SHORT CIRCUIT CALCULATIONS REVISITEDiiee.org.ph/wp-content/uploads/2014/10/Short-Circuit-Calculations.pdfSHORT CIRCUIT CALCULATIONS REVISITED FORTUNATO C. LEYNES Chairman Professional

Short Circuit CalculationsIIEE Presentation

SHORT CIRCUIT CALCULATIONSREVISITED

FORTUNATO C. LEYNESFORTUNATO C. LEYNESFORTUNATO C. LEYNESFORTUNATO C. LEYNESChairman

Professional Regulatory

Board of Electrical Engineering

Professional Regulation Commission


Short Circuit (Fault) Analysis

FAULT-PROOF SYSTEM not practical neither economical faults or failures occur in any power system

In the various parts of the electrical network under short circuit or unbalanced condition, the determination of the magnitudes and phase angles Currents Voltages Impedances


Application of Fault Analysis

1. The determination of the required mechanical strength of electrical equipment to withstand the stresses brought about by the flow of high short circuit currents

2. The selection of circuit breakers and switch ratings

3. The selection of protective relay and fuse ratings


4. The setting and coordination of protective devices

5. The selection of surge arresters and insulation ratings of electrical equipment

6. The determination of fault impedances for use in stability studies

7. The calculation of voltage sags caused resulting from short circuits



8. The sizing of series reactors to limit the short circuit current to a desired value

9. To determine the short circuit capability of series capacitors used in series compensation of long transmission lines

10. To determine the size of grounding transformers, resistances, or reactors



Per Unit Calculations


Three-phase Systems

Φ

−

Φ

−

=

×=

3

2

3

2

),(

1000),(

MVAbase

kVvoltagebaseZ

kVAbase

kVvoltagebaseZ

LLB

LLB


Per Unit Quantities

)(

)(

)(

)(

B

pu

B

pu

B

pu

ZimpedanceBase

impedanceactualZ

kVVoltageBase

kVvoltageactualV

ICurrentBase

currentactualI

=

=

=


Changing the Base of Per Unit Quantities

( )

( )

( )

][

][

][

2

][

][

][

2

][][

][

2

][

][

)(

1000

1000)(

1000

)(,

newB

newpu

new

new

newB

old

oldoldpu

old

old

oldpu

Z

ZZ

kVAbase

kVbaseZ

kVAbase

kVbaseZZ

kVAbase

kVbase

ZimpedanceactualZ

Ω=

×=

×=Ω

×

Ω=


Changing the Base of Per Unit Quantities

( )

( )

( )

][

][

][

2

][

][

][

2

][][

][

2

][

][

)(

1000

1000)(

1000

)(,

newB

newpu

new

new

newB

old

oldoldpu

old

old

oldpu

Z

ZZ

kVAbase

kVbaseZ

kVAbase

kVbaseZZ

kVAbase

kVbase

ZimpedanceactualZ

Ω=

×=

×=Ω

×

Ω=


kVA Base for Motors

Induction < 100 hp1.00

Synchronous 1.0 pf0.80

Induction > 1000 hp0.90

Induction 100 < 999 hp0.95

Synchronous 0.8 pf1.00

hp ratingkVA/hp


SYMMETRICAL COMPONENTS


1cV 1aV

1bV

Positive Sequence

2aV

2cV

2bV

Negative Sequence

Unbalanced PhasorsZero Sequence

000 cbaVVV ==

cV

aV

bV

0cV

1cV

2cV

0bV

2bV

1bV

1aV

2aV

0aV


Symmetrical Components of Unbalanced Three-phase Phasor

( )

( )

( )cbaa

cbaa

cbaa

aVVaVV

VaaVVV

VVVV

++=

++=

++=

2

2

2

1

0

3

1

3

1

3

1

2

2

10

21

2

0

210

aaac

aaab

aaaa

VaaVVV

aVVaVV

VVVV

++=

++=

++=


In matrix form:

Symmetrical Components of Unbalanced Three-phase Phasor

=

2

1

0

2

2

1

1

111

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V

=

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


Sequence Networks

Power System Short Circuit Calculations


Fault Point

The fault point of a system is that point to which the unbalanced connection is attached to an otherwise balanced system.


