Now, let us continue where we left off in the previous part in part
3b. In the bonus that we will calculate the voltage
quantities. Now in the previous section for part
3b, we calculated the phase current quantities for a SLGF on a 13.8 kV
system. Now, this is the bonus
step, in this step, we will calculate the phase voltage
quantities at the point of the fault on a 13.8 kV
system. Now, we do that with a very very similar
process – right, and what we will do first is we will go back to our faulted sequence network and redraw it
again. In this figure, we want to focus on the positive, negative, and zero sequence voltages that are shown on the Low Voltage
13.8kV fictitious
bus on the positive, negative, and zero sequence
networks… and we want to determine the sequence voltage quantities by hand and from the sequence voltage
quantities, we will calculate the actual phase voltage
quantities. So this is our objective in this
section. Now, before we move forward, let’s first decide a base value of
voltage for the calculations and because the voltage at the point of the fault is 13.8 KV, we will select that as our base value
below. Vbase is equal to 13.8 kV
divided by root 3.
Now, to calculate positive sequence voltage we will now apply KVL in the positive sequence
loop. So here we go, The positive sequence voltage will equal to the sum of the
voltages across the voltage source minus the voltage drop due to the positive sequence
impedance, that is 1 angle 0 minus
j0.15 (which is the sum of the positive sequence generator and transformer
impedance) multiplied by the per unit positive sequence
current that is flowing to the 13.8kV fictitious bus which is –j2.5
pu. The –j2.5 pu
was calculated in the previous part 3b
section. That gives us 0.625 pu
Now, it needs to be converted back to the actual voltage
quantities by multiplying with the base value of 13.8
KV divided by square root
3. So that gives us 4.98 kV at the angle of 0
degrees. Similarly, now that we have calculated
the positive sequence voltage quantities, now we will calculate the negative sequence voltage
quantity. Similarly, we will apply KVL
again in the negative sequence and zero sequence
networks to calculate the per unit voltage and then multiply by the
voltage base to get the actual voltage
quantities. Since there is no voltage source in the negative sequence
loop, it will be considered
0. So the negative sequence voltage is equal
to zero minus j0.15 pu (which is the transformer and generator
impedance) times –j2.5 pu.
(which is the current that is going through the 13.8 KV fictitious
bus – negative
sequence. That is equal to -0.375 pu.
Okay. In terms of the actual voltage quantities that’s equal to 2.988 kV
at the angle of 180
degrees. And that voltage will entirely consist of the voltage
drop across the negative sequence
impedance that is j0.15 * -j2.5,
again that gives us -0.375 per
unit and we will multiply accordingly to determine the 2.988 kV
at the angle of 180
degrees. This is the negative sequence
voltage. For Zero sequence voltage, we do the
following: Zero sequence voltage is equal to 0 minus j0.10 times the
–j2.5 which is equal to –j0.25 pu.
Here, the impedances will be
–j0.10 only because the transformer pu impedance is only considered and the generator impedance which j0.05
must be excluded from the loop due to the open
delta… and we already know that zero sequence current will
NOT flow through the HV side which strengthen this point even
more. Furthermore, like the negative
sequence, we don’t have a voltage source for the zero
sequence that is why the voltage source equals
zero… and again, we get for the zero sequence voltage
quantity. And then we multiplied with the base value to get the actual voltage quantities which is
1.992 kV at the angle of 180
degrees. That’s a zero sequence voltage quantity Now, so we have calculated the positive sequence voltage
quantity, the negative sequence voltage quantity and the zero sequence voltage
quantity. Now, let us determine using our familiar transformation
equations to get the phase voltage quantities at the point of the fault on a 13.8 KV
bus. Plugging in these calculated
values of the three sequence components and the “a”
operator. Here’s what we will
get. Solving the math with a scientific calculator will lead us to the following
answers. The voltage at the point of the fault on phase A
is equal to 0 kV
The voltage at the point of the fault on phase B is equal to 7.52 kV
at the angle of negetive 113 degrees. The voltage at the point of the fault on phase C will equal 7.52 kV at the angle of 113
degrees. Now, this is an interesting
result. Now let’s redraw the original drawing and let’s talk about
this.This is very interesting, logical and very intuitive
result, because we know that the fault was one line to ground
fault on phase A. Now the point of the fault is very clear that the phase A voltage will drive to
zero and which is something very common and that is something, we should totally
expect.
However, phase B and phase C voltage magnitudes will have a slight
depression in
voltage. however, During normal and balanced operating
conditions, we should expect phase voltages to be
13.8kV divided by square root of
3 which equals 7.96 kV
during phase A to ground
fault, however at the point of the fault, Phase B and C voltages are 7.52 kV
which is very
interesting, because there is a slight depression on the unfaulted
phases. Another interesting thing to
note is that the angles of Phase B and C voltages are
opposite. They are equal in magnitude but the angles are
opposite which makes sense because Phase A voltage is driven to zero and
again, at the point of the fault, and the remaining two phases will need to balance
out. So phase voltages are goanna be
equal but the angles are goanna be
opposite. Now, In the next video we will walk through a
similar calculation for the line to line
fault. Now, we hope that you find this video tutorial
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