Reflecting a Line to Line Fault on HV Side
In this video we will see how we can actually reflect the a Line to Line fault
current quantities which occur on the Low voltage Y-grounded side of the transfromrer
but reflect the same quantities on to the high voltage delta side
So let’s first have a recap of our video Part 3c-1
in which we calculated the fault current values for a line to line fault:
For the Line to Line fault type,
we found that the positive sequence fault current is equal to –j3.33 per unit
and also the negative sequence current is equal to j3.33 per unit.
And the zero sequence current equaled zero
which means their was no zero sequence current.
That can be expressed as 3.33 angle -90 degrees
and 3.33 angle 90 degrees respectively
in polar form. It’s also important to note the subscript –
the lower case indicates fault current on LV side
while the upper case will represent fault current on the HV Side.
We now proceed by reflecting these per unit current values
on the high voltage side and then we will multiply them
with the appropriate base values too find the actual ampere quantities.
So, Given that we have a Dy1 or DAB transformer connection,
to determine the HV positive sequence quantities
we will shift the LV positive sequence component by +30 degrees
and that can be done by multiplying the LV fault current by a value of 1 angle + 30 degrees
and that gives us 3.33 at the angle of -60 degrees.
In the same way,
we will shift the negative sequence component by -30 degrees,
and that can be done again by multiplying the LV fault current with a value
of 1 angle -30 degrees and again that gives us a negative sequence current of 3.33 at the angle of positive 60
degrees. Now, Remember that for the zero sequence current, Line to line faults are faults
that are un-symmetrical in nature and are not connected to ground
which means that,
zero sequence current will not be able to flow from ground into the fault,
and this explains as to why we don’t have zero sequence current on both the LV side
as well as the HV side of the transformer.
Therefore zero sequence current is equal to zero.
We need to use the base values to convert the per unit quantities
to ampere quantities. However,
since we are working on the HV side of the transformer
we divide base value of power which is 30 MVA,
by the HV side voltage of 115 kV to get a base value of 150.6 Amps
And then just multiply out per unit quantities with the base value
to get ampere symmetrical component quantities:
We get 501.5 Amps angle -60 degrees as the positive sequence current
and 501.5 Amps angle +60 degrees as the negative sequence current.
Back to our familiar
transformation equations, they can now be transformed to phase current quantities.
Again we would remove zero sequence current from these equations since they are non-existent to simplify our calculations.
These are the equations that we get
Now, plugging in the values of a to be 1 angle 120 degrees,
a square to be 1 angle 240 degrees
and positive and negative sequence components to be 501.5 angle -60
and 501.5 amps angle +60 degrees respectively.
We get, Line Current A equals to 501.5 angle 0 degrees,
Line Current B = 1003 angle 180
and IC equals to 501 angle 0 degrees.
Now these values,
line up very well with current flow path on the HV side.
The fault current on phase b is twice the fault current on phase A and phase C,
We can say that the fault current coming from the source on line B gets distributed evenly between
line A and line C hence they are exactly half the line B current.
It also leads us to the fact that a line to line fault on the wye side resembles a
line to ground fault on the delta side because there are two small currents on phase A and C,
which is a characteristic of a line to line fault with the exception
that the zero sequence current is absent –
now this characterization was found in Blackburn’s book.
Furthermore, this type of characteristic also makes it difficult for relays to distinguish between
the two types of faults.
Even if we tried to calculate the zero sequence current using Line A, B, and C current quantities on the HV side,
well still get zero. Because I zero is equal to one third of IA + IB + IC
and plugging those values in which is 501 amps at the angle of 0 degrees
plus Line B current of 1003 amps at the angle of 180 degrees
plus line C current of 501 amps at the angle of 0 degrees,
and multiply that one third gives us zero.
Next, we will calculate the sequence voltages with our reflected sequence current components quantities.
To do that, we go back to our faulted sequence network diagram for a line to line fault
and apply KVL in each of the connected sequence networks.
We can calculate the positive sequence voltage by considering the voltage of the positive sequence source
1 angle 0 and the impedance of the source which is j.05 pu.
The positive seq. voltage for a 1LG fault is simply the source voltage minus the voltage drop
across the positive sequence impedance.
For the positive sequence voltage, we need to consider the reflected positive sequence current on the HV side,
which is 3.33 angle minus 60 degrees.
So our equation becomes, the following:
We get a value of 0.85 angle -5.5 degrees per unit.
Similarly, the negative sequence voltage is the source voltage minus the voltage drop across
the negative sequence impedance of the high voltage side .
For the negative sequence current,
we need to consider the reflected negative sequence current on the HV side which is 3.33 angle 60 degrees.
So our equation simply becomes the following:
We get a value of 0.1665 angle 150 degrees per unit.
Zero sequence voltage will be zero because the zero sequence current was also zero.
We need to set up a base value to convert per unit voltage quantities to actual line to ground voltages quantities.
V base can therefore be taken as the following:
Now Multiplying the components with the base value we get the values as shown below:
The positive sequence voltage is equal to 57 kV angle -5.5 degrees
and the negative sequence voltage is equal to 11.05 kV angle -30 degrees
Our three legacy transformation equations will then transform the sequence components to actual one line to ground voltage quantities.
Again since the zero sequence component is non existent we can eliminate that directly
from our equation.
Plugging in the calculated sequence component values we get the following:
We finally have the actual phase-to-ground voltages,
Va is equal to 67.3 kV at the angle of minus 9.45 degrees,
Vb is equal to 48.52 kV angle -133 degrees
and Vc is equal to 57.1 kV angle 125 degrees.
For a 1LL fault, we clearly see that the HV line A voltage is the one
that is least depressed while HV Line B and C voltage are more depressed.
Again, that is probably due to the fact that we have twice the amount of line current
going through line B which is coming from the source and then equally distributed on line A and line C
returning back to the source .
Now this was a very complicated video tutorial but I hope that this video was useful and
enlightening for students or professionals.
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