Understanding c rat loading

forgive me if this has been covered somewhere but Im having trouble working out what c rate of load Im putting on my batteries.
The reason Im asking about this is because in the controls of my inverter there are cut off voltages depending on cyber C load rate.

I understand that .2c is light load and as the number get higher this is heavier loads.

So Im trying to work out what voltage to put beside each C rating for the cut off.

Its a 48 volt system I understand what the setting is for just not what load equates to each C rate.

Any help in this is much appreciated

Voltage being compared to C rating isn't actually put together in this application.
The C rating is how many Amps you are charging/discharging. C rating is done based on the capacity of the cell (hope I'm getting this right). So, if you had a 2200mAh cell, then 1C would 2.2Amps. Most laptop cells are rated around 1C. High drain cells can be as much as 10C or higher. So, a high drain of 2200mAh would 22Amp from one cell.

So, you build you battery for the voltage you require. Then you build the parallel setup for the capacity you need. If you use cells that are rated at 1C and the cells are 2200mAh, then that pack theoretically could handle 2.2Amps / cell. So, if you had 20 cells in a pack, then you could do 2.2 * 20 = 44Amps.

But most builders build their packs so the cells handle around .2 - .5C. This allows for variations of C ratings on the cells. You can mix 1C, 2C, .7C, etc all together and stay safe. Note, this is not the only reason for building these in this manner.

I see but that does not really answer my question. If you have one cell and measure its voltage then put a load on it its voltage will drop. If you then remove the load its voltage will rise again instantly. Now if you then put a heavier load on the cell its voltage under this load will be less than its voltage when it was under the lighter load.

Now my battery system is 48 volt and Im running a 14s with 36 cells in each pack. Im running my packs at 4.1 volt when fully charged so the Victron multi switches to absorption for 1 hour at 57 volts then to float after that and stops charging completely at 57.4 volts.
As well as the 5 kWh battery I built my self I also have 2 other bought batteries that came with the original system when I bought it all earlier this year and those batteries are both rated at 35amp continuous draw so at 100 amp discharge which is the max that the Victron inverter can pull each battery should never see more than 33.33 amp discharge rate.

Now to the point. ........ the Victron multi needs to know what voltage the batteries are fully discharged
So it can tell when to turn off the inverter and start pulling from the grid again. It does this depending on what voltage it sees from the batteries and what load the batteries are under. If the load is very light so .2 c it will shut off the inverter at 42volts if the load is higher it will alow the voltage to drop lower than 42volts before switching off.
So I needed to understand what amp of current was .2c .25c 1c 2c and 5c so I could set up the voltage cut off settings

Ohhh, I didn't get that by your first post.

Figure it this way, what is your lowest you want your cell to be on discharge? 3.4, 3.2, 3.0, 2.8V???? Most stop at 3.4. Now, this is the voltage under load, not the recovery bounce back voltage. It is quite possible that the batteries can bounce back and the victron turn them back on again starting a cycling issue. You keep this from happening by putting the kick-in voltage higher till it can't do this.

So, if you have 3.4 as the kick-out voltage, then 3.4 * 14 = 47.6V. If you want 3.2V, then that'd be 44.8V. However, there isn't a whole lotta life left in a cell beyond 3.2 to go any lower. And there's not much between 3.4 and 3.2, either. It tapers off rather quickly.

You need to check ur voltage drop on max load. We have no clue of that. I have my low volt set at 3v. No use going higher as last resort. My inverter have software that changed based on weather. It can be all from 20 to 50% soc and then batrium does end cut at 3v

Perhaps not the answer you wanted

I have my alow inverter to restart set at 4 volt above the cut off voltage, as I dont see much point in salowing the inverter to start drawing from the batteries again until there is a useable amount of charge built up. I have the cut off at the lowest c rating set to 3 volt per cell so 42 volts. Now if I have the next higher c rating cut off set to the same 42 volts and at the point of cut off 42 volts is reached the voltage of the battery will bounce back much higher than if the cutoff wa reached when at a light load so my thinking is that the batteries are not flat.
The higher the load when the cutoff is reached the higher the voltage the batteries will bounce back to.

Ive seen some where some one had the high load cut off set as low a 2.6 volt saying that the voltage returned to 3.1 volt after cut off so that constituted that the battery was at the point where he concidered it was flat.
But he shows no evidence or correct findings to prove he was correct. Thats why I want to understand the c rating and know what load equates to each c rating

You don't want to do that. If your battery collapses under load to a very low voltage like 2.6V then either your are putting too much load on a battery that isn't suitable for this load or the cells are worn out and can't sustain this current anymore. Lowering the cutoff voltage isn't the solution. If this is a problem for your application you have to change the battery or make it bigger. Choosing cutoff voltages based on load is a bad idea. The cells will automatically have a longer runtime with a light load compared to a heavy load with the same cutoff voltage.

So what should I set the cutoff voltages to for each c rating in my settings. All the same for all loading or what ????
The cells Im using are brand new and are these
LG INR18650-MG1 2850mAh - 10A

Yes, if this feature can't be disabled then I would set them all to the same voltage!

Most stop at 3.4. Now, this is the voltage under load, not the recovery bounce back voltage. It is quite possible that the batteries can bounce back and the victron turn them back on again starting a cycling issue. You keep this from happening by putting the kick-in voltage higher till it can't do this.

Click to expand...

