High Apparent Power and Poor Battery Performance

[NickBester]

We have the following off-grid setup for our small Airbnb in a remote location:

4x JA Solar 365W Panels
1x Kodak OG-Plus 5.48 Inverter
2x Pylontech UP5000 Batteries

We should have a very low power usage (LEFs, energy efficient fridge, no stove, gas geyser (battery operated), etc) however we use approximately 300W/h or about 7.2kW/day.

We are trying to figure out why we have such a high usage as we use approximately 40% of our batteries overnight (sunset to sunrise) which just doesn’t seem correct.

When monitoring the system on watch power, the apparent power is very high compared to the actual power, as shown below as an example reading:

AC Apparent Power: 207VA
AC Actual Power: 40W
Load Percentage: 4%
 

Any idea as to what the issue may be?

[zsde]

Hello Nick,
Not to forget that your Inverter's internal power usage is not part of what you see as actual power or load. That in itself is typically around ~45 - 60W.
When I disconnect my Solar system I can monitor the battery percentage usage per hour and also on the graph I can see the draw from the battery, even though there is zero actual power draw. Pics attached
Do you have any battery chargers for example for gate motor or alarm system that regularly recharges the 12V batteries?

One way of testing your systems internal battery usage is to disconnect all loads for one night or for a couple of hours if you can, and then check what your SOC is after the no load period. 

Edited July 5, 2022 by zsde

[Bobster.]

1 hour ago, NickBester said:

We should have a very low power usage (LEFs, energy efficient fridge, no stove, gas geyser (battery operated), etc) however we use approximately 300W/h or about 7.2kW/day.

That's not that a large figure. Is there anything else drawing from the system? TV? Decoder? Wifi? Security system?

 

1 hour ago, NickBester said:

We are trying to figure out why we have such a high usage as we use approximately 40% of our batteries overnight (sunset to sunrise) which just doesn’t seem correct.

If you meant that overnight you go from 100% to 60% state of charge then, again, I'd say that's not so bad on 10 kw/h of batteries. My house uses nearly twice what yours does in total, but we do most of the "heavy lifting" during the day - pool pump, water heating, dishwasher etc - to make full use of PV. At night it's just TVs, wifi, security, fridges, some LED lights. Varies between 250 and 700 W (the latter is peak and short lived). We usually get down to 50% or so remaining SOC.
 

[WannabeSolarSparky]

3 hours ago, NickBester said:

4x JA Solar 365W Panels
1x Kodak OG-Plus 5.48 Inverter
2x Pylontech UP5000 Batteries

those 4 panels are never going to correctly/fully charge those 2 batteries, are you using the grid to topup charge keep full those batteries?
That would use a chunk of power on its own everyday.

[Scorp007]

17 hours ago, WannabeSolarSparky said:

those 4 panels are never going to correctly/fully charge those 2 batteries, are you using the grid to topup charge keep full those batteries?
That would use a chunk of power on its own everyday.

Luckily we don't need to charge lithiums full like lead acids. It does seem it might be better to keep in a 20-80% sweet spot. I have some laptop lithiums that are now 12yrs in use and hardly ever charged them full. It's in a torch which one normally wants to use so at times they just get a splash and dash charge.

[Coulomb]

21 hours ago, NickBester said:

When monitoring the system on watch power, the apparent power is very high compared to the actual power, as shown below as an example reading:

AC Apparent Power: 207VA
AC Actual Power: 40W

That's because of the large smoothing capacitor that is effectively in parallel with the inverter's output. It causes a large apparent power load (hundreds of VA) while consuming almost no power (it won't get very hot, despite all that current, because the phase of the current is nearly 90° out of phase with the voltage). In other words, a good capacitor has a very low power factor. Most of the power drawn will be imaginary, not real. This is normal and to be expected. When the load gets higher, the power the apparent power at the input and output will still be some 170 VA different because of that capacitor, but relative to the output, it will be much less significant. Most loads are unity power factor or inductive (e.g. motors), so this will actually slightly improve your whole house power factor. That doesn't affect you much, since you only pay for real power consumed, not imaginary power.

[0012]

21 hours ago, NickBester said:

Any idea as to what the issue may be?

