Caps instead of batteries?

[PEdoubleNIZZLE]

Is it possible to use large (or many small, and in parallel) capacitors instead of batteries for a wind system? How do I figure out the "amp-hours?" (I don't have my reference book with me). I know that I would probably have to regulate the charge voltage since caps can be charge at any voltage up to their rating.

[Ampman354]

capacitors don't hold the charge as long as you need them to, to be a battery bank.

[orochi8]

If you are going to use capacitors for most anything on this scale, http://www.maxwell.com/ultracapacitors/ is going to be your best bet. Either way though, I really dont think you are going to be able to afford to use them on the scale neccesary to hold a reasonable amount of potential, at practical cost. It might be worth considering as a means to extend battery life, but I think a charge controller will do that well enough with only minimal losses in power, and far less complication.

-Orochi

[inode buddha]

If you're interested in playing with that idea, I would use both batteries and caps. I think the caps would be good for starting large inductive loads if you parallel them with batteries.

[wooferhound]

I remember a website somewhere that said that a

One Farad capaciter equals 2.5 amphours . . .

[maker of toys]

one feature of capacitors (even the ultra caps) is that the voltage on the capacitor varies much more with state of charge than a chemical battery. . .  which makes using electronic devices problematical, to say the least.  (about the only things that wouldn't care about such wide swings in supply voltage would be incandecent lighting and heater coils.

score me in the 'use'm to start big loads' camp. . . though I can't think of anything that I'd run on battery or battery/inverter power that would need capacitive help that doesn't already have capacitors built in.

one thing to remember about the ultra-capacitors is that they have much higher 'esr' (effective series resistance' than regular capacitors;  they don't react to fluctuating loads well . . . you'd want some conventional capacitors as filter caps for your alternator or generator input to the cap bank.  

Also, ultracaps tend to be electrically and mechanically fragile and can be damaged by high-rate charge or discharge.  (true, so can batteries, but not as easily)

[maker of toys]

ok, so me and the calculator tried to work that out. . . caution, I get pedantic below. . . and I've got a cold, so it's probable that I bobbled the math somewhere.

let's start with power. . .

1. amp x 1 volt  =  1 watt.
   
1. watt for 1 second  = 1 joule.
   
   

there are 3600 seconds in 1 hour.

so 2.5 amp hours is 9000 joules per volt.

therefore a 12 volt, 2.5 amphour battery has a 'nameplate' storage of 108,000 joules.

of which we can probably get about 54,000J as useful energy.

a capacitor stores energy as:  joules = (volts^2)*Capacitance/2

manipulating:

(((2 * Joules)/ Capacitance)^0.5) = volts

and, voila, a 1 Farad capacitor would have to be charged to 330 volts to give the same 54,000 J energy storage.  

Using the 2.7 V working voltage for the caps listed in the link above. . .

(2*Joules)/(V^2)  =Capacitance

which calls for 14,800 Farads;  probably more, as that's assuming complete discharge of the capacitors, and very few kinds of electronics can be expected to function below about 1.2 volts . . . and we're not given to using electrical RE for primary heating.

kinda an broccoli-to-bananas comparison, yeah?

I could whip out some calculus and figure the kind of capacitor bank you'd need to store 54kJ with the voltage only varying between, say 15 and 10.5 volts. . . but you'd all be asleep by then.

[tecker]

  I you want to pop for the caps with voltages close to device working voltage you can

work with a fast charge fast discharge srerio . The super caps that are out there are limited to 2.5 t0 5.5 volts . so you need charge distribution as well as discharge equalization circuits in between series caps . Ill post some data I've acumulated.in the form of links . They have really come of age.

[finnsawyer]

You shouldn't scare people.  You don't need calculus to solve your little example.  Simply use the equation for energy.  The change in energy delE as the capacitor discharges is:

   delE = 54,000 = 1/2xCx(V1^2 - V2^2) = .5xCx(15x15 - 10.5x10.5) = 57.375xC

Or:  C = 941 farads.

To get to a working voltage of 15 volts using the 2.7 volt capacitors you need to put six "banks" in series.  Capacitors in series divide, (by six here), so each bank must have 5,646 farads.  Since there are six banks you need a total of 33,876 farads of the 2.7 volt capacitors.  Cheers!  

[maker of toys]

true-

I worked it a little different, but came to the same conclusion after I posted that.

what can I say. . . it was late and I don't think too fast with a stuffed-up head.

-Dan

[thunderhead]

For batteries or capacitors, you can draw a graph of the voltage against the amount of charge stored.  For a 12v 100Ah lead-acid battery, the curve starts at maybe 10v, with virtually no charge stored, and then rises to just over 12v, with 360,000 coulombs stored.  For the battery this curve is steep at both ends, but quite flat in the middle: and there are two curves, one for when you are putting energy in, and one for taking energy out, separated by about a volt-and-a-half.

With a capacitor, the "curve" is a straight line, starting at zero volts and zero coulombs, and in the case of a (huge!) 1F capacitor, reaching 12 coulombs at 12 volts.

Even if you could get a capacitor rated to store 360,000 coulombs at 12 volts (it would be a 30,000F capacitor - capacitors of a millionth of this size are still as big as a torch battery) the curves would meet at 12v, and again somewhere on the "flat" bit of the battery curve.  

This matters because the total energy stored is equal to the area under the curve.  (Don't worry about why - it's integral calculus.)  When the battery is 50% discharged, the output voltage is maybe 11 volts, but the capacitor is only 6 volts.  There is a lot of paper between the left hand side of the battery curve and the left hand side of the capacitor "curve".  All that acreage of paper represents energy the battery can store, that the capacitor can't.

Another way to think of it is that the energy stored in the battery is approximately the average voltage times the total charge - call it 100Ah times 11v, or 1.1kWh.  The same is true for the capacitor, but the average voltage is now 6v, and the energy only 0.6kWh.

To make things worse, several people here pour scorn on my notion of building an invertor that could deal with the extra voltage range produced by Edison batteries.  Not without reason, either - power electronics is not an easy subject to dabble in, which is why most people buy their invertors, instead of building them.  That problem would be nothing compared to the voltage range to get all the energy out of a capacitor.  My 9-cell Edison battery's curve goes from 14.85v fully charged to 9v fully discharged, but your capacitor is at 0v fully discharged.  

Getting that last bit of charge out would be a real struggle.  You certainly couldn't do it with shop-bought invertors, and I think you'd really need to put your thinking head on to design something to do it efficiently.