fundamentals of power system modeling · 2019. 1. 10. · • per-unit calculation is more...

FUNDAMENTALS OF POWER SYSTEM MODELING

1

FORTUNATO C. LEYNESMMBA, PEE, I IEE Fellow , APEC Engineer

ASEAN Chartered Prof. EngineerAsst. Professor, Department of Electrical EngineeringFaculty of Engineering, UNIVERSITY OF STO. TOMAS

43rd ANNUAL NATIONAL CONVENTIONINSTITUTE OF INTEGRATED ELECTRICAL ENGINEERS OF THE PHILS., INC.

SMX CONVENTION CENTERNOVEMBER 16, 2018

OUTLINE OF PRESENTATION

MODELS AND SIMULATIONSPOWER SYSTEM MODELING – SHORT HISTORYPOWER SYSTEM SIMULATIONPER UNIT CALCULATIONSSYMMETRICAL COMPONENTSSEQUENCE IMPEDANCESSEQUENCE NETWORKS

2

MODELS AND SIMULATIONS

WHAT IS A MODEL?o A MODEL OF A SYSTEM IS ANYTHING AN

“EXPERIMENT” CAN BE APPLIED IN ORDER TO ANSWER QUESTIONS ABOUT THE SYSTEM;

o INSTEAD, SIMPLIFIED EXPERIMENTS ARE APPLIED INTO THE SYSTEM;

o THUS, WE HAVE A “SIMPLIFIED SYSTEM” THAT REFLECTS THE REAL SYSTEM.

3Ref.: Power System Simulation

Associate Prof., DocentKTH Royal Institute of TechnologyStockholm, Sweden


THERE ARE MANY TYPES OF MODELS – IN ENGINEERING, WE MAINLY DEAL WITH TWO TYPES:o PHYSICAL MODEL: A PHYSICAL OBJECT THAT

MIMICS SOME PROPERTIES OF A REAL SYSTEM TO HELP US ANSWER QUESTIONS ABOUT THE SYSTEM.

o MATHEMATICAL MODEL: A DESCRIPTION OF THE SYSTEM WHERE THE RELATIONSHIPS BETWEEN VARIABLES OF THE SYSTEM ARE EXPRESSED IN MATHEMATICAL FORM - THE FORM: EQUATIONS!




oMODEL KNOWLEDGE IS STORED IN BOOKS AND HUMAN MINDS WHICH COMPUTERS CANNOT ACCESS - THIS MEANS THAT EQUATIONS NEED TO BE TRANSLATED INTO COMPUTER READABLE FORM - THE FORM: COMPUTER PROGRAMS.

oTHE ARTIFACTS REPRESENTED BY MATHEMATICAL MODELS IN A COMPUTER ARE CALLED VIRTUAL PROTOTYPES (IN MOST INDUSTRIES AT LEAST).



MODELS AND SIMULATIONSWHAT IS SIMULATION?

oSIMULARE FROM LATIN, MEANS TO PRETEND. A SIMULATION IS AN EXPERIMENT PERFORMED ON A MODEL.

oWE FOCUS ON MODELS THAT CAN BE WRITTEN IN COMPUTER-REPRESENTABLE FORMS.

oHENCE, WE PERFORM NUMERICAL EXPERIMENTS BY PERFORMING COMPUTATIONS IN A COMPUTER.




THE VALUE OF SIMULATION IS COMPLETELY DEPENDENT ON HOW WELL THE MODEL REPRESENTS THE REAL SYSTEM REGARDING THE QUESTIONS TO BE ANSWERED!



WHY DO WE DEVELOP MODELS AND PERFORM SIMULATIONS?

