Electrical – Power System Faults Problem

faultpower-engineeringthree phase

A sustained three-phase fault occurs in the power system shown in the figure (See Link). The current and voltage phasors during the fault (on a common reference), after the natural transients have died down, are also shown. Where is the fault located?
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This question was asked in GATE Exam 2015, and I was unable to do it. I have no clue how to solve this problem. The question paper in which this question was asked can be Downloaded from here for verification of problem : gate.iisc.ernet.in/gate-answer-2015/EE_S05.pdf

Best Answer

The answer lies in the size and angles of the vectors.

(The size is actually not necessary at all, but it helps.)

The reason why you can't depend on the magnitude (size) of the vectors is because you don't know the impedance values, nor the production or voltage level on each side.

Yes, you'd normally have the same voltage, but it's possible (and frequently so), that generators in the same grid produce power at different voltages (but reasonably close), for instance 20kV and 22kV. This would be problematic if the generators were close, but not if they're 100km apart (and a line can easily be 100km).

The reference direction for all currents is from the buses. So, a positive angle means the current goes from the bus to the cable. A negative angle means there is current coming from the cable into the bus.

Knowing this:

  • We see that there is current coming into bus 1, through the second cable (I2). This eliminates fault location R. (There won't be current coming from the fault).
  • We see that these is current going out from both buses (I1, I3 and I4). This eliminates the buses as potential fault location, and leaves us with Q and S.

The current I4 is positive, meaning there is current going from bus 2 to bus 1 through that cable. This makes fault location S highly unlikely (unless the cable impedance is dependent on the distance from the buses, i.e. a very thin cable close to bus 2 and thick cable close to bus 1).

We can't use the current levels as indicators, since we don't know the production levels. We could use the levels if the production and voltage level was guaranteed to be equal.

We're not yet finished. we have rules out all but two, and know that S is unlikely (so we're fairly sure it must be Q).

Now, how can we make sure we're right?

The voltage vectors.

We see that the vector V1 is shorter than V2, so that pretty much tells us all we need to know. It must be Q.


Now, what if we only knew the angles, not the magnitude?

All the first arguments still stands, we only used the magnitude for the voltage vectors.

You should know that there is approximately no resistance (R), only reactance (X) in an arc.

The vector V1 is closer to being 90 degrees than V2. That means the resistance (R) between bus 1 and ground is lower than the resistance (R) between bus 2 and ground. The resistance is due to the cable, so higher resistance means a longer cable.

In essence: It must be Q.