R and L from Z1, Z2 and Z0 1

[AusLee]

Hi,

I have a cable:
Z1=Z2=0.9 + 2j
Z0 = 19.+11j

how do I go by and calculate R and L?

 

[Skogsgurra]

It may be obvious to you what you mean by Z1 etc. But it would be a lot better if you define the meaning of those quantities. Like the earlier question (voltage drop), it would be good if you present all relevant facts. When you have done that, you may find out the answer all by yourself. R seems to be given already and L you get by dividing the imaginary part by w.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

 

[deltawhy]

http://en.wikipedia.org/wiki/Symmetrical_components

 

[AusLee]

Hi, yes that's right, Z1 = positive sequence, Z2 = negative sequence and Z3 = zero sequence.

So with the fomulas from deltawhy, and for the particular case where Z1=Z2, the "real" values are:
Za = Z0 + 2 x Z1
Zb = Z0 - Z1
Zc = Z0 - Z1 = Zb

So for the particualr case:
Za = 20.8 + 15j
Zb = Zc = 18.9 + 9j

how do you go from Za, Zb, Zc to just Z = R + jLw per phase (which should be the same for all phases, right?)

 

[magoo2]

Something is wrong with your real value equations.

Also Gunnar told you about getting L from X. His w term is 2 * pi * f.

 

[AusLee]

Hi magoo2

ok i got the part where i go from L to X, but I need to get to either L or X first, so far I have only Za, Zb and Zc. How do these 3 figures relate to a one value, Z?

 

[healyx]

What are you trying to achieve AusLee? Are you using these impedances to calculate various fault currents? Is Za, Zb and Zc supposed to represent the actual phase impedances? If so, that doesn't look right - symmetrical components is based on the premise that all phases must see the same impedances.
If you are trying to find a ground fault, you need Z1, Z2 and Z0 (usually can assume Z1=Z2). If you are trying to find the maximum 3 phase fault you only need Z1.

You kind of need to answer what you are trying to achieve before we can distill it down into a simple Z = R + jX format.

 

[ijl]

The calculation of Z1, Z2, and Z0 for a symmetrical line section is described in textbooks (see Grainger & Stevenson, for example). For the calculation one needs the self-impedances Zaa, Zbb, Zcc of all phases, the self-impedance Znn of the neutral line, and the mutual impedances Zab, Zbc, Zca between the phases, and the mutual impedances Zan, Zbn, Zcn between the neutral and the phases. For a symmetrical line the parameters are the same for all phases, Zaa=Zbb=Zcc, and Zab=Zbc=Zca, and Zan=Zbn=Zcn. Thus, four parameters Zaa, Zab, Zan, Znn are needed in order to calculate Z1, Z2, and Z0.

The reverse calculation of Zaa, Zab, Zan, Znn from Z1, Z2, Z0 is obviously possible only when one additional condition for the parameters is given.

When Zaa etc are known, then R is the real part of Z, and 2*pi*L is the imaginary part of Z, as already stated by others.

 

[AusLee]

Thanks, I worked it out as ijl said. It is not possible to go the way back unless you know the actucal construction of the cable and the distance between the conductors.

Cheers.

 

[veritas]

AusLee – it probably would have been more helpful if you specified more precisely at the start which R and L you are after? If it was the positive and zero phase sequence ones then that’s pretty straightforward since you are given Z1 and Z0.

However, if you are after the original phase impedances then it gets much trickier. In actual fact you cannot work back from Z1, Z2 and Z0 to the original Z11, Z12, etc. unless there are no earth conductors AND the line is transposed. Once you have earth conductors the best you can hope for is getting the values of the equivalent phase impedance matrix where the diagonal elements are Zaa-eq and the off-diagonal elements are Zab-eq.

With say two neutral conductors the 3x3 phase impedance matrix ZP is derived from something like ZP = ZA – ZB*ZD-1*ZC. The original self-impedances and mutual impedances between the phases itself are contained in the ZA matrix.

ZP-eq is derived from ZP by averaging out the diagonal and off-diagonal entries since transpositioning is always assumed.

Zaa-eq = (Zaa + Zbb + Zcc)/3 = all the diagonal entries of ZP-eq
Zab-eq = (Zab + Zbc + Zac)/3 = all the off-diagonal entries of ZP-eq

This ensures that the sequence matrix is a decoupled one. The averaging out means your original information is lost unless you have no earth conductors and the line is transposed as mentioned before.

Hope this helps or as clear as mud?

 

[veritas]

AusLee - sorry I need to be a bit more attentive to my terminology. With ZP = ZA – ZB*ZD^-1*ZC I mean matrix operations so it should be [ZP] = [ZA] - [ZB]*[ZD]^-1*[ZC] where [ZD]^-1 is the inverse matrix of [ZD].

Also, I never quite answered your original question. The sequence impedance matrix is derived from the phase impedance equivalent matrix, [ZS] = [A]^-1*[ZPeq]*[A] where [A] = (1 1 1, 1 a a^2, 1 a^2 a).

Working backwards get, [ZP] = [A]*[ZS]*[A]^-1.