Generator Fault - Leakage Reactance 3

[tonyflair]

I'm trying to determine various faults on a synchronous generator. I received information from the manufacturer -> ie. Xd, X'd, X''d, Xq, X''q ...... The one piece of information that was added that I was not used to seeing was a leakage reactance.

Here are some typical numbers that I have obtained,
X1(leakage) = 0.08 p.u., X2(negative sequence) = 0.22,
X0(zero sequence) = 0.07, X''d = 0.12

When I'm considering a fault (ex. line-to-line), does that leakage reactance come into play in my calculation ? Normally I would treat the short circuit current (p.u) as,
Vpu/(X''d + X2). Or is the leakage reactance already included in the other reactances (i.e. negative sequence) ?

Should I subtract the leakage reactance from the others in my calculations ?

Thanks in advance,

Tony

 

[hvcad]

The leakage reactance Xl is already included in Xd, Xd', Xd", Xq, Xq", etc. The normal steady state reactance of the machine Xd, is made up of two components the leakage X1 in series with the Xd mutual Xmd, i.e Xd = X1 + Xmd. Similarly for the transient reactance Xd' = X1 + Xmd*Xf/(Xmd+Xf) where Xf is the reactance of the field. A similar equation can be written for the sub-transient reactance. Again, similar equations can be written for the q axis.

The negative sequence reactance is approximately equal to (Xd" + Xq")/2 Again the leakage reactance is already included in the value you have.

So you can calculate your phase to phase fault currents using Xd" and X2 only.



Dr K S Smith
Mott-MacDonald, Power Systems Division
Glasgow, Scotland.