2. A review of:
Nature of 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 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
6. Fault at Voltage Zero
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 Voltage Zero 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
12. 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
18. 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
19. Short Circuit Calculations
Example 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
20. 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 be
derived as the last calculation
Fault Calculations
Short Circuit Calculations
Per Unit System
21. 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 Calculations
Per Unit System
22. √3 x kV L-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
Per Unit 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
Per Unit 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
Per Unit 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
Per Unit 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
Per 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
29. Short Circuit Calculations
Symmetrical 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
34. Symmetrical Components
Network 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
35. Symmetrical Components
Network 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
36. Short Circuit Calculations
Symmetrical Components
Network Representations
I2
Z2
Negative
Sequence
I0
Z0
Zero Sequence
1 pu I1
Z1
Positive
Sequence
Fault Calculations
39. Short Circuit Calculations
Symmetrical Components
Phase to Ground Fault
1 pu I1
Z1
PositiveSequence
X
X
I2
Z2
NegativeSequence
X
I0
Z0
ZeroSequence
Fault Calculations
40. Short Circuit Calculations
Symmetrical Components
Open Conductor Condition
1 pu
X
X
X
Z2
Z1
Z0
I1
I2
I0
PositiveSequence
NegativeSequence
ZeroSequence
Fault Calculations
41. Short Circuit Calculations
Symmetrical Components
Transformer Representations
Grounded Wye - Grounded Wye
L
Z1 or Z2
H
N1 or N2
H L
Z0
N0
Fault Calculations
43. Z1
or Z2
H L
N1 or 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
Symmetrical Components
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
Symmetrical 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 2.86 pu
Ia
= I1
+ I2
+ I0
= 1.05puA
I25k V = 1.05 x 2309 = 2424A
Fault Calculations
46. Short Circuit Calculations
Symmetrical 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 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