Fault Calculations Tutorial

Fault Calculations and Selection of Protective Equipment Ralph Fehr, Ph.D., P.E. University of South Florida – Tampa Senior Member, IEEE [email protected] Wednesday, March 22, 2006 8:00AM – 3:00PM Seminole Electric Cooperative, Inc. 16313 North Dale Mabry Hwy. Tampa, Florida

Symmetrical Components

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 2

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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 5

IC

IA

IB ω

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Physical Example of Vector Components

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Calculation of Moment d

F

Moment = Force × Perpendicular Distance M=F×d IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 8

Calculation of Moment d θ

F

Moment = Force × Perpendicular Distance M≠F×d IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 9

Calculation of Moment d FH θ

FV F

Moment = Force × Perpendicular Distance M = FV × d IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 10

Calculation of Moment d

FV = F cos θ

FH θ

FH = F sin θ

FV F

FH + FV = F ⎛ FH ⎞ θ = tan ⎜⎜ ⎟⎟ ⎝ FV ⎠ −1

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 11

Application of Symmetrical Components to a Three-Phase Power System

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 12

IC

IA

IB ω

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IC

IA

IB ω A-B-C Sequencing IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 14

IB

IA

ω

IC

A-C-B Sequencing IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 15

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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 16

I C1

ω

I A1 I B1 Positive Sequence Components

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I B2 ω

I A2

I C2 Negative Sequence Components

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I A0 ω

ω

I B0

I C0

ω

Zero Sequence Components

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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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 20

The a operator

−1 3 = 1 /120° a≡ + j 2 2 a2 = 1 /240°

a3 = 1 /360°

Recall the j operator j2 = 1 /180° = −1

j3 = 1 /270° = −j

j4 = 1 /360° = 1

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 22

I C1 = a I 1

ω

I A1 = I 1 I B1 = a 2 I 1

Positive Sequence Components

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I B2 = a I 2 ω

I A2 = I 2

I C2 = a 2 I 2

Negative Sequence Components

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 24

I A0 = I 0 ω

I B0 = I 0 ω

I C0 = I 0 ω

Zero Sequence Components

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Recall the original constraint equations:

IA = IA0 + IA1 + IA2 IB = IB0 + IB1 + IB2 IC = IC0 + IC1 + IC2 Rewrite them using the a operator to 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 26

Unbalanced Phasors and their Symmetrical Components

I0

IC

IA

a I1

I0

a2 I 2 IB I0

a I2

ω

I2 I1

a2 I 1

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 27

⎡I A ⎤ ⎡1 1 ⎢ I ⎥ = ⎢1 a 2 ⎢ B⎥ ⎢ ⎢⎣ I C ⎥⎦ ⎢⎣1 a

⎡1 1 ⎢1 a 2 ⎢ ⎢⎣1 a

−1

1 ⎤ ⎡I A ⎤ ⎡1 1 a ⎥⎥ ⋅ ⎢⎢ I B ⎥⎥ = ⎢⎢1 a 2 a 2 ⎥⎦ ⎢⎣ I C ⎥⎦ ⎢⎣1 a ⎡I 0 ⎤ ⎡1 1 ⎢ I ⎥ = ⎢1 a 2 ⎢ 1⎥ ⎢ ⎢⎣I 2 ⎥⎦ ⎢⎣1 a

1⎤ a ⎥⎥ a 2 ⎥⎦

⎡I 0 ⎤ ⋅ ⎢⎢ I1 ⎥⎥ ⎢⎣I 2 ⎥⎦ −1

1 ⎤ ⎡1 1 a ⎥⎥ ⋅ ⎢⎢1 a 2 a 2 ⎥⎦ ⎢⎣1 a

1⎤ a ⎥⎥ a 2 ⎥⎦

⎡I 0 ⎤ ⋅ ⎢⎢ I1 ⎥⎥ ⎢⎣I 2 ⎥⎦

−1

1 ⎤ ⎡I A ⎤ a ⎥⎥ ⋅ ⎢⎢ I B ⎥⎥ a 2 ⎥⎦ ⎢⎣ I C ⎥⎦

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 28

⎡I 0 ⎤ ⎡1 1 ⎢ I ⎥ = ⎢1 a 2 ⎢ 1⎥ ⎢ ⎢⎣I 2 ⎥⎦ ⎢⎣1 a

⎡I 0 ⎤ ⎡1 1 ⎢ I ⎥ = 1 ⎢1 a ⎢ 1⎥ 3 ⎢ ⎢⎣I 2 ⎥⎦ ⎢⎣1 a 2 I0 =

−1

1 ⎤ ⎡I A ⎤ a ⎥⎥ ⋅ ⎢⎢ I B ⎥⎥ a 2 ⎥⎦ ⎢⎣ I C ⎥⎦

1⎤ a 2 ⎥⎥ a ⎥⎦

⎡I A ⎤ ⋅ ⎢⎢ I B ⎥⎥ ⎢⎣ I C ⎥⎦

1 (I A + I B + IC ) 3

( (

1 I1 = IA + a IB + a 2 IC 3 1 I2 = IA + a 2 IB + a IC 3

) )

