EMF, electrical signal and shielded cables

[elkumbo]

Hi

I have a practical problem/question:

I need to wire 2 PLC-s for communications (RS485 signal) and I am planning to use shielded twisted pair cable (Cat5 FTP). The signal cables are quite short - approx. 6 m.

One possible problem that could arise: between the 2 PLC-s, there is a buried power cable (perpendicular to the signal cable) - 2 x 1250 kW @ 400 VAC (I = 6,25 kA), buried approx. 0,7 m beneath the ground.

In short:
My main question would be, How strong will be the max EF (electric field) on the signal cables (signal cables run near the ground)?
Can I use the shielded twisted pair cable here? I know that twisted pair shielded cables are much more resistant to electric(/magnetic) fields, but is there a maximum allowed field in which cable manufacturers guarantee lossless operation? (should be propably in the cable datasheet...)



The longer version:
I tried to reintroduce myself with electromagnetic fields, but I got a little confused. Here is what I was thinking of:
I consider the strongest Electric field to be in the nearest point from the signal cable to the power cable, which is directly above the cable. For simplicity, I am concidering the power cable to be a uniformly charged rod with the total lenth of L. I am concidering the section of the signal cable directly above the power cable as a point P. The distance between point P and line L is b = 0,7 m. Since I have no detailed knowledge about the power cable, I assume that L >> b. If the projection of point P to the line L is marked with point O, and we mark both ends of the line as A and B, then I also assume that AO == BO >> b, which means I have to consider only the component of the electric field which is perpendicular to line L.

(The mathematical derivation of the formulas can be found here: http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ContinuousChargedRod.xml" )

$$E=\frac{\lambda}{4\pi\epsilon_{o}b }\left(\frac{AO}{\sqrt{AO^{2}+b^{2}}}+\frac{BO} { \sqrt{BO^{2}+b^{2}}}\right)$$ ,
where $$E$$ is the electric field,
$$\lambda$$ the line charge of the uniformly charged rod,
$$\epsilon_{0}$$ the electric constant,
$$b$$ the shortest distance between the signal cable and the power cable ( PO )
$$AO$$ the length of the power cable which spans to the left of the projection of point P on the line L
$$BO$$ the length of the power cable which spans to the right of the projection of point P on the line L


Since AO >> b and BO >> b, we can simplify the equation to this:
$$E=\frac{\lambda}{2\pi\epsilon_{o}b }$$


The line charge of the uniformly charged rod is calculated:
$$\lambda=\frac{Q}{L}$$
where $$Q$$ is the total charge of the rod and $$L$$ the total length of the rod


Since $$Q=I*t$$ where $$I$$ is the current that flows in the rod during the time $$t$$
we can derive this formula:
$$\lambda=\frac{Q}{L}=\frac{I*t}{L}=\frac{I}{c}$$ where $$c$$ is the speed of light.


From this I can start calculating:
$$\lambda=\frac{6250}{3*10^{8}} = 2,083*10^{-5} \left(\frac{C}{m}\right)$$
$$E=\frac{2,083*10^{-5}}{2\pi*8,854*10^{-12}*0,7}=534,98\left(\frac{kV}{m}\right)$$

Are these results reasonable? It seems quite a lot to me...
It has been a while I operated with these formulas and I wanted to ask if somebody could please check my approach and calculations. Please point out what I might be doing wrong.

phew... took me almost an hour to complete this thread...
Anyway, I hope someone can help or comment on my problem.

Cheers!

 

[uart]

The ground (which they're buried under) will mean that interference from electric field should be a non-issue. So the magnetic field from the buried cables is the most likely source of any potential interference.

One of the most important factors regarding the amount of magnetic field the cables will produce (at the ground surface) is the distance of separation between the current and return path (or between the phases). If all three phases are bundled properly (and there's no zero sequence component) then magnetic field will be small.

Also be wary of making ground loops in your shield inter-connection, as that's a major way that stray external fields can couple into a circuit.