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![Vmaxi(t) =
Z
[ sin(ωt + α – θ) – sin (α - θ)e – R t / L
]
Vm [ sin(ωt + α ) L
di
dt
] = + 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](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-3-320.jpg)
![Assume L is constant and R = 0
Time
e(t)
Time of fault
occurance
α
α - θ = 90º the fault occurs at V oltage zero:
[sin (ω t + α – θ) – sin (α - θ)]= sin (ω t + 90º) – sin (90º) = cos (ω t) -1
α = θ the fault occurs at V oltage m ax:
[sin (ω t + α – θ) – sin (α - θ)] = sin (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](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-4-320.jpg)
![Transient & Subtransient Reactance
Vmax [sin(ωt + α )]
jXd"
jXd'
jXd R
Faulti(t)
Fault Calculations](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-8-320.jpg)
![Short Circuit Calculations
Example 1– Equivalent Impedance at 25 kV
√3
Z25 = Z115 x
[25/
115/√3
]2
0.53
2.80
11.2
59.5
25kV115 kV
System —
Xfrm —
Fault Calculations](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-18-320.jpg)
![Short Circuit Calculations
Symmetrical Components
Network Equations
Ia = I1+I2+I0 I1 = 1 [Ia+aIb+a2
Ic]
Ib = a2
I1+aI2+I0 I2 = 1 [Ia+a2
Ib+aIc]
Ic
= aI1
+a2I2
+I0
I0
= 1 [Ia
+Ib
+Ic
]
3
3
3
Fault Calculations](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-32-320.jpg)
![Short Circuit Calculations
Symmetrical 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° I2 = 0
a2=1∠ 240°
∠ 120°a=1
Fault Calculations](https://image.slidesharecdn.com/fault-fundamentals-rev-080212-160117051103/85/Fault-Calculations-33-320.jpg)
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Fault Calculations
- 1.
- 2. A review of: Natureof Short Circuit Currents Fault Types Per Unit Quantities Symmetrical Components Calculation Examples Fault Calculations Fault Calculations
- 3. Vmaxi(t) = Z [ sin(ωt+ α – θ) – sin (α - θ)e – R t / L ] Vm [ sin(ωt + α ) L di dt ] = + 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
- 4. Assume L isconstant and R = 0 Time e(t) Time of fault occurance α α - θ = 90º the fault occurs at V oltage zero: [sin (ω t + α – θ) – sin (α - θ)]= sin (ω t + 90º) – sin (90º) = cos (ω t) -1 α = θ the fault occurs at V oltage m ax: [sin (ω t + α – θ) – sin (α - θ)] = sin (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
- 5. Fault Calculations e(t) i(t) Time offault occurance Time i L is constant an R = 0 Fault at Voltage Maximum
- 6. Fault at VoltageZero e(t) i(s)= steady state current Time of fault occurance Time L constant and R = 0 i(t)= transient current i(f) = i(s) + i(t) i Fault Calculations
- 7. Fault at VoltageZero R ≠0 e(t) i(s)= steady state current Time of fault occurance Time i(t)= transient current i(f) = i(s) + i(t) α - θ = 90º i L is constant and R 0≠ Fault Calculations
- 8. Transient & SubtransientReactance Vmax [sin(ωt + α )] jXd" jXd' jXd R Faulti(t) Fault Calculations
- 9. 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
- 10. Variation of Currentwith Time During a Fault Fault Calculations
- 11. Variation of GeneratorReactance During a Fault Fault Calculations
- 12. Transient and Subtransient Reactances Instantaneousunits 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
- 13. X X ZΦ ZΦ ZΦ G BΦCΦ AΦ Short Circuit Calculation FaultTypes – Single Phase to Ground Fault Calculations
- 14. X X ZΦ ZΦ ZΦ G BΦCΦ AΦ Short Circuit Calculations FaultTypes – Line to Line Fault Calculations
- 15. Short Circuit Calculations FaultTypes – Three Phase ZΦ ZΦ ZΦ G BΦCΦ AΦ X X X Fault Calculations
- 16. Short Circuit Calculations Example1– System Impedance 59.5 Ω 1.56 Ω 11.2 Ω Transformer 115kV √3 25kV √3 Fault Calculations
- 17. + V1 115kV √3 + V2 Z2Z1 V1A1 =V2A2 V1A1 = V1 2 Z1 V2A2 = V2 2 Z2 Z1 = Z2 x V1 2 V2 2 Short Circuit Calculations Example 1– Equivalent Impedance Fault Calculations
- 18. Short Circuit Calculations Example1– Equivalent Impedance at 25 kV √3 Z25 = Z115 x [25/ 115/√3 ]2 0.53 2.80 11.2 59.5 25kV115 kV System — Xfrm — Fault Calculations
- 19. Short Circuit Calculations Example1– 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
- 20. Convert all systemparameters to a common base All components at all voltage levels are combined Transformers become “transparent” to calculations Operating system current and voltage values can be derived as the last calculation Fault Calculations Short Circuit Calculations Per Unit System
