Advice on installing an inverter

[chris w]

Of course it is. Hydrogen contains a huge amount of potential energy. In order to make that energy available we are electrolysing water. In order to do so requires a huge amount of energy. This we know because we can get a huge amount of energy back by burning the hydrogen. Therefore conservation of energy is intimately related.

 

Gibbo

"Conservation of Energy" relates to the impossibility of getting more energy out of a system than is put in. (ie: energy cannot be created or destroyed). It doesn't put a limit on the actual efficiency providing that efficiency is less than 100% (ie: it disallows perpetual motion machines). So however great or not is the potential energy of hydrogen, it must follow the conservation of energy as must any potential fuel. It has no practical relevance to charging at 14.2v or 14.8v.

 

The true efficiency of electrolysis is actually quite high and not low as you imply. It is in the region of 50-80% (dependent upon the composition of the electrodes). That would imply that a lot of water would be electrolysed in a battery if charged beyond the gassing point. That is not borne out in practice by me or others at 14.8v.

 

Chris

Edited May 24, 2009 by chris w

[Gibbo]

"Conservation of Energy" relates to the impossibility of getting more energy out of a system than is put in. (ie: energy cannot be created or destroyed). It doesn't put a limit on the actual efficiency providing that efficiency is less than 100% (ie: it disallows perpetual motion machines). So however great or not is the potential energy of hydrogen, it must follow the conservation of energy as must any potential fuel. It has no practical relevance to charging at 14.2v or 14.8v.

 

You amaze me at times with your inability to see the most simple things and with the way you get competely sidetracked into totally unrelated matters. Efficiency is unrelated to the matter at hand. We know we can get a huge amount of energy from burning hydrogen and turning it into water. Therefore, as a consequence of the conservation of energy, without looking anything up or doing any sums, we can know that we have to put a huge amount of energy into water to turn it into hydrogen and oxygen.

 

The true efficiency of electrolysis is actually quite high and not low as you imply. It is in the region of 50-80% (dependent upon the composition of the electrodes). That would imply that a lot of water would be electrolysed in a battery if charged beyond the gassing point. That is not borne out in practice by me or others at 14.8v.

 

Chris

 

I never mentioned the "efficiency" of electrolysis. I never implied it was low. I said absolutely nothing relating to the subject. You did. So you are aguing with yourself.

 

The amount of energy required to electrolyuse water is huge. That has got nothing to do with efficiency. You can say it implies what you want. Go and look it up.

 

We've already had this thread and done the sums. It's old. You, apparently, missed it.

 

Gibbo

[Theo]

Right ho.

 

Point 1 which Gibbo so eloquently explained

 

Gassing the batteries reduces the charging current because of the layer of tiny bubbles on the plates. This increases the internal resistance so the charging current falls off or the voltage rises or a combination of the two.

 

Point 2

 

The gassing voltage lowers as the state of charge increases.

 

Could you use this information to detect the actual gassing voltage (once the batteries had been charged so far that the current dropped below the maximum that the charge source could deliver) at any time during the charging process? If you could then you could design an alternator controller to remain just below the gassing voltage for most of the time. I suppose it would need to take brief excursions from this to make further measurements of the gassing voltage.

 

If you could hold the charging voltage just on the gassing voltage would you not have the ideal maximum charge rate?

 

Nick

Edited May 25, 2009 by Theo

[Gibbo]

Could you use this information to detect the actual gassing voltage (once the batteries had been charged so far that the current dropped below the maximum that the charge source could deliver) at any time during the charging process? If you could then you could design an alternator controller to remain just below the gassing voltage for most of the time. I suppose it would need to take brief excursions from this to make further measurements of the gassing voltage.

 

Yes you're on exactly the right lines. I've been working on it for many years on and off. Along with many others around the globe. The idea goes back a long, long time. The technology is well known and proven (which is why I find it so highly amusing that a boater with a meter denies the possibility). There's no practical way to do it. I've done it in the lab (as have others), but translating that to a real world device isn't so easy.

 

There are patents already issued for it so for anyone to deny it is just daft.

 

The biggest problem is the expense of the equipment. It effectively combines a very wide ranging mains powered charger (or alternator controller) along with an amphour counter and a device similar to a solar PV MPPT regulator (but working in reverse) and several battery temperature sensors (which in themselves are a problem to get working reliably). No one would buy it because of the cost.

