Inductance from nearby 25kV line

If you don't have any current flow in the shield, there will be no shielding of magneting fields. But since there will be no induced current anyway from the perpendicular cable, it doesn't make any difference.

Right, but of course in real life nothing works like in theory, and the cables could very well be off perpendicular. In this case, would grounding the shield at one or two locations be optimal, considering the shield will technically already be "grounded" by being placed in the ground.

Thanks

Unless the cable is unjacketed with a bare shield, burying the cable in the ground will not ground the shield.

If you ground both ends, current in a parallel line (or the parallel component of a line that is skewed) will induce a current in both the conductor and the shield. The current induced in the shield will induce an opposing current in the conductor, thereby acting to shield the conductor from induced currents. I doubt that the shielding effect for a 25 kV line would be significant even if the cables were parallel instead of perpendicular.

25 kV is not a transmission line. It's a distribution line. Why do you ask about grounding methods?

Do your calcs despite the theory.

Sounds like a student post to me.

magoo:
been out of school for years. Yes, 25kV is technically a distribution voltage but the nomenclature isn't important, just the solution. You think this type of question would arise in an undergrad power course? Think again.
I ask about grounding methods because I have never had this specific issue arise before, and I do not know the typical method to mitigate induced currents in adjacent conductors.
Here's where I got the basic mutual inductance calculation:


jghrist: What I was thinking was a typical grounding mat you see in substations near the junction boxes and hand operators for the disconnects. Those are non-insulated aluminum meshes. Like I said before I am sure the induced currents will be near zero, but as it is non-shielded control cables I need to make sure there is zero interference. I was thinking the Faraday shield arrangement and just did not know the method of grounding the shield mat.

Any insight is appreciated.

Thanks

jghrist: What I was thinking was a typical grounding mat you see in substations near the junction boxes and hand operators for the disconnects. Those are non-insulated aluminum meshes. Like I said before I am sure the induced currents will be near zero, but as it is non-shielded control cables I need to make sure there is zero interference. I was thinking the Faraday shield arrangement and just did not know the method of grounding the shield mat.

Click to expand...

Now I'm confused. I thought you had two power cables. What is the aluminum mesh? Is the cable that you are concerned about a non-shielded control cable? What do you mean by a Faraday shield? A bare aluminum mesh placed above the cable? First of all, it will corrode and be nothing but aluminum oxide in a short while. Second, how do you ground one end of something that is continuously grounded? A mesh above a cable will not shield it from magnetically induced currents. If you want zero induced current in your control cable, put it in steel conduit.

Thanks for the input.
Yes, there is a shielded power cable but I am more worried about the non-shielded control cable running along the same line. And yes, I was also wondering how you ground something that is continuously grounded, seeing as I would think one end bonded to copper clad rods will have a lower ground impedance. And yes, now that I do a little more reading on it, zero induced currents would require a continuous closed loop of conductive shield.
What I mostly need to figure out is the proper formula to calculate the induced current in the control lines (but I have mostly found scalar equations).

Thanks for your input though,



Regards,

edit: distribution voltage actually 33kV, although I assume induced currents will be not much larger.

IEEE is a very high level of information and the indicated article is also a very important issue and treated in a high level of mathematics of course.
"Mutual inductance calculations between underground power cable with earth return and underground telecommunication Cable"
I think I did once a telecommunication cable of 3600 pairs 0.3 mm conductor copper, foam polyethylene insulated. I don't think one need to waist so much copper instead of fiber glass, and fiber glass is not sensible to magnetic field.
Let's you have an old fashioned cable not shielded [nevertheless usually the pairs or quarts are individually shielded and an overall shield is provided too] and you have to protect it against a parallel laid medium voltage [single core cables, I presume].
The solution indicated by jghrist to put the control cable through a steel conduit is very good but it is too late as the cable is already laid on, I understand.
You may try to use the very sophisticated theory from the above article in order to calculate the induced voltage. In my opinion you can calculate the induced voltage for a single control pair and if it is negligible that means the problem is not in this corner.
You may, also, provide a shield as a thin reinforced concrete envelope connected both ends to the grounding system-the connection has to be insulated.
But, since the 25/33 KV cable must be shielded the electric field is confined so only concern it is the magnetic field which depends on current intensity only. If short time of short circuit is also a concern, you can take this in you calculation.
The induction voltage depends also upon the clearance between power conductors and number of parallel conductors and the distance to the control cable.
As the return path through the ground [See Carson theory] is involved, the calculation could be a bit more complicated.
You may use the theory for overhead line against static wire for instance or for induced voltage in a pipe line. For instance see:
For two parallel loops:
L12=miuo*Length/4/pi()*ln(((Dp-h)^2+b^2)/(h^2+b^2)*((Dp+h)^2+b^2)/(h^2+b^2)) [H/m]
Where:
h-vertical distance b=horizontal distance if miuo=4*pi()/10^7 H/m then Length=m
Dp=658.5*sqrt(roxf) Carson formula for the earth return 'image' conductor. f=50-60 Hz ro=earth resistance ohm.m