As I mentioned here (with the exception that it's probably 3.2V cutoff, not 3.4V; or as daromer said he's 3V even), you set your kick-in voltage higher above the bounce back of the batteries.
If your cutoff is 3.2V, and the cell bounces back to 3.5V, then set your cut-in voltage at 3.6V. That way it won't kick in until the batteries are starting to be charged. You will need to find out where that bounce back voltage is at. Being new cells, they will probably perform uniformly across the pack without much, if any, deviation from cell to cell.
So test a few cells to find out what the bounce backs are under different loads. Do a .5C, 1C, 1.5C, 2C. But always start with a freshly charged cell and let it deplete down to your cutoff voltage. Once the cutoff voltage is reached, disconnect the load and see where the cell bounces to. Once you do this with the various loads, you'll get to know how your whole string will perform under different conditions. Then you can set your parameters on your inverter and charge controllers.

Thats kind of what I was thinking but I still need to know the answer to my original question.
What load is .25c
What load is .5c
What load is 1c
There must be a way to work out what each c load is depending on voltage and capacity or c loadings must be a set amount of amp draw ??
And there must be a reason why Victron put in the different c loading cut offs. And this must be because under greater loads the battery voltage will drop lower before its flat.

Surly a cell is deemed to be discharged when its voltage reads 3 volt at zero load ??

0.25C is your total capacity *0.25

If you got 20Ah then its 20*0.25 = 5A

1C = 20A if you got 20Ah

And since voltage drops vary depending on the load. Call it C or call it A it doesnt really matter in real life. Victron states C since they dont know how big battery bank you got. The C load compensation is just so you in worst case can go a little bit further. Especially on systems where you have undersized wires and such where you can easily see 1-3V voltage drop on max load. Thats my 5 cents.

Oh, yes, I thought this was clear by now. A C-rating is always related to your capacity, it is the complete discharge of a battery over a certain time.

Right. There is a setting so that you can tell the Victron program your battery size in Ah I know that the two batteries that I got with the system are 50 amp hour batteries each so now if I work out the amp hour of the battery I built and add that to the other batteries at least that will be one setting thats correct.

Then I can work on getting the C loading cut off set up. I guess I will have to do it by trial and error.
Perhaps if I set the lowest load cut off to 42 volt which should be 3 volt on each single pack then see what voltage the batteries bounce back to after they cut off at that light load I could then play with the other Cut offs so that when it cuts off the voltage bounces back to the same voltage as it does when on the very light load ????

That is probably not possible. The idle voltage or resting voltage is the lowest after a low current discharge. To get back to this, very low, voltage you have to discharge the cells far beyond the reasonable threshold.

Example: 14S battery, 42.0V empty, 51.8V nominal, 58.8V fully charged
Now imagine the battery comes fresh off the charger. Since you have new cells lets assume they keep their end of charge voltage of 4.2V so the battery voltage is still 58.8V. You have 36 of the MG1 in parallel so the maximum continuous discharge for your battery is 360A and the total capacity is 102.6Ah.

0.2C: 20.52A
0.25C: 25.65A
1C: 102.6A
2C: 205.2A
5C: 513.0A

One thing becomes obvious immediately, your battery can't to a 5C discharge. 2C is a normal to high load, 1C is a normal load, 0.2C is a low load.
Now you put a load of 20.52A on the battery. That's 570mA per cell. Their voltage will now drop, because of the load, from 4.20V to 4.10V immediately. They will run like this until the cutoff voltage is reached and the cells will bounce back to their idle voltage. Let's say we cut off a 3.00V and they return to their idle voltage of 3.10V.
If you would put a high load of 205.2A on the battery instead then we have 5.7A per cell. Their voltage will drop not to 4.1V but 3.9V instead and run through the discharge from there on. They will hit the cutoff voltage sooner and then get back to 3.3V idle.

Any load works like an offset on the cell voltage. And it is not necessarily the same on the high end and on the low end, this is just an example. To reach the same idle voltage after a low current discharge and a high current discharge you would have to discharge further on the high current discharge because the offset is bigger. But this is not possible because the offset is bigger. It is a bit of a causality dilemma like chicken or egg. This is the reason why cells don't have a capacity as such, but a capacity at a given discharge current. The low voltage cutoff is always the same but the offset is bigger on the bigger loads, so the capacity is lower on bigger loads than on small ones.

Yeah, what he said

The voltage drop is generally alot higher at high and low voltage comparing to the middle of what I have seen here. Perhaps depends on cells.

Dont forget that at high load you will see the drop in wires, contacts and everything between as well.

Right I think I see now. Or at least I starting to see.
Thanks for the hard figures dark raven. That really helps.
There are a couple of other things to take into account here. One is that my victron multi can only draw a max continuous current of 100 amps so it looks like I can only draw about 1c max any way. I think the higher c loadings are for when you have more than 1 multi in parallel as its possible to connect up to 6 of these together.
The other bit to concider is that as well as the 5 kWh battery I have built with the brand new cells I have the other two 2.5 kw batteries that came with the system and each of those Is rated for a 50 amp hour constant draw. So if Im right here even at my maximum draw of 100 amps each of the three batteries should only see 33.33 amp hour draw so at my maximum draw the battery I built should never see more than around .3c load. Correct me if Im wrong ????

Not quite, it doesn't work like that. Do you have these three batteries connected in parallel? They will share the current and not necessarily even. Will be more like 2:1:1 in this case.
So yes, it will never see the full current, but how much can vary on the circumstances. But it probably doesn't matter anyway, at least not up to the last amp. Lets assume 50A, so about 0.5C is the maximum.

Yess all three batteries are in parallel as its a 48 volt system.
Each battery has its own pair of cables connecting to a single pair of solid copper bars with 8 mm holes in them so that I have plenty of room for adding more batteries in the future. Then there is a single pair of heavy duty cables connecting the whole lot to the victron multi.