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Transcript of Network Calculations
ADVANCED POWER SYSTEM
2-NETWORK CALCULATIONS
1
Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
ADVANCED POWER SYSTEM
2-NETWORK CALCULATIONS
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
NETWORK CALCULATIONS
The steady state analysis of an interconnected power system during normal conditions is considered when calculating networks. The power system is assumed to be operating under balanced condition and can be represented by a single-line diagram. The power system network contains hundreds of buses and branches with impedances specified in per-unit on a common MVA base. Power flow studies, commonly referred to as load flow, are essentials of power system design and analysis. Load flow studies are necessary for planning, economic operation, scheduling and exchange of power between utilities. Load flow study is also required for many other analysis such as transient stability, dynamic stability, contingency and state estimation.
Network equations can be formulated in a variety of forms. However, node
voltage method is commonly used for power system analysis. The network equations which are in the nodal admittance form results in complex linear simultaneous algebraic equations in terms of node currents. The load flow results give the bus voltage magnitude and phase angles and hence the power flow through the transmission lines, line losses and power injection at all buses.
NODE EQUATIONS The junctions formed when two or more pure elements (R, L and C or an ideal
source of voltage or current) are connected to each other at their terminal are called nodes. Systematic formulation of equations determined at nodes of a circuit by applying Kirchhoffโs current law is the basis of some excellent computer solutions of power system problems. Usually it is convenient to consider only those nodes to which more than two elements are connected and to call these junction pointsโ major nodes.
In order to examine some features of node equations, always begin with the
one-line diagram of a simple power system. In order to further understand node equations, consider the power system diagram given below.
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a
b
c
4
1
3
2
In the figure above, generators are connected through transformers to
high tension buses 1 and 3 and supply a synchronous motor load at bus 2. For the purpose of analysis, all machines at any one bus are treated as single machine and represented by a single emf and series reactance.
Next is to draw the reactance diagram of the one-line diagram, thus
4
1
3
2
Ea
0
Eb
Ec
XaXT1
Xc XT3
XbXT2
X13
X34
X23
X14
X24
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4
1
3
2
Ea
0
Eb
Ec
Za
Zc
Zb
X13
X34
X23
X14
X24
Za=Xa+XT1
Zc=Xc+XT3
Zb=Xb+XT2
Where:
Nodes are indicated by dots, but numbers are assigned only to major nodes.If
the circuit is redrawn with the emfs and the impedances in series connecting them to the major nodes replaced by the equivalent current sources and shunt admittances, the result is the circuit below.
4
1
3
2
I1
0
I2
I3
Ya
Yb
Yc
Yf
Yd
Yg
Ye
Yh
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Single-subscript notation will be used to designate the voltage of each bus with respect to the neutral taken as the reference node 0. Applying Kirchhoffโs current Law, we have
At node 1
๐ผ1 = ๐1๐๐ + ๐1 โ ๐3 ๐๐ + ๐1 โ ๐4 ๐๐
At node 2 ๐ผ2 = ๐2๐๐ + ๐2 โ ๐3 ๐๐ + ๐2 โ ๐4 ๐๐
At node 3 ๐ผ3 = ๐3๐๐ + ๐3 โ ๐1 ๐๐ + ๐3 โ ๐2 ๐๐ + ๐3 โ ๐4 ๐๐
At node 4 0 = ๐4 โ ๐1 ๐๐ + ๐4 โ ๐2 ๐๐ + ๐4 โ ๐3 ๐๐
Rearranging these equations will yield to; At node 1
๐ผ1 = ๐1๐๐ + ๐1 โ ๐3 ๐๐ + ๐1 โ ๐4 ๐๐
๐ฐ๐ = ๐ฝ๐ ๐๐ + ๐๐ + ๐๐ โ ๐ฝ๐๐๐ โ ๐ฝ๐๐๐
At node 2
๐ผ2 = ๐2๐๐ + ๐2 โ ๐3 ๐๐ + ๐2 โ ๐4 ๐๐
๐ฐ๐ = ๐ฝ๐ ๐๐ + ๐๐ + ๐๐ โ ๐ฝ๐๐๐ + โ๐ฝ๐๐๐
At node 3
๐ผ3 = ๐3๐๐ + ๐3 โ ๐1 ๐๐ + ๐3 โ ๐2 ๐๐ + ๐3 โ ๐4 ๐๐
๐ฐ๐ = โ๐ฝ๐๐๐โ๐ฝ๐๐๐ + ๐ฝ๐ ๐๐ + ๐๐ + ๐๐ + ๐๐ โ ๐ฝ๐๐๐
At node 4
0 = ๐4 โ ๐1 ๐๐ + ๐4 โ ๐2 ๐๐ + ๐4 โ ๐3 ๐๐
๐ = โ๐ฝ๐๐๐ โ ๐ฝ๐๐๐ โ ๐ฝ๐๐๐ + ๐ฝ๐ ๐๐ + ๐๐ + ๐๐ In the above equations, it is apparent that the current flowing into the network
from the current sources connected to anode is equated to the sum of several products.
