The Peukert myth exposed

[nicknorman]

Lots of people talk about Peukert because they heard it from the bloke in the pub, whilst not really understanding what it means. Sounds impressive though! The general idea is that if you discharge a lead acid battery faster, it has less capacity. Similarly although not called "Peukert", when the battery is cold it has less capacity. But that is wrong.

 

The battery has a certain amount of chemicals in it (lead, sulphuric acid etc) and when they react, a certain number of electrons can mobilised. Newtons 4th law of thermodynamics says that electrons cannot be created or destroyed, but they might be dozing. (made that bit up). They certainly don't go on holiday just because they are being asked to move faster (high discharge rate) or they are a bit chilly.

 

So why do people think Peukert affects SoC? It's to do with the definition of capacity. To measure capacity you discharge your battery at the specified rate (C20 or whatever) and measure the Ah until it reaches the definition of "flat" which is normally 10.5v. Then you stop. If you discharge it faster, or colder, you get fewer Ah before the battery reaches the notional definition of "flat". So they are right after all? No, of course they are not!

 

Having discharged at the faster/cooler rate to the terminating 10.5, if you then stop and wait, those dozing electrons will get around to migrating to the plates and after a while, the voltage recovers and more discharging can take place. It is not that the electrons have been miraculously transported out of the battery to sunny Spain, they are just resting a bit and will get around to doing their job given a bit more time.

 

Put it another way, take a 100Ah battery (at C20 rate), discharge it at 50A for an hour. How much capacity is left? Well if you carry on discharging at 50A you will find it is flat long before you have got your 100Ah. So you could say the remaining capacity was much less than 50%. But hang on, stop discharging after the first hour, go and have several cups of tea, maybe dinner and come back 9 hours later. Now continue to discharge at the specified rate (5A) and Lo! you will get your 100Ah out. So the capacity was never lost, it was simply temporarily unavailable because the reaction couldn't proceed fast enough. Given more time, the capacity was still there. So if the battery monitor had applied Peukert, it would have given a falsely low SoC after the first hour. Of course the "time to run" display would need to apply Peukert, since if the fast discharge rate had been maintained to the bitter end, the battery would have been notionally "flat" long before the 100Ah had been extracted.

 

Interestingly when I first realised all this, an internet search only showed up one supporting piece. That was over 10 years ago I think, and now there is plenty of support on the internet. I guess I started a trend (maybe not). This series of 3 videos is quite interesting and he does eventually get to the point, if you can stand the accent, the slow delivery and the wretched Youtube adverts...

 

 

[Gibbo]

Wonderful. Clearly described, sciencifically detailed. So, lets try to pin it down to the shortest possible description we can get away with. One simple sentence/paragragh each. Is that fair?

 

I maintain, if you discharge a lead acid battery at a high discharge rate, you will get less total energy from the battery than if you had dischared at a lower rate. But you didagree with that. Am I understanding you correctly?

[nicknorman]

3 minutes ago, Gibbo said:

Wonderful. Clearly described, sciencifically detailed. So, lets try to pin it down to the shortest possible description we can get away with. One simple sentence/paragragh each. Is that fair?

 

I maintain, if you discharge a lead acid battery at a high discharge rate, you will get less total energy from the battery than if you had dischared at a lower rate. But you didagree with that. Am I understanding you correctly?

No, I totally agree with that. Watch the videos, he explains the difference between coulombic efficiency and voltage efficiency - and hence energy efficiency.

 

Actually I going to tweak your proposal slightly, you get less USEFUL total energy from the battery. You still get all the energy but some of it goes to heating the battery and hence the room, which isn’t useful.

[Gibbo]

Ah right, slow down there, energy is energy, whether you consider it useful or not. So pin it down properly.

 

I have maintained from day one, that if you discharge a lead acid battery at a high discharge current, you will get less total energy from that battery. You either agree with that or you don't.

Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.

Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.

Waiting...

 

 

 

 

 

This is why I stayed away for ten years. Totally pointless arguments with armchair experts who have absolutely no idea what they are talking about, no track record of design products and whose only evidence is youtube videos.

 

It's the flat earth argument all over again. Pointless.

 

Tara. x

[Guest]

This debate is meaningless unless the current forum expert in everything @IanD contributes.....

 

 

[Arthur Marshall]

1 hour ago, Gibbo said:

 

It's the flat earth argument all over again. Pointless.

