Fault Analysis in Power Systems Part 3a

Welcome to Part 3A of our series on fault analysis in power systems, where we will discuss three line-to-ground faults. In the previous parts, we covered per-unit systems and sequence network diagrams. In Part 3 of this series (Part 3A, 3B, 3C, and so forth), we will explain the math calculations by hand for current and voltage quantities for various types of faults, including three line-to-ground faults, one line-to-ground faults, line-to-line faults, and two line-to-ground faults.

In this tutorial, we will focus on the first example we introduced in the previous parts. We have a three-phase synchronous generator with a short-circuit capability of 600 MVA short-circuit connected to a 115 kV bus. A delta-wye transformer rated at 30 MVA connects the 115 kV bus to a 13.8 kV bus and has a positive and zero sequence impedance of 10. Step 1 of our calculations asks us to convert the system into per-unit values, which we already did in Part 2B of our series. Let’s redraw the same networks.

Next, we move to Step 2, which requires us to identify the type of fault involved. In this case, we have a three line-to-ground fault, which is a symmetrical fault type. This means it’s a balanced fault that produces only positive sequence quantities. No negative or zero sequence quantities are generated in this fault type.

Moving on to Step 3, we create an unfaulted sequence network from the per-unit converted system in Step 1. These are individual sequence network diagrams that were converted in Part 2C of the series. Let’s draw the same networks again: positive sequence network, negative sequence network, and zero sequence network. Refer to Part 2C for more details.

In Step 4, we draw the faulted sequence network, which interconnects based on the type of fault. Since we have a three line-to-ground fault that produces only positive sequence quantities, we will focus on the positive sequence network. The positive sequence network diagram represents the faulted network.

Moving on to Step 5, we hand-calculate the sequence network quantities. As we analyze the faulted sequence network diagram, we can see that we simply need to divide the source voltage (in per-unit values) by the total positive sequence impedance (also in per-unit values). After dividing, we obtain the positive sequence current.

In Step 6, we calculate the three-phase current and voltage values by converting the per-unit quantities to phase quantities. This step involves complex calculations, so understanding the fundamentals of symmetrical components is critical. The base value of current is calculated by dividing the base value of power (30 MVA) by the voltage on the bus (13.8 kV) multiplied by the square root of three. Then, we multiply the per-unit values of the sequence components by the base value current to obtain the phase current values in amperes.

To summarize the six steps, we looked at the system and converted it into per-unit values, created the unfaulted sequence network diagrams, identified the fault type, eliminated the negative and zero sequence components for a balanced three-line fault, converted the per-unit values to sequence component values in amperes, and converted the sequence component values to phase current values for the three-phase system.

If we wanted to calculate the voltage quantities, it follows a similar process with six steps. However, for a perfectly balanced three-phase fault, we would expect zero quantities for phase A, B, and C voltage quantities at the point of the fault on the 13.8 kV bus.

In Step 7, we can easily reflect the current and voltage quantities on the high voltage side of the transformer. However, we will cover that step in another fault analysis.

In the next video, we will calculate a line-to-line fault and follow the same methodology.