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--..--..-------.-.-.---------.-~:l~..~
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been dccul,ted on ,elf-';;;;.-';ObO~
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can be held responsible if anything is wrong or i
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is improper information or insufficient
I information provided in it.
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IPower
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System Analysi~ and Stability
-, .1-·
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1
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16th Sem EE
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15
--------.--.-.,,-~
..---..---- ..---.-.-.-----.- ..------.---.----.----
..-.----.-.- ..--- ...--'~~·;·fo~:"'···
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._._
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e-Notes
f'..J
by
?
Dr.M.S.Ravillrakasha,
._-----'-
Malnad Collge of Engg., Hassan
r
:.
a
~
---'--
I
I
I
CONTENTS(CHAPTER_WISE)
Subject: POWERSYSTEMANALYSIS AND STABILITY .
Code: EE61
SUBJECT EXPERTS:
.
Chapter 1
No.of Hrs.: 35
REPRESENTATIONOF POWER SYSTEMS
:
Dr. MS Raviprakasha. Professorof E&EE. MCE. Hassan
Dr. GK Purushothama. Professorof E&EE. MCE. Hassan
Dr. K Umarao, Prof. & HOD of E&EE. RNSIT. Bangalore
LECTURE SCHEDULE:
Dr. MS Ravlprakasha: Chapters # 1. 2 & 4 (Introduction. Representationof Power systems.
Sym. 3-phasefaults & Unsym. Faults- Part I): (14 Hrs.) (Plus 3 ccincludlngSessions)
Dr. GK Purushothama: Chapter # 3 (Sym. Components& Unsyrn,Faults-Part II): (13 hrs.)
Dr. K Umarao: Chapter # 5 (Stability Studies): (08 Hrs.)
OneLine Diagram
Table of Symbols
Sample ExampleSystem
ImpedanceDiagram
Reactance Diagram
Examples
Chapter 2
PROGRAMME SCHEDULE:
SYMMETRICAL THREE PHAS5 FAULTS
14 Hrs. (8 + 6 Hrs.) of Class by Dr. MS Raviprakasha,MCE. Hassan:
13 Hrs. (9 + 4 Hrs.) of Class by Dr. GK Purushothama.MCE, Hassan:
B Hrs. of Class by Dr. H Umarao, RNSIT. Bangalore
3 Classes per week: Man (2 pm), Thu (11 am) & Fri (11 am)
.•
Plus 3 concluding sessions to discuss on solutions to the recent question papers of VTU.
VIEEE class on the selected subject of PSA&S (EE61).
Transients on a Transmission line
Short Circuit of Unloaded Syn. Machine
Short Circuit Reactances
Sort Circuit Current Oscillogram
Short Circuit of a Loaded machine
Examples
Iv
.
Chapter 3
I
.'
:
SYMMETRICAL COMPONENTS
,
What are SymmetricalComponents?
Resolutionof components, Neutral Shift
Phase shift in Y-lJ.Transformer Banks
Power.in terms of Symmetrical Components
Sequence Imps. & Sequence Networks
Sequence Networks of system elements
Examples
:
f
PS- I
,
,
.
I
Chapter 4
UNSYMMETRICAL FAULTS
,
~-""'"
"'' ' ,"'',,. "
I
,
..
----'1-
r
I
. .
,
'
----l
d
I
Review o'i Syrn.'Tletrical Components
.
i
I
I:CTRICAL
,
POWERSYSTEMS·
E STATE·OF·THE-ART:
Preambleto UnsymmetricalFault Anatysls
L-G, L-L. L-L-G & 3-Phase Faults
Faults on Power Systems
of Fault Impedance
Op n Conductor Faults
Ex pies
I
EffE
,
I
<;
I
To Beg/n With ......
No.
SUB·TOPICS
TOPICS
Cha[!ter 5
- SLD/OLD
STABILITY STUDIES
1.
Representation
systems
ot Power
• Rel;lct.llmp. Diagram
- per unit Systems
Basic Stabilityterms
Swing Equation & Swing Curve
PowerAngle Equation
Equal Area Criterion
Determinationof Stability of a System
Examples
- Generation - Machines
- Transmission Trans2.
I
I
;KE;!<I!;Q
Em!;RN
.
Electric Power System
- Distribution
formers
- Utilization- Tariffs
QF QU;§TIQN EAP!:B
I
~
One question each on chapter 1 and 2 And Two questions each on chapters 3. 4 and 5.
Five questions oUiot 8 are to be answered in full.
I
Note:
- Sym. Faults
TEXTS/REFERENCES:
1
- Sym. Components
1. WD Stevenson, Elements of Power System Analysis. MH.
3.
Fault studies
- Seq. Imps. I Networks
2. IJ Nagrath and DP Kothari, modern Power System analysis, TMH ..
- UnsymmetricalFaults
3. Hadi Sadat: Power System Analysis, TMH
4. GL Kusic: Computer aided PSA, PHI
PREREQUISITE SUBJECTS:
I
I
1. DC and' Synchronous Machines
-SSS, TS, DS
2.
Transmission and distribution
3. Transformers and induction machines
- Angle Stability
4.
System Stability
• Solution of Equations
- EAC, Clarke's Diagram
5.
""'''''''''!I'lf
:I
I
I
I
Linear Graph Theory
(Linear Equations)
• Incidence Matrices
,
,----
.s.
r--
.
_2
- Framesof Reference
Present Scenario ......
,
- Singularl NS Transformations
No.
• Network Matrices
i
TOPICS
SUB-TOPICS
• Node Elimination
• ZeusBuilding
• Importanceof VArs
- CompensationDevices,
7.
Reactive Power
Management
- Buses,YBUS Advs., Loadsflow equations
I
6.
Power Flow Studies (NL
Equations)
I
Sizing, Placement,
I
Desig'n,Optimality,
• Iterative Methods
• VAr Dispatch
- GS. NR. FDLF & DCLF
• VAr Co-ordination
:
I
-
- Optimality
i
,:
8.
Gen. Expansion Planning
- Load Prediction: Short,
Medium and Long Term
I
!
,
Forecasting
.
fI::.
I
I
i
-EMS: EMC, SLDC.RLDC
9.
Operation and Control
,
-ALFC. Voltage Control
-Tie..JinePower Control
I
10.
System
- Requirements
Reliability
- Methods
~
- Unit Commitment
11.
Economic Operation
- Parallel Operation
- Optimal Load Dispatch
- Constraints
_.
-
-
-
.
i
.,
I~ ,"',"
,.,
' OM' ~"r
II
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,------
_______________________________
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11
- UD Table Storage
-J'-\01'OR.
@
_; &eneyabl ®
',,<a..~10'l)n)e~ " 'Z.-\l'l:Y\6..t~
- AI. Expert Systems
20.
Recent Computer
Applications
~t
'.~''''\~''~''''1\'t-2.o'oNe.'( c.:i'(u..i\:- 'c'lo._,~'1
- ANN. Genetic Algorithms
3~
- Fuzzy Logic
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<k.~"i\1a.v.hoJ. : ~
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Figure 1. TABLE OF SYMBOLS FOR USE ON SLDS
Example system
CHAPTER 1
REPRESENTATION OF POWER SYSTEMS
[CONTENTS: One line diagram. impedance diagram, reactance diagram, per unit quantities, per unit
impedance diagram, formation of bus admittance & impedancematrices,examples]
1.1 One Line Diagram
In practice. electric power systems are very complex and their size is unwieldy. It is very difficult to
represent all the components of the system on a single frame. The complexities could be in terms
of various types of protective devices, machines (transformers, generators, motors, etc.), their
connections (star, delta, etc.), etc. Hence, for the purpose of power system analysis, a simple
single phase equivalent circuit is developed called, the one line diagram (OLD) or the single line
diagram (SLD). All SLD is thus, the concise form of representing a given power system. It is to be
noted that a given SLD will contain only such data that are relevant to the system analysis/study
under consideration. For example, the details of protective devices need' not be shown for load
flow analysis nor it is necessary to show the details of shunt values for stability studies,
'
i
I
I
Symbols 1sad tor SLD
,.
'.
Various symbols are used to represent the different parameters and machines as single phase
equivalents on the SLD,. Some of the important symbols used are as listed in the table of Figure 1.
L
~_
___j
Consider for illustration purpose, a sample example power system and data as under: Generator
1: 30 MVA, 10.5 KV, X"= 1.6 ohms, Generator 2: 15 MVA, 6.6 KV, 'X"= 1.2 ohms, Generator 3:
25 MVA,
6.6 KV, X"= 0.56 ohms, Transformer 1 (3-phase): 15 MVA, 33/11 KV, X=15.2
ohms/phase on HT side, Transformer 2 (3-phase): 15 MVA, 33/6.2 KV, X=16.0 ohms/phase on
HT side, Transmission Line: 20.5 ohms per phase, Load A: 15 MW, 11 KV, 0.9 PF (lag); and Load
B: 40 MW, 6.6 KV, 0.85 PF (laq), The corresponding SLD incorporating the standard symbols can
be shown as in figure 2.
I..n
~---'--'-------------'--"----.-
_----------
._
.
.----
.-----------
__
..u~
_
j
I
I
Figure 2. SAMPLE SYSTEM OLD
IMPEDANC~DIAGRAM
It is observed here, that the generators are specified in 3-phase MVA, L-L voltage and per phase Y_,
equivalent impedance, transformers are specified in 3-phase MVA, L-L voltage transformation ratio and per
phase Y-equivalent impedance on anyone side and the loads are specified In 3-phase MW, L-L voltage
and power factor.
1.2 Impedance Diagram
The impedance diagram on single-phase basis for use under balanced conditions can be easily
drawn from the SLD. The following assumptions are made in obtaining the impedance diagrams.
Assumptiors:
,
'1. The sindle phase transformer equivalents are shown as ideals with impedances on appropriate side
(LV/HV),
.
2, The magnetizing reactancesof transformersare negligible,
3, The generators are representedas constantvoltage sourceswith series resistance or reactance,
4. The transmission lines are approximatedby their equivalent n-Models,
,
5. The loads are assumed to be passive and are represented by a series' branch,of resistance or
reactance and
.
6. Since the balanced conditions are assumed, the neutral grounding impedances do not appear in the
impedance diagram.
Example system
As per the list of assumptions as above and with reference to the. system of figure 2, the
impedance diagram can be obtained as shown in figure 3.
•
,-----------1.3 Reactance Diagram
With some more additional and simplifying assumptions, the impedance diagram can be simplified
further to obtain the corresponding reactance diagram. The following are the assumptions made.
Normally the nominal voltage of lines and equipment is known along with the complexpower rating in MVA.
Hence, in practice, the base values are chosen for complex power (MVA) and line voltage (KV). The
chosen base MVA is the same for all the parts of the system. However, the base voltage is chosen with
reference to a particular section of the system and the other base voltages (with reference to the other
sections of the systems, these sections caused by the presence of the transformers)are then related to the
chosen one by the turns-ratio of the connecting transformer.
Additional assumptions:
»0
,
»
)0
The resistance is often omitted during the fault analysis. This causes a very negligible error since.
resistances are negligible
Loads are Omitted
Transmission line capacitancesare ineffecUve &
Magnetizingcurrents of transformersare neglected.
If I, is the base current in kilo amperes and Vb, the base voltage in kilovolts, then the base MVA is, S.
(V,I.). Then the base values of current & impedance are given by
Base current (kA), Ib
=
= MVAb/KVb
(1.1)
Example system
Base Impedance, Zb. (Vol'.)
as per the assumptions given above and with reference to the system of figure 2 and figure 3, the
reactance diagram can be obtained as shown in figure 4.
= (KV.2 I MVA.)
(1.2)
r
Hence the per unit impedanceis given by
Zpu " ZohmslZb
"Zohms (MVAJKV.2)
(1.3)
In 3-phase systems, KVb is the line-to-line value & MVAb is the 3-phase MVA. [1-phase MVA = (1/3) 3phase MVAj.
•
•
Figure 4. REACTANCE DIAGRAM
Changing the base of a given pu value:
It is observed from equation (3) that the pu value of impedance is proportionaldirectlyto the base MVA and
Note: These impedance & reactance diagrams are also refered as the Positive SequenceDiagrams!
-Networks.'
.
i~versely to the square of the base KV. If Zpunew is the pu impedance required to be calculatedon a new
set ofbase values: MVAbnew & KVbnew from the already given per unit impedanceZpuold, specified on
the old set of base values, MVAboid & KVbold , then we have
1.4 Per Unit Quantities
during the power system analysis, it is a usual practice to represent current, voltage, impedance,
power, etc., of an electric power system in per ~nit or percentage of the b.as~~r reference value of
the respective quantities. The numerical per Unit (pu) value of any quantity IS.ItSratio to a chosen
base value of the same dimension. Thus a pu value is a normalized quantity with respect to the:
chosen base value.
'.
Definition: Per Unit value of a given quantity is the retia of the ectusl value in any given'unit to the base:
value in the same unit. The percent value is 100 times the pu value. Both the pu 'and percentage methods
are simpler than the use of actual values. Further, the main advantage in ,using the pu system of
computationsis that the result that comes out of the sum, product,quotient, etc. of two or more pu values is
expressed in per unit itself.
In an electrical power system, the parametersof interest include the current, voltage, complex power (VA),
impedance and the phase angle. Of these. the phase angle is dimensionless and the other four.quantities.
can be described by knowing any two of them. Thus clearly, an arbitrary choice of any two base values will
evidently fix the other base values.
Zpunew =
Z.:1d (MVA.....'MVA.. "') (KV"""KV,,"")2
(1.4)
On the other hand, the change of base can also be done by first converting the given pu
impedance to its ohmic value and then calculating its pu value on the new set of base values.
Merits and Demerits of pu System
Follo"Yingare the advantages and disadvantages of adopting the pu system of computations in electric
power systems:
Merits:
. l> The pu value is the same for both t-phase and & 3-phase systems
l> The pu value once expressed on a proper base, will be the same when refereed to either
side of the transformer. Thus the presence of transformer is totally eliminated
)0 The variation of values is in a smaller range 9nearby unity). Hence the errors involved in pu
computationsare very less.
---------------Usuallythe nameplateratingswill be marked in pu on the base of the name plate ratings, etc.
Demerits:
1
:.. If proper bases are not chosen,then the resulting pu values may be highly absurd (such as 5.8 pu, - i
18.9 pu, etc.). This may cause confusion to the user. However, this problem can be avoided by
.selectirlQthe base MVA near the high-ratedequipment and a convenient·base KV in any section of
the system.
~z~
~------------,
Of the various network matrices refered above, the bus admittance matrix (Yaus) and the bus
impedance matrix (Zeus) are determined for a given power system by the rule of inspection as
explained next.
Rule of Inspection
Consider the 3-node admittance network as shown In figureS. Using the basic branch relation: I =
(YV), for all the elemental currents and applying Kirchhoff's Current Law principle at the nodal
points, we get the relations as under:
1.5 pu Impedance {Reactance Diagram
for a given power system with all its data with regard to the generators, transformers, transmission
lines, loads, etc., it is possible to obtain the corresponding impedance or reactance diagram as
explained above. If the parametric values are shown in pu on the properly selected base values of
the system, then the diagram is refered as the per unit impedance orreactance diagram. In
forming a pu diagram, the following are the procedural steps involved:
.
At node 1: I,
=y,v, + Ya (V,-Va) + Ys (V,
- V2)
At node 2: 12 =Y2V2 + Ys (V2-Va) + Ye (V2 - VI)
At node
•
3: 0 = Ya (Va-V,) + YN3 + Ys (V3 - V2)
(1.8)
1. Obtain the one line diagram based on the given data
2. Choose a common base MVA for the system
3. Choose a base KV in anyone section(Sectionsformed by the presence of transformers)
4. Find the base KV of all the sections present
.
5. Find pu values of all the parameters:R,X, Z, E, etc.
6. Draw the pu impedance!reactancediagram.
1.6 Formation Qf Y~~ID.!§
Figure 5. EXAMPLE SYSTEM FOR FINDING YBUS
The performance equations of a given power system can be considered
of reference as discussed below:
In three
.
different frames'
These are the performance equations of the given network in admittance form and they can be
represented in matrix form as:
Frames of Reference:
(Y,+Ya +Y6)
-Y6
-Y6
(Y2+YS +Ys)
Bus Frame of Reference: There are b independentequations (b = no. of buses) relating the bus vectors of:
currents and voltages through the bus impedancematrix and bus admittance matrix:
-Ya
EBus = ZBUS IBUS
-Ys
(1.9)
In other words, the relation of equation (9) can be represented in the form
IBUS = YBUS Esus
(1.S)
Branch Frame of Reference: There are b independent equations (b = no. of branches of a selected Tree
sub-graph of the system Graph) relating the branch vectors of currents and voltages through the branch
impedance matrix and branch admittance matrix:
.
laus = Y BUS EBus
(1.10)
W~ere, YBUS Is the bus admittance matrix, leus & EBus are the bus current and bus voltage vectors
respectively.
