Solar series battery charging problem

[jim hardy]

The only possible way I can reconcile this is either it has to do with the chemical reactions that do or can take place in a battery (perhaps something that can happen when a cell has a certain type of failure?) or that my meter was not reading properly for some odd reason (bad meter, bad leads, poor connection?) when I connected it between the two batteries in series.

yeah, sounds to me like something opened up on ya.
I've had batteries break an internal connection. One car wouldn't start unless i pushed down on the post to make the internal connection again. It'd work the dome light and radio i guess through electrolyte but voltage collapsed with even the windshield wipers turned on.

We learn a lot in troubleshooting, among the most valuable is how little we know. I had to refresh my battery chemistry to keep up with this thread.

Here goes

a battery cell has two plates one of pure lead and the other of lead oxide. They are immersed in sulfuric acid which is a mix of hydrogen ions and sulfate ions.
The pure lead plate is the negative one
the lead oxide plate is the positive one
As the battery discharges,
the lead plate turns into lead sulfate and pushes electrons out into the external circuit, one by one, oops two by two
the lead oxide plate .turns into lead sulfate and pulls electrons in from the external circuit one by one also two at a time
no electrons get 'stored' or hide out inside the battery
Here's pictures from http://ecee.colorado.edu/ecen4517/materials/Battery.pdf
that i annotated a little bit

that allows pure lead there to become lead sulfate and push electrons out into the world.

And over at the positive plate, almost instantaneously,
two electrons (but not the same two) enter from external circuit per Kirchoff's Current Law.

That allows the positive plate to pull electrons in from the outside world.

Were electrons allowed to accumulate or deplete inside the battery it would soon have an excess or shortage of them and rise to lightning-bolt potential with respect to its surroundings. That's how lightning works in clouds.


Sulfate ions in solution get replaced by water molecules. Electron balance is maintained.


Of course for charging just revere the directionsold jim

 

[BernieM]

Sounds like what I had in mind on battery chemistry too.

OK so you have a charging situation now, and the amount of current will be limited to the total resistance in the battery between the positve and negative posts. So each cell will appear as a resistor and since there are 6 cells, the resistance will be about equal in a normally operating battery so you have 6 resistors with value X. So the resistance of the battery is 6X and using ohms law, I=E/R which means the result will be I=V/6X. Now as a cell charges, the electrolyte regains sulfur ions, and becomes more conductive. In the process of gaining electrons, the only place from which to take those electrons is the current flowing into the battery. So now the cells become more charged, we have a new resistance internally in the battery, and now we have I=E/6Y (Y for the new resistance of the cell.) Putting real values in, let's just say X = 10 and Y = 5. So to begin with we had 100 ohms resistance (yes I know it's actually called impedance) and a voltage of a partially discharged battery at say 11.0V So I = 11/60 = .18A current flow into the battery. Later when it is charged up more we have I = 12.0V/30 or .4A current flow. So current flow increases as the battery gets charged, as well as the voltage increasing. This means as the batteries get charged up they are also producing more heat then. For me that translates to more heat at the end of the long hot summer day, at the same time the day has gotten the hottest. Bad news.

What I still do not know is, since the cells are in series, whether the first cell will consume more electrons than the next cell, and it more than the one after it, or if somehow all cells will somehow magically get the same current as stated early on in this thread.

If you look at the cells as individual batteries connected by a liquid connection whose resistance changes with state of charge of the battery, then from the + post to the first cell we see 1X resistance. From the first cell to the 2nd cell we see 1X resistance. But from perspective of the first post, we see 2X resistance to get to the 2nd cell. So the current, as I see it, is highest to cell one, and 1/2 that to cell two. But as cell one charges up, the resistance between the + post and cell 2 decreases, allowing then more current to arrive at the 2nd cell.

Are you seeing my confusion here and why I can't reconcile that all cells see the same amount of current at the same time?

