Battery Fault

[Dr Bob]

2 hours ago, nicknorman said:

So in the case of the voltage check, you just have to leave it longer for the true open circuit voltage to become apparent.

Since boat batteries are not normally fully discharged at a constant rate, that measure of capacity is fairly irrelevant. Certainly if the desire is not to go below 50% for cycle life reasons, there is no need to take temperature into account to decide how much AH can be taken out.

That's exactly what I was getting at. At lower temps you will get to the 12.2v( or whatever you lower voltage) at rest sooner, so you do need to pay attention to temp, unless you leave it for a longer period to recover. If like me, your voltage is never 'at rest' - inverter always on- then the voltage will be slightly depressed as well. Nick is right that temp will not affect ahrs taken out.

[WotEver]

8 minutes ago, Dr Bob said:

Nick is right that temp will not affect ahrs taken out.

But Nick gets hung up on the phrase ‘is lost’. He argues the same about Peukert. 

Yes, if you leave a cold battery long enough it will recover. Yes, if you leave a quickly discharged battery long enough it will recover. If you don’t have the time to let it recover because you need to continue with your life then that energy is unavailable to you for that discharge cycle and is effectively ‘lost’ until the next time. 

[nicknorman]

52 minutes ago, WotEver said:

But Nick gets hung up on the phrase ‘is lost’. He argues the same about Peukert. 

Yes, if you leave a cold battery long enough it will recover. Yes, if you leave a quickly discharged battery long enough it will recover. If you don’t have the time to let it recover because you need to continue with your life then that energy is unavailable to you for that discharge cycle and is effectively ‘lost’ until the next time. 

What you say is true, and it would be relevant if we were in the habit of fully discharging our batteries at a relatively high rate all the way from fully charged to flat. But we aren’t! Most folk are at worst discharging at an average of ~ the C50 rate (ie more than 2 days to flat) and since batteries in general are specified at the C20 rate you are already well above rated capacity even with a cold battery and even with some brief periods of high discharge (followed by long periods of minimal discharge).

(”we” in the above being your average narrowboater)

Edited November 24, 2017 by nicknorman

[smileypete]

On 11/23/2017 at 23:11, WotEver said:

But Nick gets hung up on the phrase ‘is lost’. He argues the same about Peukert. 

Yes, if you leave a cold battery long enough it will recover. Yes, if you leave a quickly discharged battery long enough it will recover. If you don’t have the time to let it recover because you need to continue with your life then that energy is unavailable to you for that discharge cycle and is effectively ‘lost’ until the next time. 

Only if you discharge the batt to 10.5V.

Otherwise it's just a distraction that confuses people.

Edited November 25, 2017 by smileypete

[WotEver]

2 hours ago, smileypete said:

Only if you discharge the batt to 10.5V.

I think you need to think about that statement. Nick says the same (perhaps that’s where you got it from?) and he’s wrong. 

[nicknorman]

32 minutes ago, WotEver said:

I think you need to think about that statement. Nick says the same (perhaps that’s where you got it from?) and he’s wrong. 

I am never wrong.

 

Well, only when I'm wrong. Which I'm not in this case!

[jddevel]

15 minutes ago, nicknorman said:

I am never wrong.

 

Well, only when I'm wrong. Which I'm not in this case!

My wife adds you just didn`t hear me correctly and mis-understood!!!!

[WotEver]

27 minutes ago, nicknorman said:

I am never wrong.

 

Well, only when I'm wrong. Which I'm not in this case!

That’s impossible because I’m never wrong either. 

Except when my wife tells me I am because she never is  

 

But you are, about this, as we’ve discussed before. 

[nicknorman]

50 minutes ago, WotEver said:

That’s impossible because I’m never wrong either. 

Except when my wife tells me I am because she never is  

 

But you are, about this, as we’ve discussed before. 

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.

 

Its all very well saying I am wrong but unless you come up with an alternative theory to explain your version of events, people aren't going to believe you any more than I believe our local vicar when he talks about god.

Edited November 25, 2017 by nicknorman

[WotEver]

Here we go again. 

A nice simple example. 

Boater has a 200Ah (C20) battery. He always starts his engine as his battery approaches 50% SoC. He’s somewhat obsessive about this  

Day 1 he discharges his battery at an average C20 rate and uses 100Ah before he has to start his engine to prevent the batteries going below 50% SoC. 

Day 2 he has relatives visiting and with inverter use, constant pump, fridge being opened every 5 minutes, laptops on charge etc he discharges his batteries at an average C5 rate. He finds that he has to start his engine after using only 82Ah. He’s ‘lost’ 18Ah that day. 

You can stand on the bank and explain to him that they’re not really lost and if he rests his batteries for a few hours he can regain that missing energy if you like. In the mean time he’s started his engine so as to get on with his life. 

