Sections: Conditions | Network | Example | Problem Set

Conditions

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

V_A = 0 V_B = V_C I_A = 0 I_B = -I_C

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

In matrix form:

I_A0 = 0 I_A1 = -I_A2 I_A = 0 I_N = 0

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

In matrix form:

V_A1 = V_A2 = V_A0 = 0

In matrix form:

0 = V_f - I_A1Z₁ - I_A1Z₂

V_f = I_A1 [Z₁ + Z₂]

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Sections: Conditions | Network | Example | Problem Set

Network

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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 line-to-line voltages for subtransient conditions when a line-to-line 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 + Z₂ ] = 1 + j0 / [ j0.25 + j0.35 ] = -j1.667 p.u.

I_A1 = I_A2 = -j1.667 p.u. I_A1 = -I_A2

I_A = I_A1 + I_A2 + I_A0 = 0 p.u.

I_B = a²I_A1 + aI_A2 + I_A0 = a²(-I_A2) + aI_A2 + I_A0
I_B = (-a²+ a)I_A2 + I_A0 = (j1.732)I_A2 + I_A0
I_B = (j1.732)(-j1.667 p.u.) = -2.866 +j0 p.u.

I_B = -I_C = -(-2.866 +j0 p.u.) = 2.866 + j0 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 = -2.866 p.u. x 837 amperes = 2416 /180° Ampere.
I_C = 2.866 p.u. x 837 amperes = 2416 /0° Ampere.

Symmetrical components of voltage from point a to ground are:
V_A1 = V_A2 = V_f - I_A1Z₁ = 1.0 - (-j1.667)(j0.25) = 1 - 0.417 = 0.583 p.u.
V_A2 = 0.583 p.u.
V_A0 = 0 p.u.

line-to-line voltages:
V_A = 2V_A1 + V_A0 = 2(0.583) = 1.166/0° p.u.
V_B = V_C = a²(0.583) + a(0.583) = (a²+a)0.583 p.u. = (-1)(0.583) p.u. = -0.583 p.u.

Line-to-line voltages:
V_ab = V_A - V_B = 1.166 - (-0.583) = 1.749 /0° p.u.
V_bc = V_B - V_C = -0.583 - ( -0.583) = 0 p.u.
V_ca = V_C - V_A = -0.583 - 1.166 = -1.749 = 1.749/180° p.u.

The equivalent line-to-line voltages, would be:
V_ab = (1 /Ö3) 1.749 x 13.8 kV /0° = 13.94/0° kV.
V_bc = 0 kV.
V_ca = (1 /Ö3) (-1.749) x 13.8 kV = 13.94/180° kV.

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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 line-to-line 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 line-to-line 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 line-to-line 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 line-to-line fault occurs at its terminals. Find the subtransient current in the faulted phase.

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