Fault Calculation Steps

7/30/2019 Fault Calculation Steps

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Fault CalculationsandSelection ofProtective Equipment

Wednesday, March 22, 2006

8:00AM 3:00PM

Seminole Electric Cooperative, Inc.

16313 North Dale Mabry Hwy.

Tampa, Florida

Ralph Fehr, Ph.D., P.E.University of South Florida TampaSenior Member, [email protected]


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

Symmetrical Components


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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.



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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.



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Charles Fortescues Theory of Symmetrical

Components, first published in 1918, proves

that any set of unbalanced voltage or current

phasors belonging to a three-phase system canbe resolved into three sets of components,

each of which is balanced.



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A

ICI

BI


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Physical Example ofVector Components



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F

d

M = F d

Calculation of Moment

Moment = Force Perpendicular Distance



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d

F

M F d




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M = FV d


d

VF F

HF



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d

VF F

HF

FV = F cos

FH = F sin

FH + FV = F

=

V

H1

F

Ftan



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Application of SymmetricalComponents to a Three-PhasePower System



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AICI

BI


14/100

A-B-C Sequencing


I B

I CAI


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A-C-B Sequencing


II

B

CI

A


16/100

Fortescues theory shows that three sets of

balanced components are required to represent

any unbalanced set of three-phase phasors.



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Positive Sequence Components


I A1

C1I

B1I


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Negative Sequence Components


A2IB2I

C2I


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Zero Sequence Components


A0IB0I

C0I


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The constraint equations for the symmetrical

components require the sum of the three

components for each unbalanced phasor toequal the unbalanced phasor itself.

IA = IA0 + IA1 + IA2

IB = IB0 + IB1 + IB2

IC = IC0 + IC1 + IC2



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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



22/100

Using the a operator and the symmetry of the

sequence components, we can develop asingle-phase equivalent circuit to greatly

simplify the analysis of the unbalanced system.

We will start by expressing the sequencecomponents in terms of a single phases

components. We will use Phase A as the phase

for developing the single-phase equivalent.



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Positive Sequence Components


B1I

A1I

I C1

1= I

= a I1

= a I 12


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Negative Sequence Components


I A2 = I 2I B2 = a I2

2C2 = a II 2


25/100

Zero Sequence Components


A0I = I 0= IB0I 0

= II C0 0


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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



27/100

Unbalanced Phasors and their Symmetrical Components


I A

I 1

I 2

0I

CI

a I1

2a I 2

I 0

BI0I

2a I1a I

2


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=

2

1

0

2

2

C

B

A

I

II

aa1

aa1111

I

II

=

2

1

0

2

2

1

2

2

C

B

A

1

2

2

I

I

I

aa1

aa1

111

aa1

aa1

111

I

I

I

aa1

aa1

111

=

C

B

A

1

2

2

2

1

0

I

I

I

aa1

aa1

111

I

I

I



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=

C

B

A

1

2

2

2

1

0

I

I

I

aa1

aa1

111

I

I

I

=

C

B

A

2

2

2

1

0

I

I

I

aa1

aa1

111

3

1

I

I

I

( )CBA0 III3

1I ++=

( )C2BA1 IaIaI31I ++=

( )CB2A2 IaIaI3

1I ++=



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( )CBA0 III3

1I ++=

( )C2

BA1 IaIaI3

1

I ++=

( )CB2A2 IaIaI3

1I ++=

Summary of Symmetrical Components

Transformation Equations

IA = I0 + I1 + I2

IB = I0 + a2 I1 + a I2

IC = I0 + a I1 + a2 I2



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Workshop #1Symmetrical Components


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


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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

I1I2

I0

a I1

a I22 I0

1a I2

2a II0


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I0 = 0.7 /300I1 = 1.2 /10

I2 = 0.3 /167

Find Ia, Ib, and Ic

I1

2

0

I

I

Ia

Ic

a I1

2a I2

I0

I0

1a I2

2a I

Ia = 1.2827 /345

Ib = 2.0209 /271

Ic = 0.5749 /112


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Electrical Characteristics ofthe Sequence Currents



