Peukerts Law and evaluating true SoC

[VanHausen]

Hi folks,

So as a follow up to my other thread about my admittedly poorly-matched van solar system, I've got a question I felt would be relevant to a wider scope.


Owing to other design factors, I draw (un-ideal) currents around the 150A mark, often for up to 30mins at a time. Obviously with this sort of amperage, Peukerts Law becomes a big issue - hence why you dont use 12V systems for 2000W +, if you have a choice (I didn't). This got me thinking about something which admittedly most folk with better-matched voltages probably don't have to consider.

Hypothetically speaking, lets sa you have a battery rated for 100Ah @ 10A, but owing to Peukert, only rated to provide 50 Ah @ 50A. This would mean, then, drawing 50A for half an hour would put you in the 50% SoC zone you ideally shouldn't cross.

But you've only pulled .5 h x 50A = 25Ah out of a "100Ah" battery - or at least, 100Ah if you're only drawing 10A. SO - does this mean that after pulling your 25Ah out at 50A, can you then pull ANOTHER 25 Ah out (so long as you do it at 10A) before reaching 50% SoC

OR

Should you prioritize the stated capacity for the highest load used in the discharge cycle (in this case, 50Ah @ 50A), thus making 25Ah equivalent to 50% SoC?

Discuss? What is considered 'best practice'?

Thanks all!

[Wy_White_Wolf]

Consider:

If you have a car that has a 10 gallon tank and is rated to get 20 miles to the gallon you should be able to drive 200 miles. But if you have a lead foot and only get 10 miles to the gallon you have nothing left in the tank after 100 miles.

Your battery works about the same as the gas tank.

WWW

[bcroe]

Its more about voltage. Batteries have resistance (like everything else). You have to raise the
charging voltage to get current flowing through the resistance; the extra voltage used (multiplied
by the current flowing) results in lost energy not stored in the battery. Discharging the battery, the
resistance is still there and results in more lost energy (that was stored in the battery).

If you can operate the battery at lower currents, the voltage loss will be less, and a larger fraction
of the original energy will be delivered. Bruce Roe

[VanHausen]

Consider:

If you have a car that has a 10 gallon tank and is rated to get 20 miles to the gallon you should be able to drive 200 miles. But if you have a lead foot and only get 10 miles to the gallon you have nothing left in the tank after 100 miles.

Your battery works about the same as the gas tank.

I see what you're saying...but I feel the analogy seems somewhat off.

Yes, in much the same way as driving faster uses more L/hr, bigger wattage equals more current drawn. But driving fast doesn't make your gas tank any smaller, whereas larger currents seem to do exactly that.

The question is, what " tank" size do you use to calculate 50% SoC having been used, when you draw from your battery at both high and low current?

Its more about voltage. Batteries have resistance (like everything else). You have to raise the
charging voltage to get current flowing through the resistance; the extra voltage used (multiplied
by the current flowing) results in lost energy not stored in the battery. Discharging the battery, the
resistance is still there and results in more lost energy (that was stored in the battery.

If you can operate the battery at lower currents, the voltage loss will be less, and a larger fraction
of the original energy will be delivered. Bruce Roe

I'm sorry but I do not quite understand what you are trying to say, insofar as how it relates to the question.

[sensij]

I see what you're saying...but I feel the analogy seems somewhat off.

Yes, in much the same way as driving faster uses more L/hr, bigger wattage equals more current drawn. But driving fast doesn't make your gas tank any smaller, whereas larger currents seem to do exactly that.

The question is, what " tank" size do you use to calculate 50% SoC having been used, when you draw from your battery at both high and low current?

You are missing the analogy. Think about the mileage projection that your car gives you for a full tank... let's say it is 400 mi. That projection is based on some assumption of your driving speed and conditions. If you drive faster, you end up with only 300 mi before you have to fill the tank. It makes your gas tank *seem* smaller, but clearly, it didn't actually change size.

