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Daft question / proving a point

Featured Replies

Hello :)

daft question but trying to see if I am actually right.

 

if we have a 5000wh battery (lithium) and we have a 1000watt AC load. Then the battery will last for 5hours ? Is this correct basic math. Sure we can take 15% off for losses but for now is this the correct way to see how long something will last ? 

34 minutes ago, Dylboy said:

Hello :)

daft question but trying to see if I am actually right.

 

if we have a 5000wh battery (lithium) and we have a 1000watt AC load. Then the battery will last for 5hours ? Is this correct basic math. Sure we can take 15% off for losses but for now is this the correct way to see how long something will last ? 

Correct mostly. I would take another 10% off to in fact only discharge 90% of said capacity.

8 hours ago, Dylboy said:

Hello :)

daft question but trying to see if I am actually right.

 

if we have a 5000wh battery (lithium) and we have a 1000watt AC load. Then the battery will last for 5hours ? Is this correct basic math. Sure we can take 15% off for losses but for now is this the correct way to see how long something will last ? 

That's broadly correct if you want to fully discharge the battery. The catch is that many batteries don't allow this, and it's seldom reccomended. 

Usually the BMS will shut the battery down when it is 90% discharged. So think of that 5kwh battery as effectively being a 4.5 kwh battery. 

  • Author

Awesome thanks gents.

Yes agreed with the 90% etc. 

 

This install does not have a BMS as the battery is not communicating with the inverter so it's ok school with voltages. 

Having the battery guy argue a bit with how to calculate run time. So just wanted to make sure I was not mistaken in my math too horribly.

I have the logs from Sunsynk showing the data and battery lasted 2hours at a discharge average of 2100 watts.. it's a 7 KWH. Should have lasted 3.3 to 100% so definitely should of made the 2hr30 of loadshedding before the batteries shutdown.

 

On 2022/11/03 at 8:08 AM, Dylboy said:

This install does not have a BMS as the battery is not communicating with the inverter so it's ok school with voltages.

 

Surely the battery has a BMS. It will still be taking care of most of the charge and discharge parameters even without communication.

 

On 2022/11/03 at 8:08 AM, Dylboy said:

Having the battery guy argue a bit with how to calculate run time. So just wanted to make sure I was not mistaken in my math too horribly.

you'd have to qualify your AC load, if its a purely resistive load, then yes, 1kW load would require the battery to give up that + the inverters 70 to 100W own consumption, but if the load is not purely resistive, then the 1kW load could really be munching 1.6kVA possibly... PF needs to be specified for the AC load...

On 2022/11/03 at 8:08 AM, Dylboy said:

This install does not have a BMS as the battery is not communicating with the inverter so it's ok school with voltages.

I'm sure there's a BMS. You just don't have comms between that BMS and your inverter. But there will be one, if only to protect the batteries against things like too high a draw or to low an DOD.

Edited by Bobster.
sppeling

8 minutes ago, Kalahari Meerkat said:

you'd have to qualify your AC load, if its a purely resistive load, then yes, 1kW load would require the battery to give up that + the inverters 70 to 100W own consumption, but if the load is not purely resistive, then the 1kW load could really be munching 1.6kVA possibly... PF needs to be specified for the AC load...

He says he's working from the logs supplied by the inverter. So what would they show? The theoretical load, or the actual load?

Surely if I have a 10kWh battery, and load as reported by the inverter is a steady 0.5kW, I can expect 20 hours out of the battery? OK... 18, because the BMS will shut the batteries down when DOD reaches 90%.

  • Author

Ye sorry it does have a BMS but no comms.

 

Going to build a CAN or 485 cable as I need to start learning that and the pin outs etc. 

But at the moment my concern is the battery is not lasting as calculated.

Battery discharge of 2100watts and it does like 2 mins before the 2 hour mark where battery voltage was at 48V and then died. 

7000w / 2100 should be 3.3... give in losses and not discharge we should have reached the 2hr:30 mark.

