KINGDOM OF SAUDI ARABIA Ministry Of High Education
Umm Al-Qura University College of Engineering & Islamic Architecture
Department Of Electrical Engineering Engineering
Power System Analysis The Impedance Model And Network Calculations
2011/2012 1432/1433
Dr Houssem Rafik El Hana Bouchekara
1
1
THE IMPEDANCE MODEL AND NETWORK CALCULATIONS ............................................. 3
1.1
THE BUS ADMITTANCE AND IMPEDANCE MATRICES ............................................................... 3
1.2
THEVENIN'S THEOREM AND
1.3
MODIFICATION OF AN EXISTING
1.4
DIRECT DETERMINATION OF
1.5
CALCULATION OF
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................................................................................. ........................................................................... ...... 19
ELEMENTS FROM
Dr Houssem Rafik El Hana Bouchekara
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............................................................. .................................... ......................... 23
1
THE IMPEDANCE MODEL AND NETWORK CALCULATIONS ............................................. 3
1.1
THE BUS ADMITTANCE AND IMPEDANCE MATRICES ............................................................... 3
1.2
THEVENIN'S THEOREM AND
1.3
MODIFICATION OF AN EXISTING
1.4
DIRECT DETERMINATION OF
1.5
CALCULATION OF
.................................................................................... ......................................................... ........................... 6 ........................................................................... ............................................................. .............. 12
................................................................................. ........................................................................... ...... 19
ELEMENTS FROM
Dr Houssem Rafik El Hana Bouchekara
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............................................................. .................................... ......................... 23
1
THE IMPEDANCE MODEL AND NETWORK CALCULATIONS
The bus admittance matrix o f a large-scale interconnected power system is typically very sparse with mainly zero elements. In previous chapter we saw how
is constructed
branch by branch from primitive admittances. It is conceptually simple to invert the bus impedance matrix
but such direct inversion is rarely employed when the
systems to be analyzed are large scale. In practice, the triangular factors of
to find
is rarely explicitly required, and so
are used to generated elements of
only as they are
needed since this is often the most computationally efficient method. By setting computational considerations aside, however, and regarding
as being already
constructed and explicitly available, the rower system analyst can derive a great deal of in sight. This is the approach taken in this chapter. The bus impedance matrix can be directly constructed element by element using
simple algorithms to incorporate one element at a time in to the system representation. The work entailed in constructing
is much greater than that required to construct
the information content of the bus impedance matrix is far greater than that of shall see, for example, that each diagonal element of
but We
reflects important characteristics
of the entire system in the form of the Thevenin impedance at the corresponding bus. Unlike the bus impedance matrix of an interconnected system is never. sparse and contains
zeros only when the system is regarded as being subdivided into independent parts by open circuits. In Chap.12, for instance, such open circuits arise in the zero-sequence network of
the system. The bus admittance matrix is widely used for power-flow analysis. On the other
hand, the bus impedance matrix is equally well favored for power system fault analysis. Accordingly, both
and
have important roles in the analysis of the power system
network. In this chapter we study how to construct
directly and how to explore some
of the conceptual insights which it offers into the characteristics of the power transmission network.
1.1 THE BUS ADMITTANCE
AND IMPEDANCE
MATRICES
In Example 7.6 we inverted the bus admittance matrix the bus impedance matrix
. By definition
and called the resultant
(1)
and for a network of three independent nodes the standard form is
Since
is symmetrical around the principal diagonal,
(2)
must also be
symmetrical. The bus admittance matrix need not be determined in order to obtain in another section of this chapter we see how
Dr Houssem Rafik El Hana Bouchekara
may be formulated directly.
