Re: Battery management: Blue Sea ACR vs Pro Split R zero voltage drop isolator
Now you have me paranoid U.W. and short of running down to the boat with my turkey baster, I'm being lazy today, spot me on this "Big.W."
"RC is 140", RC is defined then as 140 Minutes @ a 25 Amp discharge rate.
Also then, 140 min * 60 = 8400 Sec's @ 25 amps
charge(q) in coulombs = current(i) * time(t)
or
q=i*t
and since time(t) = 8400 sec's and current(i) = 25 amps, the charge in the battery = 25 amps * 8400 sec's or 210000 amp sec's(as)
or 210000 coulombs(q)........if you like
To convert sec to hours, divide by 3600 (60 *60)
so amp sec(as) converted to amp hours(ah) = as/3600
therefor, 210000as/3600 = *58.33ah (#1)
*(Gross rule of thumb in the hood is RC/2 is approx. equal to amp hour(ah), so this checks)
My laptop power supply says 1.5 amps(which I'll use since it was running fairly warm and auto temp sensing/proportional cooling fan in the inverter was whirling away), so 120V * 1.5 amps = 180 Watts
My inverter is 80% efficient(on a good day), so 1/.8 = 1.25
It then requires approx 225 Watts(180*1.25) @ 12 Volts
225 Watts/12 Volts = 18.75 Amps (#2)
From #1 and #2,
58.33 ah/18.75 Amps = 3.1
My discharge rate is about C/3.1
Then using.....
which is taking into account the intrinsic losses during discharge of lead acid's using the Peukert Law, and
IF(it should be) my battery follows this C/3 curve, based on what we see, I'm thinking the batt. is still okay?
i.e., at my discharge rate of C/3 and when I eventually arrived @ 11.75V(⌂t = ?, hard to use a linear approx. since it behaves non-linear like my ex), I still should have 90%'ish remaining (down to 9.5 volts anyway):der: