Advanced Power Systems

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WEEK 19: POWER SYSTEMS: DOUBLE LINE-TO-GROUND FAULTS

Sections: Conditions | Network | Example | Problem Set

Conditions

Conditions at Fault. The following are the conditions for double line-to-ground fault for power system:

V_B = 0 V_C = 0 I_A = 0

Current. The following are the current relationships during a double line-to-ground fault for power system:

I_A = 0 = I_A1 + I_A2 + I_A0I_N = 3I_A0 I_N = I_B + I_C

Voltage. The following are the voltage relationships during a double line-to-ground fault for power system:

In matrix form:

V_A / 3 = V_A1 = V_A2 = V_A0 = V_f - I_A1Z₁

Multiplying the above matrix with the following matrix:

will give:

Multiplying both sides by the matrix [ 1 1 1 ] and
knowing that I_A = 0 = I_A1 + I_A2 + I_A0, thus

Sections: Conditions | Network | Example | Problem Set

Network

Sections: Conditions | Network | Example | Problem Set

Example

A salient pole generator is rated 20MVA, 13.8 kV and has a direct-axis subtransient reactance of 0.25 unit. The negative- and zero-sequnce reactances are 0.35 and 0.10 per unit. The neutral of the generator is grounded. Find the subtransient current in the generator and the double line-to-ground voltages for subtransient conditions when a double line-to-ground fault occurs at the generator terminals with the generator operating unloaded at rated voltage. Neglect resistance.

Solution:
Base: 20MVA, 13.8kV, V_f =1.0 p.u.
Internal voltage = terminal voltage at no load.

I_A1 = V_f / [ Z₁ + (Z₂Z₀)/(Z₂+Z₀)]
I_A1 = 1.0 + j0.25 / [ (j0.35)(j0.10)/(j0.35+j0.10) ] = -j3.05 p.u.

V_A1 = V_A2 = V_A0 = V_f - I_A1Z₁
V_A1 = V_A2 = V_A0 = 1.0 - (j3.05)(j0.25) = 1.0 -0.763 = 0.237 p.u.

I_A2 = -V_A2 / Z₂ = -0.237/j0.35 = j0.68 p.u.

I_A0 = -V_A0 / Z₀ = -0.237/j0.10 = j2.37 p.u.

I_A = I_A1 + I_A2 + I_A0 = -j3.05 +j0.68 +j2.37 = 0 p.u.

I_B = a²(-j3.05) + a(j0.68) + j2.37 = -3.23 +j3.555 = 4.80/132.3° p.u.

I_C = a(-j3.05) + a²(j0.68) + j2.37 = +3.23 +j3.555 = 4.80/47.7° p.u.

I_N = I_B +I_C = (4.80/132.3°) + (4.80/47.7°) = (-3.23 +j3.555) + (3.23 +j3.555) = j7.11 p.u.

I_N = 3I_A0 = 3(j2.37) = j7.11 p.u.

Base Current = MVA / [Ö3) x kV ] = 20,000 / [Ö3) x 13.8 ] = 837 amperes.

Subtransient Current
I_A = 0 p.u. x 837 amperes = 0 Ampere.
I_B = 4.80/132.3° x 837 amperes = 4017/132.3° Ampere.
I_C = 4.80/47.7°. x 837 amperes = 4017/47.7° Ampere.

Double Line-To-Ground voltages:
V_A = V_A1 + V_A2 + V_A0 = 3V_A1 = = 3(0.237) = 0.711/0° p.u.
V_B = V_C = 0 p.u.

Double Line-To-Ground voltages:
V_ab = V_A - V_B = 0.711 - 0 = 0.711/0° p.u.
V_bc = V_B - V_C = 0 p.u.
V_ca = V_C - V_A = 0 - 0.711 = -0.711 = 0.711/180° p.u.

The equivalent Double Line-To-Ground voltages, would be:
V_ab = (1 /Ö3) 0.711 x 13.8 kV /0° = 5.66/0° kV.
V_bc = 0 kV.
V_ca = (1 /Ö3) (-0.711) x 13.8 kV = 5.66/180° kV.

Sections: Conditions | Network | Example | Problem Set

Problems

Solve the following problems
1. A 60-hz turbogenerator is rated 500MVA, 22kV. It is Y-connected and solidly grounded and is operating at rated voltage at no load. It is disconnected from the rest of the system. Its reactances are X" = X₂ = 0.15 and X₀ = 0.05 per unit. Find the ratio of the subtransient line current for a Double Line-To-Ground fault to the subtransient line current for a symmetrical three-phase fault.

2. Determine the ohms of inductive reactance to be inserted in the neutral connection of the Generator in Problem no. 1 to limit the subtransient line current for a Double Line-To-Ground fault to that for a three-phase fault.

3. How many ohms of resistance in the neutral connection of the generator in Problem No. 1 would limit the subtransient line current for a Double Line-To-Ground fault to that for a three-phase fault?

4. A 100-MVA 18-kV turbogenerator having X" = X₂ = 20% and X₀ = 5 % is about to be connected to a power system. The generator has a current-limiting reactor of 0.162W in the neutral. Before the generator is connected to the system, its voltage is adjusted to 16kV when a double line-to-ground fault develps at terminal b and c. Find the initial symmetrical rms current in the ground and in line b.