This is the Fourth video in Part 3 of the series on fault analysis in power system.
Fault analysis in power systems Part 3d – 1. This is the Fourth video in Part 3 series
fault analysis in power system and In this video we will see how phase currents and voltages are
calculated for a double line to ground fault on the 13.8kV
bus which will involve phase b, phase c, and ground
So Let’s continue with our example which we introduced in the previous part:
We have a three phase synchronous generator having a short circuit capability of 600 MVAsc connected
to a 115 kV bus. A delta – Wye transformer rated at 30 MVA connects the 115 kV bus
to the 13.8 kV bus and has a positive and zero sequence impedance of 10 % at 30
MVAStep 1 of our calculation asks us to convert our system into perunit values
and that is exactly what we did in part 2b of our series.
Here we will draw the same network again:
Next we move to step 2,
which asks us to identify the type of fault involved.
By drawing a simple diagram with transformer connection
and the type of fault, we can see how the current will flow for a 2LG fault on phase b & c on the 13.8kV system.
A Double line to ground fault which is an unsymmetrical fault type (meaning an unbalanced fault).
This type of fault will produces all sequence quantities, positive, negative and zero sequence components
(current and voltage quantities)
and In part 3d-1, we will of-course hand calculate all phase and sequence current quantities quantities.
In Part 3d-2 we will calculate Phase and sequence voltage quantities.
Next we move to step 3,
which asks us to make unfaulted sequence networks
from the per unit converted system in Step 1.
These are the individual sequence network diagrams which we covered in part 2c
of the series,
Here we redraw the same networks:
And then, in Step 4,
we will draw the faulted sequence network,
Given that the fault is a double line to ground fault
we will connect the positive and negative sequence networks,
as well as the zero sequence network in parallel.
This is discussed in great detail in our video on double line to ground faults in Topic 2
which is the Derivations of Sequence Network Diagrams of our series.
Here we will just redraw the parallel connected faulted networks.
Now this type of fault is an unsymmetrical fault type and it will have a zero sequence network
incorporated in the faulted sequence network with the positive and negative sequence networks connected as well.
To simplify the network and to make it easy for us to
relate the formulas with the network,
we draw it again, but this time it’s a lot simpler.
And simplifying it even more we see that we have three impedances in parallel,
with the sequence fault currents, positive negative and zero,
flowing through each branch respectively.
With this network in mind, to calculate the positive sequence fault current
we first take the zero sequence and negative sequence branch impedances in parallel.
So we use our handy dandy formula that we learned and we start with 1 over R is equal
to 1 over j.15 + 1 over j .10,
well this can be simplified even further by 1 over R is equal to j.10 + j.15
divided by j.10 times j.15,
and this could be simplified even more by R is equal to j.10 times j.15 divided by j.10 + j.15.
Now this is just a way for us to combine parallel resistor or impedances together.
After taking these two impedances in parallel we find that they are in series with the positive sequence impedance,
Now we add that in to the picture,
so R is equal to j.15 which is the positive sequence impedance
plus the negative and zero sequence impedances connected in parallel
together that we solved earlier.
That done, we have successfully lumped the entire network impedance
into one single impedance and now,
dividing the voltage with this impedance we will get the positive sequence current value:
So the positive sequence current value is equal to 1 at angle of zero degrees per unit
divided by j.15 which is the positive sequence impedance plus the negative and zero sequence impedances
in parallel.
Well that simply gives us –j4.76 per unit as the positive sequence fault current,
and now we just have to simplify that you just have to multiply that with our base value of current,
the Base value of current being 1255 Amps that we have seen in the earlier videos:
That simply gives us 597 amps at the angle of negative 90 degrees.
Now this is the positive sequence fault current values in ampere form.
For the negative sequence fault current we cannot follow the same process
because there is no voltage source in the negative sequence branch,
instead we see that the positive sequence fault current will flow forward into the negative sequence branch
and zero sequence branch as shown in the diagram below.
With the difference that direction will be opposite.
Given that the positive sequence current flows into both the zero sequence and negative sequence branches,
current divider rule can be applied to find the current in negative and zero sequence branches
accordingly.
So the negative sequence current is equal to negative of the positive sequence current
times the zero sequence impedance
divided by the zero sequence impedance
plus the negative sequence impedance,
This is just our simple current divider rule that we use in all sorts of different places
including fault alysis in power systems
as shown
So, the positive sequence current with the reversed direction,
j4.76 is multiplied by the ratio of impedances of zero and negative sequence branches which is simply j0.1 divided by
j0.25. This gives us a value of j1.90 per unit
and after multiplying that with the base value of current,
1255 amperes we get negative sequence current in amperes
is equal to 2391 angle -90 amperes.
Now the positive sequence currents was equal to 5997 amps
so the negative sequence current and zero sequence current
should be simply a ratio of the positive sequence current
and so we could simply calculate the zero sequence current accordingly
but will continue to use the current divider rule to calculate the remaining fault current which is the zero sequence current.
So the zero sequence current is equal to the negative of positive sequence current
multiplied by the negative sequence impedance
divided by zero sequence impedance
plus negative sequence impedance
So, j4.76 per unit multiplied by the ratio of j0.15 and j0.25 per unit,
gives as a per unit value of j2.857 per unit.
And if we multiply that with the base we get 3586 amps at the angle of 90 degrees.
And we can see that the positive sequence impedance
is equal to the negative sequence impedance
plus the zero sequence impedance
which all matches up and agrees to each other and it alingn with the diagram that we drew
before.
Now, proceeding to step 6, we have to convert the sequence component values to phase current values
by our three legacy transformation equations:
which is shown below.
We proceed by plugging in the values of a operator,
and our calculated sequence current values as shown
And what we get by simply letting our calculators do the math is that,
The current on phase a is simply equal to 0,
the current on phase b is equal to 9025 amps at the angle of 143 degrees and
the current on phase c is equal to 9025 amps at the angle of 36.58 degrees.
So this is our hand calculated solution for fault current values
on a double line to ground fault
Now, This is quite a unique answer answer especially if we compare it with our simple diagram
that we show below. In this diagram, we see that phase b & c are involved
as well as ground making a 2LG fault.
From our hand calculations,
its clear that phase b & c currents are equal in magnitude but not quite opposite in direction.
Phase B has an angle of 143 degrees
where phase C has an angle of 36.58 degrees.
There is non-conformity of angles and it is probably due to the fact that ground is involved.
But when we draw the diagram below,
we show the fault current flowing towards the point of the fault,
is represented as flowing towards the point of the fault, but when we draw the diagram below we show that fault current flowing towards the
point of the fault, this representation may not entirely be accurate because the angles are different
and they are not quite opposite of each other,
but this is the best way that we can think of fault current flow.
Also, this is a classical 2LG fault analysis.
In real world systems however,
the magnitude of the faulted phase may not be equal and the angles may be entirely different
for a 2LG.
But generally speaking,
this is a good way to think about 2LG fault analysis.
In the next part, part 3d-2,
we will calculate the voltage quantities at the point of the fault.
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