TO REDUCE THE LIFETIME COST OF A SYSTEM.

o IN REQUIREMENTS: TRADE-OFF STUDIESo IN TEST AND DESIGN: FEWER PROTO –

TYPESo IN TRAINING: AVOID ACCIDENTSo IN OPERATION: ANTICIPATE PROBLEMS



POWER SYSTEM SIMULATION – SHORT HISTORY

1929 – THE NEED FOR COMPUTATIONAL AIDS LED TO THE DESIGN OF A SPECIAL PURPOSE ANALOG COMPUTER (AC NETWORK ANALYZER), AN OUTGROWTH OF THE DC CALCULATING BOARDS USED IN THE VERY EARLIEST POWER SYSTEM ANALYSIS

LATE 1940S – THE EARLIEST APPLICATION OF DIGITAL COMPUTERS TO SOLVE POWER SYSTEM PROBLEMS WAS USED

MID 1950S – LARGE-SCALE DIGITAL COMPUTERS BECAME AVAILABLE




BACK IN THE 60S & 70S, ALL SCIENTIFIC COMMUNITIES WERE IN THE SAME CONDITION: MOST SOFTWARE WAS OPEN SOURCE DE FACTO AND WAS SHARED AMONG EXPERTS IN THE AREA.

SOFTWARE FOR POWER FLOW AND TRANSIENT STABILITY BECAME AVAILABLE AROUND MID 60S.

PROGRAMS RAN IN MAINFRAMES, GE AND WESTINGHOUSE WERE THE MAIN SERVICE PROVIDERS.

LARGE COMPANIES THAT HAD MAINFRAMES (FOR BILLING) STARTED LOOKING INTO USING THEM FOR POWER SYSTEM STUDIES.




BY THE LATE 60S MANY UTILITIES IN THE USA HAD DEVELOPED THEIR OWN POWER FLOW AND STABILITY PROGRAMS: PHILADELPHIA ELECTRIC CO. (PECO) AND BPA'S BECAME WIDELY USED PROGRAMS FOR PLANNING.

THESE PROGRAMS AND THEIR SOURCE CODE WERE FREELY GIVEN AWAY (THE TERM "OPEN SOURCE" DID NOT EXIST YET), AND THE BPA SW WAS IN THE PUBLIC DOMAIN BECAUSE IT WAS DEVELOPED BY A US GOV’T ENTITY.




BPA AND PECO HAD WELL-KNOWN GROUPS OF POWER ENGINEERS WHO DEVELOPED, MAINTAINED AND IMPROVED THE SW THROUGHOUT THE 70S AND INTO THE 80S.

OTHER POWER COMPANIES THAT USED THESE SOFTWARE, DID NOT HAVE THEIR OWN GROUPS TO SUPPORT IT WHILE BPA AND PECO COULD NOT PROVIDE THE MUCH NEEDED TECHNICAL SUPPORT.

THUS, VENDORS OF PLANNING SW WHO COULD PROVIDE SUCH USER SUPPORT ALSO THRIVED IN PARALLEL.




BY THE LATE 80S EVEN PECO AND BPA DECIDED TO DISBAND THEIR IN-HOUSE EXPERTISE IN SW DEVELOPMENT AND THE USE OF THESE PACKAGES DWINDLED.THERE ARE FEW TRACES OF THESE PROGRAMS LEFT, EXCEPT FOR THEIR MENTION IN THE TECHNICAL LITERATURE FROM THOSE DAYS.

PRESENT – THE DIGITAL COMPUTER IS AN INDISPENSABLE TOOL IN POWER SYSTEM PLANNING WHILE DIFFERENT POWER SYSTEM ANALYSIS SOFTWARE ARE AVAILABLE IN THE MARKET



POWER SYSTEM SIMULATIONPOWER SYSTEM SIMULATION SOFTWARE'S ARE A CLASS OF COMPUTER SIMULATION PROGRAMS THAT FOCUS ON THE OPERATION OF ELECTRICAL POWER SYSTEMS. THESE TYPES OF COMPUTER PROGRAMS ARE USED IN A WIDE RANGE OF PLANNING AND OPERATIONAL SITUATIONS FOR:

ELECTRIC POWER GENERATION - NUCLEAR, CONVENTIONAL, RENEWABLECOMMERCIAL FACILITIESUTILITY TRANSMISSIONUTILITY DISTRIBUTIONRAILWAY POWER SYSTEMSINDUSTRIAL POWER SYSTEMS

14

POWER SYSTEM SIMULATIONAPPLICATIONS OF POWER SYSTEM SIMULATION INCLUDE:

LONG-TERM GENERATION AND TRANSMISSION EXPANSION PLANNING

SHORT-TERM OPERATIONAL SIMULATIONS

MARKET ANALYSIS (E.G., PRICE FORECASTING)

THESE PROGRAMS TYPICALLY MAKE USE OF MATHEMATICAL OPTIMIZATION TECHNIQUES SUCH LINEAR PROGRAMMING, QUADRATIC PROGRAMMING, AND MIXED INTEGER PROGRAMMING.

