Gel Cell Condition Check?

[exposure]

Hi All,

I have three Prevailer 186 Ah (558 Ah total) Gel Cell batteries that make up my house bank. FYI, I have a separate 86Ah gel starter battery. They were installed on my boat in '94. They have sat at the dock on float charge for 99% of thier life. I am interested in assessing their condition.

This weekend, I turned off the charger and ran a constant 6.75 amp load of house lighting. The time ran was 29 hours 21 minutes (29.35 hours). This works out to 198.1 Ahours. I have an ample power system monitor and it showed 194.8 Ah consumed. The load varied a little, probably due to the flourescent lights and temperature changes, so I'll go with the 194.8. This should be 35% of power consumed, or 65% remaining.

During this time the voltage went from 13.00 to 12.03 Volts. The voltage measurements are at the batteries, with the 6.75 amp load present. I have not been able to find a capacity verses vlotage chart for gel cells. The 12.03 volts seems a little low for 65% remaining.

Can anyone shed some "light" on these numbers?

Thanks, Woody

[GordMay]

“...During this time the voltage went from 13.00 to 12.03 Volts. The voltage measurements are at the batteries, with the 6.75 amp load present. I have not been able to find a capacity verses vlotage chart for gel cells. The 12.03 volts seems a little low for 65% remaining ...”

An open circuit voltage of 12.3 V represents a state of charge approximating 60% (40% discharged) at 80 degrees F. A fully charged battery would have an OCV of about 12.8V.

Effective internal resistance (Re) is the apparent opposition to current within a battery that manifests itself as a drop in battery voltage (Ed) proportional to the discharge current (I). Its value is dependent upon battery design, state-of-charge, temperature, and age.
Re = Ed ÷ I = (13.0 - 12.3V) ÷ 6.75A = 0.7 ÷ 6.75 = 0.1037 Ohms

The internal resistance of a fully charged battery in good condition is about 0.10 ohms.

Seems about right to me.

Rick ???

Gord

[exposure]

Gord,

From your reply I am guessing that internal resistance rises with age. So the .1037 of my bank vs .1 for a new battery is pretty good? That would only be a 3.7% increase. I am sure some of this must vary by manufacturer, etc.

It sounds like I should have measured the voltage with little or no load to get the proper voltage to compare to the charts. BTW, where did you get your information of Voltage vs. Charge State for a Gel Cell? I have not found it anywhere.

Thanks, Woody

[exposure]

Gord, Rereading your reply, did you calculate internal resistance correctly? should it be...

Re=(Vopen-Vload)÷Iload?

[GordMay]

Woody (Exposure):
I apologize for my cursory reply, which was superficial and misleading. I had hoped that Rick would jump in and correct and augment my post with his expert knowledge and profound understanding of the subject.

Some of my errors:

~ I accepted 13V value for a Gel Cell Open Circuit Voltage (OCV), whereas the OCV for a fully charged & rested Gell Cell (or AGM) is approximately 12.8V (2.13VPC), with no load - although I’ve also seen 12.6V cited (2.10VPC).

~ I missread your Loaded Voltage as 12.3V, whereas you actually stated 12.03V. Also, I didn’t point out the “Uncertainty Factor” inherent in any Voltage reading.

~ I misstated the Internal Resistance (RE) as 0.10 Ohms. The actual RE would be more like 0.004 Ohms for a single Battery. In this case we have (3) Batteries wired in parallel, which would equate to an equivalent Internal Resistance (Re) of about 0.00133 Ohms* (see Note 1).

Standard "flooded" lead acid batteries state of charge (SOC) is approximately as follows*:
12.6 V = 100%, 12.4 = 75%, 12.2 = 50%, 12.0 = 25%, 11.7 = 0%.

Sealed AGM or Gel VRLA batteries have a state of charge (SOC) as approximately follows* (sorry, I don’t recall my source for this):
12.8 V = 100%*, 12.6 = 75%, 12.4 = 50%, 12.0 = 25%, 11.8=0%

* These are very rough approximations, and cannot be assumed absolutely reliable.

