Gibbo
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Posts posted by Gibbo
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Long story but I got dragged back in by a private message. So... two points to make.
65watt hours gone.
65 watthours gone.
I know they're the same, but someone here seems to have no problem counting things twice.
And I'm gone again.
Nick, you are wrong. Tara.
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3 minutes ago, rusty69 said:
With all due respect to you sir Gibbo, how can he possibly argue about that without knowing what other useless devices you own.
You don't know Nick very well if you think he can't argue about that
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14 hours ago, dmr said:
So are there two levels of Peukert?
1 A genuine loss of amp-hour capacity due to rapid discharge
2 A ficticious loss of capacity due to the on-load volt drop when discharging down to 10.5volt.
Basically, yes. Well, very close. Remove "amp-hour" and it's probably the most concise summary so far.
9 hours ago, nicknorman said:Anyway, I think I may have killed Gibbo, which is a shame.
LOL No such luck I'm afraid. My central heating broke. Not the best time of year for it. I had to deal with that.
I think the Peukert argument has been done to death now, so...
You meantioned 3D printers. I actually got one last year. Printed a Benchy, then a bigger one. I've not used it since. Most useless device I've had for a long time. Care to argue about that?
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1 hour ago, nicknorman said:
Nick, you are wrong.
65 watt hours. Not in heat.
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Yet more irrelevant waffling. 65 missing watt hours.
Not heat. Converted into lead suphate. Takes energy.
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7 minutes ago, nicknorman said:
You're waffling again. I am in no way arguing against conservation of energy. But a battery cannot be thought of as an energy store in the way a capacitor can.
Google Nernst. I think you're in for a surprise.
I have never scoffed at degrees. Though clearly they don't always benefit their owners with wisdom.
I have shown quite clearly that higher discharge rates result in a permanent loss of battery capacity, which you clearly stated was a temporary illusion of some sort, and that the apparently lost energy would become available again given a suitable rest period. That is clearly wrong as this thread has shown.
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3 minutes ago, nicknorman said:
The missing 65 watt hours have been dissipated as heat into the battery.
No. They absolutely have not. They were simply never created in the first place. Due to...
Peukert's effect. i.e. slow ion migration resulting in less ions available at the plate/electrolyte interface.
The Nernst equation defines the voltage generated in a cell and it is proportional to the log of the quantity of ions at the cell interface. Note "log". Hence why Peukert's equation is exponential. They are intimately connected.
It has absolutely nothing (well, close to nothing. Close enough that it can be ignored for these purposes) to do with wasting energy into heat.
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8 minutes ago, nicknorman said:
Assuming that by total energy you mean electrical energy from the terminals AND heat energy released into the battery (and thence the surroundings) then YES.
And YES to the SoC.
Throughout the two discharges battery A will have spent it at an average terminal voltage of 12.4 volts. Battery B will have been at an average of 11.1 volts.
The load on the battery A will have benefitted from (converted, consumed, dissipated, call it whatever you want, it makes no difference). 12.1V x 5A x 10 hours = 620 watt hours of energy.
The load on the battery B will have "benefitted" from 11.1V x 50A x 1 hour = 550 watt hours.
Both batteries started at 100% SoC. You agree that they both ended the discharge in the same SoC.
Where are the missing 65 watt hours?
Your schoolboy error appears to be you believing an amp hour to be a unit of energy. It is not.
This thought experiment proves beyond any doubt that higher discharge currents result in less available energy from the battery. The effect is permanent and cannot be "got around" by waiting a bit.
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So Nick...
I have stated all along that as the discharge current is increased, one gets less total energy from the battery. You maintain that is incorrect. Yes?
You maintain that battery A and battery B are in exactly (or at least extremely close) the same SoC. Yes?
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And you are absolutely certain of this? Or are you going to start backpedalling later?
That's to Nick.
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Not refuting anything. You, however, appear rather confused.
Take two identical batteries. Battery A and battery B. Identical batteries. Same make, model, condition, batch etc. 100Ahr. Peukert's exponent of 1.2 Both batteries fully charged.