The fact that all the energy is available if only he used it more slowly is totally irrelevant to him. That day he has ‘lost’ nearly 20% of the energy he had available the previous day, simply by discharging at a higher rate, in total accordance with Peukert and without dropping to 10.5V. 

Note that I use the word ‘lost’ in inverted commas each time because as we both know the energy is there within the battery, just not available to him on this discharge cycle. 

[smileypete]

50 minutes ago, WotEver said:

The fact that all the energy is available if only he used it more slowly is totally irrelevant to him. That day he has ‘lost’ nearly 20% of the energy he had available the previous day, simply by discharging at a higher rate, in total accordance with Peukert and without dropping to 10.5V.

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

The 10.5V cutoff is just to minimise the chance of a weak cell going into reversal.

How does Mr Boater know when his batts have reached 50%?

Edited November 25, 2017 by smileypete

[nicknorman]

1 hour ago, WotEver said:

Here we go again. 

A nice simple example. 

Boater has a 200Ah (C20) battery. He always starts his engine as his battery approaches 50% SoC. He’s somewhat obsessive about this  

Day 1 he discharges his battery at an average C20 rate and uses 100Ah before he has to start his engine to prevent the batteries going below 50% SoC. 

Day 2 he has relatives visiting and with inverter use, constant pump, fridge being opened every 5 minutes, laptops on charge etc he discharges his batteries at an average C5 rate. He finds that he has to start his engine after using only 82Ah. He’s ‘lost’ 18Ah that day. 

You can stand on the bank and explain to him that they’re not really lost and if he rests his batteries for a few hours he can regain that missing energy if you like. In the mean time he’s started his engine so as to get on with his life. 

The fact that all the energy is available if only he used it more slowly is totally irrelevant to him. That day he has ‘lost’ nearly 20% of the energy he had available the previous day, simply by discharging at a higher rate, in total accordance with Peukert and without dropping to 10.5V. 

Note that I use the word ‘lost’ in inverted commas each time because as we both know the energy is there within the battery, just not available to him on this discharge cycle. 

How is he measuring SoC? If with a decent AH counting gauge, they correctly don’t apply peukert because it’s not relevant to SoC, only to “time to run at the current rate of discharge”. So your example is fallacious. His gauge will show the same SoC regardless of how fast the charge is taken out.

Think about it. If an AH gauge applied Peukert, at the end of a fast discharge it would show a lower reading than it would have done for the same AH taken out slowly. But then if the battery was rested the SoC reading would have to increase again. So which SoC reading was correct, the one just after the fast discharge, or the one hours later after no further discharge? It just can’t work like that. This is why AH counting gauges don’t use Peukert for SoC display, only for time to run.

SoC is a measure of the percentage of chemicals left unreacted. Peukert doesn’t come into that. Nor does battery temperature.

And let’s bear in mind other methods of measuring SoC:

Specific gravity? Unaffected by Peukert. 

Resting voltage? Unaffected by Peukert, though it would be necessary to let the battery rest for longer to get the true rested voltage following a fast discharge or with a cold battery.

Put simply, I can’t think of any means of measuring intermediate SoC that would be affected by Peukert issues.

Edited November 25, 2017 by nicknorman

[Dr Bob]

15 minutes ago, nicknorman said:

SoC is a measure of the percentage of chemicals left unreacted. Peukert doesn’t come into that. Nor does battery temperature.

 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. Looking at tables/graphs on the internet, it looks like a 20% reduction in capacity is seen dropping from 25degC to 5degC. 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. 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.

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. 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.

[nicknorman]

4 minutes ago, Dr Bob said:

 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.

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

[WotEver]

2 hours ago, smileypete said:

How does Mr Boater know when his batts have reached 50%?

With a SmartGauge

2 hours ago, nicknorman said:

How is he measuring SoC?

With a SmartGauge

[WotEver]

2 hours ago, nicknorman said:

If an AH gauge applied Peukert, at the end of a fast discharge it would show a lower reading than it would have done for the same AH taken out slowly. But then if the battery was rested the SoC reading would have to increase again. 

Yes it would, which is exactly what SmartGauge does. 

2 hours ago, nicknorman said:

So which SoC reading was correct, the one just after the fast discharge, or the one hours later after no further discharge?

Both of them of course. At the time they were displayed. 

Peukert isn’t an imaginary phenomenon and it doesn’t only occur when the battery is flat. The effect is real and can be observed from the moment you start discharging the battery.  This is why battery manufacturers offer discharge curves at C20, C10 etc. 

2 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.

Which has precisely zero to do with a real life boater who doesn’t sit around waiting for his battery voltage to stabilise over 4 hours. 

