Battery Fault

[WotEver]

13 minutes ago, nicknorman said:

Peukert is not an imaginary phenomena. As you say, it is why there are different specified capacities according to discharge rate.

So if we only take half of the available capacity whilst discharging at a high rate we have ‘lost’ some energy in that cycle. 

QED

[nicknorman]

5 minutes ago, WotEver said:

Yes, that is how it functions whenever it ‘gets it wrong’. It waits for the SoC to ‘catch up’ with its reading.

The occurrence I’m referring to was when you logged SG against your Ah counter (can’t remember which one you have). I guess it would have been a couple of years ago. 

So are you saying that by applying Peukert (and since it doesn’t measure current, it can’t do so directly) it gets it wrong and then has to freeze for a while? If so, perhaps it would be better if it didn’t get it wrong in the first place!

[WotEver]

Just now, nicknorman said:

So are you saying that by applying Peukert (and since it doesn’t measure current, it can’t do so directly) it gets it wrong and then has to freeze for a while? If so, perhaps it would be better if it didn’t get it wrong in the first place!

No, I am saying that by applying Peukert it makes an assumption that you are intending to continue to discharge at the high rate and reflects this in its display. When it sees you reduce your usage it self-corrects. 

[nicknorman]

Just now, WotEver said:

So if we only take half of the available capacity whilst discharging at a high rate we have ‘lost’ some energy in that cycle. 

QED

Careful, you lose credibility if you suddenly start talking about Energy. Peukert is nothing to do with Energy. Check the dimensions of the equation if you don’t believe me! Of course, as I said earlier, if you discharge fast you “waste” energy due to the reduced terminal voltage. But that is not Peukert.

Just now, WotEver said:

No, I am saying that by applying Peukert it makes an assumption that you are intending to continue to discharge at the high rate and reflects this in its display. When it sees you reduce your usage it self-corrects. 

It can’t apply Peukert directly as it doesn’t measure current. Current is the only parameter in Peukert’s equation (apart from time). How Gibbo dealt with a sudden dip in voltage caused by a high current drain, we can only guess at. I am not convinced he got it wrong but we’d have to do specific testing.

[WotEver]

4 minutes ago, nicknorman said:

I am not convinced he got it wrong but we’d have to do specific testing.

You did. With graphs and everything

5 minutes ago, nicknorman said:

It can’t apply Peukert directly as it doesn’t measure current.

You and I both know that it applies Peukert by interpreting voltage fluctuations. 

[nicknorman]

Just taking a different tack to break the stalemate, why do we want to know discharging SoC? Presumably to stop discharging too much? If I discharge really fast and apply Peukert, I’ll get to 50% quite soon. But I’ve only used 30% of the chemicals. Do I really need to not discharge any more, or can I continue because I’ve used less than 1/2 the chemicals up? And what is the difference between continuing the fast discharge now, vs waiting a while, resting so that the SoC reverts to 70%, and then fast discharging some more?

[WotEver]

12 minutes ago, nicknorman said:

If I discharge really fast and apply Peukert, I’ll get to 50% quite soon. But I’ve only used 30% of the chemicals. Do I really need to not discharge any more, or can I continue because I’ve used less than 1/2 the chemicals up? And what is the difference between continuing the fast discharge now, vs waiting a while, resting so that the SoC reverts to 70%, and then fast discharging some more?

I honestly don’t know enough about battery chemistry to know the answers to any of those questions. 

But your argument is convincing at present 

I shall research some more. 

Edited November 25, 2017 by WotEver

[Dr Bob]

1 hour ago, nicknorman said:

Long post to pick apart, so I've done it in-line in red.

You wrote:

Tony thinks that capacity is 'lost' and you dont. You mentioned chemistry in a post a few hours ago and that is the key if we can understand it.