3
and
The impedance elements of
on the principal diagonal are called driving-point
impedances of the buses, and the off-diagonal elements are called the transfer impedances of the buses. The bus impedance matrix is important and very useful in making fault calculations, as we shall see later. In order to understand the physical significance of the various impedances in the matrix, we compare them with the bus admittances. We can easily do so by looking at the equations at a particular bus. For instance, starting with the node equations expressed as
(3)
we have at bus of the three independent nodes
If
and voltage bus is
and
(4)
are reduced to zero by shorting buses and to the reference node,
is applied at bus so that current
enters at bus , the self-admittance at
(5)
Thus, the self-admittance of a particular bus could be measured by shorting all other buses to the reference node and then finding the ratio of the current injected at the bus to the voltage applied at that bus. Figure 1 illustrates the method for a three-bus reactive network. The result is obviously equivalent to adding all the admittances directly connected
to the bus, which is the procedure up to now when mutually coupled branches are absent. Figure 1 also serves to illustrate the off-diagonal admittance terms of equation obtained by expanding equation (3) equation (3) is
from which we see that
Thus, the mutual admittance term
(6)
(7)
is measured by shorting all buses except bus
to the reference node and by applying a voltage
At bus the
at bus , as shown in Figure in Figure 1. Then, 1. Then,
is the ratio of the negative of the current leaving the network in the short circuit at node
to the voltage
since
. The negative of the current leaving the network at node is used
is defined as the current entering the network. The resultant admittance is the
negative of the admittance directly connected between buses and , as we would expect since mutually coupled branches are absent.
Dr Houssem Rafik El Hana Bouchekara
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Figure 1: Circuit for measuring
.
and
.
We have made this detailed examination of the bus admittances in order to
differentiate them clearly from the impedances of the bus impedance matrix. Conceptually, we solve equation (3) equation (3)by by premultiplying both sides of the equation by
and we must remember when dealing with
that
to yield (8)
and are are column vectors of
the bus voltages and the currents entering the buses from current sources, respectively. Expanding equation (8) equation (8) for a network of three independent nodes yields
From equation (10) equation (10) we see that the driving-point impedance
(9) (10) (11)
open-circuiting the current sources at buses and and by injecting the source current at bus . Then,
Figure 2: Circuit for measuring
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is determined by
(12)
and and
.
Figure 2 shows the circuit described. Since sources connected to the other buses whereas
is defined by opening the current
is found with the other buses shorted, we
must not expect any reciprocal relation between these two quantities. The circuit of Figure 2
also enables us to measure some transfer impedances, for we see from equation (9) that with current sources
and
open-circuited (13)
and from equation(11)
(14)
Thus, we can measure the transfer impedances bus and by finding the ratios of
and
to
and
by injecting current at
with the sources open at all buses except
bus . We note that a mutual admittance is measured with all but one bus short-circuited and that a transfer impedance is measured with all sources open-circuited except one. Equation (9) tells us that if we inject current into bus with current sources at
buses and open, the only impedance through which conditions, equations (10) and (11) show that expressed by
flows is
. Under the same
is causing voltages at buses and
(15)
It is important to realize the implications of the preceding discussion, for sometimes used in power-flow studies and is extremely valuable in fault calculations.