By observing the elements of the bus admittance matrix, Yaus of equation (9), it is observed that
the matrix elements can as well be obtained by a simple inspection of the given system diagram:
EBR = ZeR IBR
(1.6)
Loop Frame of Reference: There are b independentequations (b = no. of branches of a selected Tree subgraph of the system Graph) relating the branch vectors of currents and voltages through the branch
impedance matrix and branch admittance matrix:
.
(1.7)
Diagonal elements: A diagonal element (YII) of the bus admittance matrix, Yaus. is equal to the
sum total of the admittance values of all the elements Incident at the b,us/node i,
Off Diagonal elements: An off-diagonal element (Y;i) of the bus admittance matrix, YBUS, is equal to
the negative of the admittance value of the connecting element present between the buses I and j,
if any,
This is the prlnctple of the rule of inspection. Thus the algorithmic equations for the rule of
inspection are obtained as:
Y;;= EY;j
Y;j
=
-y;J
= 1,2,
n)
(j = 1,2,
n)
(j
(1.11)
'--------------------------------------~--~-------j
..
..
.
'
r-~:
I
i ='1 .2, ... ~', n = no. of buses of the given system, Y;j is the admittance of-element connected
between buses i and j and y;;is the admittance of element connected between bus i and ground
(reference bus).
I
Bus impedance matrix
In cases where, the bus impedance matrix is also required, then it cannot be formed by direct
inspection of the given system diagram. However, the bus admittance matrix determined by the
rule of inspection following the steps explained above. can be inverted to 'obtain the bus
impodanc~ matrix, since the two matrices are inter-invertible. Note: It is to...be noted that the rule
of inspecti1n can be applied only to those power systems that do not have any mutually coupled
elements.
.
.
rI
1.7 Examples
I PROBLEMS ON RULE OF INSPECTION:
Problem #1: Obtain the bus admittance matrix for the admittance network shown aside by the rule of
inspection
(I)
<.tl
YlIU$=
l
16 -8 -4
~
8 24 -8
-4
-8 16
L~
(
•
4-
-P'S6
® '.
J4-
-dO.'iS
(-M."", "1\"')
..
""I.e """Ie, . i
~'hhvl.>.) •
'.
CD'.., 4(f??~
.
'-,...
S
"
4
@>
.
.
Problem #2: Obtain YBUS and Zeus matrices for the Impedance network Sh:o";"naside by the rule of
Inspection.Also. determine Yeusfor the reduced network after eliminatingthe eligltlle unwanted node. Draw
the resulting reduced system diagram.
-9.8 5 4
Y.us
=
j
5 -16 10
4 10 -14
,
L
I """"''''''II'',r
~~
~
x, =X" = j 0.1 (100/50)
(6.6/6.31)'
Xllou "j 20 (100/1102)
j 0.1652 pu.
E,
= 1.0LO"
•
=
pu, Em (6.6/6.31)
= j 0.219 pu.
[~.., ......;o._e".~are:
= 1.045'::0°pu
I
Thus Ihe pu reactance diagram can be drawn as shown in figure P2(b).
= 120.23 KV in the transmission line circuit and
:::,(115/13.2)
., 120.23 (13.26/115) = 13.8 KV in the motor circuit.
I
Cetcutetion of ou velues:
x~= j 0.15 pu.
x~,= j 0.2 (30/20)
(12.8/13.8)2
= j 0.516 pu.
= j 0.2 (30/10)
(12.8/13.8)2
= j 0.2581 pu .
. Xm2
.)(" =X" = j 0.1 (30/35)(13.2/13.8)'
Figure P2(b). Per Unit Reactance Diagram
Xli,. = j SO(30/120.232)
Eg
Problem #3:
A 30 MVA. 13.S KV. 3-phase generator has a sub transient reactance'of 15%. The generator supplies 2
motors through a step-uptransformer - transmissionline - step-down'transformer Ilrrangement. The motors
have rated inputs of 20 MVA and 10 MVA at 12.SKV with 20% sub transient reactance each. The 3-phase
transformers are rated at 35 MVA, 13.2 KV-6/115 KV-Y with 10 % leakage.reacttmce.The line reactance
Is SOohms. Draw the equivalent per unit reactance diagram by seleqting 'the g~nerator ratings as base
values in the generator circuit.
.
= 1.0'::0°pu;
= j 0.0784 pu.
= j 0.17 pu.
Em1 = Em> = (6.6/6.31) = 0.93'<:0°pu
Thus the pu reactance diagram can be drawn as shown in figure P3(b).
The one line diagramwith the data is obtained as shown In figure P3(a).
,
Figure P3(b). Per Unit Reactance Diagram
""'"", '2.0""'11;
(I'l.:L.D
'5 \<\.")
·r.
Problem #4:
\0 ~".\-
A 33 MVA. 13.8 KV, 3-phase generator has a sub transient reactance of 0.5%. me generator supplies a
motor through a step-up transformer - transmission line - step-downtransformer arrangement.The motor
has rated input of 25 MVA at 6.6 KV with 25% sub transient reactance. Draw the equivalent per unit
impedance diagram by selecting 25 MVA (34)).6.6 KV (LL) as base values in the motor circuit. given the
transformer and transmission line data as under:
Figure P3(a). OLD of the given system
bank: three single phase units•.connected e,-y. each rated 10 MVA, 13.2/6.6 KV with
7.7 % leakage reactance and 0.5 % leakage resistance;
Step up transformer
. Selection of base quantities:
Transmission
line: 75 KM long with a positive sequence reactance of 0.8 ohm! KM and a resistance of 0.2
ohm! KM; and
30 MVA. 13.S KV in the generator circuit (Given);
---~'-----------------------,--
___J
"'Wlr
"
L~' ~~
~
1---- - -.- --.----
I
----- ..._..
_24___.
...
.
.
--I
i
Step down transformer bank: three single phase units, connected 6-Y, each rated 8.33 MYA, 110/3.B8 KV
with 8% leakage reactance and 0.8 % leakage resistance;
I
Solution_;
The
0119
••
I
.
1
line diagram with the data is obtained as shown in figure P4(a),
. f]_9_l!~~ I?~(~J.:
I
Per Unit Reactance Diagram
!
1.8 Exercises for Praclice
problems
3-phase ratings of transformers:
1. Determine the reactances of the three generators rated as follows on a common base of 200 MVA, 35
KV: Generator 1: 100 MVA, 33 KV, sub transient reactance of 10%: Generator 2: 150 MVA, 32 KV, sub
transient reactance of 8% and Generator 3: 110 MVA, 30 KY, sub transient{eactance of 12%.
T,: 3(10) = 30 MYA, 13.2166.4-./3 KY = 13.2/115 KY, X = 0.077, R = 0.005 pu.
T,: 3(8.33) = 25 MYA, 1101 3.98-./3 KY
= 110/6.8936
KY, X = 0.08, R = 0.008 pu,
[Answers:
XG, = j 0.1778, Xg, = j 0.089, Xg, = j 0.16 all in per unit]
,
Selection of base quantities:
25 MYA, 6.6 KV in the motor circuit (Given); the voltage bases in other sections are: 6.6 (110/6.8936)
105.316 KV in the transmission line circuit and 105.316 (13.21115) = 12.09 KV in the generator circuit.
Calculation of pu values:
Xm ,- j 0.25 pu: Em= 1.0LOo pu.
=,
2. A 100 MVA, 33 KV, 3-phase generator has a sub transient reactance Of 15%. The. generator supplies 3
motors through a step-up transformer - transmission line - step-down transformer arrangement. The motors
have rated inputs of 30 MVA, 20 MVA and 50 MVA, at 30 KY with 20% sub transient reactance each. The
3-phase transformers are rated at 100 MVA, 32 KV-A /110 KY-Y with 8 % leakage'l'eactance. The line has
a reactance of 50 ohms. By selecting the generator ratings as base values in the generator circuit,
determine the base values in all the other parts of the system. Hence evaluate the corresponding pu values
and draw the equivalent per unit reactance diagram.
[Answers:
XG = j 0.005 (25/33) (13.8/12.09)2 = j 0.005 pu: E. = 13.8/12.09 = 1.414LOO pu..
Z" = 0.005 + j 0.077 (25/30) (13.2/12.09),
Z,2 = 0.008 + j 0.08 (25/25) (110/105.316)'
Z"c.. = 75 (0.2+j 0.8) (25/105.316')
I
I
= 0.005 + J 0.0765 pu. (ref. to LV side)
3. A 80 MVA, 10 KV, 3-phase generator has a sub transient reactance of 10%. The generator supplies a
motor through a step-up transformer - transmission line - step-down transformer arrangement. The motor
has rated Input of 95 MVA, 6.3 KV with 15% sub transient reactance. The $tep-up 3-phase transfonmer is
rated at 90 MYA, 11 KV-Y /110 KV-Y with 10% leakage reactance. The 3~phase step-down transformer
consists of three single phase Y-A connected transformers, each rated at 33.33 MVA, 68/6.6 KY with 10%
leakage reactance. The line has a reactance of 20 ohms. By selecting the 11 KV, 100 MVA as base values
in the generator circuit, detenmine the base values in all the other parts of the system. Hence evaluate the
corresponding pu values and draw the equivalent per unit reactance diagram.
= 0.0087 + j 0.0873 pu. (ref. to HV side)
= 0.0338 + j 0.1351 pu.
Thus the p1reactance diagram can be drawn as shown in figure P4(b).
L_~ .
-'" , "'rI'lr
XG = j 0.15, Xm, = j 0.551, Xm, = j 0.826, Xm3= j 0.331, E.,=1.0 LOO, Em,= Em' = Em' =
0.91 LOO,X" = X12 = j 0.0775 and X'no = j 0.39 all in per unit)
[Answers:
XG=j 1.103, x, =j 0.165, Eg,=0.91 LOo, Em= 1.022LOo, X" =jO.ll,
all in per unit]
= j 0.17
----.J
X12 =jO.114and
Xlin•
r-·---·--_·, . --
_
2(;
..
I
4. For the three·phase system shown below, draw an impedance diagram expressing all Impedances in per
unit on a common base of 20 MVA, 2600 V on the HV side of the transformer. Using this Impedance
diagram, find the HV and LV currents.
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Which of the following loss In a transformer Is zero even at full load?
a)
b)
Eddy current
Hysteresis
c)
Core loss
d)
Friction loss
..
,
5. The transmission line conductors are transposed to
tiiJ ·
(p.lt~~~~
d)
a)
Balance the current
b)
Obtain different losses
c)
Obtain same line drops
d)
Balance the voltage
, •
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,
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Figure E3. OLD of the. Example system
[Ans.: l(b), 2(a), 3(b), 4(d), 5(c)l
[Answers:
s, = 20 MVA; Vb=2.6 KV (HV) and 0.2427 KV (LV); VI=1.0LOO,XI = j 0.107, Z~bl. = 0.136
+j 0.204 and Z'o,' = 5.66 + j 2.26,1 = 0.158 all in per unit, I (hv)« 0.7 A and I (Iv) = 7.5 A]
I
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Qbjective tYol! 9yestions
1. Under
t
~
load conditions the current in a transmission line is due to.
a)
b)
Corona effects
Capacitance of the line
c)
Back flow from earth
d)
None of the above
2. In the short transmission line which of the following is used?
a)
b)
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c)
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T - Model
Both (a) and (b)
d)
None of the above
3. In the short transmission line which of the following is neglected?
I' R loss
a)
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c)
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;----------------------------------------------------------------------~------------------------------------,
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lIQ
C-H-A--'P"'TE-R--2------------------1
SYMMETRICAL
THREEPHASEFAULTS
[CONTENTS:
Preamble, transients on a transmission line, short circuit o( an unloaded synchronous
machine- short circuit currents and reactances, short circuit of a loaded machine, selection of circuit
breaker ratings, examples]
Figure 1. Short Circuit
II
2,1 Preamhle
in practice, any disturbance
in the normal working conditions is termed as a FAULT. The effect of fault is to
load the device electrically
by many times greater than its normal rating and thus damage the equipment
involved. Hence all the equipment
overloading
in the fault line should be protected from being overloaded.
damage
Consider,
v = Vm sin (cot+a)
= iR
and
i=
+ L (dildt)
1m sin
(mt+a-9)
Where
In order to protect the ~quipment during faults, fast acting circuit breakers are put in the lines. To design the
rating of these circuit breakers or an auxiliary device, the fault current has to be predicted. By considering
the equivalent per unit reactance diagrams, the various faults can be analyzed to determine the fault
parameters. This helps In the protection and maintenance of the equipment.
-
Thus
(2'1)
i,
= [Vm/Z] sin
on a transmission
(2.~
IR + L (dVdt) .. 0
(RIL + d/dt)1 = 0
In order to solve the equation (5), consider the complementary
V.l
(2.6)
(2.5)
function part of the solution as:
CF= C,
e('
Where, (= UR) is the time constant and C, is a constant given by the value of steady state current at t = O.
Thus we have,
C, =-is(O)
=-
2.2Transients
«(l)t+a-9)
Consider the performance equation of the circuit of figure 1 under circuit as:
i.e.,
Faults can be symmetrical or unsymmetrical faults. In symmetrical faults, the fault quantity rises to several
times the rated value equally In all the three phases. For example, a 3-phase fault - a dead short circuit of
all the three lines not involving the -ground. On the other hand, the unsymmetricaUaults
may have the
connected fault quantities In a random way. However, such unsymmetrical faults can be analyzed by using
the Symmetrical Components. Further, the neutrals of the machines and equipment mayor may not be
grounded or the fault may occur through fault impedance. The three-phase fault involving ground is the
most severe fault among the various faults encountered in electriC power systems.
Line.
Thus the line can be modeled by a lumped R-L series circuit. Let the short circuit take place at t=O. The
In general,
anc
on an Unloaded
parameter, a controls the instant of short circuit on the voltage wave. From basic circuit theory, it is
observed that the current after short circuit is composed of the two parts as under: 1=i. +1" Where, i. is the
steadystate current and i, is the transient current. These component currents are determined as follows.
involves the increase of current up to 10-15 times the rated value. In a few cases, like Ihe
opening or closing of a circuit breaker, the transienl voltages also may overload the equipment
them.
Transients
[Vm/Z] sin (o.-El)
; [Vm/Z] sin (S-a)
line
(2.7)
Similarly the expression for the transient part is given by:
us
Now, let
Consider a transmission
line of resistance
R and inductance
L supplied by an ac source of
voltage v, such that v = Vm sin (mt+«) as shown in figure 1'. Consider the short circuit transient on this
transmission line. In order to analyze this symmetrical 3-phase fault, the following assumptions are made:
I
I
l>
l>
l>
The supply is a constant voltage source,
The short circuit occurs when the line is unloaded and
The line capacitance is negligible.
I
l
"""" '''''''Wlr
i, ; -is(O) e(-v'l
;
[V..,IZI sin (9-0.) e(·RILI'
Thus the total current under short circuit is given by the solution of equation (1) as [combining equations (4)
and (8)),
i ;i, +1,
= [v2V/Z]
II
I
(2.8)
sin (et+c-a) ,+ [v2V/Z) sin (0-0.) e(-RlL~ (2.9)
I
Thus, ~-the
sinus:dal steady state curre~:':::
the symmetrical short. circuit current and i, is the
unidirectional value called as the DC off-set current. This causes the total current to be unsymmetrical till
the transient decays, as clearly shown in figure 2.
,
.,
I
Under steady state short circuit conditions, the armature reaction in synchFonous generator produces a
demagnetizing effect. This effect can be modeled as a reactance, X. in series with the Induced emf and the
leakage reactance, X, of the machine as shown in figure 3, Thus the equivalent reaciance Is given by:
x, = X. +X,
(2.13)
Where Xd is called as the direct axis synchronof,lS reactance of the synchrondus machine. Consider now a
sudden Ihree-phase short circuit of the synchronous generator on nO-load.IJhe machine experiences a
transient in all the 3 phases, finally ending up in steady state conditions.
Figure 3. Steady State Short Circuit Model
Figure 2. Plot of Symmetrical short circuit current, i(t).
The maximum momentary current, immthus corresponds to the first peak. Hence; if the decay in the
transient current during this short interval of time is neglected, then we have (sum of the two peak values);
imm= [V2VIZj sin (B-a) + [V2v/Z]
(2.10)
now, since the resistance of the transmission line is very small, the impedance angle
approximately equal to 900• Hence, we have
d=
[V2V/Zj cos a + [V2V/Zj
Immediately after the short circuit, the symmetrical short circuit current is limited only by the leakage
reactance of the machine. However, to encounter the demagnetization of the armature short circuit current,
current appears in field and damper windings. assisting the rotor field winding to sUflain the air-gap flux.