I see two inputs to that - one of which I dont understand. Firstly (the one I do understand) there is the 'how many molecules will react?'. The rate of reaction will be very dependent on access of one molecule to the other molecules in the plates and deep in the plates. This is likely a steric problem and the heterogenous nature of the solid/liquid interface will effect if the incoming molecule (or will it be ion?) can get to a site and react. Steric hinderance is always worse at lower temps so it is very likely as temperature is reduced, the most sterically hindered sites will not react and so the overall capacity will effectively be reduced.  I am not a chemist but I am confident that its not a case of the chemicals no longer reacting at all. Just reacting more slowly.  Looking at tables/graphs on the internet, it looks like a 20% reduction in capacity is seen dropping from 25degC to 5degC. That is because the standard way to measure capacity is to discharge to a terminating voltage, normally 10.5v. That voltage will be reached earlier at low temperatures due to sluggish reaction. As Tony says, this capacity is lost and cannot be got back until the battery heats up and those difficult to get to sites are ''reactive" again. So what you are saying is that if a cold battery is discharged to a certain SoC, and then warmed up, suddenly the Soc is higher? I suppose you can define SoC how you like but that is not a normal way of doing it. If that is all correct it would also mean that those sterically hindered sites are likely never to react as the battery is usually only drained to 50% and these sites will be the last to give up their 'charge'. I would guess therefore that these sites would also never be sulphated.

The second input is the change in internal resistance as temperature drops. This will change the fate of the electrons produced and hence change the rate of reaction at the 'slow' sites. I have little experience of resistance vs rate of reaction but it is likely that it changes the activation energy for each site - either reducing the starting energy or increasing the required energy. Note: for a reaction to take place, ie the chemical combination leading to the release of an electron, enough thermodynamic energy has to be provided to drive the reaction. In practice it could be that the low temperature reduces the rate at which the electron is taken from the site and that would push the equilibrium reaction back to the starting materials and therefore require more energy to make the reaction happen. Dont confuse charge/Soc/AH with anything to do with power or energy. The dimensions of charge are not those of power or energy. Without doubt a fast discharge or cold discharge results in less energy being extracted from the battery because the terminal voltage is lower, and of course energy = charge taken out x voltage. But that has nothing to do with remaining SoC or capacity. As I said, AH has no concept of energy or power.

Overall, it sounds to me that the 20% of capacity is just not available unless you drain it at a slower rate - but on the boat we are creatures of habit so the rate at which we use it is the same and that 20% is lost until the temperature rises. This is true if you want to discharge the battery to flat. But generally we try not to do that! The fact that the 20% is never used however means the user sees no difference in their use of the batteriess unless they try to work out capacity and then it will all depend on what the voltage at rest vs temp curve looks like. There is virtually no rest voltage vs temperature coefficient. As I said, a cold or fast-discharged battery will take longer to reach its rested voltage but since that process is asymptotic its just a matter of how long you can be bothered to wait and how accurate you want the result.

......................

A few comments.

Para 1. I agree that ALL of the potential sites will slow down and that is why the least accessible will not be available as they will be 'too slow' ie too hindered etc. I dont think I mentioned SoC as I dont really know how to define it. What I said is that the slowest 20% of the sites will not be available for reaction so the capacity is reduced by 20%. If you have a 100Ahr battery and take 50Ahr out at 5degC then there may only be 30Ahr left. I am not saying what the SoC is. If the battery is heated to 25deg C you will then have 50Ahr left. Again I am not saying the SoC goes up or down as I dont know the definition of SoC.

Para 2. You are confusing my 'energy' with your 'energy'. Apologies. When I talk about energy and equilibrium reactions, it is the energy needed to force that reaction in a forward or reverse direction - so is thermodynamic energy needed to make the reaction go. This is related to activation energy for reactions to take place and the energy state of the unreacted molecules and reacted molecules. It is not the energy that is the output of the process - ie the stream of electrons (their potential or current etc) which I think is what you are referring to.

