Transporting and Accessing Charge at EHV, Part III

Dissectible, fast-switching supercapacitors?

The main reason for the very low current calculated last time is that the dissectible capacitor analysed had a very small capacitance.

We could do far better, if we could obtain a capacitor having three essential requirements:—

1)  It must have a large capacitance — the bigger the better — in a reasonably small size.

2)  It must still be dissectible.

3)  It must still be able to be charged and discharged rapidly — the quicker the better.

Fig.3 D-cell sized supercapacitors.
Image from http://www.maxwell.com/products/ultracapacitors/d-cell-series

Nowadays capacitors of very high capacitance are readily available, such as the supercapacitors shown in Figure 3, which have a rated value of 310 farad at 2.7 volts, in a D-cell size (for version BCAP0310 P270 T10).

A D-cell has a nominal diameter of 33.2 mm, and a nominal height of 61.5 mm. So it has an external surface area of 8154.9 mm², i.e. the same as a sphere of about 25.5 mm radius.

So, requirement 1) above can certainly be met now. Unfortunately, as far as I know, supercapacitors currently cannot meet the other two requirements. They are not dissectible, nor can they be charged/discharged rapidly.

However, also as far as I know, there are no fundamental scientific or technological reasons preventing capacitors from existing that would meet all three of the above requirements. In what follows I'll assume that such capacitors will become available in the not-too-distant future.

Calculations for a fast-switching, dissectible supercapacitor

I'll now analyse, in broad outline, a seven-stage cycle of operation for a hypothetical fast-switching, dissectible supercapacitor, being used to transport charge into a dome already at high voltage.

1) Charge a capacitor rated as above to its full voltage. Then its stored energy is:—

U_E = ½CV² = ½ × 310 × 2.7² = 1129.95 joules

Its stored charge is:—

q = CV = 310 × 2.7 = 837 coulombs

Note that q is the charge on either plate, not on both plates combined (which is +q +(-q) = 0). Also, while the capacitor is assembled, the charges on its plates are bound, and contained within its can, i.e. its outer metal container. 

2) Physically transport the capacitor into the inside of a metal dome, already at a voltage of say 800kV. The capacitor can is itself initially uncharged, so no energy is expended against the dome's voltage gradient during this transport.

3) [This step is not essential, but would probably occur incidentally, so the required energy should be checked] Charge the capacitor can to the dome voltage of 800kV.

The can (comparable to a sphere of about 25.5mm radius) has a capacitance of approximately:—

C = 4πε₀R = 4π × 8.854 × 10^-12 × 0.0255 = 2.837 pF

The energy required to charge this to 800kV is:—

U_E = ½CV² = ½ × 2.837 × 10^-12 × 800,000² = 0.908 J (i.e. a negligibly small energy expenditure).

4) Dissect the capacitor, and deliver charge from the plate of the desired polarity to the dome. Recall that, provided the capacitor is dissected to the extent that the charges on its plates are no longer bound, all of the 837 coulombs of initial charge must go to the dome at 800kV, so there would be an energy gain of:—

U_E = ½qV = ½ × 837 × 800,000 = 334,800,000 J

In order to dissect the capacitor, force will need to be exerted over distance, which would generally follow a curve of shape depending on the capacitor's construction. I shall not attempt to analyse this in detail, but I doubt that any such force vs distance curve, when integrated, could substantially exceed the energy required to initially charge the capacitor.

If so, we would gain 334,800,000 joules for an expenditure of not much more than 1130 joules, negligible by comparison. This is an energy gain ratio approaching 300,000 : 1.

5) Neutralise unwanted charge on the other plate of the capacitor, e.g. by connecting it via a wire to Earth.

6) Reassemble the capacitor (still inside the dome). Since both plates are now uncharged, there is no energy loss or gain in this step.

7) Remove the capacitor from the dome. Again there is essentially no energy loss or gain in this step.

The next cycle of operation can then begin.

A possible variation

As a variation to the above cycle, it might be possible (and preferable) to keep the capacitor components always within the dome, as shown and described last time, rather than transporting the assembled capacitor as in steps 2) and 7).

Power output

If a cycle of operation could be completed in one second, the power output would be nearly 335 megawatts. If 100 cycles could be completed in one second, the power output would be nearly 33.5 gigawatts.

This enormous 33.5 GW output, from 100 Hz operation, seems improbable, but nevertheless the above analysis does indicate that a rapidly dissectible capacitor of large capacitance at very high voltage could still deliver a very worthwhile power output, even at substantially lower-frequency operation.

Transporting and Accessing Charge at EHV, Part II

Fig. 2.  Charge transfer to a dissectible capacitor within a high-voltage dome

Current carried in metal-wire conductors

Figure 2 above shows in schematic form one method of accessing charge that has been brought into the interior of a dome already at very high voltage. The charge is brought from an external source (the battery B) as a current in ordinary metallic conductors, which are initially connected to the plates of a dissectible capacitor within the dome.

Dissectible capacitor

The dissectible capacitor is an old concept, dating back to the Leyden Jars of the mid 18th century. The version shown in Figure 2 consists of two metallic plates of area A separated by a distance d, with a dielectric of constant κ and thickness b between the plates. (b must be sufficiently less than d to avoid charge transfer by corona discharge from the plates to the dielectric at any time. This unwanted effect usually does occur when Leyden Jars of traditional design are dissected — for example see the MIT video at https://www.youtube.com/watch?v=9ckpQW9sdUg).

