Is this actually an ultracapacitor?

Yes, that's an ultracap. He mechanically lowers "C" as the cap discharges, regulating voltage.

I spoke to a guy working on ultracaps for EV use--he said they're getting 25F/cm^3 with aerogels, at some fraction the weight of LiIon, 4v, and miniscule ESR.

For a 50KWhr pack (about 300 miles range), that works out to C = 22.5e6 F fitting in 900 L, if I calculated correctly.

--
Cheers,
James Arthur

There is attractive force between the plates. If the plates have to be moved, this takes power... wonder if that is significant? (I can barely remember the equations :-)

Likewise I'd expect any movement of electrolyte to consume power. ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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I love to cook with wine.     Sometimes I even put it in the food.

Hi Jim,

If the capacitor is disconnected from the external circuit when the plates are moved, it's straightforward enough to figure out from conservation of energy. Since C = QV, and Q is constant, if C increases then V must decrease to compensate and vice versa, so you just have

1/2*C_1*V_1^2 = 1/2*C_2*V_2^2 +/- energy added/removed to the system. Obviously the power depends on how quickly you do it.

If the capacitor is connected to a resistive load and is discharging while you're moving the plates it is a bit more complicated; I think it's like the capacitor discharging differential equation in Electronics

101 except you have a changing capacitance that's a function of time, dC/dt, which for a parallel-plate capacitor is a function of (a known quantity) dD/dt, i.e. how fast you're moving the plates. It's a system of equations, but it's a pretty simple system and I think it's solvable in closed form.

Once you know the voltage as a function of time while you're moving the plates you can again use conservation of energy to figure out whatever quantity you're interested in.

Hi Jim,

If the capacitor is disconnected from the external circuit when the plates are moved, it's straightforward enough to figure out from conservation of energy. Since C = QV, and Q is constant, if C increases then V must decrease to compensate and vice versa, so you just have

1/2*C_1*V_1^2 = 1/2*C_2*V_2^2 +/- energy added/removed to the system. Obviously the power depends on how quickly you do it.

If the capacitor is connected to a resistive load and is discharging while you're moving the plates it is a bit more complicated; I think it's like the capacitor discharging differential equation in Electronics

101 except you have a changing capacitance that's a function of time, dC/dt, which for a parallel-plate capacitor is a function of (a known quantity) dD/dt, i.e. how fast you're moving the plates. It's a system of equations, but it's a pretty simple system and I think it's solvable in closed form.

Once you know the voltage as a function of time while you're moving the plates you can again use conservation of energy to figure out whatever quantity you're interested in.

Sure. The true capacitor equation is...

I = d/dt(C*V) = V*dC/dt + C*dV/dt

(Substitute charge as suits you. If I had a mind to, I could derive the force situation. I think I'll have a glass of wine instead ;-)

I've had occasion to use E = d/dt(L*I) in real world situations but he not the capacitor version. ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.

Actually, Jim, it is dC/dt(C*V)

After all, capacitance changes with voltage.

Capacitive reactance changes with voltage.

Might be interesting to charge at low voltage (available) and discharge at a higher voltage (usable)........

Claims to have compensated at a 35W / 6A dischage rate, mechanically.

RL

Oh, yes, movement of the plates in a ultracap would take a SIGNIFICANT amount of power - might as well as jump out of the car and push to get better efficiency. I think the equation is (not looking) U=0.5*C*V*V ..

Look at it this way:

Take a capacitor of one farad, charged to one volt. Charge is one coulomb. Energy is 1/2 CV^2 = 0.5 joules.

Now decrease the capacitance to 0.5 farad by whatever means. Charge is still one coulomb. PD is now 1/0.5 = 2 volts. Energy is 1/2 CV^2 = 0.5 *

0.5 * 2^2 = 0.25 * 4 = 1 joule.

The extra half joule must have come from somewhere. Energy must have been put in.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Perzackly.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

That's what I don't get-- that high-dielectric fluid would be stuck like glue to the plates as long as it is charged-- it wouldn't just drain off-- and the force * distance to remove it is the missing energy.

On Wed, 25 Jul 2012 11:46:07 -0700, Jim Thompson K

If a person believes in the law of conservation of energy, then, yes, it's very significant.

If Q is charge, V is voltage, C is capacitance, and E is energy, we have:

Q = CV

and

E = CV^2/2

If by some means (mechanical in this case) we decrease the capacitance to half its starting value while no load is connected, it's not too hard to see that the capacitor will contain double the energy it had before we decreased the capacitance. This is how parametric amplifiers work.

The energy has to come from somewhere. In this case it came from the mechanical work that had to be done to decrease the capacitance.

The article says an "electromechanical system" was used to "lift the core of the device out of the electrolyte solution". What kind of electromechanical device? An electric motor perhaps? Where does the energy to run that motor come from? It seems to me that that energy is added to the capacitor. Why not just use that energy directly (assuming it's an electric power source running the motor)?

You don't get reviews in "scientific" journals, or money from the gullible that way.

Looks like they just re-invented the Electrophorus.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Oh really? I really think it is something else that has to do with material, thermo and structure. But who am I?

Jamie

Electronics

system

solvable

the

whatever

That is a property of the dielectric (or in this case physical movement = of the plates a wholly different situation), not static plate vacuum = physics. Trying to solve that for constant voltage must be interesting.

?-)

Check; just do not strain family jewels..

Actually, who cares where the energy comes from, if it can be converted, stored and reclaimed so easily and obviously. Question is, how much is lost in the physical operation of resizing?

RL

Who is this "John asS"? Dumbass or is it dumbshit... the equation IS d by dt of the product C*V

Don't you understand the partials I showed or are you AlwaysWrong? ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.