Maybe I'm not reading your message right Rick, but here's how it makes sense to me
Quote:
Originally Posted by Rick  Evaluating a 12 Volt flooded-cell battery of unknown history and quality
To do this you would need to fully charge the battery (making sure that a full 14.4V or more is reached during acceptance charging @ or around 25 C). You would then measure each cell separately, obviously the total has meaning as well yet it is the VARIATION in cell-to-cell readings that reveals the need for equalization or evidence of permanent sulphatation (sulphation is not permanent, sulphatation is).
The three things to determine: 1. That the battery is fully charged, and if so that it does not need equalization or that equalization does not yield a reversal of sulphation cell-to-cell. 2. The approximate Amp-hour rating. 3. The battery internal cell resistance.
The internal resistance is a very important and little acknowledged attribute of any battery, lead-acid or otherwise). It is the internal resistance multiplied by a heavy load current which results in a “voltage burden” to your system and directly increases losses in the energy stored by the battery that you cannot use.
Notice that the fully charged terminal voltage is not on the list because it will be sufficiently high (about 12.7V without surface charge) when the other conditions are met. Surface charge may be removed by applying around a 5% of the Amp-hour rating in Amps and noting the terminal voltage. It will linearly decrease until it becomes constant. Remove the load at that time. Note that most nominal 12V flooded-cell batteries have a specific gravity that results in a 12.66 Volt terminal reading yet many of you notice readings of 12.8Volts or so....this is due to a lingering surface charge after the application of a charge voltage. Similarly many gel-cell and AGM batteries exhibit terminal voltages of 13 Volts or so even though the readings drop to 12.9 Volts (AGM and gell-cell batteries have higher specific gravities than do most flooded-cell batteries and, therefore, the terminal voltage is higher. Terminal voltage, Vt = 6(0.85 = SG) where SG is the specific gravity of the electrolyte for a 12 Volt lead-acid battery (yes, this also includes AGM and gell-cell batteries).
Note that state-of-charge means little to you. It is state of capacity that is important and must be inferred from a relatively simple test. Given two batteries in separate boxes that do not allow you to inspect other than terminal voltage: They both read 12.65 Volts. Which one has a greater capacity? One cannot know. What good does it do to have a degraded start battery that exhibits a so-called full state of charge. One day it will easily start the engine. The next day it will not because it may have degraded from a 1 Amp-hour state of capacity to a 3/4 Amp-hour in one day's use and may not start the engine. To be sure, that same battery began as a 75 Amp-hour battery and for 6 years always started the engine. One day the final denouement occurs to one's dismay.
Because it is hard on the life of any lead-acid battery to fully discharge it in order to make an attempt to determine capacity it is not advised to do so unless one has many batteries available to waste before choosing a similar one that has not been discharged. What WILL be revealing, though, is to discharge the battery at the theoretical 20 hour rate for an hour (at least 1/2 hour), let the battery recover for at least an hour and measure the standing voltage and use a "look-up" table to determine approximate state-of-charge and using simple algebra calculate the capacity from that information. At the end of the hour observe the terminal voltage under load and then the voltage immediately after the load is removed. Note the current just before removing the load. Calculate the internal resistance by dividing the difference in voltage by the difference in current (as an example for a 180 Amp-hour rated battery the current difference will be 180 divided by 20 minus zero...zero because after removing the load the current should be zero). |
At this point as I read it, we're supposed to be calculating internal resistance.
Let's say no load voltage is 12.7 v and load voltage is 12.5 v with a 20 amp load, then (12.7-12.5)/(20-0)= 0.01 ohms. An older battery will have higher internal resistance, and will have a greater voltage drop with a load. This will make the battery less useful, particularly with higher current loads.
To rephrase what Rick said above, plot volts versus current, the slope of the line is the internal resistance of the battery. Since you can't put your voltmeter across the internal resistance to measure the voltage drop across it, you measure the voltage at the terminal at two different loads and calculate the slope.
i=internal
T=terminal voltage
V(B)=internal battery voltage (constant as is R(i))
V(i)=voltage across internal resistance
V(T)= voltage at terminal (or across external resistor)
slope or R(i) = (V(i1)- V(i2))/(I1-I2)
V(B)=V(T)+V(i)
solve for V(i)
V(i)=V(B)-V(T)
substitute
R(i)= (V(B)-V(T1))-(V(B)-V(T2))/(I1-I2)
R(i)=(V(T2)-V(T1))/(I1-I2)
Notice how V(T2) is over I1
Quote:
Originally Posted by Rick For this test note as well the initial change in voltage from no load to the 20 hour rated load applied. Calculate that initial resistance as well making sure that initially there is no surface charge on the battery which will make the results lower (better) than the actual value.
So, the 20 hour rated current is 9 Amps in this example. Using this test you will notice that the terminal voltage drops linearly for awhile then settles down to a constant value. I would expect that voltage to be approximately 12.5X Volts or better for a flooded-cell 180 Amp-hour “real” battery.
The approximate capacity, C = (measured test Amp-hours)/(decimalSOC), where |
Once again, maybe I'm reading it wrong, but I would put C = (measured test Amp-hours)/(1-decimalSOC) I put a 9 amp load for one hour, so 9 amp-hrs consumed. If I read a voltage at the end of that period of say 12.63, which my state of charge chart (not presented in this thread) says is a state of charge of 95%, then C=9/(1-0.95) = 180 amp-hour capacity at full charge. In this case my battery is still as new. Now use this battery for 3 years (or 3 months) and after 9 amp-hours, you read 12.42 volts, which state of charge chart says 80%. C=9/(1-0.80) = 45 amp-hour capacity at full charge. As Rick states below, the 9 amps out of this old battery is much greater than the c20 rate, so redo the test 45/20=2.25 amps and you will get a better capacity. John Quote:
Originally Posted by Rick the decimalSOC is the decimal value (0 to 1) of the state of charge shown by a chart showing state of charge versus voltage. Now to be more accurate one should divide C by 20 to get the 20 hour capacity current discharge rating and repeat the test at that current to recalculate a refined value of C.
Obviously the use of an accurate battery monitor is almost essential in making such tests.
AGM and gel-cell batteries require similar tests yet require slightly different observations and tests. More on that later. |