Fault Calculations
2.3 Use of the Bus Impedance Matrix Matrix methods are ideally suited to use in explaining computer solutions of short circuit problems. If the size of the network exceeds about three or four busses, hand computation becomes excessively laborious and prone to error. The computer solves the problem by setting up a matrix representation of the whole system, then calculating the Thevenin impedance at every bus. 2.3.1 Review of matrix algebra
A matrix is an N by M array array of numbers, each number may be real or complex. The rows of the matrix run horizontally across the array and the columns run vertically down the array. The ijth element is on row i and column j of the array. If N equals M, the matrix is square. A 3 x 3 matrix is shown below:
1 2 3 A 4 5 3 3 3 9 The transpose of a matrix is obtained by interchanging its rows with its columns.
a B d
b e
c
f
a d BT b d c f A matrix of size N x 1 is a column vector, and its transpose is a row vector. We take vectors to be column vectors by default.
I1 I I 2 I 3 The sum of two matrices is defined as the the sum of the corresponding elements. Note that both matrices must have the same size if they are to be added. A matrix may be multiplied by a scalar by multiplying every element of the matrix by that scalar. Matrix multiplication is defined under certain conditions: C=AB if A is p x n, and B is n x q, then C exists and is of dimension p x q:
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C. W. Brice
August 2002 n
ci, j
a i, k b k , j k 1
where i ranges from 1 to p and j ranges from 1 to q. Notice that A B does not equal B A in general, or even for the special case of p = q. Matrix multiplication is not commutative, so the order of the matrices in a product is significant. Matrix inversion is possible for square matrices whose determinants do not vanish. A-1 A = A A-1 = Identity matrix (ones on the diagonal and zeros off the diagonal). Methods for calculating the inverse of a matrix are well known. We will look at one method a little later. 3.2 Bus admittance matrix
The bus admittance matrix Y bus is formed easily by inspection. To see this, write node equations for an example system shown in Figure 2.3.1. 1
2
-j4
I2 -j5
-j3
I1 -j0.1 3
-j0.5 ADMITTANCES IN PER UNIT
Figure 2.3.1 Sample system for formation of admittance and impedance matrices. Writing equations enforcing Kirchhoff's current law at each node gives: I1 = (V1 - V2) (-j4) + (V1 - V3) (-j5) I2 = (V2 - V1) (-j4) + (V2 - V3) (-j3) + V2 (-j0.1)
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Fault Calculations 0 = (V3 - V1) (-j5) + (V3 - V2) (-j3) + V3 (-j0.5) Collect terms: I1 = -j9.0 V1 + j4.0 V2 + j5.0 V3 I2 = j4.0 V1 - j7.1 V2 - j3.0 V3
0 = j5.0 V1 + j3.0 V2 - j8.5 V3 Put in matrix form: I = Y bus V
where Y bus is the matrix of coefficients that is formed by the following rules: Yii = Sum of all admittances connected to bus i Yij = Negative of the sum of all admittances directly connecting busses i and j.
Note that busses that are not directly connected by a line or transformer will have a zero element in the corresponding location of the Ybus matrix. In most power transmission networks, each bus is connected to about 3 or 4 other busses (at most), and so only about 3 off-diagonal elements will be non-zero and the rest will be zero. In a system of hundreds or thousands of busses, almost all elements of the matrix will be zero. Such a matrix is called a sparse matrix, and methods for exploiting sparsity are often used in programming large matrix methods. These methods conserve computer memory by storing only the non-zero elements and conserve run time by not multiplying elements that are zero. The methods of sparsity programming are beyond our present scope, but it is useful to know that the methods exist, and a general understanding of the ideas is important. The bus admittance matrix is also called the nodal admittance matrix or the short-circuit admittance matrix. If the network were subjected to a test in which all nodes except one were shorted to ground, and the one exceptional node had a current of 1 per unit injected into it, the elements of Ybus could be measured as the reciprocals of the bus voltages: Yij = Ii / V j where Vk = 0 for k not equal to j.
The bus admittance matrix is easily formed from a list of line and transformer impedances. It is likewise easy to modify to reflect changes in the network caused by switching operations. It is not particularly easy to use in calculations of short-circuit currents, however. 2.3.3 Bus impedance matrix
The bus impedance matrix Z bus is the matrix inverse of the bus admittance matrix. Z bus = Y bus-1
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August 2002
The bus impedance matrix can be interpreted as the open-circuit impedance matrix, since each element is equal to the ratio of a bus voltage to a current injection, with all other current injections equal to zero: Zij = Vi / I j with Ik = 0 for k not equal to j.
