Re: Battery Replacement

[Lee Hart <leeahart@earthlink.net>]

Michael Haseltine wrote:
> Could you go throught the peukert business again? I have only the
> most rudimentary understanding of it and don't see how it relates
> here. I don't know the peukert for these various batteries, so what's
> the difference between the 875 and the 105 and how does that effect
> max current?

In theory, C = I x T, where C is a battery's capacity (in amp-hours), I
is the current (in amps), and T is the time (in hours). Thus, a 100
amp-hour battery should deliver:

  1 amp  for 100 hours = 100 amp-hours
 10 amps for  10 hours = 100 amp-hours
100 amps for   1 hour  = 100 amp-hours

But 100 years ago, Peukert noticed that the actual amp-hour capacity of
a lead-acid battery depends on the discharge current. For example, a
real battery might deliver:

  1 amp  for 100 hours = 100 amp-hours  
 10 amps for 6.3 hours =  63 amp-hours
100 amps for 0.4 hours =  40 amp-hours

He made lots of measurements, graphed the data, and figured out a simple
equation to estimate the REAL capacity. Peukert's equation is Cp = I^p x
T where Cp is the Peukert capacity (in amp-hours), I is the discharge
current (in amps), p is the peukert exponent, and T is the time (in
hours).

If the Peukert exponent = 1, it is the same as the equation for a
perfect battery. But if p=1.2, then you will get the performance quoted
above.

Every real battery has a Peukert capacity slighly larger than the quoted
capacity at the 20-hour rate, and a Peukert exponent slightly larger
than 1, to account for its loss of capacity at high discharge currents.
-- 
Lee A. Hart                Ring the bells that still can ring
814 8th Ave. N.            Forget your perfect offering
Sartell, MN 56377 USA      There is a crack in everything
leeahart_at_earthlink.net  That's how the light gets in - Leonard Cohen