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Fault Calculationsand
Selection ofProtective Equipment
Wednesday, March 22, 20068:00AM – 3:00PM
Seminole Electric Cooperative, Inc.16313 North Dale Mabry Hwy.
Tampa, Florida
Ralph Fehr, Ph.D., P.E.University of South Florida – Tampa
Senior Member, IEEEr.fehr@ieee.org
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 2
Symmetrical Components
Most power systems are designed as balanced systems.
Due to the symmetry of the problem, a single-phase equivalent approach can be taken to simplify the calculation process.
When the voltage and current behavior is calculated for one of the phases, the behaviors on the other two can be determined using principles of symmetry.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 3
But when the system phasors are not balanced, the single-phase equivalent approach cannot be taken.
This means that either
1. a three-phase solution must be found,
or
2. the unbalanced phasors must be resolved into balanced components so the single-phase equivalent method can be used.
☺
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 4
Charles Fortescue’s Theory of Symmetrical Components, first published in 1918, proves that any set of unbalanced voltage or current phasors belonging to a three-phase system can be resolved into three sets of components, each of which is balanced.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 5
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 6
AICI
BIω
Physical Example ofVector Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 7
F
d
M = F × d
Calculation of Moment
Moment = Force × Perpendicular Distance
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 8
Calculation of Moment
d
F
θ
M ≠ F × d
Moment = Force × Perpendicular Distance
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 9
Calculation of Moment
M = FV × d
Moment = Force × Perpendicular Distance
d
VF F
HFθ
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 10
Calculation of Moment
d
VF F
HFθ
FV = F cos θ
FH = F sin θ
FH + FV = F
⎟⎟⎠
⎞⎜⎜⎝
⎛= −
V
H1
FFtanθ
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 11
Application of Symmetrical Components to a Three-Phase
Power System
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 12
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 13
AICI
BIω
A-B-C Sequencing
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 14
I B
ω
I CAI
A-C-B Sequencing
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 15
II
ω
B
CI
A
Fortescue’s theory shows that three sets of balanced components are required to represent any unbalanced set of three-phase phasors.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 16
Positive Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 17
I A1
C1I
B1I
ω
Negative Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 18
A2IB2I
C2I
ω
Zero Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 19
A0IB0I
C0Iωω
ω
The constraint equations for the symmetrical components require the sum of the three components for each unbalanced phasor to equal the unbalanced phasor itself.
IA = IA0 + IA1 + IA2
IB = IB0 + IB1 + IB2
IC = IC0 + IC1 + IC2
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 20
The a operator
a ≡23
21 j+
− = 1 /120°
a2 = 1 /240° a3 = 1 /360°
j2 = 1 /180° = −1 j3 = 1 /270° = −j j4 = 1 /360° = 1
Recall the j operator
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 21
Using the a operator and the symmetry of the sequence components, we can develop a single-phase equivalent circuit to greatly simplify the analysis of the unbalanced system.