1.2
THEVENIN'S THEOREM AND
is
The bus impedance matrix provides important information regarding the power
system network, which we can use to advantage in network calculations. In this section we examine the relationship between the elements of
and the Thevenin impedance
presented by the network at each of its buses. To establish notation, let us denote the bus
voltages corresponding to the initial values voltages
to
of the bus currents I by
. The
are the effective open-circuit voltages, which can be measured by
voltmeter between the buses of the network and the reference node. When the bus currents are changed from their initial values to new values are given by the superposition equation
where
, the new bus voltages
(16)
represents the changes in the bus voltages from their original values. (a)
shows a large-scale system in schematic form with a representative bus Ⓚ extracted along with the reference node of the system. Initially, we consider the circuit not to be energized
Dr Houssem Rafik El Hana Bouchekara
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are zero. Then, into bus Ⓚ a current of
so that the bus currents
and voltages
(or
in per unit) is injected in to the system from a current source
per unit for
amp
connected to the reference node. The resulting voltage changes at the buses of
Figure 3:
Original network with bus and reference node extracted. Voltage at bus current entering the network. Thevenin equivalent circuit at node ·
the network, indicated by the incremental quantities
to
is caused by
, are given by
with the only nonzero entry in the current vector equal to
in row
column multiplication in equation (17) yields the incremental bus voltages
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(17)
. Row-by-
Ⓚ
Ⓚ
(18)
which are numerically equal to the entries in column current
of
multiplied by the
. Adding these voltage changes to the original voltages at the buses according to
equation(16) yields at bus Ⓚ
(19)
The circuit corresponding to this equation is shown in Figure 3 (b) from which it is evident that the Thevenin impedance
where
at a representative bus of the system is given by (20)
is the diagonal entry in row and column of
With set equal to 2,
this is essentially the same result obtained in equation (12) for the driving-point impedance at bus of Figure 2. In a similar manner, we can determine the Thevenin impedance between any two buses ⓙ and Ⓚ of the network. As shown in Fig. 8.4(a), the otherwise dead network is energized by the current injections
.at bus ⓙ and
the bus voltages resulting from the combination of these two current injections by we obtain
ⓙ
to
in which the right-hand vector is numerically equal to the product of
added to the product of
at bus Ⓚ. Denoting the changes in
and column
of the system
(21)
and column
. Adding these voltage
changes to the original bus voltages according to equation (16), we obtain at buses ⓙ and
Adding and subtracting (23), give
,
Dr Houssem Rafik El Hana Bouchekara
in equation (22), and likewise,
8
(22)
,
(23)
in equation
Since
is symmetrical,
equals
(24) (25)
and the circuit corresponding to these two
equations is shown in Fig. 8.4(b), which represents the Thevenin equivalent circuit of the system between buses ⓙ and . Inspection of Fig. 8.4(b) shows that the open-circuit voltage from bus to bus ⓙ is
Figure 4: Original network with circuit:
, and the
current source
short-circuit connection;
at bus
and
impedance
;
between buses
Thevenin equivalent and
impedance encountered by the short-circuit current Figure 4
at bus
.
from bus to bus ⓙ in
is evidently the Thevenin impedance.
This result is readily confirmed by substituting and (25) and by setting the difference
(26)
in equations (24)
between the resultant equations equal to
zero. As far as external connections to buses ⓙ and are concerned, Figure 4
represents the effect of the original system. From bus ⓙ to the reference node we can trace Dr Houssem Rafik El Hana Bouchekara
9
the Thevenin impedance
and the open-circuit voltage
; from bus
to the reference node we have the Thevenin impedance
the open-circuit voltage
; and between buses and ⓙ the Thevenin impedance of
equation(26) and the open-circuit voltage
impedance is given by
and
is evident. Finally, when the branch
is connected between buses ⓙ and of Figure 4
, the resulting current
(27)
We use this equation in Sec. 8.3 to show how to modify
when a branch
impedance is added between two buses o f the network.
Example 1 A capacitor having a reactance of 5.0 per unit is connected between the reference node and bus ④ of the circuit of Examples 7.5 and 7.6. The original emfs and the corresponding external current injections at buses and ④ are the same as in those examples. Find the current drawn by the capacitor.
Solution The Thevenin equivalent circuit at bus ④ has an emf with respect to reference given
by
per unit, which is the voltage at bus ④ found in Example
7.6 before the capacitor is connected. The Thevenin impedance Example 7.6 to be
per unit, and so Figure 5
at bus is calculated in
follows. Therefore, the
current leap drawn by the capacitor is
Example 2
If an additional current equal to
per unit is injected into the
network at bus ④ of Example 7.6, find the resulting voltages at buses and ④.
