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Zbus
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The Impedance Model and Network Calculation
• The Bus Admittance and Impedance Matrices
• Thevenin’s Theorem and Zbus
• Modification of An Existing Zbus
• Direct Determination of Zbus
The Bus Admittance and Impedance Matrixes 1
Properties:
1. It is symmetrical.
2. It is a highly dense matrix .
3. Zbus matrix is very useful in the fault analysis.
1bus bus
Z Y
1. The impedance elements of Zbus on the principal diagonal are called driving-point impedances of the buses
2. Off-diagonal elements are called the transfer impedances of the buses
The Bus Admittance and Impedance Matrixes 2
Analyze the elements of Y bus:
A three bus example:
+
-
1
3
1I
2I
3I 2V
02
112 31
| VVVIY
02
222 31
| VVVIY
0ijY If i and j are not directly connected
2
busI Y V
2 21 1 22 2 23 3I Y V Y V Y V
If V1 and V3 are reduced to zero by shorting buses 1 and 2 to the reference node and V2
is applied at bus 2…
1 11 1 12 2 13 3I Y V Y V Y V
The Bus Admittance and Impedance Matrixes 3
Analyze the elements of Z bus
02
222 31
| IIIVZ
02
332 31
| IIIVZ
02
112 31
| IIIVZ
Zij means bus i is isolated from bus j
+
-
1 2
3
3I 2V1V3V
+
-
+
-
1I 2I
busV Z I
1 11 1 32 2 33 3V Z I Z I Z I
2 21 1 22 2 23 3V Z I Z I Z I
3 31 1 32 2 33 3V Z I Z I Z I
Thevenin’s Theorem and Zbus 1
From the definition for Thevenin’s equivalent impedance, at bus 2
22Z
iiZ
jiijZ
It equals to in Z bus
----- driving -point impedance (Thevenin’s
equivalent , impedance at bus i)
------ transfer impedance
+
-
1 2
3
3I 2V1V3V
+
-
+
-
2I1I
1 3
20
2
|ii th I IV
Z ZI
Figure 8.3
3V
2V
1V
nV
kV
kIReference
OriginalNetwork
busZ
3
2
1
0
n
k
(a)
kI
Originalnetwork
busZ+
-
okV
th kkZ Z
Reference0
k
kV
+
-
(b)
Figure 8.3
th kkZ Z
Thevenin impedance between two buses j and k of the network.
Figure 8.4
3V
2V
1VjV jI
Reference
OriginalNetwork
busZ
3
2
1
0
k
j
(a)
kV kI
(b)Figure 8.4
jI
Reference
OriginalNetwork
busZ
0
k
j
kI+
+
0kV0jV
kk kjZ Z
jj jkZ Zjk kjZ Z
k
j
k
jbZscI scI
(c) (d), 2th jk jj kk jkZ Z Z Z
sc j kI I I 0 0
,
k j k jb
th jk b b
V V V VI
Z Z Z
Thevenin’s Theorem and Z bus 4
Example 8.1A capacitor having a reactance of 5.0 per unit is connected between the reference node and bus of a given circuit . The original emfs and the corresponding external current injection at buses and are same as in those examples. Find the current drawn by the capacitor .Solution :
43
4
4
0
-j1.43028135738 110 74660. .
+
-
V4
5.0j
Z4
0
j0.69890
0.94866 20.7466
+
-
V4
44Z
capI 5.0j
(a)
Figure 8.5
capI
07466.20
40V
4V
4V
(b)
0.94866 20.7466 0.22056 69.25345.0 0.69890capI pu
j j
Thevenin’s Theorem and Z bus 6
Example 8.2If an additional current equal to is injected into the network at bus of Example 7.6, find the resulting voltage at bus , , , and . 1 2 3
4
0.22056 69.2534
0.73128 0.69140 0.61323 0.636770.69140 0.71966 0.60822 0.641780.61323 0.60822 0.69890 0.551100.63677 0.64178 0.55110 0.69890
bus
j j j jj j j jj j j jj j j j
Z
1 14 0.22056 69.2534 0.63677 0.14045 20.7466capV I Z j
2 24 0.22056 69.2534 0.64178 0.14155 20.7466capV I Z j
3 34 0.22056 69.2534 0.55110 0.12155 20.7466capV I Z j
4 44 0.22056 69.2534 0.69890 0.15415 20.7466capV I Z j
Thevenin’s Theorem and Z bus 7
1
2
3
4
0.96903 18.4189 0.14045 20.7466
0.96734 18.6028 0.14155 20.7466
0.99964 15.3718 0.12155 20.7466
0.94866 20.7466 0
1.10938 18.7135
1.10880 1
.15415 20.