15

MOST COMMON POWER SYSTEM STUDIES

LOAD FLOW STUDIESSHORT-CIRCUIT STUDIESSTABILITY STUDIESINSULATION COORDINATIONSYSTEM PROTECTION COORDINATIONELECTROMAGNETIC TRANSIENTSHARMONIC ANALYSISMOTOR-STARTING STUDIESCABLE AMPACITY STUDIESGROUND MAT STUDIESARC FLASH ANALYSIS

16

TIME DOMAIN OF POWER SYSTEM DYNAMICS



PER UNIT CALCULATIONS

18


ADVANTAGES OF USING PER UNIT CALCULATIONS

• VALUES IN PER UNIT QUANTITIES ARE MUCH EASIER TO HANDLE

• IMPEDANCES BEING REFERRED TO ONE SIDE OF THE TRANSFORMER DUE TO TRANSFORMATION RATIO IS NOT A PROBLEM

• MANUFACTURERS SPECIFY THE IMPEDANCES OF THEIR EQUIPMENT IN PERCENT (OR PER-UNIT) USING THE NAMEPLATE RATING OF THE EQUIPMENT.

19


ADVANTAGES OF USING PER UNIT CALCULATIONS (CONT’D):

• THE PER-UNIT IMPEDANCES OF ELECTRICAL EQUIPMENT OF THE SAME TYPE BUT DIFFERENT RATINGS USUALLY LIE WITHIN A NARROW RANGE. THIS MAKES THE DETECTION OF AN ERRONEOUS IMPEDANCE DATA EASY. ALSO, IF THE IMPEDANCE OF A PARTICULAR EQUIPMENT IS NOT KNOWN, IT IS ACCEPTABLE FOR MOST STUDIES TO SELECT FROM A RANGE OF TABULATED TYPICAL VALUES.

20



• PER-UNIT REPRESENTATION YIELDS MORE RELEVANT INFORMATION AND EASILY CORRELATED DATA.

• NETWORK CALCULATIONS ARE THE SAME FOR SINGLE-PHASE AND THREE-PHASE SYSTEMS. THERE IS LESS CHANCE OF MIX-UP BETWEEN PHASE AND LINE VOLTAGES, SINGLE-PHASE AND THREE-PHASE POWERS, AND PRIMARY AND SECONDARY VOLTAGES.

21



• PER-UNIT CALCULATION IS MORE CONVENIENT TO USE WHEN THE SOLUTION REQUIRES A DIGITAL COMPUTER

POWER SYSTEM COMPONENTS, I.E., GENERATORS, TRANSFORMERS, TRANSMISSION LINES, ETC. ARE MODELED WITH PER UNIT IMPEDANCES IN THE DIFFERENT POWER SYSTEM APPLICATIONS LIKE LOADFLOW, SHORT CIRCUIT, POWER SYSTEM STABILITY, ELECTROMAGNETIC TRANSIENTS, ETC.

22

CHOICE OF PER-UNIT VALUES

• CHOOSE ANY TWO OF THE ELECTRICAL PARAMETERS. IN GENERAL, THE BASE VOLT-AMPERES AND BASE VOLTAGE ARE CHOSEN.

NOTE: For actual power systems, equipment are rated in kilovolts, kVA or MVA. Thus, the bases are often expressed in kV and MVA or kVA.

• CALCULATE THE BASE IMPEDANCE AND BASE CURRENT

NOTE: The base MVA or kVA will also serve as base for true/real power and reactive power. The base Z will also be used as base for resistance and reactance.