Solution
The voltage changes at the buses due to the additional injected current can be
calculated by making use of the bus impedance matrix found in Example 7.6. The required impedances are in column 4 of
. The voltage changes due to the added current injection
at bus ④ in per unit are
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Figure 5: Circuit for Examples 8.1 and 8.2 showing:
Thevenin equivalent circuit; .
phasor diagram at bus
By superposition the resulting voltages are determined from equation (16) by adding these changes to the original bus voltages found in Example 7.6. The new bus volt ages in per unit are
Since the changes in voltages due to the injected current are all at the same angle
shown in Figure 5
and this angle differs little from the angles of the original voltages, an
approximation will often give satisfactory answers. The change in voltage magnitude at a .bus may be approximated by the product of the magnitude of the per-unit current and the magnitude of the appropriate driving-point or transfer impedance. These values added to the original voltage magnitudes approximate the magnitudes of the new voltages very closely. This approximation is valid here because the network is purely reactive, but it also provides a good estimate where reactance is considerably larger than resistance, as is usual
Dr Houssem Rafik El Hana Bouchekara
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in transmission systems. The last two examples illustrate the importance of the bus impedance matrix and incidentally show how adding a capacitor at a bus causes a rise in bus voltages. The assumption that the angles of voltage and current sources remain constant after connecting capacitors at a bus is not entirely valid if we are considering operation of a power system. We shall consider such system operation in Chap. 9 using a computer powerflow program.
1.3 MODIFICATION
OF
AN EXISTING
In Sec. 1.2 we see how to use the Thevenin equivalent circuit and the existing
to solve for new bus voltages in the network following a branch addition without having to develop the new
Since
examine how an existing
is such an important tool in power system analysis we now
established buses. Of course, we could create a new modifying
may be modified to add new buses or to connect new lines to and invert it, but direct methods of
are available and very much simpler than a matrix inversion even for a small
number of buses. Also, when we know how to modify
directly. We recognize several types of modifications in which a branch having i mpedance
is added to a network with known
we can see how to build it
The original bus impedance matrix is identified as
, an N N matrix.
In the notation to be used in our analysis existing buses will be identified by numbers
or the letters
and
network to convert denoted by
. The letter
or
will designate a new bus· to be added to the
to an (N + 1) X (N + 1) matrix. At bus the original voltage will be
the new voltage after modifying
will be
, and
will
denote the voltage change at that bus. Four cases are considered in this section.
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
is injected at the new bus. The
Figure 6: Addition of new bus
Voltage
connected through impedance
at the new bus is equal to
Dr Houssem Rafik El Hana Bouchekara
then,
12
to existing bus
.
(28)
We note that the column vector of currents multiplied by the new
will not alter
the voltages of the original network and will result in the correct voltage at the new bus ⓟ.
CASE 2. Adding
from a new bus ⓟ to an existing bus
The addition of a new bus ⓟ connected through
to an existing bus with
injected at bus ⓟ will cause the current entering the original network at bus to become the sum of 6.
. injected at bus plus the current
coming through
The current
by the voltage
, as shown in Figure
flowing into the network; at bus will increase the original voltage just like in equation(19); that is,
(29)
and
will be larger than then new
by the voltage
. So,
(30)
and substituting for
, we obtain
We now see that the new row which must be added to
Since
(31)
In order to find
is
must be a square matrix around the principal diagonal, we must add a
new column which is the transpose of the new row. The new column accounts for the increase of all bus voltages due to
, as shown in equation(17). The matrix equation is
Dr Houssem Rafik El Hana Bouchekara
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(32)
Note that the first and the first
elements of the new column are the elements of column
CASE 3. Adding
elements of the new row are the elements of row
from existing bus to the reference node
of
of
by connecting an impedance
from an existing bus to
the reference node, we add a new bus ⓟ connected through
to bus . Then, we short-
To see how to alter
circuit bus ⓟ to the reference node by letting equation as equation (32) except that
equal zero to yield the same matrix
is zero. So, for the modification we proceed to
create a new row and new column exactly the same as in Case 2, but we then eliminate the row and
column by Kron reduction, which is possible because of the zero in
the column matrix of voltages. We use the method developed in equation(7.50) to find each element
in the new matrix, where
CASE 4. Adding
between two existing buses ⓙ and
(8.33) To add a branch impedance
in
(33)
between buses ⓙ and already established
, we examine Figure 7, which shows these buses extracted, from the original
network. The current
flowing from bus to bus ⓙ is similar to that of Figure 4. Hence,
from equation (21) the change in voltage at each
Figure 7: Addition of impedance
Dr Houssem Rafik El Hana Bouchekara
between existing buses
14
and
.