Thus during the initial part of the short circuit, there is mutual coupling between stator, rotor and damper
windings and hence the corresponding equ!valent circuit would be as shown in figure 4, Thus the
equivalent reactance is given by:
'
e, can be taken to be
(2.11)
r'
Xd' = X, +[1/X. + 1IX, + 1lXdW
(2.14)
Where 'XJ" is called as the sub-transient reactance of the synChronous machine. Here, the equivalent
resistance of the damper winding is more than that of the rotor field winding. Hence, the time constant of
the damper field winding is smaller. Thus the damper field effects and the eddy currents disappear after a
few cycles.
This value is maximum when the value of a is equal to zero, This value corresponds to the short circuiting
instant of the voltage wave when it is passing through zero, Thus the final expressIon for the maximum
momentary current is obtained as:
i",,,, = 2 [V2V/Zj
(2.12)
Thus it is observed that the maximum momentary current is twice the maximum value of symmetrical short
circuit current. This is refered as the doubling effect of the short circuit current du'ring the symmetrical fault
on a transrnjssion line.
.
2,3 Short circuit of an unloaded synchronous machine
Figure 4. Model during SUb-transient Period of Short Circuit
2.3.1 Short Circuit Reactances
L,
:"'n:" ._"",', .,'~
II'
-'-'-- _
T----------------------------~~~--------------~
f
~e~ words, X.Wgets open circuited from the model of Figure 5 to yield the model as shown in figure 4.-l
us the equivalent reactance is given by:
Xi
= X, +[l/X.
+ 1IX,
r'
(2.15)
Where Xi is called as the transient reactance of the synchronous machine. Subsequently, X, also g"ts
open circuited depending on the field winding time constant and yields back the steady state model of
figure 3.
r~
'ori"" ••
currents and rea etances
three zones of period are rei ats .. under " .. rms of tho
interceptson the OSCillogram(oa, ob and oc are the y-Intercepts as Indicated In flgure 6):
RMS value of the steady state current" I .. [oa/v2) .. [E./Xo)
RMS value of the transient current .. I' = [ob/v2] .. [EA']
RMS value of the sub tr~nsient current = I = [oclv2)
= [EJX,1
(2.16)
2.4 short circuit of a loaded machine
In the analysis of section 2.3 above, it has been assumed that the machine o~eratesat no load prior to the
occurrenceof the faull. On similar lines, the analysis of the fault occurring on a loadedmachinecan also be
considered.
Figure 5. Model during transient Period of Short Circuit
Thus the machine offers a time varying reactance during short circuit and this value of reactance varies
from initial stage to final one such that: Xd > Xd' > Xd'
Figure 7 gives the circuit model of a synchronous generator operating under steady state conditions
supplying a load current I, to the bus at a terminal voltage V,. E, is the induced emf under the loaded
conditions and X, is the direct axis synchronousreactance of the generator.
2.3.2 Short Circuit Current Oscillogram
Consider the oscillogram of short circuit current of a synchronous machine upon theoccurrence of a fault
as shown in figure 6. The symmetrical short circuit current can be divided into three·zones: the initial sub
transient period, the middle transient period and finally the steady state period, The corresponding
reactances, Xd," Xd' and Xd respectively, are offered by the synchronous machine during these time
periods.
Figure 7. Circuit models for a fault on a loaded machine.
Also shown in figure 7, are the circuit models to be used for short circuit current calculationswhen a fault
occurs at the terminals of the generator, for sub-transientcurrent and transient cutrent values. The induced
emf values used in these models are given by the expressions as under:
Eg = VI + j ILX.; Voltage behind syn. reactance
=
Eg'; V, + j ILX" Voltage behind transient reactance
/
. ' .L
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.
<l\l
vlLL..u.,
t<.J.£.
Figure 6. SC current Oscillogram of Armature Current.
"'"''''''''''11'''
E,"= V, + j ILX;
= Voltage behind subtr. Reactance
(2.17)
.
The synchronousmotors will also have the terminal emf values and reactances. However,then the current
direction is reversed. During short circuit studies, they can be replaced by circuit models similar to those
shown in figure 7 above, except that the voltages are given by the relations as under:
Em = V, - j 'LX. = Voltage behind syn. reactance
Z = R + jeol
Em'= V, - j ILX.'= Voltage behind translant reactance
Em"= V, - j 'LXd'= Voltage behind subtr. Reactance
(2.18)
= 5 + j (1001<)(0.1)
= 5 + j 31.416 ohms.
Thus we have: Zmag=31.8113Ohms; 9=80.957° and 't=LlR=0.1/5=0.02
The short circuit current is given by:
I
seconds.
i(t) = [V",/Z] sin (Olt+a-9) + [Vm/Z] sin (8-a) e.(RlL)1
The circuit models shown above for the synchronousmachines are also very useful while dealing with the
short circuit of an interconnectedsystem.
= [100/31.8113]
[sin (1001tt+150-80.9570) + sin(80.9570-150) e~vo.021
= 3.1435 sin(314.16 t - 65.96) +2.871 e-501
Thus we have:
=
i) imm 3.1435 + 2.871 e-501
2.5 Selection of cIrcuit breaker ratings
For selection of circuit breakers, the maximum momentary current is considered.corresponding to its
maximum possible value. Later, the current to be interrupted is usually taken as symmetrical short circuit
current multiplied by an empirical factor in order to account for the DC off-set current. A value of 1.6 is
usually selected as the multiplyingfactor.
where t is the time instant of maximum of symmetrical short circuit current. This instant occurs at
(314.16 te - 65.96°) = 90°; Solving we get, t = 0.00867 seconds so that imm= 5 Amps.
ii) imm= 2Vm/Z
= 6.287 A;
for a=O, and 8=90°
(Also, imm= 2 (3.1435)
iii) DC offset current = [Vm/Z] sin (tl-a) eJR/L)1
= zero, if (8-a) = zero, i.e., 8
Normally, both the generator and motor reactances are used to determine the momentary current flowing
on occurrence of a short circuit. The interrupting capacity of a circuit breaker is declded by x." for the
generators and Xi for the motors.
.
= maximum if
= a,
(8-a) = 90°, i.e., a = 8 - 90°,
or
a
or
t
= 6.287 A)
= 80.9
70
a = _ 9.0 30.
2.S Example~
Problem #2: A 25 MVA, 11 KV, 20% generator is connected through a step-up transformer- T, (25 MVA,
11/66 KV, 10%), transmission line (15% reactance on a base of 25 MVA, 66 KV) and step-down
transformer-To(25 MVA. 66/S.6 KV, 10%) to a bus that supplies 3 Identicalmotors in parallel (all motors
rated: 5 MVA, 6.6 KV, 25%). A circuit breaker-A is used near the primaryof the transformer T, and breakerB is used near the motor M,. Find the symmetrical currents to be interruptedby circuit breakersA and B for
a fault at a point P, near the circuit breaker B.
Problem #1: A transmissionline of inductance 0.1 H and resistance 5 Q Is suddenly short circuited at t "
0, at the far end of a transmission line and is supplied by an ac source of voltage v = 100 sin (100rrt+1So),
Write the expression for the short circuit current, itt). Find the approximate value of the first current
maximum for the given values of a and A. What is this value for o=O, and 9=90°7 Wtiat should be the
instant of short circuit so that the DC offset current is (i)zero and (ii)maximum7
Solution:
Solution:
Consider the SLD with the data given in the problem statement. The base values are selected as
under:
6MvA
'.6
\<.~
2.-; oj,
Figure
I Consider
P1.
the expression for voltage applied to the transmission system given by
v = Vm Si~(Olt+a) = 100 sin (100nt+150)
Thus we get:
Vm= 100 volts; f = 50 Hz
and
Consider the impedance of the circuit given by;
'~·"'··""!I'lr
a
Selection of bases:
'1 v, = 66
Sb = 25 MVA (common); Vb = 11 KV (Gen. circult)- chosen. so that the~
and Vb = 6.6 KV (Motor circuit).
Pu values:
= 150.
Xg=jO.2pu, XI1=X'2=jO.1
pu; Xm,=Xm2=Xm,=jO.25(25/5)=j1.25pu; X';n.=jO.15pu.
KV (line circuit)
Since the system is operating at no load. all the voltages before fault are 1 pu. Considering t=l
reactanced;"om W< th thetaultsat P, we have:
'
I
Thus the prefault voltage at the motor bus; V, = 0.4/0.44 = 0.909LOo.
I'
Short circuit current fed to the fault at motor bus (Ir = YV);
I, "[0.125 + 0.258],' + 2.0 }B.909 = [20.55 pu] [1000/("./3(0.4))]
!
= 20.55 (1.312) KA = 26.966 KA.
,
.
Problem #4: A generator·transformer unit Is connected to a line through a circuit breaker. The unit ratings
are: Gen.: 10 MVA, 6.6 KV, '><4" 0.1 pu, '><4'.= 0.2 pu and Xd 0.8 pu; and Transformer: 10 MVA, 6.9/33
KV, X, = 0.08 pu; The system Is operating on no-load at a line voltage of 30 KV, when a three-phase fault
occurs on the line just beyond the circuit breaker. Determine the following:
=
=
(i) Initial symmetrical RMS current in the breaker.
(il) Maximum possible DC off-set current in the breaker.
Figure
P2(b)
(iii) Momentary current rating of the br~aker.
(iv) Current to be interrupted by the breaker and the interrupting KVA and
Current to be interrupted by circuit breaker A = 1.0 1j[0.2 >0.1+0.15+0.1]
(v) Sustained short circuit current In the breaker.
~ - j 1.818 pu = - j 1.818 (25/[-./3(11)]) =. j 1.818 (1.312) KA = 2.386 KA
Solution:
And Current to be interrupted by breaker B = 1/j1.25 = - j 0.8 pu
,,- jO.8 (25/[-.13(6.6)])= - jO.8 (2.187) KA" 1.75 KA .•
Problem #3: Two synchronous motors are connected to a large system bus through a short line. The
ratings of the various components are: Motors(each)= 1 MVA. 440 volts. 0..1·pu reactance; line of 0.05 ohm
reactance and the short circuit MVA at the bus of the large system is 8.at 440 volts. Calculate the
symmetrical short circuit current fed into a three-phase fault at the motor 'bus when the motors are
operating at 400 volts.
Consider the base values selected as 10 MVA, 6.6 KV (in the generator circuit) and 6.6(33/6.9)
31.56 IW(in {he transformer circuit). Thus the base current is:
Ib
= 10 I [,13(31.56)] = 0.183 KA
The pu values are:
pu: V, = (30/31.6)
Xc/' = 0.1
pu, Xd'
= 0.95LOo pu.
= 0.2
pu and Xd
Maximum possible DC off-set current = 2 (0.9277)
Momentary current rating
The base, values are selected as
.
= 0.8
pu; and XT,
= 0.08 (6.9/6.6)2 = 0.0874
Initial symmetrical RMS current = 0.95LOo 1 [0.1 + 0.0874J = 5.069 pu = 0.9277 KA;
Solution:
Consider the SLD with the data given in the problem statement.
unde~
=
= 1.312 KA;
= 1.6(0.9277) = 1.4843 KA; (assuming 60% allowance)
~urrent to be interrupted by the breaker (5 Cycles)
= 1.1(0.9277)
= 1.0205 KA;
Interrupting MVA = 3(30) (1.0205) = 53.03 MVA;
Sustained short circuit current In the breaker
M
.1No.. V'-",e
44(9V
. ~:OA.\.~
2.7 Exercises for Practice
( e\,,'-l~\'\-c.(\;•
4~r::)"l)
",.""""
:II.'I!' ,.,~
PROBLEMS
1· The one line diagram for a radial system
Figure P3.
S, = 1 MVI>.;Vb = 0.44 KV (common)- chosen so that Xm(each)=jO.1 pu, Em."
= j 0.258 pu and Xlarge-system -= (1/8) = j 0.125 pu.
= 0.95LOo (0.183) 1 [0.8 + 0.0874]
= 0.1959 KA .
i.OLOo. X'io."jO.05 (1/0.44')
network consists of two generators. rated 10 MVA, 15% and 10
MVA. 12.5 % respectively and connected in parallel to a bus oar A at 11 KV. Supply from bus A is fed to
bus B (at 33 KV) through a transformer T, (rated: 10 MVA, 10%) and OH line (30 KM long). A transformer
T, (rated: 5 MVA, 8%) is used in between bus B (at 33 KV) and bus C (at 6.6 KV). The length of cable
,
,
-.--
__
.99._.__
.
.
~
.
_
running from the bus C up to the point of fault, F is 3 KM. Determine the current and line voltage at 11 kV
bus A under fault conditions, when a fault occurs at the paint F, given that Za."e = 0.135 + J 0.08 ohml kM
and ZoH.~".= O.<y + j 0.36 ohm/kM.
[Answer: 9.62 kV at the 11 kV bus]
a)
b)
2. A generator
(rated: 25MVA, 12. KV, 10%) supplies power to a motor (rated: 20. MVA, 3.8 KV, 10%)
c)
through a step-up transformer (rated:25 MVA, 11/33 KV, 8%), transmission line (of reactance 20 ohms) and
a step-down transformer (rated:20 MVA, 33/3.3 KV, 10%). Write the pu reactance diagram. The system is
loaded such that the motor is drawing
15 MW
3. A synchronous
d)
at 0.9 leading power factor, the motor terminal voltage being
3.1 KV. Find the sub-transient current in the generator
[Answer: I," = 9.337 KA; 1m" = 6.9 KAJ
and motor for a fault at the generator
bus.
4.
generator feeds bus 1 and a power network feed bus 2 of a system. Buses ·1and 2 are
connected through a transformer
1):0.25; Transformer:O.12
and a line. Per unit reactances
and Line:O.28. The power network
of the components
are:
Generatortbus-
is represented, by a generator
4, A synchronous generateor, rated 500 KVA, 440 Volts, 0.1 pu sub-transient reactance is supplying a
passive load of 400 KW, at 0.8 power factor (lag). Calculate the initial symmetrical RMS current for a threephase fault at the generator terminals.
[Answer: S.=0.5 MVA; V.=0.44 KV; load=O.8L-36.9°;
OBJECTIVE
5.
Zn
3Zn
An ineffective value
An infinite bus-bar should maintain
a)
Constant frequency and Constant voltage
b)
Infinite frequency and Infinite voltage
c)
Constant frequency and Variable voltage
d)
Variable frequency and Variable voltage
with an
unknown reactance in series. With the generator on no-load and with 1.0 pu voliage at each bus. a three
phase fault occurring on bus-t causes a current of 5 pu to flow into the fault. Determine the equivalent
reactance of the power network.
[Answer: X = 0.6 pu]
Zn2
Voltages under extra high voltage
a)
1KV & above
b)
11KV & above
1.=0.656 KA; Ir=6.97 KA] :
c)
132 KV & above
d)
330 KV & above
TYPE QUESTIONS
are
[Ans.: 1(c), 2(b), 3(c), 4(a), 5(d)]
1.
When a 'l-phase supply is across
magnetic field produced is
a 1-phase winding,
the nature of the
CHAPTER 3: SYMMETRICAL COMPONENTS
a)
Constant in magnitude and direction
b)
Constant in magnitude and rotating at synchronous
c)
Pulsating in
d)
Rotating in nature
(CONTENTS:
Introduction, The a operator, Power in terms of symmetrical components,Phaseshift in V-l:. transformer banks,
2.
3.
nature
s.i
The damper windings are used in alternators to
a)
Unsymmetrical series Impedances, Sequence impedances, Sequence networks, Sequence networks of an
unloaded generator, ~quence networks of elements, Sequence networks of power system!
speed
Reduce eddy current loss
b)
Reduce hunting
c)
Make rotor dynamically balanced
d)
Reduce armature reaction
The neutral path impedance Zn is used in the equivalent sequence network·
models as
l~-----------------------------------~~-------------
INTRODUCTION
Rower systems are large and complex three-phase systems. In the normal operating conditions, these systems are in
balanced condition and hence can be represented as an equivalent single phase system. However, a fault can cause
the system to become unbalanced. Specifically, the unsymmetrical faults: open Circuit, LG, LL, and LLG faults cause
the system to become unsymmetrical. The single-phase equivalent system method of analysis (using SLD and the
reactance diagram) cannot be applied to such unsymmetrical systems. Now the question is how to analyze power
systems under unsymmetrical conditions? There are two methods available for such an analysis: Kirchhoff's laws
method and Symmetrical components method.
The method of symmetrical components developed by C.L. Fortescue in 1918 is a powerful technique for analyzing
unbalanced three phase systems. Fortescue defined a linear transformation from phase components to a new
components called symmetrical components. This transformation
represents
set
of
an unbalanced three-phase system by-
I
____
.
~1n1
~---------------,
I
102
a set of three balanced three-phase systems. The symmetrical component method is:a modeling technique that
permits systematic analysis and design of three-phase systems. Decoupling a complex three-phase network into
three simpler networks reveals complicated phenomena in more simplistic terms.
