As it happens, I do enjoy rowing and because it's difficult to get a whole days outboarding in a reasonable weight of battery (lead acid cyclic batteries especially) then rowing was already a part of the deal (and / or sailing if the weather and space allowed etc).
Like the other day, we outboarded for an hour, I rowed for an hour and outboarded for an hour again.
The idea for a real trip was to row out and outboard back, hence why I need a way of monitoring the battery charge status as closely as possible so I know how far to row out. ;-)
Very clear statement of what you can find in this type of problem. Never mind what's claimed to be the theoretical equation, make sure you get the data in the actual region you are interested in, and then fit to that. If a flexicurve works, use it. Occam's razor.
On page 6 are graphs of voltage against percentage discharge for currents C1, C5, C20. (I assume that's another way of saying C/1, C/5, C/20, i.e.
21A, 5.40A, 1.60A respectively from reference 'C').
I magnified the graph, added extra grid lines, and estimated that at 50% the voltages are 11.70V, 12.05V, 12.13V respectively. These are similar to the values in your first posting:
11.75V @ 93A (12.4V initial)
12.05V @ 18.2A (12.7V initial)
12.10V @ 4.65A (12.75V initial) The "initial" voltages correspond to those shown on the graph at zero current, but I agree with you that they are irrelevant. The corresponding currents for three batteries in parallel would be (from reference 'C') 63A, 16.2A, 4.8A, for C/1, C/5, C/20. Similar to your first posting except your 93A should be 63A.
From my graph readings of C1 and C20, 21A minus 1.60A gives an extra voltage drop of 12.13V minus 11.70V. If we assume that the effective internal resistance is the same for all currents, it would be V/I = 0.43V/19.4A =
0.022 ohms.
From my readings of C1 and C5, V/I = 0.35V/15.6A = 0.022 ohms. From my readings of C5 and C20, V/I = 0.10V/3.8A = 0.026 ohms.
You say you will be drawing between 0A and 10A from each battery, and typically less than 5A. That's typically near the C/5 rate of 5.40A, so I would assume that the internal resistance of your arrangement is 0.026/3 =
0.0087 ohms i.e. 8.7 milliohms ignoring lead resistance.
The notional source voltage at the 50% discharge point is, from my C5 reading, 12.05V + 0.026 x 5.40A = 12.19V.
So if you know your current and voltage, you can work out when your notional source voltage has dropped to 12.19V, which indicates the 50% discharged point.
I think you should plot a graph of the speed of your normally-laden boat in still water against current drawn. This will help decide how fast you can go back before the battery runs out, especially if the relationship is non-linear. 25.6Ah gives 5 hours at 5A, or 2.5 hours at 10A, your maximum current.
As an aside, your reference 'B' shows a plot of lifetimes in terms of number of cycles versus depth of discharge. I multiplied the cycles by the depths, to give an indication of how the total Ah given by the battery over its life depends on the depth of discharge:
5700 x 0.1 = 570
2100 x 0.25 = 525
1000 x 0.5 = 500
600 x 0.8 = 480
450 x 1 = 450
This seems to indicate that the life doesn't depend very much on depth of discharge. The plot says that they are "Based on BSI 2-hour Capacity", and I found an explanation of the 100% discharge version here:
Quite ... and I'm not sure all of it is supportive? ;-(
Agreed.
(agreed, but only by a bit). ;-)
That's the first one in volts so a bit of a loner.
Yes, I believe you are correct (they are the sums resolved etc).
Ok.
Ok. (And my way was more 'rough and ready' than yours). ;-)
Ok.
I was simply using the typical rating which is often at C/20 (x) 3 Dave?
I'm happy to assume that with you but I think it does change in practice and hence we get Peukert's effect?
Ok (and whilst most of the leads are copper and 20mm^2, they will have resistance etc).
Sounds like a reasonable number. ;-)
Yeahbut, that in itself isn't the solution I was hoping for Dave ... it would be to be able to calculate that critical voltage at any current within the likely range?
Quite ... and back to the formula that would allow me (or an Arduino) to calculate that on the fly. ;-)
That's interesting Dave, thanks. ;-)
And the battery they reference there are very close to the spec of mine as they are the 'Solar use' MK Gel jobbies.
So, Dave, thank you very much for you in depth analysis of the above and do you think what you have gleaned so far would allow you to create a formula to calculate the 50% DOD thresholds for the 0-30A current range?
Funnily enough, whisky-dave posted a link elsewhere to a similar spec / capacity lead acid battery where they show the sort of graph I think we are considering:
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I'm looking at 'Discharge characteristics at various currents' chart although it's not particularly clear to what DOD / voltage?
