Simultaneous Fault Examples

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Simultaneous Fault Examples

pu := 1 MVA:= 1000kW a:= 1ej 120 deg A012 1 1 1 1 a2 a 1 a a2

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:= 13.8kV 100MVA ZS1 = j0.3 pu V = 1.0pu BUS 1 BUS 2 ZL1 = 0.8 pu 85o ZR1 = j0.3 pu RelayR1 24kV 100MVA RelayR2 V = 1.0pu 13.8kV: 230kV 100MVA X = 10% X/R = 15 -Yg BUS S 230:24kV 100MVA X = 10% X/R = 15 Yg- BUS R on 100MVA and 230kV ZR2 = ZR1 ZR0 = j1.0 pu ZS2 = ZS1 ZS0 = j1.0 pu ZL2 = ZL1 ZL0 = 3ZL1 Z2 = Z0 = Z1 Z2 = Z0 = Z1 VB1 := 13.8kV VB2 VB1 230 kV 13.8kV

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 := VB2 = 230 kV VB3 VB2 13.8 kV 230kV

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 := VB3 = 13.8 kV

Local source (no change of base needed):

· · Remote source (no change of base needed):

ZS1 := j 0.3 pu ZR1 := j 0.3 pu

ZS2 := ZS1 ZR2 := ZR1

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Transmission line: ·

ZL1 := 0.8pu e j 85 deg ZL1 = (0.0697+0.797i) pu ZL2:= ZL1

ZL0 := 3 Z L1 ZL0 = (0.2092+2.3909i) pu

Transformer between BusS and Bus1 (no change of base needed) · θT1 atan 15 1

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:= XT11 := 0.1pu mag_ZT11 X_T11 sin

(

θT1

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:= ZT11 mag_ZT11 e jθT1  := ZT11 = (0.0067+0.1i) pu

- Negative and zero sequence impedances equal to positive sequence impedance: ZT12:= ZT11 ZT10:= ZT11

Transformer between BusR and Bus2 (no change of base needed) · θT2 atan 15 1

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:= XT21 := 0.1pu mag_ZT21 X_T21 sin

(

θT2

)

:= ZT21 mag_ZT21 e jθT2  := ZT21 = (0.0067+0.1i) pu

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Positive sequence per unit, per phase, equivalent circuit: BusS 1.0 Bus1 Bus2 ZL1 Src1 _Src2 BusR 1.0 ZR1 ZT21 ZS1 ZT11 +30 deg +30 deg N1

Negative sequence per unit, per phase, equivalent circuit:

BusS Bus1 Bus2

ZL2 BusR ZR2 ZT22 ZS2 ZT12 -30 deg -30 deg N2

Zero sequence per unit, per phase, equivalent circuit:

BusS Bus1 BusR

ZR0 ZT20

Bus2

ZS0

ZT10 ZL0

Define Zbus matrices, but the fault bus needs to be added as a function of M for now. ·

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YBus1(M) 1 Z_S1 1 Z_T11 + 1 - e j 30 deg Z_T11 0 0 0 1 - e-j30deg Z_T11 1 Z_T11 1 M Z _L1 + 0 0 1 -M Z L1 0 0 1 1-M ( ) Z _L1 1 Z_T21 + 1 - e -j30deg ZT21 1 -1-M ( ) Z L1 0 0 1 - e j 30 deg Z_T21 1 ZT21 1 ZR1 + 0 0 1 -M Z _L1 1 -1-M ( ) Z _L1 0 1 M Z L1 1 1-M ( ) Z L1 +

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:= YBus2(M) 1 Z_S2 1 Z_T12 + 1 - e -j30deg Z_T12 0 0 0 1 - ej 30 deg Z_T12 1 Z_T12 1 M Z _L2 + 0 0 1 -M Z L2 0 0 1 1-M ( ) Z _L2 1 Z_T22 + 1 - e j 30 deg ZT22 1 -1-M ( ) Z L2 0 0 1 - e -j30deg Z_T22 1 ZT22 1 ZR2 + 0 0 1 -M Z _L2 1 -1-M ( ) Z _L2 0 1 M Z L2 1 1-M ( ) Z L2 +

