# Lead Acid AH Capacity



## Jan (Oct 5, 2009)

Lead acid batteries are rated, as far as I have seen, at only 0,1C. The 100 Ahrs are only available if you draw 10 amps. If you increase the amps the Ahrs you get will drop significantly. And there is Peukert. Peukert really loves Lead acid batteries. Peukert will let the voltage sag if you increase the amps. Not a nice guy imho.

I can not confirm that you can only discharge to 50%. That seems a bit concervative to me.


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## rtwelder (Aug 11, 2010)

So are you saying discharging more than 50% is stil safe on lead acid batteries?

Won't sulfation start and slowly/gradually destroy the lead plates?


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## rtwelder (Aug 11, 2010)

or maybe I can just ask,

if I get 1200WH out of a 12V 100AH, would the battery be at 0% charge?

Please enlighten me on this.


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## Jan (Oct 5, 2009)

Every charge and discharge will dammage your lead acid batteries. That's why you can only do it a couple of hundred times. If you discharge them less, or charge them lower, they will last longer. But not forever. All I want to say, it's not they'll die on you as soon you'll dischage them to 20% or so. They only last even less longer. Notice the 'even'.

The original question about the Whrs you might get, is still unaswered, because it all depends on the amount of amps you're planning to draw from them.


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## major (Apr 4, 2008)

rtwelder said:


> or maybe I can just ask,
> 
> if I get 1200WH out of a 12V 100AH, would the battery be at 0% charge?
> 
> Please enlighten me on this.


Hi rtw,

The rated Ah from the battery manufacturer is typically given for a discharge down to a voltage level of like 10.8 volts for a 12 volt battery. So you can get all the Ah from the battery without it going completely flat to zero volts. But that rated Ah number is typically for a 20 hour rate. So any faster discharge will yield fewer Ah, via Peukert.

Read this: http://www.diyelectriccar.com/forums/showthread.php?t=669&redir_from=668 Section: http://www.diyelectriccar.com/forums/showthread.php?t=11709 

Regards,

major


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## Dink (Jun 3, 2010)

Jan said:


> Lead acid batteries are rated, as far as I have seen, at only 0,1C. The 100 Ahrs are only available if you draw 10 amps. If you increase the amps the Ahrs you get will drop significantly.


What is this C rate? I've seen different discharge rates for different types of batteries, and different rates for the same type?


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## Jan (Oct 5, 2009)

Dink said:


> What is this C rate? I've seen different discharge rates for different types of batteries, and different rates for the same type?


Every battery's performance depends of the load. The load can be translated to the amount of amps that are drawn from the battery. And the performance is the voltage the battery gives, and the amount of energy it can deliver.

So a given discharge curve is only valid at a certain current level. This current leven (the amount of amps) is usualy given with the constant C. Where C stands for the Ah the cell is normaly capable of. Because a 100AH li-ion cells acts the same as a 50Ah, or 200AH cell, if you fill in these ahrs for C.

Li-ion is usaly rated at 1C. And you'll mostly notice that at 2C the discharge curve isn't much shorter.

Lead acid cells are usaly rated at 0.1C. And you will get a lot less at 1C. Major states with his 20hrs, that 0.05C is also often used. That are very little amps.


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## dexion (Aug 22, 2009)

the rated capacity is usually 20 hour rate. Ie discharging it for 20 hours at a low fixed rate (5amps) will result in 100ah. These numbers are fairly useless to an ev except as a starting place to then get the correct ev capacity.

So, a typical discharge rate for an ev is 1c (or 1 hour full to flat.)
Giving that is your discharge rate (100amps) you may get about 45ah out of the cells you need to further restrict discharge (ah out of the pack) to get useful life from the cells. Say pull 25ah out of the cells at a 1C rate would give a useful life in the hundreds of cycles depending on the quality of the cells and the charger, rest time etc.

If you have a fairly light ev and drive conservatively as well on flat roads at 45mpg in the correct gear (there are a few variables  ) that means around 25 miles on a 144V pack of 12 12volt cells with a 200ish cycle life. YMMV


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## major (Apr 4, 2008)

Dink said:


> What is this C rate? I've seen different discharge rates for different types of batteries, and different rates for the same type?


Good question, Dink. It is a way to normalize discharge rates (current or amps) so batteries (or cells) of different capacities can be compared.

