Battery Fault

[WotEver]

14 minutes ago, nicknorman said:

Now who does that remind me of???

Yourself, obviously

[frahkn]

16 minutes ago, nicknorman said:

We have a young chap in the gliding club, he did a degree in aeronautical engineering and is doing a PhD in computational fluid dynamics. He is pretty clever and a good pilot, and a nice guy. BUT he is a creationist. He firmly believes that the world started up a few thousand years ago, created by god in 7 days etc. He knows he is out of kilter with the rest of us but there is no swaying him. He just knows that he is right and any attempt at debate (and we have tried plenty!) is just met with blind faith that he is right regardless of any evidence. He actually seems to relish the challenge of having his views rubbished!

Now who does that remind me of???

I "just know" that both of the two highlighted phrases cannot be correct. If either is right then the other is wrong (both, of course, may be wrong).

I "just know" that I'm right here.

[nicknorman]

25 minutes ago, WotEver said:

So if there is less energy, in the new Physics Invented By Nick there is still the same charge?

 

charge x voltage = energy

same charge x less voltage = less energy. I couldn’t really put it any simpler.

Lets start a new thread about the flatness of the earth. Or otherwise.

2 minutes ago, frahkn said:

I "just know" that both of the two highlighted phrases cannot be correct. If either is right then the other is wrong (both, of course, may be wrong).

I "just know" that I'm right here.

I agree it is entirely irrational. But then as we have discovered on this thread, there are plenty of irrational people around! Doesn’t necessarily mean they arent clever in every other respect though.

Lots of clever people believe in god. Humans are not binary when it comes to rationality.

[WotEver]

1 hour ago, nicknorman said:

charge x voltage = energy

same charge x less voltage = less energy. I couldn’t really put it any simpler.

Absolutely. We don’t disagree on that, never have. 

1. Peukert notes a loss of charge. 

2. Some of that charge is lost through reduced energy in the battery through heating and other effects. 

3. Some of the charge is lost through slow dispersion. 

4. Peukert, unconcerned with why the charge is lost, writes an equation which demonstrates the results he has observed in 1). 

Nick gives a new meaning to Peukert’s observations and tries to convince us all that the world is flat. 

[nicknorman]

Sigh.

2. No charge can be lost. Kirchov’s first law. Charge does not have any intrinsic energy. I definitely give up now.

Edited November 29, 2017 by nicknorman

[bizzard]

The world is flat. The southern edge is in Benfleet creek, between the mainland and Canvey Island. Between beer and lager.

Edited November 29, 2017 by bizzard

[WotEver]

1 hour ago, nicknorman said:

No charge can be lost. Kirchov’s first law. Charge does not have any intrinsic energy.

That’s what I keep saying!

Charge is an analogue of energy. If there is less energy there will be less charge.

Peukert noted less charge. That was partly because there was less energy. But you would have us believe that Peukert noted the total losses, calculated how much of that was due to wasted energy without telling anyone, and then created an equation to show only the dispersion bit. 

Apart from being nonsense it’s an absurd assertion. 

[Dr Bob]

1 hour ago, nicknorman said:

Sigh.

2. No charge can be lost. Kirchov’s first law. Charge does not have any intrinsic energy. I definitely give up now.

Nick, I think this where you are going wrong. I think we need to define what 'charge' is. It does have 'energy' but maybe not energy that is easily defined in terms of volts and amps. Again it goes back to battery chemistry. Each reactive site can combine PbO and Pb with a HSO4- ion and release 2 electrons. Those 2 electrons have a charge and will sit on the surface of the plates giving a voltage potential. Now, not all the reactants will react as it is a reversible reaction and once enough molecules have reacted to give a voltage potential of 12.7V (in a six cell battery) with the energised electrons still sitting on the surface with nowhere to go, the forward reaction will reverse and some of the energised electrons will provide the energy to create the Pb and PbO again. Once fully charged the battery will sit there with only a proportion of the active molecules on the right hand side of the equation with the 2 electrons each.

What I dont understand is what percentage of the active molecules are actually over to the right and so sitting with energised electrons ready to go. The total 'charge' in the battery is not just the electrons ready to go but also the 2 electrons per site that will be released once the ones that are there start to flow. You cannot therefore relate total charge to voltage potential.

