Fault Analysis in Power Systems, Part 4c

In this video, we will explore how to reflect a line-to-ground fault current value, which occurs on the low voltage side of the transformer, on the high voltage side. Let’s have a recap of our video in Part 3B-1, where we calculated the fault current and voltage values for a single line-to-ground fault on the 13.8 kV bus, or the low voltage side of the transformer. We also saw the sequence component diagram for the fault type.

For a one-line to ground fault, we found that the positive sequence current was -j2.5 per unit, and the negative and zero sequence currents were also -j2.5 per unit. These values can be expressed as 2.5 angle -90 degrees in polar form. It’s important to note that the lowercase subscript indicates fault current values on the low voltage side, while the uppercase represents fault current on the high voltage side.

Now, let’s reflect these per unit current values on the high voltage side by multiplying them with the appropriate base values to find the actual values. Given that we have a delta-Y transformer connection (dy1 or dab), we shift the low voltage positive sequence components by +30 degrees and the negative sequence components by -30 degrees. However, the zero sequence component will be non-existent on the high voltage side due to the delta side of the transformer trapping the zero sequence current within itself.

To convert the per unit quantities to ampere quantities, we divide the base value of power (30 MVA) by the high side voltage (115 kV) to get a base value of 150.6 amps. Multiplying the per unit quantities with the base values, we find the positive sequence current to be 377 amps at an angle of -60 degrees, and the negative sequence current to be 377 amps at an angle of -120 degrees.

Using the transformation equations, we can transform these sequence component current values to phase current quantities. Removing the zero sequence current, the line current A is calculated to be 625 amps at an angle of -90 degrees, line B current is zero, and line C current is 625 amps at an angle of 90 degrees. These values align with the current flow path on the high voltage side, where the fault current will only flow on lines A and C for a low voltage one-line to ground fault. This is due to the specific delta transformer connection.

On the high voltage side, line A and line C current quantities are 180 degrees apart, which supports the zero sequence current calculations. Even if we try to calculate the zero sequence current using phase A, B, and C current quantities on the high voltage side, we still get zero. This is because I0 is equal to one-third of IA plus IB plus IC, and plugging in line A current and line C current, we get zero.

Next, let’s calculate the sequence voltage with the reflected sequence current components. Applying Kirchhoff’s Voltage Law (KVL) in each of the connected sequence networks, we can calculate the positive sequence voltage by subtracting the voltage drop across the positive sequence impedance (0.05 per unit) on the high voltage side from the positive sequence source voltage (1 angle 0 degrees). Taking into account the reflected positive sequence current on the high voltage side (2.5 angle -60 degrees), we find the positive sequence voltage to be 0.89 per unit at an angle of -4 degrees.

Similarly, the negative sequence voltage is calculated by subtracting the voltage drop across the negative sequence impedance on the high voltage side from the negative sequence source voltage. Considering the reflected negative sequence current on the high voltage side (2.5 angle -120 degrees), we find the negative sequence voltage to be 0.125 per unit at an angle of 150 degrees. The zero sequence voltage will be zero since the zero sequence current was also zero.

Setting up a base value to convert the per unit voltages to actual line-to-ground voltages, Vbase is taken as 115 kV divided by the square root of three. Multiplying the components with the base value, we find the positive sequence voltage to be 59.3 kV at an angle of -4 degrees, and the negative sequence voltage to be 8.3 kV at an angle of 150 degrees.

Using the three legacy transformation equations, we can transform the sequence components to actual one-line to ground phase voltages. Since the negative sequence component is non-existent, we can eliminate it directly from our equation. Plugging in the calculated sequence component values, we find that the actual phase-to-ground voltages are VA = 52 kV at an angle of 0 degrees, VB = 66.3 kV at an angle of 120 degrees, and VC = 60.5 kV at an angle of 108 degrees for a one-line to ground fault.

We observe that the high voltage line A voltage is the most depressed since the fault occurred on line A on the low voltage side. There is also some depression in the high voltage line C voltage. However, we don’t see any significant depression in the high voltage line B voltage. This aligns with the current flow on the high voltage side, where the high voltage fault current flows on lines A and C for a low voltage one-line to ground fault.

The reason why the fault current flows on lines A and C on the high voltage side is due to the specific type of delta connection mentioned in Part 4A of the series, which can also be visually seen in the diagram. It also makes sense that line A and line C current quantities are 180 degrees apart, supporting the zero sequence current calculations. On the high voltage side, the largest voltage depression occurs on line A, followed by line C, while line B does not see any significant depression.

Thank you.