Para 3. It is clear that the voltage under load will be depressed at lower temperatures. My problem here is that MY batteries are never at rest with an inverter and fridge running so I always estimate a true 'rest' voltage from my loaded voltage (either -1.7A or 5.2A with the fridge either running or not). This loaded voltage will change with temp hence my last sentance. Sorry to confuse you. What I dont understand is why is the rest voltage not lower at lower temp if the whole battery chemistry slows down? The minute you put a meter on it, the sites react slower so why not a lower value - or is this just the surface charge being the same? I guess this is the case as there must be a 'charge' on the plates so it stops the equilibrium reaction progressing. The voltage measured is therefore always 12.72V (or whatever) regardless of temperature.

 

 

[nicknorman]

21 minutes ago, Dr Bob said:

 

A few comments.

Para 1. I agree that ALL of the potential sites will slow down and that is why the least accessible will not be available as they will be 'too slow' ie too hindered etc. I dont think I mentioned SoC as I dont really know how to define it. What I said is that the slowest 20% of the sites will not be available for reaction so the capacity is reduced by 20%.  I’m no chemist but I think it’s wrong to say that a particular chunk of chemicals suddenly decides they’re not going to react, whilst some identical chemicals at the same temperature decide they will. IMO the chemicals will react, given a bit of time. If you have a 100Ahr battery and take 50Ahr out at 5degC then there may only be 30Ahr left. I am not saying what the SoC is. If the battery is heated to 25deg C you will then have 50Ahr left. Again I am not saying the SoC goes up or down as I dont know the definition of SoC. SoC can be defined however you want it. In the case of a UPS battery, it is useful to define SoC as the remaining percentage of AH at the present current drain, so that the light go out when it reaches 0%. But for a boat I think this definition is not useful since the rate of discharge is so variable. Better to know the remaining AH that can be extracted at the typical average discharge rate, which is very slow.

Para 2. You are confusing my 'energy' with your 'energy'. Apologies. When I talk about energy and equilibrium reactions, it is the energy needed to force that reaction in a forward or reverse direction - so is thermodynamic energy needed to make the reaction go. This is related to activation energy for reactions to take place and the energy state of the unreacted molecules and reacted molecules. It is not the energy that is the output of the process - ie the stream of electrons (their potential or current etc) which I think is what you are referring to. OK fair enough. But you still get 2 electrons migrating for each molecule of chemicals. Their energy and hence terminal voltage may be lower, but that is not the issue in question.

Para 3. It is clear that the voltage under load will be depressed at lower temperatures. My problem here is that MY batteries are never at rest with an inverter and fridge running so I always estimate a true 'rest' voltage from my loaded voltage (either -1.7A or 5.2A with the fridge either running or not). This loaded voltage will change with temp hence my last sentance. Sorry to confuse you. What I dont understand is why is the rest voltage not lower at lower temp if the whole battery chemistry slows down? The minute you put a meter on it, the sites react slower so why not a lower value - or is this just the surface charge being the same? I guess this is the case as there must be a 'charge' on the plates so it stops the equilibrium reaction progressing. The voltage measured is therefore always 12.72V (or whatever) regardless of temperature. Battery chemistry is very complicated and I certainly don't understand why there is no significant thermal coefficient. But then there are lots of other weird things about batteries that I don't understand either!

 

I

Edited November 25, 2017 by nicknorman

[Dr Bob]

15 minutes ago, nicknorman said:

Just taking a different tack to break the stalemate, why do we want to know discharging SoC? Presumably to stop discharging too much? If I discharge really fast and apply Peukert, I’ll get to 50% quite soon. But I’ve only used 30% of the chemicals. Do I really need to not discharge any more, or can I continue because I’ve used less than 1/2 the chemicals up? And what is the difference between continuing the fast discharge now, vs waiting a while, resting so that the SoC reverts to 70%, and then fast discharging some more?

I think this is the sensible approach and very similar to the temperature issue. Forget SoC for the discussion. If you discharge fast then you 'loose' some capacity. If you discharge in the cold you lose some capacity, but you have used less than half the chemicals if you are defining SoC as 50%. We then need to define what SoC means (and you two do not seem to agree) and be able to relate it to our batteries at whatever the temperature ie the question I ask each morning  is 'how are my batteries doing - which is code for what do I think their current capacity is?