23

24

SINGLE-PHASE SYSTEMS

LN

LNB

LNB

B

LNB

kVvoltagebasekVAbasekVvoltagebaseZ

kVvoltagebasekVAbaseI

IcurrentbaseVvoltagebaseZ

,

1000,,

,,

,

1

1

SINGLE-PHASE SYSTEMS

25

11

11

1

21

2

,,

,

1000,

kVAbasekVArPowerBasekVAbasekWPowerBase

MVAbasekVvoltagebaseZ

kVAbaseVkvoltagebaseZ

LNB

LNB

26

THREE-PHASE SYSTEMS

LL

LLB

LLB

B

LLB

kVvoltagebasekVAbasekVvoltagebaseZ

kVvoltagebasekVAbaseI

IcurrentbasekVvoltagebaseZ

,3

10003/,

,3

,10003/,

3

3

27

THREE-PHASE SYSTEMS

33

33

3

23

2

,,

),(

1000),(

kVAbasekVArPowerBasekVAbasekWPowerBase

MVAbasekVvoltagebaseZ

kVAbasekVvoltagebaseZ

LLB

LLB

PER UNIT QUANTITIES

28

)(

)()(

)(

Bpu

Bpu

Bpu

ZimpedanceBaseimpedanceactualZ

kVVoltageBasekVvoltageactualV

ICurrentBasecurrentactualI

Quantities in percent are per unit ×100.

PER UNIT QUANTITIES

29

)()(

)()(

Bpu

Bpu

kVAPowerBasekVArpowerreactiveactualQ

kVAPowerBasekWpowertrueactualP

TRANSFORMER EQUIVALENT IMPEDANCE IN P.U. SYSTEM

Zp Zs

IpVp Vs

n:1

n = transformation ratio

Is

30

p

s

s

p

spsp

II

VV

n

valuesratedIIVV ,,,

TRANSFORMER EQUIVALENT IMPEDANCE IN P.U. SYSTEM

31

p

pBp

Bp

sp

Bp

eqpppu

speqp

IV

Zwhere

ZZnZ

ZZ

Z

ZnZZ

,

2

2

2,

since,

nZ

ZthenIV

Z BpBs

p

pBp

Bs

sp

Bs

eqsspu

sp

eqs

Z

ZnZ

ZZ

Z

ZnZ

Z

2

2

p

p

p

p

p

p

s

sBs I

VnIn

VInnV

IVZ 22

1/

Bp

sp

Bp

sp

spu ZZnZ

nZ

ZnZ

Z2

2

2

ppuspu ZZ

CHANGING THE BASE OF PER UNIT QUANTITIES

32

][][

][

2][

][

][

2][][

][

2][][

][

)(

1000

1000)(

1000)(,)(,

newBnewpu

new

newnewB

old

oldoldpu

old

oldoldBoldpu

ZZZ

kVAbasekVbase

Z

kVAbasekVbaseZ

Z

kVAbasekVbase

ZimpedanceactualZ

ZimpedanceactualZ

CHANGING THE BASE OF PER UNIT QUANTITIES

33

][

2][

][

2][][

][ 1000

1000

new

new

old

oldoldpu

newpu

kVAbasekVbasekVAbasekVbaseZ

Z

][

][

2

][

][][][

][

][

2

][

][][][

old

new

new

oldoldpunewpu

old

new

new

oldoldpunewpu

MVAbaseMVAbase

kVbasekVbase

ZZ

kVAbasekVAbase

kVbasekVbase

ZZ

kVA/hp hp rating

1.00 Induction < 100 hp

1.00 Synchronous 0.8 pf

0.95 Induction 100 < 999 hp

0.90 Induction 1000 hp

0.80 Synchronous 1.0 pf34

kVA BASE FOR MOTORS

SYMMETRICAL COMPONENTS

35

BALANCED THREE-PHASE SYSTEM

THE FOLLOWING ARE THE BASIC CHARACTERISTICS OF BALANCED POLYPHASE SYSTEMS:

1) THE MAGNITUDES OF THE VOLTAGES AND CURRENTS IN EACH PHASE ARE EQUAL.