8.8764
1.12071 15.9539
746
V
V
V
V
1.1026 81 20.7466
Modification of An Existing Zbus 1
How an existing Z bus may be modified to add new buses or to connect new lines to established
buses?
Case 1:
Adding from a new bus to the reference node
The addition of the new bus connected to the reference node through without a connection to any of the buses of the original network cannot alter the original bus voltages when a current Ip is injected at the new bus.
bZ p
bZ
Modification of An Existing Zbus 2
The voltage at the new bus is equal to .
p
N
bp
N
II
II
ZVV
VV
2
1
0
02
01
0000
00
......... origZ
...pV bpZI
Increase the rank of the Z bus
)(newbusZ
Modification of An Existing Zbus 3
Case 2 : Adding from a new bus to an existing bus p
bZ
kI
pI
k
0
p
Reference
Original network with bus and the reference node extracted
k
Go next slide
bZ k
The current Ip flow into the network at bus will increase the original voltage by the voltage :
k0
kV kkpZI
Modification of An Existing Zbus 4
kkpkk ZIVV 0
pV kV bpZI
origZ bV
will be larger than the new by the voltage :
bpkkpkp ZIZIVV 0
Substituting for kV
kV 0
The new now which must be added to in order to find is
Modification of An Existing Zbus 5
Since must be a square matrix a round the principal diagonal , a new column which is the transpose of the new row must be added .
p
N
bkkkNkk
Nk
k
k
p
N
II
II
ZZZZZZ
ZZ
VV
VV
2
1
21
2
1
2
1
origZ
...
... ...
...
p
p
)(newbusZ
busZ
Increase the rank of the Z bus
Modification of An Existing Zbus 6
b
innhorighinewhi ZZ
ZZZZ
kk
)1()1()()(
Case 3 : Adding exiting bus to the reference mode1 . Add a new bus connected through to bus 2 . Short circuit bus to the reference mode by letting equal zero to yield the same matrix equation as in case 2 expect that is zero .3 . Eliminate the (N+1) row and (N+1) column .Each element in the new matrix :
pV
bZ
hiZ
p
p
k
k
bZ
pV
Modification of An Existing Zbus 7
Case 4 : Adding between two exiting buses and
Extract these buses from the original network . bZ kj
bZkI
iI
0 Reference
Original network with bus and the reference node extracted
bI
bi II
bk II
j
k
kj
Modification of An Existing Zbus 8
The change in voltage at each bus cased by the injection at bus
and at bus is given by
The matrix equation
bIaI
bhkhih IZZV )( j
k
k
b
N
k
j
bb
N
k
i
II
II
I
ZV
VV
V 11
0 (row j - row k) of
( col . j - col . k) oforigZ
origZ
origZ
Modification of An Existing Zbus 9
The coefficient of in the last row of previous matrix is denoted by
The new row is column minus column of with in the (N+1) row .The new row is the transpose of the new column .Eliminating the (N+1) row and (N+1) column each element in the new matrix is
bI
j origZk bbZ
)(newhiZ
bjkkk
iNNhhinewhi ZZZZ
ZZZZ
jj
2)1()1(
)(
bjkkkjjbjkthbb ZZZZZZZ 2,
Direct Determination of Zbus 1
1 . Start by writing the equation for one bus connected through a branch impedance to the reference as
11 IZV a2 . Add a new bus connected to the first bus or to the reference node . For instance, if the second bus is connected to the reference node through , the matrix equations :
2
1
2
1
00
II
ZZ
VV
b
a
3 . Add other buses and branch following the procedures described in previous section .
