chris w

Not having them in a closed box is not a BSS fail. 1. Your batteries are required to be prevented from moving (eg: a closed box or strapping) and 2. Must be covered to prevent something falling on the terminals. (eg: a closed box or a covering like rubber mats). So, a closed box is sufficient, but not necessary. The Peukert Factor quoted by battery manufacturers, tells you how many hours you can discharge continuously, at a certain current, before the battery voltage drops to (usually) 10.5v. There is no ambiguity in that. I use 3 Victron Battery Monitors, on my boat, to monitor various battery banks, including the SOC. The SOC indication, for the main batteries, in my experience, tracks their actual SOC based on (no load) voltage really well. On recharge, the battery monitor resets the SOC to 100% based on tail current. Tail current is the only sure way to know that your batteries are really charged.

Nick Out of courtesy to you, I did watch the 3 YouTube videos you highlighted. First of all, my initial reaction is that what the author has produced is very much a “Janet & John do batteries; chapter 1.” It’s extremely basic. He also lets himself down, by his saying, “I studied High School chemistry so I guess that makes me qualified to talk about this subject.” (!!) However, having said that, I don’t disagree substantially with what he is saying about the chemistry. As a graduate in Electronics, and with over 30 years of boat owning, I feel I’m at least as qualified as the next guy to espouse my views. There is much made in these videos as, in this discussion, of leaving a “discharged” battery to rest and then being able to recover more capacity from it. I agree this is the case. However, that doesn’t mean Peukert’s Law is incorrect as the video author asserts. When we use batteries in a high discharge situation, such as with an inverter, Peukert’s Law will give an extremely good guide to the amount of capacity we can draw out of the battery, before, say, the inverter cuts off. We don’t, however, in real use, think “I’ll now rest the battery for 6 hours and then I can drain it some more.” We recharge it, as soon as is practicable. In the real-world, Peukert is a very good predictor of useable SOC and that’s why it’s important to ensure one does set the correct Peukert Factor in AH meters etc. I suspect the author of the videos does not have much practical experience in using batteries in real-life situations. He’s just trying to belabour a theoretical point that has no practical use.

I’ll take Sundays, as I’m an atheist.

Nick You clearly didn’t read my post correctly. Particularly that sentence, after my calculations, where I specifically stated, “However, this is not what Peukert is about. My calculations were to show that it can never be the case that all the energy in a battery is available as useful energy, whereas you were pushing the opposite view.

Well, you’ve awakened a third old man. I’m back from my time travels and….. I agree with Gibbo! Furthermore, there is another loss which hasn’t been considered quantitatively. Take the 2 identical lead acid batteries in the earlier example. One discharging at 5A for 10 hours and one discharging at 50A for 1 hour. Let’s assume the batteries’ initial internal resistance is 5mohm (a typical fully charged figure), though the actual figure is not relevant to the argument. The energy lost through heat in the first battery will be: i^2Rt = (5^2)(0.005)(10) = 1.25Wh The energy lost through heat in the second battery will be: i^2Rt = (50^2)(0.005)(1) = 12.5Wh ie: the second battery will dissipate 10 times the energy compared to the first battery, owing to its internal resistance. This energy (in both cases) is unrecoverable. Further, the internal resistance of the second battery will actually increase at a faster rate than the first battery as it becomes discharged more quickly. So, in reality, the second battery will dissipate even more energy owing to its internal resistance’s increase. However, this is not what Peukert is about. Peukert is analogous to a car, whose fuel consumption is quoted by the sales person, when asked, as “30mpg”. Now, we all understand implicitly that means, driven steadily, our fuel consumption should be as quoted. We also understand (implicitly), that if we were to drive the car for an extended period at 80mph (where permitted!), we would not expect to obtain 30mpg. Peukert is a mathematical way of describing those “high speeds” in a battery.

I'm still the only one with a working paralleler with auto-switch off

Time will tell.....................

I have 4 of the "Waterbuoys" (as seen on Dragon's Den). I haven't had cause to use one in anger yet but the end caps have all stayed in on mine and one got wet in the rain and didn't go off. (Maybe it never will ). A tenner each. Chris

If we listed all your errors and mistakes on here Pete, we'd have to start a special section just to keep it manageable (that would include your latest paralleler circuit of course, which STILL isn't working). You didn't earn the epithet of "Gibbo's glove-puppet" for nothing.

Er............... that's because it was designed for MY boat that does have an alternator controller. I designed it for ME and published it because I thought others might be interested. The other difference is that it works. Chris That's after a total of 12 posts from you. Once you're into long trousers, you can start to debate with the big boys, who've spent a lifetime in electronics. Till then, keep sucking your lolly.

If you rae not willing to "tinker", you shouldn't be playing around anyway with your electrical systems. A lead-acid battery is a deadly weapon in the wrong hands. We know nothing yet as to whether Smelly's current circuits work or not. I predicted the cap would blow and indeed it did. Chris

Warm "Pol Roger '96 Winston Churchill" ........... to what is the world coming? Chris

The 3 most common causes, apart from the man-made hair problem listed above are probably: 1. Blown diode = new rectifier = £15 on Ebay 2. Faulty regulator = new regulator = £10 on Ebay 3. Worn brushes after long service = might as well replace the whole regulator (which usually includes the brush assembly = £10 on Ebay At the end of the day, a whole new alternator can be had on Ebay for around £40 - £70 depending on which day you look. Chris

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

I agree that my version needs soldering and the removal of the regulator. But Smelly doesn't seem phased by these operations based on his dialogue. My circuit also allows paralleling batteries for starting....................and, the other issue,........ it works. Chris

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

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

Why all the faffing around with capacitors and light bulbs? The way I did it was so simple..... my switching circuit works immediately (see original diagram) with no problems at all and using the regulator output transistor collector of the alternator with the PDAR connected, to achieve auto switch-off, works beautifully (thanks for Snibble's idea). You seem to understand what needs to be connected where, in this regard Smelly, as you correctly described the mods needed somewhere above. It works, it's not difficult for you, you have all the bits... just go do it - no issues and switches off everytime at 14.35v. Chris

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

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

It's all down to the contract you strike with them and who owns which bits when. Get a watertight contract and you're covered. If THEY don't like strict terms (with penalty clauses on time of delivery) then walk away. Chris

You will need to charge your batteries for an extra HOUR EVERY DAY just to make up for the inefficiencies of a 240v fridge and an inverter. And, if your inverter fails then your fridge fails too. It's 12v everytime, unless you don't have the money. Chris

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

Measure the continuity (resistance) between the D+ connection on the regulator and the brushes. You want the brush that is NOT connected to D+. This is known as the "field brush". I would swap out the regulator bolts for M3.5 screws of about 25-30mm length. They are so much easier to remove as compared to the very thin walled socket needed to remove the dstandard bolts. Chris

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