2) THE PHASE DISPLACEMENTS OF THE VOLTAGE AND THE CURRENT IN EACH PHASE ARE ALSO EQUAL.

3) THE MUTUAL REACTIONS BETWEEN THE PHASES ARE REPRESENTED BY THE EQUIVALENT SELF-IMPEDANCES OF EACH PHASE BECAUSE OF SYMMETRY.

4) THE SOLUTION OF ONE PHASE YIELDS THE SOLUTION OF OTHER PHASES AND THE TOTAL SOLUTION.

36


IN DEALING WITH NORMAL OR NEAR NORMAL OPERATION OF POWER SYSTEMS, THE SLIGHT UNBALANCES ARE IGNORED AND THEREFORE, BALANCED OPERATION IS ASSUMED, I.E., BALANCED LOADS, BALANCED GENERATOR OUTPUTS, AND BALANCED LINE/TRANSFORMER PARAMETERS

37


Zm

ZmZm

Zm

ZmZm

Ia

Ic

IbEa

Ec

Eb

ZL

ZL

ZL

38c

b

a

mLsmm

mmLsm

mmmLs

c

b

a

III

ZZZZZZZZZZZZZZZ

EEE cba

bca

acb

cba

IIIIIIIIIIII 0

Zs

Zs

Zs

mbmamcLcscc

mcmambLbsbb

mcmbmaLasaa

ZIZIZIZIZIEZIZIZIZIZIEZIZIZIZIZIE


39

c

b

a

mLs

mLs

mLs

c

b

a

III

ZZZZZZ

ZZZ

EEE

200020002

The foregoing gives us a very simple single-phase solution!

120120

2

ac

ab

mLs

aa

IIII

ZZZEI

UNBALANCED POLYPHASE CIRCUITS

• SAME SIMPLIFICATION AS IN BALANCED SYSTEMS IS NOT POSSIBLE

40

Z

Z

Z

ZF

BalancedSource

BalancedLoad

UNBALANCED POLYPHASE CIRCUITS

• CLASSICAL METHODS OF ANALYSIS USING KIRCHHOFF S LAWS AND SIMULTANEOUS EQUATIONS ARE VERY DIFFICULT TO SOLVE AND OFTEN IMPOSSIBLE

• UNBALANCED REACTIONS BETWEEN PHASES• WHERE ROTATING MACHINES ARE INVOLVED, IT

IS NECESSARY TO INTRODUCE IMPEDANCES RELATING THE STATOR AND ROTOR CIRCUITS

41

UNBALANCED POLYPHASE CIRCUITSALTERNATIVE SOLUTION METHODS

SYMMETRICAL COMPONENTS

ALPHA, BETA, ZERO COMPONENTS (POPULARIZED BY EDITH CLARKE OF GENERAL ELECTRIC)

POSITIVE-PLUS-NEGATIVE, POSITIVE-MINUS-NEGATIVE, ZERO COMPONENTS

ONLY THE METHOD OF SYMMETRICAL COMPONENTS WILL BE DISCUSSED

42

SYMMETRICAL COMPONENTSCHARLES LEGEYT FORTESCUE DISCUSSED IN HIS 114-PAGE PAPER “METHOD OF SYMMETRICAL COORDINATES APPLIED TO THE SOLUTION OF POLYPHASE NETWORKS”, WHICH WAS PUBLISHED IN 1918 BY THE THEN AIEE [NOW IEEE]), THAT ANY SET OF N UNBALANCED VECTORS CAN BE REPRESENTED BY N SETS OF BALANCED VECTORS.• BALANCED SYSTEM CAN BE SIMULATED WITH

SINGLE PHASE PARAMETERS. EASIER TO ANALYZE AND COMPUTE.

• THREE PHASE UNBALANCED VECTORS THREE BALANCED “SEQUENCE VECTORS.”

