The multiple string battery riddle.

[phil.g00]

I have inherited the care of a battery bank, lets call it 12 strings of 4 x 12V 100Ah sealed, (it's actually 11 now, but that's by the by).

It used to be a 24V system to the battery cabling is over-rated, but its paid for and I want to use or re-use to solve my problem.

The batteries are mounted on 3 shelves above one another, 16 batteries /shelf. 

So there are 4 strings of 4 batteries each per shelf.

Each battery string is cabled back with 25mm2  Cu back to busbars (on the left of each shelf, so effectively the busbars are about 1.5m in length down the height of all shelves on the left).

Re-positioning the above hardware ( batteries or busbars) is not an option.

One of the improvements I'd like to make is to fit a BMV shunt.  The cabling between the busbars has to broken to be in series with all the current. 

 

Problem 1. A 500A BMV shunt is going to be smothered with 12 x 25mm2 cables attached at least one-side.

 

At the same time, I am conscious that  the cable lengths from the furthest battery string is far longer ( 12 battery thicknesses away) than the near battery string.

Problem 2. I want to improve on the different cable length situation, at least certainly not make it worse.

 

I think I have a rather novel solution to both issues, I 'll provide it later, it has to do with how the battery connections are made.

I've been pondering this for while, so for a bit of fun I thought I'd put it out there as a sort or riddle.

Lets frame the question better and cut it down just considering a single 16 battery shelf. ( 4 strings)

And lets say, for even 1 shelf that even 4 lugs at 25mm2 cabling will overwhelm the shunt. (You'd actually probably get away with that, but this is my riddle, so my rules).  

How do I make the putting the shunt in circuit possible while at the same time make sure that each string  has the same length of cabling back to the busbars?

Unnecessary, coils of cabling to lengthen are wasteful and not allowed. Shortest amount of cabling so that the maximum amount of cabling can be re-purposed.

I have 50mm2 at my disposal.

There you go, have fun. 

 

 

 

 

[Ingo]

You mean like this?

 

Edited January 10, 2019 by Ingo

[phil.g00]

Presently, the batteries are connected worse than this, just the first 4 and back the second 4 and back and so on.

Clue : With the solution they can still remain in adjacent groups of 4.

I think for simplicity just think of each battery as being 48V. The number of batteries just serves to emphasize the physical width in real life.

The real world difference compared to your drawing, (which is an improvement on present) is:

Both busbars are physically on the left which means it isn't physically 1 hop forward from the end to negative, its actually 3 hops worth of cable back.

So if you physically compare red and black as an entire circuit loop there would a physical difference in  cable length. We want all batteries to be faced with identical charging voltages, ergo the volt drops in the cabling must be identical.

Next issue, the shunt wont fit all four 25mm2 cables (R,Bl,G&Bk) cables,  it has to be in series with all the current.

Clue: I can go thicker, but only up to maximum of two cable lugs/shunt side.

( That's the rules of this riddle, in my real scenario I have 3 times this amount of cabling whilst I might not be this strictly limited, it is still a realistic limitation).

There are two solutions I can think of, but one has more losses.

Edited January 10, 2019 by phil.g00

[phil.g00]

I have raise to the bar here, so I involved google.

This gentleman;s solution achieves what I want every battery is faced with the same cable loop length and I can thicken the the supply cables or double up on them and the shunt will handle it.

Solid B+.

But, I think there is a better answer.

 

 

[DeepBass9]

What is the whole installation powering, just a household?

[phil.g00]

No, but irrelevant to my riddle.

[phil.g00]

OK, give up?

[phil.g00]

[Guest]

Why did you not say so?

Had this pic for years now ... 

 

[phil.g00]

Not the same circuit.

Look again.

[Guest]

9 minutes ago, phil.g00 said:

Not the same circuit.

Look again.

What am I missing? Where it is 2 x 12v, make it 4 x 12v?

Pic I posted can have 8 x 48v banks. The success lies in the wiring, to not over-strain a battery.

Got that from a guy called Carribocoot on another forum.

