Advice on installing an inverter

[chris w]

The sound of silence is because I am currently enjoying the sound of waves lapping on a beach and 30degC temperatures. Even the Forum is not a priority at the moment.

 

The article calculated 1AH per 0.33cc...... I calculated 3AH per 1cc.................er......I believe those figures are identical

 

You keep shooting yourself in the foot, because if you were correct with your 237KJ figure then it requires 10 times LESS energy to electrolyse water - so your HUGE is actually HUGE/10 and since your original HUGE was actually SMALL, it's now miniscule.

 

Chris

Edited May 27, 2009 by chris w

[RLWP]

Bump

 

The sounds of silence.....

 

 

Gibbo

 

Hello Darkness, my old friend...

 

Richard

[Gibbo]

The sound of silence is because I am currently enjoying the sound of waves lapping on a beach and 30degC temperatures. Even the Forum is not a priority at the moment.

 

The article calculated 1AH per 0.33cc...... I calculated 3AH per 1cc.................er......I believe those figures are identical

 

You should be getting quite used to being wrong and backpedalling by now. You've done an awful lot of it recently.

 

You calculated 1 amp.hour at 14.8 volts.

 

The article calculated 1 amp.hour at about 1.2 volts.

 

You are roughly one order of magnitude in error. As the many references I have posted confirm. You have also "proved" free energy with your broken sums.

 

You keep shooting yourself in the foot, because if you were correct with your 237KJ figure then it requires 10 times LESS energy to electrolyse water - so your HUGE is actually HUGE/10 and since your original HUGE was actually SMALL, it's now miniscule.

 

Chris

 

It is correct that 237kJ of energy is required to electrolyse 1 mole of water. Several references and other posters have confirmed it.

 

You state that 1 amp.hour of overcharge at 14.8 volts will electrolyse 0.3cc of water.

 

The article states that 1 amp.hour of overcharge at about 1.2 volts will electrolyse 0.3cc of water.

 

Therefore the energy required by you (the figure that you calculated and wrote on this forum) to electrolyse 1 mole of water is 2.76MJ

 

The energy required by the article (and me and everyone else) is 237kJ.

 

So you say that 2.76MJ goes into electrolysing water for a given overcharge (which would electrolyse 11 moles of water), but in actual fact it is 237kJ (as confirmed by the article that you can't understand). So you think 10 times more energy goes into electrolysing water than actually does in reality. You are roughly 10 times too high. So in actual fact the amount of water lost is roughly 10 times less than the figure you calculated.

 

Fancy a self professed mathematician being in error 10 fold!

 

How embarrassing for you.

 

Gibbo

 

PS: It would have been much more honourable to admit your error when you first realised it (which I reckon you must have done by now unless you have learning difficulties) rather than try to hide it. It would have also resulted in less embarrassment for you.

Edited May 28, 2009 by Gibbo

[carlt]

The sound of silence is because I am currently enjoying the sound of waves lapping on a beach and 30degC temperatures. Even the Forum is not a priority at the moment.

You spending the fruits of your libel action already?

 

I've not even received the summons yet.....

[Gibbo]

You spending the fruits of your libel action already?

 

I've not even received the summons yet.....

 

Anyone who can do a quick bribe will probably find out that his IP address hasn't changed and he is, in point of fact, still in Accrington:)

 

Gibbo

[carlt]

Anyone who can do a quick bribe will probably find out that his IP address hasn't changed and he is, in point of fact, still in Accrington

 

Gibbo

Exactly!

 

Clicky

[Wozzap]

Advertisement removed.

Edited May 29, 2009 by Liam

[chris w]

You calculated 1 amp.hour at 14.8 volts.

 

The article calculated 1 amp.hour at about 1.2 volts.

You are roughly one order of magnitude in error. As the many references I have posted confirm. You have also "proved" free energy with your broken sums.

 

 

 

It is correct that 237kJ of energy is required to electrolyse 1 mole of water. Several references and other posters have confirmed it.

 

You state that 1 amp.hour of overcharge at 14.8 volts will electrolyse 0.3cc of water.