Of course the earth's flat. You discharged it too fast...

[blackrose]

Gibbo is BACK! 

 

Well he was, he's probably gone away for another 10 years...

Edited November 9, 2023 by blackrose

[rusty69]

....... But now called Tara. 

[magnetman]

How does one pronounce Peukert? 

 

I am tempted to say 'Puke-urt' 

 

 

[Tacet]

9 hours ago, nicknorman said:

Lots of people talk about Peukert because they heard it from the bloke in the pub, whilst not really understanding what it means. Sounds impressive though! The general idea is that if you discharge a lead acid battery faster, it has less capacity. Similarly although not called "Peukert", when the battery is cold it has less capacity. But that is wrong.

 

The battery has a certain amount of chemicals in it (lead, sulphuric acid etc) and when they react, a certain number of electrons can mobilised. Newtons 4th law of thermodynamics says that electrons cannot be created or destroyed, but they might be dozing. (made that bit up). They certainly don't go on holiday just because they are being asked to move faster (high discharge rate) or they are a bit chilly.

 

So why do people think Peukert affects SoC? It's to do with the definition of capacity. To measure capacity you discharge your battery at the specified rate (C20 or whatever) and measure the Ah until it reaches the definition of "flat" which is normally 10.5v. Then you stop. If you discharge it faster, or colder, you get fewer Ah before the battery reaches the notional definition of "flat". So they are right after all? No, of course they are not!

 

Having discharged at the faster/cooler rate to the terminating 10.5, if you then stop and wait, those dozing electrons will get around to migrating to the plates and after a while, the voltage recovers and more discharging can take place. It is not that the electrons have been miraculously transported out of the battery to sunny Spain, they are just resting a bit and will get around to doing their job given a bit more time.

 

Put it another way, take a 100Ah battery (at C20 rate), discharge it at 50A for an hour. How much capacity is left? Well if you carry on discharging at 50A you will find it is flat long before you have got your 100Ah. So you could say the remaining capacity was much less than 50%. But hang on, stop discharging after the first hour, go and have several cups of tea, maybe dinner and come back 9 hours later. Now continue to discharge at the specified rate (5A) and Lo! you will get your 100Ah out. So the capacity was never lost, it was simply temporarily unavailable because the reaction couldn't proceed fast enough. Given more time, the capacity was still there. So if the battery monitor had applied Peukert, it would have given a falsely low SoC after the first hour. Of course the "time to run" display would need to apply Peukert, since if the fast discharge rate had been maintained to the bitter end, the battery would have been notionally "flat" long before the 100Ah had been extracted.

 

Interestingly when I first realised all this, an internet search only showed up one supporting piece. That was over 10 years ago I think, and now there is plenty of support on the internet. I guess I started a trend (maybe not). This series of 3 videos is quite interesting and he does eventually get to the point, if you can stand the accent, the slow delivery and the wretched Youtube adverts...

 

 

When your car/boat won't start and the battery goes flat, such that it is incapable of turning the engine over,  wait a few minutes and it recovers sufficiently for another spin.

[MtB]

Surely the dichotomy revolves around the rather arbitrary 10.5v definition of 'flat'. 

 

I'm sure other definitions could be dreamed up that suit either point of view.

 

I hold that the very existence of Perkwotsit at all illustrates the 10.5v definition of flat is flawed.

 

 

[nicknorman]

9 hours ago, Gibbo said:

Ah right, slow down there, energy is energy, whether you consider it useful or not. So pin it down properly.

 

I have maintained from day one, that if you discharge a lead acid battery at a high discharge current, you will get less total energy from that battery. You either agree with that or you don't.

Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.

Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.

Waiting..

This is why I stayed away for ten years. Totally pointless arguments with armchair experts who have absolutely no idea what they are talking about, no track record of design products and whose only evidence is youtube videos.

 

It's the flat earth argument all over again. Pointless.

 

Tara. x

 

Waiting waiting... for me to have my beauty sleep, which is very important!

You say that if you discharge a lead acid battery at a high discharge current, you will get less total energy from that battery. I disagree. That would be a clear breach of the first law of thermodynamics. Can you be clear - are you refuting the first law of thermodynamics?

 

The battery stores a certain amount of (chemical) energy, and that energy can be released, although it takes a little time to do so. What is true is that the rate at which the energy can be dispensed will slow down a lot if the intial discharge is fast. But given enough time, all the energy can be released even if to start with, that rate was very high.