As used in this context, C = Capacity. Which is Ampere hours or Ah. An ampere hour is a unit of charge. 1 Ah = 3600 As (Ampere seconds). 1 As = 1 coulomb. A coulomb is the basic unit of electric charge and equal to 6.242×10^18 electrons. Battery people like to use Ampere hours.

This capacity (C) is the amount of charge the battery can store and deliver under specified conditions. In the battery industry, a condition of 20 hours has been the established for the standard time to measure this discharge from the battery or cell. So the Ampere hour rating specified for a battery is given and holds true if you discharge it at a constant rate and it takes 20 hours to discharge it.

Now we a talking about charge and time. And current (Amperes or A) enter into the mix. Ampere is defined as one coulomb per second, which is a rate of charge. We talked about C as being the charge, so C rate is a rate of charge, so is current. C rate is just a way for battery guys to normalize current. Instead of talking about their battery at 100 Amps, they call it 5C (for a 20 Ah battery as an example). 

So whenever you see a charge rate (C rate) on a battery just take that number and multiple it times C (capacity) for that particular battery and you get the current in Amperes. A 5C rate for a 100 Ah battery is 500 A. A 5C rate for a 200 Ah battery is 1000 A. A C/20 rate for a 100 Ah battery is 5 A. 

Hope that explains it for ya 

At least that it how I see it.

major


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## Jan (Oct 5, 2009)

major said:


> Hope that explains it for ya


With three times, the same issue, in different words explained, should be enough.

Another question: Peukert's law only addresses the capacity of the battery in Ahrs. As far as I know. The voltage sag, which is also a serious issue, especially for lead acid, isn't part of his law? Isn't it?


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## major (Apr 4, 2008)

Jan said:


> Another question: Peukert's law only addresses the capacity of the battery in Ahrs. As far as I know. The voltage sag, which is also a serious issue, especially for lead acid, isn't part of his law? Isn't it?


No Jan, Peukert says nothing about voltage. Sag is typically associated with the internal resistance of the battery or cell. And sag is the voltage drop from the open circuit voltage of the battery due to current times the internal resistance. So Vb = Voc - I * R, where Vb is actual voltage at the battery terminals, Voc is the battery voltage open circuit (zero current), I is current, and R is the internal resistance. So sag = I * R.

With most batteries or cells, there will also be a change in voltage associated with the State of Charge (SOC). This isn't accounted for in sag. But may be accounted for in the open circuit voltage. And this change in voltage w/r/t SOC is different for different chemistries. A bit difficult to quantify.

This I suppose relates back to the actual energy you can actually store and get out of a battery  And that is difficult to determine unless you actually test the darn thing under the exact conditions you intend to use it. So for most purposes, we use the Peurkert adjusted charge times the nominal voltage. It will be optimistic. And low resistance batteries will do better w/r/t energy. The low resistance battery or cell will also be the better w/r/t power.

If you know the internal resistance or sag for your battery you can use that and calculate the actual battery voltage under load. Then use that actual voltage instead of the nominal voltage to figure your energy.

major


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## tomofreno (Mar 3, 2009)

One of the best sources I found for the Peukert Effect is here:
http://www.smartgauge.co.uk/peukert2.html
There is also a calculator there to estimate battery capacity for different discharge currents. I used the equations from the reference to create the attached graph for some common lead acid batts.
View attachment Peukert graph.pdf


Peukert's equation does take voltage sag into account since it is basically a statement of conservation of energy. The reason the capacity is lower at higher discharge currents is because of higher energy dissipation internally in the battery at higher currents and power. This is also what causes the voltage sag. You could of course pull out just as much charge as you put in (conservation of charge), but not at as high a voltage level, and energy per charge, as you could at lower discharge currents, so you hit the 10.8V major mentioned with less charge removed from the battery at higher discharge currents compared to lower currents. You get less energy per charge at higher currents because you are dissipating more energy internally in the battery.


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## Jan (Oct 5, 2009)

major said:


> w/r/t


Ah, ok. Voltage sag is dependent on internal resistance. And w/r/t means with respect to? Those shortcuts are sometimes very difficult for a non native english speaker, Major.

Do you know if the IR is a constant? Or is it also dependent on load and/or SOC?


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## major (Apr 4, 2008)

Jan said:


> Ah, ok. Voltage sag is dependent on internal resistance. And w/r/t means with respect to? Those shortcuts are sometimes very difficult for a non native english speaker, Major.
> 
> Do you know if the IR is a constant? Or is it also dependent on load and/or SOC?


w/r/t w/r/t  yes, with respect to or with regards to.