Note 1:
When resistors are connected in parallel their combined resistance (Re -or- equivalent resistance) is less than any of the individual resistances.
There is a special equation for the combined resistance of two resistors R1 and R2 in parallel:
Re = (R1 × R2) ÷ (R1 + R2)
Hence, for two parallel connected Batteries, each having an assumed internal resistance of 0.10 Ohms, the total internal resistance can be calculated:
Re = (0.1 x 0.1) ÷ (0.1 + 0.1) = 0.01 ÷ 0.2 = 0.05 Ohms
or exactly half the Internal Resistance of a single Battery.

For more than two resistors connected in parallel a slightly lengthier equation must be used. This adds up the reciprocal of each resistance to give the reciprocal of the combined resistance, Hence Re:
1/Re = 1/R1 + 1/R2 + 1/R3 ...
or
Re = 1÷ [ (1/R1) + (1/R2) + (1/R/3) ... ]
Hence, for our three parallel connected Batteries, each having an assumed internal resistance of 0.004 Ohms, the total expected (equivalent) internal resistance can be calculated:
Re = 1 ÷ [ (1 ÷ 0.004) + (1 ÷ 0.004) + (1 ÷ 0.004) = 1 ÷ (250.0 + 250.0 + 250.0)
Re = 1 ÷ 750.0 = 0.00133 Ohms

As you correctly pointed out - the Effective Resistance is:
Re = (Vopen-Vload) ÷ I Load
*Ed = Voltage Drop = (Vopen - Vload)
Hence:
Re = (12.80 V - 12.03 V) ÷ 6.75 A = 0.77 V ÷ 6.75 A = 0.114 Ohms

This is about 63* to 85 times higher than I’d have expected (0.114 ÷ 0.00133) !!!
*If we assume Fully Charged OCV @ 12.6 V.

Back to your original question:
“... I have an ample power system monitor and it showed 194.8 Ah consumed. The load varied a little, probably due to the flourescent lights and temperature changes, so I'll go with the 194.8. This should be 35% of power consumed, or 65% remaining ...”

1. Your assumed Depth of Discharge (DOD) is 35%, only if the Battery Bank was fully charged when you began your test. Only if so - you’re correct in calculating your State of Charge (SOC) at 65%.

2. Were you to fully disconnect all loads (at this point), you would then expect to measure an Open Circuit Voltage (OCV) of something like 12.5V (or 12.3V, if the 12.6V more accurately represents your 100% SOC)

My Conclusions:
a. You did not start your experiment with a fully charged battery bank.
b. You may not actually have 12.03 Volts (loaded @ 6.75A) at the end of 29.35 hours
and / or
c. You may have a dying Battery (or bank). You didn’t indicate the number of cycles on the batteries, but they are 10 years old. I wouldn't be surprised if they are nearing end of useful life.

and

d. We need RICK’s help here !!!

Some good References:

A couple of Rick’s excellent postings:
“Understanding a model for battery charge acceptance:”
http://cruisersforum.com/showthread....&threadid=1442
and
“How "fast" you can reliably charge your battery:”
http://cruisersforum.com/showthread....&threadid=1432

Gel Handbook Part 1 (Basic principles and design overview)
http://www.sonnenschein.org/PDF%20fi...dbookPart1.pdf
Gel Handbook Part 2
http://www.sonnenschein.org/PDF%20fi...dbookPart2.pdf

FWIW,
Gord

[Rick]

Sorry that I've not been able to long on and answer for awhile!

You are not going to like my answer to the subject line. For any lead-acid battery one cannot know with any great precision just what is the capacity WITHOUT actually discharging the battery and making the measurement. In addition, one must realize that the capacity measured will be dependent upon the ratio of load current to current at rated capacity. Most of us are using batteries having a capacity rated at a 20 hour rate because many deep-discharge applications, like sitting on the hook and having charge/discharge cycle times near 24 or 48 hours, are best handled with designs at or near the 20 hour rate. Industry applicatons use other battery designs of 4, 8, 100 or other hour values.

