I have a 24v system comprising:

4x12v 135amp Domestic batteries.

They are connected as 2 pairs of batteries wired in parallel to give a 2 x
24v and then the 2 pairs are connected in series.

2x12v engine starter batteries connected in series

My question. when I connect these to the boat, should I take the positive
and negatives feeds from the same battery/ battery pair or one from each?
Does it matter?'

Nigel wrote:
I have a 24v system comprising:

4x12v 135amp Domestic batteries.

They are connected as 2 pairs of batteries wired in parallel to give a 2 x
24v and then the 2 pairs are connected in series.

2x12v engine starter batteries connected in series

My question. when I connect these to the boat, should I take the positive
and negatives feeds from the same battery/ battery pair or one from each?
Does it matter?'





It may not matter much, Nigel, but the preferred way of doing it is to
run one feed to each series pair, rather than both to the same series pair.

That is intended to ensure both series pairs "see" the same load and
receive the same charging voltage.

Make sense?

Chuck

Hi Nigel,

When you have batteries arranged this way you need to be very careful
that they charge and discharge in balance. Otherwise you end up with
voltage variations between the pairs that leads to early failure (I
know - it seems odd). Your positive lead should connect first to one
pair and the negative first to the other pair.

Here's a diagram of correct wiring from the high end german battery
supplier - Sonnenschein.

http://www.sonnenschein.org/Really%20FAQ's.htm

Carl

Carl wrote:
Hi Nigel,

When you have batteries arranged this way you need to be very careful
that they charge and discharge in balance. Otherwise you end up with
voltage variations between the pairs that leads to early failure (I
know - it seems odd). Your positive lead should connect first to one
pair and the negative first to the other pair.

Here's a diagram of correct wiring from the high end german battery
supplier - Sonnenschein.

http://www.sonnenschein.org/Really%20FAQ's.htm

Carl

That is a common but unfounded myth.

With appropriate jumpers between the parallel batteries, the voltage
difference caused by cables entering/exiting from the same side are
less than 0.001 volts which is far less than internal battery
incongruities and will have zero effect on the battery life.

If the two batteries were separated by a considerable distance, say 10
feet, and the jumpers between them were smaller than the starter motor
cables, in that case one battery would tend to be used slightly more
than the other but only during heavy loads. As soon as the load is
removed, the batteries will equalize within a few seconds.

Andina Marie

On 2 May 2006 06:21:06 -0700, "Andina Marie" wrote:


Carl wrote:
Hi Nigel,

When you have batteries arranged this way you need to be very careful
that they charge and discharge in balance. Otherwise you end up with
voltage variations between the pairs that leads to early failure (I
know - it seems odd). Your positive lead should connect first to one
pair and the negative first to the other pair.

Here's a diagram of correct wiring from the high end german battery
supplier - Sonnenschein.

http://www.sonnenschein.org/Really%20FAQ's.htm

Carl

That is a common but unfounded myth.

With appropriate jumpers between the parallel batteries, the voltage
difference caused by cables entering/exiting from the same side are
less than 0.001 volts which is far less than internal battery
incongruities and will have zero effect on the battery life.

If the two batteries were separated by a considerable distance, say 10
feet, and the jumpers between them were smaller than the starter motor
cables, in that case one battery would tend to be used slightly more
than the other but only during heavy loads. As soon as the load is
removed, the batteries will equalize within a few seconds.

Andina Marie
I completely and utterly disagree!!!!!

Richard

Nb "Pound Eater" Parkend G+S

On Wed, 03 May 2006 11:55:39 GMT, chuck wrote:

Electricky Dicky wrote:
On 2 May 2006 06:21:06 -0700, "Andina Marie" wrote:


Carl wrote:

Hi Nigel,

When you have batteries arranged this way you need to be very careful
that they charge and discharge in balance. Otherwise you end up with
voltage variations between the pairs that leads to early failure (I
know - it seems odd). Your positive lead should connect first to one
pair and the negative first to the other pair.

Here's a diagram of correct wiring from the high end german battery
supplier - Sonnenschein.

http://www.sonnenschein.org/Really%20FAQ's.htm

Carl

That is a common but unfounded myth.

With appropriate jumpers between the parallel batteries, the voltage
difference caused by cables entering/exiting from the same side are
less than 0.001 volts which is far less than internal battery
incongruities and will have zero effect on the battery life.

If the two batteries were separated by a considerable distance, say 10
feet, and the jumpers between them were smaller than the starter motor
cables, in that case one battery would tend to be used slightly more
than the other but only during heavy loads. As soon as the load is
removed, the batteries will equalize within a few seconds.

