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Wittenham

another set of electrical questions

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I have found this forum to be invaluable in getting my head [partly...] around the magic that is boat electrics.  A few more questions to help advance the cause...

  • how much energy does it take to charge a 500 amp/hour battery bank from zero [please don't ask...] to charged?
    • I understand that there is a loss in the transfer from generator to the batteries
    • and that the batteries may no longer hold 500 amp/hours
    • but presumably there can be numbers put in to guesstimate the adjustment for the effect of those
    • presumably there is a mathematical formula that shows this?
  • if i knew the number from the previous question and there actually is a formula, i assume i could back into my energy usage by calculating how long it took to get my charger to shout 'float'?
  • how much energy comes out of a honda eu220i?  Does the answer change if it is running on LPG instead of petrol?
    • I am assuming that the amount going into the batteries via the Victron charger is less than the generator can put out.  If so, where does the extra energy go?

 

Happy to help out with answers from my specialty subject [as soon as I figure out what it is].  I suspect there is not much call here for names of 1970s/80s professional ice hockey players in North America....

 

 

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To put 500Ah into a battery will need about 550Ah.  The battery for reasons of chemistry will only take charge at the rate it wants, which is usually a lot less than the rate alternator/charger is capable of.  So if you could put in a constant 50A it would take 11 hours.  But it will take maybe 50A to start with but quickly drops, so obviously it will take a lot longer.  Just a guess, but maybe the best part of a day to get down to a current of about 5A at say 14.4v, at which point I would say the battery bank was just about full.

 

lpg has less calories than petrol, so max power output on lpg is a bit lower.

From memory the Honda gennie can do something like 1.8kW continuous.

 

if you don’t take full power from a gennie, where does it go?   Nowhere as it never got used in the first place.  A bit like a car going at 70mph up a hill will use more fuel  than 70mph on the flat.  Same a a gennie, if there is less load, it uses less fuel.

Edited by Chewbacka

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13 minutes ago, Wittenham said:

how much energy does it take to charge a 500 amp/hour battery bank from zero [please don't ask...] to charged?

500Ah or roughly 6kw (at 12v)

 

13 minutes ago, Wittenham said:

I am assuming that the amount going into the batteries via the Victron charger is less than the generator can put out.

The batteries dictate how much charge the charger puts out.

The battery will start start with a low voltage and a high current, as the battery starts to charge the voltage increases and the current drops, this contibue to alter until with a virtually 'full' battery the charger will be supplying 14.4 volts and 2 amps.

 

You can work on a charger actually achieving (probably) about 20%-25% of its output charging a battery.

So, VERY roughly it would take (probably) about 36-48 hours continuous to recharge your 500Ah bank.

Edited by Alan de Enfield

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2 minutes ago, Alan de Enfield said:

500Ah or roughly 6kw (at 12v)

 

The batteries dictate how much charge the charger puts out.

The battery will start start with a low voltage and a high current, as the battery starts to charge the voltage increases and the current drops, this contibue to alter until with a virtually 'full' battery the charger will be supplying 14.4 volts and 2 amps.

 

You can work on a charger actually achieving (probably) about 20%-25% of its output charging a battery.

So, VERY roughly it would take (probably) about 36-48 hours continuous to recharge your 500Ah bank.

Pedant alert - 6kWh or 6kW for one hour.

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11 minutes ago, Alan de Enfield said:

500Ah or roughly 6kw (at 12v)

Probably more like 600Ah, or roughly 8.4kWh. 

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You charger may very well shout float long before the batteries are fully charged. The reason you have not fund a formula is because there is not one but you could find the answer for YOUR boat by monitoring the charging current (Amps) and also keep turning the charger off and on again when it goes into float so it stays in acceptance charge. Then, while in bulk charge, keep charging at acceptance voltage (14.2V+) until the  until the current has fallen to about1 to 2% of battery capacity so say 5 to 10 amps. 5 is better.

 

However as the batteries age and the temperature alters you are likely to keep getting different results so not sure how useful the info is.

 

I agree with Wotever, assuming a 10% charge inefficiency seem wrong to me, more like 20 to 30%

  • Greenie 1

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12 hours ago, Tony Brooks said:

[snip] Then, while in bulk charge, keep charging at acceptance voltage (14.2V+) until the  until the current has fallen to about 1 to 2% of battery capacity so say 5 to 10 amps. 5 is better.  [snip]

Which monitoring devices would give me this information?  I have installed a Victron BMV 700 which tells me a voltage number [& probably something else, if i ever read the manual].  Sitting in the pile of 'to do' is a Smartgauge.  I have a vague recollection that someone on this forum in the past told me to put in both these monitoring devices, but i will need to look it up [& then figure out what they are trying to tell me].