43

SYMMETRICAL COMPONENTS –THREE PHASE SYSTEM

• POSITIVE-SEQUENCE COMPONENTS CONSISTING OF THREE PHASORS EQUAL IN MAGNITUDE DISPLACED FROM EACH OTHER BY 120° IN PHASE AND HAVING THE SAME SEQUENCE AS THE ORIGINAL PHASORS

• NEGATIVE-SEQUENCE COMPONENTS CONSISTING OF THREE PHASORS EQUAL IN MAGNITUDE, DISPLACED FROM EACH OTHER BY 120° IN PHASE AND HAVING A PHASE SEQUENCE OPPOSITE THAT OF THE ORIGINAL PHASORS

44

SYMMETRICAL COMPONENTS• ZERO-SEQUENCE COMPONENTS CONSISTING

OF THREE PHASORS EQUAL IN MAGNITUDE AND WITH ZERO PHASE DISPLACEMENT BETWEEN EACH OTHER

• THE UNBALANCED PHASOR IS EQUAL TO THE PHASOR SUM OF THE SYMMETRICAL COMPONENTS OF EACH PHASE, I.E.,

021

021

021

cccc

bbbb

aaaa

VVVVVVVVVVVV

45

46

1cV 1aV

1bVPositive Sequence

2aV

2cV

2bV

Negative Sequence

Zero Sequence Unbalanced Phasors

000 cba VVV

cV

aV

bV

0cV

1cV

2cV

0bV

2bV1bV

1aV

2aV

0aV

47

Function:Any phasor that is multiplied by the operator a is rotated counterclockwise by 120°. This is shown by the phasor diagram on the right:

1

a2

a

-a

-a2

-1

aX

X

OPERATOR aDEFINITION:

The operator a is a phasor with a magnitude equal to unity with an angle of 120°, i.e., a=1/120°

LETTER EXPRESSION

POLAR FORM

RECTANGULAR FORM

a 1201 866.0500.0 j 2a 2401 866.0500.0 j 3a 13601 01 j

12 aa 0 0 12 aa -1 -1

23 1 aaaa 2401 866.0500.0 j 48

OPERATOR a

EQUALITIES OF OPERATOR a

SYMMETRICAL COMPONENTS OF UNBALANCED THREE-PHASE PHASOR

49

(3)Eqn (2)Eqn (1)Eqn

210

210

210

cccc

bbbb

aaaa

VVVVVVVVVVVV

1cV 1aV

1bVPositive Sequence

11

12

1

ac

ab

aVVVaV

2aV

2cV

2bV

22

2

22

ac

ab

VaVaVV

Negative Sequence Zero Sequence

000 cba VVV


50

cbaa

cbaa

cbaa

aVVaVV

VaaVVV

VVVV

22

21

0

313131

22

10

212

0

210

aaac

aaab

aaaa

VaaVVVaVVaVV

VVVV

In summary:


51

2

1

0

2

2

11

111

a

a

a

c

b

a

VVV

aaaa

VVV

In summary (matrix form):

c

b

a

a

a

a

VVV

aaaa

VVV

2

2

2

1

0

11

111

31

POWER INVARIANCE OF SYMMETRICAL COMPONENTS

52

***ccbbaa IVIVIVS

*22

*11

*00 333 aaaaaa IVIVIVS

Substituting the symmetrical components of the voltages and currents, collect terms, and with 1 + a + a2 = 0, the process yields:

POWER SYSTEM MODELING(SEQUENCE IMPEDANCES)

53

POWER SYSTEM MODELING

54

The sequence impedance of each power system element must be shown in per unit value.

