st. vincent pallotti college of engineering & technology ...secure site...
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St. Vincent Pallotti College of Engineering & Technology
Department of Electrical Engineering
Computer Application in Power System
By Dr. Nitin K. Dhote
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Unit Syllabus Details
1
Formation of network matrices by singular
transformation
Connected graph, orientred graph, tree and links
Basic cut sets and basic loops
Incidence matrices
Primitive Network
2Algorithm for formation of network matrices
Algorithm for formation of Z bus and Ybus without mutual
coupling
3
Three phase networkThree phase balanced network element with balanced and
unbalanced excitation
Incidence and network matrices for 3 phase element
Algorithm for formation of three phase bus impedance matrics
UNIT1- Q1 & Q2 UNIT2-Q3&Q4 UNIT3&5-Q5&Q6UNIT4&6-Q7&Q8 ; SOLVE 4Q , EACH QUESTION CARRY 20 MKS
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3
4
Load Flow StudiesG.S. method and N.Raphson method, fast decoupled
technique
Representation of tap changing transforemer
Phase shifting transformer
Elementaryload flow equations
5
Short circuit studiesshort circuit calculations using bus impedance matrix for
balanced and un balanced fault
Algorithm for short circuit studies
6
Transient stability studies
Modelling of synchronous machine
Numerical solution of swing equation by modified Euler,s
method
Numerical solution of swing equation by Runge Kutta
method
algorithm for transient stability study
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SR. NO. TITLE OF BOOKS / EDITIONAUTHOR /
PUBLICATION
NO. OF
COPIES
AVAILABLE
IN LIBRARY
1Computer methods in
power system analysisStagg and Ele Abaid,
TMHNil
2Electric Energy system
theory and introductionOile J. Elgard,TMH 24
3Elements of power
system analysisW.D.stevenson,TMH 13
4Computer analysis of
power systemR.N.Dhar Nil
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STUDY OF INTERCONNECTIONS OF POWER SYSTEM COMPONENTS
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1.Graph:
2. Elements:
3. Nodes:
4. Connected Graph:
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1.Graph:
2. Elements:
3. Nodes:
4. Connected Graph:
Ϩ Ϩ
Ϩ
G1
G2
G3
1 23
4
1
2
3
45
7
6
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CONNECTED GRAPH
Total elements--7
Total nodes--5
Ground node-- Ref.node
Total buses --4
1 2 3 4
0
1 2 3
456
7
Ground Node
Ϩ Ϩ
Ϩ
G1
G2
G3
1 23
4
1
2
3
45
7
6
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Each element is assigned a direction .
1 2 3 4
0
1 2 3
456
7
Ground Node
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Branches :
0
4321
1
2
3
4
Tree 1
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0
4321
1
3
4
6
Tree 2
0
4321
1
2
3
5
Tree 3
11
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Links :
Rules:
1 2 3 4
0
1 2 3
456
7
Ground Node
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0 1 2 3 4
Ele =1 1 -1
2 1 -1
3 1 -1
4 -1 1
5 1 -1
6 1 -1
7 1 -1
Branches
Links
Nodes
1 2 3 4
0
1 2 3
456
7
Ground Node
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Rules:a = 1; If element is directed away from the node.a = -1; If element is directed towards the node.a = 0; If element is not connected to the node.
1 2 3 4
Ele =1 -1
2 -1
3 -1
4 -1 1
5 1 -1
6 1 -1
7 1 -1
Branches
Links
Buses
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Ele =1
2
3
4
5
6
7
Branches
Links
1 2 3 4
-1
-1
-1
-1 1
1 -1
1 -1
1 -1
Ref. Buses
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Basic Cut set:
Rules to form Basic Cut set: a) Circular or closed loop cut.b) Should cut only one branch , may cut one or more
links.c) Should not cut any element twice.d) Direction is given by branch.
( Centrally inward or outward)
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Ground Node
1 2 3 4
0
1 2 3
456
7
AB
C
D
A : Basic Cutset ( Branch 1)B : Basic Cutset ( Branch 2)C : Basic Cutset ( Branch 3)D : Basic Cutset ( Branch 4)
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b = 1; If element is directed in same direction of basic cutset.
b = -1; If element is directed in opposite direction of basic cutset.b = 0; If element is not connected to the basic cutset.
