Headway cell temperature

[oofnik]

I have my 12 cell Headway pack in a sealed box with very little heat transfer ability. I pulled 9.5 AH in about an hour on a hard test run the other day. The cells got up to about 130 F; kind of uncomfortable to touch. Even though LiFePO4 is very stable I'm thinking this is too high and not very healthy for the cells to be reaching this temp on a regular basis.
I will probably cut out a section of my box and replace it with aluminum sheet metal to act as a sort of heat window. I'm also considering building a 12s1p A123 pack to help the Headways with peak discharge. I don't see any issue putting two different LiFePO4 batteries in parallel, do you?

 

[swbluto]

If they're not balanced so that they don't run out at the same time, you might have problems if you're continuously discharging with over-discharging or having one "not work" in whatever way(BMS cutoff, for example), but if you set it up correctly(Like say an LVC of some sort), you should be able to disconnect it manually without having those pesky BMS cut-outs and just continue with the ride. Also, depending on the relevant capacity/internal-resistance ratios, you might have possible over-charging cross-currents, but those can be prevented by diodes.

Other than that, I'm not seeing any immediate problems.

 

[oofnik]

I'm planning on paralleling the A123s at the cell level, so they will be individually protected by the BMS low voltage cut-off.

 

[swbluto]

Oh, okay.

Just as a thought experiment, let's say you have a cell at 3V and another at 2v. There's a switch between the 2V and 3V cell that basically turns the "parallel" connection on and off and it's initially turned off and the cell's voltage sensor is connected closest to the 3V cell. What would the voltage sensor read when the switch was turned on and, thus, they were instantly put into parallel contact? Would moving the voltage sensor "electrically much closer" to the 2V cell significantly decrease the reading?

Let's say both cells have an equal internal resistance.

(My reasoning goes that the output voltage of the 3V cell would decrease until the difference between the output voltage and 2v create a current determined the 2v cell's internal resistance. Eh, it's hard to express it seems. But, basically, it wouldn't be able to read the 2V cell's true voltage and moving it closer shouldn't create a significant change in the voltage reading.)

But, in the real world, what's the impact of this kind of scenario? *thinks hard*

a123s discharge before the headways, but as long as the headways don't have an output voltage less than 2V, the a123s shouldn't go below 2v as the headways would be charging the a123s at that point. The a123, at that point, however might have significant charging currents and the headways might have correspondingly significant discharging currents which might mean shortened lives but it might not be a significant impact as it could possibly be short lived and the cells wouldn't heat by that much, which is the true reaper of cells.

Anyways, if you wanna, you can get an idea of how much the a123s will discharge at and the headways will discharge at by treating them as distinctly different batteries and putting their data in a spreadsheet I created. It's in the "mixing chemistries for most economic performance" thread or some thread like that at the end. Since we have test data for the headway's internal resistance and the a123's internal resistance, it shouldn't be hard to calculate the necessary data.

 

[monster]

put a fan over the hole

 

[oofnik]

put a fan over the hole

Huh? What hole? :?

Anyway swbluto I think you're overcomplicating the situation. Maybe reading up on some electrical theory will help you conduct your thought experiment.

The internal resistance of batteries acts just the same as putting multiple resistors in parallel: the resistance decreases. Say the headways are 12 x 9 mOhm = 108 mOhm (Doctorbass's numbers), the A123s are 12x 16 mOhm = 192 mOhm (numbers pulled out of my @#%). Equivalent parallel resistance = 1 / (1/108 + 1/192) = 69 mOhms. Yes, there might be some insignificant balance currents between the cells, but nothing to get worried about unless you parallel one completely discharged cell with one hot off the charger. But I'm not going to do that.

 

[dnmun]

internal resistance in these lifepo4 batteries is not the same as resistance in a passive device, such as a resistor. the internal resistance is a measure of how hard it is for the battery pack to deliver current, with the output potential reduced by the limitations for the electrolyte and internal construction to allow ion transport from the cathode to the anode.

 

[oofnik]

You're right. It's not exactly the same. You explained the reasons why internal resistance exists, but as far as calculation of the effective internal resistance of parallel / series cells in a battery pack, I'm fairly certain that the same resistance laws apply.

 

[swbluto]

put a fan over the hole

Huh? What hole? :?

Anyway swbluto I think you're overcomplicating the situation. Maybe reading up on some electrical theory will help you conduct your thought experiment.

:lol: . If only you knew how much I've read... If you did, I would take that as an insult. Anyways, I try to think of *all* possible angles in worst case scenario analysis, even if that means some later-to-be-discovered not so relevant detours, so "over-complication" is inevitable when comparing to the "average case". Just take whatever thoughts appear to be relevant to your case and apply it. You did ask about possible problems if I remember correctly...

Anyways, my thought experiment actually turned out to be a true result according to my SPICE simulation. But, as later explained, that wouldn't be a problem because the LVC would either trip due to the other cells also going under(thus the sensed voltage) or the charged cells would be charging the depleted cells before it would actually go under LVC. So, your only concern appears to be cross currents, which might be high under certain circumstances, but depending on your relevant circumstances, it might not be a problem. It's good you've done the calculations for your situation in verifying it isn't.

(If you're curious, the sensed voltage would actually approach the ratio of the internal resistances of the different voltaged cells in between their voltages. Example, 2 V 100mOhm and 3 V 200 mOhm approaches 2.33 sensed volts. The actual resistance between the different cells and the sensor will deviate this result, but by a small amount.)

 

[liveforphysics]

With cells of identical chemistry put into P at a cell level, when at rest, all cells in P will reach the same state of charge.

During discharge, things are different.

For a continous discharge situtation, the A123s would be taking the brunt of the load, but the non-dynamic cell voltage difference would soon reach an equilibrum, and the cells would be shareing the current load better.

As long as you have a duty cycle with adquate rest periods, the state of charge would stay balanced between the headways and the A123.

 

[swbluto]

With cells of identical chemistry put into P at a cell level, when at rest, all cells in P will reach the same state of charge.

During discharge, things are different.

For a continous discharge situtation, the A123s would be taking the brunt of the load, but the non-dynamic cell voltage difference would soon reach an equilibrum, and the cells would be shareing the current load better.

As long as you have a duty cycle with adquate rest periods, the state of charge would stay balanced between the headways and the A123.

Oh yes, that's right. Since the output voltage of the a123s would be equal to the output voltage of the headways, that greatly simplifies the analysis as it'll depend on the internal resistances at that point. Basically, at the same state of charge, the a123s will be putting out nearly twice as much current as the headways. As the state of charge decreases, the a123s would slow down their contribution until "finally stopping"(Not technically accurate, but it approaches that state). During the rest periods or sufficiently low-current periods, they would charge and rebalance as liveforphysics explains. This will probably shorten the cycle life of your a123s compared to "ideal results", but it would probably not be noticeable in the real world.

(Or, wait, what kind of load currents are you outputting? If it's something like 100 amps, then the a123s would be delivering approximately 2/3rds of that initially which would be hard on the 2.2 ah a123s in the steady state condition. If it's just 100 amps pulsed, then 30c shouldn't be a problem.)

 

[oofnik]

If only you knew how much I've read... If you did, I would take that as an insult.

Sorry, didn't mean it like that!
Anyway I think we've analyzed the situation pretty well. As long as there is sufficient time for the cell voltages to balance between the A123s and Headways I should have no problems. Just need to keep the wires between cells nice and fat.

As soon as I get my current limiter board made, I will probably end up setting it at 90 amps max. This shouldn't happen for more than 3-4 seconds at a time during hard acceleration.