Internal resistance (maybe use Ri instead of IR (IR is often short for amps * resistance))....anyway, internal resistance is typically taken as a constant for a particular battery or cell. However it does actually vary w/r/t SOC (but not a lot, IMO) and w/r/t temperature and also, maybe, w/r/t battery age. I don't think internal resistance varies w/r/t load. So the same ohmic value at 100A and 300A, but obviously a higher voltage drop, or sag at 300A. And internal resistance has almost always tested higher than the manufacturer's specification. I think this is because I load test it and they use some standardized impedance testing machine.

BTW, IMO = in my opinion.

BTW = by the way.

 = smiley face.

Regards, 

major


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## Jan (Oct 5, 2009)

My first stupid thought was: Wtf, watts divided by resistance by time?


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## tomofreno (Mar 3, 2009)

Oh, I misunderstood your question. Yes, as major said if you want to know the actual energy you will get from a battery at a given discharge current you need to know the (sagged) voltage at that current level, since energy per charge is qV, where q is the charge on an electron and V is the actual terminal voltage as the electron flows out.


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## major (Apr 4, 2008)

Jan said:


> watts divided by resistance by time?


That's a good one  Just one of those abbreviated days for me.


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## EVfun (Mar 14, 2010)

Jan said:


> Another question: Peukert's law only addresses the capacity of the battery in Ahrs. As far as I know. The voltage sag, which is also a serious issue, especially for lead acid, isn't part of his law? Isn't it?


Well... Peukert defined "discharged" as 1.75 volts per cell for his equation. 

Pk = I^p * T

Where Pk is the Peukert's capacity (the capacity at the 1 amp rate.)
I is amps
p is Peukert's exponent 
T is time, either minutes or hours can be used and the resulting Pk will be either the cell amp minutes or amp hours. 

Golf cart lead is typically capable if providing about 50% the amp hours at EV rates, compared to the 20 hour rate.

Good AGM batteries are capable of providing about 60% of their stated 20 hour capacity, but will last a lot longer if you limit it to about 40%. 

The amp hours you don't get aren't "lost." They are still available, but only at a slower rate. This, combined with the extremely variable discharge rate of an EV, make it a little tough to define state of charge.


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## yarross (Jan 7, 2009)

Jan said:


> Lead acid batteries are rated, as far as I have seen, at only 0,1C. The 100 Ahrs are only available if you draw 10 amps. If you increase the amps the Ahrs you get will drop significantly. And there is Peukert. Peukert really loves Lead acid batteries. Peukert will let the voltage sag if you increase the amps. Not a nice guy imho


Uhm Peukert effect is not related to voltage sag. It's related to loss of charge (due to side reactions on electrodes) under high loads.


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## EVfun (Mar 14, 2010)

yarross said:


> Uhm Peukert effect is not related to voltage sag. It's related to loss of charge (due to side reactions on electrodes) under high loads.


This is a common misunderstanding. The capacity that isn't available at the higher discharge rate isn't lost. I challenge you to find out for yourself, like I did back in 1999. 

I did 2 tests that are easily repeatable:

1. Discharge your battery at a high rate while recording the amp hours removed. Then recharge your battery at any rate you want while recording the amp hours returned. You will find out that the returned amp hours are only about 5 to 10 percent higher than what you removed. I did this test by discharging Optimas at 185 amps for 10 minutes and finding that they where fully charged after pushing about 33 amp hours back into them. 

2. Discharge your battery at a high rate until the voltage drops to 1.75 volts per cell. Remove the load and connect a 20 hour rate load to continue discharging. You will see the battery voltage is fine and the battery will put out the total number of amp hours expected before the sag gets back to the point defined as dead (1.75 vpc.) This is best done with a battery with a few cycles on it because a brand new battery's capacity is generally slightly less than rating (they grow in capacity for the first few cycles.)

These tests are going to provide clear enough results that checking a current shunt every 3 minutes and using that to calculate amp hours in 0.05 hour increments will provide the answer. It won't be as accurate as using logging software, but since the difference between "loss" and "not available" is about a 50% difference in amp hours it will still be clear.


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## yarross (Jan 7, 2009)

EVfun said:


> This is a common misunderstanding. The capacity that isn't available at the higher discharge rate isn't lost.