Andina Marie

I completely and utterly disagree!!!!!

Richard

Nb "Pound Eater" Parkend G+S

The actual voltage drop will depend on actual wire size and length. In
the case of #4 wire, which is probably a worst case example, the voltage
drop due to a one foot length difference (two feet round trip) would be
on the order of 0.040 volts at 100 amps. That difference is really
insignificant since it represents only initial charging current, and
ultimately the battery will be charged and discharged at a much lower
current. At 1.0 amps and #0 wire, the voltage difference is 0.0002 volts
(two feet). In the end, the difference should be less than differences
in the characteristics of individual batteries, as was stated.

With what specific part do you completely and utterly disagree, Richard?

Chuck
The bit about "myth" and the inference that there is no need to
connect "cross corners".
If you have the time take a look at
http://www.smartgauge.co.uk/batt_con.html
This article discusses the situation with facts.
The article is UK based so metric cable sizes. To assist our colonial
friends, 35mm2 is slightly larger than 2AWG.
I will be interested to see comments.

Richard

Nb "Pound Eater" Parkend G+S

Electricky Dicky wrote:
On Wed, 03 May 2006 11:55:39 GMT, chuck wrote:


Electricky Dicky wrote:

On 2 May 2006 06:21:06 -0700, "Andina Marie" wrote:



Carl wrote:


Hi Nigel,

When you have batteries arranged this way you need to be very careful
that they charge and discharge in balance. Otherwise you end up with
voltage variations between the pairs that leads to early failure (I
know - it seems odd). Your positive lead should connect first to one
pair and the negative first to the other pair.

Here's a diagram of correct wiring from the high end german battery
supplier - Sonnenschein.

http://www.sonnenschein.org/Really%20FAQ's.htm

Carl

That is a common but unfounded myth.

With appropriate jumpers between the parallel batteries, the voltage
difference caused by cables entering/exiting from the same side are
less than 0.001 volts which is far less than internal battery
incongruities and will have zero effect on the battery life.

If the two batteries were separated by a considerable distance, say 10
feet, and the jumpers between them were smaller than the starter motor
cables, in that case one battery would tend to be used slightly more
than the other but only during heavy loads. As soon as the load is
removed, the batteries will equalize within a few seconds.

Andina Marie

I completely and utterly disagree!!!!!

Richard

Nb "Pound Eater" Parkend G+S

The actual voltage drop will depend on actual wire size and length. In
the case of #4 wire, which is probably a worst case example, the voltage
drop due to a one foot length difference (two feet round trip) would be
on the order of 0.040 volts at 100 amps. That difference is really
insignificant since it represents only initial charging current, and
ultimately the battery will be charged and discharged at a much lower
current. At 1.0 amps and #0 wire, the voltage difference is 0.0002 volts
(two feet). In the end, the difference should be less than differences
in the characteristics of individual batteries, as was stated.

With what specific part do you completely and utterly disagree, Richard?

Chuck

The bit about "myth" and the inference that there is no need to
connect "cross corners".
If you have the time take a look at
http://www.smartgauge.co.uk/batt_con.html
This article discusses the situation with facts.
The article is UK based so metric cable sizes. To assist our colonial
friends, 35mm2 is slightly larger than 2AWG.
I will be interested to see comments.

Richard

Nb "Pound Eater" Parkend G+S

Thanks for the reply, Richard.

A couple of points.

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

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

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

On Fri, 05 May 2006 15:42:34 GMT, chuck wrote:

Snip
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.

Granted the OP was asking re 2 x 24v in parallel which is no different
to 2 x 12v in parallel for the calcs

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


Ok with this simplistic calculation you are now agreeing that there is
a difference!

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.

Come on, you did a calculation, now you are trying to ignore it by
changing variables, you have seen that there is a difference. If you
had done the calc with 3 batteries you would have seen a greater
difference. Therefore surely always connecting a set cross corners is
the correct method, whether it be 2 3 or 4, all it takes is a couple
of feet of cable.

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

As it happens I am replacing 4 x 230Ah (24v 460Ah set) batteries next
week. These are currently wired off one end. I will do a test with a 2
or 3kw load after installation, then again after I change to cross
corners. Interestingly it is the current first pair that have failed
after 5 years ;-)

BTW just to be really really pedantic, you suggested in another post
that 0.040A over 24hrs was 1AH. Sorry no, its more like 0.24Ah
assuming a 20hr 100Ah battery. Do not forget that at low currents Mr
Peukert is your friend ;-)

Thanks for the discussion, nice to chat with no flaming and snide
comments.

Best regards

Richard
Edelec - Design and Manufacture of Control Systems
Boat Electrical repairs and Installations.