We will start by expressing the sequence components in terms of a single phase’s components. We will use Phase A as the phase for developing the single-phase equivalent.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 22
Positive Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 23
B1I
A1I
I C1
1= I
= a I1
= a I 12
ω
Negative Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 24
I A2 = I 2I B2 = a I2
2C2 = a II 2
ω
Zero Sequence Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 25
A0I = I 0= IB0I 0
= II C0 0ω
ω
ω
Recall the original constraint equations:
IA = IA0 + IA1 + IA2
IB = IB0 + IB1 + IB2
IC = IC0 + IC1 + IC2
Rewrite them using the a operatorto take advantage of the symmetry:
IA = I0 + I1 + I2
IB = I0 + a2 I1 + a I2
IC = I0 + a I1 + a2 I2
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 26
Unbalanced Phasors and their Symmetrical Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 27
I A
I 1
I 2
0I
CI
a I1
2a I 2
I 0
BI
0I
2a I1a I2
ω
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
2
1
0
2
2
C
B
A
III
aa1aa1111
III
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−−
2
1
0
2
2
1
2
2
C
B
A1
2
2
III
aa1aa1111
aa1aa1111
III
aa1aa1111
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
C
B
A1
2
2
2
1
0
III
aa1aa1111
III
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 28
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
C
B
A1
2
2
2
1
0
III
aa1aa1111
III
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡⋅
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
C
B
A
2
2
2
1
0
III
aa1aa1111
31
III
( )CBA0 III31I ++=
( )C2
BA1 IaIaI31I ++=
( )CB2
A2 IaIaI31I ++=
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 29
( )CBA0 III31I ++=
( )C2
BA1 IaIaI31I ++=
( )CB2
A2 IaIaI31I ++=
Summary of Symmetrical ComponentsTransformation Equations
IA = I0 + I1 + I2
IB = I0 + a2 I1 + a I2
IC = I0 + a I1 + a2 I2
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 30
Workshop #1Symmetrical Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 31
Ia = 0.95 /328°
Ib = 1.03 /236°
Ic = 0.98 /92°
Find I0, I1, and I2
I0 = 0.7 /300°
I1 = 1.2 /10°
I2 = 0.3 /167°
Find Ia, Ib, and Ic
Workshop #1Symmetrical Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 32
Ia = 0.95 /328°
Ib = 1.03 /236°
Ic = 0.98 /92°
Find I0, I1, and I2
I0 = 0.1418 /297°
I1 = 0.9634 /339°
I2 = 0.1622 /191°
Ia
cI
bI
I1I2I0
a I1
a I22 I0
1a I2
2a II0
Workshop #1Symmetrical Components
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 33
I0 = 0.7 /300°
I1 = 1.2 /10°
I2 = 0.3 /167°
Find Ia, Ib, and Ic
I1
2
0
II
Ia
Ic
a I1
2a I2
I0
I0
1a I2
2a I
Ia = 1.2827 /345°
Ib = 2.0209 /271°
Ic = 0.5749 /112°
Electrical Characteristics of the Sequence Currents
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 34
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 35
O
y DA
x
I
L
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 36
I
t
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 37
Bottom WireTop Wire
Middle Wire
t = T
I
t
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 38
O
I0 DA
LI0
0I
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 39
3 I
0I
0
0I
I0
O
DA
L
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 40
O3 I
0I D0 A
LI0
0I
VN = (3 I0) × ZN
VN = I0 × (3 Z(3 ZNN))
The Delta-Wye Transformer
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 41
Ic
Ia
Ib
IA
IC
IB
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 42
Assume Ia = 1 /0o, Ib = 1 /240o, and Ic = 1/120o.
Ic
IaIb
Ic
Ia
Ib
IB = Ib – Ic = 1 /240o – 1 /120o = /270o3
IC = Ic – Ia = 1 /120o – 1 /0o = /150o3
The Delta-Wye Transformer
IA
IB
IC
Ia
Ib
IcIA = Ia – Ib = 1 /0o – 1 /240o = /30o3
Workshop #2Non-Standard Delta-Wye Transformer
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 43
Given that Ia = 1/0o, Ib = 1/240o, and Ic = 1/120o, find IA, IB, and IC.