Discharge battery A at 5 amps for ten hours. Discharge battery B at 50 amps for one hour.
Now wait a few hours for things to settle. This removes your internal confusion regarding resting time and recovery etc. Wait however long you wish/desire.
Are battery A and battery B now in the same SoC or different? It is a straight forward question. A simple question. It deserves a simple answer.
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22 minutes ago, Tacet said:
When your car/boat won't start and the battery goes flat, such that it is incapable of turning the engine over, wait a few minutes and it recovers sufficiently for another spin.
That doesn't quantify anything.
Nick
Stop all the waffling and A4 pages of irrelevant words.
If a LA battery is discharged at a higher rate (current) you will get less total energy from it. This is the crux of the matter. To disagree with it points to some sort of fundamental schoolboy error.
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Ah right, slow down there, energy is energy, whether you consider it useful or not. So pin it down properly.
I have maintained from day one, that if you discharge a lead acid battery at a high discharge current, you will get less total energy from that battery. You either agree with that or you don't.
Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.
Oh, you've gone quiet. Have you suddenly realised the bloke who had spent 35 years designing battery monitors is correct, and the bloke who meddled with them here and there and just come to the realisation that he's horribly wrong but is now struggling to accept that reality? Oh I hope you don't feel too embarrassed.
Waiting...
This is why I stayed away for ten years. Totally pointless arguments with armchair experts who have absolutely no idea what they are talking about, no track record of design products and whose only evidence is youtube videos.
It's the flat earth argument all over again. Pointless.
Tara. x
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Wonderful. Clearly described, sciencifically detailed. So, lets try to pin it down to the shortest possible description we can get away with. One simple sentence/paragragh each. Is that fair?
I maintain, if you discharge a lead acid battery at a high discharge rate, you will get less total energy from the battery than if you had dischared at a lower rate. But you didagree with that. Am I understanding you correctly?
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Ifully agree with the first sentence, but when you say "illegal", exactly er...how? Would it result in a visit from a lawman:"You are under arrst on suspicion of having an un-bonded vessel on this waterway". Did you REALLY mean illegal?
Yes, but I qualified it with "probably".
The ABYC says the hull should be bonded.
The RCD says the hull should be bonded.
The BMEA says the hull should be bonded.
The relevant ISOs say it should be bonded.
If you don't bond it, it could kill someone. It's highly likely that the owner could be prosecuted for man slaughter. There's no defence of "But I didn't know" because all the regs/guidelines/standards say it should be bonded. Exactly how that relates to "illegal" I don't know because I'm just an engineer not a lawyer. Hence the "probably".
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Thanks again. I think I have now discovered that our system is one of the 'earth not bonded to hull' types.
So it's a floating electrocution machine. Probably illegal as well as dangerous.
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Yes that one can be somewhat unpredictable (which is why I suggest using one with a "diode test" position if possible)
Indeed. Proper digitals use a test voltage below 0.25 volts for the resistance range to avoid forward biasing any semi junctions. Cheap ones use whatever the battery voltage happens to be less a bit.
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If not, as a minimum you can just test for continuity (eg by using a battery and a small bulb).
Or combine the two methods to get a better reading...
Power a bulb through it, then measure the voltage drop across it whilst it's powering the bulb. A 10 watt bulb is about best. Then use a digital multimeter on the ohms range. It should read extremely high (above several tens of kOhms). Some cheap digital meters give duff readings on this last test.
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I read the thread title as "Is Carlt crazy" in which case the answer is obvious
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I was stuck in Crick for 4 months in 2003 due to closures.
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I'm curious as to my change of name ?
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That's the first report of a death of someone I don't know that genuinely saddened me personally. I have no idea why.
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I found that, you would think whoever designed it would have thought of that
Apparently he did. Apparently it's on the website. Somehwere. Apparently. Or perhaps in the manual. Apparently he can't remember.
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Use 1mm sq cable and SmartGauge is fine up to 10 metres distance. It will go further with thicker wire but it's a bit of a pig to get it into the connectors.
Lithium Batteries installation
in Boat Equipment
Posted
Quantum mechanics now?
I'm not getting involved. It's utter nonsense.