In words, the bus impedance matrix is a collection of driving-point and transfer impedances. The diagonal elements are the driving-point impedances and are equal to the Thevenin impedance at that point in the network. In comparison to Ybus, the Zbus matrix is much more difficult to compute and to change to reflect switching changes in the network. Almost every element of Zbus is not zero, so it is a full matrix (not sparse). Practice problem: Form the Ybus matrix from the data lists given below. The first list gives line reactances in per unit on a 100 MVA base, and the second gives shunt admittances (connected from a bus to neutral) in per unit on the same base. Line number
Bus numbers
Impedance [per unit]
Admittance [per unit]
1
1-2
0 + j 0.20
- j 5.0
2
1-3
0 + j 0.25
- j 4.0
3
2-3
0 + j 0.40
- j 2.5
4
3-4
0 + j 0.10
- j 10.0
Bus Number
Admittance [per unit]
1
0 - j 0.80
2
0 - j 0.80
3
0 - j 0.00
4
0 - j 0.60
Solution: Element
Value [per unit]
(1,1)
-j (5.0 + 4.0 + 0.8) = -j 9.8
(1,2)
j 5.0
(1,3)
j 4.0
(2,1)
j 5.0
(2,2)
-j (5.0 + 2.5 + 0.8) = -j 8.3 -2.3.4-
Fault Calculations (2,3)
j 2.5
(3,1)
j 4.0
(3,2)
j 2.5
(3,3)
-j (4.0 + 2.5 + 10) = -j 16.5
(3,4)
j 10.0
(4,3)
j 10.0
(4,4)
-j (10.0 + 0.6) = -j 10.6
Note that the (1,4), (4,1), (2,4) and (4,2) elements are 0. j4.0 0.0 j9.8 j5.0 j5.0 j8.3 j2.5 0.0 Ybus j4.0 j2.5 j16.5 j10.0 0.0 j10.0 j10.6 0.0
3.4 Network reductions and partial inversion
Suppose that the Y bus matrix has been formed, and we find that one of the busses has no current injections (no generator and no load modeled as a constant current, although load modeled as a constant impedance may be present). Suppose that the voltage at that bus does not need to be calculated. We may wish to simplify the network by eliminating that node from the equations (much the same way a wye-delta network reduction would). Let the bus be numbered k: Yij' = Yij - (Yik Ykj) / Ykk for i and j not equal to k.
Then the kth row and column of the matrix may be struck out. This is called the Kron reduction of the matrix, and may be derived by writing node equations, then solving the kth one for Vk in terms of the other voltages, and substituting the result in the other equations. It assumes that Ik = 0. The same result may be obtained by network reductions such as the wye-delta transformation. Shipley's partial inverse is obtained by a similar approach, except that we no longer assume that any of the current sources are zero, and the kth row and column are not eliminated. Instead, solve for the kth voltage: Vk = [Ik - j (Ykj V j)] / Ykk
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August 2002
Now the equations can be written in a hybrid matrix form. For example, a 4x4 matrix with k = 4 would give: I1 = Y11' V1 + Y12' V2 + Y13' V3 + K 14' I4 I2 = Y21' V1 + Y22' V2 + Y23' V3 + K 24' I4 I3 = Y31' V1 + Y32' V2 + Y33' V3 + K 34' I4 V4 = K 41' V1 + K 42' V2 + K 43' V3 + Z44' I4
Where Yij' = Yij - (Yik Ykj) / Ykk K ik ' = Yik / Ykk K kj' = - Ykj / Ykk Zkk ' = 1 / Ykk
This process is called pivoting on the kth axis of the matrix. If the resulting matrix is processed in a similar fashion, pivoting on the n ext axis, the result is I1 = Y11" V1 + Y12" V2 + K 13" I3 + K 14" I4 I2 = Y21" V1 + Y22" V2 + K 23" I3 + K 24" I4 V3 = K 31" V1 + K 32" V2 + Z33" I3 + Z34" I4 V4 = K 41" V1 + K 42" V2 + Z43" I3 + Z44" I4
Where Y11" = Y11' - (Y13' Y31') / Y33' Y12" = Y12' - (Y13' Y32') / Y33' Y21" = Y21' - (Y23' Y31') / Y33' Y22" = Y22' - (Y23' Y32') / Y33' Z44" = Z44' - (K 43' K 34') / Y33' K 13" = Y13' / Y33' K 31" = - Y31' / Y33' K 23" = Y23' / Y33' K 32" = - Y32' / Y33' Z43" = K 43' / Y33' Z34" = - K 34' / Y33'
Note that since we are performing the same operations as those of the Kron reduction, the Y elements at any step are the elements of the bus admittance matrix for the reduced
