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 > Fun with batteries --- some simulations

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LScamper

Los Alamos NM

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Posted: 12/31/11 05:50pm Link  |  Print  |  Notify Moderator

Thanks Salvo, that make perfect sense.

I now have a discharge model that includes Peukert n=1.3 and resistance / SOC. The model matches SOC vs voltage curves to within about .1V > .2V across the plot at any current.

Next year we will see if it is of any use to anyone. Even if it is not I learned a lot about batteries that I did not really need to know.


Lou


LScamper

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Posted: 01/03/12 10:42am Link  |  Print  |  Notify Moderator

Much improved model includes Peukert and resistance that varies with SOC. Peukert predicts that a battery will behave like it has more capacity with a low current draw that at high current. I have added that to this model. Batteries have different Peukert numbers that greatly influence how they behave. I used 1.3 as suggested by SmartGauge. I think that this data is very close to real conditions but your mileage may vary.

Note:
It was pointed out correctly that I reported some data to 3 decimal places and the accuracy of the model is not that good. There is a big difference between accuracy and precision. The data can be very precise and very inaccurate. Some of the data that will follow is reported to 2 decimal places. This is done to show trends. For instance the load voltage changes only about 1.5V. Saying that the voltage is 11V and the next data point is 11V, which may still not be accurate, does not tell you anything. Saying 11.52V and the next 11.55V tells you that the voltage is increasing a little and is useful. It may not be as accurate as the numbers imply but the trend is accurate and something that should be shown.


Parameters that I used for this simulation:
Battery capacity is 232 Amp Hour.
Two batteries in parallel for a total of 464 AH.
Interconnecting cables = .773 mOhms each including contact resistance. This is 1.5’ of #4 wire (like a pre-made cable from the auto parts store.
The two load cable resistances are 2 mOhms each --- about .4 mOhms contact resistance, 6.4 feet #4 cable = 1.6 mOhms for a total of 2 mOhms each cable.
Used constant current load 116 A--- could use constant power load that would be more like an inverter or could use a resistor.

Balanced test:
Load cables connected to battery A positive post and battery B negative post.
Start of test, 15 Min, 1 Hour, 2 hours
Load voltage: 11.46V, 11.36V, 10.97V, 11.09V
Battery terminal voltage: 11.94V, 11.84V, 11.44V, 10.56V

Results Battery A, Battery B
Battery current at start of test: 58A, 58A
Battery current at 2 hours: 58A, 58A
Amp hours out at 15 min: 14.5AH, 14.5AH
Amp hours out at 1 hour: 58.2AH, 58.2AH
Amp hours out at 2 hour: 116AH, 116AH
State of Charge at 15 min: 89.9%, 89.9%
State of Charge at 1 hour: 59.6%, 59.6%
State of charge at 2 hours: 19.5%, 19.5%



Unbalanced test 1:
Load cables connected to battery A positive post and battery A negative post.
Start of test, 15 Min, 1 Hour, 2 hours
Load voltage: 11.46V, 11.37V, 10.97V, 10.08V
Battery terminal voltage: 11.94V, 11.84V, 11.44V, 10.56V

Results Battery A, Battery B
Battery current at start of test: 61.7A, 54.3A
Battery current at 15 min: 60.8A, 55.2A
Battery current at 1 hour: 58.8A, 57.2A
Battery current at 2 hour: 57.5A, 58.5A
Amp hours out at 15 min: 15.3AH, 13.7AH
Amp hours out at 1 hour: 61.7AH, 57.3AH
Amp hours out at 2 hour: 118.1AH, 113.9AH
State of Charge at 15 min: 89.1%, 90.7%
State of Charge at 1 hour: 57.6%, 61.7%
State of charge at 2 hours: 17.4%, 21.5%

Unbalanced test 2 (heavy interconnect cables .517 mOhm, 1.5’ 2/0 wire):
Load cables connected to battery A positive post and battery A negative post.
Start of test, 15 Min, 1 Hour, 2 hours
Load voltage: 11.48V, 11.38V, 10.98V, 10.1V
Battery terminal voltage: 11.96V, 11.85V, 11.46V, 10.57V

Results Battery A, Battery B
Battery current at start of test: 60.5A, 55.5A
Battery current at 15 min: 59.9A, 56.1A
Battery current at 1 hour: 58.5A, 57.5A
Battery current at 2 hour: 57.6A, 58.4A
Amp hours out at 15 min: 15.0AH, 13.9AH
Amp hours out at 1 hour: 59.6AH, 56.7AH
Amp hours out at 2 hour: 117.4AH, 114.6AH
State of Charge at 15 min: 89.4%, 90.4%
State of Charge at 1 hour: 58.3%, 61.1%
State of charge at 2 hours: 18.1%, 20.8%

It seems that there is little difference between balanced and unbalanced connection with a two-battery system. Differences in batteries most likely would have more influence on the balance than the connection method. I would still connect in a balanced manner, it cost nothing and you have the best chance of things being balanced.

