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Sterling-NASA -Victron


Glynn

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I have fitted all of the above and my personal preference both as an installer and a user would be the Victron unit.  Make sure you have the right charger for your batteries and it is set up correct.  Different battery types can and should be charged at slightly different rates.  The previous notes about charging at 15 volts quite rightly should only be done as an equalization charge but for standard lead acid batteries you should be in the 13.5 to 14.2 volts. As you start to get towards 14.4v onwards for prolonged charging, the battery will start to gas and you will smell sulphuric gas (Eggy farty smell).
excessive charging will kill the battery prematurely so a balanced Battery/Charger % is preferred to smoothly charge the battery and not aggressively charge/boil them.

 

Hope this helps?

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On 01/10/2017 at 08:57, nicknorman said:

if the battery is left to recover with the heavy load removed, most of its remaining capacity will return after perhaps 20 hours' resting

What a lovely theory. It’s just a shame that the majority of boaters don’t have the luxury of reducing their load for a 20 hour period, or indeed anywhere close to that. In the usual period available to most boaters (a handful of hours before they need to recharge) the ‘lost’ capacity doesn’t have nearly enough time to be recovered and remains lost for that cycle. 

17 minutes ago, mchancox said:

The previous notes about charging at 15 volts quite rightly should only be done as an equalization charge

Unless it’s a semi-traction (or better) battery during winter in which case 15V+ should be the norm. 

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18 minutes ago, mchancox said:

for standard lead acid batteries you should be in the 13.5 to 14.2 volts.

If that voltage is your daily charge voltage you will eventually suffer with irretrievable sulphation. Your regular charge voltage for wet lead acid batteries should ideally be no lower than 14.4V (in the summer and higher in the winter). 

The above assumes regular use.

If the boat spends most of its life on shore power then you can afford to be gentle with the charge voltages as it will be spending many many hours on float. 

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Smart chargers will regulate the charge voltage depending on the battery type. Off shore there are all types of alternator enhancing products that will increase voltages and currents. Being kind to your batteries is key and getting the right battery bank installed for the purpose also. 

Semi traction came quite a demand from some of my customers with several years of problem free off shore power. Design it right from the start. Happy boating. 

Edited by mchancox
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2 hours ago, WotEver said:

What a lovely theory. It’s just a shame that the majority of boaters don’t have the luxury of reducing their load for a 20 hour period, or indeed anywhere close to that. In the usual period available to most boaters (a handful of hours before they need to recharge) the ‘lost’ capacity doesn’t have nearly enough time to be recovered and remains lost for that cycle. 

Not really. The point is that if you discharge the battery faster than the 20hr rate, you lose capacity. Most boaters, except those with totally knackered batteries' don't use 100% capacity in 20 hrs. More like 50% capacity or less. So the average rate for most people is less than the 20hr rate. If I put the electric kettle on and discharge at 200A (the 1.5 to 2 hr rate) for 5 mins, then continue the normal discharge at slower than the 20hr rate, I have not lost any excess capacity due to peukert as a result of the short term fast discharge. But if you apply peukert blindly to that circumstance, of course you would think that excess capacity has been lost. Wrong!

This is why Peukert is not relevant to mid-charge SoC, only to "time to run to flat" at the current discharge rate.

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6 hours ago, mchancox said:

I have fitted all of the above and my personal preference both as an installer and a user would be the Victron unit.  Make sure you have the right charger for your batteries and it is set up correct.  Different battery types can and should be charged at slightly different rates.  The previous notes about charging at 15 volts quite rightly should only be done as an equalization charge but for standard lead acid batteries you should be in the 13.5 to 14.2 volts. As you start to get towards 14.4v onwards for prolonged charging, the battery will start to gas and you will smell sulphuric gas (Eggy farty smell).
excessive charging will kill the battery prematurely so a balanced Battery/Charger % is preferred to smoothly charge the battery and not aggressively charge/boil them.

 

Hope this helps?

Are you sure - most standard leisure and slightly better batteries are now lead calcium NOT lead antinomy and they gas at a higher voltage. In fact one engine miniseries fitted 14.8 volt regulators.

Personally I would be happy with lead calcium at about 14.5 to 14.6 volts.

That smell is hydrogen sulphide and it usually occurs when a cell is shorting out and overheating & gassing.

 

 

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3 hours ago, nicknorman said:

Not really. The point is that if you discharge the battery faster than the 20hr rate, you lose capacity. Most boaters, except those with totally knackered batteries' don't use 100% capacity in 20 hrs. More like 50% capacity or less. So the average rate for most people is less than the 20hr rate. If I put the electric kettle on and discharge at 200A (the 1.5 to 2 hr rate) for 5 mins, then continue the normal discharge at slower than the 20hr rate, I have not lost any excess capacity due to peukert as a result of the short term fast discharge. But if you apply peukert blindly to that circumstance, of course you would think that excess capacity has been lost. Wrong!