Ib
aI
cI
BI
AI
CI
IC aI
IA
IB
cI
bI
Ib
cII acIbII a
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 44
Workshop #2Non-Standard Delta-Wye Transformer
Ia = 1/0o
Ib = 1/240o
Ic = 1/120o
IA = Ia – Ic = 1/0o – 1/120o
= /330o3
IB = Ib – Ia = 1/240o – 1/0o
= /210o3
IC = Ic – Ib = 1/120o – 1/240o
= /90o3Ia
Ib
Ic
IA
IC
IB
Sequence Networks
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 45
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 46
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
One-Line Diagram
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 47
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Utility M1 M2 M3G
Positive-Sequence Reference Bus
Positive-Sequence Reactance Diagram
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 48
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Positive-Sequence Reactance Diagram
M3M1
1
T1 M2 T2
Utility G
Positive-Sequence Reference Bus
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 49
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Positive-Sequence Reactance Diagram
2
T3 M3
Positive-Sequence Reference Bus
1
T1 M2M1 T2
Utility G
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 50
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Negative-Sequence Reactance Diagram
2
T3 M3
Positive-Sequence Reference Bus
1
T1 M2M1 T2
Utility G
Negative
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 51
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Negative-Sequence Reactance Diagram
2
T3 M3
1
T1 M2M1 T2
Utility G
Negative-Sequence Reference Bus
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 52
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Zero-Sequence Reactance Diagram
2
T3 M3
1
T1 M2M1 T2
Utility G
Negative-Sequence Reference BusZero
+3Xn
Adjust Topology
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 53
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Zero-Sequence Reactance Diagram
M3
2
1
T3
Zero-Sequence Reference Bus
M1T1
Utility
M2 T2
GXn+ 3 Xn
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 54
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Zero-Sequence Reactance Diagram
M3
2
1
T3
Zero-Sequence Reference Bus
M1T1
Utility
M2 T2
GXn+ 3 Xn
Connection AlterationConnection AlterationGr. Gr. WyeWye NoneNoneWyeWye Open Open CktCkt..Delta Open Delta Open CktCkt. AND. AND
Short to Ref. BusShort to Ref. Bus
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 55
T1 T2
T3
M1 M2
M3
GUtilityXn
1
2
Zero-Sequence Reactance Diagram
+ 3 Xn
2
T3
1
T1 M1 M2 T2
M3
Zero-Sequence Reference Bus
Utility GXn
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 56
Workshop #3Sequence Networks
Draw the positive-, negative-, and zero-sequence networks for the one-line diagram on the left.
T2
G
Xn
M2
Xn
1
M1
2
T3
Utility
T1
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 57
Workshop #3Sequence Networks
Positive-Sequence Reference Bus
2
T1
1
T3
M1 T2
Utility G
M2
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 58
Workshop #3Sequence Networks
T3 M2
2
Utility
T1
1
M1 T2
Negative-Sequence Reference Bus
G
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 59
Workshop #3Sequence Networks
Zero-Sequence Reference Bus
1
T1
Utility
2
T3
M1 T2
M2
G
+3Xn
Xn+3Xn
Thevenin Reduction of Sequence Networks
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 60
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 61
Positive-Sequence Reactance Diagram
Bus 1 TheveninEquivalent
Bus 2 TheveninEquivalent
[(T1+Utility) ║ M1 ║ M2║ (T2+G)] ║ (T3+M3)
M3 ║ {T3+ [(T1+Utility) ║ M1 ║ M2║ (T2+G)]}
2
T3 M3
Positive-Sequence Reference Bus
1
T1 M2M1 T2
Utility G
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 62
Positive-Sequence Reactance Diagram
TheveninEquivalent
X1
+
Pre-faultVoltage
FaultLocation
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 63
Negative-Sequence Reactance Diagram
Bus 1 TheveninEquivalent
Bus 2 TheveninEquivalent
[(T1+Utility) ║ M1 ║ M2║ (T2+G)] ║ (T3+M3)
M3 ║ {T3+ [(T1+Utility) ║ M1 ║ M2║ (T2+G)]}2
T3 M3
1
T1 M2M1 T2
Utility G
Negative-Sequence Reference Bus
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 64
Negative-Sequence Reactance Diagram
TheveninEquivalent
Location
X2
Fault
_
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 65
Zero-Sequence Reactance Diagram
+ 3 Xn
2
T3
1
T1 M1 M2 T2
M3
Zero-Sequence Reference Bus
Utility GXn
Bus 1 TheveninEquivalent
Bus 2 TheveninEquivalent
T1
T3 + T1
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 66
Zero-Sequence Reactance Diagram
TheveninEquivalent
X0
FaultLocation
0
Types of Fault Calculations
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 67
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 68
First-Cycle or Momentary
First-cycle fault calculations are done to determine the withstand strength requirement of the system components at the location of the fault.
It is the maximum amplitude of the fault current ever expected (worst case).
It requires use of the subtransient reactances of rotating machines, and includes induction motors.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 69
Contact-Parting or Clearing
Contact-parting fault calculations are done to determine the interrupting rating of the protective devices at the location of the fault.
