Introduction to Fault Analysis Part 3d – 2,
Line to Ground Faults. In this video we will consider the sequence voltage components
and phase voltage quantities for a two line to ground fault
on the 13.8 kV side of the delta wye transformer.
This is a continuation of Part 3d – 1,
Now, for this we have to look back at our faulted sequence network:
Having a closer look at the network diagrams and paying close attention to where the positive,
negative and zero sequence voltage quantities are actually measured or calculated which is the 13.8 kV side of the transformer
we can clearly see that all three sequence networks are connected
to the same node,
and therefore by way of Thevenin’s theorem
and basic circuit analysis,
we know that all sequence voltages will be exactly the same.
Now let’s calculate the positive sequence voltage:
So the positive sequence voltage is calculated by simply taking the source voltage which is at 1 per unit angle 0 degrees
minus the voltage drop of the sum of the impedances
in positive sequence network.
And so that is the sum of the impedances is j.15
and the current flowing through the positive sequence network is –j4.7619 per unit
and what we get is .285 per unit at the angle of zero degrees.
Now we know that all of the sequence components voltages will equal .285 per unit.
And we can multiply that by the base value of 13.8 kV divided by the square root of three
to get the voltage quantity which we get as 2.27 kV at the angle of zero degrees.
Now as a full and final step the sequence voltage quantities need to be transformed to a phase voltages quantities
and we could do that by plugging them in into the legacy equations which we show below:
Now, since the positive negative and zero sequence voltages are all equal,
we know that phase a voltage will equal 3 times the positive sequence voltage
and that is equal to 3 times 2.27 kV
at the angle of zero degrees
which gives us 6.829 kV orc6.8 kV at the angle of zero degrees.
And when we calculate the phase b and phase c voltage quantities,
it happens to be reduced to zero. And this makes total sense especially by looking at the earlier
diagram that we drew for double line to ground fault on the 13.8 kV system,
we know that phases b and c
were the faulted phases and
we also know that the voltage at the point of the fault will be driven to zero.
And so it makes total sense that the voltage at the point of the fault is simply equals to zero.
So there you have it.
Now we have calculated the fault current values as well as the voltage values
for a two line to ground fault and in the previous videos
we have calculated the line to ground faults,
the line to line faults, the three phase faults so forth.
Now, In the next video, In Part 4 of our series,
we will see how we can reflect fault calculations over the high voltage side
given that the fault is at the Low voltage side of the transformer
and this is going to be very very unique a lot of the times it is quite confusing for us because
there are special things that we need to consider in order to analyze the fault values
on the high voltage side from sequence components point of view,
so this is gonna be covered in Part 4.
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