–
bus ⓗ caused by the injection
at bus ⓙ and
which means that the vector column from column of
at bus is given by (34)
of bus voltage changes is found by subtracting
and by multiplying the result by
. Based on the definition
of voltage change, we now write some equations for the bus voltages as follows: (35)
And using equation (34) gives
(36)
Similarly, at buses ⓙ and
(37)
(38)
we need one more equation since
is unknown. This is supplied by equation (27),
which can be rearranged in to the form
From equation (37) we note that column of bus currents ; likewise,
equals the product of row
of
of equation (38) equals row, of
1. Upon substituting the expressions for
and
(39)
and the
multiplied by
in equation (39), we obtain
(40)
By examining the coefficients of equations (36) through (38) and eq.(40), we can write the matrix equation
(41)
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in which the coefficient of
in the last row is denoted by
(42)
The new column is column minus column of
with
The new row is the transpose of the new column. Eliminating the
in the
row.
row and
column of the square matrix of equation(41) in the same manner as previously, we see that each element
in the new matrix is
(43)
We need not consider the case of introducing two new buses connected by
because we could always connect one of these new buses through an impedance to an existing bus or to the reference bus before adding the second new bus. Removing a branch.
single branch of impedance
removed from the network by adding the negative of
between two nodes can be
between the same terminating
nodes. The reason is of course, that the parallel combination of the existing branch ( the added branch
amounts to an effective, open circuit.
Table 1 summarizes the procedures of Cases 1 to 4. TABLE.1 Modification of existing
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) and
Example 3 Modify the bus impedance matrix of Example 7.6 to account for the connection of a
capacitor having a reactance of 5.0 per unit between bus ④ and the reference node of the circuit of Fig.7.9. Then, find
using the impedances of the new matrix and the current
sources of Example 7.6. Compare this value of Dr Houssem Rafik El Hana Bouchekara
17
with that found in Example 2.
Solution.
We use equation (32) and recognize that that subscript
, and that
is the
matrix of Example 7.6,
per unit to find
The terms in the fifth row and column were obtained by repeating the fourth row and column of
and noting that
Then, eliminating the fifth row and column, we obtain for (33)
from equation
and other elements in a similar manner to give
The column matrix of currents by which the new bus voltages is the same as in Example 7.6. Since both
is multiplied to obtain the new
and
are zero while
and
are
nonzero, we obtain
as found in Example 2.
It is of interest to note that
may be calculated directly from equation (27) by
setting node ⓙ equal to the reference node. We then obtain for
since
is already calculated in Example 1.
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and
1.4 DIRECT DETERMINATION OF We could determine
by first finding
and then inverting it, but this is not
convenient for large-scale systems as we have seen. Fortunately, formulation of direct building algorithm is a straightforward process on the computer.
using a
At the outset we have a list of the branch impedances showing the buses to which
they are connected. We start by writing the equation for one bus connected through a branch impedance
to the reference as
(44)
and this can be considered as an equation involving three matrices, each of which has one row and one column. Now we might 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
we have the matrix equation
and we proceed to modify the evolving
(45)
matrix by adding other buses and
branches following the procedures described in Sec.8.3. The combination of these procedures constitutes the
building algorithm. Usually, the buses of a network must be
renumbered internally by the computer algorithm to agree with the order in which they are to be added to Example 4
Determine
as it is built up.
for the network shown in Figure 8, where the impedances labeled 1
through 6 are shown in per unit. Preserve all buses. Solution .