Power system is modeled by an impedance diagram representing the correct sequence network models (positive-, negative-, or zero-sequence)

The power system can better be described through a single-line diagram (SLD)

SEQUENCE IMPEDANCES

55

SEQUENCE IMPEDANCESDEFINITION:

56

1

11

a

a

IVZ

2

22

a

a

IVZ

0

00

a

a

IVZ

Negative-sequence impedance (Z2)

Zero-sequence impedance (Z0)

Positive-sequence impedance (Z2)

Sequence impedances of most power system components, i.e., rotating machines, transformers, etc., except transmission/ distribution lines, are generally expressed in percent or per unit based on equipment ratings (kV and kVA or MVA)

SYNCHRONOUS MACHINESMANUFACTURES PROVIDE THE FOLLOWING DATA:

• ARMATURE RESISTANCE• DIRECT-AXIS REACTANCES• QUADRATURE-AXIS REACTANCES• NEGATIVE-SEQUENCE REACTANCE• ZERO-ZERO REACTANCE

ARMATURE RESISTANCE IS USUALLY VERY SMALL COMPARED WITH THE REACTANCES, HENCE, GENERALLY NEGLECTED FOR SHORT CIRCUIT CALCULATIONS. THE REACTANCES, ON THE OTHER HAND, ARE REFERRED TO THE DIRECT-AXIS AND QUADRATURE-AXIS. THE DIRECT-AXIS REACTANCES ARE COMMONLY USED IN SHORT CIRCUIT CALCULATIONS.

57

SYNCHRONOUS MACHINES –SEQUENCE IMPEDANCES

58

2''''

2qd xx

x

Negative-sequence impedance (salient-pole machines)

Zero-sequence reactance is smaller than the positive-sequence reactance

Positive-sequence impedanceXd = direct-axis synchronous reactanceX d = direct-axis transient reactanceX d = direct-axis subtransient reactance

Xd > X’d > X”d

TYPICAL SYNCHRONOUS GENERATOR PARAMETERS*

59

Turbo-Generators(solid rotor)

Water-Wheel Generators(with dampers)**

Synchronous Condensers Synchronous Motors(general purpose)

Low Ave. High Low Ave. High Low Ave. High Low Ave. HighReactances (in p.u.)

xd 0.95 1.10 1.45 0.60 1.15 1.45 1.50 1.80 2.20 0.80 1.20 1.50

xq 0.92 1.08 1.42 0.40 0.75 1.00 0.95 1.15 1.40 0.60 0.90 1.10

x'd 0.12 0.23 0.28 0.20 0.37 0.50*** 0.30 0.40 0.60 0.25 0.35 0.45

x'q 0.12 0.23 0.28 0.40 0.75 1.00 0.95 1.15 1.40 0.60 0.90 1.10

x"d 0.07 0.12 0.17 0.13 0.24 0.35 0.18 0.25 0.38 0.20 0.30 0.40

x"q 0.10 0.15 0.20 0.23 0.34 0.45 0.23 0.30 0.43 0.30 0.40 0.50

xp 0.07 0.14 0.21 0.17 0.32 0.40 0.23 0.34 0.45

x2 0.07 0.12 0.17 0.13 0.24 0.35 0.17 0.24 0.37 0.25 0.35 0.45

x0* 0.01 0.10 0.02 0.21 0.03 0.15 0.04 0.27Resistances (in p.u.)

ra(dc) 0.0015 0.0050 0.0030 0.0200 0.0020 0.0150r(ac) 0.0030 0.0080 0.0080 0.0150 0.0040 0.0100

r2 0.0250 0.0450 0.0120 0.2000 0.0250 0.0700

Time constants(in seconds)

T'd0 2.80 5.60 9.20 1.50 5.60 9.50 6.00 9.00 11.50

T'd 0.40 1.10 1.80 0.50 1.80 3.30 1.20 2.00 2.80

T"d = T"q 0.02 0.035 0.05 0.01 0.04 0.05 0.02 0.02 0.05

Ta 0.04 0.16 0.35 0.03 0.15 0.25 0.10 0.10 0.30Source: Kimbark [19]. Used with permission from the publisher

* x0 varies from about 0.15 to 0.60 of x"d, depending upon winding pitch**For water-wheel generators without damper windings, x0 is a listed and

x"d = 0.85x'd, x"q = x'q = xq, x2 = (x'd + xq)/2

***For curves shwoing the normal value of x'd of water-wheel-driven generators as a function of kilovolt-ampere rating and speed

*Analysis of Faulted Power System- P. M. Anderson

REACTANCE VALUES FOR INDUCTION MOTORS

60

Subtransient X” (pu)

Induction Motor above 600V 0.17

Induction Motor below 600V 0.25

TRANSFORMERS

61

POSITIVE- AND NEGATIVE- SEQUENCE REACTANCE OF TRANSFORMERS

THE POSITIVE- AND NEGATIVE-SEQUENCE REACTANCES OF TRANSFORMERS ARE EQUAL, REGARDLESS OF THE CONSTRUCTION OF THE TRANSFORMER.