 

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3 minutes ago, Wittenham said:

Which monitoring devices would give me this information?  I have installed a Victron BMV 700 which tells me a voltage number [& probably something else, if i ever read the manual].  Sitting in the pile of 'to do' is a Smartgauge.  I have a vague recollection that someone on this forum in the past told me to put in both these monitoring devices, but i will need to look it up [& then figure out what they are trying to tell me].

 

The BMV should give you the charging voltage and charging current if correctly installed. If the charging voltage drops down to something like 13.6V then the charger has dropped into float. Turn it off and back on. When the charging current has dropped to around 5A with the voltage above 14.2V, then you can call the batteries ‘full’. 
 

The smartgauge will tell you the state of charge of your batteries at a glance when you are using them (ie after fully charging them). You should aim to stop discharging (using) them before the SoC drops much below 50%. Sooner is better. Although it’s usually pretty accurate the Smartgauge shouldn’t be relied upon to tell you when your batteries are fully charged when you’re charging them - use the current and voltage as above. 

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6 minutes ago, WotEver said:

The BMV should give you the charging voltage and charging current if correctly installed. If the charging voltage drops down to something like 13.6V then the charger has dropped into float. Turn it off and back on. When the charging current has dropped to around 5A with the voltage above 14.2V, then you can call the batteries ‘full’. 
 

The smartgauge will tell you the state of charge of your batteries at a glance when you are using them (ie after fully charging them). You should aim to stop discharging (using) them before the SoC drops much below 50%. Sooner is better. Although it’s usually pretty accurate the Smartgauge shouldn’t be relied upon to tell you when your batteries are fully charged when you’re charging them - use the current and voltage as above. 

That is very clear, thank you.

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We have a Mastervolt amp-hour counting meter that calculates charge efficiency on each discharge/charge cycle. In other words, the ratio of charge taken out to charge needing to be put back in again to reach fully charged. It has varied between 92% and 94% with different types of batteries over the past 8 years since I installed it.

 

So I will confidently say that to put 500AH back into your batteries will require ~500/93% = 537 AH. Of course that presumes that having fully discharged your batteries, their capacity remains as it was. In reality it has probably reduced so less than 537AH is the answer.

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8 hours ago, nicknorman said:

We have a Mastervolt amp-hour counting meter that calculates charge efficiency on each discharge/charge cycle. In other words, the ratio of charge taken out to charge needing to be put back in again to reach fully charged. It has varied between 92% and 94% with different types of batteries over the past 8 years since I installed it.

 

So I will confidently say that to put 500AH back into your batteries will require ~500/93% = 537 AH. Of course that presumes that having fully discharged your batteries, their capacity remains as it was. In reality it has probably reduced so less than 537AH is the answer.

 

That rather supposes the Ah actually converted to stored energy is correct and on those meters in the majority of cases I would not be relying on it. However I suspect in your particular case you have the thing sussed so is likely to be more accurate than most but I still would not trust it for that sort of reading.

 

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1 hour ago, Tony Brooks said:

 

That rather supposes the Ah actually converted to stored energy is correct and on those meters in the majority of cases I would not be relying on it. However I suspect in your particular case you have the thing sussed so is likely to be more accurate than most but I still would not trust it for that sort of reading.

 

Firstly, AH isn’t directly converted to energy! Ah is charge and thus, without voltage, doesn’t carry any energy. Of course the reality is that in order to push AH into a battery it has to be at a voltage and thus does convey energy. Because the voltage to charge a battery is significantly higher than that obtained when discharging, energy efficiency is much less than charge efficiency. But that wasn’t the question!

 

AH pumped into a battery have only two destinations, either to convert the lead sulphated to lead and acid (ie charge the battery) or be used to dissociate water into hydrogen and oxygen (ie cause “gassing”). So when the charge efficiency is 93%, the other 7% is lost to gassing.
Well ok a very knackered battery with a lot of sediment at the bottom can lose AH via that path, but such batteries are beyond useful life as they will self discharge quickly via the same route.