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Fault Calculations network. The K elements are equivalent current and voltage distribution factors. The current distribution factors show how current sources at busses that are eliminated must be distributed to the remaining busses. The partially inverted matrix is also a legitimate matrix that may be useful in some calculations, particularly if some bus voltages are known and some bus currents are known. The idea is to form the bus admittance matrix then partially invert it to move all the known quantities over to the right-hand side of the matrix. Then a simple matrix multiplication gives the solution for the unknown quantities. After performing a similar pivoting operation on the second axis and then on the first, the result is: V1 = Z11 I1 + Z12 I2 + Z13 I3 + Z14 I4 V2 = Z21 I1 + Z22 I2 + Z23 I3 + Z24 I4 V3 = Z31 I1 + Z32 I2 + Z33 I3 + Z34 I4 V4 = Z41 I1 + Z42 I2 + Z43 I3 + Z44 I4
The matrix of coefficients is the Z bus matrix, so this algorithm can be used to invert a matrix. A FORTRAN program for the partial inversion of a matrix follows. This subroutine may be called N times, pivoting on each diagonal element one at a time, to invert the matrix. Note that the original matrix is replaced by the partial inverse at each step, so the original matrix is destroyed by the subroutine. SUBROUTINE PINV(A, N, K) C
Computes the partial inverse of an N x N array A.
C
Pivots on element (K,K). A is an N x N complex matrix.
C
The original A array is replaced by its partial inverse. COMPLEX A(10,10) A(K,K) = 1 / A(K,K) DO I = 1,N IF (I .NE. K) THEN A(I,K) = A(I,K)*A(K,K) END IF END DO DO J = 1,N IF (J .NE. K) THEN DO I = 1,N IF(I.NE.K) THEN A(I,J) = A(I,J)-A(I,K)*A(K,J) -2.3.7-
C. W. Brice
August 2002
END IF END DO A(K,J) = -A(K,J)*A(K,K) END IF END DO RETURN END
Practice Problem: inverting Y bus.
For the 4-bus problem just given, calculate the Z bus matrix by
Solution: Using the program just given, the results of each step are:
k = 4:
j4.000 0.000 j9.800 j5.000 j5.000 j8.300 j2.500 0.000 j4.000 j2.500 j7.066 0.9434 0.000 0.9434 0.0943 0.000
k = 3:
j7.5356 j6.4152 j6.4152 j7.4155 0.5661 0.3538 0.3338 0.5340
k = 2:
. j19858 0.8651 0.8722 0.8228
k = 1:
j0.5036 j0.4357 j0.4392 j0.4143
0.5661 0.5340 0.3538 0.3338 j01415 . j01335 . j01335 . j0. 2203
0.8651 0.8722 0.8228 j01349 . j0.0477 j 0.0450 j0.0477 j01584 . j 01494 . j0.0450 j01494 . j 0. 2353 j0.4143
j0.4357
j0.4392
j0.5117
j0.4277
j0.4035
j0.4277
j0.5415
j 0.5108
j0.4035
j0.5108
j 0.5762
2.3.5 Direct formation of Zbus
If there are shunt impedances, such as load impedances or generator internal impedances, the reference bus will be the system neutral. If there are no shunt impedances, the reference bus is one of the system busses. All bus voltages are determined with respect to the reference bus. The reference bus will be bus number 0.
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Fault Calculations The method adds the impedances to the system one at a time, starting with any impedance that connects a bus to the reference. The impedances may represent lines, loads modeled as constant impedances, transformers, or generator impedances. Despite this, we call them all "lines" just to have a short description. Let the first line connect bus k to the reference. Vk = Zline0-k Ik
Add bus k to the list of busses already included in the system (this list helps us keep track, so we know if a bus is already in the system or not). Next find a line that connects to a bus that is already included (so far this is only bus k and the reference). There are several cases: a) a radial line from a new bus to the reference bus b) a radial line from a new bus to an existing bus c) a line from an existing bus to the reference bus (this creates a loop in the network) d) a line from one existing bus to another existing bus (this also creates a loop). Case a) a radial line form a new bus to the reference bus: This creates a new axis on the matrix, which means that a new row and column are formed, corresponding to the new bus k.
