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Lithium batts and Ah counters


Col_T

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2 minutes ago, MoominPapa said:

Below are three graphs, First is Soc (%) for the last four days, battery current, and battery voltage: you can see the Dr Bob is right.

 

Second is the same, but with an estimate of open-circuit voltage by linear regression, rather than actual terminal voltage.

 

Third show the discharge in Ah by Ah counting, an estimate based on voltage, and the best estimate output from a Kalman filter which is using both the other sources of data,

 

 

Raw-voltage.png

OCV.png

est-SOC.png

Dont start bringing graphs into it MP, Bob you, and I know we are right ? My gauge starts going out of sync very quickly so I have just ignored it for the last six months and worked on the voltage

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28 minutes ago, peterboat said:

Dont start bringing graphs into it MP, Bob you, and I know we are right ? My gauge starts going out of sync very quickly so I have just ignored it for the last six months and worked on the voltage

But I like my Kalman filter :)

 

MP.

 

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1 hour ago, peterboat said:

Dont start bringing graphs into it MP, Bob you, and I know we are right ? My gauge starts going out of sync very quickly so I have just ignored it for the last six months and worked on the voltage

Ah, but now I've not got anything to do looking after my lead acids, I need something to faff around with.

Here's my graph for May-June.

I use a rasp pie to log all BMV data, a reading ever 5 secs or so. I filtered the data to show all records where Amps were zero. The points under the red ring are outliers ie they are too high in voltage as if the solar is charging at 10, 15 or 20A then goes into 'float', the current drops to zero but the voltage takes 5 mins or so to decay to the 'at rest' value so a trail of higher voltage points. The light blue points were data from August overlaid and fit perfectly.

Note my system has 660Ahrs of  lead acids in the hybrid set up so voltage rarely gets below 12.7A.

Whilst you would think you can use voltage more easily with LA's to predict SoC, it is actually more difficult as even under a slight load ie 2-5A, the voltage is depressed significantly ie 0.1-0.2V, so more difficult to estimate. With Li's, this sort of load will give a 0.01 -0.02V depression so easy to use this graph to estimate Amp Hrs out  when seeing small loads.

I agree with you Peter.......and with MP.

 

If you read info on tinternet, you would think voltage doesnt vary between 10 and 90% (outside of the knees) but it does on my Thunderskys. It is easy and robust to predict amp hrs out (or SoC) from voltage.

Screen Shot 2019-08-28 at 19.44.24.png

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21 hours ago, MoominPapa said:

Interesting: I'd always wondered where the extra charge goes. Though I suspect it's not as simple as you describe - there must be ways by which electrons can travel between the electrodes without passing through the external circuit, and those paths may be favoured at high discharge rates. 

 

MP.

No, electrons don’t pass between the electrodes without passing through the external circuit with the exception of the very slow self-discharge that LAs suffer from. It is certainly no higher at rapid discharge. Think about the energy that would be released as heat if that charge allegedly lost, was actually discharged within the battery at 12v potential. Of course a battery does get a bit warm under heavy discharge, but that is due only to ohmic losses.

 

Anyway, more to the point there is a paper somewhere on the Internet where they discharged a battery at a high rate until it was flat after relatively few AH delivered (as per Peukert). Then they let it rest, so it recovered. Then they discharged it at the same high rate again until it was flat, recovering more AH. They repeated until pretty much all the AH that would have been recovered at a slow discharge, was recovered. Absolutely proving that Peukert doesn’t describe AH that are lost, merely AH that are temporarily unavailable, but become available if left to rest for a while.

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6 minutes ago, nicknorman said:

No, electrons don’t pass between the electrodes without passing through the external circuit with the exception of the very slow self-discharge that LAs suffer from. It is certainly no higher at rapid discharge. Think about the energy that would be released as heat if that charge allegedly lost, was actually discharged within the battery at 12v potential. Of course a battery does get a bit warm under heavy discharge, but that is due only to ohmic losses.

 

Anyway, more to the point there is a paper somewhere on the Internet where they discharged a battery at a high rate until it was flat after relatively few AH delivered (as per Peukert). Then they let it rest, so it recovered. Then they discharged it at the same high rate again until it was flat, recovering more AH. They repeated until pretty much all the AH that would have been recovered at a slow discharge, was recovered. Absolutely proving that Peukert doesn’t describe AH that are lost, merely AH that are temporarily unavailable, but become available if left to rest for a while.

The implication of this is that Peukert's equation is irrelevant for, on average, small discharge rates, which would imply that LA works better than it actually does, in my experience. Actually, I suspect what happens is that peukert "constant" isn't, and increases as SoC is reduced. To fully discharge a LA battery therefore needs the same long, slow, asymptotic current curve as fully charging it does. (But in mirror image, as it were.)

 

It's interesting to speculate why this effect is so much less visible in LiFePo4 cells. The mobility of Lithium ions in FePo4 is not that great, but  the big difference is that  the active material is in a very thin layer of tightly controlled depth, spread over a large area of current-collecting foil. The reason this works is that the charged and discharged forms of the active material have similar physical properties, so you can expect such an arrangement to be stable over many cycles. LA uses the "spongy lead" form because that's the only physical arrangement that can be expected to survive the transition between lead and lead sulphate and back again, not because it provides good access for ions in the electrolyte to the electrode material.

 

 

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