Rating of wet-cells based on discharge time.

[EEAOC]

I need help on finding the some resources or formulas on calculating or evaluating the rating of wet cells based on its discharge time. For example if a wet-cell discharges in 20 hours at 4.5 volts when fully charged and at the same time it takes 10 hours to discharge at 7.5 Volts. How do you rate the voltage of this cell. Is it 7.5 volts or between 4.5 and 7.5 volts? I cannot find any EE books that cover wet cells or this topic. Thanking you in advance.

 

[rbulsara]

Battery manufacture's have published charts for this. Its non-linear.

Ask C&D or Enersys or the like.

 

[nukeman]

Not really sure of your question, batteries are normally rated in amp hours not voltage. You can specify the "voltage rate" at any voltage. Normally, you would use the minimum voltage that the load could handle and then use the vendor supplied discharge curves to determine the discharge rate to whatever end voltage you have chosen. In your example is the discharge the current the same?

You might want to check into a book called Storage Batteries by George Vinal which has a lot of good information about wet cells. Also, do a web search on Peukert's equation which relates the discharge current to the time to discharge in order to calculate capacity at different discharge rates.

Hope this helps.

 

[rbulsara]

EEAOC:

I think you have confused volts with amperes. Did you mean 20 hrs at 4.5A and 10 hrs at 7.5A? That would make more sense. For the purpose of rating the voltage is assumed constant or a nominal value is designated. In reality votage does drop to some extent and certain low voltage the battery is considered dead.

 

[rbulsara]

Also if you have noticed, the battey capacity is rated in AH (ampere-hours). So the product of Amps and the hours remains relatively constant for a given battery. Higher the amps, lesser the back up time and vice versa.

 

[fsmyth]

Looks like the feller read "volts" when he should'a
read "amps".

A /cell/ will have a particular open-circuit voltage
according to its chemistry. It will have a maximum
current delivery based on its construction and size.
The amp-hour rating is determined at a particular
current drain for a specific time to a specified
terminal voltage. Any of these variables may change
according to chemistry, construction, or vendor test
methods.
Determine your load (in amps). Determine your maximum
run time required before the next charging cycle.
Multiply by 0.8 - this will be your required amp-hours.
Select the cell to reach or exceed this; more is better.
Stack cells until you reach your operating voltage
(a "battery"). Set a maintenance routine, and follow it.
<als>

 

[zacky]

fsmyth, any specific reason to muliply by the factor 0.8

 

[fsmyth]

Yeah, to give the batteries a break.
And to compensate for overly optimistic marketing.
(Ever dealt with audio power amps????)
I try not to discharge below 75% on lead-acid
and similar chemistries - the useful life drops
dramatically if over-discharged constantly.
NiAl can stand a bit more (different electrolyte).

NiCads are different.
NiMH are even more different.

<als>

 

[zacky]

If you want to give the battery a break, then you should select a higher AH battery and not smaller.
Then you must multiply by the factor (1/.08)= 1.25

 

[fsmyth]

Zacky, multiplying the run-time by 0.8 effectively
prevents over-discharge on wet lead-acid batteries.
This is a very general rule; specifics will change
with duty cycle, charge rate, reserve capacity, etc.
<als>

 

[rbulsara]

fsmyth:

zacky is right also.

You need to divide by 0.8 "the calculated load" to arrive at AH size of the battery. To determine effective runtime of given battery you multiply rated AH by 0.8.

This is just a margin of safety, nothing scientific about it.