1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.
-
Upload
tracy-singleton -
Category
Documents
-
view
225 -
download
2
Transcript of 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.
![Page 1: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/1.jpg)
1
EE6501POWER SYSTEM ANALYSIS
![Page 2: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/2.jpg)
2
UNIT I
INTRODUCTION
![Page 3: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/3.jpg)
3
Power system network
![Page 4: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/4.jpg)
4
SINGLE LINE DIAGRAM
It is a diagrammatic representation of a power system in
which the components are represented by their symbols.
![Page 5: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/5.jpg)
5
COMPONENTS OF A POWER SYSTEM
1.Alternator
2.Power transformer
3.Transmission lines
4.Substation transformer
5.Distribution transformer
6.Loads
![Page 6: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/6.jpg)
6
MODELLING OF GENERATOR AND SYNCHRONOUS MOTOR
1Φ equivalent circuit of generator 1Φ equivalent circuit of synchronous motor
![Page 7: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/7.jpg)
7
MODELLING OF TRANSFORMER
2 2 1
1 1 2
' 201 1 2 1 2
' 201 1 2 1 2
E N IK
E N I
RR R R R
KX
X X X XK
=Equivalent resistance referred to 1o
=Equivalent reactance referred to 1o
![Page 8: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/8.jpg)
8
MODELLING OF TRANSMISSION LINE
Π type T type
![Page 9: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/9.jpg)
9
MODELLING OF INDUCTION MOTOR
'
'
'
1( 1)r
S r
S r
Rs
R R R
X X X
=Resistance representing load
=Equivalent resistance referred to stator
=Equivalent reactance referred to stator
![Page 10: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/10.jpg)
10
2 2
1000
b bb
bb
kV kVZ
KVAMVA
per unit=actual value/base value
Let KVAb=Base KVA
kVb=Base voltage
Zb=Base impedance in Ω
![Page 11: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/11.jpg)
11
Changing the base of per unit quantities
Let z = actual impedance(Ω)
= base impedance (Ω)
bZ
. 2 2
* bp u
b b b
b
Z MVAZ ZZ
Z kV kV
MVA
Let , ,
, ,
&
&b old b old
b new b new
kV MVB
kV MVB
represent old base values
represent new base values
![Page 12: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/12.jpg)
12
,. , 2
,
. , ,
2
,
,. , 2
,
2
, ,. , . , 2
,,
*(1)
*(2)
*(3)
* *
b oldp u old
b old
p u old b old
b old
b newp u new
b new
b old b newp u new p u old
b oldb new
Z MVAZ
kV
Z MVAZ
kV
Z MVAZ
kV
kV MVAZ Z
MVAkV
![Page 13: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/13.jpg)
13
ADVANTAGES OF PER UNIT CALCULATIONS
The p.u impedance referred to either side of a 1Φ transformer is same
The manufacturers provide the impedance value in p.u The p.u impedance referred to either side of a 3Φ
transformer is same regardless of the 3Φ connections Y-Y,Δ-Y
p.u value always less than unity.
![Page 14: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/14.jpg)
14
IMPEDANCE DIAGRAM
• This diagram obtained by replacing each component by their 1Φ equivalent circuit.
Following approximations are made to draw impedance diagram
1. The impedance b/w neutral and ground omitted.
2. Shunt branches of the transformer equivalent circuit
neglected.
![Page 15: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/15.jpg)
15
REACTANCE DIAGRAM
It is the equivalent circuit of the power system in which the various components are represented by their respective equivalent circuit.
