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I have an AGM battery that is rated at 12V 225Ah. Using the 50% "rule", the battery should be able to deliver 12 * 225 * 0.5 = 1350 watt hours = 1.35 KWh. However, when the battery is under load, the voltage drops to as low as 10.5V, where many power inverters are set to warn the user. I don't know what the average voltage is during a capacity test, but it slowly drops from as high as 13V (with a very light load at the beginning of the test), to around 10.5V if the load is just right such that it is heavy enough to dip the battery that much at the tail end of the load test.

So my question is, do battery manufacturers expect us to use the nominal voltage when calculating the usable capacity of that battery or the average voltage during the capacity test? I don't want to overestimate this 50% rated capacity (100% usable capacity) and overdischarge my battery. The difference is about 6.25% if I use the average voltage (around 11.25V) vs. the rated voltage (12V). For example, 11.25V (ave) * 225Ah * 50% = 1.265 KWh, not the 1.35 KWh we calculated using the 12V rating. Which is more correct? is it more correct to just use the watt hour rating by the manufacturer and not worry about voltage sag under load, as that is already accounted for by the different capacity ratings such as 10 hour, 20 hour... drain rates?

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    \$\begingroup\$ By default, the capacity of a lead acid battery is given assuming that it is discharged with a constant current over a 20 hour period. So your 225 Ah battery should be able to deliver 11.25 Amps for 20 hours. If you discharge at 20 Amps, the Amp hour capacity will be less. If you discharge at 5 Amps the Amp-hour capacity will be greater. The energy delivered to the load should be calculated based on the average voltage during actual discharge in your conditions. Different discharge rates will result in different amounts of energy delivered. Energy delivered to load depends on the load. \$\endgroup\$
    – mkeith
    Jan 26 '20 at 19:45
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It depends. Manufacturers use the C20 (20-hour), C10 (10-hour) or C5 (5-hour) capacity ratings as a standard way to express the total amount of Ah a battery can deliver when completely discharged in a given time, under given circumstances.

Note that this value says nothing about the battery voltage, only the current delivered over time. The battery voltage during this time will vary between about 12.7 Volts (fully charged) to 10.8 Volts (empty). The detailed behaviour should be looked up in the manufacturer's datasheet.

Generally, the slower an AGM battery (or any lead acid battery) is discharged, the higher the total amount of Ah it can deliver.

I.e. the total amount of energy delivered by a battery is highly dependent on the circumstances.

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    \$\begingroup\$ Good answer. I might have used a bold font for "in the manufacturer's datasheet" and underlined it twice. \$\endgroup\$ Jan 26 '20 at 19:48
  • \$\begingroup\$ @ElliotAlderson, Agree, done! \$\endgroup\$
    – StarCat
    Jan 26 '20 at 20:21
  • \$\begingroup\$ The AGM battery will deliver about the same power regardless of the discharge rate, provided the finish discharge rates are similar. For example, if a 12V 100Ah battery is drained at 50A, it will not last for 2 hours, but it will likely last for at least 1 hour. The remaining 50Ah of capacity can be extracted at a slower rate (for example, 2.5Ah for 20 additional hours). If you instead take a fully charged 12V 100Ah battery and discharge it for 40 hours at 2.5A load, you will get the same 100Ah of capacity. There is very little difference in ACTUAL capacity with mixed load vs. light load. \$\endgroup\$
    – David
    Jan 26 '20 at 21:22
  • \$\begingroup\$ To clarify, suppose I want to extract 50Ah out of a 12V 100Ah fully charged battery. If I start with a 50A load, it will "beep" at 10.5V soon into the drain test (perhaps at 1/2 hour), then I can just reduce the load and continue draining, perhaps at 25A, then again it will eventually beep for low voltage warning, then I reduce the load to 10A.... Eventually I will get the 50A out of the battery. If I let this battery sit for maybe 1 hour after the test, it should be in about a 50% SoC. Similarly, if I had instead ONLY used the 10A rate, it too would be at about 50% Soc after 50Ah drained. \$\endgroup\$
    – David
    Jan 26 '20 at 21:39
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    \$\begingroup\$ @David You need to cite some evidence for your statements about load vs. capacity. See Fig. 2 on batteryuniversity.com/index.php/learn/article/…. Discharging to 1.7V/cell at 1C rate gives about 1/10 the run time as discharging at 0.2C rate. That sounds like a factor of 2 difference in available capacity, which I would say is quite significant. \$\endgroup\$ Jan 26 '20 at 22:04
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I have very seldom seen a lead acid battery that provided a capacity in energy terms (kWh, Joules, etc). Usually lead acid batteries are rated by coulomb capacity (amp-hours). If just one number is given for capacity, then it is assumed to be the capacity derived from a 20 hour test. (C/20 discharge rate). So for a 225 Amp-hour battery, if the only thing the manufacturer tells you is that the capacity is 225 Amp-hours, then that means the battery should be able to deliver 11.5 Amps for 20 hours continuously when it is new. There is no default voltage endpoint defined for zero state of charge, but 10 to 11 Volts is probably what is used for the test. Depending on battery type, discharging to 0% can be very stressful for the battery and cause it to wear out extremely fast.