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At any node, one product is the voltage of the node times the sum of all the admittances which terminate or connected on that node. This product accounts for the current that flows away from the node if the voltage is zero at each other node. Each other product equals negative of the voltage at another node times the admittance connected directly between the othe node and the node at which the equation is formulated.
The standard form for the four independent equations in matrix form is,
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
=
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
The symmetry of the equations in this form makes them easy to remember, and
their extension to any number of nodes is apparent. The order of the Y subscripts is effect-cause; that is, the first subscript is that of the node at which the current is being expressed and the second subscript is that of the voltage causing this component of current.
The Y matrix is designated Ybus and called the BUS ADMITTANCE MATRIX.
Y11, Y22, Y33 and Y44 - is called the self-admittances or driving point admittancesat the nodes
- is equal to the sum of all admittances terminating on the node identified by the repeated subscripts.
Other admittances - is called the mutual admittances or transfer admittances of
the nodes, and each equals the negative of the sum of all admittances connected directly between the nodes identified by the double subscripts.
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BUS ADMITTANCE MATRIX To clearly illustrate the bus admittance matrix, let us consider the below
Starting with the three node equation expressed in the form of matrix as
๐ฐ๐๐ฐ๐๐ฐ๐
=
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
And expanding this to equations, we have
๐ฐ๐ = ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐ ๐ฐ๐ = ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐ ๐ฐ๐ = ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐ + ๐๐๐๐ฝ๐
If V1 and V2 are reduced to zero by shorting nodes 1 and 3 to the reference node
and current I2 is injected at node 2, the self- admittance at node 2 is
๐๐๐ = ๐ฐ๐๐ฝ๐
๐ฝ๐=๐ฝ๐=๐
Remember that self-admittance of a particular node could be measured by shorting all other nodes to the reference and then finding the ratio of the current injected at the node to the voltage resulting at that node. The result is obviously equivalent to adding all the admittances directly connected to the node, as has been our procedure up to now.
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To find the mutual admittance, at node 1 and node 3 the equation obtained are
๐๐๐ = ๐ฐ๐๐ฝ๐
๐ฝ๐=๐ฝ๐=๐
And
๐๐๐ = ๐ฐ๐๐ฝ๐
๐ฝ๐=๐ฝ๐=๐
Thus, the mutual admittances are measuredby shorting all nodes except node 2
to the reference node and injecting a current I2 at node 2. Then Y12 is the ratio of the negative of the current leaving the network in the short circuit at node 1 to the voltage V2. The negative of the current leaving node 1 is used since I1 is defined as the current entering the network. The resultant admittance is the negative of the admittance directly connected between node 1 and 2, as we would expect. BUS IMPEDANCE MATRIX The Inverse of the bus admittance matrix is known as the bus impedance matrix Zbus. It is conceptually simple to invert Ybus to find the bus impedance matrix Zbus, but such direct inversion is rarely employed when the systems to be analyzed are large scale. In practice, Zbus is rarely explicitly required and so the triangular factors of Ybus are used to generate elements of Zbus only as they are needed since this is often the most computationally efficient method. By setting computational considerations aside, however, and regarding Zbus as being already constructed and explicitly available, the power system analyst can derive a great deal of insight.
The bus impedance matrix can be directly constructed element by element using simple algorithms to incorporate one element at a time into the system representation. The work entailed in constructing Zbus is much greater than that required constructing Ybus, but the information content of bus impedance is far greater than that of Ybus. We shall see, for example, that each diagonal element of Zbus reflects important characteristics of the entire system in the form of the Thevenin impedance at the corresponding bus. Unlike Ybus, 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.
