Wednesday, Jan 29, 2014 at 18:35
I understand your point Leigh.
A couple of different perspectives can be taken on this.
1. An amp hour is an amp hour no matter whether it comes from a 100Ah battery or a 1000 Ah battery, so the Ah actually consumed is what it is.
Watthours will be different due to the voltage variations over the runtime.
2. Peukert tells us though that the faster we take it out, the less we can take out, so that gives us a situation where, even though we have taken 5.4Ah from a 100Ah (C20 rate) battery, but at a high discharge current, in this case 0.65C, we can end up with less than the expected (100 - 5.4 = 94.6Ah) capacity remaining in the battery.
Math can help us here, as the formula:- "Capacity reduction in Ah = time in hrs x (discharge current x Pk)" tells us how much capacity we will lose from the battery in the time for which the discharge is occurring.
Pk is Peukert's constant, and while we don't know it for this particular battery, it will be fairly safe to pick the high end for an AGM battery of say 1.15.
As he said it was a battery pack then I assumed it would be either AGM or Gel.
You could use 1.25 as a constant for a gel battery if you prefer, but that will only change the answer by about 0.5Ah
The discharge current is 65 amps
The time is 5 minutes or 0.083 of an hour
So capacity = T x (current x Pk)
capacity = 0.083 x (65 x 1.15)
capacity = 6.2 Ah
Not a lot of difference from my original figure of 6Ah.
100 - 6.2 = 93.8Ah remaining
If the battery was a flooded type, then the lost capacity could be as high as 9Ah as you suggested,
Either way, the remaining capacity will still be better than 90Ah, assuming it was initially in good condition and fully charged.
Remember that the pump is only running for about 5 minutes.
By transposing that same formula you would find that drawing 65 amps out of that battery would give you 1.34 hrs to flat, or an effective capacity of 75Ah.
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