This is why Peukert is not relevant to mid-charge SoC, only to "time to run to flat" at the current discharge rate.

No, I disagree. If you have a 100Ah battery (C/20) then you would expect to be able to use 50Ah before getting to 50% SoC. but if you’ve discharged the battery at above C/20 for part of that time and at C/20 for the rest of the time then, due to Mr Peukert, you will find that you’ve reached the 50% marker after using only say 45Ah. So for that cycle that 5Ah is indeed ‘lost’. 

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Acknowledging discussion about the separate subject of Peukert I do think that many folk misunderstand the so called rule of 'don't discharge to less than 50% SoC' thinking that discharging to anything higher than 50% SoC is somehow in safe territory and will look after their batteries.

There's nothing special about 50% SoC other than a rough figure for a balance between battery replacement costs against fuel used to recharge, 51% SoC is hardly any better than 49% SoC. 

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6 minutes ago, nb Innisfree said:

Acknowledging discussion about the separate subject of Peukert I do think that many folk misunderstand the so called rule of 'don't discharge to less than 50% SoC' thinking that discharging to anything higher than 50% SoC is somehow in safe territory and will look after their batteries.

There's nothing special about 50% SoC other than a rough figure for a balance between battery replacement costs against fuel used to recharge, 51% SoC is hardly any better than 49% SoC. 

Whilst I agree that there is nothing special about the 50% SoC figure, it is more to do with getting a reasonable "lifetime energy capacity" out of a battery rather than fuel saving (although I acknowledge how long it takes to put the last 20% of chsrge into a battery)

The more deely you discharge a battery the fewer cycles you will get out of it, and the less "lifetime energy capacity". 50% typically represents the best compromise of lifetime capacity against battery costs.

This link explains it.

http://all-about-lead-acid-batteries.capnfatz.com/all-about-lead-acid-batteries/lead-acid-battery-fundamentals/lead-acid-cycle-life-or-performance-explained/

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32 minutes ago, nb Innisfree said:

There's nothing special about 50% SoC other than a rough figure for a balance between battery replacement costs against fuel used to recharge

 

18 minutes ago, cuthound said:

50% typically represents the best compromise of lifetime capacity against battery costs

I think you’re both saying the same thing guys :)

Whilst on the subject, it’s often said on here that “many folk misunderstand the so called rule of 'don't discharge to less than 50% SoC' thinking that discharging to anything higher than 50% SoC is somehow in safe territory and will look after their batteries.” but I’m not sure that ‘many folk’ actually do think that, do they?

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5 minutes ago, WotEver said:

 

I think you’re both saying the same thing guys :)

Whilst on the subject, it’s often said on here that “many folk misunderstand the so called rule of 'don't discharge to less than 50% SoC' thinking that discharging to anything higher than 50% SoC is somehow in safe territory and will look after their batteries.” but I’m not sure that ‘many folk’ actually do think that, do they?

Well, they don't after they have wrecked their first set of batteries :o

For long battery life it is essential to fully recharge every discharge AND to minimise the time the battery is left in a discharged condition without recharging it.

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Just now, cuthound said:

For long battery life it is essential to fully recharge every discharge AND to minimise the time the battery is left in a discharged condition without recharging it.

And that’s it in a nutshell. 

Sadly it’s much, MUCH quicker to write than it is to do ;)

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47 minutes ago, WotEver said:

No, I disagree. If you have a 100Ah battery (C/20) then you would expect to be able to use 50Ah before getting to 50% SoC. but if you’ve discharged the battery at above C/20 for part of that time and at C/20 for the rest of the time then, due to Mr Peukert, you will find that you’ve reached the 50% marker after using only say 45Ah. So for that cycle that 5Ah is indeed ‘lost’. 

No, no charge is "lost", this is the point that you don't seem to grasp even though I've mentioned it before!

It's all about two things:

1/ the definition of battery capacity and 

2/ the nature of the chemicals and their reaction in the battery.

For 1/ the battery capacity is determined by steadily discharging the battery until the terminal voltage drops to nominally flat, typically 10.5v. If you discharge the battery faster, you extract less charge before the voltage reaches 10.5v. The degree can be determined by Peukert.