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:=

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YBus0(M) 1 Z_S0 0 0 0 0 0 1 Z_T10 1 M Z _L0 + 0 0 1 -M Z L0 0 0 1 1-M ( ) Z _L0 1 Z_T20 + 0 1 -1-M ( ) Z L0 0 0 0 1 Z_R0 0 0 1 -M Z _L0 1 -1-M ( ) Z _L0 0 1 M Z L0 1 1-M ( ) Z L0 +

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:= ZBus1(M) := YBus1(M)-1 ZBus2(M) := YBus2(M)-1 ZBus0(M) := YBus0(M)-1

Simultaneous AG and BC faults 30% of the way from Bus 1 to Bus 2

I1 +I2 = 2 I 0 V1 = V2 1 -2 V0 = Vf := 1.0pu No load M:= 0.3

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I1 V_f ZBus1(M) 4 4, 1 Z_Bus2(M) 4 4, 1 ZBus0(M) 4 4,

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1₄ +

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1 -+ := I1 = 2.055 pu arg I

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1 = -86.871deg I2 -I1 Z_Bus0(M) 4 4,

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4 ZBus2(M) 4 4, Z_Bus0(M) 4 4,

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4 +

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 := I2 = 0.55 pu arg I

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2 = 91.613 deg I0 I₁ 2 Z_Bus2(M) 4 4, ZBus2(M) 4 4, Z_Bus0(M) 4 4,

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 := I0 = 0.752 pu arg I

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0 = -86.316deg I1 +I2

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-2 I 0 = 0 V1 Vf I1 ZBus1(M) 4 4,  -:= V1 = 0.211 pu arg V

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1 = -1.196deg V2 -I2 ZBus2(M) 4 4,  := V2 = 0.211 pu arg V

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2 = -1.196deg V0 -I0 ZBus0(M) 4 4,  := V0 = 0.423 pu arg V

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0 = 178.804 deg

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V₀ -2 -V1 = 0 VABC A012 V0 V1 V₂

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 := VABC   0 0.634 0.634

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= arg VABC 1

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= 178.804 deg arg VABC 2

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= 178.804 deg Very close to ATP IABC A012 I0 I1 I₂

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 := IABC   2.257 2.256 2.256

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= arg I

(

ABC

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  -86.316 177.191 -2.809

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deg  =

SLG or LL alone (without simultaneous):

SLG · I0slg V_f Z_Bus1(M) 4 4, +ZBus2(M)4 4, +ZBus0(M)4 4,

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:= I0slg = 0.752 arg I

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0slg

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= -86.316deg IA_slg := 3 I 0slg _I

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LL · I1LL V_f Z_Bus1(M) 4 4, +ZBus2(M)4 4, := IABCLL A012 10-15 I1LL I1LL

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 := IABCLL   0 2.256 2.256

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= arg I

(

ABCLL

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  0 177.191 -2.809

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deg  = IABCLL-IABC 0.145 - +2.253i 0 0

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= I_{A_slg} 0 0

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IABCLL + -IABC 0 0 0

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=

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VS BUSS V BUS1 R1 I BUSF V V SAT Y V BUS2 R2 I Y_SAT BUSR V VR IA IC I I B I I A I V

ATP System Model

·

Fault Currents (versus time): ·

(f ile Simult_AG_BC.pl4; x-v ar t) c:BUSFA -IA c:BUSFB -IB c:BUSFC -IC

0.00 0.02 0.04 0.06 0.08 [s] 0.10 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 [A]

Note that: IC = -IB ·

IA = 2.257A θA:= -86.24deg

IB = 2.257A θA:= -177.1deg

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Voltages at the fault point: ·

(f ile Simult_AG_BC.pl4; x-v ar t) v :BUSFA v :BUSFB v :BUSFC

0.00 0.02 0.04 0.06 0.08 [s] 0.10 -0.9 -0.6 -0.3 0.0 0.3 0.6 0.9 [V] VAG = 0V VBG = 0.6339V θVB := 178.9deg VCG = 0.6339V θVC := 178.9deg