Definition of Sequence Networks

Positive-sequence Network

Ea1 = Thevenin’s equivalent voltage as seen at the fault point

Z1 = Thevenin’s equivalent impedance as seen from the fault point

1111 ZIEV aaa −=

Z1

Ea1

Ia1

Va1

+

-


Negative-sequence Network

Z2 = Thevenin’s equivalent negative-sequence impedance as seen at the fault point


222 ZIV aa −=Z

2

Ia2

Va2

+

-


Zero-sequence Network

Z0 = Thevenin’s equivalent zero-sequence impedance as seen at the fault point


000 ZIV aa −= Z0

Ia0

Va0

+

-


Sequence Network Models ofPower System Components



Synchronous Machines(Positive Sequence Network)

djxZ ''1 =


Synchronous Machines(Negative Sequence Network)

2

''''2

qd xxjZ

+=

Where:

Z2

Ia2

Va2

+

-

x”d = direct-axis sub-transient reactance

x”q = quadrature-axis sub-transient reactance


Synchronous Machines(Zero Sequence Network)

Solidly-Grounded Neutral

jX0

I0

n

ground


Impedance-Grounded Neutral


jX0

I0

n

3Zg

ground


Ungrounded-Wye or Delta Connected Generators


jX0

I0

n

ground


Two-Winding Transformers(Positive Sequence Network)

Standard

Symbol

P S

ZPS

Equivalent

Positive-seq. Network

P S


Three-Winding Transformers(Positive Sequence Network)

tsst

tppt

spps

ZZZ

ZZZ

ZZZ

+=

+=

+= ( )

( )

( )psstptt

ptstpss

stptpsp

ZZZZ

ZZZZ

ZZZZ

−+=

−+=

−+=

2

1

2

1

2

1

PT

Standard

Symbol

Zp

P

ZS

S

Zt

T

Equivalent Positive-Sequence

Network

S


Transformers(Negative Sequence Network)

The negative-sequence network of two-winding and three-winding transformers are modeled in the same way as the positive-sequence network since the positive-sequence and negative-sequence impedances of transformers are equal.


Simplified Derivation of Transformer Zero-Sequence Circuit Modeling

P Q

S1 = 1 and S3 = 0or

S2 = 1 or S4 = 0

Grounded wye

S1 = 0 and S3 = 0or

S2 = 0 and S4 = 0

Ungrounded wye

S1 = 0 and S3 = 1or

S2 = 0 and S4 = 1

Delta

(Thanks to Engr. Antonio C. Coronel, Retired VP, Meralco, and former member, Board of Electrical Engineering)



S1 = 1S2 = 1S3 = 0S4 = 0

P Q

Z

S1

S3 S4

S2

Grounded wye – Grounded wye



S1 = 1S2 = 0S3 = 0S4 = 0

P Q

Grounded wye – Ungrounded wye



S1 = 1S2 = 0S3 = 0S4 = 1

P Q

Grounded wye – Delta



S1 = 0S2 = 0S3 = 0S4 = 0

P Q

Delta – Delta


Transformers(Zero-Sequence Circuit Model)

TransformerConnection

Zero-SequenceCircuit Equivalent

P Q ZPQ

P Q

P Q ZPQ

P Q



QP ZPQ

P Q

P Q ZPQ

P Q







P Q

R

ZP

ZQ

ZR

P

R

Q

P Q ZPQ

P Q






ZP ZQ

ZR

P

R

Q

P Q

R

QP

R

ZP

ZQ

ZR

P

R

Q




Transmission Lines(Positive Sequence Network)

ra jXL

Z1


Transmission Lines(Negative Sequence Network)

The same model as the positive-sequence network is used for transmission lines inasmuch as the positive-sequence and negative-sequence impedances of transmission lines are the same


Transmission Lines(Zero Sequence Network)

The zero-sequence network model for a transmission line is the same as that of the positive- and negative-sequence networks. The sequence impedance of the model is of course the zero-sequence impedance of the line. This is normally higher than the positive- and negative-sequence impedances because of the influence of the earth’s resistivity and the ground wire/s.



Classification of Power SystemShort Circuits


Shunt Faults

Single line-to-ground faults

Double line-to-ground faults

Line-to-line faults

Three-phase faults


Series Faults

One-line open faults

Two-line open faults


Combination of Shunt andSeries Faults

Single line-to-ground and one-line open

Double line-to-ground and one-line open faults

Line-to-line and one-line open faults

Three-phase and one-line open faults


Single line-to-ground and two-line open faults

Double line-to-ground and two-line open faults

Line-to-line and two-line open faults

Three-phase and two-line open faults

Combination of Shunt andSeries Faults


Balanced Faults

Symmetrical or

Three-Phase Faults


Derivation of Sequence Network Interconnections

Boundary conditions:

fccfbbfaaF

cba

ZIVZIVZIVV

III

−=−=−=

=++ 0 Eq’n (1)

Eq’n (2)

