Man, this Peukert stuff is still confusing people! :-)
Bill Dube wrote:
The internal resistance of a battery does not change its Ahr
capacity. Not even a tiny amount.
That is correct. AMPhours are not affected by internal resistance.
WATThours are, because internal resistance causes a voltage drop.
Jan Steinman wrote:
No, but it *does* determine how much of the battery capacity gets
turned into waste heat, no? Assuming voltage stays fairly constant,
the power dissipated by internal resistance rises with the square
of the current drawn. That's energy that isn't going into moving
you forward.
That's also correct. The "perfect" cell voltage may still be 3.2v, but
the terminal voltage drops due to internal resistance. For example, you
get 3.2v at 0 amps, and 3.0v at 100 amps. Then that cell has an internal
resistance R = (3.2v-3.0v)/100a = 0.002 ohms. At 100 amps, about P =
0.2v x 100a = 20 watts of heat is being produced inside the cell.
Just be aware that this is not a true resistor, and not constant. It is
the combined net effect of all the internal processes; mechanical,
chemical, state of charge, temperature, aging, etc.
David Nelson via EV wrote:
No. Don't confuse charge capacity with energy capacity. None of the
charge capacity gets turned into heat, only some of the energy
capacity gets turned into heat. That is why what Bill said is correct
and why I monitor SOC (state of charge) of my batteries and not SOE
(state of energy) to determine how much I have left in my battery.
I think Jan has it right. He's not confusing amphours with watthours.
You need to decide what you want to know; then pick your measurements to
give you that information. If you want to know how many amphours are in
the battery, then measure amphours.
But suppose you want to know how far you can go at a given speed (i.e. a
given load). That is a much more complex calculation, because it
involves many factors besides amphours. Peukert is just a "short cut" to
make a rough guess that is better than using amphours alone.
Also, to assume voltage stays fairly constant with changing current is
way off the mark of what happens. In fact, it is that very change in
voltage at the terminals that accounts for the energy dissipated as
heat internally in the cell since the current throughout the circuit
is the same.
Yes. And, the voltage sag is what causes the apparent reduction in range
at higher currents. The actual capacity of the battery hasn't changed;
but the *useable* capacity decreases at higher currents.
Is this discussion still useful, or just putting everyone to sleep?
--
The storage battery is one of those peculiar things which appeals to
the imagination, and no more perfect thing could be desired by stock
swindlers. Just as soon as a man gets working on the secondary battery,
it brings out his latent capacity for lying. -- Thomas A. Edison
--
Lee Hart's EV projects are at http://www.sunrise-ev.com/LeesEVs.htm
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