On Tue, 19 Sep 2017 13:49:30 +0100, "Dave W"
wrote:
"T i m" wrote in message
.. .
On Fri, 15 Sep 2017 14:04:56 +0100, "Dave W"
wrote:
snip
Please state the type no. of the battery;
They are MK 8GU1H
http://www.mkbattery.com/images/8GU1H.pdf
http://www.mkbattery.com/documents/9350MK_GEL_v7_r2.pdf
http://www.mkbattery.com/gel_specs.php?model=8GU1H
http://www.mkbattery.com/documents/1...I&O)_MK_r1.pdf
if more than one how they are
connected;
3, in parallel.
what current range you will be using.
0 to 30A (typically 15A).
Cheers, T i m
There's a great deal to mull over in the references you've given.
Quite ... and I'm not sure all of it is supportive? ;-(
I will
label them A, B, C, D in order.
Reference 'A' gives:
Current, Capacity
C/100, 36Ah
C/20, 31.6Ah
C/5, 25.6Ah
C is the current that would fully discharge the battery in 1 hour. C/5 is
the current that would fully discharge the battery in 5 hours.
Agreed.
Reference 'B' gives:
Current, Nominal Capacity
5 Hr Rate, 26.8Ah (different from reference 'A')
(agreed, but only by a bit). ;-)
20 Hr Rate, 31.6Ah
Reference 'C' gives:
Current, Time to fully discharge:
21A, 1h (i.e. C/1)
11.9A, 2h (i.e. C/2)
5.40A, 5h (i.e. C/5)
3.10A, 10h (i.e. C/10)
1.60A, 20h (i.e. C/20)
0.36A, 100h (i.e. C/100)
Reference 'D' gives:
% Charge, Open Circuit Voltage after 24 hours
100%, = 12.85V
75%, 12.65V
50%, 12.35V
25%, 12.00V
0%, 11.80V
That's the first one in volts so a bit of a loner.
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').
Yes, I believe you are correct (they are the sums resolved etc).
I magnified the graph, added extra grid lines, and estimated that at 50% the
voltages are 11.70V, 12.05V, 12.13V respectively.
Ok.
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)
Ok. (And my way was more 'rough and ready' than yours). ;-)
The "initial" voltages correspond to those shown on the graph at zero
current, but I agree with you that they are irrelevant.
Ok.
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?
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?
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?
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?
Cheers, T i m