Fault Calculations
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Transcript of Fault Calculations
Fault CalculationsFault Calculations
A review of:
Nature of Short Circuit Currents
Fault Types
Per Unit Quantities
Symmetrical Components
Calculation Examples
Fault Calculations
Fault Calculations
Vmaxi(t) =Z
[ sin(ωt + α – θ) – sin (α - θ)e – R t / L]
Vm [ sin(ωt + α ) L didt
] = + Ri
Vmax [sin(ωt + α )]
L R
Faulti(t)
sin (α - θ)e – R t / L = Transient current
sin (ωt + α – θ) = Steady state current
Fault Calculations
Short Circuit Currents
Assume L is constant and R = 0
Time
e(t)
Time of faultoccurance
α
α - θ = 9 0 º th e fau lt occu rs a t V o ltag e ze ro : [s in (ω t + α – θ ) – s in (α - θ )]= s in (ω t + 9 0 º) – s in (9 0 º) = co s (ω t) -1
α = θ th e fau lt o ccu rs a t V o ltag e m ax : [s in (ω t + α – θ ) – s in (α - θ )] = s in (ω t)
Vmax I(t) = [ sin (ωt + α – θ) – sin (α - θ)] Z
Vmax I(t) = [ sin (ωt + α – θ) – sin (α - θ)e – R t / L] Z
Fault Inception Angle
Fault Calculations
Fault Calculations
e(t)i(t)
Time of faultoccurance
Time
i
L is constant an R = 0
Fault at Voltage Maximum
Fault at Voltage Zero
e(t)
i(s)= steadystate current
Time of faultoccurance
Time
L constant and R = 0
i(t)= transientcurrent
i(f) = i(s) + i(t)
i
Fault Calculations
Fault at Voltage Zero R ≠0
e(t)
i(s)= steadystate current
Time of faultoccurance
Time
i(t)= transientcurrent
i(f) = i(s) + i(t)
α - θ = 90º
i
L is constant and R 0≠
Fault Calculations
Transient & Subtransient Reactance
Vmax [sin(ωt + α )]
jXd"jXd'jXd R
Faulti(t)
Fault Calculations
Asymmetrical Fault Current
Time
i''max
R = 0 and L is not constant
i'max
imax
Total Asymmetrical Current
DC Component + AC Component
Fault Calculations
Variation of Current with Time During a Fault
Fault Calculations
Variation of Generator ReactanceDuring a Fault
Fault Calculations
Transient and Subtransient Reactances
Instantaneous units are set with short circuit currents calculated with subtransient reactances, that result in higher values of current.
Time delay units can be set using the same values or the transient reactance, depending on the operating speed of the protection relays.
Transient reactance values are generally used in stability studies.
Fault Calculations
X
X
ZΦ
ZΦ
ZΦ
GBΦCΦ
AΦ
Short Circuit CalculationFault Types – Single Phase to Ground
Fault Calculations
X
X
ZΦ
ZΦ
ZΦ
GBΦCΦ
AΦ
Short Circuit CalculationsFault Types – Line to Line
Fault Calculations
Short Circuit CalculationsFault Types – Three Phase
ZΦ
ZΦ
ZΦ
GBΦCΦ
AΦX
X
X
Fault Calculations
Short Circuit CalculationsExample 1– System Impedance
59.5 Ω1.56 Ω
11.2 Ω
Transformer
115kV√3
25kV√3
Fault Calculations
+
V1115kV
√3+
V2 Z2Z1
V1A1 = V2A2 V1A1 =V1
2
Z1
V2A2 =V2
2
Z2
Z1 = Z2 x V12
V22
Short Circuit CalculationsExample 1– Equivalent Impedance
Fault Calculations
Short Circuit CalculationsExample 1– Equivalent Impedance at 25 kV
√3Z25 = Z115 x[25/115/√3
]2
0.53
2.80
11.2
59.5
25kV115 kVSystem —
Xfrm —
Fault Calculations
Short Circuit CalculationsExample 1– Fault Calculation at 25 kV
2.80 Ω 1.56 Ω0.53 Ω25kV√3
25kV
√3 x ΣZ
I fault