  • Author
16 hours ago, Kalahari Meerkat said:

you'd have to qualify your AC load, if its a purely resistive load, then yes, 1kW load would require the battery to give up that + the inverters 70 to 100W own consumption, but if the load is not purely resistive, then the 1kW load could really be munching 1.6kVA possibly... PF needs to be specified for the AC load...

Ahhh ok yes, he does ask what the loads are. There will be power factor for sure as not much is restive if anything. I will try get behind those calculations then and add in PF to estimate run time. 

 

Would 0.85 be an ok starting point ?

Not quite sure how to do the math yet but will try 

8 hours ago, Dylboy said:

Ye sorry it does have a BMS but no comms.

 

Going to build a CAN or 485 cable as I need to start learning that and the pin outs etc. 

But at the moment my concern is the battery is not lasting as calculated.

Battery discharge of 2100watts and it does like 2 mins before the 2 hour mark where battery voltage was at 48V and then died. 

7000w / 2100 should be 3.3... give in losses and not discharge we should have reached the 2hr:30 mark.

Sound like damaged cells or a large imbalance between highest and lowest voltage cells.

Have you left them for 24hours or more at full charge to see if it improves?

Edited by I84RiS

  • Author

Yes so I also seem to think balancing. I asked as to when they should be balanced and he stated 3 cycles, it did 3 according to the display but I could then see each cell was not all the same. I am not sure if tolerances but still. 

 

I did notice the following a day a bit better balance which is good. 

 

I also added in some Sonoff smart stuff to swich off fridges and it at least got the batteries to last the time gap of 2hr15 (they came back on early). 

Today in what ever gaps I had was reading about power factor and actually that we need to take that in when calculating run time of batteries. I didn't know this as thought the inverter took care of it and then we could work with direct watts but turns out we need to account for reactance.

 

So when I see 2100watts I need to add 0.80 which is 2625 and the the inverters own consumption of I believe 400watts (Sunsynk 5kw).

This then brings it to 3025. 

7000/3025 =2.3 but we can't use all 7000wh so 20% is 5600 so then 

5600/3025 =1.85... just shy of 2hr make which is what I got.

 

I assume the battery cut off at 20% and also it's a 7100wh but I use 7000wh.

 

Anyway all this is very interesting and now applying reactance for battery capacity is a new thing for me and intresting. So far it checks out to a degree. Also it's not always 0.80 powerfactor so that will also dictate more or less backup time.

 

But I must add I am still learning so if anyone can add to or dismiss anything please do so as then we can all learn :)

 

23 minutes ago, Dylboy said:

So when I see 2100watts I need to add 0.80 which is 2625 and the the inverters own consumption of I believe 400watts (Sunsynk 5kw).

Ok, slow down there, sunshine... the inverter should not self consume more than 100W, I've seen it vary from 70W on the minimum to probably just about 100W...

As for the PF, its a pity the inverter does not give us that figure based on the load, since then one could consider doing power factor correction, maybe one should consider this a project, do PF measurements and dynamic PF correction...

What you should base your calculations on is the actual current draw * the actual voltage of the battery at the time of current measurement and then you could work out how much energy the battery provided for the time period in question, of course both of those values will vary over the period you'd want to see what it was actually giving you.

As for unbalanced cells, top balancing is generally what one would want, I think, the problem is...
1. Does the BMS actually do cell balancing (some do, some (most?) don't, or if they do they do it poorly)
2. Under which conditions will it do cell balancing? You may need to Absorb at, let's say 55.4V and then Equalize at the same Voltage for 30 minutes or more to let the BMS do cell balancing and then Float at 54.5V or so thereafter.
3. The BMS will normally not care whether you're down to 20% charge remaining or 5%, usually it will disconnect the power, because one or more of the cells has reached a low Voltage threshold. If the cells are badly unbalanced then you may never see even close to the rated power out of the battery...

Edited by Kalahari Meerkat
speeling

  • Author

Ah if that's the case then great, I did think 400watts was massive.

I went to the logs in hope there was some power factor but not so I agree it would be great to see that. 

I will do so, didn't think to look at a certain time and then take that data to get power. Thank you.