[WotEver]

Been thinking.

Getting back to...

5 hours ago, nicknorman said:

You have never answered the fundamental question about what happens to the chemicals. They react, you get a fixed number of electrons regardless of their temperature. Or they don't react, no electrons. But the chemicals are thus still available. The only thing that changes is how readily / quickly they react, especially the chemicals buried deeper in the plates. By your theory the chemicals have somehow reacted without giving as many electrons. That is impossible according to physics and chemistry as we know them.

Is it?  The only thing?

So please explain how the battery has warmed up during a heavy discharge. 

3 hours ago, smileypete said:

Whether you pull out 50% of available capacity with an LED bulb or high power inverter, the recovered resting voltage will be the same.

I accepted this as read earlier but I’m unable to find any corroboration of this claim. 

So where do you get this from, Pete?

Edited November 25, 2017 by WotEver

[nicknorman]

19 minutes ago, Dr Bob said:

I think this is the sensible approach and very similar to the temperature issue. Forget SoC for the discussion. If you discharge fast then you 'loose' some capacity. If you discharge in the cold you lose some capacity, but you have used less than half the chemicals if you are defining SoC as 50%. We then need to define what SoC means (and you two do not seem to agree) and be able to relate it to our batteries at whatever the temperature ie the question I ask each morning  is 'how are my batteries doing - which is code for what do I think their current capacity is?

But you haven't lost capacity, it is just temporarily unavailable! You have only lost it if you are in a time critical application such as a UPS. But yes to some extent that argument does boil down to how one chooses to define SoC. A gold standard is the specific gravity of the acid and that doesn't suffer from any of the temporary effects you mention. I therefore suggest that its better to choose other definitions of SoC to match.

 

17 minutes ago, WotEver said:

Been thinking.

Getting back to...

Is it?  The only thing?

So please explain how the battery has warmed up during a heavy discharge. 

I accepted this as read earlier but I’m unable to find and corroboration of this claim. 

So where do you get this from, Pete?

A battery warms due to wasted energy during a rapid discharge, due to the battery's resistance. Of course discharging fast is wasteful of energy, but that is not the point in question. Once again I remind you that AH, SoC etc does not have any dimensions of power or energy and so this is a red herring.

 

Time for bed, round 2 sometime later!

Edited November 25, 2017 by nicknorman

[WotEver]

6 minutes ago, nicknorman said:

A battery warms due to wasted energy during a rapid discharge, due to the battery's resistance. Of course discharging fast is wasteful of energy, but that is not the point in question. Once again I remind you that AH, SoC etc does not have any dimensions of power or energy and so this is a red herring.

How can it possibly be a red herring? Energy is lost by heating the battery, therefore the point that you and Pete keep making (that the battery will recover to the same state it would have been in had the discharge been slower) patently incorrect. 

Now you’re inventing an energy source to heat the battery. 

A ‘red herring’ that exists, forms part of Peukert’s equation, that can be calculated and measured. That’s an interesting concept of a red herring. 

[nicknorman]

8 minutes ago, WotEver said:

How can it possibly be a red herring? Energy is lost by heating the battery, therefore the point that you and Pete keep making (that the battery will recover to the same state it would have been in had the discharge been slower) patently incorrect. 

Now you’re inventing an energy source to heat the battery. 

A ‘red herring’ that exists, forms part of Peukert’s equation, that can be calculated and measured. That’s an interesting concept of a red herring. 

Not quite in bed yet. Have a look at the dimensions of Peukert’s equation. There is nothing about power or energy in it. Therefore Peukert can have nothing to do with why a battery warms when you discharge it fast. Basic physics! You did do dimensional analysis at school/college/uni, didn’t you? If not you may be struggling to understand the concept. Look it up!

if we can’t get beyond this fundamental issue, we can’t progress. Remember (for the nth time) AH, SoC, current etc has no concept of power or energy. Peukert is only about current and time.