Reactance diagram can be obtained after omitting all
resistances & capacitances of the transmission line from
impedance diagram
![Page 16: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/16.jpg)
16
REACTANCE DIAGRAM FOR THE GIVEN POWER SYSTEM NETWORK
![Page 17: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/17.jpg)
17
PROCEDURE TO FORM REACTANCE DIAGRAM FROM SINGLE DIAGRAM
1.Select a base power kVAb or MVAb
2.Select a base voltage kVb
3. The voltage conversion is achieved by means of transformer kVb on LT section= kVb on HT section x LT voltage rating/HT voltage rating
4. When specified reactance of a component is in ohms
p.u reactance=actual reactance/base reactance
specified reactance of a component is in p.u
2
, ,. , . , 2
,,
* *b old b newp u new p u old
b oldb new
kV MVAX X
MVAkV
![Page 18: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/18.jpg)
18
p.u. calculation of 3 winding transformer
Zp=Impedance of primary windingZs’=Impedance of secondary windingZt’=Impedance of tertiary windingShort circuit test conducted to find out the above 3 impedances
![Page 19: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/19.jpg)
19
'
' '
' '
1
21
21
2
p ps pt st
s ps st pt
t ps pt st
Z Z Z Z
Z Z Z Z
Z Z Z Z
psZ
ptZ
= Leakage impedance measured in 1o with 2o short circuited and tertiary open.
= Leakage impedance measured in 1o with tertiary short circuited and 2o open.
'stZ = Leakage impedance measured in 2o with tertiary short
circuited and 1o open and referred to primary
![Page 20: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/20.jpg)
20
PRIMITIVE NETWORK It is a set of unconnected elements which provides information
regarding the characteristics of individual elements. it can be represented both in impedance & admittance form
![Page 21: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/21.jpg)
21
BUS ADMITTANCE(Y BUS) MATRIX
Y BUS can be formed by 2 methods
1.Inspection method
2.Singular transformation
Y BUS = 11 12 1
21 22 2
1 2
n
n
n n nn
Y Y Y
Y Y Y
Y Y Y
![Page 22: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/22.jpg)
22
INSPECTION METHOD
For n bus system
Diagonal element of Y BUS
Off Diagonal element of Y BUS
ij ijY y
1
n
ii ijj
Y y
![Page 23: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/23.jpg)
23
SINGULAR TRANSFORMATION METHOD
Y BUS =
Where [y]=primitive admittance
A=bus incidence matrix
TA y A
![Page 24: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/24.jpg)
24
ALGORITHM FOR FORMATION OF THE BUS IMPEDANCE MATRIX
• Modification of Zbus matrix involves any one of the following 4 cases
Case 1:adding a branch impedance zb from a new bus p to the reference bus
Addition of new bus will increase the order the Zbus matrix by 1
(n+1)th column and row elements are zero except the diagonal
diagonal element is zb
,
0
0original
bus newb
zZ
z
![Page 25: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/25.jpg)
25
Case 2: adding a branch impedance zb from a new bus p
to the existing bus q
Addition of new bus will increase the order the Zbus matrix by 1
The elements of (n+1)th column and row are the elements of
qth column and row and the diagonal element is Zqq+Zb
Case 3:adding a branch impedance zb from an existing bus p to
the reference bus
The elements of (n+1)th column and row are the elements of
qth column and row and the diagonal element is Zqq+Zb and
(n+1)th row and column should be eliminated using the following
formula
( 1) ( 1),
( 1)( 1)
1,2... ; 1, 2..j n n kjk act jk
n n
Z ZZ Z j n k n
Z
![Page 26: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/26.jpg)
26
Case 4:adding a branch impedance zb between existing buses h and q
elements of (n+1)th column are elements of bus h column –
bus q column and elements of (n+1)th row are elements of
bus h row – bus q row the diagonal element=
and (n+1)th row and column should be eliminated using the following
formula
2b hh qq hqZ Z Z Z
( 1) ( 1),
( 1)( 1)
1,2... ; 1, 2..j n n kjk act jk
n n
Z ZZ Z j n k n
Z
![Page 27: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/27.jpg)
27
UNIT II
POWER FLOW ANALYSIS
![Page 28: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/28.jpg)
28
BUS CLASSIFICATION
1.Slack bus or Reference bus or Swing bus:
|V| and δ are specified. P and Q are un specified, and to be calculated.
2.Generator bus or PV bus or Voltage controlled bus:
P and |V| are specified. Q and δ are un specified, and to be calculated
3.Load bus or PQ bus:
P and Q are specified. |V| and δ are un specified, and to be
calculated
![Page 29: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/29.jpg)
29
ITERATIVE METHOD
The above Load flow equations are non linear and can be solved by following iterative methods.
