Fault Analysis in Power Systems Part 3b-1

Welcome to Part 3B of our series on fault analysis in power systems, where we will hand-calculate the current and voltage quantities for various types of faults. In this video, we will focus on a single line-to-ground fault on Line A of the 13.8 kV system. Let’s dive into our previous example.

We have three synchronous generators 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 the 13.8 kV bus and has an impedance of 10% at 30 MVA. In Step 1, we converted the system into per-unit values, as we did in Part 2B of our series. Let’s redraw the same network.

Before we begin, let’s draw the system for a single line-to-ground fault on the 13.8 kV bus. At the point of the fault, we see that the fault current goes to ground and returns through the grounded system on the Y-side of the transformer neutral, creating a circulating loop. We also note that the fault current is not present on Phase B and C lines on the 13.8 kV system. Due to the delta-wye grounded transformer connection, fault current is only present on Phase A and C sides of the transformer, but not on Phase B.

Moving on to Step 2, we identify the type of fault involved. In this case, we have a single line-to-ground fault, which is an unsymmetrical fault, meaning it is an unbalanced fault type. This type of fault produces all sequence quantities: positive, negative, and zero sequence components for current and voltage quantities.

In Step 3, we create an unfaulted sequence network from the per-unit converted system. Let’s redraw the same networks: positive sequence network, negative sequence network, and zero sequence network. Please refer to Part 2C for more details.

In Step 4, we draw the faulted sequence network. Since the fault is a single line-to-ground fault, we connect all three sequence networks in series. We have explained this in detail in a separate video on the derivations of sequence network diagrams. Let’s redraw the series-connected faulted networks. The topmost sequence is the positive sequence network, easily identified by the voltage source. The middle network is the negative sequence network, similar to the positive sequence network but without a voltage source. The bottom network is the zero sequence network, which also does not have a voltage source. However, we account for the transformer in the zero sequence network as it is open on the high-voltage side and short-circuited on the low-voltage side.

Moving on to Step 5, we hand-calculate the sequence network quantities. Since we have an unbalanced network, we calculate all three sequence components: positive, negative, and zero sequence. In this sequence network, we observe that all impedances are connected in series in a single loop, with only one voltage source, which is 1 per unit at an angle of 0 degrees, representing the positive sequence. Therefore, we can deduce that all three sequence component current values will be equal and can be calculated by dividing the voltage by the sum of the impedance for the entire loop. The sum of the impedance for the entire loop is J0.05 + J0.10 per unit (positive sequence impedance) + J0.05 + J0.10 per unit (negative sequence impedance) + J0.10 per unit (zero sequence impedance). For the zero sequence impedance, we don’t consider the J0.05 per unit generator impedance as it represents an open circuit for the zero sequence network. The total impedance of the entire loop is J0.40 per unit. The positive sequence current in this loop is equal to the pre-fault voltage (1 per unit at an angle of 0 degrees) divided by J0.40 per unit, resulting in -J2.50 per unit.

Since this is a per-unit value, we need to find the base of the current by multiplying the base value of power (30 MVA) by the voltage on the bus (13.8 kV) divided by the square root of 3. The base value of current is 1255 amps. Multiplying this with -J2.50, we get the actual ampere quantity for the positive sequence current, which equals 3138 amps at an angle of -90 degrees. As we mentioned earlier, all sequence component values are equal, so the positive sequence current is equal to the negative sequence current, which is also equal to the zero sequence current, resulting in 3138 amps at an angle of -90 degrees for all three components.

In Step 6, we convert these sequence component values to phase current values by plugging in the sequence components and the A-operators with the familiar equations. We find that for Line A current on the 13.8 kV side, we get 9413 amps at an angle of -90 degrees. For Phase B line current on the 13.8 kV side, we get 0 amperes. And for Phase C line current on the 13.8 kV side, we also get 0 amperes. This aligns with what we assumed in our video on single line-to-ground fault derivations, where only the phase shorted to ground is supposed to carry the short-circuit current. In this case, the faulted phase is Phase A with 9413 amps at -90 degrees

, while the remaining phases, Phase B and C, have no fault current flowing, resulting in 0 amperes.

In the next part, we will calculate the voltage quantities on the 13.8 kV side. This will be covered in Part 3C of our series.

Thank you!”