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Post by Robbo on Oct 31, 2016 11:52:33 GMT
Good write up, however when it gets pinned I would rename the subject as 'amp hour counters' doesn't mean a lot to the users who actually would find this useful. 'Battery monitors' with keywords such as 'Victron BMV 501, 502, 601, 602, 701, 702' to make it easily searchable.
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Post by Gone on Oct 31, 2016 12:37:13 GMT
Charging a battery takes energy, heating also takes energy, and you can only use it once, so electrical energy converted to heat is therefore not available to be converted into 'chemical' energy - ie charging the battery. So if you think 'energy' rather than volts and amps, then heat in this case is wasted energy. Sure. However an AH counter counts AH in and out. Since AH has no concept of heat or energy, wasted heat or energy has no bearing on AH counting. Again I implore to bear in mind the fundamental difference between charge efficiency and energy efficiency. Just been walking the dog, beautiful warm day here, had a quick read through the posts - this site is more addictive than the other. Anyway I was not sure if this was answered to everyone's satisfaction, so here goes. Current flows through the shunt where it is measured and converted by the meter into amp hours. So all the current flowing into the battery is counted by the meter. The very action of measuring the current with a shunt does use some energy in the form of a voltage drop - but just a few mV. Then the current flows into the battery where it will drive various chemical reactions, all of which absorb energy. There is the reaction that we want which charges the battery and we get the energy back when we discharge the battery. Some of the energy is converted into heat and if the voltage is high enough some of the energy is lost in converting water into hydrogen and oxygen. The amp hours meter has a fiddle factor in it's counting algorithm to allow for these losses. In other words you have to put one and a bit amphours in to the battery to increase the count by one.
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Post by Telemachus on Oct 31, 2016 12:50:34 GMT
Sure. However an AH counter counts AH in and out. Since AH has no concept of heat or energy, wasted heat or energy has no bearing on AH counting. Again I implore to bear in mind the fundamental difference between charge efficiency and energy efficiency. Just been walking the dog, beautiful warm day here, had a quick read through the posts - this site is more addictive than the other. Anyway I was not sure if this was answered to everyone's satisfaction, so here goes. Current flows through the shunt where it is measured and converted by the meter into amp hours. So all the current flowing into the battery is counted by the meter. The very action of measuring the current with a shunt does use some energy in the form of a voltage drop - but just a few mV. Then the current flows into the battery where it will drive various chemical reactions, all of which absorb energy. There is the reaction that we want which charges the battery and we get the energy back when we discharge the battery. Some of the energy is converted into heat and if the voltage is high enough some of the energy is lost in converting water into hydrogen and oxygen. The amp hours meter has a fiddle factor in it's counting algorithm to allow for these losses. In other words you have to put one and a bit amphours in to the battery to increase the count by one. Not really, again confusing power/energy with current/charge. Yes, energy is lost whilst charging but that is nothing to do with AH lost. Bear in mind that AH is a bunch of electrons. They can't just disappear (there is no such thing as "current drop", only "voltage drop"). Each electron in the circuit has to go somewhere. Most of it goes to pumping the chemical reaction that simplistically is lead sulphate -> lead and sulphuric acid. Any electrons that aren't used up in that process, are used up electrolysing the water into hydrogen and oxygen. Electrons have to move between the plates in order for current to flow in the battery. The only way they can do that is via the electrolyte. Esoteric quantum effects aside, there is no means of transporting electrons between the plates other than via ions in the electrolyte. Thus the AH used to charge the battery go to the chemical reaction, and the gassing. No electrons disappear into the ether! Therefore the CEF is only <100% due to the inevitable gassing that occurs during charge. I'm not sure whether gassing (dissociating the hydrogen and water) intrinsically produces heat but even if it does, of course other mechanisms (resistive losses, and maybe the primary chemical reaction) are the predominant ones.
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Post by tonyb on Oct 31, 2016 12:50:48 GMT
Right - deleted the repeat - no idea how that got there, cut and paste from Word. Altered the title but its seems "Why battery state of charge meters can not be relied upon (for comment)" is too long so dropped some of it. While I am inclined to accept the possibility that Thelemachus's argument about it not being amps that cause the heat I do not think pursuing that will be productive as far as the target audience is concerned and is only likely to cause confusion. Others on here can discuss it amongst themselves but I think as an EASILY UNDERSTOOD explanation of why the things are not accurate it stands (even if it may not be 100% scientifically correct). I suppose that what we should be saying that the heat is produce by voltdrop but that hardly explains why the counters get out of synch.
