Running late again. I always was bad with leaving stuff until the last moment then finding it took more work than I expected. Hey, somewhere in the world it is still Sunday night!
Even though superconducting cables have zero electrical resistance, they still have a critical current. Put too much current through and it will stop superconducting, and if it's not properly controlled this can result in damage. Along with the operating voltage, this will determine the maximum power that can be put through the cable.
Starting assumptions this time.
- Circular cross section to the cable. It's the obvious shape.
- The operating voltage is limited only by the insulation of the cable.
- The superconductor is Type-II. Type-II superconductors can carry current through their whole bulk, and as such have a well-defined critical current density in Amps per unit area. Type-I superconductors confine current to their surface and thus the critical current depends on a wire's circumference not its cross-sectional area. They also have much smaller critical currents.
- The superconductor has zero electrical resistance. This isn't a tautology; a phenomenon called flux creep allows Type-II superconductors to have low but non-zero resistance while still superconducting!
I = π(fr)2Jc = πf2r2Jc
The maximum voltage is limited by the dielectric strength, d, of the insulator. This is the electric field strength, measured in volts per metre, above which the insulator will stop insulating and current will discharge across it in a phenomenon called dielectric breakdown, permanently damaging the insulator if it is solid. The dielectric breakdown of air is familiar to us all as lightning.
The electric field strength across the insulator is simply given by the operating voltage of the superconductor inside divided by the width of the insulator. Maximum operating voltage is when this equals the insulator's dielectric strength.
d = V / (1-f)r
V = d(1-f)r = dr - dfr
The maximum power that can be put through the wire, which I'll call the power capacity, is then simply given by voltage times current. Multiplying the equations for current and voltage together, we get
P = πf2r3Jcd - πf3r3Jcd
P = πr3Jcd(f2 - f3)
If f is zero, there is no superconductor and obviously no current can flow. If f is one, there is no insulator so the superconductor can't be at any voltage and no power can be transmitted. For some f between 0 and 1, f2 - f3 will be at a maximum, and thus the power capacity will be a maximum. This will tell us how much of the cable's overall radius should be superconductor, and how much insulator, for best results. To find this maximum, we differentiate P with respect to f.
DP/Df = 2πr3Jcdf - 3πr3Jcdf2
DP/Df = πr3Jcd(2f - 3f2)
(Capital Ds have been used for differentiation to avoid confusion with the lowercase d used for the dielectric constant)
When DP/Df = 0, P is at a maximum, minimum, or point of inflection.
0 = πr3Jcd(2f - 3f2)
0 = -3f2 + 2f)
0 = 3f2 - 2f
Using the quadratic formula
The solution at 0 is of no interest, P is 0 there. The remaining solution, at f = 2/3, must therefore be where P is at its maximum. There,
V = dr/3
I = 4πr2Jc/9
P = 4πr3Jcd/27
For a worked example, we shall determine the necessary radius of a cable capable of carrying 15 terawatts, the average power consumption of all of humanity, using contemporary to near-future materials.
Jc = 109 Am-2. Or 105 Acm-2. Various groups have reported values of this order of magnitude for different superconductors.
d = 108 Vm-1. The value for Teflon, among the best insulators.
P = 1.5 x 1013 W, as mentioned.
r = 0.069 m
A wire 14 centimetres across could carry the entire world's power consumption; testament to the capabilities of superconductors.
As ever, the equations have consequences, which are in fact what we're mainly interested in. The key factor is that the power capacity of the wire scale with the radius cubed, not squared, because a wider cable not only carries more current but also operates at higher voltage. A corollary of this is that it's better to use one big cable than lots of small ones with the same combined mass.
A second notable consequence is that there are gains to be had not just by improving superconductor technology but also by making better insulators.
A third consequence comes from a bit of knowledge about superconductors. Jc is temperature dependent; at the superconductor's critical temperature, the warmest it can superconduct, it approaches zero, while at lower temperatures it is higher. Thus, even given a much-vaunted room-temperature superconductor, for maximum power transmission it would still require cryogenic cooling.
If the first assumption is broken, you need more insulation to carry a given amount of power. It becomes rather a case of why would you do that?
If the second assumption is violated, and other factors limit the voltage, then for large cables the two-thirds superconductor, one-third insulation ratio will not be held, the insulation can be thinned due to the reduced voltage. Consequently the power capacity will then scale approximately with r squared, not r cubed. The drawback to using multiple smaller cables will be lessened, though not wholly eliminated as they will require more insulation than a single large one.
If the third assumption is violated the situation would considerably change. For this to happen though would require a novel Type-I superconductor with an unprecedently high critical current, or else something entirely unanticipated.
If the fourth assumption is violated we may have to consider power losses in the wire that would create inefficiency, as well as the implications for cooling system of heating in the wire. The exact impact of these factors I am unsure of, though qualitatively I expect larger cables would be required to carry the same current.
For some background information, the Open University has a free e-text on superconductivity. It focuses primarily on Type-I superconductors, which were historically discovered earlier and are better understood but have fewer practical applications than Type-II superconductors.
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