Phase to ground fault question
- Thread starter bengaltiger14
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First, both the transformer and generator need to be on the same base for their pu values to be combined. The solution chooses to change the reactance of the transformer to the same base as the reactance of the generator, so it goes from 0.04 to 0.0533 after that.
Next, the last sentence is used to find the zero sequence and negative sequence reactance values for both the generator and the transformer from the positive sequence values, e.g., X0 = X1 / 2 and X2 = X1 / 3.
Since the fault is on the far side of the transformer, the sequence component Thevenin-equivalent reactances for the generator and transformer are in series, X1eq = XG1+XT1, etc. (If the fault was between the generator and transformer, the equivalent for each sequence impedance would be the generator and transformer reactances in parallel.)
As it is a SLG fault, the Thevenin-equivalent networks are connected in series for fault current evaluation, so all reactance elements are added in the denominator (X0eq+X1eq+X2eq) to calculate the equivalent impedance seen during the fault.
P.S. I don't like the way the solution denominator adds all the generator reactances first and then adds all of the transformer reactances - even though the math comes out the same, I think that's a poor solution methodology. I think it's better to find the Thevenin-equivalents I mentioned above for each of the three sequences first based on the fault location, then add the Thevenin-equivalents since the analysis is for a SLG fault.
Next, the last sentence is used to find the zero sequence and negative sequence reactance values for both the generator and the transformer from the positive sequence values, e.g., X0 = X1 / 2 and X2 = X1 / 3.
Since the fault is on the far side of the transformer, the sequence component Thevenin-equivalent reactances for the generator and transformer are in series, X1eq = XG1+XT1, etc. (If the fault was between the generator and transformer, the equivalent for each sequence impedance would be the generator and transformer reactances in parallel.)
As it is a SLG fault, the Thevenin-equivalent networks are connected in series for fault current evaluation, so all reactance elements are added in the denominator (X0eq+X1eq+X2eq) to calculate the equivalent impedance seen during the fault.
P.S. I don't like the way the solution denominator adds all the generator reactances first and then adds all of the transformer reactances - even though the math comes out the same, I think that's a poor solution methodology. I think it's better to find the Thevenin-equivalents I mentioned above for each of the three sequences first based on the fault location, then add the Thevenin-equivalents since the analysis is for a SLG fault.
Last edited:
bengaltiger14
New member
Thanks for making this make more sense.