So, for a 200 Amp-hour battery rated at a 20 hour discharge the "rated" discharge current is ONLY at 200 divided by 20 equals 10 Amps. A discharge test with load currents BELOW 10A will result in slightly higher capacity values while discharge rates ABOVE 10A will result in lower capacity values. The values will be significantly lower as the ratio of discharge current to the "rated" current increases due to the non-linear Peukert effect. Becuase of the Peukert effect it is generally wise to "size" a battery bank so that your average continuous discharge current is equal to or less that the 20 hour discharge current rate in order to have confidence that you will not exceed your desired discharge floor on a cyclical basis. One can still use heavy intermittent loads like baking potatoes for 8 to 12 minutes in a 1200 Watt (input power) microwave or use a windlass without the engine running which puts your operation into the 'Peukert zone" without ill effects as long as your design floor is sufficiently high (like 50 % depth of discharge or less for cycling).

Back to the issue: Even if you actually completely discharge your battery and make the capacity measurement you still cannot know just what you will get ON THE NEXT DISCHARGE CYCLE. BTW, I do not recommend that you make discharge measurements unless you are merely rating a battery type because you never know what will happen on the next cycle after making such a test...so why bother? These tests are for rating batteries only, they are NOT for YOUR battery.

So, what's the point of using a battery monitor if you cannot really know just what is the actuall capacity? Here are several reasons. In other threads we've discussed how battery voltage taken "on the fly" will not give valid state of charge indication equitable to the published values for a particular battery. It is far more accurate to compare the number of Amp-hours (or prefererably kilo-Watt-hours, if you know the diference, as alternately indicated with some monitors) missing with the theoretical rated capacity value. Or, even better, merely use the percent state-of-charge indication that many monitors provide.

A good monitor will allow you to make performance checks as indicators of battery condition. Battery conditon might be thought of as a way to de-rate your battery's initial rated capacity value. The two best performance checks to make are for internal resistance and charge acceptance.

Internal resistance performance checks are NOT made using open-circuit voltages in the calculation. You can read in other threads, like Gord indicated, about the effects of surface charge on open-circuit voltage readings. The surface charge effects MUST be separated in order to make valid internal resistance performance checks. Use Gord's formulas for resistance. The important thing is to use a starting voltage which does not include the surface charge effect. You know this if the battery is not full and has not been recently on the charger. Place the battery under a load somewhere near the rated discharge current and THEN note the start voltage. After a period of time when the voltage has changed significantly note that voltage (and current, of course) and terminate the test (unless you are merely using a normally functioning load like when you are microwaving potatoes). The internal resistance is the CHANGE in voltage divided by the AVERAGE current values from beginning to end of the time during which the values are noted. If you compare resistance values made at the SAME point of indicated state of charge you will have an indication of degradation (time degradation or loss of capacity degradation).

It is true that good batteries might make a 10 year useful life and, if so, you can make a linear degradation calculation with time starting with a reasonable rated value and degrading to zero at 10 years. You then. after the first year, reset the rated capacity to a value 10% lower than when new and keep subtracting that same value every year thereafter.

A battery charge-acceptance performance check should also be made. This is easy to note if you realize that, to be valid, comparative checks must be for the same battery temperature and from, or at, the same apparent state-of-charge AND at the same charge voltage. Here you must have a monitor capable of indicating apparent depth of dischargeto do this with validity. Merely note the state of charge, the charge current accepted and the voltage. This is, of course, easier to do when first reaching an acceptance set voltage rather than during bulk charging. If, after several discharge cycles, you note that when you reach 14.4 V from an apparent state of charge of 70 %, for example, instead of seeing the usual 60 Amps you see 45 Amps you KNOW that the battery is failing to recover lost capacity and that the apparent state of charge is in error and needs to be revised downward.

This is a very manifold subject to cover in a few paragraphs...how are we doing with understanding anything useful?

Regards,
Rick

[Rick]

In the paragraph discussing measurement of battery internal resistance insert "average" after "change" (regarding voltage readings), i.e.; the resistance is calculated by using the average voltage change divided by the average current change. In addition, I did not state well that in order to truly discern an internal resistance measurement there must be a significant load. Truly a single measurement consisting of one voltage and one current at the SAME time will give you a resistance value, it is more accurate to take several measurements in a short period of time at a known state of charge and make an average.