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 29

Summary of Symmetrical Components Transformation Equations

IA = I0 + I1 + I2 IB = I0 + a2 I1 + a I2 IC = I0 + a I1 + a2 I2

1 (I A + I B + IC ) I0 = 3 1 IA + a IB + a 2 IC I1 = 3 1 IA + a 2 IB + a IC I2 = 3

(

)

(

)

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Workshop #1 Symmetrical Components

Ia = 0.95 /328° Ib = 1.03 /236°

Find I0, I1, and I2

Ic = 0.98 /92° I0 = 0.7 /300° I1 = 1.2 /10°

Find Ia, Ib, and Ic

I2 = 0.3 /167° IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 31

Workshop #1 Symmetrical Components Ia = 0.95 /328° Ib = 1.03 /236°

Find I0, I1, and I2 a2 I2 I0 a I Ic

Ic = 0.98 /92°

1

I0 = 0.1418 /297° I1 = 0.9634 /339° I2 = 0.1622 /191°

2

a I1 a I2 I0 Ib

I1 I2 Ia I0

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 32

Workshop #1 Symmetrical Components I0 = 0.7 /300° I1 = 1.2 /10° I2 = 0.3 /167°

Find Ia, Ib, and Ic a2 I2 a I1

I0 Ic

Ia = 1.2827 /345°

I1

Ib = 2.0209 /271° Ic = 0.5749 /112°

I2

Ia

I0

a2 I 1 a I2

I0

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 33

Electrical Characteristics of the Sequence Currents

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 34

I x

y

L O A D

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I t

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 36

I

Middle Wire Top Wire Bottom Wire

t t=T

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I0 I0

I0

L O A D

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I0 I0 3 I0 I0

L O A D

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 39

I0 I0 3 I0 VN = (3 I0) × ZN

I0

L O A D

VN = I0 × (3 ZN)

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 40

The Delta-Wye Transformer

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 41

The Delta-Wye Transformer IB

Ib

Ib

Ia

Ic

Ib

IA

IC

Ic

Ia

Ic

Ia

Assume Ia = 1 /0o, Ib = 1 /240o, and Ic = 1/120o.

IA = Ia – Ib = 1 /0o – 1 /240o IB = Ib – Ic = 1

/240o –

IC = Ic – Ia = 1

1

/120o –

=

/120o =

1

/0o =

3 /30o 3 3

Ic

IA

IC Ia

/270o

/150o

Ib IB

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 42

Workshop #2 Non-Standard Delta-Wye Transformer IB

Ib

IA

Ic

IC

Ia

Given that Ia = 1/0o, Ib = 1/240o, and Ic = 1/120o, find IA, IB, and IC.