As my three values for resistance are so similar, I don't see how Peukert's effect comes into it.
OK, to make a formula from my paragraph: Let Vr = Voltage at your measurement point, Let Vo = 12.19V, Let Ir = Total Current from the three batteries, Let Rt = 0.026/3 = 0.0087 ohms, plus whatever your lead resistance is,
Vr = Vo - (Ir x Rt)
As it seems the resistance is roughly independent of current, the formula will do for all currents. However if you ever want to find Vr at degrees of discharge other than 50%, you will have to work out appropriate values of Vo and Rt for each point.
Yuasa don't say, but I think the 100% discharge must be to that dotted line. Also the plots are too small to be able to work out source voltage and resistance.
I would like to add that I think you should charge your batteries individually rather than in parallel, otherwise they may not share the current equally.
Sorry I'm lost mate. If we were to be judging the worst case capacity for my real world load, wouldn't that be 30A (electric outboard on full speed) spread across 3 batteries and so would be 10A each? I'm not sure where the 21A has come from?
Ok but it does ITRW though doesn't it?
Understood.
Ok. Some of the other explanations suggest 10.5 volts (for a nominal
12V LA battery) is what is used to go with any capacity testing. Therefore, the higher the current the more the voltage will be depressed and hence the sooner that 10.5V cutoff point is reached.
Again, listening to people who do this all the time it seems like it can even come down to how you wire all the batteries together. e.g. Say you have 5 x 12V, 100Ah batteries wired in parallel using busbars, you take the +Ve off one end and the -Ve off the other as that they minimises the tiny losses (size being dependent on load and busbar resistance etc) that would work the batteries on the ends of the bus more than those in the middle.
And alternative way of wiring them is use a star so that all the wires between each battery pole and a common connection point are identical (length and gauge and so hopefully resistance), therefore spreading the load across all the batteries equally.
I believe the guidance from the battery manufacturers is:
1) All the batteries connected in parallel should *ideally* be of the same make, model (so capacity), age and cycles.
2) All batteries to be connected together should first be fully charged so that the float voltages are as similar as possible before they are joined together.
3) Once so joined, they should then be left joined together for all subsequent charge and discharge cycles.
I think the logic re parallel charging is that even if the batteries are slightly different (as they are bound to be), the battery with the deepest level of discharge (smallest capacity) will depress the charge voltage more than the others (because it will be taking more of the current) and as it charges to match the level of the others the charge current will balance out across all batteries equally.
If one battery reaches it's full charge condition before the others it's terminal voltage will be higher than the others (would be if isolated) and so the current would tend to flow though those batteries that are still a lower terminal voltage (if you were to disconnect them) and so again, the charge balances out to a reasonable degree.
The third thing is to ensure the maximum charge 'bulk phase' current is kept to such a level that it can't overdrive any one battery and similar for the 'absorption phase'.
That said, those who say live on boats and rely on a good 12V supply for a lot of things regularly generator / bank charge their paralleled batteries (often AGM) at a pretty high rate (as they don't have the luxury of time etc). ;-)
Cheers, T i m
p.s. If you like logic and maths, you might find this interesting:
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There is even a calculator you can plug values into but whilst it seems to work for me, I don't understand the outcome: ;-)
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And I can nearly follow this explanation. ;-)
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The following seems to reflect my understanding of it all and agrees that by using Peukerts in any discharge level calculations can lead to a more realistic display of charge status than when not using it (with limitations etc).
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For me the bottom line is still that 'if you discharge a lead acid battery at an increasing rate the effective capacity *will* decrease.
It is my firm belief that within a pretty strict range of parameters a formula could be found and used in a realtime display of the predicted voltage cutoff point. I just don't have the sort of brain that can handle all the variables, rules and formulae to work out what that is. ;-(
You are bored aren't you ... but I don't know. Do you have the answer to my question?
What, and sit there and wait ~12 hours or so in an open dinghy in the water?
Well with out rowing dinghy the electric outboard is all we have, other than the oars.
That will teach you to eat half cooked food! ;-)
I'm guessing you either really don't understand much or haven't been following the thread at all? ;-(
I appreciate, as a Brexiteer you don't need any facts, you just need an inkling of an idea about something to go jumping off to all sorts of conclusions. ;-(
Here, try thinking about this (or get 'nurse' to explain it to you) maybe I don't have a boat that is like your campervan, maybe I don't have the same needs with my boat as you do with your campervan and what I'm doing is all about both trying to eeek out the small amount of energy I have within the largest mass of battery we can manage, handle and transport and measuring something that is both dynamic and non-linear at the same time?