Consider a set
L
1'..
yo,
three-phase unbalanced voltages designated as V"
v; .and V,. According
to Forte>cue theorem,
V,,2
l{l
these phase voltages can be resolved into following three sets of components.
Positive-sequence components, consisting of three phasors equal in magnitude; displaced from each other by
120° in phase, and having the same phase sequence as the original phasors, designated as Vol, VbI, and Vol
Negative-sequence components, consisting of three phasors equal in magnitude,·<jisplaced from each other by
120° in phase, and having the phase sequence opposite to that of the original phasors, designated as V." V", and
Vc~
3. Zerosequence components, consisting of three phasors equal in magnitude, and with zero phase displacementfrom each other, designated as V.o,VbO, and V,.
Since each of the original unbalanced phasors is the sum of its components, the orlgina] phasors expressed in.
I:
I
I
i
I
I
I
Figure 3.1
terns of their components are
Graphical addition of symmetrical components
To obtain unbalanced phasors.
3.2 THE OPERATOR
I
I
'a'
(3.1)
The relation between the symmetrical components reveals that the phase disPlacemtnt among them is either 120°
I
or
(l.Using this relationship,
only three independent components is sufficient to dete mine all the nine components.
The svnthesis of a set odthr~e unbalanced phasors from the three sets of syrnmetrica] components is shown in
~or this purpose an operator which rotates a given phasor by 120° in the positive dire
t=igurel,
~seful. The letter 'a' is used to designate such a complex operator of unit magnitude with an angle of 1200• It is
defined by
,\.,:
v..,..r
,..
:..•...
:...•...y•·..
~ .• _,
v"
J
~
-<v.,
a
=
I LI20'
=
Ion (counterclockwise) is very
-0,5 + ) 0.866
(3,2)
Vl'~
If the operator 'a' Is applied to a phasor twice In succession, the phasor Is rotated through 2400• Similarlv, three
V"
:-. ~;iv..~e"".
-.': 'tGm~'Oll':lntt
NI.'t:~I,...e·-=nQ,JGI\c.e
.:::cmpQ!'(nh
lIl1'O":~eqlMll\Ct'·':
c.DirlpOrajl"" ':~::
successive applications of 'a' rotate the phasor through 360°.
•
To reduce the number of unknown quantities, let the symmetrical components of
v; and V, can be expressed as product
of some function of the operator a and a component of V•.Thus,
V",,=V,o
Using these relations the unbalanced phasors can be written as
.'
I
·----------------.
~~03~
.
~
__ ~
__
1M
I ".'_'" ..'""
Since the sum of three balanced voltages Is zero, the zero-sequence component voltaae In a balanced three-phase
1
Vb:;; V.a + a 2Vtl + a V,l
/
Vc=-Vao+ lIVIl+
f
a2V'2
system is always zero, Further, the sum of Une voltages of even an unbalanced three-phase system Is zero and hence
the corresponding zero-sequence component of line voltages,
(3.3)
NUMERICAL EXAMPLES
lin
,I
matrix form,
I
I
1 II!'
aJ ll'' ' ,.
a' I
.t
1
II
lI
. (~.4)
1'","
I
I
I
,,0
Let
Vp =
[ v'""1J';
rV"flj
Vs = lv", ;
Ve
V,,2
[I
1 E~;tmple I : The line currents in a 3-ph 4 -wire system are la = 100<30°; Ib = 50<300°; Ie = 30<1800.
Fihd the symmetrical components and the neutral current.
/
I
Solution:
,
laO = 1/3(la + Ib + Ie)
= 27,29 < 4.7° A
lal = 1/3(1a+ alb + a2lc) = 57.98 < 43.3° A
la2 = 1/3(la + a2 Jb + a Ic) = 18.96 < 24.9° A
111 = la + Ib + Ie = 31aO = .81.87 <4.7° A
Example 2: The sequence component voltages of phase voltages of a 3-ph system are:
100<00 V; Val = 223.6 < -26.60 V; Va2 = 100 <1800 V. Determine the phase voltages.
A = 1 a'
1 a
VaO =
(3:5)
Solution:
Va = VaO + Val + Va2
• 223.6 <-26.60 V
Vb " VaO+ a2Val + a Va2 "213 < -99.90 V
The Inverse of A matrix is
Vc = VaO+ a Val + a2 Va2 = 338.6< 66.20 V
A-' =
+[:1
a
(3.6)
a'
With these definitions, the above relations can be written as
Example J: The two seq. components and the corresponding phase voltage ofa 3-ph system are VaO=1<60° V; VaI=2<0° V; & Va = 3 <0° V. Determine the other phase voltages.
,
Solution:
Va = VaO + Val
Vp=A V.;
V, =A·'V.
(3,7)
Va2
= Va -
+ Va2
VaO- Val
= 1 <60· V
Vb = VaO+a2Val + a Va2 ~ 3 < -120· V
Thus the symmetrl!al
components of V..
v, and
V, are given by
Vc= VaO +0 Val +a2 Va2=0 V
Example 4: Determine the sequence components if la = I0<60° A; Ib =10<-60° A; Ic; 10 <180° A.
Solution:
(3,8)
laO
..
= 1/3(la
+ Ib + Ic)
=0A
la1 = 1/3(la + a Ib + a2lc)
= 10<60· A
la2· 1/3(la + a2 Ib+ a Ic)
"OA
•
.
..
llJ!>_10~
•__
___;_------
_. __
Observation: If the phasors are balanced, two sequence components will be zero.'
-The powe;ln a three-phase system can be expressed In terms of symmetrical components of the associated voltages
and currents. The power flowing into a three-phase system through three lines a, band c is
Example 5: Determine the sequence components if Va = 100 <300 V;
100
<_90" V.
i :
I
s = P + j Q = V. I;
+ VbI; + V, I, •
(3.9)
I
Solution:
VaO = 1/3(Va + Vb + Vc)
Val
= 1/3(Va
'
=0 V
+ a Vb + a2Vc)
where V. , Vb and V, are voltages to neutral at the terminals
system in the three lines. In matrix form
= 0V
Va2 = 1/3(Va + a2 Vb + a Vc) = 100<30° V
Observation:
,tJ-mm:J
If the phasors are balanced, two sequence components will be zero.
S-[,_
Example 6: The llrre b of a 3-ph line feeding a balanced Y-Ioad with neutral grounded Is:op~n resulting In line
currents: la • 10<0° A & Ic = 10<120° A. Determine the sequence current components.'
.
'.
Thus
Solution:
5 = [A
Ib =0 A.
laO = 1/3(la + Ib + Ic)
VI' [AI]'
= 3.33<60° A
lal = 1I3(la + a Ib + a2lc)
= 6.66<0°
la2 = 1/3(la + a2 Ib + a Ic)
= 3.33<-60° A
Example 7: One conductor of a 3-ph line feeding a balanced
Using the reversal rule of the matrix algebra
A
delta-load is open. Assuming that line c Is open, if
current in line a is 10<00 A, determine the sequence components of the line currents.
Noting that A' = A and
a and a'
are conjugates,
Solution:
Ic = 0 A; la = 10<0° A.
laO = 1/3(la + Ib + Ie)
f
~ Ib = 10<120° A
S:::::
= 0A
lal = 1i3(la + a Ib + a2lc)
= 5.78<-30° A
la2 = 1/3(la + a2 Ib + a Ie)
= 5.78< 30° A
[v"o
",,1
3.3 POWER IN TERMS OF SYMMETRICAL COMPONENTS
~
,,,{:
I
a'
a
:.][:
I
a
a'
~f:J
Ia2
or, since A' A* is equal to 3U where U Is 3x3 unit matrix
Note: The zero-sequence components of line currents of a delta load (3-ph 3-wire) system arezero.
L
and
s=3ko
~
~
V"I
I,. lb. and I, are the currents flowing into the
Thus the complex three-phase power Is given by
x
R _=".R"',,","·,rll1fO'---(
T;;
(3.10)
C __ :::.::-H"
~
..,...,.
. i-;;*'
I.
Here, 3V,ul'Q, ~V"I., and 3V.,I" correspond to the three-phase power 'delivered to the zero-sequence system,
Pos.tive-s~querIC~system, and negative-sequence system, respectively. Thus, t_he -totai three-phase power in the
(a)
unbalanced system Is equal to the sum of the power delivered to the three sequence systems representing the
three-phase system.
Wiring diagram
ill
C2
Al
042
3.4 PHASESHiFTOF COMPONENTSIN Y-Il TRANSFORMERBANKS
Cl
01
lJ2
Positive sequence
The dot convention Is used to designate the terminals of transformers. The dots are placed at one end of each of the.
Negativesequence
ib J Voltas. COmpOl\~nts
winding on the same iron core of a transformer to indicate that the currents flowing from the dotted terminal to the
unmarked terminal of each winding produces an mmf acting in the same direction in the magnetic ctrcuit, In that:
case, the voltage drops from dotted terminal to unmarked terminal in each side of the windings are in phase.
Figure 3,2 Wiring diagram and voltage phasors of a Y-I!. trarJformer
With Y connection on HT side.
The HT terminals of three-phase transformers are marked as Hl, H2 and H3 and the correspcnolng LT side terminals
are marked Xl~X2 and X3. In Y-Y or I!.-I!. transformers, the markings are such that voltages to neutral from terminals
Consider a Y- fl transformer as shown in Figure a. The HT side terminals H1, H2, and H3 are connected to phases A, B,
H1, H2, and H are In phase with the voltages to neutral from terminals Xl, X2, and X3,:respectively, But, there will
be a phase shlf (of 30') between the corresponding quantities of the primary and secon(lary sides 01 a star-delta (or
delta-star) transformer. The standard for connection and designatjon of transformer banks Is as follows:
linked magnetically (by being wound on the same core). In Figure a winding AN is the phase on the Y-side which Is
xi. X2
linked magnetically with the phase winding bc on the I!. side. For the location of the dots on the windings VA. is In
~hase with Vb,. Following the standards for the phase shift, the phasor diagrams for the sequence components of
'1.
The HT side tenmlnals are marked as H1, H2 and H3 and the corresponding LT side terminals are marked
and X3.
2.
The phases in the HT Side are marked in uppercase letters as A, B, and C.. Thus ior the sequence abc, A is
connected to Hl, B to H2 and eta H3. Similarly, the phases in the LT side are mark.ed:in lowercase letters as a, b
and c.
3. The standard for designating the terminals Hl and Xl on transformer banks requires that the positive-sequence
voltage drop from H1 to neutral lead the positive sequence voltage drop from Xl to Deutral by 30' regardless of
the type of connection in the HT and l T sides. Similarly, the voltage drops from H2 to neutral and H3 to neutral
an~ C, respectively and the phase sequence Is ABC. The windings that are drawn In parallel directions are those
~oltages are shown In Figure b. The sequence component of V.", is represented as VA' (leaving subscript 'N for
convenience and all other voltages to neutral are similarly represented. The phasor diagram reveals that VA' leads
V" by 30°. This will enable to designate the terminal to which b is connected as Xl. Inspection of the positivesequence and negative-sequence phasor diagrams revels that Vol leads VA' by 90' and Vol lags VA'by 90'.
lead their corresponding values, X2 to neutral and X3 to neutral by 3rt.
From the dot convention and the current directions assumed in Figure a, the phasor diagram for the sequence
components of currents can be drawn as shown in Figure c. Since the direction spectfled for I. in Figure a is away
from the dot in the winding and the direction of I" is also away from the dot in its winding, I.and I" are lSrt out of
phase. Hence the phase relation between the Y and I!. currents is as shown in Figure'c. From this diagram, it can be
seen that I" leads IAI by 90° and I" lags I., by 90°.
components on the two Sidesof the transformer gives:
L.__
"~'I"'"'Wlr
.
~~
__
...,....--------_
Summarizing these relations between the symmetrical
,
I
r------ -------------------~~Q----------~~
•
I
I
~'"
I
-. itr..,
r'b>. 'I'
"ai-""""",' ,
(vab +a ,""'.
Since Vanl = Vab1<-300 and Van2
Van1
= Vab2<300
= 0.985<73.6°-30°
= 0.985<43.6° pu (L-L base)
F;osit;vp. ~sequence
componc.nts
l-,1el!at'."ews~uene~· .
compOnents
.
Van2 = 0.235<220.3°+30°
= 0.235<250.3° pu(L-L base)
Figure 3.3 Current phasors of Y-A transformer with Y connection Oil·tiT side.
Since each resistor is of 1.0<0 pu. Impedance,
lan1 = (Vanl/Z) = 0.985<43.6° pu.
lal
=
+]
1':'1
lan2 = (Van2/Z) = 0.235<250.3° pu.
(3.11)
Where each volt,e
and current is expressed in per unit. Although, these relations are obtained for
Y- A transformer
with Y connection in the HT side, they are valid even when the HT side Is connected in A a.ndthe LTside in Y.
~he directions are tve for currents from supply toward the delta primary and away from the Y-slde toward the load.
1he HT side line to neutral voltages are
NUMERICALEXAMPLES
VAl = - j Val = 0.9'85<-46,4°
VAl = +j Va2 = 0.235<-19.7°
Example 8: Three identical resistors are Y-connected to the LTY-side of a delta-star transformer. The voltages at the
resistor loads are [vab] = 0.8 pu., IVbcl=1.2 pu., and IVcal=1.0 pu. Assume that the: neutral of the load is not
connected to the neutral of the transformer secondary. Find the line voltages on the HT .Ide of the transformer.
VA.·=VAl +VA2= 1.2<-41.3° pu,
VB1 = a2VAl
and
VB2 = a VA2
VB = VB1 + VB2 = 1<180° pu.
Solution:
VCl = a VAl
Assuming an angle of 180° for Vca, find the angles of other voltages
Vab = 0.8<82.8° pu
and
VC2 = a2VA2
VC = Vel + VC2 = 0.8<82.9° pu.
T~e HT side line voltages are
Vbc = 1.2<-41.4° pu
Vca = 1.0<180° pu
VAB = VA-VB
= (1/3) VAB
The symmetrical components of line voltages are
VBC = VB-Vc
= 2.06<-22.6° pu. (L-N base)
= 1.19<-22.6° pu. (L-L base)
= 1.355<215.8° pu. (L-N base)
= (1/3) VBe = 0.782<215.8° pu. (L-L base)
VabO = 1/3 (Vab +Vbc + Veal = 0
Vabl
= 1/3
VCA = VC-VA
= 1.78<116.9° pu. (L-N base)
(Vab +aVbc + a2Vca) = 0.985<'73.6°V
,
f---F-ig-U-re-3,-s-Se-q-u-e-nc-e-i-m-p-eJ.d~an.:1.c-e-s-o-f-a-Y--c-o-n-n-ect-ed--Io-a'-d-·.
~---.
---.------
I
z,. ~ j1 n ~'__";" •• '~Q'
I NUMERICALEXAMPLES
i
---,-'--'-_- __
Z'Gt().$eQuonce:ne~ork.
.
I
ro""
__,!
3Z. • iJO n ~L
I
t ~-
~'--'-~:
,
Example9:'A Y-connectedsourcewith phasevoltages Vag = 277<0°,Vbg = 260<-1200andVcg = 295<1150is applied
to a balancedf1.loadof JO<4000/phasethrough a line of impedance1<85°0. The neutra! bf the sourceIssotldlv
grounded. Draw the sequencenetworks of the systemand find sourcecurrents.
Solution:
VaO= 15.91<62.110V
Val = 277.1<-1.70V
Va2= 9.22<216.70 V
Y eq. of f1.load= 10<4000/phase
Zline = 1<850O.
+
Zneutral = 0
Vn
z. - ~+ il00
laO= 0<00 A
la1 = 25.82<-45.60A
la2 = 0.86<172.80 A
1,-
v,
1,.13. +./4I/Jt~iJO)
P06itive-seqycncv
Iletwork
.7.454LZUr
n
la = 25.15<-46.80A
Ib = 25.71<196.40A
Ic
1, • 1', - 7.454~
= 26.62<73.80 A
n
3.6 SEQUENCE IMPEDANCES AND SEQUENCE NETWORKS
The impedance of a circuit to positive-sequencecurrents alone is called the Impedanceto pOsitive-sequencecurrent
or simply positive-sequence impedance, which is generally denoted as Z,. Similarly, the impedance of a circuit to
negative-sequencecurrents alone is called the Impedanceto negative-sequencecurrent or simply negative-sequence
impedance, which is generally denoted as Z,. The impedanceof a circuit to zer{)-sequeneecurrents alone1scalledThe
-
-
impedance to zero-sequence current or simply zero-sequence Impedance, which is generally denoted as
Zo.
1
In the
analysis of an unsymmetrical fault on a symmetrical system, the svrnmetrtcal components of the unbalanced
currents that are flowing are determined. Since In a balanced system, the components' currents of one sequence
The negative- and zero-sequence networks are composed of only the respective sequence impedances as there is no
corresponding sequence emf. The reference bus for the positlve- and neaative-Seqtence networks Is the neutral of
the generator.