It is a reduced amplitude of the fault current anticipated at clearing time (worst case).
It requires use of the transient reactances of rotating machines, and excludes induction motors.
Short-Circuit Fault Calculations
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 70
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 71
Three-Phase Fault
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 72
Line-to-Ground Fault
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 73
Double Line-to-Ground Fault
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 74
Line-to-Line Fault
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 75
Workshop #4Short-Circuit Fault Calculations
The Thevenin-equivalent sequence reactances at a given bus are:
X1 = 0.032 p.u.X2 = 0.029 p.u.X0 = 0.024 p.u.
Find the fault currents at that bus for a (1) three-phase, (2) line-to-ground, (3) double line-to-ground, and (4) line-to-line fault.
The base current is 1.5 kA.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 76
Workshop #4Short-Circuit Fault Calculations
Three-Phase Fault
IA = 31.25 /-90o p.u. = 46.9 /-90o kA
IB = 31.25 /150o p.u. = 46.9 /150o kA
IC = 31.25 /30o p.u. = 46.9 /30o kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 77
Workshop #4Short-Circuit Fault Calculations
Line-to-Ground Fault
IA = 35.29 /-90o p.u. = 52.9 /-90o kA
IB = 0
IC = 0
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 78
Workshop #4Short-Circuit Fault Calculations
Double Line-to-Ground Fault
I0 = 12.124 /90o p.u.
I1 = 22.157 /-90o p.u.
I2 = 10.033 /90o p.u.
IA = 0
IB = 33.29 /147o p.u. = 49.9 /147o kA
IC = 33.29 /33o p.u. = 49.9 /33o kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 79
Workshop #4Short-Circuit Fault Calculations
Line-to-Line Fault
I0 = 0
I1 = 16.39 /-90o p.u.
I2 = 16.39 /90o p.u.
IA = 0
IB = 28.4 /180o p.u. = 42.6 /180o kA
IC = 28.4 /0o p.u. = 42.6 /0o kA
Open-Circuit Fault Calculations
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 80
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 81
One-Line-Open Fault
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 82
Two-Lines-Open Fault
X/R Ratio at Fault Location
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 83
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 84
The X/R ratio at the point of the fault determines the rate of fault current decay.
The larger the X/R ratio, the more slowly the fault current decays.
Small X/R Large X/R
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 85
Determination of the X/R ratio requires the construction of a positive sequence resistance network.
The X (positive-sequence reactance) and R must be determined separately at the fault location. Then the resistance is divided into the reactance to give the X/R ratio.
A SINGLE IMPEDANCE DIAGRAM COMBINING R AND X CANNOT BE USED! It will undercalculate the actual X/R ratio.
Selection of Protective Equipment
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 86
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 87
Protective devices are always sized for the highest possible fault current at the location where the device will be installed – this is NOTalways a three-phase fault!!
RMS symmetrical fault current is used to determine all protective device ratings.
With the exception of power circuit breakers, protective devices are sized based on a multiplying factor to account for X/R ratios that exceed the manufacturer’s assumptions.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 88
Power Circuit Breakers
Power circuit breakers are specified by a Close-and-Latch rating.
RMS Close-and-Latch rating = 1.6 × RMS symmetrical fault current
Crest Close-and-Latch rating = 2.7 × RMS symmetrical fault current
Example: If the maximum fault current is 23.5 kA23.5 kA,the required RMS close-and-latch rating is 1.6 × 23.5 = 37.6 37.6 kAkARMSRMS
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 89
Fused Low-Voltage Circuit Breakers
94RXfor251
1e2MF
RX2
bkrfusedLV ./.
)//(
>+
=π−
Example: Maximum fault current = 27.5 kA27.5 kAX/R at fault location = 7.87.8
1011251
1e2MF
872
bkrfusedLV ..
./
=+
=π−
So, the fused low-voltage circuit breaker must be rated at least 27.5 × 1.101 = 30.3 kA30.3 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 90
Molded-Case Circuit Breakers
( ) 66RXfor292
1e2MFRX
bkrcasemolded ./.
