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Considering a monopolar HVDC system, with the ground as the return path.

I was wondering if the current path is physically speaking like the red one or the green one in the picture? I think it is like the green one, although if there are living beings in the middle, they can be thunderstruck by the current. The green one considers that the earth has a charge.

(It's strange to say that nobody has explained me this thing until now, and I'm a student at Politecnico di Milano, I'm following the Smart grid path and I am in the fourth year. Unluckily, these such things usually are not explained.) Do you know the answer?

enter image description here

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    \$\begingroup\$ It's like the red one. \$\endgroup\$ – Andy aka Mar 20 at 11:57
  • \$\begingroup\$ @Andyaka are you sure? Is it not dangerous for living beings in the middle of that path? \$\endgroup\$ – Samuele Benito Di Gioia Mar 20 at 12:05
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    \$\begingroup\$ Yes I'm sure and yes, it can be dangerous under certain circumstances and that is why return conductors are used or the earth return is used in remote and mainly uninhabited areas. \$\endgroup\$ – Andy aka Mar 20 at 12:20
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    \$\begingroup\$ Samuele, the red and the green drawings are identical: unless you assume that the whole current is used to charge a large capacitor made of "planet earth", the charges must somehow travel. \$\endgroup\$ – Marcus Müller Mar 20 at 13:09
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    \$\begingroup\$ @MarcusMüller Exactly like that coaxial you describe. It also gives favorable, if you can call it that, hard short circuit to low impedance ground (0 V) via the cable and not local earth (dirt) if someone digs into it or drop and drag an anchor across it. \$\endgroup\$ – winny Mar 20 at 13:27
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It's the red one.

And actually you can have big potential differences on the ground. In the case for example of very high voltage powerlines fallen in the ground, you can actually get electrocuted by just walking close to them, as the potential difference between your two feet could be high enough to create a path through you.

For electric charge to flow you need a close circuit of some sort and a potential differential. The ground being a potential path.

One could imagine the ground to be ionized to some extent, as your green drawing, but that would be minimal.

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  • \$\begingroup\$ "For electron to flow ..." Please don't reinforce the notion that all electric current is electron flow. There are enough confused contributors to the site already. In the ground it is quite likely to be positive ion flow as well. \$\endgroup\$ – Transistor Mar 20 at 19:08
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I think it is like the green one, although if there are living beings in the middle, they can be thunderstruck by the current. The green one considers that the earth has a charge.

We generally assume that the charge of Earth is neutral. For a DC link as you describe current leaving the ground at the feed end and travelling through the transmission line must enter the ground at the other end.

As you probably suspect already, the current will not travel in a direct path back to the feed end but will spread out a bit like the magnetic lines of flux on the bar magnet and iron filings experiment.

A bigger engineering problem is that the electrolysis of any ground connections. With an AC electrode there might be some chance of acceptable lifespan but for DC it will be eaten away very quickly.

enter image description here

Figure 1. Consideration of ground return and metallic conductor return is given in Ireland's Grid West Project HVDC Technology Review section 2.2.2.1.

This report also points out problems with corrosion of third-party buried metal pipelines and magnetic saturation of transformers. It might be worth a read. (It might be worth my while reading it too as it passes close to my home.)

My understanding is that for the reasons listed above that a return conductor is a better engineering solution. This may initially seem wasteful but there's a benefit: the voltage can be split into a symmetrical supply arrangement. If, for example, a 1 MV DC is required then the supply would be configured as ±500 kV. The savings come in the reduced insulation costs and, presumably, easier handling during installation.

enter image description here

*Figure 1. Route map with options.

At the moment a project to lay the 575 km Celtic Interconnector between Knockraha substation in Cork, Ireland, to the La Martyre substation in Finistère, France, is proposing ±320 kV or ±500 kV for transmission. Capacity will be 700 MW and will cost €1 billion. (The Republic of Ireland's peak demand is in the order of 5000 MW so this is > 10% of that.)

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  • \$\begingroup\$ Thank you transistor! I already know a lot of these things, my question was another, but you answered me anyway and I appreciate your help! Thank you! \$\endgroup\$ – Samuele Benito Di Gioia Mar 20 at 18:36

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