 

The article states that 1 amp.hour of overcharge at about 1.2 volts will electrolyse 0.3cc of water.

 

 

Gibbo

Yeah.... and YOU calculated it at 0.3v, so you must be mortified that YOUR own article doesn't agree with you.

 

I still stand by my calculations. But even if we take your (erroneous) calculations of 237KJ of energy, do you not understand that this is NOT a huge amount of energy as you keep wrongly stating. It is a tiny amount of energy (it's just a large sounding number to the layman).

 

You keep going on about HUGE amounts of energy, I don't think that you understand Joules.

 

273KJ is the equivalent of burning 6 grams of coal. Go on work it out, if you can.

 

Ergo, running a charger at 14.8v with wet lead-acid batteries does NOT consume HUGE amounts of energy in electrolysis whether one uses my figure or your 10 times smaller figure.

 

Go figure! (Now the beach is calling).

 

Chris

Edited May 30, 2009 by chris w

[Gibbo]

Yeah.... and YOU calculated it at 0.3v, so you must be mortified that YOUR own article doesn't agree with you.

 

You've quite clearly got very serious comprehension problems. I calculated it at 0.3 volts per cell.

 

In case you don't know, there are 6 cells in a 12 volt battery.

 

And that is why the article agrees almost perfectly with my calculated figures but is 1000% different from yours.

 

I also explained very clearly (both in the thread and in PMs to you) that the actual gassing voltage varies as an inverse function of the SG so I chose 0.3 volts as a rough average.

 

And therefore, as usual, the rest of your post is completely irrelevant so I haven't even bothered reading it. As usual you jump on something and waffle on, but again, as usual, the bit you try to jump is a result of your own inabaility to comprehend things.

 

Your numbers are in error ten fold.

 

Battereis actually lose one tenth the amount water you calculated because your sums are broken.

 

Gibbo

[chris w]

So you're BACKPEDALLING now on your "HUGE amounts of energy are needed". I note that no mention of the word "HUGE", with regard to the amount of energy needed, has crept into your latest missives.

 

Let's use again your figure of 273KJ - how is that HUGE Gibbo? That's the equivalent of burning one teaspoon of diesel. Huh!!!!!!!!??????????????

 

The reason so little water is lost is because the effect is negligible. Try studying the philosophy behind "Occam's Razor"

 

Chris

[Cosmic]

(as I pointed out before, none of this has anything to do with installing an inverter)

I still stand by my calculations. But even if we take your (erroneous) calculations of 237KJ of energy...

Gibbo didn't calculate this figure - he looked it up. As he said, several other people have confirmed it.

 

I posted the calculation, and I repeat it here:

 

The reaction is

 

2H₂O -> 2H₂ + O₂

 

The bond energies (correct to 3 sf) are

 

H-O : 459 kJMol^-1

H-H : 432

O=O : 494

 

So the total energy is 459*4 - 494 - 432*2 = 478 kJ

 

But that is for two moles of H₂O,

 

so the energy per mole is 239kJ

Please explain where I have made a mistake (apart from getting involved at all )

[carlt]

Go figure! (Now the beach is calling).

Don't spend too much, in the amusement arcade.

 

I'll be seeking substantial costs and Blackpool can be pricey.

 

Try studying the philosophy behind "Occam's Razor"

14th century theory, Chris?

 

You'll be praying, next.

[Gibbo]

So you're BACKPEDALLING now on your "HUGE amounts of energy are needed". I note that no mention of the word "HUGE", with regard to the amount of energy needed, has crept into your latest missives.

 

Let's use again your figure of 273KJ - how is that HUGE Gibbo? That's the equivalent of burning one teaspoon of diesel. Huh!!!!!!!!??????????????

 

The reason so little water is lost is because the effect is negligible. Try studying the philosophy behind "Occam's Razor"

 

Chris

 

If you were even half as clever as you thihnk you are you wouldn't keep embarassing yourself in public.

 

In the scheme of things 237kJ is a huge amount of energy. Remember we are talking about batteries here not bags of coal. Oh wait, you had to stick the coal bit in to try to divert attention from your asertions that:-

 

1. That 1g of hydrogen has a volume of 1 litre.

2. It takes 2.76MJ to electorlyse 1 mole of water.

 

Do you want me to go through the enitre thread and pull up all the other c*ck ups you made in your maths?

