Fault Analysis in Power Systems Part 3b. This is the second video in Part 3 of the series on fault analysis in power
system where we will hand calculate the current and voltage
quantities for various types of
fault. In this video we will see how phase currents and voltages are
calculated for a single line to ground fault on line A of the 13.8kV
system. So Let’s get started with our previous
example. 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 an impedance of 10 % at 30
MVA. Step 1 of our calculation asks us to convert our system into per unit
values and that is exactly what we did in part 2b of our
series. Here we draw the same network
again. Now, before we begin, lets us draw the system to that we would expect for the 1LG
fault on the 13.8kV bus with the diagram
below. Now, when the fault occurs at the point of the
fault, we see that the fault current goes to
ground, then it returns from the grounded system on the wye side of the transformer neutral and makes a circulating
loop. We also see that the fault current coming
from the 115kV side of the system which is the
source. We note that fault current is not present on phase b and c lines on the 13.8kV
system and because we have a Delta-AB Wye-grounded
transformer connection, fault current is only present on Phase A & C side of the transformer but not Phase
B. Now, let us go through the calculations by
hand and prove this. Next we move to step
2, which asks us to identify the type of fault
involved. Now, as mentioned before, we have a
single line to ground fault which is an unsymmetrical fault (meaning an unbalanced fault
type). This type of fault produces all sequence
quantities, positive, negative and
zero sequence components. We talk about current and voltage
quantities. Next, we move to step no
3, which asks us to make an unfaulted sequence networks from the per unit converted system in Step
1. These are the individual sequence network diagrams which were covered in part 2c of the
series, here we will redraw the same
networks, so these are the unfaulted sequence
network. This is the positive sequence network
diagram which is indicated by a voltage
source. The negative sequence network diagram is very similar to the positive sequence
network, however it does not have a voltage
source. And the zero sequence network diagram also does not have the voltage
source, however, it does represent the Delta wye transformer with a high voltage side open and low voltage side
grounded. And then, in Step no
4, we will draw the faulted sequence
network, given that the fault is a single line to ground fault we will connect all three sequence networks in
series. This we have explained in great detail in a separate video in Topic
2: Derivations of sequence network diagrams of our
series. Here we just redraw the series connected faulted
networks. Let us describe this faulted sequence network diagram
briefly. The top most sequence is the positive sequence network which can be easily identified by the voltage
source. The middle is the negative sequence
network which is very similar to the positive sequence but does not have a voltage
source. The bottom network is the zero
sequence network which also does not have any voltage
source. However, we do account for the transformer in the zero sequence network
diagram (the open-circuit on the High Voltage side and short-circuit on the Low Voltage
side). Now let’s move towards step no
5, here we will hand calculate the sequence network quantities. Since we have an unbalanced
network we will calculate all the three sequence components, positive, negative and zero
sequence. In this faulted sequence network, we will see that all impedances are
connected in series in a single loop and there is only one voltage
source, which is 1pu angle 0
degrees in this entire loop (this comes from the positive
sequence). Therefore, it can be deduced that all three sequence components
current values and current quantities, which is positive, negative, zero sequence
currents will be equal and they can be calculated by dividing the voltage with the sum of the impedance for the entire
loop. The sum of the impedance of the entire loop will now be j0.05 + j0.10
p.u. This is the positive sequence impedance plus j0.05 + j.10
p.u. This is for the negative sequence impedance and finally plus
j0.10 only for the zero sequence
impedance. Now remember, if there is a zero
sequence impedance, there is open-circuit for the Zero Sequence
network so we cannot consider the j.05 pu generator
impedance). So the total impedance of the entire loop will be j0.40 per
unit. Positive sequence current in this loop is equal to the Prefault voltage which is 1 angle
0 degree divided by j0.40 p.u and that gives us negative –j2.5 per
unit. Now this is the positive sequence current
quantity. Since this is a per unit value we will need to find the base of the
current and multiply the per unit value with base value to find the actual current quantity in
amperes. The base value of current is equal to 30 MVA divided by square root of 3 times 13.8
KV which results to 1255
Amps Multiply that with –j2.5 and we get the actual ampere quantity of the
current. Which equals to 3138 Amperes at the angle -90
degrees. As we stated earlier that all sequence component values are
equal because the positive, negative, and zero sequence
networks are connected in
series. So the same current would flow though all three sequence
networks which means that the positive sequence current is equal to the negative sequence current which equals to
zero sequence current which equals 3138 Amps at the angle of -90
degrees. As a final step no 6, we have to convert these sequence component values
into phase current values by plugging in sequence components
values and a operators in these three familiar equations
below: What we will find is that; For line a current on 13.8kV
side, we get 9413 Amps at the angle of -90
degrees. For Phase b line current on the 13.8kV
side, we get 0, the math just happens to be that we get 0
amperes. And for Phase c line current on the 13.8kV
side, we also get
0. This is exactly the same as what we assumed in our video on single line to ground
fault derivations and intuitively only the phase shorted to
ground is supposed to carry the short circuit
current, in this case the faulted phase is phase
A which has 9413 Amps at -90
degrees. The remaining phases (meaning Phase B and Phase C) have no fault current
flowing, therefore they are
zero. So, now in the next part, what we will do is,
we actually calculate the voltage
quantities on the 13.8 Kv
side, so this will be split into the second part, Part
3b. We hope, that you find this tutorial
valuable and useful as a professional or a
student. Please support us by becoming our patron and donating through
patreon.com/generalpac. Your generous donation will help us continue making high quality power system video
tutorials.
Thank you.