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Open Roads Forum  >  Tech Issues

 > Fun with batteries --- some simulations

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KendallP

Southern Oregon

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Posted: 12/23/11 12:55pm Link  |  Print  |  Notify Moderator

Yeah, you've got plenty of obsessive friends around here who are happy to learn something. I'm also in your corner.

Smartgauge method 1 is definitely not preferred.

Unfortunately you're not set up to do it, but what I'D like to see is a test between method # 2 and # 3. With heavy interconnect cables, my money is on a very minor difference, at most.

So...

Anyone out there with 3 or more 12V batteries that's looking for a holiday project?


Cheers,
Kendall

1986 Winnebago Chieftain 22RC
Our Camper (Don't laugh...
Unlike our credit cards... she's paid for)


LScamper

Los Alamos NM

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Posted: 12/23/11 03:03pm Link  |  Print  |  Notify Moderator

Dclark wrote:

“How did you model battery voltage vs. time? Was it based on OC voltage vs. SOC or are you using output voltage vs. different load current for various SOC levels? Does battery output impedance remain constant in this simulation?”

Tom M wrote:
“1. How was state of charge determined? Voltage or specific gravity.
2. What was end of test voltage while under load?
3. Battery temperature?”

These are good questions but some not so easy to answer.

Battery impedance does remain constant in this model. I used 10 mOhms based on measurements made by BFL13 and Salvo on this forum.

Battery temperature was not modeled. The battery open circuit voltages are based on state of charge. They match a table that many seem to use on this forum at www.batteryfaq.org. The 70 degree Fahrenheit table was used.

State of Charge is calculated by measuring the current through each battery. This current is integrated over the time that the load is on. This gives total Amp hours from the battery. Say the battery current is 10 A. The load is on 10 hours. This will give a total of 100 AH. From this SOC can be calculated. It is (1-total AH used/bat AH rating)*100.
So my battery rating is 232 AH and 100 AH are used --- (1-100/232)*100 = 56.9%
SOC is 56.9%

From this SOC the program uses the battery table and looks up the voltage that corresponds to that SOC. That is the battery open circuit voltage.

Battery voltage vs. time is a lot harder to explain unless you have some electronics background. I won’t try to completely explain it. The battery voltage is determined by charging a capacitor. The voltage across a capacitor varies according to the amount of current passing through it and for how long.

The value of the capacitor is calculated by using the battery AH rating and the change of voltage across it. In this case the battery rating is 232 AH and the voltage across it is determined from the above-mentioned table. At 100% SOC the voltage is 12.643V and at 0% SOC the voltage is 11.883V.

Capacitance = battery AH rating * 3600/(V at 100% SOC – V at 0% SOC)
Capacitance = 232*3600/(12.643-11.883)
Capacitance = 1,098,947.4 farads.

I did not label this correctly in original post.
Battery voltage at load cable connection points at start of test: 11.98V
Battery voltage at load cable connection points at end of test: 11.59V
Both with load still connected.


Lou


KendallP

Southern Oregon

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Posted: 12/23/11 03:27pm Link  |  Print  |  Notify Moderator

OK...

3 115 Ah 12 batteries
2 pair of 2/0 interconnect cables

Method 1 vs. method 2

100A load or something in that realm

A real test is required to be considered any reasonable authority, but it will be fun to see what yo come up with anyway.

LScamper

Los Alamos NM

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Posted: 12/23/11 03:16pm Link  |  Print  |  Notify Moderator

KendallP wrote:

“Unfortunately you're not set up to do it, but what I'D like to see is a test between method # 2 and # 3. With heavy interconnect cables, my money is on a very minor difference, at most.”

All of this is a simulation, a mathematical model of the real world. None of it exists. There are no batteries, there are no cables and there is no load. All I have is a computer and a mouse! With this I can look at what would happen with three batteries, 4 batteries, with whatever method you want. But it takes time and some of it I’ll get to if people want it. If everyone just laughs at this it won’t get done.

LScamper

Los Alamos NM

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Posted: 12/23/11 06:07pm Link  |  Print  |  Notify Moderator

BFL13 wrote:

“So my plan to use a fifth battery as an AH pail to put AHs into the bank of four isn't going to be as fast a process as I had hoped.”