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If we invert Ybus to Zbus, by definition, we will have
๐๐๐ข๐ = ๐๐๐ข๐ โ1
And for a network of three independent nodes the standard form is
Since Ybus is symmetrical around the principal diagonal, Zbus must also be symmetrical. The bus admittance matrix need not be determined in order to obtain Zbus, and in this topic we see how Zbus may be formulated directly. The impedance elements of Zbus 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. 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
๐ฝ = ๐๐๐๐๐ฐ We must remember when dealing with Zbus that V and I are column matrices of the node voltages and the currents entering the nodes from current sources, respectively. Expanding the matrix for the three networks of three independent nodes yields to
๐ฝ๐
๐ฝ๐
๐ฝ๐
=
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐ฐ๐๐ฐ๐๐ฐ๐
Converting the matrix form to equation form, we obtain
๐ฝ๐ = ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ ๐ฝ๐ = ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ ๐ฝ๐ = ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐
ADVANCED POWER SYSTEM
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To solve for the driving-point impedances Z11, Z22, and Z33, open circuit the current sources in the remaining nodes indicated by the first subscript of impedance. The figure below shows the circuit described. Thus, for example, to determine Z22, open circuit the current sources at nodes 1 and 3 and inject the current source I2 at node 2. So,
๐๐๐ = ๐ฝ๐
๐ฐ๐ ๐ฐ๐=๐ฐ๐=๐
To enable us to measure some transfer impedances, current sources I1 and I3 should be open circuited, so we can have
๐๐๐ = ๐ฝ๐
๐ฐ๐ ๐ฐ๐=๐ฐ๐=๐
And
๐๐๐ = ๐ฝ๐
๐ฐ๐ ๐ฐ๐=๐ฐ๐=๐
Thus we can measure the transfer impedances Z12 and Z32 by injecting current at node 2 and finding the ratios of V1 and V2 to I2 with the sources open at all nodes except node 2.
ADVANCED POWER SYSTEM
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THEVENINโS THEOREM AND THE ZBUS The bus impedance matrix provides important information reagarding the power system network which we can use to advantage in network calculations. In this topic, we examine the relationship between the elements of Zbus and the Thevenin impedance presented by the network at each bus. To establish notation, let us denote the bus voltages corresponding to the initial values I0 of the bus currents of I bu V0=ZbusI
0. The voltages V1
0 to VN0 are the effective open-circuit voltages, which can be measured by
voltmeter between the buses of network and the reference node. When the bus currents are changed from the initial to the new values I0=ฮI, the new bus voltages are given by the superposition equation,
where ฮV represents the changes in the bus voltages from their original values
ADVANCED POWER SYSTEM
2-NETWORK CALCULATIONS
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Figure (a) shows network with bus k and reference node extracted voltages ฮVn, at bus n is caused by current ฮIk entering the network. (b) Thevenin equivalent circuit at node k. A large-scale system in schematic form with a representative bus k extracted along with the reference node of the system. Initially, we consider the circuit not to be energized so that the bus currents Iยฐ and voltages V0 are zero. Then, into bus k a current of ฮk amp (or ฮk per unit for Zbus 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 the network, indicated by the incremental quantities ฮVk to ฮVn, are give n by
with the only nonzero entry in the current vector equal to ฮIk in row k. Row-by-column multiplication in Eq. above yields the incremental bus voltages
which are numerically equal to the entries of k of Zbus multiplied by the current ฮIk. Adding these voltage changes to the original voltages at the buses according to equation above yields to bus k.
๐ฝ๐ = ๐ฝ๐๐+๐๐๐โ๐ฐ๐
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The circuit corresponding to this equation is shown in figure (b) from which it is evident that the Thevenin impedance Zth at a representative bus k of the system is given by
๐๐๐ = ๐๐๐ where Zkk is the diagonal entry in row k and column k of Zbus. With k set equal to 2, ths is essentially the same result obtained in Z22 equation for the driving point impedance at bus 2. In similar manner, we can determine the Thevenin impedance between any two buses j and k of the network. As shown in the figure below (a), the otherwise dead network is energized by the current injections ฮIj at bus j and ฮIk at bus k.
Denoting the changes in the bus voltages resulting from the combination of these two current injections by ฮV1 to ฮVn, we obtain
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in which the right-hand vector is numerically equal to the product of ฮIj at bus j and ฮIk at bus k of Zbus. Adding these voltages changes to original bus voltages, we obtain at buses j and k
๐ฝ๐ = ๐ฝ๐๐ + ๐๐๐โ๐ฐ๐ + ๐๐๐โ๐ฐ๐
๐ฝ๐ = ๐ฝ๐๐+๐๐๐โ๐ฐ๐ + ๐๐๐โ๐ฐ๐
Adding and subtracting ๐๐๐โ๐ฐ๐ and likewise ๐๐๐โ๐ฐ๐ in the equation above will give,
๐ฝ๐ = ๐ฝ๐๐ + ๐๐๐ โ ๐๐๐ โ๐ฐ๐ + ๐๐๐ โ๐ฐ๐ + โ๐ฐ๐
๐ฝ๐ = ๐ฝ๐๐ + ๐๐๐ โ๐ฐ๐ + โ๐ฐ๐ + ๐๐๐ โ ๐๐๐ โ๐ฐ๐
Since Zbus is symmetrical Zjk equals to Zkj and the circuit corresponding to these two equations shown in figure (b) below which represents the Thevenin equivalent circuit of the system between buses j and k.