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A B C D
Ele =1 12 13 14 15 -1 1 16 -1 17 -1 1
Branches
Links
Basic Cutsets
1 2 3 4
0
1 2 3
456
7
A B
CD
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Ground Node
1 2 3 4
0
1 2 3
456
7A
B
CD
E F
G
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A B C D
Ele
=11
2 13 14 15 -1 1 16 -1 17 -1 1
Branches
Links
Basic CutsetsE F G
1
1
1
Tie Cutsets
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g
ef
1 2 3 4
0
12
3
4
56
7
22
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c = 1; If element is directed in same direction of basic loop.
c = -1; If element is directed in opposite direction of basic loop.
c = 0; If element is not connected to the basic loop.
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e f g
Ele =1 12 1 -1 13 -1 -14 -15 16 17 1
Branches
Links
0
4321
12
3
45
6
7
f
g
e-- Basic loop ( Link5 )
f-- Basic loop ( Link6 )
g-- Basic loop ( Link7 )
e
Basic Loops
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250
4321
12
3
45
6
7
f
g
e
a
b
c
d
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Ele
=1
2
3
4
5
6
7
Branches
Links
0
4321
12
3
45
6
7
f
g
a--Open loop (Branch 1)
b--Open loop (Branch 2)
c--Open loop (Branch3)
d--Open loop (Branch 4)
e
a
b
c
d
e f g
1
1 -1 1
-1 -1
-1
1
1
1
a b c d
1
1
1
1
BASIC LOOPSOPEN LOOPS
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1 2 3 4
0
12
3
4
Ref. Node
K1
K2
K3K4
Buses: 1,2 ,3,4Ref. Node: 0
BRANCH PATH I : BUS1BRANCH PATH II : BUS2BRANCH PATH I II: BUS3BRANCH PATH I V: BUS4
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17. Branch Path Incidence Matrix K
Rules:k = 1; If branch is directed in same direction of branch path.
k = -1; If branch is directed in opposite direction of branch path .k= 0; If branch is not connected to the branch path.
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K1 K2 K3 K4
Br
=1-1
2 -13 -1 -14 -1
Branch Paths
1 2 3 4
0
12
3
4
Ref. Node
K1
K2
K3K4
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Pb.2 For the figure shown , determine matrices A, A^, B,B^, C, C^ & K. Select node 1 as reference and select elements 2&5 as a links.
1
23
4
1
2
3
4
5
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2 33 444
-1
-1
-1
1 -1
1 -1
1
3
4
2
5
Br
Link
Ref. 1
1
1
1
1
2 3
4
1
2
4
5
REF.
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1
2 3
4
1
2
4
5
REF.
A B
C
A B C
1 1
3 1
4 1
2 -1 1
5 -1 1
Br
Link
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A B C
1 1
3 1
4 1
2 -1 1
5 -1 1
Br
Link
D E
1
1
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d
1 1
3 -1
4
2 1
5
eloops
Br
Links
e
1
-1
1
1
2 3
4
1
4REF.
2
5
d
e
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a b c
1
1
1
Br
Links
1
3
4
2
5
d e
1
-1 1
-1
1
1
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K1 K2 K3
1 -1
3 -1
4 -1
BR
1
2 3
4
1
4REF.
K1 K2
K3
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Pb.3 For the Figure shown , determine incidence matrices. Select node 1 as reference and elements 1,2,4 and 6 as branches.
1
4 5
3
2
1
2
4
6
3
5
7
REF.
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1
4 5
3
2
1
2
4
6
3
5
7
ELEMENT NODE INCIDENCE MATRIX :
REF.
REF. 1
-1
1
.
1
1
2
4
6
3
5
7
2 3 4 5
1
-1
1 -1
1 -1
-1
-1 1
-1 1
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BASIC CUTSETS AND TIE CUTSETS
1
4 5
3
2
1
2
4
6
3
5
7
A
C
B
D
E
F
G
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C
B1
4 5
3
2
1
24
6
3
5
7A
D
E
F
G
AUGMENTED CUTSET INCIDENCE MATRIX
E F G
1
1
1
A B C D
1 1
2 1
4 1
6 1
3 1 -1
5 1 -1
7 -1 -1 1 -1
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BASIC LOOPS AND OPEN LOOPS
1
4 5
3
2
1
2
4
6
3
e
5
fa
b cd
7
g
a,b,c,d : Open Loopse,f,g : Basic Loops
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1
4 5
3
2
1
24
6
3
e
5
fa
b c d
7g
AUGMENTED LOOP INCIDENCE MATRIX :
1
2
4
6
3
5
7
e f g
1
-1 1
1 -1 -1
1 1
1
1
1
a b C d
1
1
1
1
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BRANCH PATHS
1
4 5
3
2
1
2
4
6
K1
K2
K3
K4
REF.