Now, if I were a Brexiteer I could just spin the bottle and go along with whatever comes up and even make that my new crusade in life but I simply can't do that.
I work on the basis 'You can manage what you can measure' whereas you work on the 'I'm not sure what it is but I want it right now'.
So, unless you can come up with whatever formula is required to calculate the real-time voltage on a battery that represents a 50% discharge level for a given current then it might be best if you went back to supporting whatever cause that spinning bottle points you at next. ;-)
You may be right John but I was just hoping that 'maths' would be able to cope with it, no matter how inaccurate the result might be ITRW.
Looking at the graphs I have seen that represent what I think I'm looking for, the range of currents I'd be dealing with seem to be in a reasonably 'linear' (as in say 'exponential') area.
If you look at the graph labeled 'Discharge characteristics at various currents' here:
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and consider the 2-10A range (= 0.33 to 0.05C when seen across the 3 x ~30Ah batteries in parallel as that battery is of a similar capacity to one of mine) then it doesn't look like there should be anything unpredictable.
Even if we only considered it as a straight line, couldn't we 'compute' the (say) 50% DOD cutoff voltage for any current within that range?
The thing is (and assuming we may now have at least a linear slope rather than just a fixed value (see my reply to John)) even if we had a graph on paper to work from I still need it in the way of a line of code (the maths part) to be able to put it to use?
Read CurrentAmps, Read CurrentVolts
If CurrentVolts CurrentAmps =< 50% DOD voltage, output alarm?
Hi Martin and thanks for the feedback. I can follow some of it. ;-)
More thinking out loud on it all .... Considering a graph of various discharge rates shows that the effective capacity does vary as a function of the discharge rate, I think it would still be worthwhile trying to accommodate it to some degree if possible (accepting all the caveats you mention etc). I agree though that whilst a linear function would be better than nothing at all, it may not reflect the situation was well as it *might*, if we could find that mathematical solution, even though it's only ever going to be an approximation at best. I was never looking for a laboratory solution here. ;-)
On that though, I can (re)evaluate the batteries capacity over time and modify the numbers accordingly and could even automatically compensate for temperature (if the typical temperature range difference we are likely to go boating would be significant etc).
So it's just a matter of (initially especially) seeing if we can make use / sense of the published data for my specific battery(ies) and if that isn't sufficient, if we can use the stats from something similar to at least start the ball rolling?
It's just knowing that there can be such a marked difference in capacity as a function of the current drawn ... and my potential maximum current (30A) *is* significant to this, even with 3 batteries in parallel (~90Ah) there is a very good chance that some massaging of the readings to predict a particular depth of discharge threshold would be worthwhile.
If I retained the original resistor based 'speed control', even I, once I guesstimated what the capacity would be for the currents typically seen at the 5 set speeds, could possibly write some code that applied the variable.
I had to google for ITRW. I admit I have not investigated Peukert's effect.
Ok, you are probably right in keeping them always in parallel.
Martin Brown has raised a very good point that I hadn't considered. Of course the capacity varies depending on the current taken, and can be predicted from the current using his formulas.
However, the voltage versus percentage discharge graph that you referenced shows none of that. I wondered how they knew where the 50% discharge point is, but I now realise it must be 50% of whatever capacity each current produces.
It looks like you will have to keep a running total of how many Ampere-Hours you have used up by calculating current x time every so often and adding to the total. Stop when you have used up half the capacity that 10A per battery gives. Martin's formula times three gives 71Ah capacity at 30A, so
36Ah represents the half discharged point. At full speed you should then have
36/30 = 1.2 hours before the battery runs out.
The question is, how many hours from this point will you get at 15A?
At 15A, Martin's formula predicts 85.5Ah capacity so the half-discharged point would be about 43Ah. I can't get my brain around this either, but it seems to me that irrespective of how you used up 36Ah, there should be at least 43Ah left at 15A i.e. 2.9 hours.
I have skimmed through these links and now my brain hurts. Your last link concludes that it's all too difficult. I may get back to you when I've done my own analysis. Regards, Dave
Ok ... and was that taken as an example or worse case etc as I'll never be drawing current at C/1?
Sorry. ;-)
Well I think that may be pivotal to the whole issue Dave, both in ITRW and mathematically re the calcs etc.
It wasn't something that sat well with me initially either but if that's what people do then so be it. ;-)
Ok.
