"Gould 0738" wrote in message
...

Rather obviously, the battery would not accept additional voltage if it
was
"fully charged" at 12.6.


I think this sentence is the core of your confusion. You can apply any
voltage you want on the battery. "Equalization" or "Reconditioning" modes
of chargers often apply 17 volts, and the battery will rise to that
potential. It will also boil and melt if you leave it there too long.

Eisboch

"Gould 0738" wrote in message
...
Where do you get this nonsense? "Open Circuit Voltage" is not the same thing
as
"Float Voltage." Did you even look at the link I provided, or are you
claiming
to know more about batteries than Trojan?


No I don't know more about batteries than Trojan.

However, if you test the batteries on your boat and you think that 12.6 means
the batteries are fully charged, that means I know more about batteries than
you.

That would appear very doubtful.

If you only test your batteries while a battery charger is runnning, then you
know very little about batteries.


I won't dispute that if a battery is removed from a charging system and
allowed
to sit around in your garage, or languish on a retailer's shelf, the voltage
will drop.

Not "lanquish," it only takes an hour or two, and can be hastened by applying a
small load. Actually, it will go below 13 Volts within minutes.



The initial drop may well be rapid, and 12.6 could be considered OK
for a partially discharged, idle battery. Put that battery back in a working
environment where it has access to charge from an alternator or a 110 charger,
and it will be "fully charged" when it achieves 2.2 volts per cell, or 13.2
volts total.

No. You're just saying that a smart charger will be at a 13.2 Volt float level
when it desides its fully charged. Take the charger off and the battery quickly
drops to 12.6.

Why do you call a "working environment" one with a charger running? Are you one
of those bozos that runs a genset all night long?



Rather obviously, the battery would not accept additional voltage if it was
"fully charged" at 12.6.

What do you mean by "accept additional voltage"? What do you think would happen
if I took a fully charged battery and put 17 Volts on it? Would it read only
13.2?



Since a "smart" charger will maintain a 13.2 voltage on a flooded, 12-volt
batery in good condition, that is the standard for full charge.

No. That's the standard for a float charge. Or are you claiming that without
the float charge the battery can't really be at 100% because it will
self-discharge some tiny amount as soon as the charger is removed? You may be
right in a pedantic way, but 99.99% is close enough to 100% for me.


If that same
charger cannot bring the voltage above 12.6, that's a likely sign that the
battery is beginning to fail.

Well, I haven't killed too many batteries, but I usually find that the Voltage
will come up on the charger, but falls too quickly when a load is applied. I'd
guess this is why people use "load testers" to test the quality of a battery.

Thanks to all of you who contributed to resolving this issue.


"Wayne.B" wrote in message
...
On Wed, 19 May 2004 23:23:28 -0400, "Eisboch"
wrote:

With the float charger attached, agreed.
Remove the float charger for an hour or so and the voltage for a fully
charged flooded battery will be 12.6 volts.

Been that way for many, many years.

============

Yes.

"Ernie" wrote in message
...
Thanks to all of you who contributed to resolving this issue.

Resolved? 8-)

According to your position on full charge, I must have three batteries that
are going bad which is not the case.



You do not have batteries going bad because they can be charged to 13.2.

The fact that they then partially discharge to 12.6 doesn't make them bad
batteries.

If your batteries could not be charged above 12.6, they would be starting to
fail.
As you noted, they can be charged to 13.2.

I think this sentence is the core of your confusion. You can apply any
voltage you want on the battery. "Equalization" or "Reconditioning" modes
of chargers often apply 17 volts, and the battery will rise to that
potential. It will also boil and melt if you leave it there too long.

17 volts is overcharged. "Equalization"
involves the short term overcharge of a battery to blow the crud off the
plates, but it is not a charge that a battery can sustain for a long period- as
you correctly note.

13.2 is the proper reading for a fully charged battery. That is the level which
any decent charger will maintain, and there is no "overcharged" condition.

There is a serious difference between a fully charged reading of 13.2 and an
equalization charge.

If you only test your batteries while a battery charger is runnning, then you
know very little about batteries.

You must have a very clean hog, with all that wash on hand.

Real life scenario for you:

When I go aboard my boat and prepare to get underway, I check the battery
voltmeter on the distribution panel. It always reads just over 13 volts. Before
I start the engine, I shut off the shorepower to the vessel. When the
shorepower is shut off, the battery charger is not "running".

Does the voltmeter take a nosedive to 12.6? No, it does not. It remains at over
13.

Yes, if I wait long enough the voltage will begin to drop.

But my fully charged batteries show 13.2 volts *without* the charger running.

Are you one
of those bozos that runs a genset all night long?

I don't run a genset at all. Propane galley boat with a diesel furnace. Two
Rolls 8D batteries. Not even equipped with a genset.

Are you one of those bozos who prefers to believe that the state of charge
represented by the .6 volts between 13.2 and 12.6 never existed, simply because
the first .6 volt bleeds off?

