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POWER SYSTEM ANALYSIS
Unsymmetrical Fault Analysis
Assoc. Prof. Dr. Saffet AYASUNDepartment of Electrical and Electronics EngineeringNide University**Sayasun
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*Fault AnalysisWe consider faults at the general three-phase bus shownTerminals abc, denoted the fault terminals, are brought out in order to make external connections that represent faults. Before a fault occurs, the currents Ia; Ib, and Ic are zero.
ProcedureSet up all three sequence networksInterconnect the networks at the point of the fault to simulate the short circuitCalculate the 012 (sequence components) currents and voltages Transform to ABC currents and voltages
Thevenin Equivalent CircuitThe sequence components of the fault currents, I0; I1, and I2, are zero before a fault occurs.Each sequence network has a Thevenin equivalent impedance. Also, the positive-sequence network has a Thevenin equivalent voltage source, which equals the prefault voltage VF.
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Example 1A single-line diagram of the power system is shown above, where negative- and zero-sequence reactances are also given. The neutrals of the generator and DY transformers are solidly grounded. The motor neutral is grounded through a reactance Xn = 0.05 per unit on the motor base. Draw the per-unit zero-, positive-, and negative sequence networks on a 100-MVA, 13.8-kV base in the zone of the generator. Reduce the sequence networks to their Thevenin equivalents, as viewed from bus 2. Prefault voltage is VF = 1.050 per unit. Prefault load current and DY transformer phase shift are neglected.
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Example 1: Solution*Sayasun*The negative sequence Thevenin impedance is (j0.21)//(j0.17+0.10+0:105+0.1)= (j0.21)//(j0.475) = j0.14562 per unit.
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Example 1: Solution*Sayasun*Thevenin impedance at bus 2 consists only of (j0.10 + 0.15)= j0.25 per unit, as seen to the right of bus 2; due to the D connection of transformer T2, the zero-sequence network looking to the left of bus 2 is open.
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Example 1: Solution*Sayasun*The positive-sequence Thevenin impedance at bus 2 is the motor impedance j0.20, as seen to the right of bus 2, in parallel with j(0.15 + 0.1+ 0.105 + 0.10)= j0.455, as seen to the left; the parallel combination is (j0.20)//(j0.455)= j0.13893 per unit.
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Example 1: Thevenin equivalent circuits*Sayasun*
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Example 2: Three-phase short-circuit calculations using sequence networksCalculate the per-unit subtransient fault currents in phases a; b, and c for a bolted three-phase-to-ground short circuit at bus 2 in Example 1
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Example 2: Three-phase short-circuit calculations using sequence networksThe sequence components of the line-to-ground voltages at the fault terminals are,
During a bolted three-phase fault, the sequence fault currents are I0=I2 = 0 and I1 = VF/Z1; therefore, the sequence fault voltages are V0 = V1 = V2 = 0, which must be true since Vag =Vbg =Vcg= 0. However, fault voltages need not be zero during unsymmetrical faults, which we consider next.
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*Unbalanced Fault Analysis
unbalanced faultssingle-line to ground 60-75%double-line to ground 15-25%line-to-line faults 5-15%
Single-Line to Ground Fault*Sayasun*
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We now transform the fault conditions given in phase domain to the sequence domain
*Sayasun*Single-Line to Ground Fault
The fault conditions can be satisfied by interconnecting the sequencenetworks in series at the fault terminals through the impedance 3ZF
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Single-Line to Ground Fault: Sequence Networks Connection *Sayasun*
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Example 3*Sayasun*Calculate the subtransient fault current in per-unit and in kA for a boltedsingle line-to-ground short circuit from phase a to ground at bus 2 in Example 1. Also calculate the per-unit line-to-ground voltages at faulted bus 2.
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Example 3: Solution*Sayasun*the sequence currents:the subtransient fault current:
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Example 3: SolutionThe sequence components of the voltages at the fault are:
Transforming to the phase domain, the line-to-ground voltages at faulted bus 2 are
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Example 4*Sayasun*
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Example 4: Solution*Sayasun*
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Example 4: Solution*Sayasun*
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Example 4: Solution*Sayasun*
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Example 4: Solution*Sayasun*
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Line-to-Line Faults*Sayasun*
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Line-to-Line Faults*Sayasun*We now transform the fault conditions given in phase domain to the sequence domain
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Line-to-Line Faults: Sequence Networks Connection *Sayasun*The fault conditions in sequence domain are satisfied by connecting the positive- and negative-sequence networks in parallel at the fault terminals through the fault impedance ZF.
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Line-to-Line Faults*Sayasun*
Fault currents:Phase current:
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Example 5*Sayasun*Calculate the subtransient fault current in per-unit and in kA for a boltedline-to-line fault from phase b to c at bus 2 in 1
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Example 5: Solution*Sayasun*
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Example 6: A line-to-line fault at bus 3*Sayasun*
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Example 6: Solution*Sayasun*
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Example 6: Solution*Sayasun*
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Example 6: Solution*Sayasun*
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Example 6: Solution*Sayasun*
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Double Line-to-Ground Faults*Sayasun*
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Double Line-to-Ground Faults*Sayasun*We now transform the fault conditions given in phase domain to the sequence domain
The positive and negative sequence networks will be connected in parallel.
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Double Line-to-Ground Faults*Sayasun*
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Double Line-to-Ground Faults: Sequence Networks Connection *Sayasun*
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Example 7 *Sayasun*For a bolted double line-to-ground fault from phase b to c to ground at bus 2 in Example 1 Calculate: the subtransient fault current in each phase,neutral fault current,contributions to the fault current from the motor and from the transmission line, (Neglect the DY transformer phase shifts.)
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Example 7: Solution*Sayasun*
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Example 7: Solution*Sayasun*
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Example 7: Solution*Sayasun*
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Example 7: Solution*Sayasun*
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Example 8: Double Line-to-Ground Fault at Bus 3*Sayasun*
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Example 8: Solution*Sayasun*
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