Fault Analysis in Power Systems Part 3d-1

In this, we will calculate the phase current and voltage quantities for a double line-to-ground fault on the 13.8 kV bus, involving phase B, phase C, and ground.

Continuing from our previous example, 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-wide transformer rated at 30 MVA connects the 115 kV bus to the 13.8 kV bus, with positive and zero sequence impedances of 10 at 30 MVA.

Step one of our calculation asked us to convert our system into per unit values, which we did in Part 2B of our series. We will draw the same network again. Moving to step 2, we identify the type of fault involved by drawing a simple diagram with transformer connections and the type of fault. We can see how the current will flow for a double line-to-ground fault on phase B and C on the 13.8 kV system. This is a double line-to-ground fault, which is an unsymmetrical fault type, meaning an unbalanced fault. This type of fault will produce all sequence quantities: positive, negative, and zero sequence components, current, and voltage quantities. In Part 3D-1, we will hand-calculate all phase and sequence current quantities. In Part 3D-2, we will calculate phase and sequence voltage quantities.

Moving to step number 3, we make unfaulted sequence networks from the per unit converted systems in step one. We redraw the same networks. In step number 4, we draw the faulted sequence network. Given that the fault is a double line-to-ground fault, we connect the positive and negative sequence networks, as well as the zero sequence network, in parallel. We simplify the network to relate the formulas with the network. We draw it again, this time simpler.

Simplifying it even more, we see three impedances in parallel with the sequence fault current (positive, negative, and zero sequence) flowing through each branch, respectively. To calculate the positive sequence fault current, we first take the zero and negative sequence branch impedances in parallel. Using formulas, we simplify it further, combining parallel resistors into a single impedance. Dividing the voltage by the impedance, we obtain the positive sequence current value.

The positive sequence current value is obtained by dividing one per unit at an angle of zero degrees by the sum of the positive sequence impedance, negative sequence impedance, and zero sequence impedance in parallel. This gives us j4.76 per unit as the positive sequence fault current. Multiplying this value by the base value of current (1255 amps), we get 5977 amps at an angle of -90 degrees.

For the negative sequence fault current, we cannot follow the same process because there is no voltage source in the negative sequence branch. Instead, we observe that the positive sequence fault current flows forward into the negative sequence branch and zero sequence branch. Applying the current divider rule, we find the negative sequence current and the zero sequence current. The negative sequence current is -j1.90 per unit (opposite direction of the positive sequence current), and when multiplied by the base value of current, it becomes -2391 amps at an angle of -90 degrees.

Similarly, using the current divider rule, we calculate the zero sequence current, which is -j2.857 per unit, resulting in 3586 amps at an angle of 90 degrees. The positive sequence impedance is equal to the sum of the negative sequence impedance and the zero sequence impedance, confirming the consistency in our calculations.

Moving to step number 6, we convert the sequence component values to phase current values using the three legacy transformation equations. After plugging in the values of the A operator and the calculated sequence current values, we find that the current on phase A is zero, the current on phase B is 9025 amps at an angle of 143 degrees, and the current on phase C is 9025 amps at an angle of 36.58 degrees.

This is our hand-calculated solution for fault current values in a double line-to-ground fault. The answer is quite unique, especially when comparing it with our simple diagram. In this diagram, we see that phase B and phase C are involved, along with ground, creating a two-line-to-ground fault. From my hand calculations, it is clear that the phase B and phase C currents are equal in magnitude but not opposite in direction. Phase B has an angle of 143 degrees, while phase C has an angle of 36.58 degrees. There is a non-conformity of angles, possibly due to the involvement of ground. The diagram representation may not entirely be accurate because the angles are different and not exactly opposite of each other. However, it provides a good way to think about two-line-to-ground fault analysis.

In the next video, Part 3D-2, we will calculate the voltage quantities at the point of the fault.

Thank you.