[phil.g00]

The object is that each battery is charged is faced with an identical voltage.

So in your case each string of 24V must be served by a equal length of cabling - in the loop, from the incoming +ve to the outgoing -ve.

So discounting the internal 3" connection on your drawing for the time being. ( Lets think of them as 24V batteries).

It's not really the same question as you has 8 strings though.

But, let me show you what I want to avoid and why your concept doesn't suit.

Your first battery on the left has a loop of:

48" on the -ve side & 32" = 80" cable in the loop back to the sources.

Now lets look at the 4th battery from the left:

32" on the -ve side & 0" on the +ve side.

That means the loop cabling for the first battery on the left is 4 foot longer than the 4th battery.

That's the specific situation I want to avoid.

[Ingo]

I tend to agree with @phil.g00, every battery in his drawing is three 'hops' away from the source.

[___]

Nice riddle and close enough... But I think 2 x 1 foot lengths is not equivalent to a 2-foot length. That lug you crimp on has a not-negligible effect on the 2x-1 foot scenario. So one could argue that the top battery is not 3-hops away... probably more like 2.1 hops :-)

Still... nice riddle!

[stoic]

...as if electricity is not complicated enough already... and now you go an add riddles to it...  

[phil.g00]

 

@plonkster,  Same number of crimps in the loop, means the same voltage across the batteries No?

Doesn't matter if that crimp is on the positive or the negative leg.

Now on to @The Terrible Triplett 's dilemma, an eight string solution anyone?

This time there might be more options, and crimp symmetry might make a difference, so solid gold, super -conductor , -273 degree Kelvin crimps everybody.

 

[___]

48 minutes ago, phil.g00 said:

Same number of crimps in the loop

Dang it, you're right. Nice solution then!

[Gabriel_2018]

Obviously, It Will be better to install balancer than not, but often, an issue is the inner unbalance of a 12 V monoblock, which has 6 cells that cant be balanced. In other words, if one cell is damaged, there's nothing to do. 

[phil.g00]

People shy away from multiple battery strings.

Lots of tales of woe, and yet connected up right I contend that they are not the bogeyman.

Connected up right, they should not cross-contaminate each other, and be only subject to the same issues as a single string.

Well, I'll leave that controversial statement out there.

Everybody knows its not true,  but I remember Y2K and Santa Claus and sometimes I doubt things.

OK gents,  back to the eight battery string?  

It is a conceivable battery configuration, and just cause something isn't googleable, doesn't mean its impossible.

 

[Coulomb]

15 hours ago, phil.g00 said:

The object is that each battery is charged is faced with an identical voltage.

Fair enough.

Quote

... That means the loop cabling for the first battery on the left is 4 foot longer than the 4th battery.

That's the specific situation I want to avoid.

It's not just the loop length you want to equalise; you also want to equalise the current through the various cables. [ Edit: see Weber's correction below. ] Your solution, which is to diagonal connect two diagonally connected pairs, does just that. But you can only do it for 2 or 4 cells/strings, theoretically for 8 but the number of cables on one terminal gets too much.

Edited January 12, 2019 by Coulomb

[phil.g00]

Nope, don't accept.

I want an answer not an excuse.

Edited January 12, 2019 by phil.g00

[weber]

These battery paralleling puzzles are fun. I've studied them extensively. Chris Gibson deserves to get credit for those diagrams posted above. He also gives your solution. So it isn't new. But good on you, since you came up with it independently. Here's his web page.

Yes, each battery must see the same voltage. But that doesn't mean that the path through each battery must have the same resistance, i.e. the same lengths of the same cross sections. Nor does it mean you have to "equalise the current through the various cables", in any way I can parse that. What is required is that the sum of the volt drops be the same along the path for every battery. i.e. the sum over all hops, along the path through each battery, of the resistance times the current for each hop.

If you make all jumpers the same length and the same cross-section (and therefore same resistance), it's easier to check any proposed solution. On your first pass, you note against each jumper, how many cells it has to carry current for. Then on a second pass, you note against each battery, the sum along the path through that battery, of the numbers you wrote on the first pass.