 

Of course this is not too relevant to real world experience because if we break down the energy into useful energy at the terminals, vs energy lost to heating the battery, then of course at a high discharge rate there is more energy wasted on heating the battery and less for useful application. This is the voltage efficiency alluded to in the video, which (true to your avatar) I'm guesing you haven't watched yet. Voltage efficiency of course having a direct bearing on (useful) energy efficiency.

 

None of which has anything to do with Peukert, which postulates about coulombic efficiency. Peukert is only relevant if you want to discharge at a high rate AND continue to discharge at that high rate until the battery is flat. And even then, if you wait a bit, all the "lost" coulombs can be recovered albeit at a much slower discharge rate.

 

15 minutes ago, MtB said:

Surely the dichotomy revolves around the rather arbitrary 10.5v definition of 'flat'. 

 

I'm sure other definitions could be dreamed up that suit either point of view.

 

I hold that the very existence of Perkwotsit at all illustrates the 10.5v definition of flat is flawed.

 

 

Yes, I would agree. Although to be fair, one has to decide that at some voltage, the battery is no longer providing useful current because the voltage is too low. Hence the need to set a nominal limit, and as you say Peukert arises only as a consequence of setting that nominal limit. It is an artificial creation.

Edited November 9, 2023 by nicknorman

[Gibbo]

22 minutes ago, Tacet said:

When your car/boat won't start and the battery goes flat, such that it is incapable of turning the engine over,  wait a few minutes and it recovers sufficiently for another spin.

 

That doesn't quantify anything.

Nick

 

Stop all the waffling and A4 pages of irrelevant words.

 

If a LA battery is discharged at a higher rate (current) you will get less total energy from it. This is the crux of the matter. To disagree with it points to some sort of fundamental schoolboy error.

[nicknorman]

13 minutes ago, Gibbo said:

 

If a LA battery is discharged at a higher rate (current) you will get less total energy from it. This is the crux of the matter. To disagree with it points to some sort of fundamental schoolboy error.

 

So you really are refuting the first law of thermodynamics. I think the conversation might as well end there.

 

And of course all this is unrelated to Peukert.

Edited November 9, 2023 by nicknorman

[Gibbo]

Not refuting anything. You, however, appear rather confused.

 

Take two identical batteries. Battery A and battery B. Identical batteries. Same make, model, condition, batch etc. 100Ahr. Peukert's exponent of 1.2 Both batteries fully charged.

Discharge battery A at 5 amps for ten hours. Discharge battery B at 50 amps for one hour.

Now wait a few hours for things to settle. This removes your internal confusion regarding resting time and recovery etc. Wait however long you wish/desire.

Are battery A and battery B now in the same SoC or different? It is a straight forward question. A simple question. It deserves a simple answer.

[nicknorman]

2 minutes ago, Gibbo said:

Not refuting anything. You, however, appear rather confused.

 

Take two identical batteries. Battery A and battery B. Identical batteries. Same make, model, condition, batch etc. 100Ahr. Peukert's exponent of 1.2 Both batteries fully charged.

Discharge battery A at 5 amps for ten hours. Discharge battery B at 50 amps for one hour.

Now wait a few hours for things to settle. This removes your internal confusion regarding resting time and recovery etc. Wait however long you wish/desire.

Are battery A and battery B now in the same SoC or different? It is a straight forward question. A simple question. It deserves a simple answer.

They are at the same SoC.

Well, within a very small and probably unmeasurable margin, a massively smaller margin than the Peukert fiction would suggest. They are a tiny bit different due to gassing of H2 and O2 which probably occurs at the faster rate more than at the slower rate.
This is all covered in the video that you haven’t dared to watch yet. And this of course is why Peukert has no place in the SoC calculation of a coulomb-measuring SoC meter.

[IanD]

18 minutes ago, nicknorman said:

 

So you really are refuting the first law of thermodynamics. I think the conversation might as well end there.

 

And of course all this is unrelated to Peukert.

No he's not, and neither are you -- there are two things going on here which I think you both agree on if you could put your boxing gloves down... 😉

 

The first is that if you discharge at high rates you get less energy *out* of the battery before it drops to a "discharged state" defined by a given voltage e.g. 10.5V. And if you stop here and wait the voltage will go back up and you can get some more energy out.

