BWAHAHAHAHA! Now its a conspiracy! Trojan, Rolls, and Calder are ganging
up
on poor Gould!

First characteristic of a losing argument, switch to personal attack.

Or, was that several posts back when you assumed I was a 'bozo" that ran a gen
set all night?

Damn, it's hard to keep the insults straight any more.

2.2 X 6 = 13.2.

On Mon, 24 May 2004 02:34:50 GMT, (Steven
Shelikoff) wrote:
The Nernst equation also explains part of why a lead-acid cell will have
a higher voltage when it's just taken off a charger and then settle to a
full-charge voltage a little bit lower and still be considered to have a
100% charge. It has to do with 1) the concentrations of the chemicals
at the plates equalizing some time after removing the charging voltage
and 2) the cooling down after charging which lowers the potential. It
nothing to do with self discharge or internal resistance.

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

Thank you. It's good to see a little high quality science interject
itself into an otherwise degenerating discussion. :-)

All the better it supports what most of us have observed. Chuck's
electrons have become damp, slugish, and not a little bit stubborn,
having lived in the Pacific North West for so long.

Thank you. It's good to see a little high quality science interject
itself into an otherwise degenerating discussion. :-)

All the better it supports what most of us have observed. Chuck's
electrons have become damp, slugish, and not a little bit stubborn,
having lived in the Pacific North West for so long.


Steve's hypothesis states that once the battery has achieved a full charge of
2.2 volts per cell, the decrease to 2.1
is due to an a change in the chemical concentration within the cell. Very
probable scenario.

Doesn't support any sort of position that the battery was never charged to
13.2, though, does it?

Merely explains with greater technical clarity than the term "self discharge"
why the voltage will eventually drop.

And that's the Pacific North WET, thank you very much. :-)

On 24 May 2004 06:11:19 GMT, (Gould 0738) wrote:

Thank you. It's good to see a little high quality science interject
itself into an otherwise degenerating discussion. :-)

All the better it supports what most of us have observed. Chuck's
electrons have become damp, slugish, and not a little bit stubborn,
having lived in the Pacific North West for so long.


Steve's hypothesis states that once the battery has achieved a full charge of
2.2 volts per cell, the decrease to 2.1
is due to an a change in the chemical concentration within the cell. Very
probable scenario.

Doesn't support any sort of position that the battery was never charged to
13.2, though, does it?

It may have been charged at 13.2, to 13.2, or whatever you want to call
it. But if you're taking the measurement right after charging, you're
reading an artificially high voltage. When the battery cools down and
the chemicals reach a constant concentration over the space inside the
battery, the voltage will drop to it's steady state full charge of well
less then 13.2 volts. The actual chemical voltage potential of the sum
of the half reactions of a lead-acid cell is between 2.05 and 2.06
volts.

The condition of a battery being fully charged is actually a chemical
condition and not a specific voltage. The battery is fully charged when
the negative plate is made fully of lead, the positive plate fully of
lead dioxide and there is no lead sulfate on either plate. The
electrolyte is sulfuric acid (around 25-35%) and the rest water. The
temperature and the concentration of the sulfuric acid can affect the
voltage of the fully charged battery. The maker of the battery will be
the one who knows what their designed full charge voltage is. They'll
also have to tell you how to measure it, i.e., what temperature, how
long to wait after charging, etc.

The voltage of the battery will drop as it discharges either through
self discharge or by being used because the electrolyte becomes diluted,
i.e., the concentration changes, when it breaks down into hydrogen and
sulfate and the hydrogen combines with oxygen to form more water. The
sulfate combines with the lead in both plates to form lead sulfate.

When the battery is fully discharged, both plates are covered in lead
sulfate and the electrolyte is almost all water. You don't want your
battery to get in that condition.

Merely explains with greater technical clarity than the term "self discharge"
why the voltage will eventually drop.

The voltage will drop eventually due to self discharge. But that's
something different then what happens in the few hours after removing
the charging voltage. The self discharge rate of lead acid batteries is
very variable. Sealed lead acid batteries lose maybe 3% of their
capacity per month. That also depends on temperature and the rate can
double if you increase the temp by like 20 degrees F.

However, you won't notice any self discharge in the few hours it takes
for the voltage to settle down after you remove the charger. I.e., if
the battery is fully charged and the float charger was at 13.2 volts and
you remove it and immediately measure the voltage as 13.2 volts, the
battery will still be fully charged (virtually no lead sulfate on the
plates and the same concentration of sulfuric acid in the electrolyte) a
few hours later when the voltage settles to a lower value.

Steve

"Steven Shelikoff" wrote in message
... a bunch of good stuff related to
batteries...

I guess that's why they call them 12 volt batteries instead of 13 volt
batteries, huh?

Eisboch

"Gould 0738" wrote in message
...
BWAHAHAHAHA! Now its a conspiracy! Trojan, Rolls, and Calder are ganging
up
on poor Gould!

First characteristic of a losing argument, switch to personal attack.

First characteristic of a losing argument: claiming all of the experts are
wrong, all of the manufacturers have a hidden agenda, and everyone is ignoring
the "science."


Or, was that several posts back when you assumed I was a 'bozo" that ran a gen
set all night?

No - you seemed to imply that not running a charging system was unnatural for a
boater. You called a "working environment" one that has a charger running.



Damn, it's hard to keep the insults straight any more.

2.2 X 6 = 13.2.


The first sign of dementia is mumbling the same nonsense over and over.

It may have been charged at 13.2, to 13.2, or whatever you want to call
it. But if you're taking the measurement right after charging, you're
reading an artificially high voltage.

No, there's nothing "artificial" about it.
That's the voltage reading at the time a battery finishes charging. As others
have said, (quoting various time periods from
"almost right away" to "overnight") the voltage eventually drops to a lower
point.

