Fault Analysis in Power Systems Part 3b-2

Now let’s continue where we left off in the previous part, Part 3B. In this bonus step, we will calculate the voltage quantities. In the previous section, Part 3B, we calculated the phase current quantities for a single line-to-ground fault on the 13.8 kV system. Now, in this step, we will calculate the phase voltage quantities at the point of the fault on the 13.8 kV system. We will follow a very similar process.

First, let’s redraw our faulted sequence network to focus on the positive, negative, and zero sequence voltages shown on the low-voltage 13.8 kV fictitious bus. Our objective in this section is to determine the sequence voltage quantities by hand and, from there, calculate the actual phase voltage quantities.

Before we proceed, let’s establish a base value of voltage for our calculations. Since the voltage at the point of the fault is 13.8 kV, we will select that as our base value, denoted as Vbase = 13.8 kV divided by the square root of 3.

To calculate the positive sequence voltage, we will apply Kirchhoff’s Voltage Law (KVL) in the positive sequence loop. The positive sequence voltage will equal 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 represents the sum of the positive sequence generator and transformer impedances, multiplied by the per-unit positive sequence current (-j2.5) flowing to the 13.8 kV fictitious bus. This calculation results in 0.625 per unit, which needs to be converted to actual voltage quantities by multiplying it with the base value of 13.8 kV divided by the square root of 3. Therefore, the positive sequence voltage is 4.98 kV at an angle of 0 degrees.

Similarly, we calculate the negative sequence voltage quantity by applying KVL in the negative sequence and zero sequence networks. Since there is no voltage source in the negative sequence loop, the negative sequence voltage is considered zero. The negative sequence voltage is -j0.15 per unit, which represents the transformer and generator impedance, multiplied by -j2.5 (the current flowing to the 13.8 kV fictitious bus in the negative sequence). This calculation results in -0.375 per unit, which, in terms of actual voltage quantities, is 2.988 kV at an angle of 180 degrees. The entire voltage consists of the voltage drop across the negative sequence impedance (j0.15 per unit) multiplied by -j2.5, resulting in -0.375 per unit. By multiplying it accordingly, we obtain 2.988 kV at an angle of 180 degrees. This is the negative sequence voltage.

For the zero sequence voltage, we follow a similar calculation. The zero sequence voltage is -j0.10 multiplied by -j2.5, resulting in -j2.5 per unit. Here, the impedances are only -j0.10, as the transformer’s per-unit impedance (j0.05) must be excluded from the loop due to the open delta. Additionally, since the zero sequence current will not flow through the high-voltage side, this strengthens the exclusion of the generator impedance. Furthermore, as there is no voltage source for the zero sequence, the voltage source is considered zero. The zero sequence voltage is -0.25 per unit, which, when multiplied by the base value, results in 1.992 kV at an angle of 180 degrees.

Now, using our familiar transformation equations, we can determine the phase voltage quantities at the point of the fault on the 13.8 kV bus. By plugging in the calculated values of the three sequence components and the A-operator, we solve the math and obtain the following results:

• The voltage at the point of the fault on Phase A is 0 kV.
• The voltage at the point of the fault on Phase B is 7.52 kV at an angle of -113 degrees.
• The voltage at the point of the fault on Phase C is 7.52 kV at an angle of 113 degrees.

This is an interesting result. Let’s redraw the original drawing and discuss it further. It is both logical and intuitive because we know that the fault was a one line-to-ground fault on Phase A. At the point of the fault, it is clear that Phase A voltage will go to zero, as expected. However, Phase B and Phase C voltage magnitudes will experience a slight depression. During normal and balanced operating conditions, we should expect phase voltages to be 13.8 kV divided by the square root of 3, resulting in 7.96 kV. However, during the Phase A-to-ground fault, at the point of the fault, Phase B and C voltages are 7.52 kV, showing a slight depression on the unfaulted phases.

Another interesting observation is that the angles of Phase B and C voltages are opposite, while their magnitudes remain equal. This makes sense because Phase A voltage is driven to zero, and at the point of the fault, the remaining two phases need to balance out. Thus, the phase voltages are equal, but the angles are opposite.

In the next video, we will walk through a similar calculation for the line-to-line fault.

Thank you!”