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Post by Telemachus on Oct 31, 2016 12:55:42 GMT
Right - deleted the repeat - no idea how that got there, cut and paste from Word. Altered the title but its seems "Why battery state of charge meters can not be relied upon (for comment)" is too long so dropped some of it. While I am inclined to accept the possibility that Thelemachus's argument about it not being amps that cause the heat I do not think pursuing that will be productive as far as the target audience is concerned and is only likely to cause confusion. Others on here can discuss it amongst themselves but I think as an EASILY UNDERSTOOD explanation of why the things are not accurate it stands (even if it may not be 100% scientifically correct). I suppose that what we should be saying that the heat is produce by voltdrop but that hardly explains why the counters get out of synch. Counters get out of synch because of CEF which as you yourself have stated, is a variable thing depending on the SoC amongst others. CEF isn't 100% due to electrons used up in gassing as opposed to converting lead sulphate to lead and sulphuric acid.
Because our batteries charge very fast in the early stages (175A alternator) the Smartgauge lags significantly during charge from say 60% to 80%. Thus the AH-counter is the much better indicator during charge. Of course the SG is the much better indicator during discharge because it effectively/intrinsically uses the actual battery capacity as opposed to the one set in the AH gauge. Our AH gauge automatically calculates the CEF and uses that during charge. The calculated CEF varies between about 93% and 95% which is not much variation, so I feel that once that correction is applied, the AH counter is a pretty accurate indicator of SoC during charge. I keep the AH capacity updated so as to match the SG during discharge (though so far, after 3 years of leisure use I have detected no reduction in the capacity of the Trojans).
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Post by Gone on Oct 31, 2016 13:16:46 GMT
Just been walking the dog, beautiful warm day here, had a quick read through the posts - this site is more addictive than the other. Anyway I was not sure if this was answered to everyone's satisfaction, so here goes. Current flows through the shunt where it is measured and converted by the meter into amp hours. So all the current flowing into the battery is counted by the meter. The very action of measuring the current with a shunt does use some energy in the form of a voltage drop - but just a few mV. Then the current flows into the battery where it will drive various chemical reactions, all of which absorb energy. There is the reaction that we want which charges the battery and we get the energy back when we discharge the battery. Some of the energy is converted into heat and if the voltage is high enough some of the energy is lost in converting water into hydrogen and oxygen. The amp hours meter has a fiddle factor in it's counting algorithm to allow for these losses. In other words you have to put one and a bit amphours in to the battery to increase the count by one. Not really, again confusing power/energy with current/charge. Yes, energy is lost whilst charging but that is nothing to do with AH lost. Bear in mind that AH is a bunch of electrons. They can't just disappear (there is no such thing as "current drop", only "voltage drop"). Each electron in the circuit has to go somewhere. Most of it goes to pumping the chemical reaction that simplistically is lead sulphate -> lead and sulphuric acid. Any electrons that aren't used up in that process, are used up electrolysing the water into hydrogen and oxygen. Electrons have to move between the plates in order for current to flow in the battery. The only way they can do that is via the electrolyte. Esoteric quantum effects aside, there is no means of transporting electrons between the plates other than via ions in the electrolyte. Thus the AH used to charge the battery go to the chemical reaction, and the gassing. No electrons disappear into the ether! Therefore the CEF is only <100% due to the inevitable gassing that occurs during charge. I'm not sure whether gassing (dissociating the hydrogen and water) intrinsically produces heat but even if it does, of course other mechanisms (resistive losses, and maybe the primary chemical reaction) are the predominant ones. This type of discussion is interesting for those involved, but confusing for those that want a simple explanation of something, which is why when you come to pin the modified first post the rest are better in another place. I think this is correct. Agreed that electrons are not lost, but to drive the electrons takes energy. The energy required to break the HO-O and HO bonds that make up water are measured in kJ/mol. An Amp being driven by one volt is one watt of power, and a Watt is also a Joule per second. So when charging your battery you are forcing (as an example) 10 amps at 14 Volts which is 140Watts. So every second you are pushing in 140Joules. To dissociate water into Hydrogen and Oxygen takes about 460kJ/mol. One mole of water is 18g. So to turn 18g of water into Hydrogen and Oxygen would take 460,000J In our example of 10A at 14v which is 140J per second, so to convert 18g of water into hydrogen and oxygen would take 460000/140=3200 seconds or almost an hour. In the battery, most of the energy is used in forcing the battery to charge.
