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%.
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.
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
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 ...
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)
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)
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.
d. We need RICK’s help here !!!
Some good References:
A couple of Rick’s excellent postings:
“Understanding a model for battery charge acceptance:”
“How "fast" you can reliably charge your battery:”
Gel Handbook Part 1 (Basic principles and design overview)
Gel Handbook Part 2