1.Gauss seidal method2.Newton Raphson method
3.Fast Decoupled method
1
*
*1
n
p pq qq
p p p P p
np p
pq qqP
I Y V
S P jQ V I
P jQY V
V
![Page 30: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/30.jpg)
30
GAUSS SEIDAL METHOD
For load bus calculate |V| and δ from Vpk+1 equation
For generator bus calculate Q from QPK+1 equation
11 * 1
1
1*Im ( )p n
k k k kp P pq q pq q
q q p
Q V Y V Y V
11 1
*1 1
1
( )
p np pk k k
p pq q pq qkq q ppp P
P jQV Y V Y V
Y V
![Page 31: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/31.jpg)
31
• Check Qp,calk+1 with the limits of Qp
• If Qp,calk+1 lies within the limits bus p remains as PV bus
otherwise it will change to load bus• Calculate δ for PV bus from Vp
k+1 equation• Acceleration factor α can be used for faster convergence
• Calculate change in bus-p voltage
• If |ΔVmax |< ε, find slack bus power otherwise increase the iteration count (k)
• Slack bus power=
1 1k k kp p pV V V
G LS S
![Page 32: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/32.jpg)
32
NEWTON RAPHSON METHOD
1
1
1
1 2
3 4
cos( )
sin( )
n
i i i j ij ij i jj
n
i i j ij ij i jj
n
i i j ij ij i jj
k sch ki i i
k sch ki i i
P Q V V Y
P V V Y
Q V V Y
J JP
J J VQ
P P P
Q Q Q
![Page 33: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/33.jpg)
33
• Calculate |V| and δ from the following equation
• If
• stop the iteration otherwise increase the iteration count (k)
1
1
k k ki i
k k ki i iV V V
ki
ki
P
Q
![Page 34: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/34.jpg)
34
FAST DECOUPLED METHOD
J2 & J3 of Jacobian matrix are zero1
4
1
4
'
''
1'
1''
0
0
i
i
JP
VJQ
PP J
QQ J V V
V
PB
V
QB V
V
PB
V
QV B
V
![Page 35: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/35.jpg)
35
1
1
k k ki i
k k ki i iV V V
This method requires more iterations than NR
method but less time per iteration
It is useful for in contingency analysis
![Page 36: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/36.jpg)
36
COMPARISION BETWEEN ITERATIVE METHODS
Gauss – Seidal Method
1. Computer memory requirement is less.
2. Computation time per iteration is less.
3. It requires less number of arithmetic operations to complete an iteration and ease in programming.
4. No. of iterations are more for convergence and rate of convergence is slow (linear convergence characteristic.
5. No. of iterations increases with the increase of no. of buses.
![Page 37: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/37.jpg)
37
NEWTON – RAPHSON METHOD
Superior convergence because of quadratic convergence. It has an 1:8 iteration ratio compared to GS method. More accurate. Smaller no. of iterations and used for large size systems. It is faster and no. of iterations is independent of the no. of buses. Technique is difficult and calculations involved in each iteration are
more and thus computation time per iteration is large. Computer memory requirement is large, as the elements of jacobian
matrix are to be computed in each iteration. Programming logic is more complex.
![Page 38: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/38.jpg)
38
UNIT III
FAULT ANALYSIS-BALANCED FAULT
![Page 39: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/39.jpg)
39
Need for fault analysis To determine the magnitude of fault current throughout
the power system after fault occurs. To select the ratings for fuses, breakers and switchgear. To check the MVA ratings of the existing circuit breakers
when new generators are added into a system.
![Page 40: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/40.jpg)
40
BALANCED THREE PHASE FAULT
All the three phases are short circuited to each other and to earth. Voltages and currents of the system balanced after the symmetrical fault
occurred. It is enough to consider any one phase for analysis.