 

The next problem is the complexity of the installation. It requires, as an absolute necessity, a shunt. And as people in the trade know, end users (and many installers) seem to be completely incapable of fitting a shunt following a simple diagram. It's not rocket science, but for some reason people nearly always get it wrong and stick wires on the wrong side of it. With a battery monitor you just get duff readings. If it was controlling this type of charger the results could be absolutely catastrophic (boiled batteries, leaked electrolyte, explosions etc). Ditto the battery temperature sensor.

 

Some of the engineers at Cruising Equipment (sister company of Heart Interface) looked at the idea about 20 years ago but management wouldn't fund development as they decided the installation issues would result in too many problems.

 

Gibbo

[chris w]

We know we can get a huge amount of energy from burning hydrogen and turning it into water. Therefore, as a consequence of the conservation of energy, without looking anything up or doing any sums, we can know that we have to put a huge amount of energy into water to turn it into hydrogen and oxygen.

 

 

 

I never mentioned the "efficiency" of electrolysis. I never implied it was low. I said absolutely nothing relating to the subject. You did. So you are aguing with yourself.

 

The amount of energy required to electrolyuse water is huge. That has got nothing to do with efficiency. You can say it implies what you want. Go and look it up.

 

We've already had this thread and done the sums. It's old. You, apparently, missed it.

 

Gibbo

From where do you keep getting this "huge amount of energy" idea by burning hydrogen and producing water?

 

The calorific value of hydrogen is 3 times that of diesel, so while it is good it is not HUGE. I think you are getting confused with the "Hydrogen bomb" which utilises nuclear fusion not the burning of hydroge, Now that IS huge!!

 

The calorific value of hydrogen is 142MJ/kg. If you rattle this number, this comes out at 39KWHr/kg or 39 watt.hours/gram.

 

At say 14v, that is at most 3AH/gram of hydrogen. Even if we assume that electrolysis is only 50% efficient (this is why your saying "efficiency isn't important" is incorrect because it has a bearing on the answer) that equates to 6AH of electrolysis to produce 1 gram of hydrogen. In fact, if you research, electrolysis is more often quoted to be higher than 80%, but let's stick with 50% for this exercise.

 

So how much water will be "used up" by electrolysis in producing that 1 gram of hydrogen. (BTW, 1 gram of hydrogen is just over a litre of gas.) Well comparing the atomic weights of oxygen and hydrogen and the fact that water is H₂O, it takes 9 grams of water (ie: 9cc in volume) to produce 1 gram (~1 litre) of hydrogen by electrolysis.

 

So, at 50% efficiency, 9cc of water will be lost in the batteries by 6AH of electrolysis. ie: 1.5cc/1AH. I lose virtually no water from my batteries over months and that is the experience too of other alternator regulator users on here. Ergo, large amounts of current at 14.8v are not going into electrolysing water and producing gas. Even if I lost one litre of water per year (which I don't) that equates to less than 700AH of "wasted" Ampere.Hours PER YEAR out of a total that is around 150AH PER DAY of charging whilst cruising.

 

So the percentage wasted by electrolysis at 14.8v is absolutely negligible and, in reality, the efficiency of electrolysis is probably much higher than I used in the maths. Further some of the water loss will be due to evaporation not electrolysis, so the whole electrolysis thing at 14.8v is a big fat red herring.

 

 

Chris

Edited May 25, 2009 by chris w

[Gibbo]

From where do you keep getting this "huge amount of energy" idea by burning hydrogen and producing water?

 

The calorific value of hydrogen is 3 times that of diesel, so while it is good it is not HUGE.

 

Compare it with the same weight of lead acid battery. I think you'll find it's huge.

 

I think you are getting confused with the "Hydrogen bomb" which utilises nuclear fusion not the burning of hydroge, Now that IS huge!!

 

No I'm not confusing it with that.

 

The calorific value of hydrogen is 142MJ/kg. If you rattle this number, this comes out at 39KWHr/kg or 39 watt.hours/gram.

 

At say 14v, that is at most 3AH/gram of hydrogen.