 

The second is that even when you do this the total energy available *outside the battery* is lower at high discharge rates because of increased losses in the battery internal resistance -- the total Ah is the same but the terminal voltage is lower at high rates, so the usable Wh is lower. This isn't breaking any laws of thermodynamics, the energy released by the chemical reaction is the same but less of it makes its way outside the battery.

 

Agreed?

Edited November 9, 2023 by IanD

[Gibbo]

And you are absolutely certain of this? Or are you going to start backpedalling later?

 

That's to Nick.

[nicknorman]

3 minutes ago, IanD said:

No he's not, and neither are you -- there are two things going on here which I think you both agree on if you could put your boxing gloves down... 😉

 

The first is that if you discharge at high rates you get less energy *out* of the battery before it drops to a "discharged state" defined by a given voltage e.g. 10.5V. And if you stop here and wait the voltage will go back up and you can get some more energy out.

 

The second is that even when you do this the total energy available *outside the battery* is lower at high discharge rates because of increased losses in the battery internal resistance -- the total Ah is the same but the terminal voltage is lower at high rates, so the usable Wh is lower. This isn't breaking any laws of thermodynamics, the energy released by the chemical reaction is the same but less of it makes its way outside the battery.

 

Agreed?

Yes to both points. But I don’t think Gibbo agrees. And by the way, the boxing gloves are the entire point of this thread (you don’t understand the history), Peukert is just a vehicle. But feel free to join in anyway!

3 minutes ago, Gibbo said:

And you are absolutely certain of this? Or are you going to start backpedalling later?

Is that directed to me? Yes I am absolutely certain of it.

Edited November 9, 2023 by nicknorman

[Gibbo]

So Nick...

 

I have stated all along that as the discharge current is increased, one gets less total energy from the battery. You maintain that is incorrect. Yes?

 

You maintain that battery A and battery B are in exactly (or at least extremely close) the same SoC. Yes?

[nicknorman]

Just now, Gibbo said:

So Nick...

 

I have stated all along that as the discharge current is increased, one gets less total energy from the battery. You maintain that is incorrect. Yes?

 

You maintain that battery A and battery B are in exactly (or at least extremely close) the same SoC. Yes?

Assuming that by total energy you mean electrical energy from the terminals AND heat energy released into the battery (and thence the surroundings) then YES.

And YES to the SoC.

[MtB]

40 minutes ago, Gibbo said:

 

That doesn't quantify anything.

Nick

 

Stop all the waffling and A4 pages of irrelevant words.

 

If a LA battery is discharged at a higher rate (current) you will get less total energy from it. This is the crux of the matter. To disagree with it points to some sort of fundamental schoolboy error.

 

My own view is one should count the Coulombs out until you can't get any more. Then it is flat.

 

Do it as fast or as slow as you like but I bet you'll get the same. 

 

Writing as a plummer, obviously. 

 

 

[nicknorman]

Just now, MtB said:

 

My own view is one should count the Coulombs out until you can't get any more. Then it is flat.

 

Do it as fast or as slow as you like but I bet you'll get the same. 

 

Writing as a plummer, obviously. 

 

 

I'd agree. But using the plumber analogy, when said plumber says "there is nothing wrong with my pipework, water is appearing at all the taps", the householder doesn't appreciate the fact that the water only drips out of the shower, they want the water to spray out of the shower, otherwise that water is pretty useless.

[magnetman]

A lead acid battery is flat if you can hold a steel ruler across the terminals. 

 

 

[Gibbo]

8 minutes ago, nicknorman said:

Assuming that by total energy you mean electrical energy from the terminals AND heat energy released into the battery (and thence the surroundings) then YES.

And YES to the SoC.

 

Throughout the two discharges battery A will have spent it at an average terminal voltage of 12.4 volts. Battery B will have been at an average of 11.1 volts.

The load on the battery A will have benefitted from (converted, consumed, dissipated, call it whatever you want, it makes no difference). 12.1V x 5A x 10 hours = 620 watt hours of energy.

The load on the battery B will have "benefitted" from 11.1V x 50A x 1 hour = 550 watt hours.

Both batteries started at 100% SoC. You agree that they both ended the discharge in the same SoC.

Where are the missing 65 watt hours?

Your schoolboy error appears to be you believing an amp hour to be a unit of energy. It is not.

This thought experiment proves beyond any doubt that higher discharge currents result in less available energy from the battery. The effect is permanent and cannot be "got around" by waiting a bit.