For 2/, the chemicals (simplistically, lead and sulphuric acid) react together to produce lead sulphate for each molecule, an electron is liberated (or is it 2, can't be bothered to check!). But the reaction isn't instantaneous and relies on the two chemicals being able to come into intimate contact. To start with, lead on the surface of the plate and sulphuric acid adjacent, is free to react swiftly. But later once those chemicals have been exhausted, lead deeper in the plate and sulphuric acid further away, need to react to pump the electrons round. It takes a while for the chemicals to diffuse towards each other and so the reaction slows, the terminal voltage decreases. The faster one tries to discharge the electrons the faster the chemicals have to react and so, due to the time it takes for the chemicals to migrate to each other, the lower the terminal voltage.

Hopefully you agree so far!

What is behind Peukert bearing in mind 1/ and 2/, is that if you discharge at say C5 you will reach the nominally "flat" voltage of say 10.5v having extracted fewer AH than if you discharged at C20. But remember that for each molecule of the starting chemicals, there is a correlated number of electrons available to be transported. Faster discharge doesn't miraculously cause electrons to disappear. It merely means that the chemicals buried deeper in the plate don't have time to migrate to meet the acid in time to keep the terminal voltage above 10.5v. But having discharged at C5 and reached 10.5v, if you then let the battery rest you will find that the voltage increases and further discharge at a slower rate is possible, recovering nearly all the apparently lost charge since the chemicals have had time to migrate.

This phenomena is often observed when one tries to start the car/boat, it won't start, the battery is now "flat" and won't turn the engine over. But walk away and come back after 30 mins and the battery has a new lease of life and is able to turn the engine over again.

So in summary, Peukert means that if you discharge at a higher rate you will reach "flat" voltage sooner. But it doesn't mean chemicals have evaporated nor that the concomitant electrons arising from the reaction have been beamed into space! They are still there to react and produce more electrons once they've had time to find each other.

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1 hour ago, nb Innisfree said:

Acknowledging discussion about the separate subject of Peukert I do think that many folk misunderstand the so called rule of 'don't discharge to less than 50% SoC' thinking that discharging to anything higher than 50% SoC is somehow in safe territory and will look after their batteries.

There's nothing special about 50% SoC other than a rough figure for a balance between battery replacement costs against fuel used to recharge, 51% SoC is hardly any better than 49% SoC. 

You are of course absolutely correct but I think your views would not be believed by the majority of boaters (and lots of other people too). Even well known battery suppliers suggest that batteries will be Damaged if they are discharged below 50%.

I discharged mine to just below 40% recently and am now so worried that I have needed to drink a bottle of Henry Westons 8.2% cider every day to calm my nerves.

...............Dave

20 minutes ago, nicknorman said:

No, no charge is "lost", this is the point that you don't seem to grasp even though I've mentioned it before!

It's all about two things:

1/ the definition of battery capacity and 

2/ the nature of the chemicals and their reaction in the battery.

For 1/ the battery capacity is determined by steadily discharging the battery until the terminal voltage drops to nominally flat, typically 10.5v. If you discharge the battery faster, you extract less charge before the voltage reaches 10.5v. The degree can be determined by Peukert.

For 2/, the chemicals (simplistically, lead and sulphuric acid) react together to produce lead sulphate for each molecule, an electron is liberated (or is it 2, can't be bothered to check!). But the reaction isn't instantaneous and relies on the two chemicals being able to come into intimate contact. To start with, lead on the surface of the plate and sulphuric acid adjacent, is free to react swiftly. But later once those chemicals have been exhausted, lead deeper in the plate and sulphuric acid further away, need to react to pump the electrons round. It takes a while for the chemicals to diffuse towards each other and so the reaction slows, the terminal voltage decreases. The faster one tries to discharge the electrons the faster the chemicals have to react and so, due to the time it takes for the chemicals to migrate to each other, the lower the terminal voltage.

Hopefully you agree so far!

What is behind Peukert bearing in mind 1/ and 2/, is that if you discharge at say C5 you will reach the nominally "flat" voltage of say 10.5v having extracted fewer AH than if you discharged at C20. But remember that for each molecule of the starting chemicals, there is a correlated number of electrons available to be transported. Faster discharge doesn't miraculously cause electrons to disappear. It merely means that the chemicals buried deeper in the plate don't have time to migrate to meet the acid in time to keep the terminal voltage above 10.5v. But having discharged at C5 and reached 10.5v, if you then let the battery rest you will find that the voltage increases and further discharge at a slower rate is possible, recovering nearly all the apparently lost charge since the chemicals have had time to migrate.

This phenomena is often observed when one tries to start the car/boat, it won't start, the battery is now "flat" and won't turn the engine over. But walk away and come back after 30 mins and the battery has a new lease of life and is able to turn the engine over again.