I believe that is correct yes. As an aside, if this was a conventional flooded battery with openable cell caps you could simply read the specific gravity of a cell to glean the current charge condition. No such possibilities with a sealed gel battery. ;-(
The problem with that Dave is that doesn't automatically compensate for the change in capacity as the current changes?
Ok.
Again, (and it still leaves me with a headache) it's all down to the capacity changing with current drawn. I think thinking of 'remaining time' is *way* advanced for this project and all I would like to do is to display the current voltage and the predicted voltage that represents the 50% DOD at the current current. ;-)
You and me both Dave!
I think it concludes it's not realistic, *all_things_considered* to expect an accurate gauge / outcome. However, I still believe it's not outside the brains of this group to come up with a formula that would be better than nothing. ;-)
Ok, and thanks again for your interest in this.
I think may help to picture yourself sitting in a 10' wooden and folding dinghy, steering via a small electric outboard motor with 3 x lead acid batteries in parallel in a box in the bow and whilst looking at a small Arduino driven meter that is displaying:
12.56 V (current battery terminal voltage)
10.8 A (current current drawn from battery)
11.75V (voltage that is calculated from the current voltage and current to represent the voltage alarm point of 50% DOD).
In theory and if you kept at the same speed especially ...
1) the voltage would initially drop as you turned the outboard on and then slowly drop over time.
2) The current to mostly stay the same (it would also drop slightly as the voltage dropped etc).
3) The predicted 50% DOD voltage to remain more or less constant (allowing for the fluctuations in the above etc).
As I mentioned elsewhere, if you had 10 points taken from a graph that represented 10 different currents and their resultant voltages (to 50% DOD) you could simply look for those 10 and display accordingly.
The trick is to not have just 10 but a completely dynamic calculation that would work with anything from 1A (realistically) and 30A. ;-)
Whilst I'm pretty sure it's not rocket science, the maths might as well be from my POV. ;-(
If only you could apply the same amount of energy to helping us resolve this (and what must be to you, 'brain as big as a planet'), simple *energy* measurement issue as trying to defend the indefensible (Linux or leaving the EU) we might all get on faster! ;-)
No, I was only correcting the 93A "the manufacturers have provided" that you gave in your first post.
Well if 'remaining time' is too advanced, what's wrong with my simple formula? It's better than nothing. It might be better to display the source voltage Vo rather than the terminal voltage Vr, and when it gets to 12.19V you have reached the 50% discharged point. Vo = Vr + (Ir x Rt) is the rearranged formula.
I can't see the need for that as my three calculations of internal resistance were all pretty similar, as would be 10 intermediate curves. You mentioned that for a liquid battery you could measure the specific gravity to find the
50% point, so surely that point will also have my predicted source voltage of
2.19V?
The display of Vo should stay fairly constant as the current is ramped up and down. If it doesn't, the assumed value for Rt can be adjusted until it does.
I suspect the motor current will be very noisy, making any digital display very erratic and difficult to read. Better to use a mechanical meter, with a big capacitor across it to smooth out wobbles. Or perhaps use software to create a running average. A meter would be easy to read in bright sunlight.
The meter could have a big mark at 12.19V, with perhaps everything below
11.5V marked in red, as that's where your discharge graph starts to drop like a stone.
If you are using a simple rheostat to adjust the motor current, that's a big waste of energy in the form of heat. Better to have an efficient switched-mode power converter to convert the battery voltage into motor voltage.
Ideally this would have current feedback to compensate for the resistance of the motor, in effect raising the voltage as more current is taken. The control knob could then be calibrated directly in propeller speed.
The feedback could be further modified if you know the relationship between boat speed and propeller speed, and the knob could be calibrated in boat speed. This would be helpful for estimating journey time.
I'll step in here (without, ahem, having read all the thread).
I think looking for a formula won't work here, but as you are using an Arduino:
Take the nominal Ah of the batteries.
Measure the current drawn.
Calculate the capacity of the battery at the current amperage drawn. This will be a fraction of the nominal capacity. Calculate the inverse ration of this capacity to the nominal capacity. (100 Ah pack, effectively 75 Ah at current X. So 100/75= 4/3.)
Multiply current amperage drawn by the time it is drawn (i.e. you measure every second, 10 Amps, so 10 Amp-seconds). Multiply this by the ratio above, and subtract from the nominal capacity. So, reduce nominal capacity by 4/3*10 As.
Display remaining capacity as a percentage. Adjust for real-world experience.
Repeat.
If i have thought this through correctly, you reduce the full charge by what you use, weighted by how much the currently drawn amperage reduces that capacity.
Alternatively, reformulate to use time-to-discharge instead of capacity.
And of course you can just recharge the battery if you run out.
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