[WotEver]

6 minutes ago, nicknorman said:

Not quite in bed yet. Have a look at the dimensions of Peukert’s equation. There is nothing about power or energy in it. Therefore Peukert can have nothing to do with why a battery warms when you discharge it fast. Basic physics! You did do dimensional analysis at school/college/uni, didn’t you? If not you may be struggling to understand the concept. Look it up!

if we can’t get beyond this fundamental issue, we can’t progress. Remember (for the nth time) AH, SoC, current etc has no concept of power or energy. Peukert is only about current and time.

Oh, this is the problem. You don’t understand Peukert. 

You’re wrong Nick, Peukert is all about lost energy. He showed how more power was removed by a heavier discharge than a lighter one. He wasn’t the first to notice it but he was the first to quantify it. And his equations take into account the energy lost due to the higher internal resistance of the battery as well as the simple dispersal limitations which are reversed by battery recovery. 

Edited November 25, 2017 by WotEver

[WotEver]

Peukert didn’t invent his calculations from basic theory which you appear to get hung up on, his equations are entirely empirical. 

[nicknorman]

So you have 50AH stored in a battery. How much energy is that? No idea!

You have 10A flowing in a circuit. How much power is that? No idea?

because charge and current have no energy or power in themselves. They only do when there is voltage involved.

take 100AH battery, fully charged. Discharge at 50 A for 1 hour. How many AH used? 50. How many left? 50.

ditto but discharge at 5A for 10 hours. You have use 50AH. You have 50 left.

BUT how much energy did you get? Well in the first case the voltage was an average of 12v so you got 600 WH

second case the voltage was an average of 12.5 so you got 625 WH. In the first case, 25WH was lost and it heated the battery.

but in both cases you have used 50 AH and you have 50AH left (if you discharge slowly).

None of that has anything to do with Peukert.

18 minutes ago, WotEver said:

Oh, this is the problem. You don’t understand Peukert. 

You’re wrong Nick, Peukert is all about lost energy. He showed how more power was removed by a heavier discharge than a lighter one. He wasn’t the first to notice it but he was the first to quantify it. And his equations take into account the energy lost due to the higher internal resistance of the battery as well as the simple dispersal limitations which are reversed by battery recovery. 

This is your fundamental problem. You don’t understand what is behind Peukert. Please write down Peukert’s equation and show where it represents energy. You won’t be able to. It doesn’t have the dimensions of energy. Don’t believe me, ask a physicist. Your argument is akin to sayin 2+ 2 = an Orange. It simply doesn’t make sense at all. 

Edited November 26, 2017 by nicknorman

[nicknorman]

9 minutes ago, WotEver said:

Peukert didn’t invent his calculations from basic theory which you appear to get hung up on, his equations are entirely empirical. 

Absolutely, and they are all about current and charge, which don’t have the dimensions of power or energy (for the n+1th time!). He showed that a battery became “flat” quicker if you discharged it faster, which of course it does. His equation has nothing to do with lost energy, only to do with lost AH.

Of course as I’ve pointed out, those AH are only temporarily lost!

We are getting nowhere. You will never believe me as you clearly don’t have the grounding in basic physics to understand my point. Please take Peukert’s equation to a physicist and ask him whether it describes a loss of energy, or a loss of charge. Perhaps you will believe someone independent.

Edited November 26, 2017 by nicknorman

[WotEver]

16 minutes ago, nicknorman said:

But as I’ve pointed out, those AH are only temporarily lost.

And this is where you’re wrong. For the n’th time!

Some of those Ah are lost completely for that cycle because the energy that could have produced them got spent heating the battery. 

Despite all your waffling about dimensions Peukert clearly demonstrates empirically that you won’t have as many Ah to play with when discharging heavily as you will when discharging gently. 

The capacity does not completely recover, no matter how many times you say it does. 

Get a battery on a bench and do some tests if you don’t believe me. Every battery monitor designer agrees with me. 