So in summary, Peukert means that if you discharge at a higher rate you will reach "flat" voltage sooner. But it doesn't mean chemicals have evaporated nor that the concomitant electrons arising from the reaction have been beamed into space! They are still there to react and produce more electrons once they've had time to find each other.

Have you done a good experiment to prove this yet?

And are you not getting bored with having the same battery conversations again and again on this forum? :D

...............Dave

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On 03/10/2017 at 10:04, Robbo said:

Think you will like, it still feels like a early stage product, but the firmware is getting updated regularly.   I've not had it long to do a proper review yet and the stuff I want (logging/history) isn't out yet :(

I hope you are correct and if that firmware ever gets good I am gonna get one of these. There is an old bit of wisdom when purchasing any software (or firmware).....

Buy it for what it does Now, not for what the salesman promises it might do in the future!

..................Dave

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3 minutes ago, dmr said:

Have you done a good experiment to prove this yet?

And are you not getting bored with having the same battery conversations again and again on this forum? :D

...............Dave

1/ Yes. We have two knackered vehicles on the airfield. Well, with knackered batteries anyway. Sometimes they won't start. Then they won't turn over. But come back after a while and they fire up. The batteries were "flat" due to the high discharge rate of the starter, but given time the "lost" charge was recovered. That, plus the logic I described above, and the fact that AH-counting gauge manufacturers agree with me, is sufficient proof IMO.

2/ Yes, but people still don't get it so it's my duty to keep going!

9 minutes ago, dmr said:

You are of course absolutely correct but I think your views would not be believed by the majority of boaters (and lots of other people too). Even well known battery suppliers suggest that batteries will be Damaged if they are discharged below 50%.

I discharged mine to just below 40% recently and am now so worried that I have needed to drink a bottle of Henry Westons 8.2% cider every day to calm my nerves.

...............Dave

 

You will be fine. Trojan say that the best level of discharge to maximise the total AH that can be taken from their batteries over their lifetime, is 40% SoC. Something which the pedlars of convention are oblivious to!

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58 minutes ago, nicknorman said:

No, no charge is "lost", this is the point that you don't seem to grasp even though I've mentioned it before!

I haven’t missed anything; on the contrary it is you who persists in repeating the same theory which isn’t borne out by the facts of real world usage.

When I (or anyone else with a grasp of the theory) use the word ‘lost’ I am referring to the inaccessibility of that bit of charge to the user during that particular discharge cycle. I thought I explained that clearly enough in my previous post. 

Why do you suppose there are different capacities quoted for C/5, C/10, C20? It’s because of the so-called Peukert effect. In none of those cases are electrons ‘lost’ but the apparent capacity of the battery varies according to the discharge rate. At higher discharge rates the battery ‘loses’ capacity. 

You don’t need to go all the way down to 10.5V to experience this, it’s just as apparent at 12.2V. 

The theory that it will all come back if you now allow the battery to rest for a few hours is nothing more than idealistic wishful thinking - instead of resting his batteries for a few hours the boater is far better off charging his batteries back up which allows him to continue using them. 

Edited by WotEver
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40 minutes ago, dmr said:

I discharged mine to just below 40% recently and am now so worried that I have needed to drink a bottle of Henry Westons 8.2% cider every day to calm my nerves.

If you drank it more slowly you’d have more to drink ;)

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1 hour ago, nicknorman said:

the chemicals (simplistically, lead and sulphuric acid) react together to produce lead sulphate for each molecule, an electron is liberated (or is it 2, can't be bothered to check!)....

Yes it’s 2. If we’re going to get into chemistry, then riddle me this, which I’ve been meaning to post for months and months...

During discharge............

The positive plate (which is actually a cathode for the purposes of the internal reaction) reduces from lead dioxide to lead sulphate. In doing so it gains two electrons from the external circuit (thus reduction) and the voltage of the reaction is 1.685 volts. This is an atomic/chemical reaction. There are no iffs or buts about it. The voltage is not negotiable.

The negative plate oxidises from lead to lead sulphate. In doing so it releases two electrons (thus oxidation) to the external circuit. The voltage of this reaction is 0.356 volts. Again, this is an atomic/chemical reaction. It’s a fixed voltage.

Add the two voltages (for each reaction) together and we get 2.041 volts. That is the natural voltage of that cell.  6 cells gives us 12.246V.

Now there either are available ions at the interface between the acid and the plates or there aren't available ions. There is no in between.

So either the plates react (thus producing those voltages) or they don't. Again, there is no in between.