 

Back to the 237KJ joules to electrolyse 1 mole of water (the longer you continue to refuse to accept that extremely well known scientific fact the longer you continue to look silly).....

 

A 12 volt, 100Ahr battery has a total available electrical energy of 4.3MJ. It takes 237kJ to electrolyse 1 mole of water (unfortunately for you that's a fact, I know you don't like it, but you're the only person on the planet who thinks it takes 2.76MJ) so a 100Ahr battery just about has sufficent energy (assuming 100% conversion and electrolysis efficiency) to electrolyse 0.33 litres of water. That's a small amount of water.

 

So yes, it iis a huge amount of energy

 

Gibbo

 

The reason so little water is lost is because the effect is negligible.

 

Now that's backpedalling !!

 

You said loads of water would be lost. I explained why loads of water isn't lost. Seems you've now changed your mind, accept that (again) I was correct and loads of water isn't in fact lost. You just haven't worked out why yet.

 

You think it's because it takes 2.76MJ to electrolyse 1 mole of water. When you finally accept that is actually out by a factor of ten you'll have to reappraise your sums. Eventually you might even get the right answer.

 

Try studying the philosophy behind "Occam's Razor"

 

How appropraite. The simplest explanation in this case is that you don't know much about electronics or batteries. Hence you keep getting things wrong.

 

Gibbo

[chris w]

A 12 volt, 100Ahr battery has a total available electrical energy of 4.3MJ. It takes 237kJ to electrolyse 1 mole of water (unfortunately for you that's a fact, I know you don't like it, but you're the only person on the planet who thinks it takes 2.76MJ) so a 100Ahr battery just about has sufficent energy (assuming 100% conversion and electrolysis efficiency) to electrolyse 0.33 litres of water. That's a small amount of water.

 

So yes, it iis a huge amount of energy

 

Gibbo

No - that's a small amount of energy. 237KJ is only 5.5% of that 4.3MJ. What you have also overlooked is that it's important to realize that the environment (at typical ambient temperature) contributes thermal energy equal to 48.7 kJ per mole to the hydrogen electrolysis process. That's a 17% contribution "for free". So a full battery can actually cause the electrolysis of 400cc of water if it is not recharged.

 

I probably put in half a litre of water a year spread over 3 x domestics and 1 x start battery. Let's ignore the start battery. So, based on your figures, (and allowing for the "free energy" which you forgot) the amount of energy from the batteries needed to electrolyse that amount is only around 5.5MJ PER YEAR, ie: 1.8MJ per battery per year or 100KJ per cell per year.

 

Your comparison of the necessary energy to the amount of energy in a battery is a red-herring because the battery is continuously charged and does not just have 4.3MJ available. The energy taken from the battery bank over a typical boating year, even by a non-liveaboard, will be in the region of 2 x 10⁹ Joules (ie: 2 BILLION Joules) per year. The 5.5MJ needed to electrolyse 0.5 litres of water over a year is just 0.3% of the total energy used.

 

Even if we say we have underestimated the amount of water needed to top up - let's say it's really 2 litres, not half a litre, the amount of energy taken from the total battery bank as compared to typical use over a year is only around 1.2%. Further, a goodly proportion of this will be due to evaporation and not electrolysis.

 

So, it's a tiny amount of energy - not a (Gibbo) HUGE amount and charging at 14.8v is not wasting energy - it's a red herring.

 

Chris

[smileypete]

charging at 14.8v is not wasting energy - it's a red herring.

 

Charging above gassing voltage IS wasting energy.

 

Whether 14.8v IS above gassing voltage depends on temperature, specific gravity (SG), and state of charge (SoC).

 

This is typically what happens:

 

If a battery is at 90% SoC and charging at gassing voltage at 20A, only 10A (50% of the 20A) is actually raising the state of charge of the battery.

 

If the charge voltage is raised above gassing voltage and current increased to 50A, then approx 25A goes into electrolysis, and of the 25A remaining, only 12.5A (50% of the 25A) is actually raising the state of charge of the battery.