The analogy is good. a 400 Ah capacity is based on certain assumptions (discharge rate). If you violate that assumption by discharging too fast, you might get only 300 Ah before needing to recharge. It makes your battery *seem* smaller, but the amount of lead and acid in it haven't changed.

A more mathematical way to think about it might be to come up with a weighted average of discharge rate, and figure out what battery capacity on the chart corresponds best to that rate. The 20 hour capacity only applies if you are discharging at the 20 hour rate. If you discharge at the 10 hour rate, the 10 hour capacity applies. If your average discharge is something between the 10 hour and the 20 hour rate, your effective capacity will fall in between those numbers as well.

[NEOH]

VanHausen ,
Its more about the energy wasted as heat.
Heat is wasted energy.
Wasted energy = I x I x R (internal resistance).
Going from 10 Amps to 50 Amps is 5 TIMES the Amps.
But 5 TIMES the amps generates 25 TIMES the heat loss because of "amps squared".

At high amps, the energy consumed by the heat is GONE.
At high amps, your battery is less efficient at delivering the Amp-Hours to the load, as some is also going towards heat.
At high amps, the internal resistance of the battery becomes significant and introduces I^2xR Heat Losses into the equation
At high amps, you battery will be warm <= that is where your "lost" AH energy is going and it is not recoverable.

[VanHausen]

edit - apologies to anyone trying to read the edit-fest that was the initial post. My copypaste had a lot of incompatible characters I had to edit out manually.

sensij Thanks, I understand all that. The notion of an 'average' capacity between two discharge rates makes sense, however...

I've just done a bit more google-fu and found a few relevant points, namely this, in which someone points out:

It's an interesting question and Ralph Hiesey at Bogart Engineering, who probably has more years testing batteries than anybody here, tells why he don't use Peukert Number or Factor in his battery monitors:

http://www.bogartengineering.com/sit...tsComments.pdf

What Ralph says is true in my experience. We got three different battery monitors on our system here - two that use Peukert compensation, and the TriMetric that doesn't. Time after time the TriMetric is the most accurate one on heavy discharges. I've seen many times when the two monitors that use Peukert Factor say the battery is 25% SOC or lower and the TriMetric says 40% SOC or so. And indeed if I check SG's the result agrees with the TriMetric, and on full recharge the TriMetric is the one that reaches 100% SOC at end amps while the Peukert compensated monitors are only showing 85% SOC or so, and have to "sync" when the battery reaches full charge.

The referenced document states (edited for length):

The reason you get less amp hours out when the amps are higher is that the higher amps drag down the volts more than a lower amp load, so the 10.5 volts endpoint is reached sooner. It is not because some of the electrons in the battery get lost or wasted when you draw current out faster.

In the past there has been a misunderstanding of Peukerts relation ...that if you removed current (amps) out of a battery faster that somehow that wasted some of the battery electrons, so they wouldnt be available anymore until the battery was recharged ... if you discharged a battery down to its endpoint voltage at a high rate of discharge, then reduced the rate of discharge, according to his description you still would have an empty battery, with no ability to deliver further energy at a lower rate. This is not true. ...

Suppose you have a battery that is supposed to be discharged after 5 hours if you discharge it at a 20 amp rate. So you start to discharge it at that rate, but after 4.9 hours you reduce the load to only 1 amp. If you had let the load go on longer at 20 amps you would have gotten only 0.1 hours times 20 amps, or 2 amp hours more from the battery. The person who was confused about the meaning of Peukerts relation would have expected that even though the load was now only a 1 amp load that it still would have only 2 amp hours to go, or only 2 hours. But because the load was reduced the voltage would bounce up- and you would be able to quite a bit more than 2 amp hours before the voltage would drop below the "discharged" point.

This all seems to indicate that higher amps draws DO NOT in fact 'consume' extra Amp Hours (more on that later) But rather, the "lesser" Ah capacity at a 10 or 5 hr rate is an arbitrary value based on the induced voltage sag, and that 100 Ah out @ 100A is the same as 100 Ah out at 10A. Ergo, if you pull notionally 50% of the effective capacity at the 10 hr rate, you can drop to the 20 Hr rate and continue discharging while still being above 50% of actual Ah capacity used...at least thats how I'm reading it

All the points made about internal resistance make sense, but I think the key to remember there is that those loss equations are in expressed in terms of energy - Joules and Watt Hours, not Ah.