SHORT CIRCUIT CAPACITY
It is the product of magnitudes of the prefault voltage and the post fault current. It is used to determine the dimension of a bus bar and the interrupting capacity of a circuit breaker.
![Page 41: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/41.jpg)
41
Short Circuit Capacity (SCC)0
2
,1
1.
,3
3.
6
3
6,
/
*10
3 * *10
F
TF
T
bT
T T p u
b
T p u
f
L b
SCC V I
VI
Z
SVSCC MVA
Z Z
SSCC MVA
Z
SCCI
V
![Page 42: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/42.jpg)
42
Procedure for calculating short circuit capacity and fault current
Draw a single line diagram and select common base Sb MVA and kV
Draw the reactance diagram and calculate the total p.u impedance from the fault point to source (Thevenin impedance ZT)
Determine SCC and If
![Page 43: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/43.jpg)
43
ALGORITHM FOR SHORT CIRCUIT ANALYSIS USING BUS IMPEDANCE MATRIX
• Consider a n bus network. Assume that three phase fault
is applied at bus k through a fault impedance zf
• Prefault voltages at all the buses are
• Draw the Thevenin equivalent circuit i.e Zeroing all voltage sources and add voltage source at faulted bus k and draw the reactance diagram
1
2
(0)
(0)
.(0)
(0)
.
(0)
busk
n
V
V
VV
V
(0)kV
![Page 44: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/44.jpg)
44
• The change in bus voltage due to fault is
• The bus voltages during the fault is
• The current entering into all the buses is zero.the current entering into faulted bus k is –ve of the current leaving the bus k
1
.
.
.
busk
n
V
VV
V
( ) (0)bus bus busV F V V
![Page 45: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/45.jpg)
45
11 1 1
1
1
. . 0
. . . . . .
. . ( )
. . . . . .
. . 0
( ) (0) ( )
( ) ( )
(0)( )
( ) (0) ( )
bus bus bus
k n
bus k kk kn k
n nk nn
k k kk k
k f k
kk
kk f
i i ik k
V Z I
Z Z Z
V Z Z Z I F
Z Z Z
V F V Z I F
V F Z I F
VI F
Z Z
V F V Z I F
![Page 46: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/46.jpg)
46
UNIT IV
FAULT ANALYSIS – UNBALANCED FAULTS
![Page 47: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/47.jpg)
47
INTRODUCTION
UNSYMMETRICAL FAULTSo One or two phases are involvedo Voltages and currents become unbalanced and each phase is to be
treated individuallyo The various types of faults are Shunt type faults 1.Line to Ground fault (LG) 2. Line to Line fault (LL) 3. Line to Line to Ground fault (LLG) Series type faults Open conductor fault (one or two conductor open fault)
![Page 48: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/48.jpg)
48
FUNDAMENTALS OF SYMMETRICAL COMPONENTS
Symmetrical components can be used to transform
three phase unbalanced voltages and currents to
balanced voltages and currents Three phase unbalanced phasors can be resolved into
following three sequences
1.Positive sequence components
2. Negative sequence components
3. Zero sequence components
![Page 49: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/49.jpg)
49
Positive sequence components Three phasors with equal magnitudes, equally displaced from one
another by 120o and phase sequence is same as that of original phasors.
Negative sequence components Three phasors with equal magnitudes, equally displaced from one
another by 120o and phase sequence is opposite to that of original phasors.
Zero sequence components Three phasors with equal magnitudes and displaced from one another
by 0o
1 1 1, ,a b cV V V
2 2 2, ,a b cV V V
0 0 0, ,a b cV V V
![Page 50: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/50.jpg)
50
RELATIONSHIP BETWEEN UNBALANCED VECTORS AND SYMMETRICAL COMPONENTS
0 1 2
0 1 2
0 1 2
0
21
22
2
2
1 1 1
1
1
1 1 1
1
1
a a a a
b b b b
c c c c
a a
b a
c a
V V V V
V V V V
V V V V
V V
V a a V
a aV V
A a a
a a
Similarly we can obtain for currents also
![Page 51: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/51.jpg)
51
SEQUENCE IMPEDANCE
Impedances offered by power system components to positive, negative and zero sequence currents.