aZ
bZ1
2
2
1
1
1
Direct Determination of Zbus 2
Example 8.4
Determine the Zbus
125.0j
2.0j
25.0j
25.1j
4.0j
25.1j
0
1 2 3
4
(1)
(2) (3)
(4)
(5)
(6)
Direct Determination of Zbus 3
11 25.1 IjV
1
1
25.11, jZbus
1
1
The establish bus with its impedance to bus , Follow CASE TWO :
5.125.125.125.1
2, jjjj
Zbus
1
1
2
2
2 1
The term j1.50 above is the sum of j1.25 and j0.25 .
25.1j
25.0j1 2
CASE ONE
25.1j
1
Direct Determination of Zbus 4
Bus with the impedance connecting it to bus is established by writing
90.150.125.150.150.125.125.125.125.1
3,
jjjjjjjjj
Zbus
1
2
2
2
1
3
3
The term j1.90 above is the sum of Z22 of the matrix being modified and the impedance Zb of the branch being connected to bus from bus .2 3
3
25.1j
25.0j
1 2
4.0j
3
CASE TWO
Direct Determination of Zbus 5
To add the impedance from bus to the reference node, follow case three to connect a new bus through and obtain the impedance matrix
25.1jZb
15.390.150.125.190.190.150.125.150.150.150.125.125.125.125.125.1
4,
jjjjjjjjjjjjjjjj
Zbus
1 2 3
3
2
1
p
p
3
p bZ
Where j3.15 above is the sum of .
bZZ 33
CASE THREE
25.1j
25.0j1
2
4.0j3
25.1j
Direct Determination of Zbus 6
After the row p and column p are eliminated , the new matrix is
75397.059524.049603.059524.078571.065476.049603.065476.075397.0
5,
jjjjjjjjj
Zbus
1 2
2
3
3
1
No change in rank
Direct Determination of Zbus 7
Some of the elements of the new matrix are calculates by :
75397.015.3
)25.1)(25.1(25.1)(11 jj
jjjZ new
78571.015.3
)50.1)(50.1(50.1)(22 jj
jjjZ new
15.3)90.1)(50.1(50.1)(32)(23 j
jjjZZ newnew
59524.0j
Direct Determination of Zbus 8
To add the impedance from bus to establish bus , follow the CASE TWO
95397.075397.059524.049603.075397.075397.059524.049603.059524.059524.078571.065476.049603.049603.065476.075397.0
6,
jjjjjjjjjjjjjjjj
Zbus
1
1
2
2
3
3 4
4
The new diagonal element is the sum of Z33 of the previous matrix and Zb = j0.20
3 420.0jZb
CASE TWO
25.1j
25.0j1
2
4.0j
3
25.1j
4
Direct Determination of Zbus 9
Finally , to add impedance between buses and ,follow the CASE FOUR
125.0jZb 2 4
15873.049603.065476.0141215 jjjZZZ
19047.059524.0.078571.0.0242225 jjjZZZ 15873.0.075397.059524.0343235 jjjZZZ
35873.095397.059524.0444245 jjjZZZ
bZZZZZ 24442255 2 125.0)}59524.0(2)95397.078571.0{( jj 67421.0j
CASE FOUR
25.1j
25.0j1
2
4.0j
3
25.1j
4
Direct Determination of Zbus 10
The 5*5 matrix is
67421.035873.015873.019147.015873.035873.015873.0
19047.015873.0
jjjjjjj
jj
6,busZ
q
q
Direct Determination of Zbus 11
After the row q and column q are eliminated, the bus impedance matrix to be determined is:
76310.066951.069659.058049.066951.071660.064008.053340.069659.064008.073109.060992.058044.053340.060992.071660.0
jjjjjjjjjjjjjjjj
Zbus
1 2
1
2
3
3
4
4
No change in rank