VanHausen ,

At high amps, the internal resistance of the battery becomes significant and introduces I^2xR Heat Losses into the equation
At high amps, you battery will be warm <= that is where your "lost" WATT HOURS is going and it is not recoverable.

Thats how I think the above statement should read. Although 100Ah out is still 100 Ah out, be it at 10A or 100A, under a 10A load you've looking at voltage sag down to only about 12.5 V or so. Compare that 100 loads of 100A, which cause sag to 10-11V, and thus a corresponding drop in P = I*V, and thus fewer WattHours out. The difference in watthours between 100Ah@100A and 100Ah@10A is what I think becomes heat due to higher internal resistance losses.

Any of this scan?

[sensij]

Any of this scan?

I think you are still missing the fundamentals here. A battery doesn't store Ah, it stores energy, which can be stated in units of Wh.

When you discharge at higher current, more of that energy is lost to heat and less of it goes to the load. The publication you cited is making the point that the losses aren't neatly described by the peukert equation in applications with variable discharge rates.

If your mental model makes more sense by treating that loss to be only in the voltage term, go with what works for you, but it isn't really right. As NEOH pointed out, the resistive energy losses are proportional to current squared, while the voltage sag is just linear with current, so it isn't correct to look at voltage sag difference of 12.5 V vs 10.5 V and call that 16% energy loss. The energy losses are higher than voltage alone will indicate (especially before you take into account that voltage isn't linear with SOC, either).

The good news is that with your AGM batteries, the IR is low and the peukert effect is much smaller than it would be for FLA, so your system will tolerate those high discharge spikes better than one which was designed less appropriately for them.

[NEOH]

I think you are still missing the fundamentals here. A battery doesn't store Ah, it stores energy, which can be stated in units of Wh.

When you discharge at higher current, more of that energy is lost to heat and less of it goes to the load. The publication you cited is making the point that the losses aren't neatly described by the peukert equation in applications with variable discharge rates.

If your mental model makes more sense by treating that loss to be only in the voltage term, go with what works for you, but it isn't really right. As NEOH pointed out, the resistive energy losses are proportional to current squared, while the voltage sag is just linear with current, so it isn't correct to look at voltage sag difference of 12.5 V vs 10.5 V and call that 16% energy loss. The energy losses are higher than voltage alone will indicate (especially before you take into account that voltage isn't linear with SOC, either).

The good news is that with your AGM batteries, the IR is low and the peukert effect is much smaller than it would be for FLA, so your system will tolerate those high discharge spikes better than one which was designed less appropriately for them.

Yes Watt-Hours is energy, not AH ( my mistake ) rephrase

At high amps, the energy consumed by the heat is GONE.
At high amps, your battery is less efficient at delivering the Amp-Hours to the load, as some is also going towards heat.
At high amps, the internal resistance of the battery becomes significant and introduces I^2xR Heat Losses into the equation
At high amps, you battery will be warm <= that is where your "lost" Watt-Hour energy is going and it is not recoverable.

Chart Below...
Let's assume C = 20AH and the knees are "square corners".
I approximated the area under the curves to calculate Watt-Hours for three discharge curves, as follows ...

At 0.05C Rate = 200 Watt-Hours = 12.5 Volts x 1 Amp x 20 Hours = Very efficient & battery is room temperature
At 1C Rate = 121 Watt-Hours = 12.1 Volts x 20 Amp x 0.5 Hours
At 7C Rate = 19 Watt-Hours = 10.2 Volts x 140 Amp x 0.0133 Hours = Not efficient & battery is very warm

If the resistance of the load and the internal resistance of the battery are equal then
50% of the battery's Watt-Hour energy is dissipated / wasted as internal heat energy.