So how can the voltage ever be higher than this? 12.7 volts is about fully charged and rested. Where does the extra 0.65 volts come from?

And... why does the voltage fall as the battery discharges?

There are no good answers out there in Google-Land that I can find. 

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7 minutes ago, WotEver said:

I haven’t missed anything; on the contrary it is you who persists in repeating the same theory which isn’t borne out by the facts of real world usage.

1/ When I (or anyone else with a grasp of the theory) use the word ‘lost’ I am referring to the inaccessibility of that bit of charge to the user during that particular discharge cycle. I thought I explained that clearly enough in my previous post. 

2/ Why do you suppose there are different capacities quoted for C/5, C/10, C20? It’s because of the so-called Peukert effect. In none of those cases are electrons ‘lost’ but the apparent capacity of the battery varies according to the discharge rate. At higher discharge rates the battery ‘loses’ capacity. 

3/ You don’t need to go all the way down to 10.5V to experience this, it’s just as apparent at 12.2V. 

4/ The theory that it will all come back if you now allow the battery to rest for a few hours is nothing more than idealistic wishful thinking - instead of resting his batteries for a few hours the boater is far better off charging his batteries back up which allows him to continue using them. 

Enumerated your paragraphs for clarity

1/ Yes, the "lost" bit of charge is indeed inacessible during that particular discharge cycle, because there isn't time for it to react before the battery is nominally "flat". Unless the discharge is paused and the necessary time given.

2/ I explained why there are different capacities for C5, C10 etc and this is because of peukert. As you say, at higher discharge rates the battery "loses" capacity. Because there isn't time for the buried chemicals to react before the battery is nominally flat (ie reaches the terminating voltage) at the higher discharge rates.

3/ Yes it is apparent at 12.2v as it is at 10.5v. A fast discharge to 12.2v will make you think you've hit 50% SoC. But go away for a while with the discharge stopped  and the apparent SoC will miraculously recover to 60% or whatever.

4/ Nothing you have said above makes your point. You have not explained where the chemicals or electrons go that would otherwise be providing the full capacity of the battery. Sure, the battery is better put on charge but that is not the point. However, it's now clear that you are copying and pasting responses from Gibbo!

And fortunately, makers of AH-counting gauges agree with me not you (or Gibbo!).

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3 minutes ago, nicknorman said:

it's now clear that you are copying and pasting responses from Gibbo!

It most certainly ain’t, sir. This is all my own work :)

1 minute ago, cereal tiller said:

Is because of Pukers effect?

That’s the one :D

4 minutes ago, nicknorman said:

4/ Nothing you have said above makes your point.

????

1/, 2/, 3/, all make my point, and you agreed with me on all 3 points. 

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1 minute ago, WotEver said:

Yes it’s 2. If we’re going to get into chemistry, then riddle me this, which I’ve been meaning to post for months and months...

During discharge............

The positive plate (which is actually a cathode for the purposes of the internal reaction) reduces from lead dioxide to lead sulphate. In doing so it gains two electrons from the external circuit (thus reduction) and the voltage of the reaction is 1.685 volts. This is an atomic/chemical reaction. There are no iffs or buts about it. The voltage is not negotiable.

The negative plate oxidises from lead to lead sulphate. In doing so it releases two electrons (thus oxidation) to the external circuit. The voltage of this reaction is 0.356 volts. Again, this is an atomic/chemical reaction. It’s a fixed voltage.

Add the two voltages (for each reaction) together and we get 2.041 volts. That is the natural voltage of that cell.  6 cells gives us 12.246V.

Now there either are available ions at the interface between the acid and the plates or there aren't available ions. There is no in between.

So either the plates react (thus producing those voltages) or they don't. Again, there is no in between.

So how can the voltage ever be higher than this? 12.7 volts is about fully charged and rested. Where does the extra 0.65 volts come from?

And... why does the voltage fall as the battery discharges?

There are no good answers out there in Google-Land that I can find. 

Ooh, scary big writing! But anyway, as you know, the exact details of how the chemicals react and the potentials thus produced, are very complicated and beyond me and this forum to understand exactly. However one should bear in mind that whilst the voltages involved are linearly variable, the charge is in increments of one electron. One electron has a specific fixed charge. Two electrons arise from the reaction of lead and acid. The lead and acid react, two electrons are moved. Or the lead and acid don't react, no electrons are moved. You can't partially move and Elton or have the chemicals reacting without the two electrons moving. There is no linear variation, it's binary.

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1 minute ago, nicknorman said:

Ooh, scary big writing!

That’s what comes from copy/paste on an iPhone where you can’t get at any font controls. I wrote that question out in Notes about a year ago. 

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