Pack / Cell compression Optimized By Using Springs.

Goldminer

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Nov 11, 2021
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My first post on the forum, so firstly a big thanks to everyone for the advice and knowledge shared. I have learned a lot, I think – what follows may prove/disprove that!

My build is at feasibility stage, but compression is an important consideration for me as space constraints require my 1P16S battery to be configured 1 x 16 cells in a single row. From the feedback on this forum, I therefore need to manage 8mm expansion across my battery pack through the charging cycle (0.5mm per cell) and maintain specified compression.

I am planning 4 bolts for compression, with springs at either end (so 8 in total), allowing each spring to manage 4mm of expansion. I would also plan to apply 300kgf at 30-40% SOC specified in the latest EVE 280K specification (attached - for which thanks, Amy Wan).

With c 150mm space each end of the pack to limit free length of each spring and prioritising as low change in load/mm as possible, 2 springs off the Lee Spring website really fit my requirements (LHL 1250A 10 & LHL 1000AB 12). Both springs have change in load of c 1.6kg/mm, so that a 4mm change per spring would increase individual spring load by c 6.5kg or 26kg across the pack. Compared to the 300kgf, this is a <10% change in compression force. This was confirmed in conversation with a technician from Lee Spring.

In psi terms, this equates to an increase from 12 to 13 psi if cells expand the full 8mm across the pack. Well below the critical level of 17 psi referenced elsewhere in this thread. They would really have to blow to get close to 17…

One other limiting factor highlighted by their technician: if the springs are deflected 80% or more from their resting state, this compromises their future performance. The 2 springs selected above under my assumptions only reach 70% deflection when 8mm cell expansion occurs, so some headroom there, too.

This spring configuration fits my purpose and space constraints, and appears to keep the cells very close to 12 psi on average - if my calcs are correct! This may be of use to other forum members.
 

Attachments

  • EVE LF280K (3 2V 280Ah) Product Specification( Version B ).pdf
    542.4 KB · Views: 18

Hedges

I See Electromagnetic Fields!
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I would also plan to apply 300kgf at 30-40% SOC specified in the latest EVE 280K specification (attached - for which thanks, Amy Wan).

I don't think the data sheet mentions compression, so relying on other sources.

The data sheet describes welding of terminals. Do you know what you'll be getting, and how you will make connections?
With 0.5mm per cell of motion, you'll want a busbar which can flex that much.
 

Goldminer

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Nov 11, 2021
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Compression per the datasheet attached above:
- "under 300kgf +/- 20kg" specified when applied to the thickness dimension of the cell at 30-40% SOC (section 3.13 on p2)
- "under the 300kgf clamp" specified for life cycles (sections 5.5/5.6 on pp3-4)

I'll be asking for welded terminals, and will most likely make crimped wire connections to eliminate the risk you highlights. Happy for any suggestions, though. I've seen some beautiful copper braided connections on a build by another forum member, but that might be above my pay grade...
 

Hedges

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Mar 28, 2020
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Ah, under "cycle life". Document wasn't searchable.

Are the welds good these days? Some people have twisted welded studs off. We've see that for round terminals, intended assembly is a busbar with round hole to fit over the terminal. Both tapped holes and welded studs were after-market hacks. There were some pictures of properly done ones.

One guy built his own braid terminals with copper pipe. I wonder about maintaining crimp force in that case.
Ring terminals can be crimped on braid. Someone did that with a braid ground strap at work.
One member measured better voltage drop with solid busbars. It is possible to have a busbar bent in a shape so it is compliant for space between studs (but maybe not misalignment.)

My system is 8x 6V AGM batteries, not lithium cells. I bought 10x 4/0 cables with crimped ring terminals for $80.

Whatever kind you use, take care of native oxide on aluminum.
 

Bob B

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Sep 21, 2019
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Ah, under "cycle life". Document wasn't searchable.

Are the welds good these days? Some people have twisted welded studs off. We've see that for round terminals, intended assembly is a busbar with round hole to fit over the terminal. Both tapped holes and welded studs were after-market hacks. There were some pictures of properly done ones.

One guy built his own braid terminals with copper pipe. I wonder about maintaining crimp force in that case.
Ring terminals can be crimped on braid. Someone did that with a braid ground strap at work.
One member measured better voltage drop with solid busbars. It is possible to have a busbar bent in a shape so it is compliant for space between studs (but maybe not misalignment.)

My system is 8x 6V AGM batteries, not lithium cells. I bought 10x 4/0 cables with crimped ring terminals for $80.

