> From: David Nelson via EV <[email protected]>
>
> On Wed, May 7, 2014 at 11:06 PM, Jan Steinman via EV <[email protected]>
> wrote:
>>> From: Bill Dube <[email protected]>
>>>
>>> The internal resistance of a battery does not change it's Ahr
>>> capacity. Not even a tiny amount.
>>
>> 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.
>
> 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.
Other than for pedantic interest, what's the difference? If some of the
battery's capacity gets turned into heat, it still isn't available to the
wheels, no?
Perhaps you can give us a mini-tutorial on the difference between SOC and SOE.
Because this EE is confused.
> With LiFePO4
> cells, however, the warmer they are the more energy efficient they
> become so it actually doesn't affect range as much as you might first
> think.
I'm getting more and more confuseder by the minute!
You just said the batteries had 100% charge/discharge efficient. Now you say
heating from internal resistance makes them even more efficient.
When they heat up to 110% efficiency, can't we just eliminate charging
altogether? :-)
> Also, to assume voltage stays fairly constant with changing current is
> way off the mark of what happens.
What I meant by "fairly constant" is that the terminal voltage changes by
millivolts as the current through the internal resistance goes from 0 to a
couple hundred amps. So yea, that change in voltage times the current times
time is the energy "lost" to heat.
----------------
:::: Jan Steinman, {link:"http://www.EcoReality.org"; textDesc:"EcoReality
Co-op" :link} ::::
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