Adding third house battery advice

[Ningaloo]

I have two 160Ah batteries (B1 and B2) in my house bank, side by side.

B1 negative goes to negative bus bar (95mm2 cable) and to B2 negative (possibly 70mm2 cable)
B1 positive goes to B2 positive (70mm2 cable)
B2 positive goes to positive bus bat (95mm2 cable)

The new (identical) battery will go behind the existing bank (about 50cm between) rather than side by side.

I was planning to take a cable from B2 negative to B3 negative and B2 positive to B3 positive AND remove the positive busbar cable from B2 and connect it to B3 so the busbar negative is from B1 and the busbar positive is from B3,

My electrician says that I can just wire the new battery direct to the bus bars (positive and negative) which are both positioned between the existing bank and the new battery.

What do the experts think?

 

[pvb]

My electrician says that I can just wire the new battery direct to the bus bars (positive and negative) which are both positioned between the existing bank and the new battery.

Sounds like that would result in shorter cables, so why not do it that way?

 

[Ningaloo]

Sounds like that would result in shorter cables, so why not do it that way?

So the sequence of cables connecting the batteries/busbar isn't important?
I understand that all points on the busbar are essentially equal by my mind is thinking I should take the bus bar connections from each "end" of the battery bank, i.e. B1 and B3.

 

[pvb]

So the sequence of cables connecting the batteries/busbar isn't important?
I understand that all points on the busbar are essentially equal by my mind is thinking I should take the bus bar connections from each "end" of the battery bank, i.e. B1 and B3.

With only 2 or 3 batteries it's less important. If you had, say, 6 batteries in parallel there are so many connections that taking the pos and neg from each end of the bank will help to balance the voltages on each battery. But, with decent connections, the effects are minimal.

 

[VicS]

So the sequence of cables connecting the batteries/busbar isn't important?
I understand that all points on the busbar are essentially equal by my mind is thinking I should take the bus bar connections from each "end" of the battery bank, i.e. B1 and B3.

The aim is to ensure that the total lengths of the connections , and therefore the resistances, are the same for each battery in the bank. This is to make sure that the loading and charging is equal across all the batteries.
It's all described in the technical info pages on the Smart Gauge website

SmartGauge Electronics - Interconnecting multiple batteries to form one larger bank

It is probably not so important in a small yacht as it is in a liveabord situation with a very substantial electrical system like the author of the above has.

 

[TernVI]

If you are lucky, the different length connections might counteract the effects of one battery being newer.

 

[PaulRainbow]

So the sequence of cables connecting the batteries/busbar isn't important?
I understand that all points on the busbar are essentially equal by my mind is thinking I should take the bus bar connections from each "end" of the battery bank, i.e. B1 and B3.

This is correct.

 

[HissyFit]

So the sequence of cables connecting the batteries/busbar isn't important?
I understand that all points on the busbar are essentially equal by my mind is thinking I should take the bus bar connections from each "end" of the battery bank, i.e. B1 and B3.

This is correct.

Hi Paul,
If I'm reading Ningaloo's post correctly, he is actually making two contradictory statements. So which are you agreeing to: the sequence being unimportant or the connections being from each end of the bank? I know what I've seen recommended and why I'd do what I'd do, but I'm not an expert (as you know).

 

[pvb]

It's all described in the technical info pages on the Smart Gauge website

SmartGauge Electronics - Interconnecting multiple batteries to form one larger bank

This stuff is regurgitated all over the internet on countless websites, but nobody ever queries the claims. Do you really believe that 4 batteries in parallel, with the connections at one end of the bank, can result in one battery providing double the current of another battery? Chris Gibson declines to publish his calculations, claiming they're "quite horrific". They could also be quite wrong.

Logically, connecting the pos and neg feeds at opposite ends of a big bank will help to balance voltages. But if generously-sized cable is used, and if the connections are properly made, there'll be very little resistance to produce a significant voltage drop. For the OP, who already has feeds at opposite ends of his 2-battery bank, adding a third battery wired directly to the busbars wouldn't have any harmful effect.

 

[TernVI]

This stuff is regurgitated all over the internet on countless websites, but nobody ever queries the claims. Do you really believe that 4 batteries in parallel, with the connections at one end of the bank, can result in one battery providing double the current of another battery? Chris Gibson declines to publish his calculations, claiming they're "quite horrific". They could also be quite wrong.

Logically, connecting the pos and neg feeds at opposite ends of a big bank will help to balance voltages. But if generously-sized cable is used, and if the connections are properly made, there'll be very little resistance to produce a significant voltage drop. For the OP, who already has feeds at opposite ends of his 2-battery bank, adding a third battery wired directly to the busbars wouldn't have any harmful effect.

The calculations are not simple.
The problem is, lead acid batteries do not naturally share current equally, through a charge/discharge cycle.
Although the differences in resistances of the cabling may be tiny, the internal resistance of the battery is very small, and the temp coefficients work against sharing equally.
Once battery A is warmer, its internal voltage drops, so it takes a bigger share of the charge current and gets warmer still.
If you have to think about loads which are not steady DC, or noisy charging voltages, it gets messier still.

So, for optimum battery life, parallelled batteries should be identical and 'see' identical circuit conditions.

 

[pvb]

Once battery A is warmer, its internal voltage drops, so it takes a bigger share of the charge current and gets warmer still.

Can you expand on that? Is its "internal voltage" different from the voltage measured at the terminals?