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BRANCH PATH INCIDENCE MATRIX K
K1
1
4 5
3
2
1
24
6K2
K3
K4
K1 K2 K3 K4
1 1
2 -1 -1 -1
4 1 1
6 -1
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Pb.4 For a particular power system, element node incidence matrix is given by
1 2 3 4
1 1 -1
2 1 -1
3 -1 1
4 1 -1
5 1 -1
Draw oriented graph, select 4 & 5 as link and node 1 as ref. node. Determine matrices B^, c^ and K.
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ORIENTED GRAPH
1
32
4
1
2
3
4
5
BASIC CUTSETS
1
32
4
1
2
3
4
5A
B
C
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47
BASIC CUTSET INCIDENCE MATRIX
1
32
4
1
2
3
4
5A
B
C
A B C
1 1
2 1
3 1
4 -1 1 -1
5 -1 1
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48
AUGMENTED CUTSET INCIDENCE MATRIX
1
32
4
1
2
3
4
5A
B
C
A B C
1 1
2 1
3 1
4 -1 1 -1
5 -1 1
D
E
D E
1
1
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49
1
32
4
1
2
3
4
5
d
e
1
2
3
4
5
d e
1 1
-1 -1
1
1
1
ab c
a b c
1
1
1
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50
32
4
1
2
3
1
K1
K2
K3
REF.
K1 K2 K3
1 -1
2 -1 -1
3 1
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51
[Ab]br X b X [KT]b X br = [Ubr]
[Al]l X b X [KT]b X br = [Bl]lXbr
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52
Pb.5 4 bus sample system is shown in fig. Consider ground as ref. and elements 4 &5 as links. Determine A^, B^,C^ and K and also verify the relation sUbr = Ab.K T ; Al.K T = Bl and Cb = -Bl T
G
G
1 2 3
4
1
2 3
45
6
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53
ORIENTED GRAPH
ELEMENT NODE INCIDENCE MATRIX
0 1 2 3 4
1 1 -1
2 1 -1
3 1 -1
6 1 -1
4 1 -1
5 1 -1
1 2 3
4
0REF.
1
2 3
45
6
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54
BUS INCIDENCE MATRIX
1 2 3 4
1 -1
2 1 -1
3 1 -1
6 -1
4 1 -1
5 1 -1
Ab
Al
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55
AUGMENTED CUTSET INCIDENCE MATRIX
1 2 3
4
0REF.
1
2 3
45
6 A
B
C
D
A B C D E F
1 1
2 1
3 1
6 1
4 -1 -1 -1 1 1
5 -1 1 1
E
F
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56
AUGMENTED LOOP INCIDENCE MATRIX
1 2 3
4
0REF.
1
2 3
45
6
f
a
b c
d
a b c d e f
1 1 1 1
2 1 1
3 1 1
6 1 -1 -1
4 1
5 1
e
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57
BRANCH PATH INCIDENCE MATRIX K
1 2 3
4
0
1
2 3
6 K4
K1 K2 K3 K4
1 -1 -1 -1
2 -1 -1
3 -1
6 -1
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58
-1
1 -1
1 -1
-1
-1
-1 -1
-1 -1 -1
-1
1
1
1
1
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1 -1
1 -1
-1
-1 -1
-1 -1 -1
-1
-1 -1 -1 1
-1 1
1 1
1
1
-1 -1
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60
Pb.6 For the power system shown , select element 1,2,6,7 as links and ground node as reference. Determine matrices A^, B^,C^ and K . Verify the relations 1. Ab. K T= Ubr 2. Al.K T =Bl and 3. Cb= -Bl T
G
G
L L
G
1 321
2
34
5 6
7
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61
ORIENTED GRAPH
1
0
323 4
51
2
6
7
REF
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62
1
0
323 4
512
67
REF
ELEMENT NODE INCIDENCE MATRIX
0 1 2 3
3 1 -1
4 1 -1
5 -1 1
1 -1 1
2 -1 1
6 -1 1