After the plates have been charged, one of them is disconnected from the source, and removed as far as possible from the rest of the device, as shown dashed, so that (most of) its charge can then be delivered through an electrical connection to the dome. As we have seen previously, by Gauss's law, this delivery of charge must occur, no matter what voltage the dome has already acquired.

After this, the movable plate is returned to its original position, reconnected to the external source and the operating cycle continues.

Force through distance

It could be objected that force must be exerted through a distance, i.e. that energy must be expended, to separate out the movable charged plate. That is correct, and no doubt there would be no net energy gain if this process was not occurring within a dome already at very high voltage. But it is, and the higher the dome voltage, the higher the net energy gain as the plate's charge is transferred to it.

Charge accumulation on fixed plate

A more serious problem involves getting rid of undesired charge that would otherwise build up on the fixed plate. This could be neutralised against Earth via switch S, once per cycle, but even so, problems could still remain in accessing the desired charge at low energy penalty. However, I'm certainly not convinced that any such problems are insoluble. Experiments would be necessary to decide this.

Example calculation

When assembled, the capacitor obeys the formula given in Figure 2, where:—

C = capacitance
κ = dielectric constant
ε₀ = permittivity constant = 8.854 × 10^-12 farad/meter
A = area of plate
d = plate separation
b = dielectric slab thickness

Substituting "reasonable" values: A = 0.01m², κ = 5.4, d = 1mm, b = 0.8mm in this formula, we get:— 

C = 255.6pF.

If the external source charges the capacitor to a voltage V of say 1kV, we would get:—

q = CV = 2.556 × 10^-7 coulombs.

If the capacitor could be charged/discharged say 100 times/sec, we would get:—

I = dq/dt = 25.56 microamps.

This is a very small current, still typical of what is usually encountered in electrostatic machines employing isolated "static" charge carriers.

Next time I'll look into the possibility of using a similar approach to this, to achieve far higher levels of current and power.

Transporting and Accessing Charge at EHV, Part I

I'll now revert to the dual topics of i) transporting charge against a voltage gradient, without necessarily paying a full energy penalty for doing so; and ii) accessing that charge at very high voltage, thus gaining net energy.

Electrostatic vs electromagnetic charge transport

There is a very simple method of transporting charge "shielded" from an external voltage gradient. It is used daily in billions of electrical devices around the world (although the shielding effect is usually incidental). It is simply the familiar current-carrying metallic wire conductor. We recall that electric current is simply the rate at which electric charge flows; i = dq/dt.

Is that all that needs to be said? These days, probably not — a lot more words are probably required. But if so, rather than use my own words, I'll again quote the foremost expert, Professor Noël Felici, on the matter of electrostatic vs electromagnetic current [Ref 1]:— 

Figure 1. Comparison of electrostatic and electromagnetic generators.
(a) Electrostatic air-blast generator. Ash particles move from the left to the right. They pick up a negative charge from the ionizing wire (A), travel in the insulating pipe (B), the walls of which carry stationary charges of opposite sign, and are discharged by the blades of the collector (C). For a significant current to be developed, the electric field near the walls must be strong, with the attendant risk of breakdown.
(b) Electromagnetic generator. The copper bar is displaced in a magnetic field (not shown) giving rise to a Lorentz force (arrow) driving electrons to the right. Their electric field is caught by interspersed positive ions, and no macroscopic electric field related to current will appear.

The caption of the above figure is exactly as Felici wrote it, except for my added emphasis on the last phrase. Again with my emphasis, he comments further:—

"To settle the matter, let us consider an "electrostatic" and an "electromagnetic" generator. In the first one (Pauthenier's air-blast machine, figure 1(a)), we have electrified ash particles moving in an insulating pipe against the pull of the field due to the voltage between terminals. In the second (figure 1(b)), mobile electrons are driven in a copper bar against a similar pull by Lorentz forces, which replace the air blast. What significant differences do we find? In both cases static charges are driven against a true static field due to the voltage on terminals by an extraneous force, and this gives rise, in modern parlance, to an e.m.f. In the "electrostatic" generator, the electric field of the charge carriers extends down to the walls of the insulating pipe and should be caught there by static charges of opposite sign carried by the surface of the walls or, better, by electrodes at cascaded potentials in belt machines. At any rate, the field of the charge carriers is macroscopic; if too strong somewhere, it entails gas breakdown and thus limits the possible current output. Besides, the ash particles are not spontaneously electrified; they need a charging (and discharging) arrangement to accomplish their turn of duty. This is not so with the "electromagnetic" (or "electrodynamic") generator. The field of the electrons drifting in the bar is not caught at its surface but at the atomic level by stationary copper ions. Although incredibly strong, it cannot cause breakdown anyway, since breakdown (in an insulator) would require the field to run over many electronic mean free paths; besides, electrons are charged particles with an enormous charge-to-mass ratio.

Thus, the difference between the two machines reduces to the fact that the field of the charge carriers does not extend so far in the second case as in the first."

Transporting charge without paying a full energy penalty

So, the field of the charge carriers (electrons) in a wire conductor does not extend far enough for those electrons to be strongly influenced by an external voltage gradient, such as necessarily exists between a dome at high voltage and Earth. This means we could indeed transport charge against the gradient without paying a full energy penalty. There is still a problem to be solved in accessing the charge, once transported. I'll look at that next time.

Ref 1: "Electrostatics and electrostatic engineering," N. J. Felici, 1967 Static Electrification Conference.