Zf

Zf

Vf

System A System B

A

B

C

Va

Vb

Vc

Ia

Ib

Ic

Zf


Z0

Ia0

Va0

+

-

Z1

Ea1

Ia1

Va1

+

-

Zf

Z2

Ia2

Va2

+

-

Zf

1

11

1

11

210

1

11

,0

Z

EII

Z

ZZ

EII

IIII

ZZ

EI

aaf

f

aaf

aaaf

f

aa

f

==

=

+==

++=

+=


Unbalanced Faults

Single Line-to-Ground Faults


0

0

==

==

faa

cb

ZIV

II

Zf

System A System B

A

B

C

Va

Vb

Vc

Ia

Ib

Ic




f

aaaa

ZZZZ

EIII

3210

1210 +++===

Z1

Ea1

Ia1

Va1

+

-

Z2

Ia2

Va2

+

-

Z0

Ia0

Va0

+

-

3Zf

10

1

21

01210

10

1210

21

210

1

210

2

3

0

33

2

0

ZZ

EII

ZZandZIf

IIIIIII

ZZ

EIII

ZZ

ZZZ

EIII

ZIf

aaf

f

aaaaaaf

aaaa

a

aaa

f

+==

==

==++==

+===

=

++===

=


Unbalanced Faults

Line-to-Line Faults


fbcbcfbb

cb

a

ZIVVVZIV

II

I

=−=−

−=

=

or ;

0

Zf

System A System B

A

B

C

Va Vb Vc

Ia

Ib

Ic




1

11

21

21

11

2

0

Z

EI

ZZandZIf

ZZZ

EI

aa

f

f

aa

=

==

++=

( ) 11

2

21

21

2

0

3

;0

aaf

aaao

aaacbf

IjIaaI

III

aIIaIIII

currentfaultThe

−=−=

−==

++=−==

Z1

Ea1

Ia1

Va1

+

-

Zf

Z2

Ia2

Va2

+

-

Z0

Ia0

Va0

+

-


]3[][

1

1]3[

1

1

1

1

1

1

21

2

3

2

3

2

3

0 if

23

with ,thus

φ

φ

fLLf

af

aaf

f

f

af

II

Z

EI

Z

E

Z

EjI

Z

ZZ

EjI

ZZ

=

=

=−=

=

+−=

=

−


Unbalanced Faults

Double-to-line Ground Fault


( )( ) gcbfcc

gcbfbb

a

ZIIZIV

ZIIZIV

I

++=

++=

= 0 Eq’n BC-1

Eq’n BC-2

Eq’n BC-3

Zf

System A System B

A

B

C

Va

Vb

Vc

Ia

Ib

Ic

Zf

Vf

Zg (I

b+I

c)




( )( )gf

gff

f

aa

ZZZZ

ZZZZZZZ

EI

32

3

02

02

1

11

+++

+++++

=

Z0

Ia0

Va0

+

-

Z2

Ia2

Va2

+

-

Zf

Zf

Zf+3Z

g

Z1

Ea1

Ia1

Va1

+

-


Negative-sequence Component:

+++

++−=

gf

gf

aaZZZZ

ZZZII

32

3

02

0

12

Zero-sequence Component:

The fault current( ) ( )

( ) ( )( ) ( ) ( )

( )0

210210

210210

2

2

1

2

0

2

2

1021

2

0

3 thus,

or ;0but

2112

2

af

aaaaaa

aaaaaaf

aaaf

aaaaaacbf

II

IIIIII

IIIIIII

IaaIaaII

IaaIIaIIaIIII

=

+−==++

+−=−+−+=

++++=

+++++=+=

+++

+−=

gf

f

aaZZZZ

ZZII

3202

2

10


( )

01

2

1

10

01

0

01

011

1

01

012

01

2

1

101

01

011

11

21

2

2

0

ZZZ

EZ

ZZ

Z

ZZ

ZZZ

E

ZZ

ZII

ZZZ

EZZ

ZZ

ZZZ

EI

ZZandZZIf

a

aaa

aaa

gf

+−

=

+

++

−=

+−=

++

=

++

=

===


01

10

01

1

01

2

1

11

01

1

01

011

1

01

110

2

33

22

ZZ

EII

ZZ

E

ZZZ

EZ

ZZ

Z

ZZ

ZZZ

E

ZZ

ZII

aaf

aa

aaa

+==

+=

+=

+

++

=

+−=

( )