NOTE:
The X-Scale is not linear ( logrithmic scale? )
Linear "Minutes Section" should be 60 times wider than the 60 second section!
Linear "Hours Section" should be 20 times wider than the stretched 60 minute section!

[Wy_White_Wolf]

...The question is, what " tank" size do you use to calculate 50% SoC having been used, when you draw from your battery at both high and low current?...

The one for the draw. Just like voltage drop goes up the more amperage you push through wire. You never get that lost voltage back. Same with batteries. The more amperage the higher the internal resistance that you have to overcome. It gets dissipated as heat loss so you never get it back just because you lowered the amp draw.

WWW

[NEOH]

Amp-Hour Capacity "recovery"...

Sometimes, a low voltage or low Amp-Hour capacity, is due the speed of the chemical reaction.
The electrolyte near the plates is depleted of sulfuric acid ions - it is a higher percentage of water.
If you have a short but high discharge rate, then let the battery rest.
After some time, the electrolyte is more evenly distributed and the Amp-Hour Capacity will recover.

[NEOH]

The one for the draw. Just like voltage drop goes up the more amperage you push through wire. You never get that lost voltage back. Same with batteries. The more amperage the higher the internal resistance that you have to overcome. It gets dissipated as heat loss so you never get it back just because you lowered the amp draw.
WWW

I do not find that to be true.
The internal resistance is fairly constant over a range of reasonable load amps.
The energy (watt-hours) lost as heat varies dramatically with increased amps, per the "amps squared"

The internal resistance does change over time.

[NEOH]

edit - apologies to anyone trying to read the edit-fest that was the initial post. My copypaste had a lot of incompatible characters I had to edit out manually.

sensij Thanks, I understand all that. The notion of an 'average' capacity between two discharge rates makes sense, however...

I've just done a bit more google-fu and found a few relevant points, namely this, in which someone points out:

This all seems to indicate that higher amps draws DO NOT in fact 'consume' extra Amp Hours (more on that later) But rather, the "lesser" Ah capacity at a 10 or 5 hr rate is an arbitrary value based on the induced voltage sag, and that 100 Ah out @ 100A is the same as 100 Ah out at 10A. Ergo, if you pull notionally 50% of the effective capacity at the 10 hr rate, you can drop to the 20 Hr rate and continue discharging while still being above 50% of actual Ah capacity used...at least thats how I'm reading it

All the points made about internal resistance make sense, but I think the key to remember there is that those loss equations are in expressed in terms of energy - Joules and Watt Hours, not Ah.


Thats how I think the above statement should read. Although 100Ah out is still 100 Ah out, be it at 10A or 100A, under a 10A load you've looking at voltage sag down to only about 12.5 V or so. Compare that 100 loads of 100A, which cause sag to 10-11V, and thus a corresponding drop in P = I*V, and thus fewer WattHours out. The difference in watthours between 100Ah@100A and 100Ah@10A is what I think becomes heat due to higher internal resistance losses.

If you don't think that a significant increase in HEAT in your battery is lost energy then
you need to explain where that energy came from to make that heat.

Peukert's Discharge Law accurately predicts the Apparent Loss of Amp-Hour Capacity under continuous heavy load.
Test it for yourself and you see that it is quite accurate.

BUT ...
Peukert's Discharge Law does not apply once the heavy load is removed.
Peukert's Discharge Law does not predict the % SOC or the % DOD.
Peukert's Discharge Law does not predict Amp-Hours needed to recharge.

There is a huge misunderstanding with what you quoted in Message #7 vs what Puekert's Discharge Law actually predicts.
Not one thing that you "QUOTED" disagrees with Peukert's Discharge Law.
The authors of those quotes wrongly equate Peukert's Discharge Law with some specific % SOC or % DOD - that is clearly their problem.

Peukert's Discharge Law never stated ... What happens after the Heavy Load is disconnected and the battery is allowed to rest.
Peukert's Discharge Law never stated ... What the % DOD is at the end of a continuous heavy discharge of the battery.