,
cause voltage drops of like sequence only and are independent of currents of O\he.r·seq~ences,currents O.fanyone
sequence may be considered to flow in an independent network composed of the generated voltages, if any. and
impedances to the current of that sequence only.
.
,
.
The cu;rent flowing in the Impedance Zn between neutral and ground Is 31'0as sho~n I~ Fig.Thus the zero-sequence
The singl.e-p:fhse.equivalent circuit consisting of the impedances to currents of any on.:e Sequence only is called the
sequence ne
current
IS
voltage drop from point a to ground, is
(-I",Zoo - 31.oZn)where Z,o Is the z~ro-sequence Impedance of the
generator. Thus the zero-sequence network, which Is single-phase equivalent circuit assumed to carry only one
phase, must have an zero-sequence impedance of Zo =
+3Zn).
(z.o
ark of that particular sequence. Thus, the sequence network corresporiding to positive-sequence
led the positive-sequence network. Similarly,· the sequence network corresponding to negative-
sequence current is called negative-sequence network, and that corresponding to zero-sequence current is called
zero-sequence network. The sequence networks are interconnected
in a particular way to represent various
from the sequence networks, the voltage drops from point a to reference bus (or ground) are given by
unsymmetrical fault conditions. Therefore, to calculate the effect of a fault by: the method of symmetrical
components, it is required to determine the sequence networks.
3.7 SEQUENCE NETWORKS OF UNLOADED GENERATOR
Vao :;:
(3.15)
Consider an unloaded generator which is grounded through a reactor as shown in :Figure. When a fault occurs.
"\!\~l~-";
..t.'C.'t\~'i>
unbalanced currents depending on the type of fault will flow through the lines. Thes~ currents can be resolved irto
their symmetrical components. To draw the sequence networks of this generator, the 'component voltages/currents,
component Impedances are to be determined. The generated voltages are of positive-sequence only as the
fo.IT'+-
generators are designed to supply balanced three-phase voltages. Hence, positive-sequence network is composed of
1:,
an emf In series with the positive-sequence Impedance. The generated emf In this network Is the no-load terminal
1.
V..1
,
~1;''''' «
voltage to neutral. which Is also equal to the transient and subtranslent voltages as the generator is not loaded. The
(b) PQsiti\fe~,,(!qut!ne~Jei\lj\)rk
reactance in this network is the subtransient, transient, or synchronous reactance, depending on the condition of
study.
~
Ie) Nea.Uv.-,oquente current paths
a
~
Reference bus
3Z.} Zn
\'",,0
Z..,
Figure 3.6 Circuit of an unloaded generator grounded through reactance.
~
(e} Z"o-$oqvonce. curr"nt paths
(f)
.Zoro.sequenc.
network
Figure 3.7 Sequence current paths in a generator and
,
f
,----------------.------
lU1DL_
~------
_
The corresponding sequence networks.
according to the SLD. All the generated
emfs are positive-sequence voltages. Hence all the per unit
reactance/impedance diagrams obtained in the earlier chapters are positive-sequence networks. The negativesequence generated emfs are not present. Hence, the negative-sequence network for a power system is obtained by
Eq. 3.15 applicable to any unloaded generator are valid for loaded generator under steady state conditions. These
omitting all the generated emfs (short circuiting emf sources) and replacing all impedances by negative-sequence
impedances from the positive-sequence networks.
relations are also applicable for transient or subtranSient conditions of a loaded generator if Eg' or Eg" is substituted
for Ea.
Since all the neutral paints of a symmetrical three-phase system are at the same potential when balanced currents
are flowing, the neutral of a symmetrical three-phase system is the logical reference point. It is therefore taken as
3,8 S"QUENCEIMPEDANCEOFCIRCUITELEMENTS
the reference bus for the positive- and negative-sequence networks. Impedances connected between the neutral of
the machine and ground is not a part of either the positive- or negative- sequence networks because neither
positive- nor negative-sequence currents can flow In such Impedances.
For obtaining the sequence networks, the component voltages/ currents and the component impedances of all the
elements of the network are to be determined. The usual elemehts of a power system are: passive loads, rotating
machines (generators/ motors), transmission lines and transformers: The posltlve- and negative-sequence
impedances of linear, symmetrical, static circuits are identical (because the impedance of such circuits is
independent of phase orier provided the applied voltages are balanced).
Zero-Sequence Networks: The zero-sequence components are the same both in mafitude and in phase. Thus, It Is
equivalent to a single-phase system and hence, zero-sequence currents will flow a Iy if a return path exists. The
reference paint for thls network is the ground (Since zero-sequence currents a e flowing, the ground Is not
necessarily at the same point at all points and the reference bus of zero-sequence networle does not represent a
ground of uniform potential. The return path is conductor of zero Impedance, which is the reference bus of the zerosequence network.).
The sequence impedances of rotating machines will generally differ from one another. This.is due to the different
conditions that exists when the sequence currents flows. The flux due to negative-sequence currents rotates at
double the speed of rotor while that the positive-sequence currents Is stationary with respect to the rotor. The
resurtant
flux due to zero-sequence currents is ideally zero as these flux components adds up.to zero, and hence the
If a circuit is Y-connected, with no connection from the neutral to ground or to another neutral paint in the clrcult,
zero-sequence reactance is only due to the leakage flux. Thus, the zero-sequence impedance of these machines is
smaller than positive- and negative-sequence impedances.
no zero-sequence currents can flow, and hence the impedance to zero-sequence current is infinite. This is
represented by an open circuit between the neutral of the Y-connected circuit and the reference bus, as shown in
Fig. 3.6a. If the neutral of the V-connected circuit is grounded through zero impedance, a zero-impedance path
(short circuit) is connected between the neutral paint and the reference bus, as shown in Fig. 3.6b. If an impedance
The positive- and negative-sequence impedances of a transmission line are identical, while the zero-sequence
Zn is connected between the neutral and the ground of a Y-connected Circuit, an Impedance of 3Zn must be
impedance differs from these. The posltive- and negative-sequence impedances are identical as the transposed
• connected between the neutral and the reference bus (because, all the three zero-sequence currents (31.. ) flows
through this impedance to cause a voltage drop of 31ec10), as shown in Fig. 3.6c.
transmission lines are balanced linear circuits. The zero-sequence impedance is higher due to magnetic field set up
by the zero-sequence currents Is very different from that of the positive- or negative-sequence currents ( because of
no phase difference). The zero-sequence reactance .is generally 2 to 3.5 times greater than the positive- sequence
reactance. It Is Customary to take all the sequence Impedances of a transformer to be Identical, although the zerosequence impedance slightly differs with respect to the other two.
3.9 SEQUENCENETWORKSOF POWERSYSTEMS
In the method ad symmetrical components, to calculate the effect of a fault on a power! system, the sequence
networks are developed corresponding to the fault condition. These networks are then interconnected depending
on the type of fault. The resulting network is then analyzed to find the fault current and other. parameters.
Positlve- and Negative-Sequence Networks: The positive-sequence network is obtained by determining all the
positive-sequence voltages and positive-sequence impedances of individual elements, and connecting them
..
A II-connected circuit can provide no return path; its impedance to zero.... quence line curr8~ts Is therefore Infinite.
Thus, the zero-sequence network is open at the II-connected circuit, as shown in Fig.3.7 i-lbwever zero-sequence
currents can circulate inside the II-connected circuit.
~
~1.1.~q
I
~
~
A-efeN-nce..:b'-O':;::'"'- __
~
__
,
;""' __
~~
.
I
I
2.
I
I
~~N
~
Z.
(<&)
i-~
I
!
l.
I
..
I
.
Rof.r"nee b'~~._
..
·[
I
i
I
.
.
Z
I
I
(b)
Case 1: Y-Y Bank with both neutral grounded:
In this case, a path through transformer exists for the zerosequence current. Hence zero-sequence current can flow In both sides of the transformer provided there is
complete outside closed path for it to flow. Hence the points on the two sides of the transformer are connected
by the zerO-sequence impedance of the transformer.
3. Case 3: Y- ~ Bank wIth grounded Y: In this case, there Is path for zero-sequence current to ground through the Y
as the corresponding Induced current can Circulate in the A. The equivalent circuit must provide for a path from
lines on the V side through zerd-sequence Impedance of the transformer to the reference bus. However, an open
circuit must exist between line and the reference bus on the A side. If there is an impedance Zn between neutral
and ground, then the zero-sequence impedance must Include 3Zn along with zero-sequence Impedance of the
transformer.
.
,
1-'----'VVYV\r---:--:-. 'N
...._..3
I
I
I
'1
.'
:.
,.
I
'":;.."
z""-S..
the other). An open circuit exists for zero-sequence current between two parts of the system connected by
the transformer bank.
4.
Case 4: Y- ~ Bank with ungrounded Y:
In this case, there Is no path for zero-sequence current. The zerosequence Impedance is infinite and is shown by an open circuit.
5. , Case 5: ~-~ Bank: In this case, there is no return path for zero-sequence current. The zero-sequence current
cannot flow in lines although it can circulate in the t!. windings.
6. The zero-sequence equivalent circuits determined for the Individual parts separately are connected according to
the SLD to form the complete zero-sequence network.
Procedure to draw the sequence networks
The sequence networks are thre1! separate networks which are the Single-phase equivalent of the corresponding
symmetrical sequence systems. These networks can be drawn as follows:
(0)
1. For the given condition (steady state, transient, or subtransient), draw the re,\ctance diagram (selecting proper
base values and converting all the per unit values to the selected base, If nece~sary).This will correspond to the
positive-sequence network.
Figure 3.8 Zero-sequence equivalent networks of Y-connected load
~eJJrenc'ebUS:
I '.'
'••...::....•........•....
I
I
f
I
'
~
.:
: : .'
:
,.:
.. '
~'."
~. '_'
- ' ..
,
,
Z
Figure 3.9 Zero-sequence equivalent networks of II -cormected load
The zero-sequence equivalent circuits of three-phase transformers deserve special attention. The different possible
combinations of the primary and the secondary windings in Y and t; alter the zero-sequence network. The five
possible connections of two-winding transformers and their equivalent zero-sequence networks are shown in
Fig.3.8, The nfworks
are drawn remembering that there will be
no
primary current when there is no secondary
current, negle ing the no-load component. The arrows on the connection diagram show the possible paths for the
zero-sequenc
current. Absence of an arrow indicates that the connection is such that zero-sequence currents
cannot flow, The letters P and Q identify the corresponding points on the connection dlagram.and equivalent CIrcuit.
1.
Case 1: Y-Y Bank with one neutral grounded:1f either one of the neutrals of a Y-Y bank is ungrounded, zerosequence current cannot flow in either winding ( as the absence of a path through one winding prevents current
I
T.,
~'
-::?(
I
.
'--"-'-'d·
Line.
X, u,X, '" 20
n .::,r
Xa=60n
4-1
100 MVA
=-".
M
t~l~
..,.
100 MVA
138 !N
X-' Co· tl.ll)
X, = 0.17
XI)
2
~~
100 MVA
138,kVA/138··kVY
X = O,lO per unit
0.06 per unit
13S-kVY/13.S-kV A
x "'" 0.10
par
100 MVlI.
13.8 kV
X· ~ 0.20
= 0.21
Xu = 0.10
X,,1:&. 0.05 por.\.
unit
'x,
~.
r
f•
·1.0..'
fi4 ~~
~
'
~~J2
C
/"., .. ,
R.I.,.nc~
bu-,--
;0,105'
,(>.15
-
J",_..,,,,,,
.
.
.
1.0 ",
Figure 3.8 Zero·sequence equivalent networks of three-phase
+
_.
ibi
transformer banks for various combinations.
2.
Determine the per unit negative-sequence impedances of all elements (if the values 9f negative sequence is not
given to any element, it can approximately be taken as equal to the positive-sequence Impedance). Draw the
negative-sequence network by replacing all emf sources by short circuit and all Impedances by corresponding
negative-sequence impedances In the positive-sequence network.
: .
3.
Determine the per unit zero-sequence impedances of all the elements and draw the zero-sequence network
corresponding to the grounding conditions of different elements.
JO.20
P05i1lve'S(lQuer,CQ
....._
'__
F.~,... 1.0 !.!X
I\OIWlJj1: .
NUMERICAL EXAMPLES
EXERCISE PROBLEM:
ham pie 10: For the power system shown in the SLD,draw the sequence networks.
For the power system shown in the SLD, draw the sequence networks.
I
L____
L. ,,·,'
.._,,__~
~------
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~--------~
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(4.2)
a power system wtthout and with fault Impedance, open conductor faults In power systems, examples,
This equation is used along with the equations i.e., conditions under fault (c.u.f.), derived to describe the fault under
consideration, to determine the sequence current '" and hence the fault current I" In terrlll of E, and the sequence
impedances, Z" 2, and Zo. Thus during unsymmetrical fault analysis of any given type of fault, two sets of equations
as follows are considered for solving them simultaneously to get the required fault parameters:
I
4.1 PREAMBLE
>
»
Equations for the conditions under fault (c.u.f.)
Equations for the sequence components (sequence equations) as per (4.2) above.
The unsymmetrical faults will have faulty parameters at random. They can be analyzed by: using the symmetrical
components. The standard types of unsymmetrical faults considered for analysis include the:following (in the order
of their severity):
4.2 SINGLELINETO GROUND FAULTON A CONVENTIONAL(UNLOADED)GENERATOR
~
>
~
»
Line-tO-<;round (L-G) Fault
Line-to-Line (L-L)Fault
Double Llne-to-Ground (L-L-G)Fault and
Three-Phase-to-<;round (LLL·G)Fault.
Further the neutrals of various equipment may be Irounded or Isolated, the faults can occur at any general point F
of the given system, the faults can be through a fault Impedance, etc. Of the various typ~s of faults as above, the 3-$
fault involving the ground Is the most severe one. Here the analysis Is considered in two:stages as under: (i) Fault at
the terminals of a Conventional (Unloaded) Generator and (ii) Faults at any point F, of a given Electric Power System
(EPS).
.
Consider now the symmetrical component
relational equations derived from. the 'three sequence networks
corresponding to a given unsymmetrical system as a function of sequence Impedance~ and the. positive sequence
voltage source in the form as under:
Figure 4.1 LG Fault on a Conventional Generator
(4.1)
These equations are refered
as;
as the sequence equations. In matrix
Form the sequence equations can be considered:
A conventional generator is one that produces only the balanced voltages. Let Ea, nd Ec be the internally
generated voltages and Zn be the neutral impedance. The fault is assumed to be on the phase'a' as shown in
figure 4.1. Consider now the conditions under fault as under:
c.u.f.:
(4.3)
Now consider the symmetrical components of the current I, with 1.=1,=0, given by:
'''wlr
..
1.
0
I
1 I"
11
~ (1/3) ~ a
/'0
.'
a' a
I.,
I
I
I
0
I~-6.
Fault phase voltage V. = 0,
7. Sound phase voltages Vb= a'Vn+aV,,+V.o; V, = aV,,+a'V.,+v.o
89'. Fault phase power: V.I; = 0, Sound pahse powers: Vbl; = 0, and V,I,' = 0,
If Z, = 0, then Zo= ZoO'
(4.4)
I,
10. If Z, = <1J, then Zo= <1J, i.e., the zero sequence network is open so that then, 1.=1.=0.
Solving (4.4) we get,
I" = I" = 1.0 = (lj3)
(4.5)
4.3 LINETO LINEFAULTON A CONVENTIONALGENERATOR
Further, using equation 14.5) in (4.2). we get,
I~:J1j ;r : :
tV"
°
0
0
z,
I"
(4.6)
Pre-multiplying equation (4.6) throughout by [1 1 1), we get,
i.e.,
Va = E, -I., (Z, + z, + 20) = zero,
Or in other words,
I" = [E,/(Z, + Z, + Zo)l
(4.7)
Figure 4.3 LL Fault on a Conventional Generator
I
J
Consider a line to line fault between phase 'b' and phase 'c' as shown in figure 4.3, the terminals of a conventional
generator, whose neutrsl ts grounded through a reactance. Consider now the conditions under tault as under:
FIgutj, 4.2 Connection of sequence networks·for LG Fault
on phase a of a Conventional Generator
c.u.f.:
I. = 0; I, = - I~ and
The equation (4.7) derived as above implies that the three sequence networks arc connected in series to
simulate a LG fault, as shown in figure 4.2. Further we have the following relations satisfied u.nderthe fault
conditions:
= = =
=
z, ...
