Electrical Power System Analysis 4. The Impedance Model And Network Calculations 6y1r6q

KINGDOM OF SAUDI ARABIA Ministry Of High Education

Umm Al-Qura University College of Engineering & Islamic Architecture

Department Of Electrical Engineering

Power System Analysis The Impedance Model And Network Calculations

Dr : Houssem Rafik El- Hana BOUCHEKARA 2011/2012 1432/1433

Dr Houssem Rafik El Hana Bouchekara

1

1

THE IMPEDANCE MODEL AND NETWORK CALCULATIONS ............................................. 3 1.1 1.2 1.3 1.4 1.5

THE BUS ITTANCE AND IMPEDANCE MATRICES ............................................................... 3 THEVENIN'S THEOREM AND .................................................................................... 6 MODIFICATION OF AN EXISTING ........................................................................... 12 DIRECT DETERMINATION OF ................................................................................. 19 CALCULATION OF ELEMENTS FROM ............................................................. 23


2

1 THE IMPEDANCE MODEL AND NETWORK CALCULATIONS The bus ittance 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 ittances. It is conceptually simple to invert to find the bus impedance matrix but such direct inversion is rarely employed when the systems to be analyzed are large scale. In practice, is rarely explicitly required, and so the triangular factors of 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 but the information content of the bus impedance matrix is far greater than that of We shall see, for example, that each diagonal element of 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 ittance 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 ITTANCE AND IMPEDANCE MATRICES In Example 7.6 we inverted the bus ittance matrix the bus impedance matrix . By definition

and called the resultant

(1)

and for a network of three independent nodes the standard form is

(2)

Since is symmetrical around the principal diagonal, must also be symmetrical. The bus ittance matrix need not be determined in order to obtain and in another section of this chapter we see how may be formulated directly.


3

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 ittances. 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 (4)

If and voltage bus

and

are reduced to zero by shorting buses

is applied at bus

so that current

and

to the reference node,

enters at bus

, the self-ittance at

is (5)

Thus, the self-ittance 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 ittances 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 ittance of

At bus

the

equation obtained by expanding equation (3) is (6)

from which we see that (7)

Thus, the mutual ittance term

is measured by shorting all buses except bus

to the reference node and by applying a voltage

at bus

, as shown in Figure 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 ittance is the

negative of the ittance directly connected between buses expect since mutually coupled branches are absent.


4

and

, as we would

Figure 1: Circuit for measuring

.

and

.

We have made this detailed examination of the bus ittances in order to differentiate them clearly from the impedances of the bus impedance matrix. Conceptually, we solve equation (3)by premultiplying both sides of the equation by to yield (8)

and we must when dealing with that and are column vectors of the bus voltages and the currents entering the buses from current sources, respectively. Expanding equation (8) for a network of three independent nodes yields (9) (10) (11)

From equation (10) we see that the driving-point impedance open-circuiting the current sources at buses at bus

and

is determined by

and by injecting the source current

. Then, (12)

Figure 2: Circuit for measuring


5

and

.

Figure 2 shows the circuit described. Since is defined by opening the current sources connected to the other buses whereas 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 and

to

and

by injecting current at

bus

and by finding the ratios of

with the sources open at all buses except

bus

. We note that a mutual ittance 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 buses

and

open, the only impedance through which

conditions, equations (10) and (11) show that

with current sources at

flows is

. Under the same

is causing voltages at buses

and

expressed by (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.

is

1.2 THEVENIN'S THEOREM AND 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 of the bus currents I by . The voltages to 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 , the new bus voltages are given by the superposition equation (16)

where

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

6

so that the bus currents

and voltages

are zero. Then, into bus Ⓚ a current of

amp

(or per unit for in per unit) is injected in to the system from a current source 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 ⓝ is caused by current entering the network. Thevenin equivalent circuit at node ⓚ·

the network, indicated by the incremental quantities

to

, are given by

(17)

with the only nonzero entry in the current vector equal to in row . Row-bycolumn multiplication in equation (17) yields the incremental bus voltages


7

Ⓚ (18)

Ⓚ

which are numerically equal to the entries in column of multiplied by the current . 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 at a representative bus of the system is given by (20)

where 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 .at bus ⓙ and

energized by the current injections

at bus Ⓚ. Denoting the changes in

the bus voltages resulting from the combination of these two current injections by we obtain

ⓙ

to

(21)

in which the right-hand vector is numerically equal to the product of and column added to the product of and column of the system . Adding these voltage changes to the original bus voltages according to equation (16), we obtain at buses ⓙ and

(22) (23)

Adding and subtracting (23), give

,


in equation (22), and likewise,

8

,

in equation

(24) (25)

Since is symmetrical, equals 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 voltage from bus

to bus ⓙ is

Figure 4: Original network with circuit:

. Inspection of Fig. 8.4(b) shows that the open-circuit , and the

at bus ⓙ and

current source

short-circuit connection;

impedance

at bus

Thevenin equivalent

between buses ⓙ and

impedance encountered by the short-circuit current Figure 4

;

from bus

.

to bus ⓙ in

is evidently the Thevenin impedance. (26)

This result is readily confirmed by substituting in equations (24) and (25) and by setting the difference 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

equation(26) and the open-circuit voltage impedance

and

and ⓙ the Thevenin impedance of is evident. Finally, when the branch

is connected between buses ⓙ and

of Figure 4

, the resulting current

is given by (27)

We use this equation in Sec. 8.3 to show how to modify impedance is added between two buses o f the network.

when a branch

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 per unit, which is the voltage at bus ④ found in Example

by

7.6 before the capacitor is connected. The Thevenin impedance Example 7.6 to be per unit, and so Figure 5 current leap drawn by the capacitor is

at bus is calculated in follows. Therefore, the

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


10

Figure 5: Circuit for Examples 8.1 and 8.2 showing:

Thevenin equivalent circuit;

phasor diagram at bus

④.