The authors of your quotes made some very erroneous assumptions about Peukert's Discharge Law.

[VanHausen]

So, a long time since I posted this thread. So for future readers sake, here's the conclusion:

The responses that came back originally didn't quite a make sense to me. But rather than argue fruitlessly around in circles, I decided to go test the theory in the real world.

Using a sacrificial battery with a 216AH C20 rating, we did a full discharge and obtained 212Ah at the cutoff V of 10.5V. Following a full recharge and equalize, we ran a truly horrifying C1 discharge, again to 10.5V. Spec sheet lists the 60 min rate as 139A . We got 130Ah - more or less a total discharge.

Here's the interesting bit:

A 24hr rest resulted in a recovery of OCV to 12.0V, at which point we re-applied the C20 load of 11A...and withdrew another ~70 Ah to 10.5V - giving a total of 201Ah. After another rest, this voltage barely recovered to 11V verifying the battery was now trulydepleted.

This battery actually recovered with a good recharge, yielding 210Ah on the next C20 discharge. So we did 3 more repeat tests, comparing discharge to 10.5 V the 10 Hr and 5 Hr rates.

The results were the same each time : discharging at a high rate to the 10.5V then allowing a rest, would result in OCV recovery, allowing further discharge at a lower rate, always giving a total yield within a few Ah of the C20 capacity



I've also since seen these results echoed in several other working setups I've helped build people over this time, including my own.

In the case of my Fullriver DC400's, I can pull roughly 120A for an hour while cooking and heating the shower, before inverter starts to complain about apparent batt V. The spec sheet lists 126A as the 2 Hr rate, meaning by rights, I've used 50% of the batteries capacity this way.....but my monitor says I've really only used about 30%, based on the 415Ah C20 capacity.

Here's the thing - turn it all off for about 20mins, and the volts jumps right back up to about 12.4v - more or less where the DC400 spec sheet says it should be for for 70% SoC. And I can keep happily running normal loads for hours after, and never see it drop under 12.1...at least not until the monitor says I've actually used 50% of that 415Ah .


So, wrapping up:

The results here confirmed what I suspected initially - Peukerts Law is describing an effective loss in capacity with increasingly heavy loads, as the induced voltage sag hastens you towards the 10.5V 'cliff'. It does not, in fact, reflect energy losses due to resistance/heating/magic - while there is indeed some energy lost to resistance, it is nowhere as large as this, which can be confirmed with a basic sanity check below**

Peukerts law CAN be used to accurately estimate your usable capacity at any CONSTANT given rate of draw. However - if you have variable loads, you're perfectly OK to figure out what capacity best corresponds to the 'typical' draw over time, and just use that. Any Ah used by intermittent heavy loads just comes off this C20 total, as the V will recover once the heavy load is withdrawn, leaving you free to continure draw at your 'typical' load

The only issue crops up is when you've used up a good portion of that C20 total, and THEN try to pull a high load. If you've already the equivalent of your C2 rating, but at a lower draw, trying to then pull C2 at 40% SoC will sag the voltage to 10.5v right away. In that sense, as far as those heavy loads are concerned, you've used all the battery can give.





**If the difference in AH yield shown by Peukerts Law was lost energy due to heating from internal resistance, then in the case of my DC 400 batteries, comparing the C20 rate (415 Ah) to the C2 Rate (240Ah), this would mean that if I used 240 Ah @ 120A for 2 hrs, versus 20A @ 12 hrs, I would have "lost" ~175Ah of charge as internal heat in the batteries alone.

That is an INSANE amount of energy. By comparison, it take about 40Ah for me to BOIL my 10L water heater. 175Ah would do that, fours times over, and then some. I'll let you do the math on the joules if you want, but suffice to say, if this WAS the case wires would be smoking and acid would be boiling. Having now personally tested this exact scenario, I can say thats simply not where the energy is going.

[Sunking]

Well your conclusions are wrong based on junk science. One fact you missed is heat losses from resistance has absolutely nothing to do with Peukert Law or Amp Hours. You are making it up.