 

I knew you'd make that mistake. That's why I said the sums had already been done in a previous thread. You need to recalculate with a different voltage. The voltage being that required to dissociate the electrolyte, not the voltage applied across the entire battery. The electrolysing voltage is in parallel with a small part of the internal resistance of the battery which is in series with the battery voltage. Try about 0.3 volts per cell which instantly drops your figure eight-fold.

 

And therefore the rest of your post is superfluous.

 

Re-do your sums. Properly this time.

 

They've already been done. Properly.

 

Gibbo

 

Edited to correct maffs.

Edited May 25, 2009 by Gibbo

[Gibbo]

The calorific value of hydrogen is 142MJ/kg. If you rattle this number, this comes out at 39KWHr/kg or 39 watt.hours/gram.

 

At say 14v, that is at most 3AH/gram of hydrogen. Even if we assume that electrolysis is only 50% efficient (this is why your saying "efficiency isn't important" is incorrect because it has a bearing on the answer) that equates to 6AH of electrolysis to produce 1 gram of hydrogen. In fact, if you research, electrolysis is more often quoted to be higher than 80%, but let's stick with 50% for this exercise.

 

So your maffs proves that 6Ahr at 14 volts produces 1 gram of hydrogen which you reckon is 1 litre?

 

Odd that. Because that's 302kJ to electrolyse 1 litre of hydrogen. Yet according to this:-

 

http://hyperphysics.phy-astr.gsu.edu/Hbase...o/electrol.html

 

It takes 237kJ to electrolyse 1 mole of water which will produce 22 litres of hydrogen. Your figures are out by a factor of 20

 

So how much water will be "used up" by electrolysis in producing that 1 gram of hydrogen. (BTW, 1 gram of hydrogen is just over a litre of gas.) Well comparing the atomic weights of oxygen and hydrogen and the fact that water is H₂O, it takes 9 grams of water (ie: 9cc in volume) to produce 1 gram (~1 litre) of hydrogen by electrolysis.

 

Again you've got problems with your maffs.

 

9 grammes of water is half a mole.

 

Half a mole of water produces half a mole of hydrogen.

 

Half a mole of hydrogen = 11 litres. Somewhat more than the 1 litre you quote.

 

So, at 50% efficiency, 9cc of water will be lost in the batteries by 6AH of electrolysis. ie: 1.5cc/1AH. I lose virtually no water from my batteries over months and that is the experience too of other alternator regulator users on here. Ergo, large amounts of current at 14.8v are not going into electrolysing water and producing gas. Even if I lost one litre of water per year (which I don't) that equates to less than 700AH of "wasted" Ampere.Hours PER YEAR out of a total that is around 150AH PER DAY of charging whilst cruising.

 

9cc of water (half a mole) will be lost by 118kJ.

 

Again assume 50% efficiency, 118kJ at the gassing voltage (let's take an average of 0.3 volts - it changes depending upon the state of charge).

 

118000J = 33 watt.hours - at 1.8 volts (the averaged gassing voltage across the 6 cells) that is 18Ahrs. At 50% efficiency that's 36Ahrs

 

See how far out your figures are?

 

In summary:-

 

You state that 6Ahrs will electrolye 9cc of water. In reality it is 36Ahrs ie 6 times more

 

You say 9cc of water will produce 1 litre of hydrogen. In reality it is 11 litres.

 

How far out do you want your maths to be?

 

Gibbo

 

Edited again to correct broken sums

 

There seems little point in even trying to get you to accept reality when your maths is so far adrift from reality. Until you get those right you've got no chance. A 600% error on one part, 2000% error on another, 1100% error on another? Come on!

 

(I'm still checking my maths and references because I'm having trouble believing that you got this so wrong but, so far, it appears that you did)

Edited May 25, 2009 by Gibbo

[AKULA]

i use a domestic washing machine i disconect the heating element i do use 10 x 135 amp batts coupled to a sterling charger using 2 110 amp alternators and a 3000 watt invert.

[Michele]

2 TV's for 4 hours each and 40W each = 320WH per day (WH = watt.hours)

Microwave for say 15 minutes per day @ 1400W = 350Wh per day

Fridge @ 120W @ 50% duty cycle for 24 hours = 1440WH per day

 

Total WH's = 2110 WH. We'll divide this by 10v rather than 12v to allow for the inefficiencies of the inverter. This gives 211AH (Ampere.Hours) per day which is about double what a "typical" narrow boater would use. It's a very high consumption.

