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I have 6 6 volt Intimidator 8AGC2M Group GC2 AGM batteries in series-parallel to make one battery bank that is ~13volts at 570ah. I'm using a Pico One battery monitor and it wants to know what the C5, C10, and C20 ratings are. How do I find them seeing as they are not stamped on the batteries.

Any help would be great!

Thanksenter image description here

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    \$\begingroup\$ As well as MKeith's answer you should find this Stack Exchange Q&A useful. \$\endgroup\$
    – Russell McMahon
    Oct 4, 2022 at 7:45
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    \$\begingroup\$ Also see here noting that only some of these are AGM batteries. \$\endgroup\$
    – Russell McMahon
    Oct 4, 2022 at 7:45
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    \$\begingroup\$ The manufacturer reall really really should be able to provide these figures !!!!!!!!!!!!! \$\endgroup\$
    – Russell McMahon
    Oct 4, 2022 at 7:45
  • \$\begingroup\$ The maker of those particular batteries is a USA based company that has been in business for many years. I would not be surprised if they have people on staff who respond to technical inquiries in a competent fashion. I have intimidator AGMs in my cars as a starting batteries. \$\endgroup\$
    – user57037
    Oct 4, 2022 at 16:47

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C20 is the capacity when discharged over 20 hours. This is the default for lead acid batteries. C10 is the capacity when discharged over 10 hours, and C5 is the capacity when discharged over 5 hours.

I found the datasheet for that battery, but it only gives C20 (which is the default for all lead acid batteries).

C20 is 190 Ah per battery, and since you have 3 batteries in parallel, it is 3 x 190 = 570 Ah. So C20 = 570 Ah. Obviously you already knew that.

The reserve capacity at 25 Amps is 409 minutes which is 6.82 hours. That means when discharged at 25 Amps, it can run for 6.82 hours. So that gives a capacity of 170 Ah per battery when discharged over 6.82 hours. That is not what you need, but it can help us. 170 * 3 = 510 Ah over 6.82 hours.

Now that we have two capacities, we can use Peukert's law to calculate Peukert's constant for your battery. Then we can use the constant to estimate C5 and C10.

I will leave it to you to search out Peukert's law.

To solve for the constant, k, we can use this equation:
k = ln(6.82 / 20) / [ln(190 / 20) - ln(170 / 6.82)]
k = 1.115

If you rearrange Peukert's law you can get the following:
C5 = C20 * (5 / 20) (k-1)/k
So this means that C5 will be 165 Ah (per cell)
C10 = C20 * (10 / 20) (k-1)/k
So this means that C10 will be 177 Ah (per cell).

These are just estimates from Peukert's law, which is not really a law. But you can try plugging them in and see how it goes. Don't forget to multiply by 3.

C5 = 165 x 3 = 495 Ah
C10 = 177 x 3 = 531 Ah

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  • \$\begingroup\$ You could also try reaching out to East Penn and see if they have the numbers you need. \$\endgroup\$
    – user57037
    Oct 4, 2022 at 4:18

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