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 43

Workshop #2 Non-Standard Delta-Wye Transformer IB

Ib

IA

Ia

Ib Ic

IC

Ic

Ib Ia

= 3 /330o Ic Ia

IB = Ib – Ia = 1/240o – 1/0o = 3 /210o

IC Ic Ia IB

IA = Ia – Ic = 1/0o – 1/120o

Ia = 1/0o Ib = 1/240o Ic = 1/120o

IC = Ic – Ib = 1/120o – 1/240o = 3 /90o

IA Ib IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 44

Sequence Networks

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 45

One-Line Diagram Utility

G

T1

Xn

T2

1 M1

M2

T3 2 M3 IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 46

Positive-Sequence Reactance Diagram Utility

G

Xn

Positive-Sequence Reference Bus

T1

T2

1

Utility

M1

M1

M2

G

M3

M2

T3 2 M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 47

Positive-Sequence Reactance Diagram Utility

G

Xn

Positive-Sequence Reference Bus

T1

T2

1

Utility

M1

G

M2 T1

T3

M1

M2

T2

M3

1

2 M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 48

Positive-Sequence Reactance Diagram Utility

G

Xn

Positive-Sequence Reference Bus

T1

T2 Utility

1 M1

M2

T1

T3

G

M1

M2

1 T3

2 M3

T2

M3

2

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 49

Negative-Sequence Reactance Diagram Utility

G

T1

Xn

Negative Positive-Sequence Reference Bus

T2 Utility

1 M1

M2

T1

T3

G

M1

M2

1 T3

2 M3

T2

M3

2

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 50

Negative-Sequence Reactance Diagram Utility

G

Xn

Negative-Sequence Reference Bus

T1

T2 Utility

G

1 T1

M1

M1

M2

T2

M2 1

T3 2

T3

M3

2

M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 51

Zero-Sequence Reactance Diagram Utility

G

T1

Xn

Zero Negative-Sequence Reference Bus

T2 Utility

G +3Xn

1 T1

M1

M1

M2

T2

M2 1

T3 2

T3

M3

2

M3

Adjust Topology

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 52

Zero-Sequence Reactance Diagram Utility

G

Xn

Zero-Sequence Reference Bus

T1

T2

G + 3 Xn

Utility

Xn

1 T1

M1

M1

M2

T2

M2 1

T3 2

T3

M3

2

M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 53

Zero-Sequence Reactance Diagram Utility

G

Xn

Zero-Sequence Reference Bus

T1

T2

G + 3 Xn

Utility

Xn

1 T1

M1

M1

M2

T2

M2 1

T3 2 Connection Gr. Wye Wye Delta

Alteration None Open Ckt. Open Ckt. AND Short to Ref. Bus

T3

M3

2

M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 54

Zero-Sequence Reactance Diagram Utility

G

Xn

Zero-Sequence Reference Bus

T1

T2

G + 3 Xn

Utility

Xn

1 T1

M1

M1

M2

T2

M2 1 T3

T3

M3

2

2 M3

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 55

Workshop #3 Sequence Networks Utility

G

Xn

T2

T1 Xn

1 M1

Draw the positive-, negative-, and zero-sequence networks for the one-line diagram on the left.

T3 2 M2

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 56

Workshop #3 Sequence Networks Positive-Sequence Reference Bus

Utility

T1

G

M1

T2

1 T3

M2

2

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 57

Workshop #3 Sequence Networks Negative-Sequence Reference Bus Utility

T1

G

M1

T2

1 T3

M2

2 IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 58

Workshop #3 Sequence Networks Zero-Sequence Reference Bus G

Utility

T1

+3Xn

M1

Xn

T2

+3Xn

1 T3

M2

2 IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 59

Thevenin Reduction of Sequence Networks

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 60

Positive-Sequence Reactance Diagram Positive-Sequence Reference Bus

Bus 1 Thevenin Equivalent Utility

G

[(T1+Utility) ║ M1 ║ M2║ (T2+G)] ║ (T3+M3)

T1

M1

M2

T2

Bus 2 Thevenin Equivalent

1 T3

2

M3

M3 ║ {T3+ [(T1+Utility) ║ M1 ║ M2║ (T2+G)]}

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 61

Positive-Sequence Reactance Diagram + Pre-fault Voltage X1 Fault Location

Thevenin Equivalent IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 62

Negative-Sequence Reactance Diagram Negative-Sequence Reference Bus Utility

Bus 1 Thevenin Equivalent

G

[(T1+Utility) ║ M1 ║ M2║ (T2+G)] T1

M1

M2

║ (T3+M3)

T2

1 T3

2

M3

Bus 2 Thevenin Equivalent M3 ║ {T3+ [(T1+Utility) ║ M1 ║ M2║ (T2+G)]}

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 63

Negative-Sequence Reactance Diagram _ X2 Fault Location

Thevenin Equivalent IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 64

Zero-Sequence Reactance Diagram Bus 1 Thevenin Equivalent

Zero-Sequence Reference Bus G + 3 Xn

Utility

Xn

T1 T1

M1

M2

T2

1 T3 2

M3

Bus 2 Thevenin Equivalent

T3 + T1

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 65

Zero-Sequence Reactance Diagram

0 X0 Fault Location

Thevenin Equivalent IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 66

Types of Fault Calculations

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 67

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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 68

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.

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 69

Short-Circuit Fault Calculations

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 70

Three-Phase Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 71

Line-to-Ground Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 72

Double Line-to-Ground Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 73

Line-to-Line Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 74

Workshop #4 Short-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) lineto-line fault. The base current is 1.5 kA. IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 75

Workshop #4 Short-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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 76

Workshop #4 Short-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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 77

Workshop #4 Short-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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 78

Workshop #4 Short-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 IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 79

Open-Circuit Fault Calculations

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 80

One-Line-Open Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 81

Two-Lines-Open Fault

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 82

X/R Ratio at Fault Location

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 83

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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 84

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.

