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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 lets continue where we left off in the previous part in part 3b. In the bonus step we will calculate the voltage quantities. Now in the previous section for part 3b we calculated the phase current quatities for a single line to ground fault on 13.8 kV system. Now this is a bonus step in this step we will calculate the phase voltage quantities at the point of the fault on the 13.8 kV system. Now we do tht with a very very similar process right and what we will do first is that we will go back to a 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 13.8 kV fictitious bus on the positive negative zero sequence network. And we want to determine the sequence voltage quantities by hand and from the sequence voltages 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.

V_Base=13.8kV/√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

V_a^((1) )=1 ∠0°-(j0.15)(-j2.5)=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.

V_A^((1) )=(0.625 pu)* 13.8kV/√3=4.98 kV∠0°

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.

V_a^((2) )=0-(j0.15)(-j2.5)=-0.375 pu

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.

V_a^((2) )=(-0.375pu)*(13.8kV/√3)=2.988 kV∠180°

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

V_a^((0) )=0-(j0.10)(-j2.5)=-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 -0.25 pu 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

V_a^((0) )=(-0.25)*(13.8kV/√3)=1.992 kV∠180°

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

V_a= 0

The voltage at the point of the fault on phase B is equal to 7.52 kV at the angle of negative 113 degrees.

V_a=7.52kV∠-113°

The voltage at the point of the fault on phase C will equal to 7.52 kV at the angle of positive 113 degrees.

V_c=7.52kV ∠113°

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.

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.

>V_phase=13.8kV/√3=7.96 kV

however, during phase A to ground fault, 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.

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