Nick, you are an electronic engineer so what is the voltage potential from a single electron and how many electrons would need to be available to give 12.7V?

Now moving on, as you connect up a circuit and start removing electrons at a constant rate (ie constant amps) you can calculate the power as you have As and a voltage. As you remove electrons, then they will be replaced by other active sites now reacting to produce 2 electrons. As each set of reactants disappears and produces 2 electrons, that 'charge' has now been used and the overall charge of the battery will decrease.

What I am not understanding is as the reactants get used up – and so charge reduces (ie unreacted active molecules are less in number) and the number of energised electrons on the surface reduces – the voltage drops in the battery – but why? It sounds like it is a function of less electrons on the plate surface hence we need to understand the relationship between number of electrons and voltage potential.

The link Chewbacca provided had the following lines:

When a current flows round a series circuit the current at any point in the circuit is the same. The same number of electrons flow past any point in the circuit every second - no electrons are lost. Although the number of electrons is always the same their energy gets less as they move round the circuit.
This energy appears as heat, light or magnetism in say an electrical heater, a light bulb or an electromagnet.

I think this explains the heat loss from internal resistance arising. Current will flow with the same number of electrons going round the internal circuit but their energy will reduce – so less voltage potential. Hence energy is lost. That brings me back to the point above – what is the voltage potential relationship to fully energised electrons?

In summary, I think a big issue in what you and Wotever are discussing is that you cannot just describe battery charge as a voltage or Amps or Ahrs etc as the majority of the energy stored in a battery is held by the constituent chemicals held in a higher state of thermodynamic energy and when allowed to mix will release that stored 'energy' as charged electrons. What I dont understand is how many electrons need to be released to generate a Volt of potential (but maybe we also need to define a current as well).

[WotEver]

9 minutes ago, Dr Bob said:

In summary, I think a big issue in what you and Wotever are discussing is that you cannot just describe battery charge as a voltage or Amps or Ahrs etc

Correct. Ah are proportional to energy at a given fixed voltage. If the energy is less, so will the observed charge be less. That’s the bit that Nick can’t see. 

And as previously agreed, a battery doesn’t store charge, it stores energy. 

[Chewbacka]

1 hour ago, Dr Bob said:

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

In summary, I think a big issue in what you and Wotever are discussing is that you cannot just describe battery charge as a voltage or Amps or Ahrs etc as the majority of the energy stored in a battery is held by the constituent chemicals held in a higher state of thermodynamic energy and when allowed to mix will release that stored 'energy' as charged electrons. What I dont understand is how many electrons need to be released to generate a Volt of potential (but maybe we also need to define a current as well).

I don't think there is a fixed number of electrons required.  The reaction that 'frees' electrons can go in either direction. the voltage that will stop the reaction is about 2.2v in a lead acid cell.  (For NiCd it is about 1.2V and for lithium about 3V)  So long as there are sites on the plates ready to chemically change there will be a voltage.  The amount of current that can be taken is dependant upon the rate at which electrons can be liberated, and so for more current we need bigger area plates.  Bigger plates (by area) means either a bigger peak current or the same current for longer.  As we increase the current, the rate of electron release increases until the voltage starts to drop, and this will encourage a faster rate of electron releases as the voltage preventing electron release on the plate surface is reduced, so more sites will release electrons, if you get my meaning.  So the bigger the current the bigger the volt drop.  If you suddenly stop taking current the reaction on the plate surface will continue releasing electrons, until the voltage rises to that required to stop further electron release - about 2.2V for lead acid.  I shall stop now before I ramble for too long and confuse myself.

 

Added - If we were talking about a capacitor then for a given capacitance it would require a fixed number of electrons to generate a voltage across the plates, the bigger the capacitance the more electrons are required to create a 1V potential.  However we are talking chemical reactions..............