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O

y DA

x

I

L


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I

t


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Bottom Wire

Top WireMiddle Wire

t = T

I

t


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O

I0

D

A

LI0

0I


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3 I

0I

0

0I

I0

O

DA

L


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O3 I

0I D

0 A

LI0

0I

VN = (3 I0) ZN

VN = I0 (3 Z(3 ZNN))


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The Delta-Wye Transformer



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Ic

Ia

Ib

IA

IC

IB


Assume Ia = 1 /0o, Ib = 1 /240

o, and Ic = 1/120o.

Ic

Ia

Ib

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 /0

o 1 /240o = /30o3


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Workshop #2Non-Standard Delta-Wye Transformer


Given that Ia = 1/0o, Ib = 1/240

o, and Ic = 1/120o, find IA, IB, and IC.

Ib

aI

cI

BI

AI

CI


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IC aI

IA

IB

cI

bI

Ib

cII acI

bII a


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


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Sequence Networks



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T1 T2

T3

M1 M2

M3

GUtilityXn

1

2

One-Line Diagram

Positive Sequence Reactance Diagram


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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

Utility M1 M2 M3G

Positive-Sequence Reference Bus

Positive-Sequence Reactance Diagram



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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2


M3M1

1

T1 M2 T2

Utility G




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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2


2

T3 M3


1

T1 M2M1 T2

Utility G

Negative-Sequence Reactance Diagram


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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

Negative Sequence Reactance Diagram

2

T3 M3


1

T1 M2M1 T2

Utility G

Negative



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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

Negative Sequence Reactance Diagram

2

T3 M3

1

T1 M2M1 T2

Utility G

Negative-Sequence Reference Bus

Zero-Sequence Reactance Diagram


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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

Zero Sequence Reactance Diagram

2

T3 M3

1

T1 M2M1 T2

Utility G

Negative-Sequence Reference BusZero

+3Xn

Adjust Topology



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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

q g

M3

2

1

T3

Zero-Sequence Reference Bus

M1T1

Utility

M2 T2

GXn

+ 3 Xn



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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

q g

M3

2

1

T3


M1T1

Utility

M2 T2

GXn

+ 3 Xn

Connection AlterationConnection Alteration

Gr.Gr. WyeWye NoneNone

WyeWye OpenOpen CktCkt..

Delta OpenDelta Open CktCkt. AND. AND

Short to Ref. BusShort to Ref. Bus



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T1 T2

T3

M1 M2

M3

GUtility Xn

1

2

q g

+ 3 Xn

2

T3

1

T1 M1 M2 T2

M3


UtilityG

Xn


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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


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2

T1

1

T3

M1 T2

Utility G

M2


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T3 M2

2

Utility

T1

1

M1 T2


G


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1

T1

Utility

2

T3

M1 T2

M2

G

+3Xn

Xn+3Xn


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Thevenin Reduction ofSequence Networks


P iti S R t Di


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Bus 1 Thevenin

Equivalent

Bus 2 Thevenin

Equivalent

[(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


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Thevenin

Equivalent

X1

+

Pre-faultVoltage

FaultLocation

N ti S R t Di


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Bus 1 Thevenin

Equivalent

Bus 2 Thevenin

Equivalent

[(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



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Thevenin

Equivalent

Location

X2

Fault

_

Zero Sequence Reactance Diagram


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+ 3 Xn

2

T3

1

T1 M1 M2 T2

M3


UtilityG

Xn

Bus 1 Thevenin

Equivalent

Bus 2 Thevenin

Equivalent

T1

T3 + T1


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Thevenin

Equivalent

X0

FaultLocation

0


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Types of Fault Calculations



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First-Cycle or Momentary

First-cycle fault calculations are done to determine

the withstand strength requirement of the systemcomponents 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.