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9 minutes ago, nicknorman said:

Firstly, AH isn’t directly converted to energy! Ah is charge and thus, without voltage, doesn’t carry any energy. Of course the reality is that in order to push AH into a battery it has to be at a voltage and thus does convey energy. Because the voltage to charge a battery is significantly higher than that obtained when discharging, energy efficiency is much less than charge efficiency. But that wasn’t the question!

 

AH pumped into a battery have only two destinations, either to convert the lead sulphated to lead and acid (ie charge the battery) or be used to dissociate water into hydrogen and oxygen (ie cause “gassing”). So when the charge efficiency is 93%, the other 7% is lost to gassing.
Well ok a very knackered battery with a lot of sediment at the bottom can lose AH via that path, but such batteries are beyond useful life as they will self discharge quickly via the same route.

And some of the energy is lost as heat, more so at higher currents, as cells do have an internal resistance.

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1 minute ago, Chewbacka said:

And some of the energy is lost as heat, more so at higher currents, as cells do have an internal resistance.

And your point is...? As I explained, AH doesn’t have any energy, so it can’t lose any! Cell internal resistance must have zero bearing on charge efficiency because the two concepts don’t share the same dimensions.

 

Obviously cell resistance and consequential energy lost as heat does have a significant bearing on energy efficiency, but that wasn’t the question.

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14 minutes ago, nicknorman said:

And your point is...? As I explained, AH doesn’t have any energy, so it can’t lose any! Cell internal resistance must have zero bearing on charge efficiency because the two concepts don’t share the same dimensions.

 

Obviously cell resistance and consequential energy lost as heat does have a significant bearing on energy efficiency, but that wasn’t the question.

My point was that you said “AH pumped into a battery have only two destinations, either to convert the lead sulphated to lead and acid (ie charge the battery) or be used to dissociate water into hydrogen and oxygen” my point was that some of the energy ‘pumped’ into a battery has a third destination, ie heat.

 

added - whilst you say AH doesn’t have energy which is true, but current can not flow into or out of a battery without a voltage difference, a battery does not store amps it stores energy.  

Edited by Chewbacka

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13 minutes ago, Chewbacka said:

My point was that you said “AH pumped into a battery have only two destinations, either to convert the lead sulphated to lead and acid (ie charge the battery) or be used to dissociate water into hydrogen and oxygen” my point was that some of the energy ‘pumped’ into a battery has a third destination, ie heat.

The question was about AH taken out vs AH to be put back in again. I say again, AH has no energy. It is simply the integral of current. Current on its own, even over time, does not have any dimensions of power and energy. Electrical power is the product of current and voltage. Energy the product of AH and voltage. But we are discussing neither voltage nor energy, only AH which, unsurprisingly, is the thing AH counters measure.

 

Some of the energy pumped into a battery is indeed lost to heat. But current is not lost to heat nor is AH. Kirchhoff’s  first law, current cannot be dropped/lost around a circuit.

 

So your point about energy efficiency is valid, but the question was about charge efficiency which is not related to energy efficiency.

 

Perhaps your confusion arises over the word “charge”. Charge the noun is the integral of current. But it is also used as a verb, eg to charge a battery. Charging a battery requires electrical power, not just electrical charge (because the charge has to be raised to a potential in order to stuff it in).

 

But let’s be clear, charge efficiency relates to the noun, not the verb.

Edited by nicknorman

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On 19/11/2019 at 09:48, nicknorman said:

AH has no energy. It is simply the integral of current. Current on its own, even over time, does not have any dimensions of power and energy. Electrical power is the product of current and voltage. Energy the product of AH and voltage. But we are discussing neither voltage nor energy, only AH which, unsurprisingly, is the thing AH counters measure.

But how does an Ah counter count, er, Amphours?  It measures the current and integrates that measurement over time. So if it measures a continuous 10A for an hour it states that 10Ah have passed. Obviously that current has a voltage pushing it, which the Ah counter doesn’t care about; nevertheless there is of necessity a power component in the circuit. So that 10Ah (coupled with the voltage) could have contributed to a charge in the battery, or heating up a cable, or creating gas.  You can’t simply say that because the Ah counter has measured 500Ah that the battery has received 500Ah; it means nothing more than an average of 50A has passed over the last 10 hours (or whatever time you’ve measured) and if the average voltage was 14V then 7kWh of power has been consumed, whatever task that energy was doing. 