Or another way to put it would be the old water pressure model, where voltage is pressure and amperage is water volume. You now have a hose with 6 holes in it, all the same size. As the water goes through the hose, some of the pressure is lost at hole #1 (but pressure is equal through the whole system unless the far end of the hose is open to the air,) as well some of the volume of water moving through the hose. Same for hole #2, etc., until at the end hole you have only the quantity of water flowing in the pipe at that location as the water that will flow through hole #6. So if one looked at it as gpm, and showed how much water was flowing in the pipe between holes, each hole draining 1GPM, before the first hole you have 6GPM going into the pipe. Between hole #1 and #2 you have 5GPM, 2#3, 4GPM, etc., until hole 6 at 1GPM.

This is driving me nuts. If you guys are right then somehow I have the wrong model of the internal workings of the battery, and all I can find on the Internet are basic explanations of batteries, nothing in depth enough to cover this adequately. Please feel free to point out the mistake I am making if you see it!

P.S. Regarding your Kirchoffs Current Law. The sum of the current leaving a node is equal to the current entering. However there is more than one path to leave the node. One path is electron capture to reverse the chemical process, and the other is conduction on to the next cell, correct?

 

[jim hardy]

I think you're grasping to hold on to an old misconception. It's human nature to do that.

If your pipe has no leaks,
every molecule of water that enters one end must exit the other end abeit at reduced pressure.

Gallons in = gallons out
divide both sides by minutes and gpm in = gpm out
but
pressure in ≠ pressure out

 

[BernieM]

Yes if the pipe has no leaks. Each cell is technically a leak, as electrons are required to put the sulfur molecule back in solution as electrolyte, right? As I said above (you may not have read it since I posted it after you responded) there are two paths for the current. One is on to the next cell, the other is capture in a chemical reaction.

 

[jim hardy]

Wow you're making me go back to high school chemistry , i hope i get this right

When working equations for chemical reactions you have to balance not only atoms but charges too.

Note that when SO4 ion plates out on negative plate

The charge in the electrolyte appears unbalanced, at first glance. It lost two negatives and no positives. There's two unwed hydrogen ions now. Mother Nature won't stand for that.

She takes care of it over at the positive plate. It's a little sneakier there...

Note it used up 4 hydrogen ions here and only one SO4.
The ionized H2SO4 molecule consumed here donated 2 hydrogens, and the one consumed over at the negative plate donated the other two.
So, both atoms and charge are conserved.

To put it in fewer words -
an electron that jumps into the lead atom at + plate causes one to jump into the electrolyte aboard an oxygen ion
and causes another one to jump out of the electrolyte over at the negative plate .
Hydrogen ions are pretty much free to migrate in water. (Like everything else, it's not quite that simple - see https://chem.libretexts.org/Core/Ph..._Bases_in_Aqueous_Solutions/The_Hydronium_Ion)

The load completes the circuit .
Without a load, only enough electrons move to establish an electric field in the battery.

Were it not so Kirchoff would be wrong.

on balancing charge in chemistry equations:
http://www.periodni.com/half-reaction_method.php?eq=PbS+H2O2=PbSO4+H2Ohope this helps
keep us posted what you find
old jim

 

[The Electrician]

BernieM, if you have your in-line ammeter connected in series with the + terminal of a battery at the top of one of the 24 volt series strings how does the measurement of the DC current compare with what your DC responsive Snapon clamp ammeter reads?

Assuming the readings are essentially the same (meaning we can trust the Snapon), without changing the setup move the Snapon clamp ammeter to the cable leaving the bottom of the 24 volt series string and get a reading of the current leaving the string. What are the two readings on the clamp ammeter?

 

[jim hardy]

Ahhh, Electrician i like your method. Make everything prove itself good especially your test equipment.

How many times have i got fooled because i didnt understand how my measuring instrument would respond to an unusual input.
Example: Some true RMS meters won't measure AC in presence of DC beyond voltage range selected - ie don't try to measure millivolts of AC riding atop ten volts of DC.

