CycleAnalyst: Tuning Watt-Hours?

[PeteCress]

I'm running a couple of DeWalt DC9360 power tool batteries through my CycleAnalyst.

I'm getting the rated 2.3 AH almost right on the money cycle-after-cycle.

But in Watt-Hours, I'm only getting sixty or so - and the batteries have "79 Wh" printed on their backs and I take that to mean they are rated for 79 watt hours. There is nothing about amp hours on the batteries, but I've seen the 2.3 figure in quite a few places.

Seems like a rather large descrepancy to me - who knows nothing electric.

Since both batteries deliver 2.3 AH time-after-time, I assume they are in good working condition.

That seems to leave some sort of tweak in CycleAnalyst to bring the Watt Hours more into line with the Amp Hours.

Or could 60 vs 79 be explained some other way?

 

[gge5]

The pack's nominial voltage should be around 36V, but under load it will sag. Still, 60 Wh divided by 2.3A is about 26V. I can't imagine why voltage would sag 25% under a reasonable load. If you cycle through the displays on the cycle analyst, what values is it giving for min and max voltage?

 

[PeteCress]

The pack's nominial voltage should be around 36V, but under load it will sag. Still, 60 Wh divided by 2.3A is about 26V. I can't imagine why voltage would sag 25% under a reasonable load. If you cycle through the displays on the cycle analyst, what values is it giving for min and max voltage?

aMax = 30.91
vMin = 26.2

I'm guessing these are between resets and not overall.

Sounds like I may be abusing these batteries. There are only 2 connected in parallel and my use is generally full throttle when I come to a hill, otherwise nothing. Full throttle seems to pull something in the high 400's watt-wise.

 

[dnmun]

capacity is measured with discharge at .1C

 

[pdf]

Off hand, I can't see why there would be an adjustment for W-hrs. I assume the CA is calculating the W-hrs by integrating the product of the instantaneous current and the voltage, in which case, it would seem that it is not possible for the W-hr and A-hr to disagree by much if at all.

You can try a discharge at a low "C" rate to see if you get a capacity in W-hr more similar to what you expect as suggested above, but it still wouldn't help explain why the energy divided by the current capacity is higher than what the apparent average voltage is. Might send an email to the CA people; they might have an explanation of why they don't agree better. They seem pretty sharp.

Of course, if you are actually getting that much voltage sag all the time (such that the min. is actually close to the average), then it might be correct after all.