Whatever kind you use, take care of native oxide on aluminum.

I either save things as .pdf .... or print to .pdf .. One of my favorite things about the PDF docs it the search function.
 

Vonglick

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Nov 7, 2021
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Here’s a picture of my fixture. I used the same springs as 100 Proof (post 272).

3/4” plywood on ends (had some spare maple plywood)
metal on both sides of springs is 3/16” x 1 1/2” steel (local metal shop)
1/4” all thread (10 ft piece from Home Depot)
1/4” id plastic tubing (Home Depot)

As 100 Proof, I used a 162 lb. weight to compress the spring and measure it’s length.5DF67CC8-DAA8-469B-9256-3F582918A7F3.jpeg
Springs we’re compressed to 1.660”
 

Solar123456

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Jun 4, 2021
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Here’s a picture of my fixture. I used the same springs as 100 Proof (post 272).

3/4” plywood on ends (had some spare maple plywood)
metal on both sides of springs is 3/16” x 1 1/2” steel (local metal shop)
1/4” all thread (10 ft piece from Home Depot)
1/4” id plastic tubing (Home Depot)

As 100 Proof, I used a 162 lb. weight to compress the spring and measure it’s length.View attachment 76012
Springs we’re compressed to 1.660”
Looks great. Where are you mounting the bms?
 

KYSolar

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Jul 17, 2021
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1640651582405.png
I have read through this thread (and a few others on the "clamping" issue) and this way of distributing the load on the cell faces has me the most puzzled. It was mentioned by noenegdod but there doesn't seem to have been any discussion about it.
My understanding is that if you have one EVE LF280K cell with a surface area to be compressed at 12psi that would be :
173.7mm X 204.6mm = 35,539.02 or 55 square inches......X 12psi = 660 pounds of force on that surface.
If you were to use 4 bolts for the clamping then you would divide 660 pounds by 4 and apply 165 pounds at each bolt, therefore applying a total of 660 pounds on that cell face. I think everyone agrees on this.
If you were to use 2 bolts for the clamping then you would divide 660 pounds by 2 and apply 330 pounds at each bolt, therefore applying a total of 660 pounds on that cell face.
If you were to use 6 bolts for the clamping then you would divide 660 pounds by 6 and apply 110 pounds at each bolt, therefore applying a total of 660 pounds on that cell face.

If you had two cells side by side and used one plate to cover both cell walls and 4 bolts to do the clamping you would need (660 X 2) or 1320 pounds of clamping force or; 1320 divided by 4 equals 330 pounds at each bolt. This is assuming there is no deflection in the plate.
If you assume there might be deflection in the plate you might consider putting two more bolts at the midway point to counteract that deflection bringing your total to 6 bolts. That would be 1320 pounds total divided by 6 equals 220 pounds at each bolt.
I don't understand how varying the pressure at different places on the clamp will create an even pressure on the cell walls.
I would be very happy if someone could explain this to me. Thank you.
By the way, I love this forum, how everyone works to be helpful and considerate and all the good information that is passed around. I certainly wouldn't be where I am without it.
 

justgary

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Jan 1, 2022
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If you assume there might be deflection in the plate you might consider putting two more bolts at the midway point to counteract that deflection bringing your total to 6 bolts. That would be 1320 pounds total divided by 6 equals 220 pounds at each bolt.
I don't understand how varying the pressure at different places on the clamp will create an even pressure on the cell walls.
I would be very happy if someone could explain this to me. Thank you.
Your first assumption is that you want to use a plate and bolt configuration that does not allow enough deflection to worry about. A gross example of too much deflection would be trying to make an end plate out of aluminum foil. Since you want a calibrated total deflection, the springs should do most of the flexing while the plate stays flat.

The second assumption is that it is the 12 PSI that is important, so you multiply that by the surface area that will be under compression. In your example, that is 2 X 55 = 110 square inches. So the force you need is 12 X 2 X 55 = 1320 pounds applied by your non-deflecting plate. Divide that by the number of spring rods you plan to use.

The third assumption is that the geometric center of your fastener pattern lies at the geometric center of the areas under compression. If it isn't, you risk applying the force unevenly. A gross example would be putting eight spring rods at the bottom of the pack and none at the top.

Once you have a non-deflecting plate with the bolt pattern centered at the geometric center of the surface area of the cells, you are good. Spring rods between the two stacks of cells simply allow you control deflection of your plates a little better. In other words, they can be a bit less beefy since the cantilever on them is half as much.
 
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