Electrical Induction between parallel conductors

[drdimmer]

THis isn't homework or coursework, I'm a bit old for that! Apologies if this has been posted in the incorrect place, I am just after advice from some electrical engineers.

Not sure where to attack this question from so any help would be appreciated. I was talking to a friend of mine who lays cables for a contractor and he was saying that in a recent job they laid pilot cables under ground next to High Voltage 11kV cables (in the same duct).

This got me interested, if a fault occurred on the HV cable then an EMF would be induced in the pilot cable running parallel to it. I'm trying to calculate approximately what the induced EMF could potentially reach if I presume the pilot cable is laid a set distance away from the HV cable.

I'm really not sure where to start with this problem apart from calculating the magnetic field produced by the HV cable under fault conditions and from there calculate the induced voltage in the adjacent cable.

The B field produced by the HV cable would be = ((u0)*I)/(2*pi*r);
Knowing that voltage induced = d(phi)/dt, = -AdB(t)/dt, would A simply be the cross sectional area of the pilot wire, as going through this process is giving me incredibly small values of which I know a much greater voltage would be induced in reality.

I'm sure I must be missing something obvious like including the length of the pilot cable somewhere or something similar.

Any help greatly appreciated

 

[Bob S]

Here is the inductance per unit length of two parallel conductors of radii a and c, and separated by a distance b:

L = [μ₀/4π]·[1 + 2 Ln(b²/ac)]

See Smythe Static and Dynamic Electricity, 3rd edition, page 342.

Bob S

 

[drdimmer]

Many thanks, am I correct in presumig that to find the induced voltage I can simply use V=L(dI/dt)?

 

[Bob S]

Many thanks, am I correct in presumig that to find the induced voltage I can simply use V=L(dI/dt)?

Yes, as long as the wavelength of the current excitation is long compared to the length of the cables. If they are comparable, then the coupling capacitance becomes important, and the coupling becomes a directional coupler. See

http://www.google.com/url?sa=t&sour...IGOzUJxAPO5AxWeAw&sig2=vZIyit52J72TUL6Krr00dg

In particular, see the papers in the references by Firestone and Oliver.

Bob S

 

[drdimmer]

Apologies again for the stupid question, using the V=L(dI/dT), would I then multiply this value by the length of the cable in metres? Presuming that the values of a,b and c used in the inductance equation you gave are in metres also?

Apologies for the questions, I have had a look in Static and Dynamic Electricity in an older edition but a lot of it seems to be going above my head!

 

[Bob S]

Apologies again for the stupid question, using the V=L(dI/dT), would I then multiply this value by the length of the cable in metres? Presuming that the values of a,b and c used in the inductance equation you gave are in metres also?!

The inductance in the formula is Henrys per meter. The value of μ₀ is

μ₀ = 4π x 10^-7 Henrys per meter.

Bob S

 

[m.s.j]

L = [μ₀/4π]·[1 + 2 Ln(b²/ac)]

That formula is correct for total inductance of two-wire transmission line and applied for induced voltage drop calculation of a two-wire system. But to find the voltage, in V/m, induced in a nearby one conductor line by the adjacent another line we must consider inductance of single conductor which is half of mentioned amount.
Indeed for one conductor of two-wire transmission line we can write:

Ls=(μ/2п).(1/4+ln D/r)

Where D is the distance of two parallel wire and r is the radios of each wire.

And for total inductance of two wire system we can to consider:

Lt=2.Ls


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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[drdimmer]

Does Lt=2Ls not presume that both conductors are of equal diameter?
Presumably each one individually can be summed to get a value of Lt if varying diameters are used.

Apologies for any misunderstanding on my behalf.

Cheers

 

[m.s.j]

No my friend, when presume that both conductors are equal diameter we can write:

a= c = r
and L = [μ0/4π]•[1 + 2 Ln(b2/r2)]= [μ0/π]•[1/4 + Ln(b/r)] which is same total inductance and should not be used for interference induced voltages calculations.


Meanwhile you are very genteel, please don't apologize, indeed we are serving us when we try to help others.

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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com