7 -1 1
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63
0 1 2 3
3 1 -1
4 1 -1
5 -1 1
1 -1 1
2 -1 1
6 -1 1
7 -1 1
BUS INCIDENCE MATRIX
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64
AUGMENTED CUTSET INCIDENCE MATRIX
1
0
323 4
512
67
REF
AB
C
DE
FG
A B C D E F G
3 1
4 1
5 1
1 1 1 1
2 1 1 1
6 -1 1 1
7 -1 1 1
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65
AUGMENTED LOOP INCIDENCE MATRIX
1
0
323 4
512
67
REF
e
d f
g
a b
c
a b c d e f g
3 1 -1 -1
4 1 1 1
5 1 -1 1 -1 -1
1 1
2 1
6 1
7 1
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66
BRANCH PATH INCIDENCE MATRIX
1
0
23 4
5
3
REF
K1
K2 K3
K1 K2 K3
3 1
4 -1
5 1 1 1
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67
PRIMITIVE ELEMENT : PRIMITIVE MATRIX :
A) IMPEDANCE FORM :
EqZpq
P QEp
epq
ipq+Vpq-
+- +
+-
-
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68
Vpq = Zpq ipq – epqVpq +epq = Zpq .ipq
V nx1 + e nx1 = Z nxn . i nx1
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69
YpqP Qipq
jpq
+ Vpq -
Ipq +jpq = Vpq . Ypq
i nx1 + j nx1 = Y nxn . V nx1 ,
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70
A) BUS REFERENCE FRAMEEBUS = [ZBUS] . IBUS
IBUS = [YBUS]. EBUS
B) BRANCH REFERENCE FRAMEEBR = [ZBR].IBR
IBR = [YBR]. EBR
C) LOOP REFERENCE FRAMEELOOP = [ZLOOP]. ILOOP
ILOOP = [YLOOP].ELOOP
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PERFORMANCE EQUATION OF PRIMITIVE ELEMENT IN ADMITTANCE FORM:i + j = [y]. V
PREMULTIPLYING EQN.1 BY AT
AT . i + AT . j = AT . [y] . V
APPLYING KCL, AT . i =0
I BUS = AT . j USING EQNS. (2) AND (3)
I BUS = AT [Y] VTRANSFORMATION FROM PRIMITIVE TO BUS REF. IS POWER INVARIANT
S = IBUS* T . EBUS = j *T .V
FROM EQ.(3) , IBUS = AT. j TAKING CONJUGATE TRANSPOSE
IBUS*T = j*T . A*
71
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72
IBUS*T = j*T . A*
A* = A IBUS
*T = j*T . AS = IBUS
* T . EBUS = j *T .V (5)FROM EQ.(5) AND EQ.(6),
j*T . A. EBUS = j *T .VV = A. EBUS
I BUS = AT [Y] VFROM EQS.(4) AND (7),
IBUS = AT[Y]A.EBUS
PERFORMANCE EQUATION IN BUS REF. FRAME IN ADMITTANCE FORM
IBUS = [YBUS] . EBUS
FROM EQNS. (8) AND (9)YBUS = AT . Y. A
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PERFORMANCE EQUATION OF PRIMITIVE ELEMENT IN ADMITTANCE FORM:i + j = [y]. V
PREMULTIPLYING EQN.1 BY BT
BT . i + BT . J = BT . [y] . V
APPLYING KCL, BT . i =0
I BR = BT . j USING EQNS. (2) AND (3)
I BR = BT [Y] VTRANSFORMATION FROM PRIMITIVE TO BRANCH REF. IS POWER INVARIANT
S = IBR* T . EBR = j *T .V
FROM EQ.(3) , IBR = BT. j TAKING CONJUGATE TRANSPOSE
IBR*T = j*T . B*
73
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74
IBR*T = j*T . B*
* = B IBR
*T = j*T . BS = IBR
* T . EBR = j *T .V (5)FROM EQ.(5) AND EQ.(6),
j*T . B. EBR = j *T .VV = B. EBR
I BR = BT [Y] VFROM EQS.(4) AND (7),
IBR = BT[Y]B.EBR
PERFORMANCE EQUATION IN BRANCH REF. FRAME IN ADMITTANCE FORM
IBR = [YBR] . EBR
FROM EQNS. (8) AND (9)YBR = BT . Y. B
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75
V + e = [Z].i
C T .V +C T . e = C T . [Z].iC -- INCIDENCE OF ELEMENTS TO BASIC LOOPS C T .V -- ALGEBRIC SUM OF VOLTAGES OF THE ELEMENTS AROUND EACH BASIC LOOP
