"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.
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.
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.
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.
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.
Regards,
Dave