Fault Analysis in Power Systems Part 3c-1

In this series on fault analysis and power systems, our objective is to hand calculate the current and voltage quantities for various faults. In this video, we will focus on the calculation of phase current and voltage quantities for a line-to-line fault on lines B and C of the 13.8 kV side of the transformer. Let’s continue with our example introduced in the previous parts.

We have a three-phase synchronous generator with a short-circuit capability of 600 MVA, short-circuited to a 115 kV bus. A delta-wye transformer rated at 30 MVA connects the 115 kV bus to the 13.8 kV bus, with positive and zero sequence impedance of 10 at 30 MVA. In Step 1, we converted our system into per unit values, as explained in Part 2B of our series. Now, let’s redraw the same network before constructing the converted sequence network diagram.

Let’s take a look at the type of fault we are dealing with: a line-to-line fault on the 13.8 kV system involving phases B and C. The fault current on Phase B flows towards the point of the fault, while on Phase C, the fault current is flowing away from the point of the fault. For a line-to-line fault, we expect the fault currents on both phases to be equal in magnitude but opposite in direction. Let’s proceed with the hand calculations to verify this result.

In Step 2, we identify the type of fault involved, which is a line-to-line fault. It is an unsymmetrical fault, meaning it is unbalanced. This type of fault produces sequence quantities: positive and negative sequence components for current and voltage. However, due to the nature of a line-to-line fault, it does not have a ground mode, and we should not expect any zero sequence current to flow.

Moving on to Step 3, we construct the sequence network from the per unit converted system. We draw the positive sequence network and the negative sequence network. However, since we have a line-to-line fault that does not have a ground mode, there is no zero sequence network.

In Step 4, we construct the faulted sequence network based on the type of fault. As the fault is a line-to-line fault, we connect the positive and negative sequence networks in parallel. For a line-to-line fault, zero sequence networks are omitted because zero sequence current cannot flow from ground into the fault. This also means that our subsequent calculations for fault current and voltage quantities will consider zero sequence components as zero.

Moving to Step 5, we hand calculate the sequence current quantities. We set up a base value and then proceed with sequence current calculations. The positive sequence current is found by dividing the pre-fault voltage (1 per unit at an angle of 0 degrees) by the total impedance in the loop, which is the sum of the positive and negative sequence impedances. Calculating this gives us -j3.333 per unit. To obtain the current value in amperes, we multiply the base value of 1255 amperes (derived from the base value of power, 30 MVA, divided by the base value of voltage at the faulted end, 13.8 kV) by the per unit value. This results in a positive sequence current of 4184 amperes at an angle of -90 degrees.

It is evident from the faulted sequence network that the positive and negative sequence currents will be equal in magnitude but flowing in opposite directions. To account for this, we give a 180 degrees phase shift to the positive sequence current to find the negative sequence current. Thus, the negative sequence current is also 4184 amperes but at an angle of +90 degrees.

As there is no zero sequence component for a line-to-line fault, we move to Step 6 to convert the sequence component values to phase current values. Using the familiar three-phase transformation equations, we simplify the equations further due to the absence of zero sequence current. After plugging in the values for the A operator (1 at an angle of 120 degrees) and the A squared operator (1 at an angle of 240 degrees), as well as the positive sequence current (4184 amperes at an angle of -90 degrees) and negative sequence current (4184 amperes at an angle of +90 degrees), we find the following:

• The fault current in Phase A is zero.
• The fault current in Phases B and C is equal in magnitude but opposite in direction.

This intuitive result can be explained by assuming that Phases B and C are shorted, resulting in no fault current in Phase A. As Phase B and C represent the faulted phases, they should have the same magnitude of fault current but opposite directions. This aligns well with the diagram we previously drew.

In the next part, Part 3C-2, we will calculate the voltage quantities for a face-to-face fault..

Thank you