I wrote "charge". Still support this statement. Charge is being removed from electrodes faster than one could expect just by integrating load current. This charge loss is not recoverable just by releasing accelerator, but this does not mean that the battery has been badly damaged and has only 50-60% of rated capacity.



EVfun said:


> I challenge you to find out for yourself, like I did back in 1999.
> 
> I did 2 tests that are easily repeatable:


Not related to charge loss because of side reactions.


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## EVfun (Mar 14, 2010)

yarross said:


> Not related to charge loss because of side reactions.


If it is a "charge loss" then how come you don't have to put it back when recharging? How come you can still get the capacity that supposedly lost to drive a load if you just use a smaller load? The reduction in available capacity at high discharge rates that was described by W. Peukert in 1897 is not loss of charge. It simply isn't available without excessive voltage sag at the desired discharge rate.


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## yarross (Jan 7, 2009)

EVfun said:


> If it is a "charge loss" then how come you don't have to put it back when recharging? How come you can still get the capacity that supposedly lost to drive a load if you just use a smaller load? The reduction in available capacity at high discharge rates that was described by W. Peukert in 1897 is not loss of charge. It simply isn't available without excessive voltage sag at the desired discharge rate.


I can say just my experience is different. I observe ~50% deficit. Maybe Optimas have lower load per area of active mass, I don't know.
Compare this to Li cells. They have negligible Peukert effect (1.03 for TS cells), but still can suffer from voltage sag. Quite different behaviour.
And last question - did you correct the cutoff voltage for sag? Maybe you haven't discharged yout batts to the same DoD when testing with different loads.


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## Sunking (Aug 10, 2009)

EVfun said:


> 1. Discharge your battery at a high rate while recording the amp hours removed. Then recharge your battery at any rate you want while recording the amp hours returned. You will find out that the returned amp hours are only about 5 to 10 percent higher than what you removed.


That only holds true for AGM batteries like the Optima you used in the test. Flooded lead acid batteries will require around 20 to 30% more.

Secondly I would have to ask is how did you determine when you reached 100% fully charged, and what algorithm did you use to recharge the batteries?


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## EVfun (Mar 14, 2010)

Sunking said:


> That only holds true for AGM batteries like the Optima you used in the test. Flooded lead acid batteries will require around 20 to 30% more.
> 
> Secondly I would have to ask is how did you determine when you reached 100% fully charged, and what algorithm did you use to recharge the batteries?


Flooded cells are less efficient than AGM batteries. My chosen charge algorithm when testing Optimas was generally 10 amps to 15.0 volts (it may have been 14.x where x was at least 7 - these tests where 12 years ago) and hold until the current dropped to 1 amp. I also did a few tests using a lower charge current. 

Charge time with a Lester charger on my one golf cart pack made it pretty clear that I was only replacing used amp hours plus a little too. I was not having to dump 200+ amp hours into them after running the pack down low. I'm sure the "plus a little" in this case was more than with the Optimas but I never followed the end of charge profile that closely. I believe the e-meter had self adjusted the CEF (charge efficiency factor) to 80-something percent but that was years ago.


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## major (Apr 4, 2008)

Just for the hell of it, here a discharge test I ran a while back.










This was a Transcontinental 75DT-525, 525 CCA, 80 RC, Pb-Acid, maintenance free, about 50 Ah rated, I think. This is just an example of how the voltage sags and how Ah and Wh are accumulated. It was run into a constant resistance load. So the Ah and Wh were calculated from the spreadsheet recording of volts, amps and time. Too bad I didn't do the same thing for the recharge  Maybe next time.

major


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## dtbaker (Jan 5, 2008)

all I can tell you for sure is that my current pack of TS-100ah rated cells produce better range than my original pack of 8v floodies rated at 185ah under 'normal' driving loads. There is also no discernable sag under load, and no sag as DOD progresses as there was with lead, and the car's performance (accel, braking, handling) are much improved with dropping 500#.


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## yarross (Jan 7, 2009)

Sunking said:


> That only holds true for AGM batteries like the Optima you used in the test.


Not sure. Got data from specs and calculated Peukert quantities for Optima Yellowtop D34 (AGM): exponent 1.138, capacity 76.30Ah. Quite high, although lower than T-105 (1.251, 435Ah) (I have used data around 1C current for curve fitting).


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