01

2

1

10

01

0

01

011

1

01

012

01

2

1

101

01

011

1

1

21

2

2

0

ZZZ

EZ

ZZ

Z

ZZ

ZZZ

E

ZZ

ZII

ZZZ

EZZ

ZZ

ZZZ

EI

ZZandZZIf

a

aaa

aa

a

gf

+−

=

+

++

−=

+−=

++

=

++

=

===


Voltage Rise Phenomenon

Single-to-line Ground Fault


Unfaulted Phase B Voltage During Single Line-to-Ground Faults


+

−−=

+

−

−=

+

−

−=

+

−−=

+−

+−+

+−=

++=

1

0

1

0

2

1

01

1

0

1

0

2

1

1

0

1

0

1

12

1

01

102

1

1

01

1

1

01

1

1

2

0

01

1

21

2

0

2

1

;R and R s,resistance neglecting

2

1

2

1

2

222

X

X

X

X

aEV

Z

Z

Z

Z

aEV

Z

Z

Z

Z

Z

ZaE

ZZ

ZZaEV

ZZZ

EaZ

ZZ

EEaZ

ZZ

EV

aVVaVV

ab

ab

aab

aa

a

a

b

aaab



Neglecting resistances R0 & R1

0

5

10

15

20

25

30

35

40

45

50

-10 -6 -3 -2.8 -2.6 -2.4 -2.2 -2 -1.8 -1.6 -1.4 -1.2 -1 0 1 2 5 9 25 45 65 85

X0/X1

Phase B

Voltage (p.u

.)


Fault MVA

10

)(

1

)3(

2

3

1

:p.u.in fault ground-to-line singefor

:p.u.in fault phasefor three

p.u.; 1.01

for where,

ZZI

ZI

MVAIMVA

SLGF

F

baseFF

aE

+=

=

×=

−

=

φ


Fault MVA

( ) ( )

( )

( ) ( )

( )

( )10

01

3

1

33

23

..3

2

.).(

:

..1

.).(

:

ZI

Z

upI

ZZ

MVAupIMVA

MVAfaultgroundtolineSingle

upI

Z

MVAupIMVA

MVAfaultphaseThree

SLGF

SLGF

baseSLGFSLGF

F

baseFF

−=

=+

×=

−−

=

×=

−

φ

φφ


Assumptions Made to SimplifyFault Calculations

1. Pre-fault load currents are neglected.

2. Pre-fault voltages are assumed equal to 1.0 per unit.

3. Resistances are neglected (only for 115kV & up).

4. Mutual impedances, when not appreciable are neglected.

5. Off-nominal transformer taps are equal to 1.0 per unit.

6. Positive- and negative-sequence impedances are equal.


Outline of Procedures for Short Circuit Calculations

1 Setup the network impedances expressed in per unit on a common MVA base in the form of a single-line diagram

2 Determine the single equivalent (Thevenin’s) impedance of each sequence network.

3 Determine the distribution factor giving the current in the individual branches for unit total sequence current.



4 Interconnect the three sequence networks for the type of fault under considerations and calculate the sequence currents at the fault point.

5 Determine the sequence current distribution by the application of the distribution factors to the sequence currents at the fault point

6 Synthesize the phase currents from the sequence currents.



7 Determine the sequence voltages throughout the networks from the sequence current distribution and branch impedances

8 Synthesize the phase voltages from the sequence voltage components

9 Convert the pre unit currents and voltages to actual physical units


CIRCUIT BREAKING SIZING(Asymmetrical Rating Factors)

Momentary Rating

Multiplying Factor = 1.6

Interrupting Rating

Multiplying Factor8 cycles = 1.0

5 cycles = 1.1

3 cycles = 1.2

1 ½ cycles = 1.5


EXAMPLE PROBLEM

In the power system shown, determine the momentary and interrupting ratings for primary and secondary circuit breakers of transformer T2.


Solution:

pux

pux

puI

puI

SLGF

F

05.0125.0210

3

125.08

1

10100

1000

8100

800

:100MVA system@ 69kV Equivalent

0

1

)(

)3(

=×−=

==

=

=

=

=φ

pux

pux

pux

pux

pux

3333.35.4

10015.0

4.5MVAor

450050000.9kVA :M2 M1,

80.075

10006.0

:T3 T3, T2,

24.025

10006.0

40.025

10010.0"

:G1

2667.030

10008.0

:T1

B

B

0

=

=

=

=×=

=

=

=

=

=

=

=

=


Solution:Positive sequence network:


Solution:

Zero sequence network:


Improper Sequence Network Models






Incorrect Sequence Network Models


QUESTIONS?


Documents

SHORT CIRCUIT CALCULATIONS REVISITEDiiee.org.ph/wp-content/uploads/2014/10/Short-Circuit-Calculations.pdfSHORT CIRCUIT CALCULATIONS REVISITED FORTUNATO C. LEYNES Chairman Professional