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This question is inspired by the single-wire earth return power transmission systems used in very rural areas (Wiki).

Consider a power distribution system with a generator in the middle, connected to two stakes in the ground far from each other. For very high frequencies, the current through the earth should attenuate rapidly as we move away from the stakes; that is, my intuition tells me that the Earth return should act as a very large capacitor. Consequently, the region in between should be "invisible" to the current. One corollary (for example) is that the impedance of the return should tend to some fixed, finite number in the large-interstake-distance limit.

At the same time, we should expect to see this in homogenous ground. The simplest model for the earth return in that case is a continuum limit of capacitors connected in series; I originally began with a more complex model. But it seems any model consisting of components connected in series does not have a coherent large-distance limit.

Suppose the earth return has impedance $Z_E(d)$. We can split this into a "left" earth return and a "right" earth return; because impedances connected in series add, we expect $$Z_E(d)=Z_E\left(\frac{d}{2}\right)+Z_E\left(\frac{d}{2}\right)$$ But for large $d$, the only difference between $Z_E(d)$ and $Z_E\left(\frac{d}{2}\right)$ is the bulk, which (we assumed above) is negligible. So $$Z_E(d)\approx Z_E\left(\frac{d}{2}\right)$$ and thus the large-distance impedance is $0$!

The only assumptions we used here are (1) additive impedance, (2) homogenous ground, and (3) negligible bulk. Is there some way to weaken (3) while preserving the locality of current around the stakes that gives a nonzero impedance?

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  • $\begingroup$ Typical power transmission frequencies (50-60 Hz) have associated wavelengths in the 1000's of km. Are you intending that the distance between the stakes is much more than the wavelength? $\endgroup$ – The Photon Apr 15 '19 at 2:50
  • $\begingroup$ @ThePhoton: Yes. This is primarily a hypothetical question. $\endgroup$ – Jacob Manaker Apr 15 '19 at 2:53
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For very high frequencies, the current through the earth should attenuate rapidly as we move away from the stakes;

Depending what you mean by "very high frequencies", this isn't correct.

At normal power transmission frequencies (50-60 Hz), the return currents will indeed spread out, but will overall tend to run from one stake to the other. You would do better to model this situation with a grid of resistors rather than capacitors.

At modestly high frequencies (in the kilohertz, maybe) the current won't spread out so much, but be concentrated beneath the wires leading from the generator to the stakes. It will "follow the path of least inductance" in the same sense we mean when we say that at low frequencies it "follows the path of least resistance". We don't mean that absolutely all of the current follows the path of least inductance, but that the current follows paths between the two stakes in inverse proportion to their inductance, resulting in most of the current being concentrated fairly near the lowest inductance path.

At actual high frequencies (megahertz or higher, and I'm assuming the cable is routed some meters above the ground) the power in the system will mostly be lost due to radiation from the cables before it ever reaches the stakes into the ground, and very little current would be seen at the stakes themselves.

my intuition tells me that the Earth return should act as a very large capacitor.

As mentioned above, at the frequencies where this matters, you'd do better to treat it as a large distributed resistor, not a capacitor.

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