[RE-wrenches] discharging Rolls batteries

Hugh hugh at scoraigwind.co.uk
Fri Jan 15 00:02:20 PST 2010


Hi Jamie,

>
>Remember, as batteries cool actual capacity is reduced, so if 200AH 
>is 50% @ 25C it is significantly more than 50% @ 5C.   Thus, you are 
>discharging more deeply.

But earlier you put it this way:

>>Regarding temperature effects on capacity, earlier responses are 
>>spot on as the lower capacity is totally as a result of slower 
>>reaction times as a result of lower temperatures.  

There is an issue here that I need to understand better.  You state 
that a battery has lower capacity in low temperatures.  Suppose you 
take a fully charged, 400 Ah battery and cool it down to -5 degrees C 
where according to our numbers it will only have 80% of its nominal 
capacity.  You then remove 160 Ah (say 10 amps for 16 hours).  It 
will then be 50% discharged.  Now warm it up again to 20 degrees or 
whatever.  My question is: will you only have 200 amphours left in it 
now?  And if so, what happened to the other 40 amphours?  Does low 
temperature operation actually lose amphours, or is it just more 
sluggish?  What is the chemical explanation for the lost amphours?

I understand batteries as a chemical process of converting amphours 
into chemical changes.  I assume that a given amount of electrical 
charge converts a given amount of lead into lead sulphate (and 
likewise) back again.  I understand that cooling will make this 
process less efficient and thereby result in a rise in charging 
voltage and a drop in discharging voltage.  But does a low 
temperature actually mean that a given amount of lead being converted 
to sulphate actually give you less amphours electrically?

(I have similar questions in relation to Peukert's equation where 
high discharge rates impact on the amphour capacity.  The capacity 
apparently 'recovers' when the discharge rate is reduced.  To what 
extent is the capacity actually lost by using high discharge rates 
and to what extent is it just a voltage effect that impacts on the 
terminal voltage, rather than the actual chemical state of the 
battery?)

I hope you can follow my descriptions.
-- 
Hugh Piggott

Scoraig Wind Electric
Scotland
http://www.scoraigwind.co.uk
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