What is the only case that makes the single line-to-ground fault bigger than the 3-phase short circuit fault?

The L-G fault current will exceed the 3Ø fault current whenever you are near the terminal of a D-Y transformer. The Delta winding effectively blocks the zero sequence impedance contributed by the source. Thus Z0 source is zero and the only Z0 contribution to the impedance is the transformer impedance. As you move out from the transformer, the zero component should quickly increase until the L-G fault current value is less than the 3Ø fault current. This will happen whether you are near a power plant or a long distance from the source.

If there is a delta-wye transformer with the wye neutral grounded, then the zero sequence model for the transformer on the wye side is the transformer impedance to the zero sequence neutral, making the zero sequence impedance smaller in value as compared to the positive sequence and negative sequence. On the delta side the zero sequence model for the transformer is an open circuit, and the single phase short circuit current will depend on the zero sequence equivalent of the network on the delta side. The highest values for the single phase short circuit will be on the wye side of the transformer near the transformer. Usually the value for the zero sequence impedance for the overhead transmission line is about 3 times the values for the positive and negative sequence, and this is why the zero sequence impedance becomes larger as the fault is located further away from the transformer.

Considering a short-circuit near to a generator, if we call Z0 as the zero sequence impedance and 3Zg as the ground return impedance, the total zero sequence impedance is (Z0 + 3Zg), and the phase-to- ground short-circuit current If-t holds as 3Vth over (2Z1+ (Z0 + 3Zg)). Assuming Zg≈ 0 and Z0= Z1, the If-t becomes equal to 3Vth over (2Z1 + Z0) which, then, becomes 3Vth over (3Z1+Z0). Assuming Z1= Z0, this relation becomes 3Vth over 3V1, which is equal to the 3-phase short-circuit. But, if Z0< Z1, this implies If-t > I3p, that is, for phase-ground short-circuits near to the generator with a solidly grounded neutral, we do have If-t greater than I3p. That is the reason the generator designers put, at least, a small reactance in the neutral of generators to make the If-t not higher than I3p, to avoid damage in the iron core and windings due to internal faults of the generator. If the generator is large neutral resistors must be used instead.

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do graphical T-C representations of curves work?

thanks!

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