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