 

Can I ask why you need 2 TV's?

There's b*gger all on anyway - we very rarely need one!

[chris w]

I knew you'd make that mistake. That's why I said the sums had already been done in a previous thread. You need to recalculate with a different voltage. The voltage being that required to dissociate the electrolyte, not the voltage applied across the entire battery.

 

 

That's nonsense. IF the voltage is as you say then the current associated with it would also be tiny and my figures would be back on track even by your reckoning.

 

The charge current, say 50A, is being driven by 14.8v NOT 0.3v. I think you make this stuff up as you go along.

 

Chris

[Gibbo]

That's nonsense. IF the voltage is as you say then the current associated with it would also be tiny and my figures would be back on track even by your reckoning.

 

The charge current, say 50A, is being driven by 14.8v NOT 0.3v. I think you make this stuff up as you go along.

 

Chris

 

Let's go over the figures again:-

 

My post shows that 1Ahr of overcharge will electrolyse 0.25cc of water.

 

Your so called expert maths shows it will electrolyse 1.5cc of water.

 

So we are 6 times adrift.

 

How come this reference............

 

http://www.azsolarcenter.com/technology/ba...es/pdfs/ch3.pdf (see page 40) shows that 1ahr of overcharge will electrolyse 0.33cc of water?

 

I say 0.25, they say 0.33

 

You say 1.5

 

Who do you suspect has got it wrong (yet again)?

 

And you say I make it up? How come I'm right and you're wrong then?

 

How's your one litre of hydrogen?

 

Are you sure you have a degree?

 

Gibbo

[chris w]

You say 9cc of water will produce 1 litre of hydrogen. In reality it is 11 litres.

 

How far out do you want your maths to be?

 

Gibbo

 

Edited again to correct broken sums

 

There seems little point in even trying to get you to accept reality when your maths is so far adrift from reality. Until you get those right you've got no chance. A 600% error on one part, 2000% error on another, 1100% error on another? Come on!

 

(I'm still checking my maths and references because I'm having trouble believing that you got this so wrong but, so far, it appears that you did)

 

Well let's analyse your dissertation just using common sense. What is 9 grams of water made out of.......er... hydrogen and oxygen. 2 atoms of hydrogen and 1 atom of oxygen per molecule of water.

 

An atom of oxygen is 16 times heavier than an atom of hydrogen (atomic weights). So, in the 9 grams of water, the hydrogen will weigh 2/18 of 9 grams and the oxygen 16/18 of 9 grams. So the hydrogen will weigh 1 gram and the oxygen 8 grams. (ie: 2 atoms of hydrogen will weigh 8 times less than 1 atom of oxygen).

 

 

Chris

Edited May 26, 2009 by chris w

[Gibbo]

Come on, just use common sense. What is 9 grams of water made out of.......er... hydrogen and oxygen. 2 atoms of hydrogen and 1 atom of oxygen per molecule of water.

 

An atom of oxygen is 16 times heavier than an atom of hydrogen so in H₂O. So, in the 9 grams of water, the hydrogen will weigh 2/18 of 9 grams and the oxygen 16/18 of 9 grams. So the hydrogen will weigh 1 gram and the oxygen 8 grams. (ie: 2 atoms of hydrogen will weigh 8 times less than 1 atom of oxygen).

 

The density of hydrogen is 0.9 grams/litre so 1 gram of hydrogen will have a volume 1/0.9 = 1.1 kitres. QED. Simples

 

Chris

 

Oh dear. Your chemistry is even worse than your electronics (and that in itself is attrocious).

 

Gibbo

 

Well let's analyse your rantings...... just use common sense. What is 9 grams of water made out of.......er... hydrogen and oxygen. 2 atoms of hydrogen and 1 atom of oxygen per molecule of water.

 

An atom of oxygen is 16 times heavier than an atom of hydrogen so in H₂O. So, in the 9 grams of water, the hydrogen will weigh 2/18 of 9 grams and the oxygen 16/18 of 9 grams. So the hydrogen will weigh 1 gram and the oxygen 8 grams. (ie: 2 atoms of hydrogen will weigh 8 times less than 1 atom of oxygen).