With clamp around meters one has to be really careful about DC.
Early electronic DC clamp-ons used an interesting technique, they had a feedback winding to which a feedback circuit applied DC to drive average flux to zero as measured by the Hall sensor. (That required power for the amplifier, my thirty year old Fluke takes four AA cells in the handle.)
What got reported is the amp-turns necessary to achieve that null, scaled of course to represent amps through the donut hole. My ancient fluke delivers 2 volts proportional to 20 amps..
It takes time to reach balance and that's why their frequency response was only out to a khz or so.

Moral of story: it's important to understand the working principle of one's test equipment. I like analog 'scopes and Simpson 260 multimeters because i understand them.

 

[BernieM]

OK, first of all let me state that my measurement between the two batteries that was zero was apparently due to a faulty connection (as you recall the measurements I initially made were using my DMM inline on 20amp scale.) I checked it a few times but I guess I wasn't getting good continuity. Using the Snap On ammeter, I get decent readings. This meter has a full scale range of 100 amps, with 0 center. So it's 100-0-100. Now when I took a reading (early in the morning so sun isn't high in the sky yet and amps are still low) I get about 2 amps at both the + and - of the series battery pair. Odd though, at the - post and the common posts where the two batteries are connected, the reading is left of zero on the scale, where the reading at the + post is right of zero. I tested it at multiple locations on the wire and the result is consistent.

 

[jim hardy]

Using the Snap On ammeter, I get decent readings. This meter has a full scale range of 100 amps, with 0 center. So it's 100-0-100. Now when I took a reading (early in the morning so sun isn't high in the sky yet and amps are still low) I get about 2 amps at both the + and - of the series battery pair. Odd though, at the - post and the common posts where the two batteries are connected, the reading is left of zero on the scale, where the reading at the + post is right of zero. I tested it at multiple locations on the wire and the result is consistent.

Very clearly worded - Thanks !

Only reason i can see for polarity reversal is
(no idea what yours looks like so take this as generic)

Usually there's an arrow and + sign someplace on the meter indicating which direction of conventional current (not electron current), up or down, will give a positive reading..

old jim

 

[The Electrician]

I assume that all your clamp ammeters can read AC, but it looks like the Snapon is the best.

This afternoon when you have a heavy load on the batteries (air conditioning?), go around and measure the AC current into each battery with the Snapon (and you could check the current out as well, but they better be the same--law of physics and all). Write them all down and let's see what the largest and smallest is.

The impedances everywhere in your bank are in the order of milliohms (batteries themselves, cabling, connectors, etc.). Any differences on the order of milliohms can lead to large differences in ripple current in the batteries.

I'll be back from lunch in a while.

 

[BernieM]

A/C amperage ranges from .8 to 1.5 amps throughout the bank.
D/C amperage is around 8 amps throughout the bank with the exception of one pair of batteries that is measuring around 25 amps charging. This is also the same pair of batteries that I got the zero amps read on at the common posts between the two batteries that I mentioned in an earlier post, that I decided must have been a poor connection which had caused the faulty reading. But this pair of batteries I expect is failing, at least the one that is connected to + in the parallel array, and it's the one with the damp top I had the picture of.
So at this time for some reason it is getting far more charge than the rest of the bank, and since there is one more negative plate than positive, the positive plate inside it is likely getting more cooked than the negative plate. As well, the positives are a paste in a bag, right? (Is this true for flooded cell?) And the negatives are solid lead.
So my bet at this point will be that if this battery fails (the one connected to + wire of the parallel array) that the other will be way undercharged when I separate the two. That's the MO I have described before.

 

[jim hardy]

A/C amperage ranges from .8 to 1.5 amps throughout the bank.

That sure sound reasonable and safe. Is that with heavy load on the inverter ?

If your inverter is imposing ripple on the bank it'll be in proportion to load and at inverter's switching frequency.

Looks to me like you're making progress .

Difficult troubleshooting proceeds 'One tiny step at a time.'

Keep on truckin' .


old jim

 

[The Electrician]

So you have a 2 to 1 variation in the ripple current around your bank.

If it were me, the next thing I would do is measure the internal AC impedance of each 12 volt battery. This isn't too hard to do.