Unless there is a battery datasheet, you really shouldn't assume too much more about the battery.

Some battery makers may provide additional specifications (for example, East Penn Deka, Trojan, Concorde, and other higher end battery makers).

One Trojan AGM battery I looked at (MOTIVE T105-AGM) provided the following specifications for capacity:

  • Discharge time at 25 Amps: 440 minutes
  • Discharge time at 75 Amps: 115 minutes
  • Amp hours at C/5: 171
  • Amp hours at C/10: 187
  • Amp hours at C/20: 217
  • Amp hours at C/100: 230
  • Energy at C/100: 1.38 kWh

As you can see this is a 6V battery, and they used the C/100 capacity and nominal voltage of 6V to measure the energy. As you increase the battery discharge rate above C/100, both the average voltage and the capacity will be reduced.

This tells you that there will be a non-linear decrease in energy delivered to the load as you increase the discharge rate.

Note also that the internal resistance for the battery is only 1.9 mOhm, and the short circuit current is 3250 Amps (both from datasheet). The internal resistance cannot explain the relationship between discharge rate and Coulombic capacity. The voltage drop due to 10.85 Amps (C/20) and 1.9 mOhm is only around 20mV. At C/100 it is only around 5mV. That tiny voltage drop difference of 15 mV cannot explain why the capacity increases from 217 Ah to 230 Ah when you go from C/20 to C/100.

Datasheet link: https://www.trojanbattery.com/pdf/datasheets/Motive_T105-AGM_DS.pdf

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First of all, let's get one thing out of the way: using manufacturer's data should always be done conservatively (as you've done), since they list the specs of their batteries at the most favourable conditions — e.g. room temperature, battery is brand new, ...

This overprovisioning alone is likely to account for more than the 6.25% you're wondering about.

So my question is, do battery manufacturers expect us to use the nominal voltage when calculating the usable capacity of that battery or the average voltage during the capacity test?

They do not expect anything, since they cannot know what you're doing.

They give their specifications expecting a constant-current load. That may well be the case - suppose you want to power a device that requires 9 volts and draws 1 amp. You may choose to use a linear regulator (likely a low-dropout one) to make those 9 volts. Then you'll be pulling constant 1A from the battery, and you won't care about the voltage as long as it is above 9 volts; voilà, the Ah rating is all you need!

You may also choose to use a DC-DC converter for this task, since that would be more efficient. How much time can you power your 9V/1A circuit? It suddenly becomes much more difficult to compute, because:

  • you need to factor in the efficiency of the converter, and it doesn't have to be constant - it may degrade as the voltage gets lower
  • the voltage of the battery becomes important, and you need to consider it varies with temperature
  • you don't have the full data. Ideally, you want to know, if on a specific State-of-Charge (%), if you draw X amps, how much will the battery voltage be?

The manufacturers try to be useful by giving you typical discharge curves, that look like this:

discharge curve of a typical lead-acid battery

however that's not the full data you need, even though you may try to interpolate. In the end you may derive a mathematical model of the battery which fits the stated data, and then, by numerical integration, you can compute the Wh/KWh you'd get in your specific scenario.

The manufacturers don't do this for you, since there's a continuum of possible scenarios.

In the end, I find it much easier to do a real-life test with the intended load.

The (average volts × average current × hours × derating factor) Wh computation is a useful approximation that gets you in the right ballpark. It's not terribly important whether you'd pick the nominal vs average voltage; it's an approximation anyway, you should not count on it.

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