Edited November 29, 2017 by Chewbacka

[Dr Bob]

4 minutes ago, Chewbacka said:

I don't think there is a fixed number of electrons required.  The reaction that 'frees' electrons can go in either direction. the voltage that will stop the reaction is about 2.2v in a lead acid cell.  (For NiCd it is about 1.2V and for lithium about 3V)  So long as there are sites on the plates ready to chemically change there will be a voltage.  The amount of current that can be taken is dependant upon the rate at which electrons can be liberated, and so for more current we need bigger area plates (ie much less steric hinderance and bigger area so ion depletion is less significant).  Bigger plates (by area) means either a bigger peak current or the same current for longer.  As we increase the current, the rate of electron release increases until the voltage starts to drop, and this will encourage a faster rate of electron releases as the voltage preventing electron release on the plate surface is reduced, so more sites will react and release electrons, if you get my meaning. Absolutely. So the bigger the current the bigger the volt drop.  If you suddenly stop taking current the reaction on the plate surface will continue releasing electrons, until the voltage rises to that required to stop further electron release - about 2.2V for lead acid. 

At this point I fully agree with your description and I think it mirrors what I have said from the chemistry angle. I agree that the bigger the current the bigger the voltage drop - but and the big BUT is why when 50% of the available sites have reacted and given up their electrons, why is the 'observed' at rest voltage now at 12.2V (ish) on a six cell LA battery? My chemistry and your description both would suggest that the at rest voltage should still be 12.7V and I think Nick in a post way way back here said just one electron would have a potential of the max value (ie 2.2v in a single cell).

What is the relationship between active electrons on the surface and the voltage at rest on the terminals - given only a small fraction of the active sites will have reacted to produce electrons. Again I ask 'what is the voltage potential of an electron. Is the lower voltage at 50% SoC is 12.2V lower than 12.7V because there are less electrons on ths surface of the plate? I havent a clue.

I shall stop now before I ramble for too long and confuse myself. I also shall stop now as I have confused myself

I have added a few phrases in red.

1 hour ago, WotEver said:

 

And as previously agreed, a battery doesn’t store charge, it stores energy. 

...and that energy is in the form of thermodynamic energy stored in the molecules that have not reacted to release their electrons.

[nicknorman]

39 minutes ago, Dr Bob said:

Nick, I think this where you are going wrong. I think we need to define what 'charge' is. Well you may be right there. Charge is well defined elsewhere by the rest of the scientific world, but not here it seems! As I mentioned earlier, some confusion may be arising because folk are looking at "charge" in relation to "charging" a battery. But that is not what the word nor the dimension/unit/property means in scientific terms, and since Mr Peukert was a scientist I think its fair to say that the "charge" in his equation is the same "charge" as used by the rest of the scientific world. Charge is the property of something that causes it to experience a force when in an electric field. It does have 'energy' but maybe not energy that is easily defined in terms of volts and amps. No, charge absolutely does not have any intrinsic energy. The Newtonian analogy of Charge is Mass. Mass as a property doesn't intrinsically instill its posessor with any energy. An item having mass that has zero velocity and is on the notional surface of the earth has zero energy. It only gains energy when it has velocity or is raised up such that it can fall back again. Voltage is the analogy of height in that one. You are supposed to have a PhD - you cannot just say "Oh, I think has energy, but not as we know it (Jim)"! Again it goes back to battery chemistry. Each reactive site can combine PbO and Pb with a HSO4- ion and release 2 electrons. Those 2 electrons have a charge and will sit on the surface of the plates giving a voltage potential. Now, not all the reactants will react as it is a reversible reaction and once enough molecules have reacted to give a voltage potential of 12.7V (in a six cell battery) with the energised electrons still sitting on the surface with nowhere to go, the forward reaction will reverse and some of the energised electrons will provide the energy to create the Pb and PbO again. Once fully charged the battery will sit there with only a proportion of the active molecules on the right hand side of the equation with the 2 electrons each.

What I dont understand is what percentage of the active molecules are actually over to the right and so sitting with energised electrons ready to go. The total 'charge' in the battery is not just the electrons ready to go but also the 2 electrons per site that will be released once the ones that are there start to flow. You cannot therefore relate total charge to voltage potential.

Nick, you are an electronic engineer so what is the voltage potential from a single electron and how many electrons would need to be available to give 12.7V? An electron only has charge, it doesn't have voltage. The relationship between voltage and charge is that of capacitance, ie V = Q/C. So if we take an item, say a battery plate at zero voltage, and add an electron, we get a voltage rise (negative of course, but lets not complicate things!) according to that equation. The capacitance of a plate is probably fairly small but then so is the charge of an electron so I dare say it takes quite a lot of electrons being pushed onto the plate to get the voltage up to 2.2v or whatever. Of course after the first electron gets to the plate, the plate's voltage starts to rise and so the next electron has to be pushed "up hill" to the plate at a slightly higher potential. This carries on with subsequent electrons having to be pushed harder (in an instant, no doubt) until the energy required to push the electron up the electric field gradient reaches the limit of what the chemical reaction can impart, then the flow of electrons stops with the voltage at 2.2v. Its like trying to build a pile of stones, eventually the pile gets so high that there isn't enough energy to add any more to the top.