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Post by Telemachus on Oct 31, 2016 13:24:52 GMT
Not really, again confusing power/energy with current/charge. Yes, energy is lost whilst charging but that is nothing to do with AH lost. Bear in mind that AH is a bunch of electrons. They can't just disappear (there is no such thing as "current drop", only "voltage drop"). Each electron in the circuit has to go somewhere. Most of it goes to pumping the chemical reaction that simplistically is lead sulphate -> lead and sulphuric acid. Any electrons that aren't used up in that process, are used up electrolysing the water into hydrogen and oxygen. Electrons have to move between the plates in order for current to flow in the battery. The only way they can do that is via the electrolyte. Esoteric quantum effects aside, there is no means of transporting electrons between the plates other than via ions in the electrolyte. Thus the AH used to charge the battery go to the chemical reaction, and the gassing. No electrons disappear into the ether! Therefore the CEF is only <100% due to the inevitable gassing that occurs during charge. I'm not sure whether gassing (dissociating the hydrogen and water) intrinsically produces heat but even if it does, of course other mechanisms (resistive losses, and maybe the primary chemical reaction) are the predominant ones. This type of discussion is interesting for those involved, but confusing for those that want a simple explanation of something, which is why when you come to pin the modified first post the rest are better in another place. I think this is correct. Agreed that electrons are not lost, but to drive the electrons takes energy. The energy required to break the HO-O and HO bonds that make up water are measured in kJ/mol. An Amp being driven by one volt is one watt of power, and a Watt is also a Joule per second. So when charging your battery you are forcing (as an example) 10 amps at 14 Volts which is 140Watts. So every second you are pushing in 140Joules. To dissociate water into Hydrogen and Oxygen takes about 460kJ/mol. One mole of water is 18g. So to turn 18g of water into Hydrogen and Oxygen would take 460,000J In our example of 10A at 14v which is 140J per second, so to convert 18g of water into hydrogen and oxygen would take 460000/140=3200 seconds or almost an hour. In the battery, most of the energy is used in forcing the battery to charge. Yes I agree with all that, hence a typical CEF of 90-something%. As to this thread, I presumed the purpose was to have a discussion to refine the OP text, which would eventually be relocated away from all the technical chit chat.
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Post by tonyqj on Oct 31, 2016 13:28:06 GMT
From a >7 year old thread on CW where electrolysis was discussed it was concluded that 239kJ of energy is required to electrolyse 1 mole of water. Several references confirmed it. So you're about double that energy in your post above, Chewy.
1Ah of overcharge at about 1.2 volts will electrolyse 0.3cc of water.
Therefore the energy required is 239kJ, no?
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Post by tonyqj on Oct 31, 2016 13:35:54 GMT
Calculations for the above (not my work): The reaction is
2H2O -> 2H2 + O2
The bond energies (correct to 3 sf) are
H-O : 459 kJMol-1 H-H : 432 O=O : 494
So the total energy is 459*4 - 494 - 432*2 = 478 kJ
But that is for two moles of H2O,
so the energy per mole is 239kJ
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Post by tonyb on Oct 31, 2016 14:05:40 GMT
This type of discussion is interesting for those involved, but confusing for those that want a simple explanation of something, which is why when you come to pin the modified first post the rest are better in another place. I think this is correct. Agreed that electrons are not lost, but to drive the electrons takes energy. The energy required to break the HO-O and HO bonds that make up water are measured in kJ/mol. An Amp being driven by one volt is one watt of power, and a Watt is also a Joule per second. So when charging your battery you are forcing (as an example) 10 amps at 14 Volts which is 140Watts. So every second you are pushing in 140Joules. To dissociate water into Hydrogen and Oxygen takes about 460kJ/mol. One mole of water is 18g. So to turn 18g of water into Hydrogen and Oxygen would take 460,000J In our example of 10A at 14v which is 140J per second, so to convert 18g of water into hydrogen and oxygen would take 460000/140=3200 seconds or almost an hour. In the battery, most of the energy is used in forcing the battery to charge. Yes I agree with all that, hence a typical CEF of 90-something%. As to this thread, I presumed the purpose was to have a discussion to refine the OP text, which would eventually be relocated away from all the technical chit chat.Exactly that. By all means chat away and delve into the intricacies of the subject but remember someone with a first class degree, a masters, and a Doctorate often make very poor school (and FE) teachers because they have never experienced the problems others have in grasping concepts - especially if the concept has to be taken on trust because the obvious physical evidence is not there. If anyone else can come up with an absolutely accurate explanation for the inefficiencies of battery charging that is easy to understand I will happily modify the post and attribute it.