V1 Zbus V j existing Vk 0 0
0 I1
0 I j line Z 0k I k
Case b) a radial line from an existing bus p to an new bus q: A current Iq into q has the same effect on the rest of the system as if it were injected into p. V pnew = V pold + Z pp Iq Vq = V pold + (Z pp + Zline p-q) Iq
So we add axis q and let Zqk = Z pk
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August 2002
Zkq = Zkp Zqq = Z pp + Zline p-q
Note that the first two cases, addition of radial lines, did not alter the existing part of Z bus, but in the third and fourth cases, addition of loop-closing lines, the whole matrix will change. Case c) addition of a line from an existing bus p to the reference bus: Think of first adding a fictitious bus q (as in case b) but then shorting it to the reference (Vq = 0) and using Kron reduction to eliminate the fictitious bus q. Zij' = Zij - Zip Z pj / (Z pp + Zline)
Case d) addition of a new line between two existing busses: Add a fictitious axis called the loop axis to represent the constraint 0 = V p - Vq + Zline Iline
V1 Zbus existing V j 0 Z L1 Z L j
Z1 L I1
Z j L I j Z L L I line
ZL-i = Z pi - Zqi Zi-L = Zip - Ziq ZL-L = Z pp + Zqq - Z pq - Zqp + Zline
Now eliminate the loop axis using Kron reduction Zij' = Zij - Zi-L ZL-j / ZL-L
This process of adding one line at a time is continued until all the lines have been added. As we mentioned earlier, the word "line" here is used to mean any impedance whether it represents a line or a transformer or a generator internal impedance. Practice Problem: Use the bus impedance matrix formation algorithm given above for this problem:
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Fault Calculations Bus list
X [per unit]
0-1
0.010
0-2
0.015
1-2
0.084
2-3
0.122
0-3
0.005
Solution: Add 0 - 1
Zline = 0.010
Add 0 - 2
Zline = 0.015 1
Z bus
2
= j 0 . 010
0. 000
1
0 . 000
0. 015
2
Add 1 - 2
Z bus
Z bus = j 0.01
Zline = 0.084
=j
1
2
L
0 . 010
0. 000
0. 010
0 . 000
0. 015
0 . 010
0. 015
1
0. 015 2 0. 109
L
Eliminate the loop axis 1 Z bus
= j 0 . 00908
Add 2 - 3
Z bus
Add 0 - 3
2 0. 00138
1
0 . 00138 0. 01294
2
Zline = 0.122
=j
1
2
3
0 . 00908
0. 00138
0. 00138
1
0 . 00138 0. 01294
0. 01294
2
0 . 00138 0. 01294
0. 13494
3
Zline = 0.005
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C. W. Brice
Z bus
August 2002
1
2
3
L
0 . 00908
0. 00138
0. 00138
0. 00138
1
= j 0 . 00138 0 . 01294 0 . 00138 0 . 01294
0. 01294
0. 01294
2
0. 13494
0. 13494
3
0 . 00138 0 . 01294
0. 13494
0. 13994 L
Eliminate the loop axis
Z bus
=j
1
2
3
0 . 00907
0. 00125
0. 000049
1
0 . 00125
0. 01174
0. 000462
2
0 . 000049
0. 000462
0. 00482
3
2.3.6 Application to network calculations (capacitor sizing)
Each diagonal element of the bus impedance matrix is the Thevenin impedance at the corresponding bus. If we desire to find the effects of a change in the network, such as the voltage rise produced by a shunt capacitor bank, we can use the Z bus matrix. For example, the system of the last practice problem has voltage at bus 2 of 0.90 per unit. Calculate the current drawn by a capacitor having a reactance of 1.5 per unit, and calculate the voltage at bus 2. Ic = 0.90 / (j 0.01174 - j 1.50) = j 0.605 per unit V3 = 0.90 - j 0.605 x j 0.01174 = 0.907 per unit
The changes in the voltage at bus 1 can be computed by multiplying the capacitor current by the Z12. In this example, the change in voltage is quite small, so this capacitor will not produce a large change in voltage.
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