C T . V = 0C T .e -- ALGEBRIC SUM OF EXTERNAL VOLTAGE SOURCE AROUND EACH LOOP
ELOOP = C T . e
ELOOP = C T .[Z]. i
I LOOP iE LOOPe
S= ILOOP * T . ELOOP = i* T .e
I LOOP* T . C T.e = i* T . e
I LOOP * T .C T = i* T
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76
I LOOP * T .C T = i* T
i = C *. I LOOP
SINCE C IS REAL MATRIX, C* = Ci = C . I LOOP
ELOOP = C T .[Z]. i (4)
ELOOP = C T .[Z]. C . I LOOP
E LOOP = Z LOOP . I LOOP
Z LOOP = C T . [Z]. C
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RHS = K . [ ]. K T
YBUS = A T [Y] A= [Y]
A K T = B , B T = K .A T
RHS = B T [Y] B= YBR
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78
RHS = Ab T [ ] AbY BR = B T [Y] B
= Ab T [Y] AbB = A K T
B T = KA T
RHS A T [Y] A Ab K T =Ubr
K T = Ab -1
KT Ab = Ubr
K Ab T = UbrAb T = K -1
Ab T K = Ubr
RHS = A T [Y] A= Y BUS
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79
1 2
4
3
0.51
0.52
50.2
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80
1 2 3
4
0.52
REF.
1 3 4
1 1
3 1 -1
5 -1
2 1
4 -1 1
0.51
0.25
4
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1 3 5 2 4
1 1 1 1
3 -1 -1
4 -1 1
1 3 5 2 4
1 2
3 2.5
5 5
2 2
4 1.25
ELEMENTNO.
1 2 3 4 5
Z 0.5 0.5 0.4 0.8 0.2
Y 2 2 2.5 1.25 5
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82
1 3 5 2 4
1 1 1 1
3 -1 -1
4 -1 1
1 3 5 2 4
1 2
3 2.5
5 5
2 2
4 1.25
1 3 4
1 1
3 1 -1
5 -1
2 1
4 -1 1
2 2.5 2
-5 -1.25
-2.5 1.25
1
1 -1
-1
1
-1 1
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83
6.5 0 -2.5
0 6.25 -1.25
-2.5 -1.25 3.75
1 2 3
4
3
1
5REF.
K1
K3
K2
K1 K2 K3
1 1 1
3 -1
5 -1
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84
1 1
-1
-1
6.5 0 -2.5
0 6.25 -1.25
-2.5 -1.25 3.75
1
-1
1 -1
4 -1.25 1.25
2.5 1.25 -3.75
-6.25 1.25
1
-1
1 -1
5.25 -1.25 1.25
-1.25 3.75 -1.25
1.25 -1.25 6.25
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85
2
41
3
1 (0.5)
2 (0.5)
5 (0.4)
3 (1.0)
0.2
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86
2
41
3
1 (
0.5
)
2 (0.5)
5 (0.4)
3 (1
.0)
0.2
2 3 4
1 -1
4 -1
2 -1
3 1 -1
5 1 -1
** MUTUAL COUPLED BRANCHES 1 & 4 ARE KEPT ADJACENT AND PLACED AT THE TOP OF THE MATRIX
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87
1 4 2 3 5
2 -1 1
3 -1 1 -1
4 -1 -1
1 4 2 3 5
1 0.5 0.2
4 0.2 0.4
2 0.5
3 1
5 0.4
Z1 0
0 Z4
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88
Y = Z -1 Y1 0
0 Y4
0.5 0.2
0.2 0.4=
-1
=
2.5 -1.25
-1.25 3.125
=0.5
1
0.4
-1
=
2
1
2.5
Y = -j
2.5 -1.25
-1.25 3.125
2
1
2.5
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89
-1 1
-1 1 -1
-1 -1
2.5 -1.25
-1.25 3.125
2
1
2.5
-1
-1
-1
1 -1
1 -1
5 -3.75 0
-3.75 6.625 -1
0 -1 3
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90
2
41
3
1
2
4K1
K3
K1 K2 K3
1 -1
4 -1
2 -1
REF.
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91
1
0.1025
0.42
0.63
4
0.1025
1 32
1
0
32
1
2 3
4
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92
1
0
32
1
2 3
4
A
B
C
A B C
1 1
2 1
3 1
4 1 1 1
1 2 3 4
A 1 1
B 1 1
C 1 1
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93
ELEMENTNO.
