ranges that this inverter can handle continuously, I have drawn a P−Q diagram ( see attached)
I used the following formula to plot the curve: Q = √(10² - P²) with P ≤ 8 kW.
1- Can you confirm if this P−Q curve, representing the loads that the inverter can handle continuously, is correct? If not, please feel free to correct it.
2- Can you explain the implications of the inverter having a power factor (PF) of 0.8 in offgrid mode ? Does this mean that if the inverter is connected to a purely resistive load with an active power of 8 kW (which, in this case, would also be equal to 8 kVA, because PF = 1), the power drawn from the battery, on the DC side, will be 8 kw/(0.8*0.96) = 10 .41 kVA?
comment: 0.96 stands for effeciency of the inverter.
Thank you so much for your assistance.
in the datasheet for the Victron Quattro 48/10000/140-100/100, the AC output current is listed as 37 A with a voltage of 230 V. However, when I calculate the apparent power using these two values, I get S=U×I=230×37=8510 VA, whereas the datasheet specifies 10,000 VA.
Could you please help me understand this apparent discrepancy?
Thank you in advance for your insights.
In off-grid mode, the PF is set by the load.
Technically, it does not make sense to talk about “inverter PF 0.8” in off-grid mode.
The power drawn on the DC side depends only on the active power and inverter losses (efficiency).
I’m sorry but I cannot help you. Technically, it does not make sense.
does your response mean that their is a mistake in the datasheet ?
No, it means the data is for electrical engineers or equally educated. Without a profound knowledge of electricity you’re not qualified to work on this equipment, endangering yourself and others.
Sorry
True power = Apparent power x Power factor
There is a footnote in the datasheet that specifies a non-linear load for that specification.
So we can say a power factor of 1.2
10,000 = 8500 x 1.2
Thank you for your response. I agree with the equation: Active Power = Apparent Power × Power Factor. However, I would like to point out that the power factor cannot exceed 1, as it represents the ratio of real power to apparent power. A power factor of 1.2 is therefore not physically possible.
Could you please reconsider your response in light of this clarification?
This is being expressed as a ratio rather the actual power factor.
(A switch between electrical electrical engineering maths to basic math.)
Yes, but at the same time, the following equation must also be true:
Apparent Power = Current x Voltage
And if Voltage is 230V and Current is max 37A, then apparent power cannot be 10,000 VA.
In off-grid mode, the ratio between the max. active power to the max. apparent power is not a power factor.
In off-grid mode, the power factor is set by the load.
The Quattro can supply any load at any power factor as long as the apparent power is max 10 kVA AND the active power is max 8 kW.
The Quattro is responsible for the power factor in one of the following cases:
A. When charging the battery from the grid or generator AC input (the Quattro is a load on the grid or generator).
B. When exporting power from the battery into the grid with ESS.
Typically a PF=1 is used, but the DNO may require grid-tied inverters to run at a slightly lower inductive power factor to support the capacitive reactive loads from the grid that would otherwise lead to increased grid voltage.
The inverter synchronises to the incoming wave form.
The grid code chosen governs the rest. That is what the grid code aproval process/certification tests.
When weak ac is checked in offgrid with generator applications it changes how the power is drawn.
The grid code gives you some default settings, that were probably the official DNO requirements at the time of the certification. But the DNO might update the requirements over time, so you can adjust some of them from VE.Configure.
That is what i am saying. The grid code used affects the way the inverter works.
It can be altered.
The inverters charge at PF1. They synchronise to gird and their respose can be programmed.