 

[TernVI]

Can you expand on that? Is its "internal voltage" different from the voltage measured at the terminals?

Its internal voltage is the electrochemical potential of the cells.
It will be different from the terminal voltage due to the ohmic resistance of the internal metalwork etc, if a current is flowing.

 

[PaulRainbow]

Hi Paul,
If I'm reading Ningaloo's post correctly, he is actually making two contradictory statements. So which are you agreeing to: the sequence being unimportant or the connections being from each end of the bank? I know what I've seen recommended and why I'd do what I'd do, but I'm not an expert (as you know).

Apologies if my reply was confusing. Here's how i always connect batteries:

 

[halcyon]

This stuff is regurgitated all over the internet on countless websites, but nobody ever queries the claims. Do you really believe that 4 batteries in parallel, with the connections at one end of the bank, can result in one battery providing double the current of another battery? Chris Gibson declines to publish his calculations, claiming they're "quite horrific". They could also be quite wrong.

Logically, connecting the pos and neg feeds at opposite ends of a big bank will help to balance voltages. But if generously-sized cable is used, and if the connections are properly made, there'll be very little resistance to produce a significant voltage drop. For the OP, who already has feeds at opposite ends of his 2-battery bank, adding a third battery wired directly to the busbars wouldn't have any harmful effect.

Agree.

A battery has a lower voltage under load, this voltage varies with battery state and current drawn. Two batteries in parallel one battery cannot carry on loosing voltage when compared with the other, one may be 0.5% lower capacity, but will balance load to balance battery voltage.

Brian

 

[HissyFit]

Can you expand on that? Is its "internal voltage" different from the voltage measured at the terminals?

No battery is perfect. Any battery can be represented by a theoretical perfect battery in series with a resistance. When no current is being drawn from a battery there is no potential being dropped across the internal resistance, so the potential that can be measured at the terminals is that of the cell. [Edit: Conversely, when current flows, i.e. the battery is under load, potential is dropped across the internal resistance thus reducing the potential measured at the battery terminals]. As batteries age, the internal resistance increases, resulting in a drop in both the charge acceptance rate and, conversely, the maximum supplied current. There is a way to calculate the resistance through the Thevenin theorem.

 

[HissyFit]

Apologies if my reply was confusing. Here's how i always connect batteries:

I thought that's what you meant. You were agreeing with his last statement. Given this arrangement and the topic of adding a new battery to the current bank, would it be of benefit to put the new battery in the middle of the pack?

 

[halcyon]

No battery is perfect. Any battery can be represented by a theoretical perfect battery in series with a resistance. When no current is being drawn from a battery there is no potential being dropped across the internal resistance, so the potential that can be measured at the terminals is that of the cell. As batteries age, the internal resistance increases, resulting in a drop in both the charge acceptance rate and, conversely, the maximum supplied current. There is a way to calculate the resistance through the Thevenin theorem.

Your are not taking in to account battery voltage under load. If say battery is 80% capacity, open circuit voltage is around 12.53 volt. But pull 20 amp out of 2 x 100 AH batteries the terminal voltage falls to around 12.25 volt, assuming each battery supplies 10 amp. Now we have 70 - 90 sq mm cable, volt drop @ 20 amp around 0.05 v for 500 mm cable link. So the biggest difference between cells is around 0.05 volt, which equals around 3% in batt capacity. Thus if one battery pulls more current, once the capacity drops by 3% both batteries are equal and will supply equal loads. So though you may have a slight inial in-balance, it will equalize out and remain so.

Charging follows the same , with both batteries reaching the final common capacity, though one may be slightly earlier.

Brian

 

[Ningaloo]

I love this forum! There is so rarely consensus which keeps it interesting.

I read the SmartGauge article and see the recommendation is to make each battery "equal" by having identical cable lengths and wiring all direct to the busbars, but in most installations this would result in longer cables.

I imagine that this is a common problem given the size and weight of LA batteries. If I had wanted two additional batteries they would have had to go on the opposite side of the boat with significant cable lengths.

I'm going with my original plan to take the + busbar feed from the additional battery.

If I rotate my third battery so the positive post is adjacent to the busbars, I can take the existing positive busbar feed from my additional (all batteries are brand new) battery. I'll need 0.5m to connect B2+ to B3+ and 1.5m to connect B2- to B3-, all 95mm2.

By the time these batteries need replacing I hope that lithium has less of a cost differential!

Thanks for all the contributions.

 

[HissyFit]

Your are not taking in to account battery voltage under load. If say battery is 80% capacity, open circuit voltage is around 12.53 volt. But pull 20 amp out of 2 x 100 AH batteries the terminal voltage falls to around 12.25 volt, assuming each battery supplies 10 amp. Now we have 70 - 90 sq mm cable, volt drop @ 20 amp around 0.05 v for 500 mm cable link. So the biggest difference between cells is around 0.05 volt, which equals around 3% in batt capacity. Thus if one battery pulls more current, once the capacity drops by 3% both batteries are equal and will supply equal loads. So though you may have a slight inial in-balance, it will equalize out and remain so.

Charging follows the same , with both batteries reaching the final common capacity, though one may be slightly earlier.

Brian

I didn't need to go into any of this to answer pvb's question. Although this is important for the wider subject, it overcomplicates things.

 

[halcyon]

I didn't need to go into any of this to answer pvb's question. Although this is important for the wider subject, it overcomplicates things.

Sorry to complicate it for you, it's just how batteries work.

Brian