KendallP

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Posted: 01/03/12 10:55am Link  |  Print  |  Notify Moderator

LScamper wrote:

I would still connect in a balanced manner, it cost nothing and you have the best chance of things being balanced.

Unless you have 3 or more batteries and must use either a buss bar or run all of the equal lengths of cable to a single point for each the positive and negative in order to comply with smartgauge method # 3.

I have three 27s and employ Smartgauge method # 2 with 2/0 interconnects. THIS method truly is no more difficult than method # 1. But, of course... it is not considered perfectly balanced. When you get a moment, I would love to see what you come up with for that scenario with both small and large cables.

As to your latest findings, I'm not surprised that the difference is that minimal. Nor am I surprised that cable size makes a difference. It has always bothered me that smartgauge failed to take this into account. To me, their report is sorely lacking in setup data to be taken seriously. YOU have done a much better job.


Cheers,
Kendall

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


KendallP

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Posted: 01/05/12 11:19pm Link  |  Print  |  Notify Moderator

Great job!

Wow!

2 things jump out at me.

1. Interconnect resistance makes a very discernible difference. No surprise. This is why I went with 9" 2/0s.

2. But with either interconnect size, with a 100A load, your model shows MUCH less of a difference for Method # 2 than smartgauge shows. It's certainly adequate for the job.

And for everyone who doesn't run the A/C unit of the inverter for a couple hours at a time, method # 2 is MORE than adequate.

Unless it's an easy and inexpensive project, I see no reason to employ method # 3 for most RV applications.

Of course we need reliable experimental data to truly put this one to bed, but at least now we have a 2nd opinion to smartguage (besides my own.) And a much more detailed data set it is too.

Again... good work!

Thanks!
.

* This post was edited 01/05/12 11:26pm by KendallP *

LScamper

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Posted: 01/05/12 10:43pm Link  |  Print  |  Notify Moderator

KendallP wrote:

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

Method 1 vs. method 2

100A load or something in that realm”

I’m a little late getting this done, trying to update model.

Parameters used for this simulation:
Battery capacity is 115 Amp Hour.
Three batteries in parallel for a total of 345 AH.
The two load cable resistances are 2 mOhms each --- about .4 mOhms contact resistance, 6.4 feet #4 cable = 1.6 mOhms for a total of 2 mOhms each cable.
Used constant current load 100 A.

Method 3:
Load cables connected to center connection of three cables each for positive and negative outputs.
Interconnecting cables = .773 mOhms each including contact resistance. This is 1.5’ of #4 wire (like a pre-made cable from the auto parts store.

Start of test, 15 Min, Time voltage reaches 10.5V
Load voltage: 11.4V, 11.3V, 1.24 hours
Battery terminal voltage: 11.9V, 11.7V, 1.54 hours

Results Battery A, Battery B, Battery C
Battery current at start of test: 33.3A, 33.3A, 33.3A
Same for all three batteries.
State of Charge at 15 min: 88%
SOC load V = 10.5V: 40%
SOC battery terminal V = 10.5V: 25%
Time SOC = 0%: 2.05 hours
Amp hours out at 15 min: 8.4 AH
Amp hours out at 1.24 hour (load V = 10.5V): 42 AH
Amp hours out at 1.54 hour (battery terminal B = 10.5V): 51AH
Amp hours out at 2.05 hour (SOC = 0): 68 AH

Method 2:
Load cables connected to battery A +post and battery C –post.
Interconnecting cables = 1.5 mOhms each including contact resistance. This what SmartGauge used.

Start of test, 15 Min, Time voltage reaches 10.5V
Load voltage: 11.3V, 11.2V, 1.12 hours
Battery terminal voltage: 11.7V, 11.6V, 1.5 hours

Results Battery A, Battery B, Battery C
Battery current at start of test: 34A, 31.9A, 34A
Battery current at 15 min: 33.9A, 32.2A, 33.9A
Battery current at 1 hour: 33.4A, 33.1A, 33.4A
Current link A+ > B+: 66A
Current link B+ > C+: 33.3A
Current link A- > B-: 34A
Current link B- > C-: 66A
State of Charge at 15 min: 87%, 88%, 87%
SOC 1 hour: 50%, 52%, 50%
SOC load V = 10.5V: 45%, 47%, 45%
SOC battery terminal V = 10.5V: 28%, 30%, 28%
Time SOC = 0%: 2.03 hours
Amp hours out at 15 min: 8.5 AH, 8 AH, 8.5 AH
Amp hours out at 1.12 hour (load V = 10.5V): 37.7 AH, 36.5 AH, 37.7 AH
Amp hours out at 1.5 hour (battery terminal B = 10.5V): 49AH, 47.8 AH, 49 AH
Amp hours out at 2.03 hour (SOC = 0): 68.1 AH, 67.1 AH, 68.1 AH


Interconnecting cables 2/0 = .517 mOhms each including contact resistance.

