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Re: (ET) Fully charged pack voltage




Ohhh, I was forgetting about the affect the concentration of sulfate and hydrogen ions would have on the Gibbs free energy of the reaction (the voltage any battery produces is the change in gibbs free energy of the reactions that take place divided by the amount of charge that has to move for that much energy to be transferred). As you keep charging above the resting SOC voltage, you should be producing more H+ and SO42- ions, which will increase the Gibbs free energy of the reaction, thereby increasing the voltage of the pack. Those hydrogen and sulfate ions gradually recombine, bringing the gibbs free energy (and therefore voltage) back down to what it should be.
        Ok, I think I'm all set now....

Thanks,
Mike

--

--------------------------------------------------------------
Michael S. Briggs, PhD
UNH Physics Department
(603) 862-2828
---------------------------------------------------------------

On Thu, 23 Apr 2009, Tim Humphrey wrote:

The act of charging causes a chemical reaction between the plates and
the electrolyte. You already know how that works.

The resultant change occurs where the plates and electrolyte are in
contact with one another. The time delay for the surface charge to
dissipate is a result of the time it takes the chemical change to
absorb deeper into the plates.


Stay Charged
Hump




-----Original Message-----
From: elec-trak-bounces cosmos phy tufts edu
[mailto:elec-trak-bounces cosmos phy tufts edu] On Behalf Of Michael S
Briggs
Sent: Thursday, April 23, 2009 10:20 AM
To: SteveS
Cc: et
Subject: Re: (ET) Fully charged pack voltage


Oh - I'm still perplexed about *why* it takes so long for the pack
voltage
to drop back down to its resting voltage after charging. I'm a
physicist,
I feel compelled to understand why things are the way they are. :)

Mike

--

--------------------------------------------------------------
Michael S. Briggs, PhD
UNH Physics Department
(603) 862-2828
---------------------------------------------------------------

On Thu, 23 Apr 2009, SteveS wrote:

It actually can take up to 24 hours to get to resting voltage. And I
don't
think you can 'hurry' it up by momentarily applying a load.

Here's a good reference:


http://www.trojanbattery.com/productLiterature/documents/UsersGuide_0109
_English_001.pdf

- SteveS

Michael S Briggs wrote:

 On Thu, 23 Apr 2009, John J Casey wrote:

 In my experience, the onboard ge charger takes my newish T-105's
up to
 44-45v over an 18 hour cycle, the last 3-5 hrs between 44.4 and
45v for
 cell
 equalization.  End the charge cycle and in a short time the
voltage
 settles
 back to 39v.Jack

 So, the fully charged pack then has a voltage of 39 V (since that is
what
 it settles to after charging).
     One thing that is troubling me - what does it take a while for
the
 battery's voltage to gradually drop from the elevated charging
voltage
 down to its resting voltage. I know that while charging the voltage
 applied will be higher than the battery's voltage due to the voltage
drop
 across the internal resistance of the battery (batteries) - but once
the
 charger turns off, I would think that the battery pack voltage would
drop
 rather immediately down to its resting voltage (the potential
difference
 that the chemical reactions taking place at the electrodes produce).
     BTW, this has some interesting information about how the
internal
 resistance of the electrolyte changes with state of charge, which
gives
 lead acid batteries their unique voltage vs. state of charge curves:

http://www.arttec.net/Solar_Mower/4_Electrical/Battery%20Charging.pdf
     The main thing I'm trying to figure out right now is what to set
my
 Landis charge controller to to make sure that the pack gets fully
charged
 (and doesn't take 3 days to get there). It would seem that the
default 38V
 setting is too low, since the pack should be higher than that when
fully
 charged.

 Thanks,
 Mike

 On Apr 23, 2009, at 8:13 AM, Michael S Briggs wrote:


       I just got a nice new pack of Trojan T-605s (slightly lower
       capacity than T-105s, with a proportionately lower price).
I'm
       trying to figure out what the voltage of a fully charged new
       pack *should* be, and I have seen very different things as
       skimming through the archives.
       My understanding is that the chemical reactions taking place
in
       a lead-acid cell produce a potential difference of 2.13
Volts,
       so a fully charged lead-acid battery measured with a very
high
       impedance voltmeter should measure that 2.13 Volts per cell,
or
       6.39V for a 6V battery, and 38.34 V for a 36V pack.
       But, I have seen statements in the archives saying that a
fully
       charged pack should have a voltage of 42V or so (various
posts
       saying anywhere from about 40V up to 44V).

       Thanks,
       Mike

       --


--------------------------------------------------------------
       Michael S. Briggs, PhD
       UNH Physics Department
       (603) 862-2828

---------------------------------------------------------------


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