Positive sequence impedance The impedance of a component when positive sequence
currents alone are flowing.
Negative sequence impedance The impedance of a component when negative sequence
currents alone are flowing.
Zero sequence impedance The impedance of a component when zero sequence currents
alone are flowing.
![Page 52: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/52.jpg)
52
SEQUENCE NETWORK
SEQUENCE NETWORK FOR GENERATOR
positive sequence network negative sequence network Zero sequence network
![Page 53: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/53.jpg)
53
SEQUENCE NETWORK FOR TRANSMISSION LINE
positive sequence network negative sequence network Zero sequence network
![Page 54: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/54.jpg)
54
SEQUENCE NETWORK FOR TRANSFORMER
positive sequence network negative sequence network Zero sequence network
![Page 55: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/55.jpg)
55
SEQUENCE NETWORK FOR LOAD
positive sequence network negative sequence network Zero sequence network
![Page 56: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/56.jpg)
56
SINGLE LINE TO GROUND FAULT
b
c
fa a
a1 a2 a0 a
a11 2 0
I 0
I 0
V Z I
I I I I / 3
I3
af
E
Z Z Z Z
Consider a fault between phase a and
ground through an impedance zf
![Page 57: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/57.jpg)
57
LINE TO LINE (LL) FAULT
a2 a1
a0
fa1 a2 a1
a11 2
1 2
I 0
I I
V V I
I I
I 0
V V Z I
I3
I I3
a
c b
fb c b
af
ab c f
Z
E
Z Z Z
jE
Z Z Z
Consider a fault between phase b and c
through an impedance zf
![Page 58: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/58.jpg)
58
DOUBLE LINE TO GROUND (LLG) FAULT
a0
a1 a2 a0
fa0
fa0 a1 a0
a11 2 0 2 0
I 0
I I I 0
V =V (I I ) 3Z I
V V V 3Z I
I( 3 ) / ( 3 )
fb c b c
b
af f
Z
E
Z Z Z Z Z Z Z
Consider a fault between phase b and c
through an impedance zf to ground
![Page 59: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/59.jpg)
59
UNBALANCED FAULT ANALYSIS USING BUS IMPEDANCE MATRIX
SINGLE LINE TO GROUND FAULT USING Zbus
Consider a fault between phase a and ground through
an impedance zf at bus k
For a fault at bus k the symmetrical components of fault current
0 1 21 2 0
V (0)I I I
3k
k k k fkk kk kkZ Z Z Z
![Page 60: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/60.jpg)
60
LINE TO LINE (LL) FAULT
Consider a fault between phase b and c through an impedance zf
0
1 21 2
I 0
V (0)I I
k
kk k f
kk kkZ Z Z
![Page 61: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/61.jpg)
61
DOUBLE LINE TO GROUND (LLG) FAULT
Consider a fault between phase b and c through an impedance zf to ground
12 0
12 0
1 12
2
1 10
0
V (0)I
( 3 )
3
V (0) II
V (0) II
3
I ( ) I I
kk f
kk kkkk f
kk kk
k kk kk
kk
k kk kk f
kk
b ck k k
Z Z ZZ
Z Z Z
Z
Z
Z
Z Z
F
![Page 62: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/62.jpg)
62
BUS VOLTAGES AND LINE CURRENTS DURING FAULT
0 0 0
1 0 1 1
2 2 2
0 00
0
1 11
1
2 22
2
( ) 0 I
( ) (0) I
( ) 0 I
( ) ( )I
( ) ( )I
( ) ( )I
i ik k
i i ik k
i ik k
i jij
ij
i jij
ij
i jij
ij
V F Z
V F V Z
V F Z
V F V F
Z
V F V F
Z
V F V F
Z
![Page 63: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/63.jpg)
63
UNIT V
STABILITY ANALYSIS
![Page 64: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/64.jpg)
64
STABILITY The tendency of a power system to develop restoring
forces equal to or greater than the disturbing forces to maintain the state of equilibrium.
