Re: Making Battery Choices (more EV idea

[bdube@boulder.nist.gov (Bill Dube')]

>> There is still one thing that I've not heard an Optima owner talk about.
>> Battery efficiency in EV service. The Peukert's losses are quite a bit less
>> than with the flooded batteries. This means that the capacity of the AGM is
>> less for a given range (the Peukert's capacity of an Optima is around 72).
>> This means that I can have respectable charging time while only charging
>> off a 120v 15amp circut.
>
>No; it doesn't quite work like this.

    Actually, it does work like that.
>
>On discharge: Peukert effects mean you can't remove as many full amp-hrs at
>high currents as you can at low currents. But the excess amp-hours are still
>in the battery. Suppose you had a battery rated 100 amp-hrs (20-hr rate), and
>50 amp-hrs (1-hr rate). You pull 50 amp-hrs out of it at the 1-hr rate; 1 hour
>later it is "dead" (10.5v under load). But, you can keep discharging it for
>another 19 hours at a much lower rate. At the end of 20 hours total it will
>have delivered another 50 amp-hrs for a total of 100 amp-hrs!

    I'm going to roll up my pants. It's too late to save my shoes.  

    Yes, you can reduce the load and get some addidional amp-hrs. No, you
won't get the full capacity. You will not get anywhere near the full
capacity at the reduced load.

    The Peukert equation models the internal workings of the battery quite
well. Most of the loss of capacity is due to internal resistance losses.
(Some is due to diffusion rate limitations, but this shows up "mostly" as
resistive losses.) The resistive losses are equal to I^2 x R. Thus, high
currents cause much greater amount of the energy to be converted to heating
the battery. This energy is lost forever and will not appear at the battery
terminals at some later time. The reduction in voltage across the terminals
due to diffusion limitations results in energy reduction during discharge.
This energy is also never recovered. Indeed, one can wait a bit for the
chemistry to settle out and draw a few more amp-hrs, but the amp-hrs drawn
while the voltage was reduced (resulting in fewer watt-hrs) are gone forever.

    A low Peukert exponent is, indeed, a good indication that the battery
has a good charge efficiency. If you can pull out energy efficiently, you
can put in energy efficiently. At a selected amp-hr capacity, the lower the
Puekert exponent, the higher the charge efficiency. (The battery capacity,
in amp-hrs, strongly influences the Peukert exponent, so, in terms of charge
efficiency, it is only useful for camparision of batteries that have about
the same amp-hr capacity.) 
>
>Batteries with a bad (high) Peukert number have the same amp-hr efficiency,
>it's just that the voltage sag becomes so bad that they are useless at the
>desired current. The loss of power efficiency comes from the voltage drop
>(internal resistance); not from the Peukert value (amp-hrs).

    For a given amp-hr capacity (C/20) the Peukert exponent is a perfect
indicator of battery efficiency. All the factors (except C/20 capacity) that
increase the Peukert exponent decrease the charge efficiency.
>
>On Charge: Lead-acid battery efficiency is about the same for all types;
>flooded, gels, and AGMs. Mainly, this is because we don't charge at high
>enough rates for Peukert effects or internal resistance to make any
>difference. Sub-hour charging; yes. Many-hour charging; ignore it.

    The charge efficiency is the ratio of watt-hrs out over watt-hrs in. The
entire cycle is included in the calculation. Indeed, very low charge rates
improve the efficiency of that portion of the cycle, but this is the Peukert
equation hard at work in the reverse fashion. The efficiency of both
protions of the cycle matter in the calculation of charge efficiency and
both are well-modeled by Peukert's equation. For a given application and
charge rate, the Peukert's exponent is an excellent judge of the charge
efficiency.

    If we switch to Coulombic efficiency (amp-hrs out / amp-hrs in) the
Peukert equation does an even better job of modeling the system. Amp-hrs get
eaten up in I^2 R losses just the same as Watt-hrs do. The reduction in
voltage due to diffusion limitations does not strongly influence the
Coulombic efficiency, (as it does the charge efficiency.)

    You can more easily measure Coulombic efficiency, but what you really
care about is charge efficiency.
             _   /|             
             \'o.O'              Bill Dube'
           =(___)=           bdube@boulder.nist.gov
              U