Fault Analysis in Power Systems Part 3c-2

In this video, we will calculate the voltage quantities for a phase-to-phase fault, continuing from Part 3C-1. Moving forward to fault voltage calculations, we observe from the fault sequence network that the positive and negative sequence voltages at the point of the fault will be equal to each other since we are measuring at the same node. As shown in the sequence network diagram, we use the same base values as before, where Vbase is equal to 13.8 kV divided by the square root of three.

Applying KVL (Kirchhoff’s Voltage Law) in the positive sequence loop, the positive sequence voltage at the point of the fault is the source voltage (1 per unit at an angle of zero degrees) minus the voltage drop across the positive sequence impedance, which is the positive sequence fault current (-j3.333 per unit) multiplied by j0.15 per unit. This gives us a value of 0.5 per unit at an angle of zero degrees. This is a per unit value, and we need to convert it to sequence voltage quantities. We do this by simply multiplying it by the base value calculated earlier, resulting in a positive sequence voltage of 3.98 kV at an angle of zero degrees.

As discussed earlier, the negative sequence voltage will also be equal to 3.98 kV at an angle of zero degrees. So, both the positive sequence and the negative sequence voltages will be equal.

Now, let’s apply our familiar transformation equations to obtain the phase voltage values. We do this by plugging in the calculated sequence components and the A operator as shown below. Since faulted phases B and C will have the same voltage levels (Vphase B = Vphase C), we only need to calculate two phase voltages: Vphase A and Vphase B. Vphase A gives us 7.96 kV, and Vphase B and Vphase C will be equal to 3.98 kV.

This is quite an interesting result, especially when considering our original drawing. We know that Phase A on the 13.8 kV system is the non-faulted phase and, therefore, will not exhibit any fault current. However, Phase B and Phase C are the faulted phases. We have previously proven that the fault current on Phase B and Phase C will be equal in magnitude but opposite in direction.

When it comes to voltage quantities at the point of the fault, we have shown by hand calculation that the phase A voltage at the point of the fault on the 13.8 kV system is the normal operating quantity, which is 13.8 kV divided by the square root of three, resulting in 7.96 kV phase voltage. However, for the faulted phases at the point of the fault (Phase B and Phase C), the fault voltages are exactly half, meaning they are 7.96 kV divided by 2, which is 3.98 kV phase voltages.

This is an interesting characteristic that is quite common for a classic line-to-line fault in power systems. These types of faults typically occur when two bare conductors or phases come into contact due to conditions such as heavy wind or geo disturbances like accidents involving vehicles or poles. They can also occur during small earthquakes that cause the conductors to move back and forth, eventually slapping each other. In such cases, it is a phase-to-phase fault, and if we analyze the fault waveforms, we would typically expect the voltages to be exactly half for the faulted phases, with fault currents on the faulted phases equal in magnitude but opposite in direction.

Thank you.