fault calculations and selection of protective equipmenthelios.acomp.usf.edu/~fehr/slides.pdf ·...

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, [email protected]

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.


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.

☺


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.



AICI

BIω

Physical Example ofVector Components


F

d

M = F × d

Calculation of Moment

Moment = Force × Perpendicular Distance



d

F

θ

M ≠ F × d




M = FV × d


d

VF F

HFθ



d

VF F

HFθ

FV = F cos θ

FH = F sin θ

FH + FV = F

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

V

H1

FFtanθ


Application of Symmetrical Components to a Three-Phase

Power System



AICI

BIω

A-B-C Sequencing


I B

ω

I CAI

A-C-B Sequencing


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.


Positive Sequence Components


I A1

C1I

B1I

ω

Negative Sequence Components


A2IB2I

C2I

ω

Zero Sequence Components


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


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


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.


Positive Sequence Components


B1I

A1I

I C1

1= I

= a I1

= a I 12

ω

Negative Sequence Components


I A2 = I 2I B2 = a I2

2C2 = a II 2

ω

Zero Sequence Components


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


Unbalanced Phasors and their Symmetrical Components


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


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⋅

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

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 ++=


( )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


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



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



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



O

y DA

x

I

L


I

t


Bottom WireTop Wire

Middle Wire

t = T

I

t


O

I0 DA

LI0

0I


3 I

0I

0

0I

I0

O

DA

L


O3 I

0I D0 A

LI0

0I

VN = (3 I0) × ZN

VN = I0 × (3 Z(3 ZNN))

The Delta-Wye Transformer


Ic

Ia

Ib

IA

IC

IB


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


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


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



T1 T2

T3

M1 M2

M3

GUtilityXn

1

2

One-Line Diagram


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2

Utility M1 M2 M3G

Positive-Sequence Reference Bus

Positive-Sequence Reactance Diagram


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


M3M1

1

T1 M2 T2

Utility G



T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


2

T3 M3


1

T1 M2M1 T2

Utility G


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2

Negative-Sequence Reactance Diagram

2

T3 M3


1

T1 M2M1 T2

Utility G

Negative


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


2

T3 M3

1

T1 M2M1 T2

Utility G

Negative-Sequence Reference Bus


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


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


M3

2

1

T3

Zero-Sequence Reference Bus

M1T1

Utility

M2 T2

GXn+ 3 Xn


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


M3

2

1

T3


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


T1 T2

T3

M1 M2

M3

GUtilityXn

1

2


+ 3 Xn

2

T3

1

T1 M1 M2 T2

M3


Utility GXn


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




2

T1

1

T3

M1 T2

Utility G

M2



T3 M2

2

Utility

T1

1

M1 T2


G




1

T1

Utility

2

T3

M1 T2

M2

G

+3Xn

Xn+3Xn

Thevenin Reduction of Sequence Networks




Bus 1 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



TheveninEquivalent

X1

+

Pre-faultVoltage

FaultLocation





[(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




TheveninEquivalent

Location

X2

Fault

_



+ 3 Xn

2

T3

1

T1 M1 M2 T2

M3


Utility GXn



T1

T3 + T1



TheveninEquivalent

X0

FaultLocation

0

Types of Fault Calculations



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.


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



Three-Phase Fault


Line-to-Ground Fault


Double Line-to-Ground Fault


Line-to-Line Fault


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.



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



Line-to-Ground Fault

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

IB = 0

IC = 0



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



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



One-Line-Open Fault


Two-Lines-Open Fault

X/R Ratio at Fault Location



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


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



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.


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


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


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


Medium-Voltage Expulsion Fuses

15RXfor521

1e2MF

RX2

fuseMV >+

=π−

/.

)/(/


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


Low-Voltage Expulsion Fuses

94RXfor251

1e2MF

RX2

fuseLV ./.

)/(/

>+

=π−


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


Current-Limiting Fuses

10RXfor441

1e2MF

RX2

fuseitinglimcurrent >+

=π−

− /.

)/(/


0661441

1e2MF

2162

fuseitinglimcurrent ..

./

=+

=π−

−

So, the current-limiting fuse must be rated at least 58.4 × 1.066 = 62.3 kA62.3 kA


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.



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.



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



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



Since X/R ≤ 10, no multiplying factor is used.

Required Interrupting Rating = 27.3 kA

10X/Rfor1.44

1e2MF

(X/R)/2π

fuselimitingcurrent >+

=−

−5.



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:



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, [email protected]

fault calculations and selection of protective equipmenthelios.acomp.usf.edu/~fehr/slides.pdf ·...

Documents