Ability to keep the machines in synchronism with another machine
![Page 65: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/65.jpg)
65
CLASSIFICATION OF STABILITY
Steady state stability Ability of the power system to regain synchronism after small and
slow disturbances (like gradual power changes)
Dynamic stability Ability of the power system to regain synchronism after small
disturbances occurring for a long time (like changes in turbine speed, change in load)
Transient stability This concern with sudden and large changes in the network
conditions i.e. . sudden changes in application or removal of loads, line switching operating operations, line faults, or loss of excitation.
![Page 66: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/66.jpg)
66
Steady state limit is the maximum power that can be transferred without the system become unstable when the load in increased gradually under steady state conditions.
Transient limit is the maximum power that can be transferred without the system becoming unstable when a sudden or large disturbance occurs.
![Page 67: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/67.jpg)
67
Swing Equation for Single Machine Infinite Bus System
• The equation governing the motion of the rotor of a synchronous machine
where
J=The total moment of inertia of the rotor(kg-m2)
=Singular displacement of the rotor
Tm=Mechanical torque (N-m)
Te=Net electrical torque (N-m)
Ta=Net accelerating torque (N-m)
2
2m
a m e
dJ T T T
dt
m
![Page 68: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/68.jpg)
68
• Where pm is the shaft power input to the machine
pe is the electrical power
pa is the accelerating power
2 2
2 2
2
2
m sm m
m msm
m m
mm a m e
t
d d
dt dt
d d
dt dt
dJ p p p
dt
![Page 69: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/69.jpg)
69
2
2
2
2
2
2
2
20
20 0
2max2
20
2
2
2
2
2
sin
m
ma m e
machinesm
m a m e
sm machine machine
a m es
s
a m e
m a
a
J M
dM p p p
dtH
M S
d p p pH
dt S S
H dp p p
dt
f
H dp p p
f dt
f fdp p p
dt H Hd
dt
fd dp
dt H dt
δ and ωs are in electrical radian
p.u
p.u
H=machine inertia constant
![Page 70: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/70.jpg)
70
Swing Equation for Multimachine System
2
2
systema m e
machinesystem machine
system
H dp p p
f dt
SH H
S
p.u
machineS =machine rating(base)
systemS =system base
![Page 71: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/71.jpg)
71
Rotor Angle Stability
• It is the ability of interconnected synchronous machines of a power system to maintain in synchronism. The stability problem involves the study of the electro mechanical oscillations inherent in power system.
• Types of Rotor Angle Stability
1. Small Signal Stability (or) Steady State Stability
2. Transient stability
![Page 72: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/72.jpg)
72
Voltage Stability
It is the ability of a power system to maintain steady acceptable voltages at all buses in the system under normal operating conditions and after being subjected to a disturbance.
The major factor for instability is the inability of the power system to meet the demand for reactive power.
![Page 73: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/73.jpg)
73
• Mid Term Stability
It represents transition between short term and long
term responses.
Typical ranges of time periods.
1. Short term : 0 to 10s
2. Mid Term : 10 to few minutes
3. Long Term : a few minutes to 10’s of minutes• Long Term Stability
Usually these problem be associated with
1. Inadequacies in equipment responses.
2. Poor co-ordination of control and protection equipment.
3. Insufficient active/reactive power reserves.
![Page 74: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/74.jpg)
74
Equal Area Criterion
• This is a simple graphical method to predict the transient stability of two machine system or a single machine against infinite bus. This criterion does not require swing equation or solution of swing equation to determine the stability condition.
• The stability conditions are determined by equating the areas of segments on power angle diagram.
![Page 75: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/75.jpg)
75
Power-angle curve for equal area criterion
multiplying swing equation by on both sides
Multiplying both sides of the above equation by dt and then integrating between two arbitrary angles δ0 and δc
![Page 76: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/76.jpg)
76
Once a fault occurs, the machine starts accelerating. Once the fault is cleared, the machine keeps on accelerating before it reaches its peak at δc ,
The area of accelerating A1
The area of deceleration is given by A2
If the two areas are equal, i.e., A1 = A2, then the power system will be stable
![Page 77: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/77.jpg)
77
Critical Clearing Angle (δcr) maximum allowable value of the clearing time and angle for the system to remain stable are known as critical clearing time and angle.