[nicknorman]

1 minute ago, WotEver said:

With a SmartGauge

With a SmartGauge

So it then boils down to how Gibbo decided to programme it. Since his thinking in this area is flawed, I’m not that confident it will be right. To test it one would have to discharge a cold battery fast for quite a while. Observe the SG’s SoC reading. Then let the battery rest for quite while. Then continue a more modest discharge. If the SG reading froze at its current indication for quite a while, we’d know that Gibbo got it wrong. But if it continued on its path of downward SoC (at a much slower rate of course) we’d know he’d got it right.

[WotEver]

2 hours ago, nicknorman said:

Put simply, I can’t think of any means of measuring intermediate SoC that would be affected by Peukert issues.

Instantaneous voltage. We have plenty of folk on here who claim to know their battery’s SoC from a voltmeter. 

[nicknorman]

Just now, WotEver said:

Instantaneous voltage. We have plenty of folk on here who claim to know their battery’s SoC from a voltmeter. 

But as you well know, instantaneous voltage is not a reflection of SoC.

[WotEver]

Just now, nicknorman said:

To test it one would have to discharge a cold battery fast for quite a while. Observe the SG’s SoC reading. Then let the battery rest for quite while. Then continue a more modest discharge. If the SG reading froze at its current indication for quite a while, we’d know that Gibbo got it wrong. But if it continued on its path of downward SoC (at a much slower rate of course) we’d know he’d got it right.

I seem to recall that you did this. Turning on the kettle for a while caused a dip in displayed SoC which gradually recovered once the heavy load was removed. 

1 minute ago, nicknorman said:

But as you well know, instantaneous voltage is not a reflection of SoC.

Yes, I know that but there are plenty of folk on here who claim they can tell their battery’s SoC from their voltmeter. “Nothing else is required” is their mantra. 

3 minutes ago, nicknorman said:

Since his thinking in this area is flawed,

I don’t believe it is. You keep referring to rested voltage, which Pete then repeats, but as I keep pointing out, the battery doesn’t get rested. It gets used and then recharged. 

[nicknorman]

11 minutes ago, WotEver said:

Yes it would, which is exactly what SmartGauge does. 

Both of them of course. At the time they were displayed. 

Peukert isn’t an imaginary phenomenon and it doesn’t only occur when the battery is flat. The effect is real and can be observed from the moment you start discharging the battery.  This is why battery manufacturers offer discharge curves at C20, C10 etc. 

Which has precisely zero to do with a real life boater who doesn’t sit around waiting for his battery voltage to stabilise over 4 hours. 

Well then your idea of SoC is different from mine. In my world, a battery not subject to charge or discharge has a SoC that doesn’t vary with time except for self-discharge. In your world a battery not subject to charge or discharge (etc) has a variable SoC depending on its recent usage. I think that is an unhelpful definition of SoC but you are entitled to your opinion! It certainly doesn’t correlate with measuring SoC by means of specific gravity.

Peukert is not an imaginary phenomena. As you say, it is why there are different specified capacities according to discharge rate. But it is a phenomena that is misapplied by some people including you.

Edited November 25, 2017 by nicknorman

[WotEver]

4 minutes ago, nicknorman said:

it is a phenomena that is misapplied by some people including you.

I respectfully disagree. 

[nicknorman]

6 minutes ago, WotEver said:

I seem to recall that you did this. Turning on the kettle for a while caused a dip in displayed SoC which gradually recovered once the heavy load was removed. 

Yes, I know that but there are plenty of folk on here who claim they can tell their battery’s SoC from their voltmeter. “Nothing else is required” is their mantra. 

I don’t believe it is. You keep referring to rested voltage, which Pete then repeats, but as I keep pointing out, the battery doesn’t get rested. It gets used and then recharged. 

1st para. No, SG tends not to increase its reading if it’s too low. It tends to just sit and wait until the actual SoC matches the indicated SoC. But I don’t remember the occurrence you’re talking about.

3rd para: I know Gibbo is your hero and he has a lot of good thought processes, but he isn’t perfect. It doesn’t matter whether the battery is rested or not, unless you intend to fully discharge it.

[WotEver]

1 minute ago, nicknorman said:

SG tends not to increase its reading if it’s too low. It tends to just sit and wait until the actual SoC matches the indicated SoC. But I don’t remember the occurrence you’re talking about.

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. 

[nicknorman]

Just now, WotEver said:

I respectfully disagree. 

As is your prerogative! But unfortunately for you, manufacturers of AH counting gauges side with me in not taking Peukert or battery temperature into account when calculating SoC. They only use it for time to run. And as I said, if your idea of SoC floats around according to prior usage pattern, I think it is woolly and not useful. SoC in my book relates to the percentage of charge taken out vs that which is available. Not vs that which is available in a short time.

of course if you are Mr Cuthound etc, Peukert is very important because he wants to use the batteries for backup power and needs to know how long they will run for beforemtheir voltage drops too low. But that isn’t how boaters use their batteries.