Electricky Dicky wrote:
On Thu, 04 May 2006 14:28:36 GMT, chuck wrote:
Snipped a lot
http://www.smartgauge.co.uk/batt_con.html
I think the article you referenced confirms the data I provided on wire
resistance. It is a bit naive, however, in assuming a constant and
uniform internal resistance among the batteries. Internal resistance
varies with the age and condition of the battery and non-linearly with
the amount of current being drawn from the battery. Variations of
internal resistance between the cells of 8% seem to be common among new,
similar batteries and the added wire resistance to the farther battery
in the case of "corners not crossed" is only about 7.5% of the battery's
assumed internal resistance. Battery manufacturers want to see internal
resistances increase by 50% before replacement under warranty,
suggesting that increases of as much as 45% are within normal variances!
See:
http://www.batteryuniversity.com/print-parttwo-42A.htm
Does internal resistance reveal battery capacity?
Real batteries are not easy to model. For example, even if there is a
difference in each battery's contribution to a particular load, the
battery providing the greater contribution will more rapidly see an
increase in its internal resistance, thus enabling the other battery to
increase its contribution.
Except in extreme cases, the method of connection is not too important.
Chuck
Chuck,
Thanks for the URL above, most interesting.
The author of the site that I quoted does not follow this group these
days, however we had a conversation and the following is copied from a
mail to me
(Start quote from mail)
With battery internal resistance of 0.01R (a typical figure for a
brand new fully charged deep cycle 100AHr battery) the figures come
out at:- (End wired)
Battery share of current
1 44.2
2 26.9
3 17.6
4 13.5
The figure used on the website of 0.02R represent a battery at about
half life. It doesn't matter what figures are used. The end result is
the same. Just the magnitude changes.
With batteries with an internal resistance of 0.05R (i.e. totally
knackered - 100 Ahr batteries unable to support a 50 amp load without
the terminal voltage falling to 10.5 volts) the figures come out:-
Battery share of current
1 29.8
2 25.6
3 22.9
4 21.6
To suggest that the internal resistance of each battery
represents a larger percentage of the total is completely irrelevant.
Each battery's internal resistance is in series with it's own battery
current and NOTHING ELSE.
The resistance of the battery interconnecting leads is in series with
different batteries and each one has hugely differing currents through
it if the batteries are not wired corner to corner.
If one battery is slightly older or more tired than the others then
it's internal resistance will be higher. This leads to that battery
producing less current during discharge and accepting less current
during charge. It is therefore worked gentler. This therefore works
the other batteries harder. This gives a natural balancing effect
between the different batteries in the bank with the result that each
battery is worked by the same amount.
This of course, is on the assumption that the battery bank is wired up
correctly. If it is wired "all from one end" then this doesn't happen
and the first battery is worked harder.
(End of quote from mail)
If you have the opportunity then I suggest doing the actual physical
test (preferably with a new battery set) to see if the numbers crunch
as above.
I am just a simple soul but it seems to me that working a battery set
"equally" is the best way to go.
Regards
Richard
Nb "Pound Eater" Parkend G+S
Thanks again for the thoughtful reply, Richard. More than anyone, and
more than ever, I need and welcome reality checks on what I say!
Starting with the end, I do agree with your sentiment, as I noted in my
initial post on this thread, although I continue to doubt that it is
very important.
The example given on the web site you referenced is really not the
situation posed by the OP. It is probably more instructive (for me at
least) to approach this from the most simple abstraction, to wit, two 12
volt batteries with different internal resistances (0.01 ohm and 0.0115
ohm), in parallel with each other and with a resistive load. The
difference in internal resistances is due to the added resistance of the
connecting wires as we have discussed. We need not know the load
resistance, but we assume the load current is 100 amperes.
Each battery has the same voltage, so setting them equal:
V1 = V2
12 - I1 x 0.01 = 12 - I2 x 0.0115
I1 = 1.15 x I2
Since I1 + I2 = 100,
I1 = 53.5
I2 = 46.5
I believe this is a pretty insignificant difference in current between
the two batteries. It is in the realm of battery-to-battery differences
in internal resistance and open-circuit voltage at full charge.
Moreover, since a 100 ampere load can reasonably considered a worst
case, in practice, and for most of their operating lives, these two
batteries will experience an even lower imbalance than that suggested by
the above analysis.
On the other hand, if I've overlooked something or made one of my
notorious arithmetic mistakes, please let me know.
Chuck