22 minutes ago, nicknorman said:

take 100AH battery, fully charged. Discharge at 50 A for 1 hour. How many AH used? 50. How many left? 50.

Nope. Not even close. Try it some time. 

Better still, apply Mr Peukert’s equation on it. 

Edited November 26, 2017 by WotEver

[nicknorman]

1 minute ago, WotEver said:

And this is where you’re wrong. For the n’th time!

Some of those Ah are lost completely for that cycle because the power that could have produced them got spent heating the battery. 

Despite all your waffling about dimensions Peukert clearly demonstrates empirically that you won’t have as many Ah to play with when discharging heavily as you will when discharging gently. 

The capacity does not completely recover, no matter how many times you say it does. 

Ok you carry on with your delusion, I’m going to sleep. Do look at dimensional analysis though. It is not waffle, it is a very useful and powerful - and simple! - concept.

in your very first sentence you equate AH and power. An instant fail of dimensional homogeneity and therefore must be wrong without having to look any further.

[WotEver]

1 minute ago, nicknorman said:

Ok you carry on with your delusion, I’m going to sleep. Do look at dimensional analysis though. It is not waffle, it is a very useful and powerful - and simple! - concept.

in your very first sentence you equate AH and power. An instant fail of dimensional homogeneity and therefore must be wrong without having to look any further.

Everyone will see through your waffle Nick. 

You've invented some battery fairies to warm it up.  

[nicknorman]

9 hours ago, WotEver said:

Everyone will see through your waffle Nick. 

You've invented some battery fairies to warm it up.  

You confuse Peukert, which is only about current and charge and has no dimensions of power or energy, with the energy efficiency of a fast vs slow discharge, the former producing less energy due to lower terminal voltage due to a combination of battery reaction speed and resistance. The lost energy goes to warming the battery.

These two issues are unrelated. Must be unrelated if you look at the dimensions, unless you think 2+2= an orange. Until you grasp this absolutely fundamental and basic point, which seems unlikely, this discussion is going no where.

I give up. This discussion reminds me of arguing on the doorstep with Jehova’s Witnesses. Once someone has blind faith in a point, no amount of rational argument is going to sway them even though they cannot produce any evidence or science to support their position, and even though there is fundamental science and evidence against their point.

Edited November 26, 2017 by nicknorman

[WotEver]

No, I understand Peukert and his observations on the capacity (the real, observed capacity) of a battery under differing loads. 

Your claim, repeated by Pete, that the battery will recover to the same SoC regardless of the discharge rate, is unsupported by the evidence. Everyone appears to know this except you two. 

I came across an interesting snippet on Gibbo’s site last night...

Some have gone further and stated that Peukert's effect is purely a result of limited electrolyte dispersal and diffusion slowing down the chemical reactions (this is really battery recovery and/or surface discharge which is covered elsewhere on this website).

The whole article makes an interesting read:

http://www.smartgauge.co.uk/peukert2.html

[rusty69]

Perhaps one of you could persuade Gibbo back, a guest appearance,  to act as umpire 

[nicknorman]

Just now, WotEver said:

No, I understand Peukert and his observations on the capacity (the real, observed capacity) of a battery under differing loads. 

Your claim, repeated by Pete, that the battery will recover to the same SoC regardless of the discharge rate, is unsupported by the evidence. Everyone appears to know this except you two. 

Nobody disputes Peukert’s results and the validity of his equation. It is only the false application of his equation that is in question. Which is odd because everyone has seen that my position is correct. Well certainly those of a certain age, before cars were perfect.

You go to start your aging car, it grinds over for a while until the battery is flat. Click click nothing happening, the battery is flat at that high discharge rate. You go away for 30mins, come back and click whirr vrooom! The car starts. This is a perfect example of the allegedly “lost” capacity being recovered.

2 minutes ago, rusty69 said:

Perhaps one of you could persuade Gibbo back, a guest appearance,  to act as umpire 

Gibbo is driving Tony’s position. It something he got wrong and could never see. He wasn’t THAT clever!