[RE-wrenches] Fwd: discharging Rolls batteries

Sun Jan 17 12:10:14 PST 2010

Julie explains the correct application of Peukert's Law.  However she 
does not explain why the battery discharge trials all end at the same 
voltage. The SOC of a battery on high current will be much higher 
than that of a battery at low current at this chosen voltage.

  If the high rate discharge ends, then the battery voltage will rise, 
and a more gentle discharge will be able to take place for a while 
before again reaching the same 'end of discharge' voltage.  I have 
not measured this (and would not do it with my own battery of course) 
but it does make me wonder about Peukert.

Julie does state that amphours are lost during discharge though, 
which is new to the discussion here.

I don't get any messages from the list at all any more, but Michael 
kept me in the loop with this one from Julie:

At 11:23 -0800 17/1/10, Michael Welch wrote:
>Forward from non-member Julie Haugh:
>- - - - -
>
>Hugh Piggot writes:
>>  According to Peukert's law (and manufacturers' data bears this out
>>  well) the capacity of a battery depends on the rate of discharge.
>>  Everyone who knows anything about batteries knows that the capacity
>>  is specified at a certain rate whether it be C20 for a 20 hour
>>  discharge or whatever. And the capacity at 100 hours (C100) is about
>>  33% higher than the capacity at the 20 hour rate. If you look at the
>>  way this is measured though, it is based on running the battery down
>>  to a chosen 'discharge limit' voltage. And I have not heard anything
>>  from Wrenches nor seen anything in the literature to suggest that the
>>  battery discharged in 20 hours has actually lost any amphours
>>  compared to the 100 hours one. So it appears to me that if you give
>>  it a rest and then start discharging it again, but now at the 100
>>  hour rate you could still get another 33% extra capacity.
>
>The amp-hours that are not produced at the higher discharge rates
>are lost until the next charge / discharge cycle.  You can't get them
>back by discharging at a lower rate after the battery is fully
>discharged.
>You can increase the remaining capacity by lowering the discharge rate,
>but that only applies to any remaining capacity, not to the energy that
>was removed.
>
>The proper way to calculate remaining capacity is to add up all of
>the Peukert-corrected amp-hours that are removed, using the Peukert
>amp-hour capacity.  Unlike regular amp-hour ratings (load for X number
>of hours that produces a given terminal voltage), Peukert is calculated
>using a 1 amp load, and then the number of hours it takes to produce
>some voltage.  Here's a Peukert calculator --
>
>http://www.smartgauge.co.uk/calcs/peukert.xls
>
>(There are other calculators on the web, but many of them seem to
>confuse "rated amp-hours" with Peukert-corrected amp-hours, or expect
>you to calculate the Peukert capacity to start with.  This spreadsheet
>allows you to work with capacity the way you normally do.)
>
>To give an example, consider a T-105 battery which is typically
>rated around 225 amp-hours at the 20 hour rate.  If it is an FLA
>battery, a good value for Peukert's exponent is 1.3.  Fill in those
>3 boxes.  To calculate the Peukert capacity, use the spreadsheet above,
>then in the pink "Discharge Rate" box, enter "1".  Under "Total Amp
>Hours Available", you'll have the Peukert capacity -- 465 amp-hours.
>That is, if you discharged the battery at a constant rate of 1 amp,
>at the end of 465 hours, the battery would be dead (ignoring all
>of the self-discharge that would have happened in 465 hours).
>
>Now, in the same pink box, enter 225 / 20 = 11.25 amps as the
>discharge rate.  You'll get 20 hours as the discharge time and
>225 amp-hours as the number of amp-hours that are available.
>
>To see an example of why this matters, consider a 112.5 amp load
>for 6 minutes (C/2 at a 10% duty cycle).  Enter "112.5" in the
>same pink discharge rate box and you see that the "hours" drops
>from 20, to 1 rather than 2.  Those "Peukert corrected amps" is
>the amount of capacity that the battery's chemistry "thinks" has
>been removed, and it can't be restored without recharging the
>batteries.
>
>Thus, you could discharge for 20 hours with a C/20 load at a 100%
>duty cycle, or 10 hours with a C/2 load at a 10% duty cycle.  If
>you can control load and duty cycle, go for higher duty cycles
>at lower loads.  Even though watts-hours per hour remains
>constant (100 watts for 1 hour, versus 1000 watts for 0.1 hour),
>as you can see, one solution allows you to run for 20 hours,
>while the other only allows you to run for 10 hours.
>
>To use the bank teller analogy, when you ask the bank teller to
>give you more money faster, they get nervous and some of the
>money falls onto the floor and they have to get more out of the
>drawer.  They aren't allowed to stop and pick the money up
>until the next day when the cash drawer is filled up again.
>
>This URL will allow you to calculate the Peukert exponent if
>you have two different discharge rates --
>
>http://www.smartgauge.co.uk/calcs/peukert_2.xls
>
>If you fill that in for the T-105, you'd see that the Peukert
>exponent is about 1.3 -- a T-105 will (typically) deliver 225Ah
>at the C/20 rate, or 75 Amps for 105 minutes.  Enter 225 for C1,
>1.75 for R2 (105 minutes / 60), and 156 for C2 (1.75 hours x 75
>amps = 131Ah).  The answer I get is 1.29
>
>>  Well now. I don't expect to get away with saying that. But why not?
>>  I can't find any evidence that it is not true.
>
>This will be covered whenever the discussion of Peukert happens.
>
>The short answer is that the chemical reactions aren't always
>productive, in the sense that the electrons go where you want
>them to go -- out the battery posts and into your loads.
>--
>Julie Haugh
>Senior Design Engineer
>greenHouse Computers, LLC // jfh at greenhousepc.com // greenHousePC on
>Skype
>
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-- 
Hugh Piggott

Scoraig Wind Electric
Scotland
http://www.scoraigwind.co.uk