Very straight forward, Lee.
-----Original Message----- From: EV [mailto:ev-boun...@lists.evdl.org] On Behalf Of Lee Hart via EV Sent: 08 May, 2014 6:56 AM To: Electric Vehicle Discussion List Subject: Re: [EVDL] Peukert # with Lithium Batteries 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 _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA) _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA)