1. I" I., I•• (1./3) [E,/(Z, +
20)]
2. Fault current I, = I. = 3101 = [3E./(Z, + Z, + loll
3. Vol = E. -I"Z, = E,(Z,+Zo)/(Z,+Z,+Zo)
4. V" = - E.Z,/(Z,+Z,+Zo)
5. V•• = - E.Zo/(Z,+Z,+Zo)
"""""'wlr
(4.8)
No.W consider the symmetrical components at the voltage V. with Vb=V" given by:
IVo·1
Vol = (1/3)
V"
I'
v,= V,
1J 1a
r' I I
b
a'
a' a
~
Vb
(4.9)
1---Solving (4.4) we get,
Vol =
l'
V" = V" = E,Z,/(Z,+Z,)
S. V,o=D,
4.
(4.10)
6.
Further, consider the symmetrical components of current I, with 1,=-1" and 1,=0; given by:
j I"1"1
r, j I
1/
= (1/3) ~
a'
• Vc z aVu+a2V'2+V,O = (a+a1)V'1 I:: _ V'I
10. Since 1,0=0,the presence of absence of neutral Impedance does not make any difference In the analysis.
a' a
I"
Fault phase voltages;V"
7. Sound phase voltage; Va = V"+V,,tV .. = 2V,,;
8. Fault phase powers are V,I; and V,I:,
9. Sound phase power: V,I: cO,
(4.11)
Solving (4.11) we get,
(4.12)
4.4 DOUBLE LINE TO GROUND FAULT ON A CONVENTIONAL GENERATOR-
Using equation (4.10) and (4.12) in (4.2), and since V.o = 0 ( 1'0being 0), we get,
Vol
o
o
0
Z,
-I"
(4.13)
Pre-multiplying equation (4.13) throughout by [0 1 -1], we get,
Or in other words,
I" = [E,/(Z, ...Z,)]
(4.14)
I
Figure 4.5 LLG Fault on a Conventional Generator
COhsider a double-line to ground fault at the terminals of a conventional unloaded generator, whose neutral is
.gr6unded through a reactance, between. phase 'b' and phase 'c' as shown in figure 4.5, Consider now the conditions
under fault as under:
q.u.f.:
Figure 4.4 Connectionof sequence networks for LL Fault on phases" & c of
a Conventional Generator
(4.15)
Now consider the symmetrical components of the voltage with V,=V,=O, given by:
The equation (4.14) derived as above implies that the three sequence networks 'are connected such that the zero
sequence network is absent and only the positive and negative sequence networks are connected in seriesopposition to simulate the LL fault, as shown in figure 4.4. Further we have the following relations satisfied under the
fault conditions:
.
1. I" = -I" = lE,/(z,
=
V.,
VOl
= (1/3)
1J 1a
a'
a' a
ro'l
II
o
I'
(4.16)
Solving (4) we get,
• Z,)] and 1'0= 0,
2. Fault current I, = I, -I, = ["3E,/(Z, • Z,J] (since I,
3.. Vol ~ E, -I"Z, = E,Z'/(Z,.Z,)
IV,ol
= (a'-a)l"
= "31,,)
(4.17)
..
,,,
~~sweha~------------------------~~-------~---------~-----~--------------'
IV"
(4.26)
= E.. -I .. Z,
r
1 So that after solving for 1., we, get,
i
I., = [ E./ Zd
(4.27)
I
Figure 4.9 Unsymmetrical faults in Power Systems
Thus, ior all the cases of unsvmrnetrtcal fault analysis conslderad above, the sequence equations are to be changed
as under so as to account for these changes:
Figure 4.8 Connection of sequence networks for 3.phase ground Fault 00 phases band
c of a Conventional Generator
o
The equation (4.26) derived as above implies that, to simulate the 3-phase groundJautt, the three sequence
networks are connected such that the negative and zero sequence networks are absent and only the positive
sequence network Is present, as shown in figure 4.8. Further the fault current, I, in case Of a 3-phase ground fault is
given by
o
0
0
Z,
I"
(4.29)
(i) LG Fault at any point F of a glrl~n Power system
I, = 1,,= I, = (EJZ,)
Let phase 'a' be on fault at F so that then, the c.uJ. would be:
It is to be noted that the presence of a neutral connection without or with a neutral" impedance,
simulated conditions in case of a three phase to ground fault.
z., will
not alter the
I. = 0; I, = 0;
and
V. = o.
Hence the derived conditions under fault would be:
I., = 1.2 = I", = (1./3)
4.6 UNSYMMETRICALFAULTSON POWERSYSTEMS
I" = [V'/ (Z, + Z, + lo)l and
I,= 31.,
(4.30)
In all the analysis 50 far, only the fault at the terminals of an unloaded generator have been consldered. However, 1
i
faults can also occur at any part of the system and hence the power system fault at any general point is also quite
I
important.
The analysis of unsymmetrical fault on power systems is done in a slrnilar' way as that followed thus far
for the case of a fault at the terminals of a generator. Here, instead of the sequence impedances of the generator,
(Ii) LLFault at any paint F of a given Power system
each and every element is to be replaced by their corresponding sequence impedances and the fault is analyzed by
Let phases 'b' and 'c' be on fault at F so that then, the c.u.I, would be:
SUitably connecting them together to arrive at tne Thevenln equivalent Impedance If that given sequence. Also, the
I, = 0; I, = - I,; and V,
internal voltage of the generators of the equivalent circuit for the positive sequence network is now V, (and not E,),
the pre-fault voltage to neutral at the point of fault (PoF) (ref. Figure 4.9).
.
I" = (V, I (Z, + Z,)] and
I, = i, = ·1,= [..,13 Vr! (Z, +Z,)]
i
i
L____
t
. "~:'lr
V,
Hence the derived conditions under fault would be:
I
I
.-<1-
a
:1
1
I
1
(4.31)
~
(ii) LLGFault at any paint F of a given Power system
Let phases 'b' and 'c' be on fault at F so that then, the c.u.f. would be:
4.7 OPEN CONDUCTOR FAULTS
I, = 0 and V. = V, = 0
Hence the derived conditions under fault would be:
Various types of power system faults occur In power systems such as the shunt type foults (lG, ll, LlG, lLlG faults)
VOl= V" = V.o = (V,/3)
and series type faults (open conductor and cross country faults). While the symmetrical fault analysis is useful in
determination of the rupturing capacity of a given protective circuit breaker, the unsymmetrical fauit analysis is
useful in the determination of relay setting, single phase switching and system stability studies.
1,,= -I"W(Z, + Z,); I., = -I,,2,/(z, + 2,) and
(4.32)
When one or two of a three-phase circuit is open due to accidents, storms, etc., then unbaiance is created and the
,symmetrical currents flow. Such types of faults that come in series with the lines are refered as the open conductor
faults. The, open conductor faults can be analyzed by using the sequence networks drawn for the system under
consideration as seen from the potnt of fault, F. These networks are then suitablYICOnnectedto simulate the given
(ii) Three Phase Fault at any paint Fof a given Power system
Let all the 3 phases a, band c be on fault at F so that then, the c.u.f. would be: .
type of fault. The following are the cases required to be analyzed (ref. fig.4.1O).
Hence the derived conditions under fault would be:
a.~:
~:~tQ
I
b
I., = [V,/2,] and I, = 1.,=1,
)lj,'
,
V<tt'
J
8
[4.33)
:-
cl
Figure 4,10 Openconductor faults.
(i) Single Conductor Open Fault: consider the phase 'a' conductor open so that then the conditions under fault are:
The derived conditions are:
V.... = V.. r = V.. O' = (V.. /3)
(4.34)
These relations suggest a parallel combination of the three sequence networks as shown In fig. 4.11.
""''''''';...;..''
.....p
Zo
~
I
L--.---
-'---------J
t~<l4
•
______________________
I
----2~'h
~
.~
_
Figure 4.11 Sequence network connections for 1..conduct.or open fault
AlI'the faults considered so far have comprised of a direct short circuit from one or two lines to ground. The effect of
I
impedance in the fault is found out by deriving equations similar to those for faults through zero valued neutral
Impedance. The connections of the hypothetical stubs for consideration of faults through fault Impedance Z, are as
shown in figure 4.13.
It is observed that a single conductor fault is similar to a LLGfault at the fault point F of the system considered.
I
I
(ii) Two Conductor Open Fault: consider the phases 'b' and 'c' under open condition' sp that then the conditions
under fault are:
I
~
I
"I(~)
The derived conditions are:
!
I,] ; I" ; 1'0;1,/3 and
(4.351
I
I
These relations suggest a series combination of the three sequence networks as shown .in fig. 4.12. It is observed
figure 4.13
that .. double conductor fault Is similar to a LGfault at the fault point F of the system consldered,
Stubs Connections for faults through fault impedance
z,.
I
l
(i) LGFault at any point F of a given Power system through Z,
I
Let phase 'a' be on fault at F through Zp so that then, the c.u.f, would be:
Ib .0;
Ie == 0;
and
V••
o.
Hence the derived conditions under fault would be:
I
I,] ; I,,; 1'0 = (1,/3)
I
Figure 4.12
Sequence network connection for 2-conductor open fault.
I
1,= 31"
(4.37)
(iii) Three Conductor Open Fault: consider all the three phases a. band c, of a 3-phase system conductors be open.
The conditions under fault are:
(ii) LL Fault at any point F of a given Power system through Z,
1,+ 1,+ I, =0
Let phases 'b' and 'c' be on fault at F through Z" so that then, the c.u.f. would be:
The derived conditions are:
V.o= V,,2:=Oand Vil
= V,
. Henc~ the derived conditions under fault would be:
(4.361
These relations imply that the sequence networks are all open circuited. Hence. in a strict analystical sense, this is
not a fault at all!
I.] ; [Vr/ (Z, + Z,+z,1l and
I, = I, = -I, = [..J3 V, / (Z, + Z,+Z,))
4.8
(4.381
(iii) LLGFiJUltat any paint F of a given Power system through Z,
FAULTSTHROUGH IMPEDANCE
Let phases 'b' and 'c be on fault at F through ZP.so that then, the c.u.f. would be:
IL.----------
-l
I·
,
i-Hence
I
I
the derived conditions under fault would be:
V,' = Vol = v: = (VJ3)
I
I
I" = {V,/ [Z,+Z,IZo.3Z,)/(Z,+Zo+3Zf)J)
I.
1,,= -1.,(lo.3Z,j/(Z,+lo+3Z,); 1,0= -1"Z,!(Z,+(lo+3Z,) and
I
Example-
I, = 31'0
a:
Athree
phase generator wtth OOM"~ terminal voltages gives the following currents wnen under fault:
1400 A for a line-to-line fault and 2200 A for a line-to-ground fault. If the positive sequence generated voltage to
neutral is 2 ohms, find the reactances o.fthe negative and zero sequence currents.
(4.39)
I
Case a) Consider the conditions w.r.t. the LL fault:
I" = (E,d(Z, + Z,))
(iv) Three Phase Fault at any pO"'t F oj a given Power system through Z,
,
.
Let all the 3 phases a, band c be on fault at F, through Zr so that the c.u.f. would be! V.= I,Z,; Hence the derived
conditions under fault would be: I... [Vr/(Z,+Z,); The connections of the sequence n~tworks for all the above types
of faults through Z, are as shown in figures 4.14 and 4.15.
.
I, = I.
= -I, = "31"
,
.";3 E,.! (Z. + Z,) or
(Z, + Z,) = ";3 E,,/ I,
i.e., 2 + Z, = ";3 [2000/1400J
Solving, we get,
Z, = 0.474 ohms.
Case b) Consider the conditions w.r.l. a LG fault:
I" = [E,d(Z, + Z,+lo)J
I, = 31..
LGFault
LLFault
Figure 4.14
Sequence network connections for
= 3 E., / (Z, + Z,+Zo)
IZ,. + Z,+Z,)
= 3 EO)/
or
I,
LG& LLfaults through impedance
i.e, 2 + 0.474 + lo = 3 (2000/2200J
Solving, we get,
lo = 0.253 ohms.
Example-2: A dead fault occurs on one conductor of a 3-conductor cable supplied y a 10 MVA alternator with
earhed neutral. The alternator has tva, -ve and a-sequence components of Impedances per phase respectively as:
(0.5+)4.7), (0.2+)0.6) and 00.43) ohms. The corresponding LN values for the cable UP to the polnt of fault are:
(0.36+)0.25), (0.36+)0.25) and (2.9+)0.95) ohms respectively. If the generator voltage at no load (e,,) Is 6600 volts
between the lines, determine the (i}Fault current, (ii)Sequence components of currents In lines and (iii)Voltages of
healthy phases.
llG Fault
Figure 4.15
3-Ph. Fault
Sequence network (onnections for
LLG& LLLGfaults through impedance
There is LG fault 011 anyone
the fault current is given by:
of the conductors.
(i) Fault current: I, = 31", = [3E,/(Z, + Z, + loll
; 3(6600/"3)/ (4.32+j7.18)
= 1364.24 LS8.97°.
(ii) Sequence components of line currents:
4.9
EXAMPLES
I" ; I" = I" = 1,/3 = 1';3
Consider the LG fault to be on conductor in phase a. Thus
. .
.
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e-Notes
..2-"""64'--
by
--'-
_
i
Dr.K.Uma Rao. HOD & E&E Engg.. RNSIT. Bangalore
..265
I
isolation by circuit breakers will cause variations in the power flows, network bus voltages and generators rotor
.
,
Subject: Power System Analysis and StabilitY
speeds. The voltage variations will ,actuate the voltage regulators In the system and generator speed variations will
:
actuate the prime mover governors; voltage and frequency variations will affect the system loads, In stable systems,
practically all generators and loads remain connected, even though parts of the system may be Isolated to preserve
portion Cpyer,dj Chapter "S on
power
Sy't,m
Stability
bulk operations. On the other hand, an unstable system condition could lead to casf'dlng outlses ind i Shutdown
of a major portion of the power system.
5.1 INTROOUCTION
5.2 CLASSIFICATION OF POWER SYSTEM STABILITY
Power system stability of modern large inter-connected systems is a major problem for secure operation of the
system, Recent major black-outs across the globe caused by system Instability, even in' very sophisticated and secure
The high complexity of stability problems has led to a meaningful classification of the power system stability Into
systems, illustrate the problems facing secure operation of power systems, Earlier, stability was defined as the ability
various categories. The classlflcatlon takes into account the main system variable in which instability can be
of a system to return to normal or stable operation after having been subjected to some form of disturbance. This
observed, the size of the disturbance and the time span to be considered for assessingstability,
fundamentally refers to the ability of the system to remain in synchronism, However, modern power systems
operate under complex interconnections,
controls and extremely stressed conditions. Further, with increased
autcrnatlon and use of electronic equipment, the quality of power has gained utmost importance, shifting focus on
to concepts of voltage stability, frequency stability,
5.2.1 ROTOR ANGLE STABILITY
Inter-area OSCillationsetc,
Rotor angle stability refers to the ability of the synchronous machines of an inte~connJcted power system to remain
The IFEE/CiGRE'Oint Task Force on stability terms and conditions have proposed the following definition in 2004:
in synchronism after being subjected. to a disturbance,
"PowerS~stem tabillty Is the ability 0/ on electricpower system, tor a given initial operatln!!condition, to regain a
(decelerating) and lasing synchronism with other generators, Rotor angle stability depends on the ability of each
state 0/ operating equilibriumafter being subjected to a physical disturbance, with most system variablesbounded,
D...' that practicallythe entire system remains intact".
steady state, there is equilibrium between the input mechanical torque and output electromagnetic torque of each
synchronous machine to maintain equilibrium
Instability results in some generators accelerating
between electromagnetic torque and mechanical torque. Under
generator, and its speed remains a constant. Under a disturbance, this equilibrium is upset and the generators
The Power System Is an extremely non-linear and dynamic system, with operating parameters continuously varying.
accelerate/decelerate according to the mechanics of a rotating body. Rotor angle stability. is further categorized as
follows:
Stability is hence, a function of the initial operating condition and the nature of the disturbance, Power systems are
continually subjected to small disturbances in the form of load changes, The system must be ,in a position to be able
to adjust to the changing conditions and operate satisfactorily. The system must also withstand large disturbances,
Small single (or small disturbance) rotor angle stability:
which may even cause structural changes due to isolation of some faulted elements,
synchroni'sm under small disturbances. In this case, the system equation can be linearized around the initial
It is the ability of the power system to maintain
operating paint and the stability depends only on the operating paint and not on the disturbance. Instability may
result in
A power system may be stable for a particular Ilarge) disturbance and unstable. for another disturbance, It is
(i) A non OSCillatoryor a periodic increase of rotor angle
impOSsibleto design a system which is stable under all disturbances. The power system is generally designed to be
1
i
stable under those disturbances which have a high degree of OCcurrence.The response to a disturbance is extremely
complex and Involves practically all the equipment at the power system. For example, a shorticircuit learling to a line
,I
I
"-r'"
'n"'1r I n
Iii) Increasing amplitude of rotor oscillations due to ins~fflcient damping.
1
I
r
_266_:
-
--1
'''7'--
r- .._.
---,
.......