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

11

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 is such an important tool in power system analysis we now examine how an existing may be modified to add new buses or to connect new lines to established buses. Of course, we could create a new and invert it, but direct methods of modifying are available and very much simpler than a matrix inversion even for a small number of buses. Also, when we know how to modify we can see how to build it directly. We recognize several types of modifications in which a branch having impedance is added to a network with known 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 . The letter or will designate a new bus· to be added to the network to convert

to an (N + 1) X (N + 1) matrix. At bus

the original voltage will be

denoted by the new voltage after modifying will be , and denote the voltage change at that bus. Four cases are considered in this section. CASE 1. Adding

will

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 ⓟ connected through impedance

Voltage

at the new bus is equal to


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

injected at bus ⓟ will cause the current entering the original network at bus the sum of

. injected at bus

plus the current

coming through

with to become

, as shown in Figure

6. The current by the voltage

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 (31)

We now see that the new row which must be added to

In order to find

is

Since 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 s for the increase of all bus voltages due to , as shown in equation(17). The matrix equation is


13

(32)

Note that the first elements of the new row are the elements of row and the first elements of the new column are the elements of column of CASE 3. Adding

from existing bus

To see how to alter

of

to the reference node

by connecting an impedance

the reference node, we add a new bus ⓟ connected through

from an existing bus to bus

to

. Then, we short-

circuit bus ⓟ to the reference node by letting equal zero to yield the same matrix equation as equation (32) except that 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 (33)

CASE 4. Adding

between two existing buses ⓙ and

(8.33) To add a branch impedance in

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


between existing buses ⓙ and ⓚ.

14

bus ⓗ caused by the injection

at bus ⓙ and –

at bus

is given by (34)

which means that the vector of bus voltage changes is found by subtracting column from column of 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 which can be rearranged in to the form

is unknown. This is supplied by equation (27),

(39)

From equation (37) we note that column of bus currents ; likewise,

equals the product of row

of equation (38) equals row,

1. Upon substituting the expressions for

and

of

of

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)


15

in which the coefficient of

in the last row is denoted by (42)

The new column is column minus column of with in the row. The new row is the transpose of the new column. Eliminating the 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 between two nodes can be removed from the network by adding the negative of between the same terminating nodes. The reason is of course, that the parallel combination of the existing branch ( ) and the added branch amounts to an effective, open circuit. Table 1 summarizes the procedures of Cases 1 to 4. TABLE.1 Modification of existing


16

Example 3 Modify the bus impedance matrix of Example 7.6 to 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 with that found in Example 2. Dr Houssem Rafik El Hana Bouchekara

17

Solution. We use equation (32) and recognize that is the that subscript , and that per unit to find

matrix of Example 7.6,

The 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

from equation

(33)

and other elements in a similar manner to give

The column matrix of currents by which the new is multiplied to obtain the new bus voltages is the same as in Example 7.6. Since both 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.


18

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 using a direct building algorithm is a straightforward process on the computer. 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

(45)

and we proceed to modify the evolving 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 as it is built up. Example 4

Determine 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

To establish bus

bus impedance matrix

with its impedance to bus


19

we follow equation(32) to write

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

Since the new bus of

is being connected to bus

of the matrix being modified and the impedance

bus

from bus

is established by writing

, the term

above is the sum

of the branch being connected to

, 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

and obtain the

impedance matrix


20

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

from bus

to establish bus ④

using equation (32), and we obtain

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

equation (41) equal 2 and 4, respectively, we obtain the elements for row 5 and column 5. Dr Houssem Rafik El Hana Bouchekara

21

in

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.

by the network

Example 5 Find between bus

of the circuit or Example 4 by determining the impedance measured and the reference node when currents injected at buses

zero. Solution:


22

,

, and ④ are

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 is not explicitly required for an application, we can readily calculate elements of as needed if the upper-and lower-triangular factors of are available. To see how this can be done, consider postmultiplying by a vector with only one nonzero element in row and all other elements equal to zero. When is an matrix, we have

(46)

Thus, postmultiplying by the vector shown extracts the have called the vector that is


23

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 of for a four-bus system. Using convenient notation for the elements of and , we have

(49)

We can solve this equation for

in two steps as follows:

(50)

Where:

(51)

By forward substitution equation(50) immediately yields


24

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 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

(52)

Where ting column from, column of the vector shown.

is the vector formed by subtract of

, and

in row

and

in row

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 in this text. Example 6 The five-bus system shown in Fig.8.9 has per-unit impedances as marked. The symmetrical bus ittance matrix for the system is given by

and it is found that the triangular factors of


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 equal the column elements of 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


26

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.


27

Problem 1 Construct the bus impedance matrix for the network given by the following figure.

Figure 10: Impedance diagram.

Solution:


28

Problem 2:


29

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

30


31

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 its 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.


32

Solution


33

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:


34

Problem 8 The system in E 4.3 is further extended and the radial system is converted into a ring system ing bus (2) to bus (4) for reliability of supply. Obtain the ZBUS. The self impedance of element 5 is 0.1 p.u Solution:


35


36

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:


37


38


39

Problem 10: Using the building algorithm construct zBUS for the system shown below. Choose 4 as reference BUS. Solution:


40


41


42


43


44


45

Problem 11: Given the network shown in Fig. E.4.1S.


46

Solution:


47

Problem 12: E 4.8 Consider the system in Fig. E.4.17. Obtain ZBUS by using building algorithm.


48

Solution:


49

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.


50


51


52


53


54


55

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