Advanced Power Systems

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WEEK 12: UNLOADED GENERATOR: SINGLE LINE-TO-GROUND FAULT

Sections: Conditions | Network | Example | Problem Set

Unloaded Generator: Single Line-To-Ground Fault

Conditions at Fault. The following are the conditions for single line-to-ground fault for unloaded generator:

I_b = 0 I_c = 0 V_a = 0

Current. The following are the current relationships during a single line-to-ground fault for unloaded generator:

In matrix form:

I_a0 = I_a1 = I_a2 = I_a / 3

Voltage. The following are the voltage relationships during a single line-to-ground fault for unloaded generator:

In matrix form:

V_a0 + V_a1 + V_a2 = -I_a1Z₀ + E_a - I_a1Z₁ - I_a1Z₂

But V_a = V_a0 + V_a1 + V_a2 = 0, thus:

I_a1 = E_a / [Z₁ + Z₂ + Z₀]

Sections: Conditions | Network | Example | Problem Set

Unloaded Generator: Single Line-To-Ground Fault: Network

If the neutral of the generator is not grounded, the zero sequence network is open-circuited, thus, Z₀ = infinite. Also, I_a1 becomes zero, as well as I_a2 and I_a0, thus no current flows in line a.

Sections: Conditions | Network | Example | Problem Set

Unloaded Generator: Single Line-To-Ground Fault: 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. Determine the subtransient current in the generator and the line-to-line voltages for subtransient conditions when a simple line-to-ground fault occurs at the generator terminals with the generator operating unloaded at rated voltage. Neglect resistance.

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

I_a1 = E_a / [ Z₁ + Z₂ + Z₀ ] = 1 + j0 / [ j0.25 + j0.35 + j0.10 ] = -j1.43 p.u.

I_a = 3I_a1 = 3(-j1.43 p.u.) = -j4.29 p.u.

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

Subtransient Current = I_a = -j4.29 p.u. x 837 amperes = j3,590 Amperes.

Symmetrical components of voltage from point a to ground are:
V_a1 = E_a - I_a1Z₁ = 1.0 - (-j1.43)(j0.25) = 1 - 0.357 = 0.643 p.u.
V_a2 = - I_a2Z₂ = - (-j1.43)(j0.35) = - 0.50 p.u.
V_a0 = - I_a0Z₀ = - (-j1.43)(j0.10) = - 0.143 p.u.

Line-to-ground voltages:
V_a = V_a1 + V_a2 + V_a0 = 0.643 - 0.50 - 0.143 = 0 p.u.
V_b = a²V_a1 + aV_a2 + V_a0 = a²(0.643) + a (-0.50) + (-0.143) = -0.215 - j0.990 p.u.
V_c = aV_a1 + a²V_a2 + V_a0 = a(0.643) + a² (-0.50) + (-0.143) = - 0.215 + j0.990 p.u.

Where a² = (-0.5-j0.866) and a = (-0.5+j0.866)

Line-to-line voltages:
V_ab = V_a - V_b = 0 - (-0.215 + j0.990) = 0.215 + j0.990 = 1.01/77.7° p.u.
V_bc = V_b - V_c = (-0.215 -j0.990) - ( -0.215 +j0.990) = 0 + j01.980 = 1.98/270° p.u.
V_ca = V_c - V_a = (-0.215 +j0.990) - 0 = -0.215 +j0.990 = 1.01/102.3° p.u.

The equivalent line-to-line voltages, would be:
V_ab = (1 /Ö3) 1.01 x 13.8 kV /77.7° = 8.05/77.7° kV.
V_bc = (1 /Ö3) 1.98 x 13.8 kV /270° = 15.78/270° kV.
V_ca = (1 /Ö3) 1.01 x 13.8 kV /102.3° = 8.05/102.3° kV.

Before the fault, the line voltages were balanced and equal to 13.8kV.
The pre-fault voltages, with V_an = E_a as reference, would be:
V_ab = 13.8/30° kV.
V_bc = 13.8/270° kV.
V_ca = 13.8/150° kV.

Sections: Conditions | Network | Example | Problem Set

Unloaded Generator: Single Line-To-Ground Fault: 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 single 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 single 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 single line-to-ground fault to that for a three-phase fault?

4. A generator rated 100 MVA, 20 kV has X" = X₂ = 20% and X₀ = 5 %. Its neutral is grounded through a reactor of 0.32 W. The generator is operating at rated voltage without load and is disconnected from the system when a single line-to-gound fault occurs at its terminals. Find the subtransient current in the faulted phase.