anyone know the answer to his one?

[bob golding]

hi all,

i am trying to get a ballpark figure for the total energy stored in a 12 volt lead acid battery over its lifetime. i am assuming 12 volt 110 amp hours with a  cycle life of 800 charge discharge cycles. this  is proberly  on the optimistic side. i have tried to work it out but am not sure of my figures. so far i estimate 1320 watts when fully charged at 110 amp hours. i know the the capacity goes down with age but not sure how much. is there an equation to calculate this. i am using a standard  so called leisure battery for my calculations. my 3 year old 110 amp hour batteries are now down to 20 amp hours after a load test at 5 amps.

cheers

bob golding

[Slingshot]

Bob,

Energy would be expressed as a power-time product, ie watt-hours, not power only (watts).  

But anyway, by your numbers the nominal battery capacity is 1320 watt-hours, or 1.32 KWH.  If you can really add/remove this amount of energy 800 times, then your total storage/release capability of the battery over it's useful life would be 1056 KWH.

In reality, if this is a typical garden-variety "deep-cycle" battery, the specs are not always as they seem.  The capacity of 110 amp-hours may be specified at a fairly low discharge rate, typically C/20, or 5-1/2 amps in this case.  Higher discharge rates will result in lower capacity before recharge is needed.  Also, if the maker specifies an 800-cycle life, it is probably referenced to a much lower depth of discharge.  If you really use all the capacity between recharge cycles, then your battery lifetime (total # of cycles) will be much less.  Finally, the capacity will also decrease with usage.  Capacity after several hundred moderately-deep cycles may be substantially less than 110 AH.

Manufacturers often show lifetime data as a function of depth-of-discharge.  It would be interesting to integrate this energy transfer vs cycles over lifetime, to see if there is an optimum depth-of-discharge that results in maximum energy transfer over lifetime.

[richhagen]

I think you have a large number of variables for which you will have difficulty finding equations to model them on.  I would probably make some basic assumptions, for example a 20% average depth of discharge relative to the new battery capacity for each cycle over the life of the battery.  Of course in reality the life expectancy of the battery will be effected by the depth of discharge distribution pattern.  If you make the assumption of 20% depth of original capacity discharge, with 800 cycles before the battery reaches end of life at the point where it can no longer cycle to this depth, then you can make some ball park calculations.  In your specific case, you may model it on a deeper discharge for a period of time and then lessen the amount of power taken out each cycle as the battery ages as your end of life discharge is only 20 out of the original 110 amp hours, but for a simplistic model.

 110 Amp Hours * .20 = 22 Amp Hours

If the average voltage during discharge is 12 Volts (of course it actually varies) then one can estimate the average Watts taken out per cycle.

 22 Amp Hours * 12 Volts = 264 Watt Hours

If the average Watt Hours removed is 264 and the number of lifetime cycles is only 800, then the total Watt Hours removed would be:  

 800 Cycles * 264 Watt Hours = 211200 Watt Hours, or 211.2 Kwh

Now if the battery cost you 40 Pounds over there, then you would end up with an added cost of just under 19 pence per Kwh.  Although IIRC that seems quite a bit higher maybe a bit more than double previous estimate calculations I did on this subject regarding the added cost of battery storage for my tiny system which were about 18 Cents U.S. based upon cheap $60 dollar marine battery, 25 Amp hour discharge and about 3 years of life cycled daily, it is an example of why battery storage is not advantageous for grid tied systems without the specific need for battery backup in the event of power outages.  Better batteries lessen the cost per KWH for storage, but the total system basically still cannot compete with the cost of grid power if storage is required.  Rich

[wdyasq]

Hello bob,

You have opened a real "can o' worms" here. It depends on a lot of variables.

One would need to get some sort of feel of how much energy was in at each particular point in time. This will mean getting a discharge curve. To get close to this one would  plot energy and time on a graph. One would start with the 12.8V (just a guess) and discharge with a constant Ohm load. As time increased and energy drained one would plot the curve and have a spreadsheet program get a time/voltage graph.

This would give one an idea of the point in time of a 'new' battery. As the battery aged, new graphs would need to be built. After all this work, one would get a new progression/regression curve and finally a master formula for the power.

This curve could then be 'Integrated' over the time span and give an approximation for your particular case.

Enjoy,

Ron

[bob g]

you might want to check out rolls/surrette website

they have alot of info and the formula to determine the answer to your question

of course it is tailored to their products but the relative info is useful for other manufactures as well,,, generally speaking.

as the others have stated there are many variables.

in general you will get more kwatt/hrs out using a specific discharge/charge regime.

if you can work within those constraints you can maximize the life of the batteries or get more kwatt/hrs out over their lifespan,but not both.

so it all comes down to what you want or need from your batteries.

for instance, keep them cold and they will last longer, with lower capacity.

keep em hotter and they will deliver more, but at a reduced lifespan

and the list goes on.

having a clear understanding of all the variables and how they effect the battery

lifespan will dictate which method you use to get what you need or want out of them.

or at the very least have a reason to believe you will

otherwise it can be a crap shoot

bob g

[bob golding]

thanks guys,

 that is what i thought. at least i can plug in the number for the dicharge as the bank is suppling a fairly constant load to an inverter.i say fairly as of course there is less load in the summer than the winter from the lights. recharging is from a wind turbine so is almost immpossable to estimate. the reason i asked is because i am trying to cost up making a vanadium redox battery. if i can tackle the technical challenges,and there are a few, i want i see if it will be worth the trouble. could come into its own when recharging from the genny in the summer as the recharge time is 1 to 1. i might just do it anyway for the challenge but thought it might be worth a least finding out. slowly sourcing all the difficult to get parts. still need a cheap source of vanadium pentoxide. there is a firm over the pond sells it but  not to the UK. and before anyone asks yes i have read the MSD on it. be ok as long as i dont breathe while handling it.

cheers

bob golding

[Jeff]

Hi Bob,

As Ron (wdyasq) said: "You have opened a real "can o' worms" here". Now that I'm in the "post comment" window, I can't see who the first one was to give you an answer. What I wanted to say though, is the first comment showed something very obvious (to me, at least). All the others back this up also...  this is a very good point to have or ass more batteries. The more batteries you have in your array, the less each one is drained. This adds to a longer life in a number of ways: 1) The less drain on each battery before re-charge, the longer the life. 2) Less drain on each battery means a quicker recovery, or recharge time. Both these reasons can get very in-depth, such as more efficient charging, and less amperage required for that recharge. To make a short-story-long...add more batteries (of equal amp-hour rating) ASAP, if not all at once.

Best of luck to ya'!