It would be good if you had more than 2 digits for your current reading. Is the Snapon clamp meter auto-ranging, or is 100-0-100 its only range? I'm sure that if the Snapon doesn't have a lower range, you probably have another clamp on with a low AC amp range.

So what you do is this. Have a handheld DMM set to AC volts range. Use test leads with sharp points on the DMM.

Clamp the clamp-on AC ammeter around the cable in the middle of a 24 volt pair; that's the cable that connects the two 12 volt batteries together. Poke the sharp tips of the probes on the DMM (set to AC volts) into the lead battery posts of one of the 12 volt batteries of the pair (do this same procedure to the other battery of the pair next. You don't have to move the clamp ammeter; it can stay clamped on that cable between the two batteries for the measurement on each battery because the ripple current in the two batteries is the same). Write down the AC ripple voltage measured by the DMM and the ripple current measured by the clamp ammeter. Divide the AC ripple voltage by the AC ripple current--that's the internal AC impedance of the battery.

Do this for all your batteries. Generally speaking, high internal impedance is bad, low is good. This measurement may help you detect batteries likely to fail.

Let us know what results you get.

 

[jocke]

You must monitor each cell of each battery for equal voltage during equalisation charge.
Keep record of all cells individually. You can see over time if there is a pattern of which cells
fail. Check temperature of each cell too. Should be equal, if not cells will not receive the same
charge and wiil fail over time. This seems to be your problem.
I suggest you design a datalogger for the purpose. For the cost of a battery you get a nice system.

 

[Tom.G]

around 8 amps throughout the bank with the exception of one pair of batteries that is measuring around 25 amps charging

and it's the one with the damp top I had the picture of.

Damp Top = Lots of gas being generated causing bubbles = Hydrolysis of H₂O = Happens during overcharging

1. Since it is one battery in a battery bank it's a battery problem, not a charging problem. (Yeah, we already knew that.)
2. Why so much current? Not charged? No, that's not it because it's outgassing.
3. Why so much current? Because one (or more) cell is shorted, reducing battery voltage.
4. With less battery voltage bucking the charging voltage, the Volts per Cell goes up in the rest of the series cells, causing higher charge current = Overcharge of remaining cells = Outgassing = Damp top.

Two tests to verify: (use both)
(A)
Measure the voltage of each battery in the string that is charging at 25A while they are charging.
I expect the 'good' one to read a higher voltage than the 'bad' (wet?) one.

(B)
Disconnect that battery string (pair) from the bank.
If you have a battery load tester, use it here. If not, you can use an ordinary old fashioned car headlight as a load, high beam is about 3Amps.
Measure and record the voltage of each of the two batteries under load.
Leave disconnected for at least 24Hrs and measure their voltages again, under load.
The 'bad' battery will have a noticeably lower voltage after 24Hrs.

Might as well replace both of the series batteries at the same time. The second one, too, was severely overcharged and is about ready to fail. If you just replace one battery the older one will soon fail, taking the new one with it.

 

[jim hardy]

I once tried to measure a car battery's per cell voltage by connecting my voltmeter negative to battery negative then dipping the positive probe tip into each cell's electrolyte . I was looking for an open cell and figured with the engine running the open cell would show higher delta-voltage . I thought perhaps it was a broken intercell connector...
That battery was so weak it was inconclusive, each cell showed close to same increment . So i figured it probably had to be something else.Well duhhhh, the alternator belt was so loose that it was slipping and alternator couldn't keep up with electrical load of the airconditioner fan and clutch.

Anyhow - that imprecise and desperate test might find a shorted cell. You can work from either battery post.

old jim

 

[The Electrician]

Ahhh, Electrician i like your method. Make everything prove itself good especially your test equipment.

How many times have i got fooled because i didnt understand how my measuring instrument would respond to an unusual input.
Example: Some true RMS meters won't measure AC in presence of DC beyond voltage (or current--added by The Electrician) range selected - ie don't try to measure millivolts of AC riding atop ten volts of DC.