δcr expression can be obtained by substituting δc = δcr in the equation A1 = A2
Substituting Pe = 0 in swing equation
Integrating the above equation
![Page 78: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/78.jpg)
78
Replacing δ by δcr and t by tcr in the above equation, we get
the critical clearing time as
![Page 79: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/79.jpg)
79
Factors Affecting Transient Stability
• Strength of the transmission network within the system and of the tie lines to adjacent systems.
• The characteristics of generating units including inertia of rotating parts and electrical properties such as transient reactance and magnetic saturation characteristics of the stator and rotor.
• Speed with which the faulted lines or equipments can be disconnected.
![Page 80: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/80.jpg)
80
Numerical Integration methods
Modified Euler’s method
Runge-Kutta method
![Page 81: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/81.jpg)
81
MODIFIED EULER’S METHOD
• Using first derivative of the initial point next point is obtained• the step
• Using this x1p dx/dt at x1
p=f(t1, x1
p)
• Corrected value is
1 0p dX
X X tdt
0 1
1
1 0
1
2
2
p
pi i
X XP
X Xci i
dx dxdt dt
X X t
dx dxdt dt
X X t
1 0t t t
![Page 82: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/82.jpg)
82
Numerical Solution of the swing equation
• Input power pm=constant
• At steady state pe=pm,
• At synchronous speed
10
1max
'
1max1
sin mp
p
E Vp
X
0
'
2max2
0
E Vp
X
![Page 83: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/83.jpg)
83
2
20
20 0
2max2
20
2
sin
a m e
m a
a
H dp p p
f dt
f fdp p p
dt H Hd
dt
fd dp
dt H dt
The swing equation
Applying Modified Eulers method to above equation
1 0t t t
1
1
i
i
pi i
pi i
dt
dt
dt
dt
![Page 84: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/84.jpg)
84
• The derivatives at the end of interval
1
11
1
0
pi
ppii
pi
a
d
dt
fdp
dt H
The corrected value
1
1
1
1
2
2
pi i
pi i
ci i
ci i
d ddt dt
t
d ddt dt
t
![Page 85: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/85.jpg)
85
Runge-Kutta Method• Obtain a load flow solution for pretransient conditions• Calculate the generator internal voltages behind transient reactance.• Assume the occurrence of a fault and calculate the reduced admittance matrix• Initialize time count K=0,J=0• Determine the eight constants
1 1
1 2
1 12 1
1 12 2
2 23 1
2 23 2
3 34 1
3 34 2
1 2 3 4
( , )
( , )
( , )2 2
( , )2 2
( , )2 2
( , )2 2
( , )2 2
( , )2 2
2 2
k k k
k k k
k kk k k
k kk k k
k kk k k
k kk k k
k kk k k
k kk k k
k k k
k
K f t
l f t
K lK f t
K ll f t
K lK f t
K ll f t
K lK f t
K ll f t
K K K K
1 2 3 4
6
2 2
6
k
k k k k
kl l l l
![Page 86: 1 EE6501 POWER SYSTEM ANALYSIS. 2 UNIT I INTRODUCTION.](https://reader036.fdocuments.net/reader036/viewer/2022062309/5697bfc21a28abf838ca5100/html5/thumbnails/86.jpg)
• Compute the change in state vector
• Evaluate the new state vector
• Evaluate the internal voltage behind transient reactance using the relation
• Check if t<tc yes K=K+1• Check if j=0,yes modify the network data and obtain the new reduced admittance matrix and set j=j+1• set K=K+1• Check if K<Kmax, yes start from finding 8 constants
1
1
k k k
k k k
1 2 3 4
1 2 3 4
2 2
6
2 2
6
k k k k
k
k k k k
k
K K K K
l l l l
1 1 1cos sink k k k kp p p p pE E j E