 

The density of hydrogen is 0.9 grams/litre so 1 gram of hydrogen will have a volume 1/0.9 = 1.1 litres. NOT 11 litres QED. Simples Even intuitively, it seems highly improbable that one would get 11 litres of hydrogen from less than 2 x teaspoons of water, as you "proved". ???????????????

 

Chris

 

Er........

 

Look it up. The density of hydrogen is 0.09 g/L

 

http://www.google.co.uk/search?hl=en&s...ity&spell=1

 

And of course, you still won't admit that you're wrong.

 

Gibbo

[chris w]

Oh dear. Your chemistry is even worse than your electronics (and that in itself is attrocious).

Look it up. The density of hydrogen is 0.09 g/L

 

Gibbo

OK...................details, details. I agree that was a slip so 1 gram of hydrogen does indeed have a volume of 11 litres....... you are correct BUT that actually supports my argument about electrolysis in batteries at 14.8v. In fact it just supported it by a factor of 10.

 

Because, if we are electrolysing so much water, as you maintain, and I don't, the whole of the bloody cut would be full of hydrogen at that rate.

Chris

 

Your "attrocious" spelling should only have one "t"

Edited May 26, 2009 by chris w

[Gibbo]

A bold statement but just that....... a statement...... no facts from you.

 

Chris

 

How come that my calculated figure for the amount of water electrloysed by 1ahr of overcharge agrees with the reference I posted but yours doesn't?

 

How come my figure for the volume of hydrgen agrees with the link I posted but yours doesn't?

 

Is the rest of the world wrong?

 

Gibbo

 

All together...................

 

"Oh it's all gone quiet over there............. Oh it's all gone quiet over there........"............

 

You should be getting quite used to this by now.

 

 

Gibbo

[chris w]

How come that my calculated figure for the amount of water electrloysed by 1ahr of overcharge agrees with the reference I posted but yours doesn't?

 

How come my figure for the volume of hydrgen agrees with the link I posted but yours doesn't?

 

Is the rest of the world wrong?

 

Gibbo

You are now erroneously using an article that isn't discussing the same point. They talk about 1AH of OVERCHARGE electrolysing 0.3cc of water.... note the use of the word OVERCHARGE........ not charging at 14.8v. If a battery is overcharged then I would also agree that water will be electrolysed. But we are NOT talking about OVERCHARGE just CHARGING. There has been no mention of OVERCHARGE. Where does that article state that charging at 14.8v will result in loads of electrolysis and "wasted" current? It doesn't mention either.

 

Jeez - you are really having to scrape the barrel on this one.

 

Chris

Edited May 26, 2009 by chris w

[Gibbo]

You are now erroneously using an article that isn't discussing the same point. They talk about 1AH of OVERCHARGE electrolysing 0.3cc of water.... note the use of the word OVERCHARGE........ not charging at 14.8v. If a battery is overcharged then I would also agree that water will be electrolysed. But we are NOT talking about OVERCHARGE just CHARGING. There has been no mention of OVERCHARGE. Where does that article state that charging at 14.8v will result in loads of electrolysis and "wasted" current? It doesn't mention either.

 

Jeez - you are really having to scrape the barrel on this one.

 

Chris

 

What do you think overcharge is?

 

You are going round and round in circles, making yourself look sillier and sillier. Each time you get proven to be wrong (which is happening so often it is becoming monotonous), you try to change the argument to something else.

 

It is exactly overcharge that we are discussing. If you can't even see that then there is no hope for you whatsoever.

 

Look up overcharge in a technical publication about lead acid batteries.

 

It is actually defined as the current drawn by the battery as a result of exceeding the gassing voltage. The overcharge power is the overcharge current multiplied by the gassing voltage (not the charging voltage). I'm amazed that you can't see that simple fact. I suspect you can now. You're just afraid to admit that, yet again, you got it wrong.

 

You really need to stop spouting off as if you know what you are talking about with electronics when quite clearly you are grappling in the dark on this subject.

 

0.9g/L my *rse

 

Even when you are shown a reference that very clearly shows you are in error and agrees with my figures you try to make out it is misinterpreted. Absolutely pathetic.