 

Yes agreed as soon as one cell is low it shuts off. I want to try go back in the week to see how the balancing is going. I do not know how often or when it does it though. The voltages are set to what he stated. Besides that I want to get a CAN in as well to then allow better communication.

I did read a note in a REVOv document about setting the voltage settings in the AGM-V page and then selecting it back to Lithium and also doing the settings there ( only if you don't use a cable) I thought that was interesting and want to test that too however the battery does appear to be reaching the voltages set.

38 minutes ago, Dylboy said:

Yes so I also seem to think balancing. I asked as to when they should be balanced and he stated 3 cycles, it did 3 according to the display but I could then see each cell was not all the same. I am not sure if tolerances but still. 

 

I did notice the following a day a bit better balance which is good. 

 

I also added in some Sonoff smart stuff to swich off fridges and it at least got the batteries to last the time gap of 2hr15 (they came back on early). 

Today in what ever gaps I had was reading about power factor and actually that we need to take that in when calculating run time of batteries. I didn't know this as thought the inverter took care of it and then we could work with direct watts but turns out we need to account for reactance.

 

So when I see 2100watts I need to add 0.80 which is 2625 and the the inverters own consumption of I believe 400watts (Sunsynk 5kw).

This then brings it to 3025. 

7000/3025 =2.3 but we can't use all 7000wh so 20% is 5600 so then 

5600/3025 =1.85... just shy of 2hr make which is what I got.

 

I assume the battery cut off at 20% and also it's a 7100wh but I use 7000wh.

 

Anyway all this is very interesting and now applying reactance for battery capacity is a new thing for me and intresting. So far it checks out to a degree. Also it's not always 0.80 powerfactor so that will also dictate more or less backup time.

 

But I must add I am still learning so if anyone can add to or dismiss anything please do so as then we can all learn :)

 

Not easy to work out the running power factor. In this pic the measurement was taken from an old style chest freezer. Works out at a PF of 0.685. As this does not draw all the time and a geyser can draw 10 X the fridge value the overall PF will be much higher. A average in 24hrs could be 0.9 but can vary a lot from installation to installation based on the motor loads.

IMG_20221106_195100.thumb.jpg.514d67d7a190bca69c77ec3659212dd7.jpg

  • Author
11 minutes ago, P1000 said:

Power factor won't affect power from the battery that much, it might increase losses, but the battery only supplies the real power part.

That is what I originally thought but now I'm extra confused as then I thought power factor needed to be added to then see how long the battery will last, but if the battery only supplies true power then we only need to use the watts and not apply power factor. Basically them 5kwh battery and a 1kw load would then last 5 hours or close to( Assuming 100% discharge and some heat losses).

 

Unless you mean the 5kwh can supply the 5kw with or without the power factor of the load taken into account?

Edited by Dylboy

  • Author

In essence which is the correct method to more accurately estimate how long the battery can last. (Basic numbers)

Option A: no power factor used.

5kwh battery, 100% DoD, 
Load: 1000watt (i.e read on back of appliance) constant load.

Answer: 5000/1000 = 5 hours. 


Option B: power factor used 

Same as above (5kwh battery,100% DoD,
Load: 1000 watt(i.e read on back of appliance) constant load.
Assumed power factor of 0.90

Answer:

1000w / 0.90 = 1111 watts.

5000kwh / 1111 = 4.5 hours. 


Which is the better one to use, or rather which is the more correct one and if power factor must even be considered in calculating estimated time duration to sustain the load.

I also discounted what the inverter power needs are which is I believe at a max of 100watts.

Also temperature I know plays a role but how much I don't know but I want to understand the baby steps first.

Option A is correct, but you still need to account for efficiency losses. Sunsynk does not give the efficiency of it's battery to AC conversion and if they did it would be load dependent. I would work on between 90 and 95% + 70W self-use.

So option A is the correct method, but option B's answer is closer.

Keep in mind that you are unlikely to find a battery that actually allows you to go down to 0% SoC (those that do are usually 10% overspec).

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