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Drake,<br>
<br>
The spreadsheet name was Peukert.xls and all it computed was the
Peukert effect - greater discharge current <-> smaller battery
capacity. Peukert's equation is based on a constant discharge current
from fully charged to fully discharged (10.5 volts at the current in
question). The misunderstanding (almost urban myth) comes when
considering what happens when a battery is subject to partial (or even
complete) discharge at a high current. The amphours accrued at a high
discharge rate actually have little effect on the total amphours that
can be delivered at a low discharge rate.<br>
<br>
Example:<br>
Brand XYZ L-16 battery battery has a capacity of 400 Ah @ C/20 or 300
Ah @ C/6.<br>
There's a 25% difference between these two capacities.<br>
>From a full battery, discharge at 50 amps until the voltage is 10.5
volts, that takes 300 Ah. <br>
The battery is functionally dead, 100% DOD, if we need 50 amps at more
than 10.5 volts.<br>
But at 20 amps the battery is not dead. The battery will still deliver
another 80 Ah for a total of 380 Ah (5% less than 400 Ah @ C20) before
the voltage gets down to 10.5 volts.<br>
<br>
The 380 Ah and 5% numbers here are just for example purposes; if you
did this test to a L-16 the total would probably be higher than 380 Ah,
but you certainly would get less than 400 Ah.<br>
<br>
Kent Osterberg<br>
Blue Mountain Solar, Inc<br>
<br>
<br>
Drake Chamberlin wrote:
<blockquote
cite="mid:20100119181758.C2AEB5F574@friskymail-a1.g.dreamhost.com"
type="cite">Kent, <br>
<br>
At 12:34 PM 1/19/2010, you wrote:<br>
<blockquote type="cite" class="cite" cite="">In the tests performed
by Dennis
Doerffel and Suleiman Abu Sharkh it is was in the five to ten percent
range.</blockquote>
<br>
In the first Excel spreadsheet, posted by Julie Haugh, the loss rates
seem a lot higher than five to ten percent. Using 220 AH battery,
for example, I used a 50 amp discharge rate. The result was that there
would be about 140 AH available. That would amount to a 37% loss of
capacity. <br>
<br>
Is it your thinking that this high result is due to the low terminal
voltage at that point, and that the remaining AH would be available
after
the battery terminal voltage had a chance to recover? Is the
attached spreadsheet just incorrect?<br>
<br>
Thanks,<br>
<br>
<br>
<x-sigsep></x-sigsep>
<p><font color="#0000ff">Drake Chamberlin<br>
Athens Electric<br>
OH License 44810<br>
CO License 3773<br>
NABCEP </font><font color="#0000ff" size="1">TM
</font><font color="#0000ff">Certified PV Installer <br>
Office - 740-448-7328<br>
Mobile - 740-856-9648 </font></p>
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