25kV√3 x 4.89ΩIF = = = 2952A
Fault Calculations
Convert all system parameters to a common base
All components at all voltage levels are combined
Transformers become “transparent” to calculations
Operating system current and voltage values can bederived as the last calculation
Fault Calculations
Short Circuit CalculationsPer Unit System
Establish two base quantities:
Standard practice is to define- Base power – 3 phase- Base voltage – line to line
Other quantities derived with basic power equations
MVA3Φ
kVL-L
Fault Calculations
Short Circuit CalculationsPer Unit System
√3 x kV L-L·baseI base =
x1000MVAbase
Z base =kV2
L-L·base
MVAbase
Fault Calculations
Short Circuit CalculationsPer Unit System
Per Unit Value = Actual QuantityBase Quantity
Vpu = VactualVbase
Ipu = IactualIbase
Zpu = ZactualZbase
Fault Calculations
Short Circuit CalculationsPer Unit System
Short Circuit CalculationsPer Unit System – Base Conversion
Zpu = ZactualZbase
Zbase = kV 2baseMVAbase
Z1base = MVA1basekV 21base X Zactual
Z2base = MVA2basekV 22base
X Zactual
Ratio • Z1baseZ2base
Z2base =Z1base x kV 21base x MVA2base
kV 22base MVA1base
Fault Calculations
Short Circuit CalculationsPer Unit System – Base Conversion
Z2pu = Z1pu x MVA2baseMVA1base
Use if equipment voltage ratings are the same as system base voltages.
Fault Calculations
Short Circuit CalculationsPer Unit System – 25 kV Base
Select MVAbase = 100
Ibase = 100 x 103
√3 x 25
Zbase = 252 = 6.25Ω100
= 2309A
Fault Calculations
Short Circuit CalculationsPer Unit System – Transformers
11.5 / 25kV Delta – Grounded Wye20 / 26.7 / 33.3 MVA Z = 9.0%
Impedance is 0.09 per unit on a 20 MVA base
Z100 pu = 0.09 x 100 = 0.45 pu20
Fault Calculations
Short Circuit CalculationsPer Unit System – Example 1
0.45 pu 0.25 pu0.08 pu25kV√3
1.0
ΣZ
I fault
1.0
0.78IF = = = 1.28 · pu · amperes
I25kV = 1.28pu x 2309A = 2960A
Fault Calculations
Short Circuit CalculationsSymmetrical Components
“Provides a practical technology for understanding and analyzing power system
operation during unbalance conditions”
Protective Relaying Principles and Applications
J. Lewis Blackburn
Fault Calculations
lb2 la2
lc2
NegativePositive
lc1
la1
lb1
lb0
la0
lc0
Zero
Short Circuit CalculationsSymmetrical ComponentsSequence Components
Fault Calculations
Short Circuit CalculationsSymmetrical Components
“a” operator
1∠ 0°
a=1∠ 120°
a2=1∠ 240°
Ib1=1a1∠ 240 = a2Ia1
Ic1=1a1∠ 120 = aIa1
Ia1=1∠ 120
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Network Equations
Ia = I1+I2+I0 I1 = 1 [Ia+aIb+a2Ic]
Ib = a2 I1+aI2+I0 I2 = 1 [Ia+a2Ib+aIc]
Ic = aI1+a2I2+I0 I0 = 1 [Ia+Ib+Ic]
3
3
3
Fault Calculations
Short Circuit CalculationsSymmetrical Component Vectors
I1=1/3 Ia
I2 = 1/3 [Ia+a2Ib+aIc]
Ia=1 ∠0
Ib=1∠ 240
Ic=1∠ 120
I1 = 1/3 [Ia+aIb+a2Ic] aIb= 1∠ 120° 1∠ 240 °x = 1 ∠ 360 °
1∠ 240° 1 ∠ 120°x = 1∠ 360 °a Ic= 2
1∠ 240° 1 ∠ 240°x = 1∠ 120 °a Ib= 2
aIc= 1∠ 120° 1∠ 120 °x = 1 ∠ 240 ° I 2 = 0
a2=1∠ 240°
∠ 120°a=1
Fault Calculations
Symmetrical ComponentsNetwork Equations
In three phase systems, the neutral current is equal to In = (Ia + Ib + Ic) and,therefore, In = 3Io.