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3 hours ago, WotEver said:

But how does an Ah counter count, er, Amphours?  It measures the current and integrates that measurement over time. So if it measures a continuous 10A for an hour it states that 10Ah have passed. Obviously that current has a voltage pushing it, which the Ah counter doesn’t care about; nevertheless there is of necessity a power component in the circuit. So that 10Ah (coupled with the voltage) could have contributed to a charge in the battery, or heating up a cable, or creating gas.  You can’t simply say that because the Ah counter has measured 500Ah that the battery has received 500Ah; it means nothing more than an average of 50A has passed over the last 10 hours (or whatever time you’ve measured) and if the average voltage was 14V then 7kWh of power has been consumed, whatever task that energy was doing. 

I’m not sure how many times I’m expected to explain the basic laws of physics without getting grumpy. This is the last time.

 

All OK until you say “So that 10Ah (coupled with the voltage) could have contributed to a charge in the battery, or heating up a cable, or creating gas”. Let’s focus on the middle point - heating up the cable. The formula for this power used to heat the cable is I^2 x R. Or if you like, VI where V is the voltage dropped across the resistance of the cable.

 

When the 10AH (let’s say 10A for 1hr) flows through the cable resistance, how many A (or AH) comes out the other side?

 

Could it by any chance be 10AH (or 10A for an hour)? If you disagree, can I suggest you take it up with Mr Kinrchhoff? Or maybe claim the Nobel Prize for the rewriting of physics?

 

No, I’m fairly sure you will agree that what comes out the other end is 10AH / 10A. You cannot “drop” current around a circuit.

 

Of course it is true that some voltage and hence energy is lost to the wire’s resistance. The produce of V and I (aka power) is reduced. But that reduction comes solely from the reduction in V. The I remains constant. Unless you want to rewrite the basic laws of physics etc etc.

 

So hopefully you now understand that no AH is lost in the wire’s resistance. It cannot be, under the basic laws of physics.

 

So now we come to the actual battery itself. Of course the vast majority of AH pass between the battery’s plates by means of the chemical reaction, which can be grossly simplified as the turning of lead sulphate into lead and sulphuric acid. One electron for each molecule of reaction (or is it 2, I can never remember). I should of course point out that 10AH / 10A that goes in through one terminal of the battery, also comes out of the battery at the other terminal. It has to, by Kirchhoff’s first law. But where it comes out, of course instead of it being at say 14.4v potential, it is now at 0v potential. That lost energy in the VI product that went into charging the battery, has come solely from the V element, none at all from the I element. But when you multiply the 10A by 0v, you of course get 0 power remaining (the “missing” power having been stored in the battery - mostly)

 

But in the above para, we have failed to take into account the small alternative path the current through the battery can take. Not all of it is transported by means of the lead sulphate / lead and acid route. A small proportion passes via an alternative mechanism which is via the conversion of water into hydrogen and oxygen, which evolve at the two plates and for each molecule allow an electron to move between those plates. This is a parallel path to the lead acid path. And so not all of the current / charge (noun)  is used to charge (verb) the battery. Some is lost to gassing. Which is why when we put charging current through a battery, only about 93% goes to the desired chemical reaction of lead sulphate into lead and acid. About 7% is wasted in the undesired conversion of water into hydrogen and oxygen.

 

In summary, your problem has been a failure to understand the difference between AH, which has no intrinsic energy, but is what (no surprise!) AH counters measure, vs energy, which is the product of AH and the voltage that was used to move that AH. Voltage is lost around a circuit and certainly decreased in circuit resistance. But current is not. Just like the hose pipe analogy where water pressure is voltage and water flow rate is current. You can put 10 litres per hour of water at 10psi into one end of a pipe. At the other end you will still get 10 litres of water per hour, but at less that 10psi. In the same way it is impossible for AH to be “used up” by resistance in a circuit, only voltage can be lost around a circuit, not current.

 

I do hope this is now clear to you.

 

Edited by nicknorman

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3 minutes ago, nicknorman said:

I’m not sure how many times I’m expected to explain the basic laws of physics without getting grumpy. This is the last time.

 

All OK until you say “So that 10Ah (coupled with the voltage) could have contributed to a charge in the battery, or heating up a cable, or creating gas”. Let’s focus on the middle point - heating up the cable. The formula for this power used to heat the cable is I^2 x R. Or if you like, VI where V is the voltage dropped across the resistance of the cable.

 

When the 10AH (let’s say 10A for 1hr) flows through the cable resistance, how many A (or AH) comes out the other side?

