[ausev] Fwd: Good sources to get started?

[Christopher Robison]

On Wed, 2007-11-28 at 11:07 -0600, Roy Holder wrote:

> yes, but 1 amp per mile per 1000 lbs at 120v is a calculation you can
> easily do in your head.

My main point here, primarily intended for those who are new to these
terms, is that amp-hours is water in a bucket. Amps is the rate of the
flow of water as you're pouring it in our out. "Amps per mile" is a
phrase that describes a rate of increase/decrease in current with
respect to distance -- an unusual measure that's not likely to be useful
in general and it's a reasonable guess it's not what you meant here.
"Amps" are a unit of current, not capacity. "1 amp-hour per mile" would
be the correct term, indicating a unit of electrical charge (an actual
count of electrons) transferred per unit of time. Again, sorry to be
picky here. I'm guessing Roy knows this but his typing is on
autopilot. :-)

Besides, I think 120Wh/mile per 1000lbs is just as easy.  :-P  It's an
interesting estimate; I would also be interested to see how it pans out
with other vehicles. Clearly aerodynamics makes a big difference, but
how much?

> The calculation you are refering to is called Peukert's Law, and it is
> actually different for every battery.  My batteries are Trojan 150's, I
> used a web page to calc peukerts ration based on the manufacturers specs.
> It actually calcs out to about 105 ah at the 1 hour rate.  I have taken the
> batts to 97 amps once and 80 to 85 amps three other times in the 18 months
> I have had them.

As Erik said, .57 is the usual rule of thumb used to calculate a 1 hour
rate from the standard 20 hour rate given by most manufacturers. Some
variation occurs between models and manufacturers; I meant that I was
surprised to see such a significant difference. Turns out I shouldn't
have been. I'll get to that in a moment. (Roy, you know the rest of this
so you can cut to the chase at the equations and conclusion below.
Pretty revealing, for me anyway.)

The Peukert equation is a means of predicting the change in energy
available from a battery as discharge current increases, due to heating
losses plus the effect of increased voltage sag at higher current, plus
undoubtedly some effect from other factors in the chemical reaction that
are beyond my education to understand. :-) These effects resulting from
the inherent characteristics of the battery further shorten the time as
current increases, before the battery reaches a "dead" threshold voltage
(1.75 for lead acid). This is beyond the decrease you'd expect if you
used a simple proportion with the higher discharge rate. So the exponent
in the equation is in effect the numerical representation of the nature
and behavior of the battery itself. 

Heating causes true capacity loss -- energy wasted. Voltage sag simply
lowers the *available* capacity at that current, and has no effect on
actual capacity. (In fact when you're driving and your battery is near
"dead" at 1.75Vpc, you can slow down, discharging at a continually
decreasing current to keep the voltage from sagging below the threshold,
in order to creep home. This allows you to use more of the capacity in
the battery.)

The less voltage sag seen in a battery, the closer it is to the ideal
Peukert exponent of 1.0. This is seen in high-end lithiums -- very low
sag, very low exponents.

The most convenient equation I've found to calculate the capacity
variation due to the Peukert effect is mentioned here, specifically the
version allowing easier translation from the 20-hour rate figure
provided by manufacturers: 

http://en.wikipedia.org/wiki/Peukert's_law

The article suggests that the typical range of exponent values seen in
lead acid batteries is 1.1 to 1.3. I had believed there was very little
difference between one lead acid battery and another, but I see I was
mistaken. Running the numbers showed a surprising difference (to me)
from one battery to another, depending on this exponent value. One could
guess this difference relates to the quality of the battery. Using their
second equation for the example exponent values of 1.1, 1.2 and 1.3, we
get multipliers of:

20/(((150*20)/150)^1.1) = .74
20/(((150*20)/150)^1.2) = .55
20/(((150*20)/150)^1.3) = .41

for the one hour rate.

These suggest a resulting capacity of 111, 82.5 and 61.4 amp-hours
respectively for these examples.

So assuming a high quality battery, a .57 multiplier is pretty
pessimistic. For an average quality battery, it's about right. Again I'm
surprised at the difference as I had just been going on the rule of
thumb without actually bothering to check it. But it's very good news
that you really get what you pay for.  Trojans are more expensive, but
the numbers, verified by your experience, suggest that they're worth it.


-- 
Christopher Robison
chris at ohmbre.org
http://ohmbre.org          <-- 1999 Isuzu Hombre + Z2K + Warp13!