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Contact-Parting or Clearing

Contact-parting fault calculations are done to

determine the interrupting rating of the protectivedevices 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.


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Short-Circuit FaultCalculations



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Three-Phase Fault


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Line-to-Ground Fault


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Double Line-to-Ground Fault


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Line-to-Line Fault


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Workshop #4Short-Circuit Fault Calculations

The Thevenin-equivalent sequence reactances at agiven 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.


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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


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Line-to-Ground Fault

IA = 35.29 /-90o p.u. = 52.9 /-90o kA

IB = 0

IC = 0


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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


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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


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Open-Circuit FaultCalculations



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One-Line-Open Fault


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Two-Lines-Open Fault


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X/R Ratio at Fault Location


The X/R ratio at the point of the fault


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p

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


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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! Itwill undercalculate the actual X/R ratio.


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Selection of ProtectiveEquipment



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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 manufacturers assumptions.

P Ci it B k


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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 = 37.637.6 kAkARMSRMS

F d L V lt Ci it B k


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Fused Low-Voltage Circuit Breakers

94RXfor251

1e2MFRX2

bkrfusedLV ./.

)//(

>+=

Example: Maximum fault current = 27.5 kA27.5 kA

X/R at fault location = 7.87.8

1011

251

1e2MF

872

bkrfusedLV .

.

./

=+

=

So, the fused low-voltage circuit breaker

must be rated at least 27.5 1.101 = 30.3 kA30.3 kA


90/100

Medium Voltage Expulsion Fuses


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Medium-Voltage Expulsion Fuses

15RXfor521

1e2MFRX2

fuseMV >+=

/.

)/(/



0381

521

1e2MF

4212

fuseMV .

.

./

=+

=

So, the medium-voltage expulsion fuse must

be rated at least 45.8 1.038 = 47.6 kA47.6 kA

Low Voltage Expulsion Fuses


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Low-Voltage Expulsion Fuses

94RXfor251

1e2MFRX2

fuseLV ./.

)/(/

>+=



1801

251

1e2MF

8112

fuseLV .

.

./

=+

=

So, the low-voltage expulsion fuse must be

rated at least 38.2 1.180 = 45.1 kA45.1 kA

Current Limiting Fuses


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Current-Limiting Fuses

10RXfor441

1e2MFRX2

fuseitinglimcurrent >+=

/.

)/(/



0661

441

1e2MF

2162

fuseitinglimcurrent .

.

./

=+

=

So, the current-limiting fuse must be rated at

least 58.4 1.066 = 62.3 kA62.3 kA

Workshop #5P t ti D i S ifi ti


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Workshop #5Protective Device Specification

1. Find both the RMS Close-and-Latch rating and the

Crest Close-and-Latch rating required for apower 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.

Workshop #5P t ti D i S ifi ti


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3. Find the required interrupting rating for a medium-

voltage fuse installed on a bus with a maximumfault 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.



96/100



2.

46.5 kA 1.111 = 51.7 kA

( )1.111

2.29

1e2MF

/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



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4.

64.8 kA 1.182 = 76.6 kA

1.1821.25

1e2MF

12/2

fuseLV =+

=

1.0211.52

1e2

MF

18/2

fuseMV=

+=

3.

46.5 kA 1.021 = 47.5 kA



98/100



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

10X/Rfor1.44

1e2MF(X/R)/2

fuselimitingcurrent >+=

5.


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http://http://web.tampabay.rr.com/usfpower/fehr.htmweb.tampabay.rr.com/usfpower/fehr.htm

which includes a link toAlex McEacherns Power Quality Teaching Toy

Dont forget the power engineering resources mentioned in this course:


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Thank you!Seminole Electric Cooperative, Inc.

16313 North Dale Mabry Hwy.Tampa, Florida

Ralph Fehr, Ph.D., P.E.University of South Florida TampaSenior Member, [email protected]

Fault Calculation Steps

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