With clamp around meters one has to be really careful about DC.
Early electronic DC clamp-ons used an interesting technique, they had a feedback winding to which a feedback circuit applied DC to drive average flux to zero as measured by the Hall sensor. (That required power for the amplifier, my thirty year old Fluke takes four AA cells in the handle.)
What got reported is the amp-turns necessary to achieve that null, scaled of course to represent amps through the donut hole. My ancient fluke delivers 2 volts proportional to 20 amps..
It takes time to reach balance and that's why their frequency response was only out to a khz or so.

Moral of story: it's important to understand the working principle of one's test equipment. I like analog 'scopes and Simpson 260 multimeters because i understand them.

This is a very good point, Jim. What I do to check this is to change to a higher range on the DMM and see if the reading remains the same. The higher range will lose a digit of resolution, but when the reading is wildly different beyond just the loss of a least significant digit, you've got a problem!

So, BernieM, occasionally while you're measuring AC ripple current, change to DC mode and see if the DC current is much larger than the AC current. Change the range of the AC mode to a higher range and see if you get essentially the same reading. Of course, if your meter is autoranging, it may be impossible to force it to a higher range, although some autoranging meters do allow this.

Also, use a different clamp ammeter and compare readings to increase your confidence that you're not getting bogus measurements.

 

[BernieM]

Damp Top = Lots of gas being generated causing bubbles = Hydrolysis of H₂O = Happens during overcharging

1. Since it is one battery in a battery bank it's a battery problem, not a charging problem. (Yeah, we already knew that.)
2. Why so much current? Not charged? No, that's not it because it's outgassing.
3. Why so much current? Because one (or more) cell is shorted, reducing battery voltage.
4. With less battery voltage bucking the charging voltage, the Volts per Cell goes up in the rest of the series cells, causing higher charge current = Overcharge of remaining cells = Outgassing = Damp top.

Two tests to verify: (use both)
(A)
Measure the voltage of each battery in the string that is charging at 25A while they are charging.
I expect the 'good' one to read a higher voltage than the 'bad' (wet?) one.

(B)
Disconnect that battery string (pair) from the bank.
If you have a battery load tester, use it here. If not, you can use an ordinary old fashioned car headlight as a load, high beam is about 3Amps.
Measure and record the voltage of each of the two batteries under load.
Leave disconnected for at least 24Hrs and measure their voltages again, under load.
The 'bad' battery will have a noticeably lower voltage after 24Hrs.

Might as well replace both of the series batteries at the same time. The second one, too, was severely overcharged and is about ready to fail. If you just replace one battery the older one will soon fail, taking the new one with it.

I checked the cells and there is little or no water missing, so this wetness is caused by a small amount of battery acid, as it never ran down the sides or got the top literally wet, merely dampened the dust on it.

A) 13.99 for the one connected to the + terminal and 14.06 for the one connected to the - terminal.

B) I have a carbon pile load tester (I think it's carbon pile anyhow.) I already did this test a week ago when I initially was discussing it here, and after a day (or was it two?) being disconnected the voltages were 12.6 and 12.6. Same. So if it hasn't failed yet. But I do believe it is in the process.

I did most of this in an earlier post (#57) where I actually answered some of these questions there.

 

[BernieM]

I once tried to measure a car battery's per cell voltage by connecting my voltmeter negative to battery negative then dipping the positive probe tip into each cell's electrolyte . I was looking for an open cell and figured with the engine running the open cell would show higher delta-voltage . I thought perhaps it was a broken intercell connector...
That battery was so weak it was inconclusive, each cell showed close to same increment . So i figured it probably had to be something else.Well duhhhh, the alternator belt was so loose that it was slipping and alternator couldn't keep up with electrical load of the airconditioner fan and clutch.

Anyhow - that imprecise and desperate test might find a shorted cell. You can work from either battery post.

old jim

I'm not sure but I think introducing a probe into a cell would create a new plate and a new battery, as well a different potential due to it being different metal (unless you used a lead probe...) which might be the reason for all the readings being about the same. I think you were measuring the potential of the probe in the electrolyte, so if there was a difference it was probably related to the strength of the electrolyte in the cells? Just a random guess.