 

Gibbo

 

Where does that article state that charging at 14.8v will result in loads of electrolysis and "wasted" current? It doesn't mention either.

 

No it doesn't say "loads" of elecrolysis. And neither did I.

 

How this came up was because I explained that all the extra ebergy resulted in no extra charge but simply electrolysed water.

 

It was you who decided that it would electrolyse loads of water and got your figures wrong by several hundred percent.

 

Remember, it wasn't me that said lots of water would be lost. YOU said it. I said the opposite and showed why. I said very little would be lost because it takes such a huge amount of energy to do so.

 

Now I have shown mathematically why that is the case and also provided references showing the same.

 

You now try to divert it.

 

You still don't accept it.

 

That's fine by mne. You can sit there believing what you want. But you're wrong. As you so often are.

 

Gibbo

[chris w]

Bt it doesn't take HUGE amounts of energy to electrolyse water. The article to which you refer does NOT mention charging at 14.8v as being wasteful of current in electrolysing water. It talks about that happening (at any voltage) once the battery is FULLY charged. We are talking about a battery that is in a state of discharge.

 

The article is not relevant to your argument therefore in this regard. The misreading error in the density of hydrogen, that I freely admitted above that I made, only serves to bolster MY argument paradoxically, because it makes no difference to the energy used to electrolyse water at all only to the amount of hydrogen produced. By your reckoning the whole engine bay must be always full of hydrogen until the engine cover is lifted.

 

Show me the unambiguous reference in the article to the bad effects of charging at 14.8v compared to 14.2v. This is similar to your argument about adaptive charging from a couple of weeks ago. Once I showed mathematically, both off-line and on-line that your argument was incorrect based on your own assumptions (not mine), you suddenly went all quiet.

 

 

Chris

Edited May 26, 2009 by chris w

[Gibbo]

Bt it doesn't take HUGE amounts of energy to electrolyse water.

 

Yes it does. It takes 237,000J per 18cc

 

The article to which you refer does NOT mention charging at 14.8v as being wasteful of current in electrolysing water. It talks about that happening (at any voltage) once the battery is FULLY charged. We are talking about a battery that is in a state of discharge.

 

The article is not relevant to your argument therefore in this regard. The misreading error in the density of hydrogen, that I freely admitted above that I made, only serves to bolster MY argument paradoxically, because it makes no difference to the energy used to electrolyse water at all only to the amount of hydrogen produced. By your reckoning the whole engine bay must be always full of hydrogen until the engine cover is lifted.

 

Show me the unambiguous reference in the article to the bad effects of charging at 14.8v compared to 14.2v. This is similar to your argument about adaptive charging from a couple of weeks ago. Once I showed mathematically, both off-line and on-line that your argument was incorrect based on your own assumptions (not mine), you suddenly went all quiet.

 

 

Chris

 

You are absolutely full of sh*t.

 

You have shown no such thing mathematically. What you have done, yet again, is get yourself buried in mathematical equations without actually understanding how they relate to reality. This seems to be a speciality of yours. That is how you ended up arguing with the world that the W terminal could have 60 volts on it. You even went so far as to state that you had measured that voltage on it in an attempt to back up your broken maths when we all, in fact, knew that you had measured no such thing. Quite simply because we all knew it was impossible.

 

From the reference quoted:-

 

"(1) The charge current at the start of recharge can be any va1ue that does not produce an average ce11 vo1tage in the battery string greater than the gassing voltage.

High recharge rates are less efficient. However I2R losses in the battery and cabling must be considered to prevent overheating. This is generally not a major

factor in renewable energy systems unless a large generator is part of the system."

 

"(2) During the recharge and until 100 percent of the previous discharge capacity has been returned, the ampere rate should be controlled to maintain the gassing

voltage, or lower, but as close to the gassing voltage as possible to minimize charge time."

 

"A key rule is that the cell gassing voltage should not be exceeded except during the finishing step of charge. The gassing voltage is the voltage at which the predominant

reaction consuming charge current is electrolysis of water in the electrolyte with evolution of oxygen at the positive plates and hydrogen at the negative plates."