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 85

Selection of Protective Equipment

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 86

Protective devices are always sized for the highest possible fault current at the location where the device will be installed – this is NOT always 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 87

Power Circuit Breakers Power circuit breakers are specified by a Close-andLatch 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 kA, kA the required RMS close-and-latch rating is 1.6 × 23.5 = 37.6 kARMS

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 88

Fused Low-Voltage Circuit Breakers

MFLV fused bkr = Example:

2 e −2 π /( X / R ) + 1 for X / R > 4.9 1.25

Maximum fault current = 27.5 kA X/R at fault location = 7.8

MFLV fused bkr =

2 e −2 π / 7.8 + 1 = 1.101 1.25

So, the fused low-voltage circuit breaker must be rated at least 27.5 × 1.101 = 30.3 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 89

Molded-Case Circuit Breakers

MFmolded−case bkr = Example:

(

)

2 e − π /( X / R ) + 1 for X / R > 6.6 2.29

Maximum fault current = 45 kA X/R at fault location = 9.2

MFmolded−case bkr =

(

)

2 e − π / 9 .2 + 1 = 1.056 2.29

So, the molded-case circuit breaker must be rated at least 45 × 1.056 = 47.5 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 90

Medium-Voltage Expulsion Fuses

MFMV fuse = Example:

2 e −2 π / ( X / R ) + 1 for X / R > 15 1.52

Maximum fault current = 45.8 kA X/R at fault location = 21.4

MFMV fuse =

2 e −2 π / 21.4 + 1 = 1.038 1.52

So, the medium-voltage expulsion fuse must be rated at least 45.8 × 1.038 = 47.6 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 91

Low-Voltage Expulsion Fuses

MFLV fuse = Example:

2 e −2 π / ( X / R ) + 1 for X / R > 4.9 1.25

Maximum fault current = 38.2 kA X/R at fault location = 11.8

MFLV fuse =

2 e −2 π / 11.8 + 1 = 1.180 1.25

So, the low-voltage expulsion fuse must be rated at least 38.2 × 1.180 = 45.1 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 92

Current-Limiting Fuses

MFcurrent −limiting fuse = Example:

2 e −2 π / ( X / R ) + 1 for X / R > 10 1.44

Maximum fault current = 58.4 kA X/R at fault location = 16.2

MFcurrent −limiting fuse =

2 e −2 π / 16.2 + 1 = 1.066 1.44

So, the current-limiting fuse must be rated at least 58.4 × 1.066 = 62.3 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 93

Workshop #5 Protective 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 moldedcase 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 94

Workshop #5 Protective Device Specification 3. Find the required interrupting rating for a mediumvoltage 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 lowvoltage 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 currentlimiting 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 a March 2006 a Ralph Fehr, Ph.D., P.E. a 95

Workshop #5 Protective Device Specification 1. Close-and-LatchRMS = 32.9 kA × 1.6 = 52.6 kA Close-and-LatchCrest = 32.9 kA × 2.7 = 88.8 kA

2. MFmolded−case bkr =

(

)

2 e − π/14 + 1 = 1.111 2.29

46.5 kA × 1.111 = 51.7 kA

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 96

Workshop #5 Protective Device Specification 3. MFMV fuse =

2 e −2π/ 18 + 1 = 1.021 1.52

46.5 kA × 1.021 = 47.5 kA

4. MFLV fuse =

2 e −2 π / 12 + 1 = 1.182 1.25

64.8 kA × 1.182 = 76.6 kA IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 97

Workshop #5 Protective Device Specification

5. MFcurrent −limiting fuse =

2 e −2π/ (X/R) + 1 for X/R > 10 1.44

Since X/R ≤ 10, no multiplying factor is used. Required Interrupting Rating = 27.3 kA

IEEE Fault Calculation Seminar a March 2006 a Ralph Fehr, Ph.D., P.E. a 98

Fault Calculations and Selection of Protective Equipment Don’t forget the power engineering resources mentioned in this course:

http://web.tampabay.rr.com/usfpower/fehr.htm which includes a link to

Alex McEachern’s Power Quality Teaching Toy

Fault Calculations and Selection of Protective Equipment Ralph Fehr, Ph.D., P.E. University of South Florida – Tampa Senior Member, IEEE [email protected]

Thank you! Seminole Electric Cooperative, Inc. 16313 North Dale Mabry Hwy. Tampa, Florida