, The first form of Instability Is largely eliminated by modern fast acting voltas. r~,ulato(s and ·the second form of
Instability Is more common. The time frame of small signal stability Is of the order' of 10-20 seconds after a
: :
disturbance.
Large disturbance voltage stability:
It refers to the systems ability to maintain steady voltages following larg.
disturbances; It requires computation
of the non-linear response of the power system to include Interaction
between various devices like motors, transformer tap changers and field current limiters. Short term voltage stability
I .......
na rotor .....
"'.,,"
" ,,~,-
,.""
.." ,."
refers to th e abllltvof the
,",W",
system tc malr.tarn
synchronism under large disturbances, such as short Circuit, line outages etc. The system response involves large
I
excursions of the generator rotor angles. Transient stability depends on both the initial operating point and the
!
I
involves dynamics of fast acting load components and period of Interest Is In the order of several seconds. Long term
voltage stability involves slower acting equipment like tap-changing transformers and generator current limiters.
• Instability is due to loss of long-term equilibrium,
disturbance parameters like locatien. type, magnitude etc. Instability Is normally in the form of a periodic angular
separation. The time frame of interest is 3-5 seconds after disturbance.
The term dynamic stability was earlier used to denote the steady-state stability
,
tn
the presence of automatic
controls (especially excitation controls) as opposed to manual controls. Since all generators are equipped with
5.2.3 FREQUENCYSTABILITY
automatic controllers today, dynamic stability has lost relevance and the Task Force has recommended against its
I
usage.
Frequency stability refers to the ability of a power system to maintain steady frequency following a severe
disturbance, causing conSiderable Imbala,nce between generation and load. Instability occurs In the form of
5.2.2 VOLTAGESTABILITY
sustained frequency swings leading to tripping of generating units or loads. During frequency swings, devices such as
under frequency load shedding, generator controls and protection equipment get activated in a few seconds.
However, devices such as prime mover energy supply systems and load voltage regulators respond after a few
!
I
Voltage stability refers to the ability of a power system to maintain steady voltages at all buses in the system after
minutes. Hence, frequency stability could be a short-term or a long-term phenomenon.
being subjected to a disturbance. It depends on the ability of the system to maintain equlllbrlum between load
I
demand .nd load supply. Instability results in a progressive fall or rise of voltages of some buses, which could lead to
II
10" of load In an area or tripping of transmission lines, leading to cascading outages. This may eventually lead to loss
5.3 MECHANICSOF ROTATORYMOnON
of synchronism of some generators.
iI
Since a synchronous machine is a rotating body, the laws of mechanics of rotating bodies are applicable to it. In
I
rh~ cause of voltage instability is usually the loads. A run-down situation causing voltage instability occurs when load
I
dynamics attempt to restore power consumption beyond the capability of the transmission network. Voltage
I
capacity of the available reactive power resources. Voltage stability is categorized into the following sub-categories:
rotation we first define the fundamental quantities. The angle
em
Is defined, with respect to a circular arc with its
center at the vertex of the angle, as the ratio of the arc length s to radius
r.
stability is also threatened when a disturbance increases the reactive power demand bevond the sustainable
I
The unit is radian. Angular velocity
Small- disturbance voltage stability: It refers to the system's ability to maintain steady voltages when subjected to
smali perturbations such as incremental changes in load. This is primarily influenced by the load characteristics and
I
II
(S.l)
wm;;--
so,
dt
the controls at a given point of time.
and angular acceleration as
wm
Is defined as
(5.2)
----
dro"
a = -d(
-
=
.---__,_.------.-------l
.-----26,Ro__
d'B"
--d-I~-
269
M=IWm
(5.3)
(5.13),
The torque on a body due to a tangential force F at a distance r from axis of rotation is given by
And the kinetic energy is given by
T
r
F
(5.41
1
KE=-[ro,-
2
The total torque is the summation of infinitesimal forces, given by
I
T = J r dF
,
..
I
2
= -Mwm
(5.14)
From (5.14) we can see that the unit of M has to be J-sec/rad.
(5.5)
The unlt of' torque is N-m. When torque is applied to a body, the body experiences angular acceleration. Each
particle experiences a tangential acceleration a
= ra , where r is the distance of the particle from axis of rotation.
5.4 SWING EQUATION:
The tangential force required to accelerate a particle of mass dm Is
dF = a dm = r a dm
(5.6)'
The laws of rotetlon developed In section.3 are applicable to the synchronous machine. From(.5.8)
Ie<' =T
The torque required for the particle is
dT = r dF = r'
or
a dm
(5.15)
(5.7)
and that required for the whole body is given by
Here T is the net torque of all torques acting on the machine, which includes the shaft torque (due to prime mover of
T = a f r'drn
1= f
=Ia
(5.8)
r'cm
It is called the moment of inertia of the boov, The unit is Kg -
a generator or load on a motor), torque due to rotational
losses (friction, windage and core 1055) and
electromagnetic torque.
(5.9)
m'. It the
torque Is assumed to be the result of a
number of tangential forces F, which act at different paints of the body
Let Tm = shaft torque or mechanical torque corrected for rotational losses
T,
= Electromagnetic
or electrical torque
Now each force acts through a distance, ds = r de", and the work done Is }: F . ds i.e.,
dW = }: F r dem= demT
For a generator Tmtends to accelerate the rotor In positive direction of rotation as shown in Fig 5.1.
the corresponding torque for a motor with respect to the direction of rotation.
(5.10)
and
T= dW
dB",
(5.11)
Thus the unit of torque may also be Joule per radian. The power Is defined as rate of dOinll work. Using (5.11)
p=dW = TdB"=Tw
dt
dt
'"
The angular momentum M Is defined as
"n'''lr
1 'I
(5.12)
It also shows
,------.-------
.-----
ll1...
--,-_~
_
r'~--------------
~2~7~
3600 rpm
(5.25)
Non condensing
.
In (5.25)
om
is expressed In mechanical radians and
mechanical units). If 0 and
I
3600 rpm
2H d'o
I --;;:
d/' '" r; '" P,,,-P,
4-3
Water wheel generator
as
J
7-4
In mechanical radians per second (the subscript m indicates
0).••,
have consistent units (both are mechanic ..1or electrical units) (5.25) can be written
0)
-,
Slow speed < 200 rpm
2- 3
High speed> 200 rpm
2- 4
Synchronous condenser
Large
pu
12S}
1:0
.
25% less for hydrogen cooled
Small
Here (0., Is the synchronous speed in electrical
radls
((0., '" (~)
same base as H. For a system with an electrical frequency
H d'(j
l!
f dl'
f
= P" = P,,,=P;
(0.""
Synchronous motor with load varying
1 and p. is acceleration power in per unit on
from 1.0 to 5.0
Hz,(5.26) becomes
2.0
(S.27)
pu
In defining the inertia constant H, the MVA base used is the rating of the machine. In a multi machine system, swing
when
(5
equation has to be solved for each machine, in which case, a common MVA base for the system has to chosen. The
is in electrical radians and
constant H of each machine must be consistent with the system base.
pu
Let
(5.28)
Gm", = Machine MVA rating (base)
G'''''m = System MVA base
when (j is in electrical degrees. Equations (5.27) and (5.28) also represent the swln~ equation. It can be seen that
the swing equation is a second order differential equation which can be written as two .flrst order differential
In (5.25), H Is computed on the machine rating G", Goo.. ;'
equations:
2H
tJ_(V '"
dll
(0.,
Multiplying (5.25) by
p_p
pu
"'.
(5.29)
G"",.
G.~.,..{f~1II
on both sides we get
(5.31)
(5.30)
·In which
(I).
(0.,
and
0 are in electrical
units. In deriving the swing equation, damping has been neglected.
2Hno ...,,,, d' (~'"'" p _ P
(() ...M
I
dt'
IJI
pu (on system base)
C'
Table 5_1 H constants of synchronous machines
Type of machine
fijine'generator
condensing 1800 rpm
H(MJ/MVA)
(5.321
9- 6
•
,
.' ~".' ,_,."'nl",r
j'l
I'"
r:::.,,",!,..,"",',
I
m,", moo"", .... m.
00:::""'" " 00",",.", .. "'" "..: ..
moo<"
equations to be solved is reduced. Machines within a plant normally swing together'. after a disturbance. Such
machines are called coherent machines and can be replaced by a single equivalent machine, whose dynamics
reflects the dynamics of the plant. The concept is best understood by considering a two machine system.
I
,w'"'
I
.2:15
H,H,
Multiplying both sides by -----
5.4.1 SWING EQUATION OF TWO COHERENT MACHINES
we get
H, +H,
2(
(0,
H, H, ) d' (0, ~o,)= p., H, - Pm,H, _
H,+H,
d t'
H,+H,
1'., H,
- p" H,
H,+H,
I
The swing equations for two machines on a common system base are:
I.e
2H, d ' ", _ P
P
~
d t' - ,,' - "pu
I
2H, d ' 0,
--= p - P pu
ro,. d t!
""
,.,
-~
(S.33)
where
(S.37)
0" = 0, -0"
Now,)',
= d, = <5
1i,1i,
H,+H,
HI'=--~
(5.34
-
~
I
(since they swing together). Adding (S.33) and (S.34) we get
_ Pm' H, - Pm' H,
p
,.,12 -
I
I
2H,q d'c5 ~ P _ P,.
ra, d t?
••
pu
the relative angle of the two machines
(S.35)
P
H, +H,
_ p" H, - p" H,
HI + H2
<.'12 -
I
Where 11..
=H, +H,
i
From (S.37) it is obvious that the swing of a machine Is associated with dynamics of other machines in the system. To
I
P,,, = P"" + P,,,,
I!
r. = p", + P",
i
be stable. the angular differences between all the machines must decrease after the disturbance. In many cases.
when the system loses stability. the machines split Into two coherent groups. swinging against each other. Each
-coherent group of machines can be replaced by a single equivalent machine and the relative swing of the two
equivalent machtnes solved using an equation similar to (S.37). from which stability can be assessed.
The relation (S.3S) represents the dynamics of the single equivalent machine.
The acceleration power Is liven bv p•• P,. - P" Hence. under the condition that p.. Is a ,o&stlnt. an ac,eleratlnl
I
machine should have a power characteristic, which would Increase P, as 6 Increases.This would reduce p. and hence
5.4.2
the acceleration and help in maintaining stability. It' on the other hand. P, decreases when 6 Increases, P, would
SWING EQUATION OF TWO NON - COHERENT MACHINES
further increase which is detrimental to stability. Therefore.
I
For any two non -boherent
machines also (5.33) and (S.34) are valid. Subtracting (S.34) from (33) we get
2 d'~, _2
I
lV.,
dt '
(0.•
d'~, = P"" - P",__ P", -P",
dt :
H,
H,
(5.36)
power-angle relationship plays a crucial role In stability.
5.5 POWER-ANGLE EQUATION:
(JP
ali
must be positive for a stable system. Thus the
~-----------------------------------------,.,------
..--------------------------------------------------------~.~----~~~-----------------------------
?7h
In solving the swing equation, certain assumptions are normally made
(I)
Mechanical power input Pm is a constant during the period of in~erest, immediately after the
The power output of the generator is given by the real part of E,I,' .
disturbance
{iiI
Rotor speed changes are insignificant.
(iii)
Effect of voltage regulating loop during the transient Is neglected i.e t~e excitation is assumed to be
I
= E, L.t5 -
t
a constant.
V. L.O·
(5.38)
R" + jx,
"
As discussed in section 4, the power-angle relationship plays a vital role In the soluticn of the swing equation.
Neglecting R.,
E, L.t5 - V. L.O·
t =--_._"
jx"
5.5.1 POWER-ANGLE EQUATION FOR A NON-SAUENT POLE MACHINE:
The simplest model for the synchronous generator is that at a constant voltage behind an impedance. This model is
called the classical model and can be used for cylindrical rotor (non-salient pole) machines. Practically all high-speed
turbo alternators are of cylindrical rotor construction, where the physical air gap around the periphery of the rotor is
uniform. This type of generator has approximately equal magnetic reluctance, regardless of the angular position of
the rotor, with respect to the armature mmf. The phasor diagram of the voltages and currents at constant speed and
excitation is shown in Fig. 5.2.
t
(5.39)
(Note- R stands for real part of). The graphical representation of (9.39) ~ called the power angle curve and It Is as
shown In Fig 5.3.
t
I
Fig S.2 Phasor diagram of a non-salient pole synchronous generator'
E, = Generator Internal emf.
p","JC
i
r
-
-_..
-""-
Fl' i
I
!
V, "Terminal voltage
e
I
I
'1-
_--_
__
..:.\__
)
= Power factor angle
---.3d
I, = Armature current
r.
Fig 5.3 Power angle curve of a non - salient pole machine
= Armature res.stance
x, = Direct axis reactance
I
-----------~~--------------+--------------
r----
E
The maximum power that can be transferred for a particular excitation is given by
ir
;d: '
at 6 = 90'.
I
I
5.5.2 POWERANGLEEQUATION FORA SALIENTPOLEMACHINE:
I
Here because of the salient poles, the reluctance of the magnetic circuit in which flows the flux produced by an
armature mmf in line with the quadrature axis is higher than that of the magnetic circuit in which flows the flux
produced by the armature mmf in line with the direct axis. These two components of armature mmf are
proportional to the corresponding components of armature current. The component of armature current producing
an mmf acting in line with direct axis is called the direct component, I,. The componerit of armature current
producing an mmf acting in line with the quadrature axis is called the quadrature axis component, I,. The phasor
diagram is shown in Fig 5, with same terminology as Fig5.4 and R, neglected.
o
cl""e.,.{:- .. )t.~, ~,
't"...J ;" {;W',", 7'
(1.):.:1>
Fig 5.4 Phasor diagram of a salient pole machine
Power output
p" ~!" case
(5.40)
From Fig 5.4,
Ed =V,sino
E -E
!,,~--y---q=1"sin(o+B)
x ,/
,
/ = E" = I " cos (0 +0)
••
0 ~-'"
.-_' "
'n"'1r
I-::ituting
I
'I
I"
«;
(5.41) :
(SAl) in (5.40), we obtain
V,~ (X,I
El' V, sin 0
p=
.
Xd
+
- «, )sin 2(5
'--2~J-x-,---
(5.42)
the relation (5.42) gives the steady state powerangle relationship for a salient pole machine, The second term does
not depend on the excitation and ii called the reluctance power component. This component makes the maximum
power greater than in the classical model. However, the angle at which the maximum power occurs is less than 900.
..
r-
I ,., '~,,,.m
5.6 POWER ANGLERELATIONSHIPIN A SMIB SYSTEM:
I
...urv
Without loss of generalit¥. many important conclusions on stability can be arrived at by considering the simple case
of a Single Machine Infinite Bus (SMIS). where a generator supplies power to an infinite bus, The concept of on
I
i
infinite bus arises from the fact that if we connect a generator to a much larger power system. it is reasonable to
I
Transient stability Is the ability of the system to remain stable under large disturbances like short circuits, line
assume that the voltage and frequency of the larger system will not be affected by control of the generator
I
outages, generation or load loss etc. The evaluation of the transient stability is required offline for planning, design
!
parameters. Hence, the external system can be approximated by an infinite bus, which Is.equivalent to an ideal
etc, and online for load management, emergency control and security assessment. Transient stability analysis deals
voltage source, whose voltage and frequency are constant. The one line diagram is shown in Fig 7.
with actual solution of the nonlinear differential equations describing the dynamics of the machines and their
controls and interfacing it with the algebraic equations describing the interconnections through the transmission
network.
Since the disturbance Is large, linearlied analysis of the swing equation (which describes the rotor dynamics) Is not
I
possible. Further, the fault may cause structural changes in the network, because of which the power angle curve
prior to fault, during the fault and post fault may be different (See example 9.8). Due to these reasons, a general
stability criteria for transient stability cannot be established, as was done In the case of stea~ state stability (namely
Ps > 0). Stability can be established, for a given fault, by actual solution of the swing equation. The time taken for the
!
. Flg.S.S SMIB System
I
fault to be cleared (by the circuit breakers) is cafled the clearing time, If the fault is cleared fast enough, the
probability of the system remaining stable after the clearance is more. If the fault persists for a longer time,
i '" "
55. the
'1""''' b. voltage is taken as reference and , is the aos" between ....
i
assumed to be connected to the infiflite bus through a lossless line of reactance
I
classical model) is given by
I
I
E, Eb
I
x,,
likelihood of instability Is increased.
d E•. 'h,
""""0< rs
The power transferred (using
Criticalclearingtime is the maximum time available for clearing the fault, before the system loses stability. Modern
circuit breakers are equipped with auto reclosure facility, wherein the breaker automatically
."
(5.43)
P=---S1I1u
Ad +X ..
sequential openings. If the fault still persists, the breakers open permanently.
reclcses
after two
• faults are transient, the
Since most
first r~closure is in general successful. Hence, transient stability has been sreatly enhanced by auto closure breakers.
and using salient pole model,
E, E; sino + ('E.' (x")(- x,) )
x" +.t..