 

"Summarizing, to decrease charge time while maintaining a consistent state-of-charge,

the following procedures should be used:

• Increase the initial charge current up to a maximum value of 25 A/100 Ah rated

5-h cell capacity.

• Use the highest setting of constant potential during the initial charge current

period without exceeding the temperature compensated cell gassing voltage."

 

According to the maths I have shown the overcharge is one part of the energy going into the battery. The total energy going into the battery is:-

 

Total energy = electrolysing energy + charge energy + heating energy

 

Therefore:-

 

Total volts * amps * time = (electrolysing voltage * total amps * time) + (total amps * [total voltage - electrolysing voltage - internal r volt drop] * time) + total current^2 * internal R)

 

Run your smartarse maths on that and it adds up correctly.

 

Yet, because you wear blinkers, you want the electrolysing energy to be the total charge voltage * total amps * time.

 

Therefore, by your broken maths:-

 

Total volts * amps * time = (total voltage * total amps * time) + (total amps * [total voltage - electrolysing voltage - internal r volt drop] * time) + total amps^2 * internal R)

 

This, by your maths: energy in = energy in + charge energy + heating energy.

 

By your own broken maths you have created overunity. A form of free energy.

 

When fully charged a wet cell battery will have a natural voltage of about 12.8 volts. Don't take my word for it. Look it up.

 

When fully charged the gassing voltage of a single cell is about 0.1 volts (remember it is highly SoC dependant and falls inversely with the electrolyte specific gravity). So that's 0.6 volts for a 12 volt battery. Don't take my word for it. Look it up.

 

12.8 volts natural voltage plus 0.6 volts = 13.4 volts. Again, don't take my word for it. Look it up.

 

Once the charge voltage reaches that voltage the battery starts to gas. Anything above 13.4 volts is wasted. It doesn't do anything other than electrolyse water.

 

If we charge that battery at 14.8 volts we have an overcharge voltage of 1.4 volts.

 

Now let's stick with this 1amp.hour figure that keeps cropping up.

 

1.4 volts (the overvoltage) * 1amp.hour = 5,040J

 

We know from the link earlier in the thread that it takes 237KJ to electrolyse 1 mole of water. 1 mole of water = 18cc

 

So 5,040J will electrolyse 5,040/237,000 moles of water = 0.021 moles.

 

1 mole = 18cc therefore 0.021 moles = 0.378cc of water.

 

That agrees very closely with the figure given in the reference of 0.336cc per amp.hour.

 

That is the reality. Those are the facts. That is what happens.

 

You state that the energy going into electrolysis is the amp.hours * charge voltage

So that would be 14.8 volts * 1amp.hour = 53,280J

 

53,280J will electrolyse 53,280/237,000 moles = 0.22 moles = 3.96cc

 

Your figure is 10 times too high. Which is what I said earlier in the thread.

 

My figure agrees with the reference I posted.

 

Yours is out by one order of magnitude.

 

I have proved it mathematically.

 

I have provided references backing up what I say.

 

Your next argument will be that we shouldn't be discussing a fully charged battery and that of course you knew all along that would be how much water gets electrolysed. That argument doesn't hold. You maintain that more water will be electrolysed not less.

 

If we take a battery at about 50% SoC the gassing voltage will be higher. It will therefore require more energy throwing at the battery in order to reach the overvoltage required to gas the battery. Therefore even less electrolysis will take place for a 1amp.hour overcharge.

 

You can try to argue all you want. You are wrong.

 

One day you might realise that your knowledge on this subject is horribly inadequate. You are punching way beyond your weight and ability.

 

Why not accept that and try to learn from it? Because all that is happening at the present time is you are being continually proved wrong and learning absolutely nothing.

 

Why does my calculation agree with the reference posted but yours is out by 1000% ?

 

Don't you think it's probably because you're wrong?

 

Gibbo

 

This is similar to your argument about adaptive charging from a couple of weeks ago. Once I showed mathematically, both off-line and on-line that your argument was incorrect based on your own assumptions (not mine), you suddenly went all quiet.

 

That is total cr*p. Just like most of your broken maths. You have done nothing of the sort. You got lost in the maths and reality because it was beyond your comprehension.

 

I'll repeat what I have previously said. Your mathematical ability is beyond question. But you have a serious problem relating it to reality. It's no good being able to formulate a great big equation if you then put cr*p into it: Cr*p in, cr*p out.