)//(
>+
=π−
−
Example: Maximum fault current = 45 kA45 kAX/R at fault location = 9.29.2
( )0561
2921e2
MF29
bkrcasemolded ..
./
=+
=π−
−
So, the molded-case circuit breaker must be rated at least 45 × 1.056 = 47.5 kA47.5 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 91
Medium-Voltage Expulsion Fuses
15RXfor521
1e2MF
RX2
fuseMV >+
=π−
/.
)/(/
Example: Maximum fault current = 45.8 kA45.8 kAX/R at fault location = 21.421.4
0381521
1e2MF
4212
fuseMV ..
./
=+
=π−
So, the medium-voltage expulsion fuse must be rated at least 45.8 × 1.038 = 47.6 kA47.6 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 92
Low-Voltage Expulsion Fuses
94RXfor251
1e2MF
RX2
fuseLV ./.
)/(/
>+
=π−
Example: Maximum fault current = 38.2 kA38.2 kAX/R at fault location = 11.811.8
1801251
1e2MF
8112
fuseLV ..
./
=+
=π−
So, the low-voltage expulsion fuse must be rated at least 38.2 × 1.180 = 45.1 kA45.1 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 93
Current-Limiting Fuses
10RXfor441
1e2MF
RX2
fuseitinglimcurrent >+
=π−
− /.
)/(/
Example: Maximum fault current = 58.4 kA58.4 kAX/R at fault location = 16.216.2
0661441
1e2MF
2162
fuseitinglimcurrent ..
./
=+
=π−
−
So, the current-limiting fuse must be rated at least 58.4 × 1.066 = 62.3 kA62.3 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 94
Workshop #5Protective Device Specification
1. Find both the RMS Close-and-Latch rating and the Crest Close-and-Latch rating required for a power circuit breaker to be installed on a bus where the maximum fault current is 32.9 kA.
2. Find the required interrupting rating for a molded-case circuit breaker installed on a bus with a maximum fault current of 46.5 kA and an X/R ratio of 14.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 95
Workshop #5Protective Device Specification
3. Find the required interrupting rating for a medium-voltage fuse installed on a bus with a maximum fault current of 46.5 kA and an X/R ratio of 18.
4. Find the required interrupting rating for a low-voltage fuse installed on a bus with a maximum fault current of 64.8 kA and an X/R ratio of 12.
5. Find the required interrupting rating for a current-limiting fuse installed on a bus with a maximum fault current of 27.3 kA and an X/R ratio of 8.
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 96
Workshop #5Protective Device Specification
2.
46.5 kA × 1.111 = 51.7 kA
( )1.111
2.291e2
MFπ/14
bkrcasemolded =+
=−
−
1. Close-and-LatchRMS = 32.9 kA × 1.6 = 52.6 kA
Close-and-LatchCrest = 32.9 kA × 2.7 = 88.8 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 97
Workshop #5Protective Device Specification
4.
64.8 kA × 1.182 = 76.6 kA
1.1821.25
1e2MF
12/2
fuseLV =+
=− π
1.0211.52
1e2MF
18/2π
fuseMV =+
=−
3.
46.5 kA × 1.021 = 47.5 kA
IEEE Fault Calculation Seminar March 2006 Ralph Fehr, Ph.D., P.E. 98
Workshop #5Protective Device Specification
Since X/R ≤ 10, no multiplying factor is used.
Required Interrupting Rating = 27.3 kA
10X/Rfor1.44
1e2MF
(X/R)/2π
fuselimitingcurrent >+
=−
−5.
Fault Calculationsand
Selection ofProtective Equipment
http://http://web.tampabay.rr.com/usfpower/fehr.htmweb.tampabay.rr.com/usfpower/fehr.htm
which includes a link to
Alex McEachern’s Power Quality Teaching Toy
Don’t forget the power engineering resources mentioned in this course:
Fault Calculationsand
Selection ofProtective Equipment
Thank you! Seminole Electric Cooperative, Inc.
16313 North Dale Mabry Hwy.Tampa, Florida
Ralph Fehr, Ph.D., P.E.University of South Florida – Tampa
Senior Member, IEEEr.fehr@ieee.org
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