Start of test, 15 Min, Time voltage reaches 10.5V
Load voltage: 11.3V, 11.3V, 1.2 hours
Battery terminal voltage: 11.8V, 11.7V, 1.54 hours

Results Battery A, Battery B, Battery C
Battery current at start of test: 33.6A, 32.8A, 33.6A
Battery current at 15 min: 33.5A, 33A, 33.5A
Battery current at 1 hour: 33.4A, 33.3A, 33.4A
Current link A+ > B+: 66.4A
Current link B+ > C+: 33.6A
Current link A- > B-: 33.6A
Current link B- > C-: 66.4A
State of Charge at 15 min: 88%, 88%, 88%
SOC 1 hour: 51%, 52%, 51%
SOC load V = 10.5V: 40%, 40%, 40%
SOC battery terminal V = 10.5V: 25%, 26%, 25%
Time SOC = 0%: 2.04 hours
Amp hours out at 15 min: 8.4 AH, 8.2 AH, 8.4 AH
Amp hours out at 1.2 hour (load V = 10.5V): 41.3 AH, 40.8 AH, 41.3 AH
Amp hours out at 1.54 hour (battery terminal B = 10.5V): 51.2AH, 50.8 AH, 51.2 AH
Amp hours out at 2.04 hour (SOC = 0): 68.2 AH, 67.9 AH, 68.2 AH

Method 1:
Load cables connected to battery A +post and battery A –post.
Interconnecting cables = 1.5 mOhms each including contact resistance. This what SmartGauge used.

Start of test, 15 Min, Time voltage reaches 10.5V
Load voltage: 11.3V, 11.2V, 1.13 hours
Battery terminal voltage: 11.7V, 11.6V, 1.5 hours

Results Battery A, Battery B, Battery C
Battery current at start of test: 39.9A, 31.9A, 28.1A
Battery current at 15 min: 38.4A, 32.3A, 29.3A
Battery current at 1 hour: 34.3, 33.1A, 32.6A
Current link A+ > B+: 60.1A
Current link B+ > C+: 28.1A
Current link A- > B-: 60.1A
Current link B- > C-: 28.1A
State of Charge at 15 min: 85%, 88%, 90%
SOC 1 hour: 45%, 52%, 57%
SOC load V = 10.5V: 38%, 46%, 51%
SOC battery terminal V = 10.5V: 22%, 30%, 34%
Time SOC = 0%: 1.92 hours
Amp hours out at 15 min: 9.8 AH, 8 AH, 7.2 AH
Amp hours out at 1.13 hour (load V = 10.5V): 41.3 AH, 36.8 AH, 34.5 AH
Amp hours out at 1.5 hour (battery terminal B = 10.5V): 52.3AH, 47.8 AH, 45.6 AH
Amp hours out at 1.9 hour (SOC = 0): 67.5 AH, 63.4 AH, 61.4 AH

Interconnecting cables 2/0 = .517 mOhms each including contact resistance.

Start of test, 15 Min, Time voltage reaches 10.5V
Battery terminal voltage: 11.8V, 11.7V, 1.54 hours

Results Battery A, Battery B, Battery C
Battery current at start of test: 35.8A, 32.8A, 31.4A
Battery current at 15 min: 35.2A, 33A, 31.9A
Battery current at 1 hour: 33.7A, 33.3A, 33.1A
Current link A+ > B+: 66.2A
Current link B+ > C+: 31.4A
Current link A- > B-: 64.2
Current link B- > C-: 31.4
State of Charge at 15 min: 86.7%, 87.9%, 88.5%
SOC 1 hour: 48.8%, 51.6%, 53%
Time SOC = 0%: 2 hours

LScamper

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Posted: 01/05/12 11:16pm Link  |  Print  |  Notify Moderator

KendallP wrote:

"Unless you have 3 or more batteries and must use either a buss bar or run all of the equal lengths of cable to a single point for each the positive and negative in order to comply with smartgauge method # 3."

For three batteries there is another way to get perfect balance. I mentioned it in my first post but nobody caught it. I'll call it method 2.5.

Method 2.5:

Connect as in method 2, load from A+ and C-.

Make link from battery B+ > C+ equal two times the resistance of link from A+ > B+.