If you've got a $3000 paycheck, spend $200 on the way home from work and only
have $2800 with which to pay your bills, would that mean that you *never* had
$3000?

If you put the battery to work as soon as it has recharged to full capacity and
before it undergoes the internal changes that lower the voltage, you can take
advantage of (some of) the additional voltage.

A battery that cannot be brought to a point above 12.6 when charging,
considering the benefits of the internal heat and the concentration of
chemicals on the plates,
is not in good shape. What would happen to the battery that barely charges to
12.6?
It will also cool down, the chemicals will also equalize, etc.......leaving you
with
what? 12.1? 12.2?

If we are going to apply the "battery cools down and chemicals equalize after
charging" caveat to the battery charged to
2.2 volts per cell, it must also be applied to
batteries charged to only 2.1.

First characteristic of a losing argument: claiming all of the experts are
wrong,

Your experts disagree with my experts.
Surely you noticed that?

No - you seemed to imply that not running a charging system was unnatural for
a
boater. You called a "working environment" one that has a charger running.

Normally a battery works in an environment where there is a charger, or an
alternator. Neither the charger nor the alternator runs all the time, but one
or the other will be running when the battery is being charged, which has a
direct bearing on a question regarding the voltage reading of a battery that
has reached a full charge state.

From your perspective, on a sailboat, that may not be true. If you don't have
an auxiliary and you're charging the battery at home in your garage, you're
probably never going to see anything above 12.6 on the boat.

On 24 May 2004 16:14:56 GMT, (Gould 0738) wrote:

It may have been charged at 13.2, to 13.2, or whatever you want to call
it. But if you're taking the measurement right after charging, you're
reading an artificially high voltage.

No, there's nothing "artificial" about it.
That's the voltage reading at the time a battery finishes charging. As others
have said, (quoting various time periods from
"almost right away" to "overnight") the voltage eventually drops to a lower
point.

If you've got a $3000 paycheck, spend $200 on the way home from work and only
have $2800 with which to pay your bills, would that mean that you *never* had
$3000?

Actually, that's not quite the way to think about it because you're
assuming something got spent. The way to think about it is, you got a
paycheck, you don't spend any and take it home. A few hours later look
at it and it says it's worth $2800. You can spend that $2800 or let it
sit under your mattress where inflation will mean that it's worth 3%
less every month (self discharge). It doesn't matter that it said $3000
when it was given to you a few hours earlier because the penalty for
spending even a single dollar is that it immediately becomes worth only
$2799. That's because even though the battery may read 13.2 volts
immediately after charging, it won't have hardly any more capacity then
it does after it settles.

If you put the battery to work as soon as it has recharged to full capacity and
before it undergoes the internal changes that lower the voltage, you can take
advantage of (some of) the additional voltage.

No, not really. There are only a fixed number of moles of molecules
available to be converted to electric power. If you put it to work as
soon as it has recharged and is still at 13.2 volts, by the time it
reaches say a 90% charge, you won't get measurably more power out of it
then if you waited a few hours.

A battery that cannot be brought to a point above 12.6 when charging,
considering the benefits of the internal heat and the concentration of
chemicals on the plates,

You can bring a battery to as high as you want when charging. You can
charge it at 100 volts and measure that across the terminals. There
will be current flowing with the battery acting as a resister. It won't
last long though. It will probably boil off so much flammable gas and
get so hot that it'll explode. But that doesn't mean you can't get 100
volts to appear across the terminals beforhand.

is not in good shape. What would happen to the battery that barely charges to
12.6?
It will also cool down, the chemicals will also equalize, etc.......leaving you
with
what? 12.1? 12.2?

If we are going to apply the "battery cools down and chemicals equalize after
charging" caveat to the battery charged to
2.2 volts per cell, it must also be applied to
batteries charged to only 2.1.

It applies to everything. But the problem is that now you're looking at
a battery that is not in good shape. Big lead sulfate crystals may have
formed that can't be broken down by normal charging. They may have even
broken loose from the lead plate and settled on the bottom of the
battery. That locks up some of the sulfate which would be converted
into sulfuric acid when charging and so the concentration of sulfuric
acid in the electrolyte never gets very high. It also takes away some
of the sponge lead on the plate that's available to be converted into
lead sulfate when discharging, reducing the capacity of the battery.

This is a different case again then looking at the settling voltage of a
fully charged battery in good condition.

Steve

Actually, I'm not sure they disagreed. All of my experts were talking about
marine batteries in a practical environment. Two are the manufacturers of the
batteries we each happen to use, the other is the acknowledged expert in
cruising boat systems. None of your "experts" ever mentioned marine batteries;
in fact, I don't think they mentioned production batteries hardly at all.
Several weren't even talking about flooded cells. One actually gave numbers
closer to mine than yours.

Also, your "experts" mention an approximate value in passing, without a
discussion of charging, measuring state of charge, and surface charge. My
experts were all talking specifically about these subjects, and were unequivocal
that your approach to measuring state of charge is flawed.



"Gould 0738" wrote in message
...
First characteristic of a losing argument: claiming all of the experts are
wrong,

Your experts disagree with my experts.
Surely you noticed that?

No - you seemed to imply that not running a charging system was unnatural for
a
boater. You called a "working environment" one that has a charger running.

Normally a battery works in an environment where there is a charger, or an
alternator. Neither the charger nor the alternator runs all the time, but one
or the other will be running when the battery is being charged, which has a
direct bearing on a question regarding the voltage reading of a battery that
has reached a full charge state.

From your perspective, on a sailboat, that may not be true. If you don't have
an auxiliary and you're charging the battery at home in your garage, you're
probably never going to see anything above 12.6 on the boat.