Some LEDs have horriffic power factor. Do a test at night. Switch all the lights off and compare W and VA readings. Switch on all the lights that are normally on, and compare W and VA readings again. That could possibly tell you whether you went too cheap on the LEDs

A 5W LED bulb can draw as much as 40VA if it is cheap and nasty

[Coulomb]

5 minutes ago, 0012 said:

A 5W LED bulb can draw as much as 40VA if it is cheap and nasty

There are two type of low power factor: due to inductance or capacitance, or due to rectifiers. Cheap and nasty appliances just rectify and smooth AC, resulting in power being drawn in bursts only at the peaks of the AC waveform.

[Scorp007]

15 minutes ago, Coulomb said:

There are two type of low power factor: due to inductance or capacitance, or due to rectifiers. Cheap and nasty appliances just rectify and smooth AC, resulting in power being drawn in bursts only at the peaks of the AC waveform.

I might as well comment on your previous post here. A low power factor from a capacitor is not a function of its quality but of its size. The higher the value the lower the power factor. What was perhaps ommited in the previous good post was done here by indicating it is in the opposite direction to the induction. That is why in electrical terms we talk of a leading (caps) and lagging (induction) power factor. Unity is with no capacitance or inductance or both are changed by to other to bring voltage and current without an angle between them.

[NickBester]

There is no grid connection (it is completely off-grid). The battery therefore provides all the power required when no solar input. The power that battery will need to provide, is that the apparent or actual power?

[GreenFields]

The battery doesn't directly do apparent power, it is a DC source feeding the inverter. Apparent power is on AC side, where you have inductive or capacitive loads that don't consume real power, they just draw current and return it it to the grid. In this case however where there is no grid power, I would wonder whether the inverter is not cycling that power in and out of the battery and causing conversion losses with each cycle. No idea really, just a question that's bothering me now.

Other than that, possibly the MPPT backfeeding into the panels at night? Maybe try to disconnect the panels and see if the overnight usage drops.

[Coulomb]

21 hours ago, NickBester said:

There is no grid connection (it is completely off-grid).

Ah, my bad. So you have a very poor power factor load (40/207 x 100% = 19%). 

Quote

The power that battery will need to provide, is that the apparent or actual power?

Mostly actual power. However, the extra current required to support the poor power factor means extra losses in the inverter, though even 207 VA is pretty trivial for a 5 000 VA inverter.

On 2022/07/06 at 1:49 AM, NickBester said:

however we use approximately 300W/h or about 7.2kW/day.

I assume you mean 300 Wh/h, or 300 W. The idle consumption of the inverter will be of the order of 50 W; that becomes significant when the loads are very low. Where is this 300 W measured or calculated? As about 6 A from the battery? Your example from Watchpower shows 40 W, plus some 50 W of idle power, and a trivial loss from the 207 VA (roughly 6 W). That's less than 100 W total, if the 40 W of real power is typical. I imagine that the fridge will consume a lot more than 40 W when it's running, say 80 W and 100 VA.

[Scorp007]

4 hours ago, Coulomb said:

Ah, my bad. So you have a very poor power factor load (40/207 x 100% = 19%). 

Mostly actual power. However, the extra current required to support the poor power factor means extra losses in the inverter, though even 207 VA is pretty trivial for a 5 000 VA inverter.

I assume you mean 300 Wh/h, or 300 W. The idle consumption of the inverter will be of the order of 50 W; that becomes significant when the loads are very low. Where is this 300 W measured or calculated? As about 6 A from the battery? Your example from Watchpower shows 40 W, plus some 50 W of idle power, and a trivial loss from the 207 VA (roughly 6 W). That's less than 100 W total, if the 40 W of real power is typical. I imagine that the fridge will consume a lot more than 40 W when it's running, say 80 W and 100 VA.

Thanks @Coulomb for listing all the details. It does seem the data needs to be supplied where and how it was measured. Just looking at the power factor for the fridge at 0.19 is worse than I have ever seen on a motor inductive load and not sure if one can have such a bad power factor if one looks at the resistance vs inductance of a motor's windings.

I do suspect some readings are suspect and needs to be confirmed. I know my 3kw Axpert has no clue how to calculate Watts vs VA as it gives the same value while running 2 fridges and an electric blanket.

[Coulomb]

18 hours ago, Scorp007 said:

I know my 3kw Axpert has no clue how to calculate Watts vs VA

Some of the smaller models run very small 8-bit microcontrollers, and I suspect that they might skip some of the calculations on those. But the models powered by TI DSPs I've been impressed with: they really do calculate RMS power, as in the square root of the average (mean) of the square of the power, and the power is instantaneous voltage times current, measured some 19,000 times per second.