- 21. Establish two basequantities: 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 Calculations Per Unit System
- 22. √3 x kVL-L·base I base = x1000MVAbase Z base = kV2 L-L·base MVAbase Fault Calculations Short Circuit Calculations Per Unit System
- 23. Per Unit Value= Actual Quantity Base Quantity Vpu = Vactual Vbase Ipu = Iactual Ibase Zpu = Zactual Zbase Fault Calculations Short Circuit Calculations Per Unit System
- 24. Short Circuit Calculations PerUnit System – Base Conversion Zpu = Zactual Zbase Zbase = kV 2 base MVAbase Z1base = MVA1base kV 2 1base X Zactual Z2base = MVA2base kV 2 2base X Zactual Ratio • Z1base Z2base Z2base =Z1base x kV 2 1base x MVA2base kV 2 2base MVA1base Fault Calculations
- 25. Short Circuit Calculations PerUnit System – Base Conversion Z2pu = Z1pu x MVA2base MVA1base Use if equipment voltage ratings are the same as system base voltages. Fault Calculations
- 26. Short Circuit Calculations PerUnit System – 25 kV Base Select MVAbase = 100 Ibase = 100 x 103 √3 x 25 Zbase = 252 = 6.25Ω 100 = 2309A Fault Calculations
- 27. Short Circuit Calculations PerUnit System – Transformers 11.5 / 25kV Delta – Grounded Wye 20 / 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 pu 20 Fault Calculations
- 28. Short Circuit Calculations PerUnit 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
- 29. Short Circuit Calculations SymmetricalComponents “Provides a practical technology for understanding and analyzing power system operation during unbalance conditions” Protective Relaying Principles and Applications J. Lewis Blackburn Fault Calculations
- 30. lb2 la2 lc2 NegativePositive lc1 la1 lb1 lb0 la0 lc0 Zero Short CircuitCalculations Symmetrical Components Sequence Components Fault Calculations
- 31. Short Circuit Calculations SymmetricalComponents “a” operator 1∠ 0° a=1∠ 120° a2=1∠ 240° Ib1=1a1∠ 240 = a2Ia1 Ic1=1a1∠ 120 = aIa1 Ia1=1∠ 120 Fault Calculations
- 32. Short Circuit Calculations SymmetricalComponents Network Equations Ia = I1+I2+I0 I1 = 1 [Ia+aIb+a2 Ic] Ib = a2 I1+aI2+I0 I2 = 1 [Ia+a2 Ib+aIc] Ic = aI1 +a2I2 +I0 I0 = 1 [Ia +Ib +Ic ] 3 3 3 Fault Calculations
- 33. Short Circuit Calculations SymmetricalComponent 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° I2 = 0 a2=1∠ 240° ∠ 120°a=1 Fault Calculations
- 34. Symmetrical Components Network Equations Inthree 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
- 35. Symmetrical Components Network Equations Thesame 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
- 36. Short Circuit Calculations SymmetricalComponents Network Representations I2 Z2 Negative Sequence I0 Z0 Zero Sequence 1 pu I1 Z1 Positive Sequence Fault Calculations
- 37. Short Circuit Calculations SymmetricalComponents Three Phase Fault 1 pu I1 Z1 Positive Sequence X Fault Calculations
- 38. Short Circuit Calculations SymmetricalComponents Phase to Phase Fault 1 pu I1 Z1 PositiveSequence X X I2 Z2 NegativeSequence Fault Calculations
- 39. Short Circuit Calculations SymmetricalComponents Phase to Ground Fault 1 pu I1 Z1 PositiveSequence X X I2 Z2 NegativeSequence X I0 Z0 ZeroSequence Fault Calculations
- 40. Short Circuit Calculations SymmetricalComponents Open Conductor Condition 1 pu X X X Z2 Z1 Z0 I1 I2 I0 PositiveSequence NegativeSequence ZeroSequence Fault Calculations
- 41. Short Circuit Calculations SymmetricalComponents Transformer Representations Grounded Wye - Grounded Wye L Z1 or Z2 H N1 or N2 H L Z0 N0 Fault Calculations
- 42. Short Circuit Calculations SymmetricalComponents Transformer Representations Z1 or Z2 Z0 N0 H L N1 or N2 H L Delta - Grounded Wye Fault Calculations
- 43. Z1 or Z2 H L N1or N2 R Z0 N0 H L 3R G Delta-Grounded Wye with Grounding Resistor Short Circuit Calculations Symmetrical Components Transformer Representations In = (Ia + Ib + Ic) In = 3I0 Fault Calculations
- 44. Short Circuit Calculations SymmetricalComponents Three Phase Fault 0.45 pu 0.25 pu0.08 pu 1 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
- 45. Short Circuit Calculations SymmetricalComponents 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 2.86 pu Ia = I1 + I2 + I0 = 1.05puA I25k V = 1.05 x 2309 = 2424A Fault Calculations
- 46. Short Circuit Calculations SymmetricalComponents 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 Low Side 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
- 47. ReferencesReferences Blackburn, 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.