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Fault Analysis in power systems Part 3c – 2. In this video we will calculate the voltage quantities for a phase to phase fault and it is a continuation of part 3c – 1.

Moving forward to fault voltage calculations, we see that from the faulted sequence network that the positive and negative sequence voltages at the point of the fault will be equal to each other because we are measuring at the same node, so positive sequence voltage and negative sequence voltage will be equal as shown in the sequence network diagram:

We use the same base value as before of V base is equal to 13.8 kV divided by square root of 3.

Now applying KVL in the positive sequence loop, the positive sequence voltage at the point of the fault is the source voltage 1 angle 0 degrees minus the voltage drop across the positive sequence impedance which is positive sequence fault current –j3.333 p.u multiplied by j0.15 p.u.

It gives us a value of 0.5 p.u angle 0 degrees, now this is a per unit value and we need to convert this value into sequence voltage quantities and we do this by simply multiplying the base value we calculated earlier which gives us the positive sequence voltage of 3.98 kV at the angle of 0 degrees.

Now as discussed earlier, the negative sequence voltage will also be equal to 3.98 kV at ∠0°. So both the positive sequence voltage and the negative sequence voltage will be equal.

Now that said, we are now left with the task of applying our three familiar transformation equations to get the phase voltage values. Now we do this by plugging in the calculated sequence components and the a operator as shown below:

Now, we have an added advantage here because since the faulted phases B and C will have the same voltage levels Vphase_b is equal to Vphase_c, we will only have to calculate two phase voltages, which is Vphase_a and Vphase_b. Of which VA gives us 7.96 kV and VB and VC will equal 3.98 kV. Now this is quite an interesting answer especially if we consider our original drawing as shown

We know that phase a on the 13.8kV system is the non-faulted phase and therefore, it will not exhibit any fault current. However, Phase b and phase c are the faulted phases. We have proved previously that phase b and phase c fault current will be equal in magnitude however it will be opposite in direction. Now when it comes to voltage quantities at the point of the fault, we proved by hand that phase a voltage, at the point of the fault on the 13.8kV system, is the normal operating quantity meaning that it’s 13.8 kV/root(3) = 7.96kV phase voltage, okay, However, for the faulted phases, at the point of the fault, and now we are talking about the phase b&c fault voltages are exactly half meaning it’s 7.96kV/2 = 3.98kV phase voltages.

Now this is an interesting characteristic which is quite common for a classic line to line fault in power systems, Most notably these types of faults occur where two bare conductors or phases slap each other due to like a heavy wind condition or geo-disturbance like accidents, vehicular accidents on a pole where bare conductor transmission lines are strung or a small earthquake that causes these conductors to go back and forth and eventually slap each other, when that happens it’s a phase to phase fault and we analyze the fault looking at waveforms we would expect typically for the voltages to be exactly half for the faulted phase and the fault currents on the faulted phases to be equal in magnitude but opposite in direction.

In the next video we will perform a similar fault calculation for a double line to ground fault which is much more complex and not as intuitive.

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Our coorporate sponsor AllumiaX,LLC published a new blog on "Fault Analysis in Power Systems". This blog covers the following topics:
  • Types of Faults in Power System
  • Steps to perform a Fault Analysis in a Power System
  • Causes of Power System Faults

For Continue Reading, CLICK HERE.

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