Make link from battery A- > B- equal two times the resistance of link from B- > C-.

This will give perfect balance same as method 3. You can make cables different length or different size to get the correct resistance.

I don't think it is needed as method 2 is almost perfect. SOC never gets more than about 1% difference. Even method 1 with 2/0 cables never gets more than about 4% difference. Batteries that are not matched will make more difference I think.

KendallP

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Posted: 01/05/12 11:24pm Link  |  Print  |  Notify Moderator

LScamper wrote:

KendallP wrote:

"Unless you have 3 or more batteries and must use either a buss bar or run all of the equal lengths of cable to a single point for each the positive and negative in order to comply with smartgauge method # 3."

For three batteries there is another way to get perfect balance. I mentioned it in my first post but nobody caught it. I'll call it method 2.5.

Method 2.5:

Connect as in method 2, load from A+ and C-.

Make link from battery B+ > C+ equal two times the resistance of link from A+ > B+.

Make link from battery A- > B- equal two times the resistance of link from B- > C-.

This will give perfect balance same as method 3. You can make cables different length or different size to get the correct resistance.

I don't think it is needed as method 2 is almost perfect. SOC never gets more than about 1% difference. Even method 1 with 2/0 cables never gets more than about 4% difference. Batteries that are not matched will make more difference I think.

Yeah, that was always my point. The difference is so negligible, just KISS it. And again... the overwhelming majority of today's RVers will never ask their batteries to offer up anywhere NEAR 100A at all... let alone long enough to run the bank down past 50%. But for the few that do... you're still covered with method # 2.

But your thought IS a cool one, though. Not sure why no one thought of that sooner.
.

* This post was edited 01/06/12 09:08am by KendallP *

KendallP

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Posted: 01/06/12 10:09am Link  |  Print  |  Notify Moderator

LScamper wrote:

KendallP wrote:
“But with either interconnect size, with a 100A load, your model shows MUCH less of a difference for Method # 2 than smartgauge shows. It's certainly adequate for the job. “

I don’t remember SmartGauge giving any numbers on three battery system.

Method 2
Current at start of test.
Battery A, B, C, D
26.6A, 23.4A, 23.4A, 26.6A

For all intents and purposes, regarding method # 2, 3 and 4, the 4 battery model that smartgauge uses equates to your 3 battery model. Some will often say that when you have 3 or more batteries in parallel, you should avoid the simple method # 2 in favor of method # 3 (or 4.)

You can clearly see that the outer batteries react equally, but more importantly, the inner batteries react equally with each other. The inner 2 batteries in the smartgauge model should correspond directly to the single, inner battery in your model.

Also... the bigger spread numbers are only for the beginning of the discharge. If one really expects unbalanced cabling to have an undesired effect on the batteries, then one should expect this spread to be maintained throughout the discharge to 50% or so. Your data clearly shows that this does not happen. AND... the recharge will likely show it evening out even more.

However, I've always said that, if it's just as easy to do so, then method 3 or 4 should be employed. For me and my 220W inverter, I wasn't able to justify the added labor and cable costs. Not by a long shot. Your model here further justifies my tack. And I believe the vast majority of RVers today... will never see any appreciable difference in battery life by opting for method 3 or 4 over method 2... or... for that matter... even your, more simple, cable sizing method.
.

* This post was edited 01/06/12 10:22am by KendallP *

LScamper

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Posted: 01/06/12 11:14am Link  |  Print  |  Notify Moderator

KendallP wrote:

"The inner 2 batteries in the smartgauge model should correspond directly to the single, inner battery in your model."

Not exactly. The two inner batteries in parallel will have two times the AH capacity and half the resistance of the single battery in the three battery system.

LScamper

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Posted: 01/06/12 10:02am Link  |  Print  |  Notify Moderator

KendallP wrote:
“But with either interconnect size, with a 100A load, your model shows MUCH less of a difference for Method # 2 than smartgauge shows. It's certainly adequate for the job. “

I don’t remember SmartGauge giving any numbers on three battery system.

Here is a shorter list of four batteries for comparison:
Interconnect 1.5 mOhms as SmartGauge
Method 1
Current at start of test.
Battery A, B, C, D
34.8A, 26.1A, 20.8A, 18.3A

Method 2
Current at start of test.
Battery A, B, C, D
26.6A, 23.4A, 23.4A, 26.6A

The small difference from SmartGauge is because I used a little different battery resistance.

Interconnect .517 mOhms
Method 1
Current at start of test.
Battery A, B, C, D
28.8A, 25.5A, 23.4A, 22.3A

Method 2
Current at start of test.
Battery A, B, C, D
25.6A, 24.4A, 24.4A, 25.6A

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