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Post by Telemachus on Oct 31, 2016 14:17:14 GMT
Yes I agree with all that, hence a typical CEF of 90-something%. As to this thread, I presumed the purpose was to have a discussion to refine the OP text, which would eventually be relocated away from all the technical chit chat.... someone with a first class degree, a masters, and a Doctorate often make very poor school (and FE) teachers If anyone else can come up with an absolutely accurate explanation for the inefficiencies of battery charging that is easy to understand I will happily modify the post and attribute it. Phew well that's alright, I don't have any of those! As I said earlier, CEF is a consequence of electrons (current, if you like) "lost" to gassing. EEF (energy efficiency factor) is lower than CEF and a consequence of differing charge vs discharge voltage, resistive losses within the battery and of course gassing. However I'm not sure it's necessary to mention EEF since it's not relevant to most folk with an AH gauge (or without one).
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Post by Gone on Oct 31, 2016 14:18:32 GMT
Calculations for the above (not my work): The reaction is 2H2O -> 2H2 + O2 The bond energies (correct to 3 sf) are H-O : 459 kJMol-1 H-H : 432 O=O : 494 So the total energy is 459*4 - 494 - 432*2 = 478 kJ But that is for two moles of H2O, so the energy per mole is 239kJ You are correct - I forgot the energy released by O+O becoming O2 and H+H becoming H2 - very embarrassed.
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Post by Gone on Oct 31, 2016 14:20:36 GMT
... someone with a first class degree, a masters, and a Doctorate often make very poor school (and FE) teachers If anyone else can come up with an absolutely accurate explanation for the inefficiencies of battery charging that is easy to understand I will happily modify the post and attribute it. Phew well that's alright, I don't have any of those! As I said earlier, CEF is a consequence of electrons "lost" to gassing. EEF (energy efficiency factor) is lower than CEF and a consequence of differing charge vs discharge voltage, resistive losses within the battery and of course gassing. However I'm not sure it's necessary to mention EEF since it's not relevant to most folk with an AH gauge (or without one). and neither do I have any of these "first class degree, a masters, and a Doctorate"
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Post by tonyb on Oct 31, 2016 14:28:09 GMT
... someone with a first class degree, a masters, and a Doctorate often make very poor school (and FE) teachers If anyone else can come up with an absolutely accurate explanation for the inefficiencies of battery charging that is easy to understand I will happily modify the post and attribute it. Phew well that's alright, I don't have any of those! As I said earlier, CEF is a consequence of electrons "lost" to gassing. EEF (energy efficiency factor) is lower than CEF and a consequence of differing charge vs discharge voltage, resistive losses within the battery and of course gassing. However I'm not sure it's necessary to mention EEF since it's not relevant to most folk with an AH gauge (or without one). Is there not an inconsistent argument there? "Resistive losses within the battery" that manifest itself - how? Would that be heat by any chance. I think it is important to give the less technical something by which to pin the losses on. So far noone seems to have come up with a simpler "pin" than heat and gassing.
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Post by Graham on Oct 31, 2016 14:43:26 GMT
Phew well that's alright, I don't have any of those! As I said earlier, CEF is a consequence of electrons "lost" to gassing. EEF (energy efficiency factor) is lower than CEF and a consequence of differing charge vs discharge voltage, resistive losses within the battery and of course gassing. However I'm not sure it's necessary to mention EEF since it's not relevant to most folk with an AH gauge (or without one). Is there not an inconsistent argument there? "Resistive losses within the battery" that manifest itself - how? Would that be heat by any chance. I think it is important to give the less technical something by which to pin the losses on. So far noone seems to have come up with a simpler "pin" than heat and gassing. Tonyb Maybe the thought that the charge has to be pushed/attracted deeper into the plates the more the battery is charged. That means work has to be done, work produce heat etc. A thought
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