2 », + XI' x" + »,
p = .----
'" 2"
Sin
(5.44)
Some common assumptions made during transient stability studies are as follows:
1. Transmission line and synchronous machine resistances are neglected. Since resistance Introduces a
damping term In the swing equation, this gives pessimlstlc results.
I
An important measure of performance Is the steady state stability limit, which Is defined .as the maximum power
2.
Effect of damper windings is neglected which again gives pessimistic results.
that can be transmitted in steady state Without loss of synchronism, to the recelvlng end, If transient analysis 'is
3.
Variations In rotor speed are neglected.
required, respective transient quantities namely Eg r
power output.
x; x;
and
are. used in (5.43) and (5.44) to calculate the
4.
Mechanical Input to the generator is assumed constant. The governor control loop is neglected. This also
leads to pesstmlsttc results.
5. The generator is modeled as a constant voltage source behind a transient reactance, neglecting the
voltage regulator action.
I
"I""''''n''',r I '1 I'"
6.
Loads are modeled as constant admittances and absorbed into the bus admittance matrix.
dO(I)
=0
.d t
The above assumptions, vastly simplify the equations. A digital computer program for transient stability analysis can
easily include more detailed generator models and effect of controls, the discussion of whlch' is beyond the scope of
present treatment. Studies on the transient stability of an SMIB system, can shed light
on some important
aspects of
o
some time after the fault is cleared In a stable system and d >
dl
stability of larger systems. One of the important methods for studying the transient stability of an SMIB system Is the
0, for a long time after the fault Is cleared in an
,
unstable system.
application of equal-area criterion.
Consider the swing equation (21)
5, 8 EQUAL- AREA CRITERION
I
Transient stability assessment of an SMIB system is possible without resorting to actual solution of the swing
equation, by a method known as equal-area criterion. In a SMIB system, if the system is unstable after a fault is
cleared, 6(t) increases indefinitely with time, till the machine loses synchronism. In contrast.jn
a stable system, 6(t)
Multiplying both sides by
reaches a maximum and then starts reducing as shown in Fig.5.6.
2 do,
dt
we get
This may be written as
!!..[(
dO)']
a t dr
~
(",.;;(;.),I<?
= 2 do p..
dt M
Integrating both sides we get
1 1\-.
8,~1'
Sf
$"1---' :
\__
as ~.-
or-=
\L_. __._~___
~-;fA.
...Il- i<t.; b"ll:~~
_~
!:D'
t;-.->
dt
do
For stability -
dt
Fig.S.6 Swing Curve (6 V, t) for stable and unstable system'
M
IPdo
(5.45)
s. •
= 0, some time after fault is cleared. This means
s
fp" do =0
Mathematically stated,
(5.46)
Th~ integral gives the area under the p. - Ii curve. The condition for stability can be, thus stated as follows: A SMIB
system is stable if the area under the P, - 6 curve, becomes zero at some value of 6. Ttlis means that the accelerating
(postnva) area under P, - 0 curve, must equal the decelerating (negative) area under P, _ Ii curve. Application of
equal area criterionior
"'I""
·'ft'''lr 1 'I
I"
several4lsturba~
is discussed next.
,I
,----------------,~----------------=--____::_
---_ .._--------_._------'-''''---------_.
__
,
..--------~
----_
.....
__
..
i--
5.9 SUDDEN CHANGEIN MECHANICALINPUT
I
Consider the SM IB system shown in Fig. 5.7,
I
I
fig.5.S
Equai area criterion;udden
change In mechanical input
lf there is a sudden step increase in input power to Pm' the accelerating' power is positive (since Pm' > Pm,) and power
angle 6 increases. With increase in 6, the electrical power P, increases, the accelerating power decreases, till at 6 •
6,. the electrical power matches the new Input Pm" The area A" during acceleration
A, =
11 given by
('(p,,,, - p.,)do
J,.
(5.47)
At b, even though the accelerating power is zero, the rotor is running above synchroMus speed. Hence, 6 and P,
increase beyond b, wherein P, < Pm, and the rotor is subjected ~o dec~eratlon, The rotor decelerates and speed
Flg.5.7
SMIB System
starts dropping, till at paint d, the machine reaches synchronous speed and 6
= 6..... The area Au during deceleration
is given by
The electrical power transferred is given by
p,.
(5.48)
= P"., sin"
By equal area criterion A, = A,. The rotor would then oscillate between 00 and om.. at its natural frequency. However,
damping forces will reduce subsequent swings and the machine finally setties down to the new steady state value 6,
Prl~I.X
E V
=-,-'x, +x"
(at paint b). Stability can be maintained only if area A, at least equal to A" can be located above Pm" The limitmg
case is shown in Flg.5.9, where A, is just equal to A,.
Under steady state Pm= p•. Let the machine be initially operating at a steady st~te angl,e:6,. at synchronous speed
I
i
w" with
a mechanical input P"", as shown in Fig.5.B( paint a).
!
I
~
I
Fig.S.9 Maximum Increase in mechanical power
I
I
L
"''1''''' "'n'i'lr I 'I
--
I ,.
__
._--l
--·
1
_286
.
r-----------
--.--.----
~~
_,
Here 6m.. Is at the intersection of P, and Pm" If the machine does not reach synchronous speed at d, then beyond d,
P, decreases
Wit
increase In 6, causing 6 to increase Indefinitely. Applying equal area criterion to Fig.S.9we get
A, =A,.
I
Acceleration
From (5.47) and (5.48) we get
a = d'8
dt'
I
= ~=166.6
om/s'
= 166.6 • ~
Substituting
p
P,,,I = Pmol)( sin 6'maJi'we get
= 166.6. ~
Equation (5.49) is a non-linear equation In 6mu and can be solved by trial and -erroror
by uslng any numerical
method for solution of non-linear algebraic equation (like Newton-Raphson, bisection. Eitd).From solution of
x ~
p
(5.49)
om.;,
=166.6'
360
can be calculated. Pml- Pm,will give the maximum possible Increase In mechanical input before the machine looses
rps Is
.L
~.
P
Pm,
Re/s'
0.15
.60 rpm/s
360
= 13.88 rpm/s
stability.
,
• Note 2e = electrical degree; 2m = mechanical degree; P=number of poles.
(c) 10 cycles =
10
- = 0.2 s
50
N, = Synchronous speed =
120 x 50
--4--
Rotor speed at end of 10 cycles = N. + a. 0.2
5.10 NUMERICALEXAMPLES
= 1500rpm
= 1500 + 13.88"
0.2.1502.776
rpm.
Ex~mple 2: Two SO Hz generating units operate In parallel within the same plant, with the following ratings:
1: ~OO MVA, 0.8 pf, 13.2 kV, 3600 rpm: H z 4 MJ/MVA;
Unit
Unit 2: 1000 MVA, 0.9 pf, 13.8 kV, 1800 rpm: H D 5 MJ/MVA.
Caiculate the equivalent H constant on a base of 100 MVA.
Example 1: A 50Hz, 4 pole turbo alternator rated 150 MVA, 11 kV has an inertia constant of 9 ·MJ/ MVA ..Find the (a)
stored energy at synchronous speed (b) the rotor acceleration if the Input mechanical power is raised to 100 MW
when the electrical luad is 75 MW, (c) the speed at the end of 10 cycles if acceleranon is assumed constant at the
~olution:
initial value.
= 4x 500 = 20 MJ/MVA
100
Solution:
= 5x 1000 =50 MJ/MVA
I
l
{ill
Stored energy = GH = 150 • 9 = 1350 MJ
(b)
p.=pm-p.
M=__(!.H
IMOf
-----I~~-~_1
=100-75=25MW
= __1350
180 x50
,
100
li.q =H, +H,=20+50=70MJ/MVA
=O.15MJ-s/2e
This is the equivalent inertia constant on a base of 100 MVA and can be used when the two machines swing
coherently.
..
______
.
~J~qn.
.:._
_
JQ'
~c;100''b5
Fig. P4 Example 4
J
.
0'4--
Solution:
JO·35
Current I = V,LB -
1.0/0° = ~~-
jl.O
'
1.0LO°
jl.O
Fig. PS Example 6
E, Lo = V,LO + jl.O(/)
=lLB+
I
w_._ I
jI.0(1.0LB-1.0LOo)
jl.O
.
0.35
(a) X= 0.4 + --2- = 0.575 pu
= cos e + J sinS + cosS + J sinS -1.0
p
• zcosa - 1+ J 2sln9
.....
E. E. _ 1.2 x 1.0 _ 2 087
X - o:57S -.
pu
When maximum power Is transferred <5 = 90'; which means real part of E = 0
:. 2 cosS-1 = 0
:.0, = sin -I ~
P"..
S = cos'! 0.5= 60'
= sin -l
(_Q:!._)
2.087
= 22.54°.
(c) P, = Pm.. cos 6. = 2.087 cos (22.54')
IE,I = 2x sin 60°= 1.732
= 1.927 MW (pu)/ elec rad.
E, = 1.732 L90° (for maximum power)
Steady state stability limit =
1.732 x 1.0 = 0.866
1.0 + 1.0
M (pu) = .!!_ = .2.3_
nf n x50
pu
Without damping S = ± j
~:
A 50 Hz synchronous generator having an internal voltage 1.2 pu,
H = 5.2 MJ/MVA and a
= 0.0331 s ~/ elec
{P; = ± j
fM'
rad
J 0.0331
1.927
= ± j 7.63 rad/sec = 1.21 Hz
reactance of 0.4 IJU is connected to an infinite bus through a double circuit line, each :line of reactance 0.35 pu. The
I
generator is dehverrng 0.8pu power and the Infinite bus voltage is 1.0 pu. Determlbe: maximum power transfer,
Natural frequency of oscillation
w, = 1.21 Hz.
Steady state operating angle, and Natural frequency of oscillation if damping is neglected.
I
example 6: In example .6, If the damping is 0.14 and there is a minor disturbance of /J. ~ = 0.15 rad from the initial
I
i
Solution: The one line diagram is shown in Fig P5.
L-----r--,-------------------------------
operating point, determine: (a) 00,(b) ~ (c) OOd (d) setting time and (el expression for
~
o.
~.------------------------------------------~
.------.-----_-L>L. __
.
:
~
,----- ...-.---------
~
(a) 0). = 11-"- =
vM
(b) ~=
(Od
1-- = 7.63 rad/sec
~ 0.0331
= 1.21 Hz
M P,
= (0. JI -(
2
~;
..L.:l"'-
---,
For the given network
=0.277
VO.033' x 1.927
= 7.63~'-(O.277)'
.--
In example 6, find the po,,:,er angle relationship
(i)
!l_J- , . = ~ C- ,2
(c)
'1.927
= 7.33 red/sec
= 1.16 Hz
(ii)
If
(iii)
If fault is cleared by line outage
a short
circuit occurs In the middle of
a line
Assume the generator to be supplying 1.0 pu power initially.
(d) Setting time =4t =4_1_ = 4x .
~(()..
(e)
6o, = 0.15
e
s:
u
1__
= 1.892 s
0.277 x 7.63
rad = 8.59'
= cos" I;= cos" 0.277 = 73.9'
s:
= o;+
(
Jss,
' e _'.,.
••
I-qSin
(0.1
+8
(I)
)
(ill
From example 6, Pm". 2.087, p. = 2.087 sin 0.
If a short circuit occurs In the middle of the line, the network equivalent can be draw as shown In Fig.
12a.
= 22.54" +
8.59
e-o.",.? .., sin (7.331+73.9" )
JI - 0.277'
= 22.54° + 8.94 e' 2.l1. sin (7.33t + 73.9')
The variation of delta with respect to time is shown below. It can be observed that-the
angle reaches the steady
state value of 22.54' after the Initial transient, it should be noted that the magnitudes of the swings decrease in a
stable system with damping .
::1---- --.
-_ ..---.-~ ----
.. . ...
...
"ni'
I
l,of
!"-::
-,~.o---~--~-·
a
Fig.P7a Short circuit in middle of line
Ih>1<J(u()Con(Je)
Fig.P6
Swing Curve for example 7
The network is reduced by converting
and P7c.
L_____
._~
~
the delta to star and again the resulting star to delta as shown In Fig P7a, P7b
..
~:
iP
p.... = 1.12x 1.0 = 1.244pu
0.9
P",o .. p.o· 1.0 .. Pm.. sIn 60.1.244 sin 60
11:,:.1 !
I~01
:. 6. = sln'
~\.~
L.,----.....
"",.
_I.:Q_
= 53.470= 0.933 rad.
1.244
Cf' i(.c
\-til
_.L.--_
...
,-
The above can be solved by N-R method
since
it is of the form f(6m .. )
= K. Applying
N-R method, at any iteration 't',
we get
v
-·
K - J(8;:;)
dJ
do""" (,T
Flg.P7b
,
The transfer reactance Is 1.55 pu. Hence,
(ThiSIs the derivative evaluated at a value of
Pm..
0 = 0,••("j
)
= 1.2 X 1.0 = 0.744
1.55
P, = 0.744 sin 6
Starting from an initial guess of lim.. between!!"
(iii) When there is a line outage
Here K
= cos 0 = 0.595.The computations
0
2
to
1[,
the above equations are solved Iteratively till A
are shown In table P8, starting from an initial guess
x = 0.4 + 0.35= 0.75
1.2 x 1.0
=1.6
0.75
Table P8
Pm.. ::: ---
Interaction
P, = 1.6 sin 0
Example 8: A generator supplies active power of 1.0 pu to an infinite bus, through a !ossless line of reactance x, =
2
1.745 -0.1407
0.626
0.22
1.965
1.965
- 0.396
0.568
- 0.068
1.897
1.897
- 0.309
0.592
- 0.0097 1.887
1.887 - 0.2963
0.596
- 0.0033 1.883
0.6 pu. The reactance of the generator and the connecting transformer is 0.3 pu, The.transient internal voltage of
the generator is 1.12 pu and infinite bus voltage is 1.0 pu. Find the maximum increase in: mechanical power that will
not cause instability.
4
o!:;~ S E.
o;~,= 1.745rad.
Since d
8,~:,~
is sufficient
by small, we can take
6"" = 1.883 rad = 107.88'
Simplifying, we get
Maximum step increase permissible is Pm. - Pm" = 1.183 -1.0 = 0.183 pu
Example 9: Transform a two machine system to an equivalent SMIB system and show
applicable to it.
~C(lnSider
equal area criterion is
,
d'o
or
M,. d/' =Pm-P"
M,M,
. M,+M,
where M•• = ---
, ,
Ihe two machine system show in Flg.P9.
P,.
= _.-E..!.,_E~,,---.sin"
This relation is Identical to that of an SMIB system In form and can be used to determine the relative swing (0, _ 0,)
'between the two machines to assessthe stability .
_ .._ ........_ .... ..
..-- ......
Fig.P9 Two machine system under steady state (neglecting lo~s;es)
~,=-P", =P"
The swing equations are
d'b'
P
-» = .:si:
P -r
_"
_,1 ~_'_"'
__sl,
dt:
'""T"""'n'I'Ir 1'1
I"
M,
M,
-
. Po vve y
h+Ovbl~W~
~+e-y
to
y ....6t- e.=VV)===h=~=c::L=b=i=\=,·=r==y=-. _
~
&~(~.
(.,()-v\.V\ec...recl
Y"e
hA
1()
yV\
1,:)0 We/{'
V\..O"
VV) &-..J
hOvv~
·b~e--vl .
b?VV\e..e·
c.oV) ve. Yse Itct
h'dY\
ate
O\A-t of
Atep'
CAA)""y0V\.-+:s
;
I0{ e
h"Oe
-vvv
l1YtA.r~CA.t
to
~lcl
by
C lAo.y?tc+e 'Y\"seo(
~fs
c(e si"( ~ble.
Oe-e¥l 'Y'a co fJ vvC S eo(
g'tr---t<WYl 1'"I "'- Y1 V> ,
W~+
~e.o-
~VY'et?..6~
~y ~
h'YD vvU S
f eo..+V\.yes
bov..\""c
.f~lllYl:i
1 he
~
to .
VlA
v~
p~+-s of lA povverr, ~+0vv\.
d y V\ 0., VV\A' G6 0f- 0.. po tN e_,y' -~ y ~t e.v-v1.
TVl e
_
-flAAc-t--
UVvl
t0e-c..o~
VV\.~~
OY
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OveN'
i+ ~
CcrYl ck.:
01..
Ce-V\.,-n YLV\a.. t tq ~TO W _
ex+~~
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11lA.Ve
ll'~
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p~ rc:t:n' 0 V\
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h' ~
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p"" e...vvo VV\evvo V\ pe (.AA...l",: oJ\.. te
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c.1~-1..f\·(2d
9t-e ~~
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~ro
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A~61~'~
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