Fault analysis and symmetrical components in Power Grids

[PainterGuy]

Hi,

I was trying to learn fault analysis in three phase system from conceptual point of view. I need your help to clarify few seemingly contradictory points in the following video:



1:
At 32: 42, phase to ground fault in introduced. It is said that line voltage Vbc is not affected as is also the case with phase voltages Vbn and Vcn. Then, it is said that Vab and Vca are affected because Van is affected. I agree with this.

After that in a segment on symmetrical components, around 46:51, the analysis of phase to ground fault (between phase A and ground) is shown. At 47:21 it is said that phase voltages on lines B and C remain unchanged and I agree with this. But then surprising it is said that at the fault point positive sequence voltage fall to two-thirds of their pre-fault value. Isn't this a contradiction?

2:
Similarly around 28:40 it was said that in a phase to phase fault, unfaulted phase voltage is not affected. In a segment on symmetrical components around 43:46 phase to phase fault between lines a and b is revisited. Around 44:39 it is said that the voltage at the fault, the unfaulted phase is approximately equal to its pre-fault value. But later it is said, in a seemingly contradictory way, that unfaulted phase voltage Vcn has been affected.

Thank you for your help and time.

 

[Baarken]

After that in a segment on symmetrical components, around 46:51, the analysis of phase to ground fault (between phase A and ground) is shown. At 47:21 it is said that phase voltages on lines B and C remain unchanged and I agree with this. But then surprising it is said that at the fault point positive sequence voltage fall to two-thirds of their pre-fault value. Isn't this a contradiction?

Take a look at the phasor diagram below. The phasors in red, Va, Vb and Vc are the phasors of the unbalanced system. Please note that its phase C that has a ground fault now (Vc = 0), instead of phase A as in your video.

The phasors Va and Vb remains unchanged.

From the picture, the unbalanced system can be represented as a vectorial sum of the positive (black), negative (purple) and zero (blue) sequence phasors.

You can see how the positive sequence component can be 2/3, and added together vectorially with the negative and zero sequence component gives the original phasor.

 

[PainterGuy]

Thank you.

Just curious, did you take the picture from some book?

 

[Baarken]

Just curious, did you take the picture from some book?

It's from some lecture notes I got from a friend some time ago, I don't think they're available online.

This link gives an introduction to symmetrical components, with equations and has a similar picture: http://www.cadickcorp.com/download/symmetricalcomponents.pdf