Q7Power System - I

Question

Q.7. Derive an expression for fault current for single line to ground fault.

Answer 1

Consider a single line-to-ground (LG) fault occurring at phase a of an unloaded generator (with internal EMF E, purely positive-sequence, and sequence impedances Z1, Z2, Z0), with the fault connected to ground through a fault impedance Zf. The boundary conditions at the fault point are: Ib = 0, Ic = 0 (the healthy phases carry no current at the fault location) and Va = Ia·Zf (the faulted phase's voltage equals the drop across the fault impedance).

Step 1 — apply the current boundary condition: using the symmetrical component transformation,

I_{a 0} = \frac{1}{3} (I_{a} + I_{b} + I_{c}) = \frac{1}{3} I_{a} (since I_{b} = I_{c} = 0)

I_{a 1} = \frac{1}{3} (I_{a} + a I_{b} + a^{2} I_{c}) = \frac{1}{3} I_{a}, I_{a 2} = \frac{1}{3} (I_{a} + a^{2} I_{b} + a I_{c}) = \frac{1}{3} I_{a}

This shows Ia0 = Ia1 = Ia2 = Ia/3 — all three sequence currents are equal, a condition that corresponds physically to connecting the three sequence networks in series with each other.

Step 2 — apply the voltage boundary condition: Va = Ia·Zf, and Va = Va0+Va1+Va2 (from the inverse symmetrical component transform). The sequence network equations for the generator (with the source only in the positive-sequence network) are: Va1 = E - Ia1·Z1, Va2 = -Ia2·Z2, Va0 = -Ia0·(Z0+3Zf') (where any additional neutral grounding impedance Zn contributes 3Zn here; if the fault-to-ground impedance Zf is the only impedance in the ground path, it is conventionally included as 3Zf in the zero-sequence branch to account for the threefold ground-return current).

Since Ia = 3Ia1 (as Ia0=Ia1=Ia2), the boundary condition Va = Ia·Zf becomes Va0+Va1+Va2 = 3Ia1·Zf:

(E - I_{a 1} Z_{1}) + (- I_{a 1} Z_{2}) + (- I_{a 1} Z_{0}) = 3 I_{a 1} Z_{f}

(using Ia1=Ia2=Ia0 throughout). Rearranging:

E = I_{a 1} (Z_{1} + Z_{2} + Z_{0} + 3 Z_{f}) ⟹ I_{a 1} = \frac{E}{Z _{1} + Z _{2} + Z _{0} + 3 Z _{f}}

Step 3 — total fault current: since Ia = 3Ia1:

I_{f} = I_{a} = \frac{3 E}{Z _{1} + Z _{2} + Z _{0} + 3 Z _{f}}

This is the standard expression for single line-to-ground fault current, showing that it depends on the series sum of all three sequence impedances (plus 3 times the fault impedance) — physically represented in the sequence-network diagram by connecting the positive-, negative- and zero-sequence networks of the system in series with each other and with the fault impedance, driven by the single positive-sequence source EMF E, which is the standard, most convenient network interconnection method for analyzing this particular (and the most commonly occurring) fault type in a power system.

Q7Power System - I

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