Ia = I1+I2+I0
Ib = a2 I1+aI2+I0
Ic = aI1+a2I2+I0
Fault Calculations
Symmetrical ComponentsNetwork Equations
The same equations apply to the voltages:
Va0 = 1/3(Va + Vb + Vc)
Va1 = 1/3(Va + aVb + a2Vc)
Va2 = 1/3(Va + a2Vb + aVc)
Fault Calculations
Short Circuit CalculationsSymmetrical ComponentsNetwork Representations
I2
Z2
NegativeSequence
I0
Z0
Zero Sequence
1 pu I1
Z1
PositiveSequence
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Three Phase Fault
1 pu I1
Z1
Positive Sequence
X
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Phase to Phase Fault
1 pu I1
Z1
Positive Sequence
X
XI2
Z2
Negative Sequence
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Phase to Ground Fault
1 pu I1
Z1
Positive Sequence
X
XI2
Z2
Negative Sequence
XI0
Z0
Zero Sequence
Fault Calculations
Short Circuit CalculationsSymmetrical ComponentsOpen Conductor Condition
1 puX
X
X
Z2
Z1
Z0
I1
I2
I0
Positive Sequence
Negative Sequence
Zero Sequence
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Transformer Representations
Grounded Wye - Grounded Wye
LZ1 or Z2
H
N1 or N2
H LZ0
N0
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Transformer Representations
Z1 or Z2
Z0
N0
H L
N1 orN2
H LDelta - Grounded Wye
Fault Calculations
Z1 or Z2
H L
N1 or N2
R
Z0
N0
H L
3R
G
Delta-Grounded Wyewith Grounding Resistor
Short Circuit CalculationsSymmetrical Components
Transformer Representations
In = (Ia + Ib + Ic)In = 3I0
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Three Phase Fault
0.45 pu 0.25 pu0.08 pu1 pu I1
I2 = I0 = 0
Ia = I1 + I2 + I0 = I1= 1.28pu A
Ib = a2I1 Ic = aI1
I25kV = 1.28pu x 2309A = 2960A
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Phase to Ground Fault
1 pu
X
X
X
Zero Sequence
0.27 pu 0.45 pu 0.95 pu
Negative Sequence
Positive Sequence
0.45 pu0.08 pu 0.25 pu
0.45 pu0.08 pu 0.25 pu
I1 = I2 = I0
I1 = 1.0 = 1pu = 0.35puA
ΣZ 2.86pu
Ia = I1 + I2 + I0 = 1.05puA
I25kV = 1.05 x 2309 = 2424A
Fault Calculations
Short Circuit CalculationsSymmetrical Components
Phase to Ground Fault
1 pu
X
X
X
Zero Sequence
0.27 pu 0.45 pu
Negative Sequence
Positive Sequence
0.45 pu0.08 pu
0.45 pu0.08 pu
I1
I2
I0
Transformer LowSide Faults
I1 = I2 = I0
I1 = 1.0 = 1pu = 0.66puA
ΣZ 1.51pu
Ia = I1 + I2 + I0 = 1.99puA
I25kV = 1.99 x 2309 =4595A
Fault Calculations
ReferencesReferencesBlackburn, J. I., Protective Relaying Principles and Applications, Marcel Dekker, Inc., copyright 1987
Blackburn, J. I., Symmetrical Components for Power Systems Engineering, Marcel Dekker, Inc., copyright 1993
ABB Power T&D Co., Protective Relaying Theory and Application, Marcel Dekker, Inc., copyright 1994
Stevenson, W.D., Elements of Power System Analysis, McGraw-Hill Book Company, Inc., copyright 1962
IEEE Std 242-1986, IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems
Cooper Power Systems, Electrical Distribution System Protection, copyright 1990, Third Edition
Fault Calculations
©2008 Beckwith Electric Co., Inc.