 

Could it by any chance be 10AH (or 10A for an hour)? If you disagree, can I suggest you take it up with Mr Kinrchhoff? Or maybe claim the Nobel Prize for the rewriting of physics?

 

No, I’m fairly sure you will agree that what comes out the other end is 10AH / 10A. You cannot “drop” current around a circuit.

 

Of course it is true that some voltage and hence energy is lost to the wire’s resistance. The produce of V and I (aka power) is reduced. But that reduction comes solely from the reduction in V. The I remains constant. Unless you want to rewrite the basic laws of physics etc etc.

 

So hopefully you now understand that no AH is lost in the wire’s resistance. It cannot be, under the basic laws of physics.

 

So now we come to the actual battery itself. Of course the vast majority of AH pass between the battery’s plates by means of the chemical reaction, which can be grossly simplified as the turning of lead sulphate into lead and sulphuric acid. One electron for each molecule of reaction (or is it 2, I can never remember). I should of course point out that 10AH / 10A that goes in through one terminal of the battery, also comes out of the battery at the other terminal. It has to, by Kirchhoff’s first law. But where it comes out, of course instead of it being at say 14.4v potential, it is now at 0v potential. That lost energy in the VI product that went into charging the battery, has come solely from the V element, none at all from the I element. But when you multiply the 10A by 0v, you of course get 0 power remaining (the “missing” power having been stored in the battery - mostly)

 

But in the above para, we have failed to take into account the small alternative path the current through the battery can take. Not all of it is transported by means of the lead sulphate / lead and acid route. A small proportion passes via an alternative mechanism which is via the conversion of water into hydrogen and oxygen, which evolve at the two plates and for each molecule allow an electron to move between those plates. This is a parallel path to the lead acid path. And so not all of the current / charge (noun)  is used to charge (verb) the battery. Some is lost to gassing. Which is why when we put charging current through a battery, only about 93% goes to the desired chemical reaction of lead sulphate into lead and acid. About 7% is wasted in the undesired conversion of water into hydrogen and oxygen.

 

I do hope this is now clear to you.

 

Which all goes to prove that Ah is a poor measurement of how much energy is being replaced in the battery. 500Ah passing through a 12V LA battery with say 12.6V at the battery posts isn’t going to do very much to charge it. 

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1 minute ago, WotEver said:

Which all goes to prove that Ah is a poor measurement of how much energy is being replaced in the battery. 500Ah passing through a 12V LA battery with say 12.6V at the battery posts isn’t going to do very much to charge it. 

Ah Is absolutely a poor measurement of energy. In fact it is no measurement of energy at all. It cannot be, as it doesn’t have dimensions of energy. I have 450AH of battery capacity, MTB has only a pathetic 225AH. How does he cope? Oh!, could it be that he has a 24v system and thus his 225AH is worth exactly the same amount of energy as my 450AH!

 

But just to mention that you scenario of passing 500AH of charge through a battery at 12.6v isn’t a feasible scenario. You can’t charge a 12v battery significantly with only 12.6v, and there is no way you could get those 500AH into that battery at that voltage.

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1 minute ago, nicknorman said:

But just to mention that you scenario of passing 500AH of charge through a battery at 12.6v isn’t a feasible scenario. You can’t charge a 12v battery significantly with only 12.6v, and there is no way you could get those 500AH into that battery at that voltage.

Sure you could. But it would take an awful long time. I’m fully aware that I used an extreme example. 

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Just now, WotEver said:

Sure you could. But it would take an awful long time. I’m fully aware that I used an extreme example. 

Well you’d have to specify the capacity of the battery. If it was a flat 500AH battery I’d say that it would never gain 500AH of charge (ie become fully charged) at 12.6v. Yes it would take some charge, but not much. The current would fall to zero long before the battery was fully charged.

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4 minutes ago, nicknorman said:

Well you’d have to specify the capacity of the battery. If it was a flat 500AH battery I’d say that it would never gain 500AH of charge (ie become fully charged) at 12.6v. Yes it would take some charge, but not much. The current would fall to zero long before the battery was fully charged.

But as you well know the current never falls to zero, just an ever decreasingly small amount... 

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1 minute ago, WotEver said:

But as you well know the current never falls to zero, just an ever decreasingly small amount... 

until quantum effects kick in... but I know what you mean! But my statement is true since the battery would never become fully charged!

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