Now moving on, as you connect up a circuit and start removing electrons at a constant rate (ie constant amps) you can calculate the power as you have As and a voltage. As you remove electrons, then they will be replaced by other active sites now reacting to produce 2 electrons. As each set of reactants disappears and produces 2 electrons, that 'charge' has now been used and the overall charge of the battery will decrease. Agree. Of course one source of confusion is that the battery's "charge" is chemical energy, not a massive capacitor-like thing that is storing charge. By mixing these two uses of the word "charge" indiscriminately, confusion is created.

What I am not understanding is as the reactants get used up – and so charge reduces (ie unreacted active molecules are less in number) and the number of energised electrons on the surface reduces – the voltage drops in the battery – but why? It sounds like it is a function of less electrons on the plate surface hence we need to understand the relationship between number of electrons and voltage potential. I don't really know but as you have said, it is a reversible reaction and something to do with the acid strength or whatever, determines the equilibrium point. As the chemical energy is used up (being careful not to say "charge"!). With less reaction energy available to push the electrons up the electric field gradient, the reaction stops at a lower voltage. As I said earlier, the number of electrons required to achieve a certain voltage depends on the capacitance of the plate.

The link Chewbacca provided had the following lines:

When a current flows round a series circuit the current at any point in the circuit is the same. The same number of electrons flow past any point in the circuit every second - no electrons are lost. Although the number of electrons is always the same their energy gets less as they move round the circuit.
This energy appears as heat, light or magnetism in say an electrical heater, a light bulb or an electromagnet.

I think this explains the heat loss from internal resistance arising. Current will flow with the same number of electrons going round the internal circuit but their energy will reduce – so less voltage potential. Hence energy is lost. That brings me back to the point above – what is the voltage potential relationship to fully energised electrons? The energy of the electrons starts out high because they have been "pumped up" to a higher potential (by the chemical reaction), and that energy reduces round a circuit because they move down a potential gradient. It is a bit like a moving mass gradually slowing down due to friction. In that analogy, voltage is analogous to speed. But as I've said there is no relationship between voltage and electrons any more than there is a relationship between mass and speed. Mass on its own, or speed on its own, do not have any intrinsic energy. Only when you have mass and speed do you get energy. So the only relationship between electrons and voltage potential is that described by the capacitance equation I mentioned earlier.

In summary, I think a big issue in what you and Wotever are discussing is that you cannot just describe battery charge as a voltage or Amps or Ahrs etc as the majority of the energy stored in a battery is held by the constituent chemicals held in a higher state of thermodynamic energy and when allowed to mix will release that stored 'energy' as charged electrons. What I dont understand is how many electrons need to be released to generate a Volt of potential Refer to the capacitance equation (but maybe we also need to define a current as well). Correct, and as I've said this is where Tony's confusion arises. He sees the word "charge" and thinks of the stored chemical energy. I see the word "charge" and I see the internationally defined scientific term - the property that makes something react to an electric field. Peukert's equation only talks of charge, measured in coulombs, AH or whatever. It does not discuss "charge" as in the chemical energy stored in a battery. It can't, because it doesn't have the dimensions of energy (it would need to include voltage to do that).

Answers in red above.

1 hour ago, WotEver said:

Correct. Ah are proportional to energy at a given fixed voltage. If the energy is less, so will the observed charge be less. But charge can't disappear or be "dropped" around a circuit so if energy is lost, that can only happen because the voltage reduces. The charge (in AH, Coulombs or whatever) cannot evaporate. You are using charge in the wrong sense. Charge is not a measure of energy. It is a measure of charge! That’s the bit that Nick can’t see. 

And as previously agreed, a battery doesn’t store charge, it stores energy. 