The branches are added in the order of their labels and numbered subscripts on will indicate intermediate steps of the solution. We start by establishing bus with its
impedance to the reference nod e and write
We then have the
bus impedance matrix
To establish bus with its impedance to bus we follow equation(32) to write
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The term
above is the sum of
and
. The elements
in the new
row and column are the repetition of the elements of row 1 and column 1 of the matrix being modified.
Figure 8: Network. Branch impedances are in per unit and branch numbers are in parentheses.
Bus with the impedance connecting it to bus is established by writing
of
Since the new bus is being connected to bus , the term of the matrix being modified and the impedance
above is the sum
of the branch being connected to
bus from bus , The other elements of the new row and column are the repetition of row 2 and column 2 of the matrix being modified since the new bus is being connected to bus . If we now decide to add the impedance
from bus to the reference
node, we follow equation (32) to connect a new bus ⓟ through impedance matrix
Dr Houssem Rafik El Hana Bouchekara
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and obtain the
where
above is the sum of
. The other elements in the new row and
column are the repetition of row 3 and column 3 of the matrix being modified since bus is being connected to the reference node through
We now eliminate row ⓟ and column ⓟ by Kron reduction. Some of the elements of the new matrix from equation (33) are
When all the elements are determined, we have
We now decide to add the impedance using equation (32), and we obtain
from bus to establish bus ④
The off-diagonal elements of the new row and column are the repetition of row 3 and column 3 of the matrix being modified because the new bus ④ is being connected to
bus . The new diagonal element is the sum of Finally, we add the impedance
of the previous matrix and
.
between buses and ④. If we let and
in
equation (41) equal 2 and 4, respectively, we obtain the elements for row 5 and column 5.
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and from equation(42)
So, employing
previously found, we write the 5 x 5 matrix
and from equation(43) we find by Kron reduction
which is the bus impedance matrix to be determined. All calculations have been rounded off to five decimal places. Since we shall again refer to these results, we note here that the reactance diagram
of Figure 8 is derived from Fig.7.10 by omitting the sources and one of the mutually coupled branches. Also, the buses of Fig.7.10 have been renumbered in Figure 8 because the
building algorithm must begin with a bus connected to the reference node, as previously remarked. The
building procedures are simple for a computer which first must determine
the types of modification involved as each branch impedance is added. However, the operations must follow a sequence such that we avoid connecting an impedance between two new buses. As a matter of interest, we can check the impedance values of calculations of Sec.8.1.
Example 5 Find
by the network
of the circuit or Example 4 by determining the impedance measured
between bus and the reference node when currents injected at buses , , and ④ are zero.
Solution:
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The equation corresponding to equation(12) is
Were cognize two parallel paths between buses and of the circuit of Figure 8 with the resulting impedance of
This impedance in series with
combines in parallel with
to yield
which is identical with the value found in Example 4. Although the network
reduction method of Example 5 may appear to be simpler by comparison with other methods of forming
such is not the case because a different network reduction is
required to evaluate each element of the matrix. In Example 5 the network reduction to find , for instance, is more difficult than that for finding
, The computer could make a
network reduction by node elimination but would have to repeat the process for each node.
1.5 CALCULATION OF
ELEMENTS FROM
When the full numerical form of
can readily calculate elements of
is not explicitly required for an application, we
as needed if the upper-and lower-triangular factors of
are available. To see how this can be done, consider postmultiplying
with only one nonzero element When
is an
in row
by a vector
and all other elements equal to zero.
matrix, we have
(46)
Thus, postmultiplying have called the vector
by the vector shown extracts the
that is
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th column, which we
Since the product of
and
equals the unit matrix, we have
(47)
If the lower-triangular matrix
and the upper-triangular matrix
available, we can write equation(47) in the form
of
are
(48)
It is now apparent that the elements in the column vector
can be found from
equation(48) by forward elimination and back substitution, as explained in Sec.7.8. If only some of the elements of
are required, the calculations can be reduced accordingly. For
example, suppose that we wish to generate
and
convenient notation for the elements of and
We can solve this equation for
of
for a four-bus system. Using
, we have
(49)
in two steps as follows:
(50)
Where:
By forward substitution equation(50) immediately yields
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(51)
and by back substitution of these intermediate results in equation (51) we find the required elements of column 3 of
If all elements of
,
are required, we can continue the calculations,
The computational effort in generating the required elements can be reduced by judiciously choosing the bus numbers.