Z Y
1 0.1025 9.756
2 0.4 2.5
3 0.6 1.67
4 0.1025 9.756
1 2 3 4
1 9.756
2 2.5
3 1.67
4 9.756
1 1
1 1
1 1
9.756
2.5
1.67
9.756
1
1
1
1 1 1
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94
19.512 9.756 9.756
9.756 12.256 9.756
9.756 9.756 11.426
0.0937 -0.034 -0.05
-0.034 0.267 -0.198
-0.05 -0.198 0.3009
1
0
32
1
2 31 2 3
1 -1
2 1 -1
3 1 -1
BR
BUSES
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95
-1 1
-1 1
-1
19.512 9.756 9.756
9.756 12.256 9.756
9.756 9.756 11.426
-1
1 -1
1 -1
12.256 -2.5 0
-2.5 4.17 -1.664
0 -1.67 11.42
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96
B
DC
G A
REF.
1
2
3
4
5
6
7
ELEMENT NO.
1 2 3 4 5 6 7
Z 0.04 0.04 0.05 0.03 0.07 0.02 0.1
Y 25 25 20 33.33 14.28 50 10
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97
1 2 3 4 5 6 7
1 25
2 25
3 20
4 33.33
5 14.28
6 50
7 10
B
DC
G A
1
2
3
4
5
6
7
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98
B
DC
G A
1
2
3
45
6
7
A
B
C
D
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99
B
DC
G A
1
2
3
4 5
6
7
A
BC
D
A B C D
1 1
2 1
3 1
4 1
5 1 1 1
6 -1 1
7 -1 -1 -1
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100
1 2 3 4 5 6 7
A 1 1 -1
B 1 -1 -1
C 1 1 1
D 1 1 -1
1 1 -1
1 -1 -1
1 1 1
1 1 -1
25
25
20
33.33
14.28
50
10
1
1
1
1
1 1 1
-1 1
-1 -1 -1
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101
49.28 10 14.28 24.28
10 85 -50 10
14.28 -50 84.28 14.28
24.28 10 14.28 57.61
B
DC
G A
1
2
3
4
A B C D
1 1
2 -1
3 -1
4 -1 1
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102
1 2 3 4
A -1
B -1
C 1 -1
D 1
-1
-1
1 -1
1
49.28 10 14.28 24.28
10 85 -50 10
14.28 -50 84.28 14.28
24.28 10 14.28 57.61
1
-1
-1
-1 1
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103
85 -50 0 -10
-50 84.28 0 -14.28
0 0 58.33 -33.33
-10 -14.28 -33.33 57.61
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104
1
0
4
32
REF.
1
2 3
6
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105
ELEMENT NO.
1 2 3 4 5 6
Z 0.1 0.2 0.22 0.43 0.3 0.12
1 2 3 6 4 5
1 0.1
2 0.2
3 0.22
6 0.12
4 0.43
5 0.3
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106
1
0
4
32
REF.
1
2 3
d
e
d e
1 1 1
2 1
3 1
6 -1 -1
4 1
5 1
1 2 3 6 4 5
d 1 -1 1
e 1 1 1 -1 1
6
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107
1 -1 1
1 1 1 -1 1
0.1
0.2
0.22
0.12
0.43
0.3
1 1
1
1
-1 -1
1
1
0.65 0.22
0.22 0.94
1.67 -0.391
-0.391 1.1553
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108
SELF MUTUAL
ELEMENT NO. Z ELEMENT NO.
Z
1 0.6
2 0.5 1 0.1
3 0.5
4 0.4
5 0.2
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2 3 4
1 -1
2 -1
3 1 -1
4 -1
5 1 -1
3
21
4
1
2
3
4
5
REF.
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A B C
1 1
2 1
3 1
4 1
5 -1 1 1
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1 2 3 4 5
1 0.6 0.1
2 0.1 0.5
3 0.5
4 0.4
5 0.2
Y1 0
0 Y4
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112
Y BUS = A T [Y] A
9.22 -0.34 -5
-0.34 4.07 -2
-5 -2 7
1 2 3 4 5
1 1.724 -0.344
2 -0.344 2.07
3 2
4 2.5
5 5
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113
Y BR = K [Y BUS] K T
Z LOOP = C T [Z] C