You already know the answer to that. Think of the charged battery as a constant voltage battery charger with a voltage set point of 12.6V. If you try to charge a low battery with a 12.6V charger you won’t get very far! It is even worse than that. As the low battery charges up the charged battery voltage goes down.

mena661

Southern California

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Posted: 12/23/11 03:27pm Link  |  Print  |  Notify Moderator

LScamper wrote:

If everyone just laughs at this it won’t get done.
There are at least 4 people in this thread that would like to see this. Would that be enough?


2009 Newmar Canyon Star 3205, Ford F53 V10
Trojan L16 6V's 740 Amp-hours


dclark1946

Richardson,TX,USA

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Posted: 12/24/11 06:32am Link  |  Print  |  Notify Moderator

A few observations:

A 10% difference in SOC for the two batteries at the end of this extreme current load does not seem to be all that bad in my opinion. You have picked the discharge time to be the exact time for the balanced batteries to hit the magic 50% SOC which does cause the heavier loaded battery to go below 50% in the unbalanced configuration. Your simulation does show that the battery that was initially supplying the most current at the beginning is beginning to decrease and the other battery is starting to supply more of its share. Under a lighter (more realistic) load the difference in SOC would be even less. If you ran the original simulation another 0.5 hour they would be even closer in current balance and the SOCs should start to converge.

If the battery impedance increases with lower SOC (suspect it might but don't know for sure)then the difference between the two batteries SOC would be even less as the lightly loaded battery would start carrying even more of the load than your simulation predicts.

Finally when you start to recharge the battery with the lower SOC will absorb more of the total current causing it to charge at a faster rate and as the total charge current decreases the resistance inbalance will become less significant so the batteries should end at at the same SOC.

I would be interested in seeing the results with one of the battery models suggested by another poster. You are off to a good start.


Dick


Dick & Karen
Richardson,TX
2007 KZ Spree 240RBS
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BFL13

Victoria, BC

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Posted: 12/23/11 06:35pm Link  |  Print  |  Notify Moderator

Well phooey. The fifth battery still has to be worth something with its usable 50 ah or so to add to the bank somehow. By driving around and topping it up, that's extra ahs right there. Meanwhile the four in the trailer are getting their dose of whatever the solar can do for them with the 130w panel (50ah?)


The thing is, the fifth battery is worth one whole day of solar at 50ah so that's like having twice my solar of one 130w panel. (50ah partially cloudy day)

So now just before supper and microwaving etc then a movie, I add the fifth battery with jumper cables. Now Peukert is happening for those big draws on five instead of four batteries, at supper microwaving, evening movie, and morning kettle and toaster, while also equalizing from the fifth battery overnight.

Then next day off in the truck recharging the fifth battery while solar works on the other four back at the trailer. This has to be better than nothing. Maybe I should have more batteries in the truck and fewer in the trailer.

I did think the fifth would equalize with the four a lot faster though. Never really though about how long that would take.


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mena661

Southern California

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Posted: 12/23/11 10:49pm Link  |  Print  |  Notify Moderator

Second alternator for charging, take two 6's with you while you're out sightseeing (or take all 4). Next day, swap them. Leave the 12 on the solar.

Cedarhill

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Posted: 12/24/11 07:41am Link  |  Print  |  Notify Moderator

Here are a few comments now that I understand what you are doing. (1) Your original post mentions a linear simulation but if you are using a capacitor to model a battery then that certainly isn't linear. That threw me off. I had an idea you were using a spreadsheet type simulation. (2) It would improve your results to use one of the lead-acid battery models you can find on line. You could model both the discharge and charging phases in one simulation. (3) The internal resistance of a lead-acid battery is not constant and will vary by a factor of 2 or so depending on the SOC. That behavior will affect on the outcome of your simulation enough to matter. (4) You reported your outputs to 5 significant digits. That implies a level of accuracy you really don't have.

LTspice is a remarkably good tool to be available for free. I worked in an office where there weren't enough commercial licenses available (Orcad) so I could use it whenever I wanted and for as long as I wanted. I, and a lot of other people, just started using LTSpice all the time and the results were just as good for simple applications like vehicle systems.

Before I stop writing, let me compliment you on your efforts. They are interesting and I can see how they would have practical applications in some situations.

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