This also shows that TTT's 8-way is wrong. Coulomb knows the correct 8-way solution, he just hasn't spelled it out, presumably because he thinks it's obvious once you've seen the correct 2-way and 4-way, and realised that the correct 4-way scheme is just the correct 2-way scheme (known as "diagonal takeoff") applied at a higher level, to two 2-ways instead of two batteries.

Edited January 12, 2019 by weber

[weber]

Some more such puzzles:

1. There is no theoretically-perfect sharing arrangement of this kind for 3 strings, so you are forced to either use Chris Gibson's method 3, or parallel some jumpers. By "of this kind" I mean where all connections are made at the cell terminals, and all jumpers have the same length and cross-section (or all jumpers in symmetrically-equivalent positions have the same length and cross-section). How can we parallel some jumpers to make a theoretically-perfect arrangement for 3 in parallel?

2. You might, by now, think that such arrangements only exist for powers of 2. But some years ago I discovered such an arrangement for 5 in parallel. I sent it to Chris Gibson and he agreed it was legit (although I'm not sure if it meets your rather strict "no link longer than it has to be" requirement, Phil). What is this arrangement for 5 in parallel?

I have not found any such arrangement for 6, 7 or 9, although I can't say I've exhausted all possibilities, as I have with 3. But since 5 and 2 both work, we can do 10 as 2 x 5, and of course any number whose prime factors are only 2 or 5.

Edited January 12, 2019 by weber

[weber]

Unfortunately, these puzzles, while fun as geometrical and mathematical exercises, abstract away too much of the real world. First, we have to recognise that most of the resistance is not in the cable, and not even in the crimp. The bolted joints generally have higher resistance than the cable or crimp, and this is highly variable. It is close to inversely proportional to the bolting pressure, and also depends on thickness of oxide layers. All cells and lugs had better be equally clean, and you had better use a torque wrench to set them all the same.

You also need to pay strict attention to the order in which the lugs are stacked on each terminal. In a simple 2-way diagonal takeoff, the takeoff lugs must be stacked on top (i.e. furthest from the terminal).

But it gets worse. The internal resistance of the cells themselves is usually much greater than that of the cables plus crimps plus bolted connections. If not, then you probably haven't designed the cabling correctly. And cell internal resistance depends strongly on temperature. Lower temperature means higher resistance, due to chemical reactions slowing down. And notice the vicious cycle here. Heat generated inside a cell is inversely proportional to its resistance P = (Vcell - Voc)²/Rint, which means the amount of heat generated increases as temperature increases, which increases the temperature further ...

The temperature profile of a cell on the end of a row will be very different from one in the middle of a row. So you either need a way to keep all your cells at the same temperature, such as the liquid circulation in the Tesla car batteries, or you need to arrange it so that every string has equal numbers of cells near the middle of the pack, and equal numbers of cells near the outside of the pack.

And then you have manufacturing variation between cells, so you have to sort them and group them correctly.

Hopefully now you understand why balancing the current between multiple strings of cells is really really difficult, and should be avoided if at all possible.

Edited January 12, 2019 by weber

[___]

2 hours ago, weber said:

The internal resistance of the cells themselves is usually much greater than that of the cables plus crimps plus bolted connections

Yup, this basically says what I was going to say. And even if that is exactly the same on day 0, it tends to diverge. Even for two-string banks. You can however compensate for it, as is explained in this paper I've posted too many times before, essentially by always overdoing it a bit so that the weaker string has time to catch up. The theory is that this slight amount of overcharge on the better string is acceptable even if it does shorten the life of that string: Because the extra capacity vs cost of replacing the whole bank vs amount by which the life is shortened is acceptable. Or so they argue. But the paper says don't do it with shallowly cycled banks.

So the accepted general mantra that parallel strings are always bad is wrong... but the generalised counter statement that it isn't is also wrong. As always, the answer is somewhere there in the middle. I contend though that from 4 strings upwards... you're pretty much always wrong :-)