Multiplus-ii with Pylontech charging efficiency

Question

question

Graham Walmsley asked Dec 9, '23

Multiplus-ii with Pylontech charging efficiency

Hi,

Running a Multiplus-ii with 12 x Pylontech US3000C batteries. When watching the numbers on the Cerbo GX screen or VRM I seem to have a lot more losses than I was expecting based on the expected efficiency.

For example, taking the screenshot above, there is 6744W coming from the grid, 626W of load and 5465W charging the battery. That means 653W is being lost (9.6%) which is a lot higher than I expected.

The only DC loads are the inverter, GX and two MPPT 250/85.

Is this expected or do I have an issue?

Thanks

Multiplus-II

img-4076.jpeg (386.0 KiB)

2 |3000

Attachments: Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

Modifications Space

Node Red Space

Answer 1 · 2023-12-09T16:44:14Z

dirk-s answered · Dec 09, 2023 at 04:44 PM

Please search in this community for efficiency curves of Multiplus 5000 and 3000. Peak efficiency for the charger is nearly 93% and at the middle of the charging power. Under full charging power it decrease below 90%. You did not write which multiplus you use. But the values seem to be nearly the same and totally ok.

Inverter efficiency is a little bit better.

3

2 |3000

Attachments: Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

Graham Walmsley commented · Dec 09, 2023 at 07:55 PM

Sorry, missed out the model. It is a 48/8000 230v.

I have searched and I can’t find an efficiency curve for the 48/8000, if anyone has a link I would appreciate it.

0 ·

Craig Chamberlain Graham Walmsley commented · Dec 09, 2023 at 11:01 PM

Here are the 5kva efficiency curves - you can probably extrapolate from this to the 8kva model. During charging from grid, I try to reduce my charge current as much as possible while still getting the required charge within my off-peak tariff period. That typically means a 48A DC charge current which is around 2.5kW AC input and 92% efficient. I can't control discharge current so easily but most of the day is spent in the 500-1000W range so 95-96% efficient. If you combine these you get 0.92 * 0.95 = 0.87 (or 87%) but that doesn't take account of battery module efficiency. I can't find efficiency figures for Pylontech LFP batteries but this Victron article states 92% as the round trip efficiency ( https://www.victronenergy.com/upload/documents/Datasheet-BMS-12-200-EN.pdf )

So if you do 0.87 (from before) * 0.92 you get 0.80 (or 80%). That's pretty much what I expect and see from my system (Multiplus-II 5kva + Pylontech US5000 x 3). It does get a bit better in summer when I charge from my MPPTs which have a higher efficiency and are sourcing free energy from the sun, but no surprise there.

HTH

Just to add, even though 80% might seem low, here in the UK we have peak electricity prices which are ~4x off-peak prices so even a 20% efficiency loss isn't a big problem.

1 ·

28639-multiplus-ii-48v-5kva-charge-efficiency.jpg (184.5 KiB)

Graham Walmsley Craig Chamberlain commented · Dec 10, 2023 at 10:51 AM

Thanks for the detailed explanation. Checking my maths I got 89.5% efficiency when charging from 20%-97% so that all seems good.

I agree about the electricity prices in the UK. One of my goals is to do time shifting to make best use of the economy 7 tariff I have.

0 ·

Answer 2 · 2023-12-09T17:16:28Z

Alexandra answered · Dec 09, 2023 at 05:16 PM

The missing power is usually heat.

Pylons being lower voltage also have higher conversion losses.

And yes the system also uses power. It's efficiency is affected by heat and load.

All in all 10% or around there is not too bad.

3

2 |3000

Attachments: Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

Graham Walmsley commented · Dec 09, 2023 at 08:03 PM

Thanks for the response. Charging has finished (at 97%) and I have been working out the losses. Based on what has been pulled from the grid (39.9kwh), used for AC loads during the charging (2.9kwh) and has gone into the battery (20%-97% - 32.8kwh) the loss is 4.2kwh or 10.5%.

The pylontech batters are 48V so I expected it to be at the ‘better’ end of the efficiency scale.

0 ·

Alexandra ♦ Graham Walmsley commented · Dec 10, 2023 at 12:32 PM

Yeah their charged voltages are 52v or around there alot of batteries 55v and 56v so more efficient.

At 97% they aren't fully charged. If they are new the first bunch are charged are worse as they get their balance right.

Depending how you have stacked and wired them as well will affect their efficiency. Particularly if you have a long DC bus.

Most of the Pylon systems with bigger banks are just under 80% efficiency including balancing heat losses and inverter efficiency

0 ·

Graham Walmsley Alexandra ♦ commented · Dec 10, 2023 at 01:07 PM

Thanks for the reply.

There are 12 US3000C modules wired in pairs to Lynx Distrutors. Max run is 3M and voltage losses appear to be less than 0.1V overall to the MP2.

I think the issue is more around the numbers being sent to the GX from the MP2 and Pylontech BMS not lining up. The BMS seems to think it is putting out 2 amps more than the MP2 says it is receiving with the same DC voltage.

With no DC loads other than the MP2 and GX/MPPT there must be an error in one number or the other.

Related to this, the AC out Volts/Amps/Watts from the MP2 just don’t add up, but that may be down to power factor.

0 ·

Answer 3 · 2023-12-09T20:11:24Z

Graham Walmsley answered · Dec 09, 2023 at 08:11 PM

Just looking at consumption from the Battery (AC is disabled using Ignore AC) and it has 479W going to AC Loads but pulling 622W from the battery, so only 77% efficient, significantly below the curve. The batteries are at 25C and the room with the Multiplus is 9C so won’t be hot.

Any ideas?

2

2 |3000

Attachments: Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

assyntdavid commented · Dec 09, 2023 at 09:42 PM

Right now my system, disconnected from the grid, indicates that the loads total 355W whilst drawing 330W from the batteries. In other words the system is creating energy from nowhere, or - more likely - there is an uncertainty associated with each reading, especially the current measurement. Another thing to factor in.

1 ·

Graham Walmsley assyntdavid commented · Dec 09, 2023 at 09:44 PM

That's the kind of efficiency I like :-)

0 ·

Answer 4 · 2023-12-10T11:17:07Z

Graham Walmsley answered · Dec 10, 2023 at 11:17 AM

I think I am getting closer to what is going on. If I compare the DC in figures to the AC out figures on the Multiplus I am seeing 93% efficiency which is about what you would expect, but as you can see below the DC amps in figure on the Multiplus doesn’t line up with the DC amps out on the Pylontech data.

I’m not getting a voltage drop (short < 3m cable run) and nothing else is on the DC side (nothing getting hot either) so I am starting to think the readings from the Pylontech BMS is wrong.

Thoughts?

img-4077.jpeg (405.8 KiB)

1

2 |3000

Attachments: Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

Craig Chamberlain commented · Dec 10, 2023 at 09:37 PM

Given that you have 12 US3000C modules in parallel, I wonder how dependable the -12.80A figure is. Presumably it is calculated by adding up all the currents from the 12 modules, but if that's the case then there will inevitably be some lag while the individual modules communicate their data to the master module and then it will send that to the Cerbo. That shouldn't matter as long as the system is in steady state conditions so if you are doing these measurements then try to ensure a nice stable AC load.

That said, the two currents really should be the same because they are being measured within a series circuit. Any losses within the battery modules themselves will manifest as a loss of CAPACITY (Volts x Amps x time) not loss of current. Any losses due to cabling and connections will manifest as a loss of voltage at the inverter compared to the battery.

You'd really need to have a current clamp meter to measure true instantaneous current.

0 ·

question