 

So of that added 30A, only 2.5A goes towards raising the state of charge of the battery.

 

cheers,

Pete.

[MoominPapa]

What you have also overlooked is that it's important to realize that the environment (at typical ambient temperature) contributes thermal energy equal to 48.7 kJ per mole to the hydrogen electrolysis process. That's a 17% contribution "for free".

 

Can you explain that a bit more? It would seem to open the possibility of a heat-pump style energy producer. Electrolyse water and get hydrogen "worth" 48.7kJ per mole more than the energy put in; feed said hydrogen to a fuel cell, producing enough electricity for the next mole, and a useful surplus. Energy comes from cooling of surroundings. I think my "thermodynamics warning light" is flashing red at this point.

 

 

MP.

[PaddingtonBear]

Heavy man! I think we have gone beyond tin hats and Anderson Shelters with this one.

[Cosmic]

I still stand by my calculations. But even if we take your (erroneous) calculations of 237KJ of energy,

If my calculation is wrong, then please explain fully what the correct figure is.

237KJ is only 5.5% of that 4.3MJ.

Why are you usung the wrong figure?

What you have also overlooked is that it's important to realize that the environment (at typical ambient temperature) contributes thermal energy equal to 48.7 kJ per mole to the hydrogen electrolysis process. That's a 17% contribution "for free". So a full battery can actually cause the electrolysis of 400cc of water if it is not recharged.

Wrong.

[chris w]

Can you explain that a bit more? It would seem to open the possibility of a heat-pump style energy producer. Electrolyse water and get hydrogen "worth" 48.7kJ per mole more than the energy put in; feed said hydrogen to a fuel cell, producing enough electricity for the next mole, and a useful surplus. Energy comes from cooling of surroundings. I think my "thermodynamics warning light" is flashing red at this point.

 

 

MP.

Your "thermodynamic warning light" is flashing because you only did "O" level physics presumably. This is big boys' stuff. No, it's not perpetual motion owing to the inefficiency of producing the electricity to electrolyse the hydrogen in the first place.

 

You need to understand the effect of increasing entropy on a closed system. Since the entropy increases in the process of electrolysis, an amount of energy can be provided from the environment at temperature T degK. The amount which must be supplied by the battery is actually the change in the "Gibbs free energy" (note: Gibbs, not Gibbo )

 

Thus the environment "helps" the process by contributing an amount equivalent to 17% of the apparent energy per mole at typical room temperature (~300deg K). The usefulness of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

 

The overall efficiency is still less than 100%, but can be very close to 100% as there is no thermodynamic limit to efficiency, as would be implied if the process were linked to a Carnot cycle.

 

Chris

Edited May 31, 2009 by chris w

[MoominPapa]

The overall efficiency is still less than 100%, but can be very close to 100% as there is no thermodynamic limit to efficiency, as would be implied if the process were linked to a Carnot cycle.

 

My electrolysis followed by fuel cell example is not a Carnot cycle, but it's still a cycle. ΔS in the electrolysis step is equal and opposite to ΔS in the recombination step, so the two balance out. I guess since Gibbo just said "energy" and not "ΔH" or "ΔG" you can have your point, but if you're taking extra energy in the electrolysis step, you have to pay it back in the energy of recombination, you can't use ΔH for one and ΔG for the other.

 

I have a feeling the fact that electrolysis is not adibatic is probably relevant here too; but you're right that my grounding in thermodynamics (though rather beyond 'O'-level, TYVM) is not good enough to work out the details.

 

MP.

[coldor]

Reading all the posts here has been edifying, mystifying and highly entertaining, great stuff!

[bowten]

Reading all the posts here has been edifying, mystifying and highly entertaining, great stuff!

Hmmmmmm,I have just had a read myself.I have a headache now.

[Guest]

Reading all the posts here has been edifying, mystifying and highly entertaining, great stuff!

 

 

... and over three and a half years old too.

[bowten]

... and over three and a half years old too.

Has Chris w left the arena?

[Guest]

Has Chris w left the arena?

 

I don't think he posts anymore...