On Tue, 19 Sep 2017 20:29:26 +0100, "Dave W"
wrote:
"T i m" wrote in message
.. .
On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:
I've snipped a lot here for clarity.
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.
I was simply using the typical rating which is often at C/20 (x) 3
Dave?
Of course, but 3 x 21A for C/1 is 63A not 93A.
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?
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.
I'm happy to assume that with you but I think it does change in
practice and hence we get Peukert's effect?
As my three values for resistance are so similar, I don't see how Peukert's
effect comes into it.
Ok but it does ITRW though doesn't it?
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.
Ok (and whilst most of the leads are copper and 20mm^2, they will have
resistance etc).
The notional source voltage at the 50% discharge point is, from my C5
reading, 12.05V + 0.026 x 5.40A = 12.19V.
Sounds like a reasonable number. ;-)
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.
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?
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.
Understood.
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.
Quite ... and back to the formula that would allow me (or an Arduino)
to calculate that on the fly. ;-)
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.
That's interesting Dave, thanks. ;-)
The plot says that they are "Based on BSI 2-hour Capacity", and I
found an explanation of the 100% discharge version he
https://www.civicsolar.com/support/i...-gel-batteries
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:
https://www.thesafetycentre.co.uk/doc/583.pdf
I'm looking at 'Discharge characteristics at various currents' chart
although it's not particularly clear to what DOD / voltage?
Yuasa don't say, but I think the 100% discharge must be to that dotted line.
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.
Also the plots are too small to be able to work out source voltage and
resistance.
Ok.
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.
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:
http://www.smartgauge.co.uk/peukert_depth.html
There is even a calculator you can plug values into but whilst it
seems to work for me, I don't understand the outcome: ;-)
http://www.smartgauge.co.uk/calcs/peukert_2.xls
And I can nearly follow this explanation. ;-)
http://preview.tinyurl.com/ycyznwaf
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).
http://www.bogartengineering.com/wp-...tsComments.pdf
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.
;-(