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.