Fault Analysis in Power Systems, Part 4b

In this video, we will explore how we can take fault current and voltage quantities that occur on the low voltage side of the transformer and reflect them on the high voltage side. Let’s have a recap of Part 3A, where we calculated fault current and voltage quantities for a three-line to ground fault on the 13.8 kV bus on the low voltage side. We found that the positive sequence current was equal to -j6.67 per unit, and the negative and zero sequence currents were equal to zero.

Now, let’s reflect these per unit current quantities on the high voltage side by multiplying them with the appropriate base values to find the actual ampere current values. Given that we have a delta-Y transformer connection (dy1 or dab), we shift the positive sequence component by +30 degrees. The positive sequence current on the high voltage side is then calculated as 6.67 per unit at an angle of -60 degrees.

However, the zero sequence and negative sequence components will be non-existent on the high voltage side, as a three-line to ground fault is a balanced, symmetrical fault with no negative or zero sequence components. We can convert these per unit values to ampere values in sequence components by setting up a base value. Dividing the power value of 30 MVA by the high voltage side of the transformer (115 kV), we get a base value of 150.68 amps. Multiplying the per unit quantities with the base values, we find that the high voltage positive sequence current is 1004 amps at an angle of -60 degrees.

Now, using the transformation equations, we can transform these sequence component current values to phase current quantities. It’s important to note that we remove the zero and negative sequence current components from these transformation equations because they do not exist on the high voltage side. Line A current is found to be 1004 amps at an angle of -60 degrees, line B current is 1004 amps at an angle of 180 degrees, and line C current is 1004 amps at an angle of 60 degrees. These current values align with the current flow on the high voltage side for a low voltage three-line to ground fault involving all three lines.

On the high voltage side, the fault current flows on lines A, B, and C of the transformer due to the specific dy1 transformer connection. The negative and zero sequence currents do not exist on the high voltage side. Even if we try to calculate them using the phase A, B, C current quantities on the high voltage side, we still get zero. This is because the three-line to ground fault is perfectly balanced on both the low voltage side and the high voltage side of the transformer.

Next, let’s calculate the sequence voltage quantities reflected on the high voltage side of the transformer. Considering the faulted sequence network diagram for a three-line to ground fault, we apply Kirchhoff’s Voltage Law (KVL) in the positive sequence network. The positive sequence voltage can be calculated by subtracting 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). Taking into account the reflected positive sequence current on the high voltage side (6.67 per unit at an angle of -60 degrees), we find the positive sequence voltage to be 0.73 per unit at an angle of -13 degrees.

Zero sequence voltage and negative sequence voltage will be non-existent since the zero sequence current and negative sequence current were also non-existent. To convert the per unit voltage to sequence voltage quantities, we set up a base value. Multiplying the base values, we find the voltage values in sequence component form to be 48.51 kV at an angle of -13.2 degrees.

Using the three legacy transformation equations, we can transform the sequence component voltage quantities to actual line-to-ground voltage quantities. It’s important to note that we remove the negative and zero sequence components from the transformation equations. Plugging in the calculated values, we find the line A to ground voltage to be 48.51 kV at an angle of -13.2 degrees, line B to ground voltage to be 48.51 kV at an angle of -133.2 degrees, and line C to ground voltage to be 48.51 kV at an angle of 106.8 degrees.

This is an interesting result, as the fault was a three-line to ground fault with equal-magnitude fault currents on each phase, displaced by 120 degrees. The voltage magnitudes also reflect this pattern, being equal on each phase but displaced by 120 degrees. We observe a voltage depression on the high voltage side of the transformer, despite the fault occurring on the low voltage side. On the high voltage side, the voltage quantities during normal operating conditions are 66 kV line to ground, but during a three-phase fault on the low voltage side, the voltage quantities are depressed to 48.51 kV.

Thank you.