 

[nicknorman]

Just as an aside, I have noticed that when our Trojans are low on water (more concentrated acid) the voltage seems a little higher for a given SoC than it is when the electrolye is "full" ie weaker acid. (It's more noticable with the Trojans because they have a fairly big reservoir of electrolyte above the plates and do use water). Therefore I would be fairly confident that the equilibrium point of the reaction is determined by acid strength. I think the voltage could be made higher with a battery at say 50% SoC by increasing the strength of the acid. But of course the maximum acid strength is a compromise - too strong and a fully charged battery would suffer from excessive plate corrosion.

[Dr Bob]

7 minutes ago, nicknorman said:

I don't really know but as you have said, it is a reversible reaction and something to do with the acid strength or whatever, determines the equilibrium point. As the chemical energy is used up (being careful not to say "charge"!). With less reaction energy available to push the electrons up the electric field gradient, the reaction stops at a lower voltage. As I said earlier, the number of electrons required to achieve a certain voltage depends on the capacitance of the plate.

Nick, that is really useful. Firstly on the point of reduced voltage. You have introduced the concept of capacitance and that does help me a great deal understand why the voltage would go down to 12.2V at 50% SoC. That will do me and as long as someone doesnt come back and knock that idea, it fits my chemistry of the model.

For the rest I think you agree that the energy stored in the battery is a combination of the chemical energy (and to be more specific this is thermodynamic energy) and the electrons. Where we differ if that I would call this energy 'charge' ie as in state of charge. Fully agree it is not an electrical charge. The word "charge" I am using is the noun derived from the verb to charge (but I am crap at english and did fail my english o level at first attempt...duh!). The problem then comes when someone says 'what charge is left in the battery' and the answer to that is either a number of Ahrs or a % number which compares the current state to empty or full, but both of these are defined by the number of chemical sites left to react and the number of electrons on the plates.

Are we in agreement?

[Dr Bob]

4 minutes ago, nicknorman said:

Just as an aside, I have noticed that when our Trojans are low on water (more concentrated acid) the voltage seems a little higher for a given SoC than it is when the electrolye is "full" ie weaker acid. (It's more noticable with the Trojans because they have a fairly big reservoir of electrolyte above the plates and do use water). Therefore I would be fairly confident that the equilibrium point of the reaction is determined by acid strength. I think the voltage could be made higher with a battery at say 50% SoC by increasing the strength of the acid. But of course the maximum acid strength is a compromise - too strong and a fully charged battery would suffer from excessive plate corrosion.

Yes. A chemist would write the equation for the reaction and from that equation say what effects reation rate. For LA's the reaction rate is first order with respect to HSO4- ion concentration (as I said earlier on) so the higher the ion concentration the higher the rate of reaction hence now using your capacitor comment, the higher the voltage.

I think we are almost there!

[nicknorman]

3 minutes ago, Dr Bob said:

Nick, that is really useful. Firstly on the point of reduced voltage. You have introduced the concept of capacitance and that does help me a great deal understand why the voltage would go down to 12.2V at 50% SoC. That will do me and as long as someone doesnt come back and knock that idea, it fits my chemistry of the model.

For the rest I think you agree that the energy stored in the battery is a combination of the chemical energy (and to be more specific this is thermodynamic energy) and the electrons. Where we differ if that I would call this energy 'charge' ie as in state of charge. Fully agree it is not an electrical charge. The word "charge" I am using is the noun derived from the verb to charge (but I am crap at english and did fail my english o level at first attempt...duh!). The problem then comes when someone says 'what charge is left in the battery' and the answer to that is either a number of Ahrs or a % number which compares the current state to empty or full, but both of these are defined by the number of chemical sites left to react and the number of electrons on the plates.

Are we in agreement?

Yes, I think we pretty much are. On the definition of "charge" the confusion is as you say. But let's be clear, shunt based SoC gauges count AH. They don't count WH. SoC is a measure of the percentage remaining AH, not the percentage remaining WH. Peukert doesn't describe a loss of WH, he only describes a loss of AH (and its temporary, of course, because given time the "buried" chemicals diffuse together and are eventually able to react).

I can well see that it would be useful to have a battery energy meter that takes into account the percentage WH remaining. AH SoC gauges are a bit misleading because of course the first 50% of the "charge" is worth a lot more in terms of energy than the second 50%. That is probably what Tony was on about with his military contracts. But the fact remains that in boaty world we are stuck with AH SoC gauges. And Peukert's equation only describes "lost" AH, not lost WH from heating etc. Which is ironic because the lost AH are not ultimately lost, whereas the lost WH are!