 

Gibbo

Edited May 26, 2009 by Gibbo

[Cosmic]

Chris and Gibbo have kindly produced a

on how to install an inverter

[honey ryder]

brilliant.

[chris w]

Yes it does. It takes 237,000J per 18cc

 

if you then put cr*p into it: Cr*p in, cr*p out.

 

Gibbo

Your quoted article (but not you) calculated 1AH per 0.33cc of water. I calculated EXACTLY the same thing even before you highlighted the article and expressed it as 3AH per 1cc of water (ie: the same thing). I then gave you a chance to be right by saying that I would only use 50% efficiency - thus increasing the AH needed per cc of water to 6AH. Your article assumes 100% efficiency.... which is actually incorrect.....but whatever.

 

So for 1cc of water, 3AH @ 14.2v = 42.6 watt.hours = 153,360 watt.secs = 153,360 Joules (since 1 Joule = 1watt.sec) an answer which your article AND I both agree on perfectly. (but not you).

 

So for your volume of 18cc (why 18cc?) of water that needs 2.76MJ of energy, an even larger number than you have erroneously calculated.

 

However, although that number SOUNDS HUGE, let's say it again, 2.76 Mega Joules of energy - WOW......., in reality it is not a large amount of energy at all, it is just a very large number because the joule is so small. It is the same energy as that produced by burning 60 grams (!!!) of coal.

 

But, you stated above that one cannot use 14.2v but must use 0.3v for some reason that you made up. So, in YOUR world the energy required should be reduced by 0.3/14.2 = only 2% of that calculated by me (and your quoted article).

 

That calculates out to be 58KJ of energy....... the same as 1.2 GRAMS of coal!!!!! Now that really is HUGE

 

So the enrgy required to electrolyse 1cc of water according to your quoted article and me is 2.76/18 MJ = 153KJ or about the same as burning about 3 grams of coal.

 

Have you calculated the number of atoms in 1cc of water? That's a really huge number!!!

 

Chris

[Gibbo]

Your quoted article (but not you) calculated 1AH per 0.33cc of water. I calculated EXACTLY the same thing even before you highlighted the article and expressed it as 3AH per 1cc of water (ie: the same thing). I then gave you a chance to be right by saying that I would only use 50% efficiency - thus increasing the AH needed per cc of water to 6AH. Your article assumes 100% efficiency.... which is actually incorrect.....but whatever.

 

So for 1cc of water, 3AH @ 14.2v = 42.6 watt.hours = 153,360 watt.secs = 153,360 Joules (since 1 Joule = 1watt.sec) an answer which your article AND I both agree on perfectly. (but not you).

 

So for your volume of 18cc (why 18cc?) of water that needs 2.76MJ

 

I'll stop you there and give you a clue..........

 

18cc = 1 mole of water.

 

1 mole of water requires 237,000J to electrolyse it (look it up - don't try to calculate it with broken sums - I've already given you a link for it).

 

NOT 2,760,000

 

Again you are 1 order of magnitude out. Almost exactly.

 

Just like all the other figures you have produced throughout this thread.

 

You keep coming back with figures that are 10 times wrong. You did it with the volume of hydrogen, you do it with the amount of energy required to electrolyse water.

 

You keep doing it.

 

Just go and look the figures up!

 

The reason you keep coming back with figures that are wrong by 10 times is because you are thinking that the charge voltage goes into electrolysis. It doesn't The electrolysing voltage does. They are roughly ten times apart.

 

I feel like Gordon Ramsey trying to explain a pancake recipe to Raymond Babbit

 

Gibbo

[Gibbo]

So for 1cc of water, 3AH @ 14.2v = 42.6 watt.hours = 153,360 watt.secs = 153,360 Joules (since 1 Joule = 1watt.sec) an answer which your article AND I both agree on perfectly. (but not you).

 

You might agree on it.

 

The article doesn't.

 

Your ability to apply your highly advanced maths to reality is crap, Chris. Absolutely crap.

 

You are one order of magnitude in error.

 

Just go and look the f*ck*ng numbers up.

 

Gibbo

[Gibbo]

Bump

 

The sounds of silence.....

 

 

Gibbo