"Gould 0738" wrote in message
...
If you only test your batteries while a battery charger is runnning, then you
know very little about batteries.

You must have a very clean hog, with all that wash on hand.

Real life scenario for you:

When I go aboard my boat and prepare to get underway, I check the battery
voltmeter on the distribution panel. It always reads just over 13 volts.
Before
I start the engine, I shut off the shorepower to the vessel. When the
shorepower is shut off, the battery charger is not "running".

Does the voltmeter take a nosedive to 12.6? No, it does not. It remains at
over
13.

Why do you care? If it was on the charger overnight it should be fully charged.
Even if the battery is near death it probabaly won't show at that point.


Yes, if I wait long enough the voltage will begin to drop.

But my fully charged batteries show 13.2 volts *without* the charger running.

Are you one
of those bozos that runs a genset all night long?

I don't run a genset at all. Propane galley boat with a diesel furnace. Two
Rolls 8D batteries. Not even equipped with a genset.

Are you one of those bozos who prefers to believe that the state of charge
represented by the .6 volts between 13.2 and 12.6 never existed, simply
because
the first .6 volt bleeds off?

So if you stop and anchor, shut off the engine, relax for an hour or so, then
check the battery and it reads 12.6, do you assume the battery is discharged?
Everyone else would say it virtually fully charged. The bottom line is that the
extra 0.6 Volts is an artifact of a recent charge, not a sign of depletion.

"Gould 0738" wrote in message
...
I think this sentence is the core of your confusion. You can apply any
voltage you want on the battery. "Equalization" or "Reconditioning"
modes
of chargers often apply 17 volts, and the battery will rise to that
potential. It will also boil and melt if you leave it there too long.

17 volts is overcharged. "Equalization"
involves the short term overcharge of a battery to blow the crud off the
plates, but it is not a charge that a battery can sustain for a long
period- as
you correctly note.

13.2 is the proper reading for a fully charged battery. That is the level
which
any decent charger will maintain, and there is no "overcharged" condition.

There is a serious difference between a fully charged reading of 13.2 and
an
equalization charge.


I understand Gould. My point was that just as the battery will take a
charge of 13.2 volts when on a float mode charger, it will also temporarily
take the 17 volt equalization charge. Neither has anything to do with the
battery's natural full charge of --- 12.6 volts.

Two final questions for you, then I will give up and let the electrons fall
as they may ...

Following your logic, if 75% of the battery charge takes place during the
bulk charge mode at 14.4 volts - why aren't you claiming that a fully
charged 12 volt battery should read 14.4 volts?

Secondly, if the bulk charge takes place at 14.4 volts, then drops to 13.5v
or whatever for the absorbsion and float modes, what happened to the .9
volts? The charger never turned off.
Hmmmm .... where did those electrons go?

The answer is that your "fully charged" value of 13.2 volts is really just
the output voltage of the charger, not the value of the battery potential by
itself.

Eisboch

Equation 8 is certainly not the whole story. I haven’t done the experiment
myself, but according to “common knowledge� lead sulfate is formed on the
“+� plate, more in agreement with equation 3 than equation 8.

It is a tangentially interesting question whether any of the hypothetical
reactions listed above play any role -- even as intermediates -- in the real
reaction.

That is only tangentially interesting, because whether or not those reactions
occur, we are still left with a major mystery: how and why does anything
containing the SO4 group attack the “+� electrode? Neutral H2SO4 could
reach the plate by simple diffusion, but it is present in fantastically low
concentration. The bisulfate ion is present in high concentration, but would
have to swim uphill against the electric field. The sulfate ion is present in
low concentration and would have to swim doubly hard uphill. That means that
when the cell is under heavy load, i.e. when there is a large field across the
electrolyte, the SO4-related reaction would come to a halt.

Let’s put in some numbers: The cell has an open-circuit voltage of 2.2 volts
are so. Suppose that we put it under heavy load, so that there is ? v = 0.4
volts “IR� drop across the electrolyte. As always, room temperature
corresponds to 25 meV (.025 electron-volts). Putting it all together: the
Boltzmann factor that tells you what fraction of the bisulfate ions manage to
climb the potential is exp(q ? v / kT) = exp(.4 / .025) = 9,000,000. So we
would expect the reaction to proceed millions of times slower when we need it
to proceed faster.

In the foregoing calculation, we ignored the effect of dielectric screening.
This may or may not have been the right thing. Argument pro: energy is
conserved. At the end of the day, to move a bisulfate ion up a hill 0.4 volts
high, you have to do 0.4 eV of work. Argument con: most of the height of the
hill is associated with the dipole layer at the edge of the water, at the place
where the electrolyte meets the plate; within the bulk of the electrolyte the
field is smaller. The ions can with relatively resonable probability get close
to the “+� plate, just outside the dipole layer. Unanswered question: if
they get that close, is that close enough?

**********************

Note the basis of calculation, opening sentence, paragraph four: 2.2 volts per
cell.


From

http://www.av8n.com/physics/lead-acid.htm