[smileypete]

22 hours ago, WotEver said:

Also completely irrelevant. 

I have no idea what you mean by ‘extra electrons’ but if you mean why has the charge reduced by more than 50% it’s because some of it has been used to warm the battery due to the rising internal resistance. Just like Battery FAQ state. 

Nah, the heat comes from lowering the terminal voltage under heavy load, NOT from mysteriously 'lost charge'

If you short a battery, all the power developed by the battery is lost as heat inside the battery, there's no charge 'lost', only that used by flowing between the shorted terminals.

[WotEver]

12 minutes ago, smileypete said:

Nah, the heat comes from lowering the terminal voltage under heavy load, NOT from mysteriously 'lost charge'

There’s nowt mysterious about using power to generate heat, Pete. I thought you’d understand that. The only two folk who keep banging on about ‘lost charge’ are you and Nick. If Energy is consumed (and it must be if the battery is getting hot) then the battery contains less energy than it started off with. Less than an Ah counter would suggest. 

So now the battery has less than 50% of the energy it started off with. How do you then equate that to 50% charge?  Nick’s faries topping it up?

[nicknorman]

5 minutes ago, WotEver said:

There’s nowt mysterious about using power to generate heat, Pete. I thought you’d understand that. The only two folk who keep banging on about ‘lost charge’ are you and Nick. If Energy is consumed (and it must be if the battery is getting hot) then the battery contains less energy than it started off with. Less than an Ah counter would suggest. 

So now the battery has less than 50% of the energy it started off with. How do you then equate that to 50% charge?  Nick’s faries topping it up?

Aha! You obviously now realise the error of your ways and have switched to talking about energy. Quite right. But a pity that Mr Peukert wasn’t talking about lost energy, only lost AH.

Oh no! You spoiled it with your last sentence, once again mixing up the word “charge” to mean energy when it doesn’t. Back to the naughty corner for you!

[WotEver]

55 minutes ago, nicknorman said:

But let's be clear, shunt based SoC gauges count AH. They don't count WH

You sure about that?

Here’s a clue - some do

1 minute ago, nicknorman said:

Aha! You obviously now realise the error of your ways and have switched to talking about energy. Quite right. But a pity that Mr Peukert wasn’t talking about lost energy, only lost AH.

Oh no! You spoiled it with your last sentence, once again mixing up the word “charge” to mean energy when it doesn’t. Back to the naughty corner for you!

So are you now changing physics so that charge has no relationship to stored energy then Nick?

[nicknorman]

Just now, WotEver said:

You sure about that?

Here’s a clue - some do

Ok fair cop, some do. But my point was that they don’t use it to calculate the SoC figure, they only use AH for that. Battery monitors that are fitted to canal boats report SoC in terms of the AH remaining that could be taken out as a proportion of the total AH that could be. Not the WH. SoC is state of CHARGE and we all know what that means!

[WotEver]

Charge = watt.hours

Energy = joules.second

Both are power over time. 

Ah are an analogue of energy, proportional at a given fixed voltage. 

[WotEver]

3 minutes ago, nicknorman said:

Battery monitors that are fitted to canal boats report SoC in terms of the AH remaining that could be taken out as a proportion of the total AH that could be. Not the WH. SoC is state of CHARGE and we all know what that means!

So if a battery is storing no energy, how much ‘charge’ does it have?  If a battery has as much energy as it can hold, how much ‘charge’ does it have? If a battery only has 40% of its original ‘full’ energy available, how much ‘charge’ does it have?

[nicknorman]

8 minutes ago, WotEver said:

Charge = watt.hours

Energy = joules.second

Both are power over time. 

Ah are an analogue of energy, proportional at a given fixed voltage. 

You can make up definitions for charge if you like, but they are not recognised by anyone else and thus it’s rather unhelpful. Charge is a fundamental property, not made up by multiplying two other things. An AH-counting SoC meter calculates based on charge. The internationally-recognised definition of charge that is, not the one just invented by you.

[WotEver]

3 minutes ago, nicknorman said:

An AH-counting SoC meter calculates based on charge.

No it doesn’t. It integrates current flow over time. It has no concept of ‘charge’ other than what the user inputs.