THE IMPEDANCE OF SINGLE-PHASE TRANSFORMERS WHEN CONNECTED IN TREE-PHASE BANK IS THE SAME

62

ZERO-SEQUENCE REACTANCE OF TRANSFORMERS

FOR THREE-PHASE SHELL TYPE TRANSFORMERS, THE ZERO-SEQUENCE REACTANCE IS EQUAL TO THE POSITIVE-SEQUENCE REACTANCE. THE SAME IS TRUE FOR AND SINGLE-PHASE TRANSFORMERS.

THE ZERO-SEQUENCE REACTANCE OF THE THREE-PHASE CORE-TYPE TRANSFORMERS IS SMALLER THAN THE POSITIVE-SEQUENCE REACTANCE DUE TO THE LEAKAGE OF ZERO-SEQUENCE FLUX TO THE TRANSFORMER TANK DURING GROUND FAULTS. 63

TYPICAL PERCENTAGE IMPEDANCES OF 50 HZ THREE-PHASE TRANSFORMERS *

64* J.P. Transformer Handbook

IMPEDANCE VALUES OF THREE-PHASE MEDIUM VOLTAGE TRANSFORMERS

65

VOLTAGE RATING kVA RATING % IMPEDANCE

2.4kV – 13.8kV 300 - 500 Not less than 4.5%

2.4kV – 13.8kV 750 – 2,500 5.75%

General PurposeLess than 600V

15 – 1,000 3% to 5.75%

65

Typical Values for X/R Ratio of Medium Voltage Transformers

X/R = 6

TRANSMISSION LINES

66

TRANSMISSION LINES –SEQUENCE IMPEDANCES

THE POSITIVE- AND NEGATIVE-SEQUENCE IMPEDANCES OF TRANSMISSION LINES ARE EQUAL.THE ZERO-SEQUENCE IMPEDANCE OF TRANSMISSION LINES IS OF HIGHER VALUE THAN THE POSITIVE-SEQUENCE IMPEDANCE DUE TO THE FACT THAT THE ZERO-SEQUENCE CURRENT MUST RETURN THROUGH THE EARTH, OR VIA THE EARTH AND GROUND WIRES, IF THERE ARE ANY.

67

SEQUENCE NETWORKS

68

69

DEFINITION OF SEQUENCE NETWORKSPOSITIVE-SEQUENCE NETWORK

EA1 = THEVENIN S EQUIVALENT VOLTAGE AS SEEN FROM THE FAULT POINT

Z1 = THEVENIN S EQUIVALENT IMPEDANCE AS SEEN FROM THE FAULT POINT

1111 ZIEV aaa

Z1

Ea1

Ia1

Va1

+

-

70

NEGATIVE-SEQUENCE NETWORK

Z2 = THEVENIN S EQUIVALENT NEGATIVE-SEQUENCE IMPEDANCE AS SEEN FROM THE FAULT POINT

NEGATIVE SEQUENCE NETWORK

DEFINITION OF SEQUENCE NETWORKS

222 ZIV aaZ2

Ia2

Va2

+

-

71

ZERO-SEQUENCE NETWORK

Z0 = THEVENIN S EQUIVALENT ZERO-SEQUENCE IMPEDANCE AS SEEN FROM THE FAULT POINT

DEFINITION OF SEQUENCE NETWORKS

000 ZIV aa Z0

Ia0

Va0

+

-

72

73

MERALCO DC CALCULATING BOARD

74Photo courtesy of Engr. Eduardo S. Gonzales, former VP of Meralco

fundamentals of power system modeling · 2019. 1. 10. · • per-unit calculation is more...

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