# Ultracap Question



## GZ1 (May 31, 2008)

Hello,

I'm curious about ultracaps. I'm guessing they're not terribly practical or this would be all over the forums but I'll ask anyway.

I see that Maxwell makes a 3000 farad ultracap rated at 2.7 volts. Trouble is I don't know how (or if) a farad relates to an amp/hr. Anyone know how many of these little wonders would I need to equal the equivalent of 1 amp/hr of charge?

At around $100/per whatever an EV's-worth would require will likely be lots more than li-ion power, but a solid-state power cell that virtually never wears out and recharges "instantly" does fuel the imagination.

Maybe a better use would be a small bank of these in front of lead cells for some instant accelleration. (I can almost imagine a cool LED "boost meter"!)

Thanks!
Greg


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## Jacob Riskin (Jun 10, 2008)

Current gen. Maxwell caps are only good for a density of about 5-7Wh/kg, compare that to LiFePO4 at around 90-110Wh/kg or Lead Acid/AGM 25-45Wh/kg.

So it would weigh 5x lead acid...or more than the car...

On the other hand, it should be possible in theory to setup a system where the caps are used for acceleration/regeneration which would increase range, acceleration, and battery cycle life. For that you only really need seconds (20-60?) of power. In theory...

In the future EESTOR has promised 200-300Wh/kg SuperDuper-capacitors which would be game changing, all this battery talk would be a moot point if they could provide these at reasonable costs.


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## ClintK (Apr 27, 2008)

GZ1 said:


> Hello,
> 
> I'm curious about ultracaps. I'm guessing they're not terribly practical or this would be all over the forums but I'll ask anyway.
> 
> ...


I was very curious as well and decided to do some investigation (with Excel of course). If you buy 12 Maxwell 48.6V Ultracap modules (www.tecategroup.com) and connect them all together, you would have:
583.2V, 13.75 F, 2338340 J, and would cost you $25,200 with a range of 2.17 miles (equivalent to about 120V with 5.5 Ah).

This assumes you've got some serious (perfect efficiency) DC to DC converters because capacitor voltage drops to 0 as they're depleted.

I'm sure there are benefits to linking regular Lead Acids with Ultracaps, but I'm not smart enough with EE and chemistry to figure it out. I do believe a couple major advancements in capacitors would be the ultimate energy storage solution for electric vehicles (see EEStor).

Btw, if anyone else has done the capacitor calculation and gotten different results please let me know!


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## Jacob Riskin (Jun 10, 2008)

Hey can you break down that math for me? I've been trying to work that out.


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## major (Apr 4, 2008)

GZ1 said:


> Hello,
> Trouble is I don't know how (or if) a farad relates to an amp/hr. Anyone know how many of these little wonders would I need to equal the equivalent of 1 amp/hr of charge?


Hi GZ1,

Use energy for comparison. E=½CV². E is in joules, or watt-seconds. C in farads. V in volts.

Regards,

major


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## Jacob Riskin (Jun 10, 2008)

So lets say you used 500F 16V modules at $750 each, lets say you wanted enough to accelerate at 20kw for 10 seconds, or 200kw total. That roughly works out to 457 caps at $15 or $2343.

Sound about right? If you did the same thing but built the modules yourself (with raw cells) then it would be about half $1200 for 11 3000F 2.5V modules. 

This is all assuming you have a fancy controller than can get the voltages all right. What I like about this is you could have a car that does 0-60 and 1/4 mile in very quick times, yet doesn't destroy the batteries doing it.

Unless my power is off, i just calculated 20kw is only 26hp, so obviously this isn't enough. So its more like 4x that. Hmm that's quite a bit more expensive. To accelerate 3000lbs to 60mph in 8 seconds is 538kw/second (with 10% loss) or more like 29 * $111 or $3252.


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## GZ1 (May 31, 2008)

Thanks for the great responses ya'll.

The cap-front-ended battery pack isn't bad. Quick 0-60 times plus plenty of battery juice for cruising. If Jacob's math is reasonable then $3200 isn't a bad price to pay for some significant snap off the line.

I start back to school in the fall...I'm gonna have to make a friend or two in the EE department and get some help with what it would take to marry batteries and caps. 

I hear ya with the EEStor thing. Virtually unlimited and short recharges for usable long-distance ranges. That wouldn't be game-changing. That would be world-changing. The tides of war and fortunes would radically shift. Oil-based transport would die out faster than CD's took over LP's in the music business. Enormous political/economic implications.

I'll check my expectations until I actually see something though. Someone will invent such a solution one day and I can only hope EEStor found the secret sauce.

Greg


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## Jacob Riskin (Jun 10, 2008)

Don't trust my math! lol...I think it was still off...can someone verify this?

i fixed a typo i had 3.5v for the one of the caps, now they all work out to be roughly the same price. 

Lets say 3000F 2.5V $111 each
To get from 0-60 in 8 seconds
1361kg*.5*26.8224^2/8=63000watts add 10% then your at 67kw
3000*.5*(2.5^2) = 9375 watt seconds 

67300/9375=7.17 * $111 = $800

So now I'm getting roughly $800 to get 0-60 in 8 seconds with 3000lbs. That's much more reasonable. You'll probably need to double that to compensate for the linear loss nature of the caps, and adjust numbers and sizes to achieve the voltages you need. 

Can someone double check this?


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## ClintK (Apr 27, 2008)

Jacob Riskin said:


> Don't trust my math! lol...I think it was still off...can someone verify this?
> 
> i fixed a typo i had 3.5v for the one of the caps, now they all work out to be roughly the same price.
> 
> ...


Kinetic Energy:
Energy (J) = .5*Mass*Velocity^2
So it takes .5*1361*26.8224^2 = 490,000 J to just move the mass (not taking into account rolling or wind resistance).

Electrical (Capacitor) Energy:
Energy (J) = .5*Capacitance*Voltage^2
Assume no losses (terrible assumption), Capacitor Energy = Kinetic Energy, so...
490,000 = .5*Capacitance*Voltage^2

You can add capacitors either in parallel (more capacitance, same voltage) or series (less capacitance, more voltage). In an EV application you would probably want more voltage, but since their total stored energy is the same I'm going to use the easier calculation (parallel):
Ctotal = C1 + C2 + Cn

The Maxwell Capacitor Modules are 500F, 16.2 V, so now our equation is:
490,000 = .5*Capacitance*16.2^2
Capacitance required = 3734 F
Round up, and you need 8 Capacitor Modules
At $750 each that's $6000 (Assuming perfect DC to DC converter, no rolling resistance, no wind resistance)

Please feel free to correct my calculations!


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## Jacob Riskin (Jun 10, 2008)

Ok, I see my mistake, I was dividing by the time for some reason. So I had 1/8th the numbers I should have. 8x$750=$6000...

Hmm that's still pretty pricey. Accounting for losses (drivetrain, controller, motor, air resistance, etc...) I'm not sure how much good 2-4 second boost would be. You could probably get a 2-3 second boost form a $3k system (30-50% loss) with an expensive controller that keeps the voltage leveled.

Still for something like the KillaCycle i would think you would be better off than batteries.


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