Fault Analysis in Power Systems, Part 4d

In this, we will explore how to reflect line-to-line fault current quantities, which occur on the low voltage Y-grounded side of the transformer, on the high voltage delta side of the transformer. Let’s have a recap of our video in Part 3C-1, where we calculated the fault current values for line-to-line faults.

For the line-to-line fault type, we found that the positive sequence current was -j3.33 per unit, the negative sequence current was j3.33 per unit, and the zero sequence current was zero. These values can be expressed as 3.33 angle -90 degrees and 3.33 angle 90 degrees, respectively, in polar form. It’s important to note that the lowercase subscript indicates fault current 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 ampere quantities. Given that we have a delta-Y (DY1 or DAB) transformer connection, we shift the low voltage positive sequence component by +30 degrees and the negative sequence component by -30 degrees. However, the zero sequence component will be non-existent on both the low voltage and high voltage sides of the transformer due to the unsymmetrical nature of line-to-line faults.

To convert the per unit quantities to ampere quantities, we divide the base value of power (30 MVA) by the high voltage side voltage (115 kV) to get a base value of 150.6 amperes. Multiplying the per unit quantities with the base value, we find the positive sequence current to be 501.5 amps at an angle of -60 degrees, and the negative sequence current to be 501.5 amps at an angle of 60 degrees.

Using the transformation equations, we can transform these sequence component current values to phase current quantities. Removing the zero sequence current, we find that line A current is 501.5 amps at an angle of 0 degrees, line B current is 1003 amps at an angle of 180 degrees, and line C current is 501 amps at an angle of 0 degrees. These values align with the current flow path on the high voltage side. The fault current on phase B is twice the fault current on phases A and C, as line B current gets evenly distributed between line A and line C. This characteristic resembles a line-to-ground fault on the delta side, with the exception that the zero sequence current is absent.

It is worth noting that this characterization was found in Blackburn’s books, and it also makes it challenging for relays to distinguish between the two types of fault currents. Even if we try to calculate the zero sequence current using line A, B, and C current quantities on the high voltage side, we still get zero, as the zero sequence current is equal to one-third of the sum of line A, line B, and line C currents.

Next, let’s calculate the zero sequence voltage with our reflected sequence current component quantities. Returning to our faulted sequence network diagram for a line-to-line fault, we apply Kirchhoff’s Voltage Law (KVL) in each of the connected sequence networks. To calculate the positive sequence voltage, we subtract the voltage drop across the positive sequence impedance (j0.05 per unit) on the high voltage side from the positive sequence source voltage (1 angle 0 degrees). Considering the reflected positive sequence current on the high voltage side (3.33 angle -60 degrees), we find the positive sequence voltage to be 0.85 per unit at an angle of -5.5 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. Taking into account the reflected negative sequence current on the high voltage side (3.33 angle 60 degrees), we find the negative sequence voltage to be 0.1665 per unit at an angle of 150 degrees. The zero sequence voltage is zero since the zero sequence current was also zero.

Setting up a base value to convert the per unit voltage quantities to actual line-to-ground voltage quantities, Vbase is taken as the following. Multiplying the components with the base values, we find the positive sequence voltage to be 57 kV at an angle of -5.5 degrees, and the negative sequence voltage to be 11.05 kV at an angle of -30 degrees. Using the three transformation equations, we can then transform the sequence components to actual line-to-ground voltage quantities. Since the negative sequence component is non-existent, we can eliminate it directly from our equations. Plugging in the calculated sequence component values, we find that the actual phase A to ground voltage is 67.3 kV at an angle of -9.45 degrees, the phase B to ground voltage is 48.52 kV at an angle of -133 degrees, and the phase C to ground voltage is 57.1 kV at an angle of 125 degrees for a line-to-line fault.

We observe that the high voltage line A voltage is the least depressed, while the high voltage line B and line C voltages are more depressed. This is likely due to the fact that we have twice the amount of line current going through line B, which is distributed equally between line A and line C, and then returning back to the source.

Thank you.