In later chapters we shall find it necessary to evaluate terms like ( involving differences between columns ⓜ and ⓝ of
If the elements of
)
are not
available explicitly, we can calculate the required differences by solving a system of equations such as
Where
(52)
is the vector formed by subtract
ting column from, column
of
, and
in row
and
in row
of the vector shown.
In large-scale system calculations considerable computational efficiency can be
realized by solving equations in the triangularized form of equation (52) while the full
need not be developed. Such computational considerations underlie many of the formal developments based on
Example 6
in this text.
The five-bus system shown in Fig.8.9 has per-unit impedances as marked. The symmetrical bus admittance matrix for the system is given by
and it is found that the triangular factors of
Dr Houssem Rafik El Hana Bouchekara
25
are
Use the triangular factors to calculate
, the
Thevenin impedance looking into the system between buses ④ and ⑤ of Fig.8.9.
Solution
Since
is symmetrical, the reader should check that the row elements of
the column elements of
equal
divided by their corresponding diagonal elements. With is
representing the numerical values of , forward solution of the system of equations
Figure 9: Reactance diagram for Example 8.6, all values are per-unit impedances.
yields the intermediate values
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Backsubstituting in the system of equations
where
represent the numerical values of
, we find from the last two rows that
The desired Thevenin impedance is therefore calculated as follows :
Inspection of Figure 9 verifies this result.
Dr Houssem Rafik El Hana Bouchekara
27
Problem 1 Construct the bus impedance matrix for the network given by the following figure.
Figure 10: Impedance diagram.
Solution:
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Problem 2:
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Problem 3
Problem 4
Problem 5 A transmission line exists between buses 1 and 2 with per unit impedance 0.4. Another line of impedance 0.2 p.u. is connected in parallel with it making it a doubl-circuit line with mutual impedance of 0.1 p.u. Obtain by building algorithm method the impedance of the two-circuit system.
Solution Dr Houssem Rafik El Hana Bouchekara
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Dr Houssem Rafik El Hana Bouchekara
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Problem 6 The double circuit line in the problem E 4.1 is further extended by the addition of a transmission line from bus (1). The new line by virtue of i ts proximity to the existing lines has a mutual impedance of 0.05 p.u. and a self – impedance of 0.3 p.u. obtain the bus impedance matrix by using the building algorithm.
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Solution
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Problem 7 The system E4.2 is further extended by adding another transmission line to bus 3 w itil 001£ in pedance of 0 .3 p.u .0 bta.in tile ZBUS
Solution:
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Problem 8 The system in E 4.3 is further extended and the radial system is converted into a ring system joining bus (2) to bus (4) for reliability of supply. Obtain the ZBUS. The self impedance of element 5 is 0.1 p.u
Solution:
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Problem 9: Compute the bus impedance matrix for the system shown in figure by adding element by element. Take bus (2) as reference bus
Solution:
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Problem 10: Using the building algorithm construct zBUS for the system shown below. Choose 4 as reference BUS.
Solution:
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Problem 11: Given the network shown in Fig. E.4.1S.
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Solution:
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Problem 12: E 4.8 Consider the system in Fig. E.4.17. Obtain ZBUS by using building algorithm.
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Solution:
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Problem 13 Form the impedance matrix of the electric network shown in the following figure by using the branch addition method.
Solution According to the node ordering, we can make the sequence table of branch adding as follows.
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