Inspection of the figure (b) hows that the open-circuit voltage from bus k to bus j is
๐ฝ๐๐ โ ๐ฝ๐
๐, and the impedance encountered by the short-circuit Isc from bus k to bus j is
evidently the Theveninโs impedance
๐๐๐,๐๐ = ๐๐๐ + ๐๐๐ โ ๐๐๐๐
This result is readily confirmed by substituting Isc = ฮIj = -ฮIk in above equations and by
setting the difference ๐ฝ๐๐ โ ๐ฝ๐
๐ between the resultant equations equal to zero. As for
external connections to the buses j and k are concerned figure (b) represents the effect of the original system. From bus j to the reference node, we can trace the Thevenin
impedance ๐๐๐ = ๐๐๐ โ ๐๐๐ + ๐๐๐ and the open-circuit voltage ๐ฝ๐๐.
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From bus k to the reference node we have the Thevenin impedance ๐๐๐ =
๐๐๐ โ ๐๐๐ + ๐๐๐, and the open- circuit ๐ฝ๐๐; and between buses j and k , the Thevenin
equation of ๐๐๐,๐๐ = ๐๐๐ + ๐๐๐ โ ๐๐๐๐ and open voltage of ๐ฝ๐๐ โ ๐ฝ๐
๐ is evident. Finally
when the branch impedance Zb is connected between buses j and k, the resulting current Ib is given by
๐ฐ๐ =๐ฝ๐
๐ โ ๐ฝ๐๐
๐๐๐,๐๐ + ๐๐=
๐ฝ๐ โ ๐ฝ๐
๐๐
Sample Problem:
1. Write in matrix form the node equations necessary to solve for the voltages of the numbered buses and Find the bus voltages by inverting the bus admittance matrix of the figure below. The emfs are Ea=1.5 cis 0, Eb = 1.5 cis -36.87, and Ec = 1.5 cis 0, all in per unit.
4
1
3
2
j1.15 j0.1Ea
0
j1.15 j0.1
Eb j1.15 j0.1
Ec
j0.25
j0.4
j0.2
j0.125
j0.2
Reqโd: a.) node equations in matrix b.) Find the bus voltages, V1, V2, V3 and V4
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4
1
3
2
-j0.8
!1
0
-j0.8
-j 4.0
-j 2.5
-j 5.0
-j0.8
!2
!3
-j 8.0
-j 5.0
Ya
Yb
Yc
Yf
Yd
Yg
Ye
Yh
Solโn:
a.) node equations in matrix
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
=
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
1. Draw the equivalent reactance diagram converted to current sources and admittances,
2. Solve for the current sources,
๐๐ = ๐๐ =1.5โ 0
j 1.25= ๐. ๐โ โ ๐๐ = ๐ โ ๐ฃ๐.๐๐ ๐ฉ๐ฎ
๐๐ =1.5โ โ 36.87
j 1.25= ๐. ๐โ โ ๐๐๐. ๐๐ = โ๐. ๐๐ โ ๐ฃ๐. ๐๐ ๐ฉ๐ฎ
3. Solve for the self admittances, Y11, Y22, Y33 and Y44. Thus,
๐๐๐ = ๐๐ + ๐๐ + ๐๐
= โ๐5 โ ๐4 โ ๐0.8 ๐๐๐ = โ๐๐. ๐
๐๐๐ = ๐๐ + ๐๐ + ๐๐
= โ๐0.8 โ ๐2.5 โ ๐5 ๐๐๐ = โ๐๐. ๐
๐๐๐ = ๐๐ + ๐๐ + ๐๐ + ๐๐
= โ๐0.8 โ ๐8 โ ๐4 โ ๐2.5 ๐๐๐ = โ๐๐๐.๐
๐๐๐ = ๐๐ + ๐๐ + ๐๐
= โ๐5 โ ๐5 โ ๐0.8 ๐๐๐ = โ๐๐๐
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4. Solve for the mutual admittances,
๐๐๐ = ๐๐๐ = ๐ ๐๐๐ = ๐๐๐ = +๐๐ ๐๐๐ = ๐๐๐ = +๐๐
๐๐๐ = ๐๐๐ = +๐๐. ๐ ๐๐๐ = ๐๐๐ = +๐๐ ๐๐๐ = ๐๐๐ = +๐๐
Hence, the node equations in matrix form is,
๐ โ ๐๐.๐โ๐. ๐๐ โ ๐๐. ๐๐
๐ โ ๐๐.๐๐
=
โ๐๐.๐ ๐๐ โ๐๐.๐
๐๐ ๐๐๐๐.๐ ๐๐
๐๐ ๐๐. ๐๐๐ ๐๐
โ๐๐๐.๐ ๐๐๐๐ โ๐๐๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
b.) Find the bus voltages, V1, V2, V3 and V4
1. Get the inverse of the bus admittance matrix, that is From,
๐ผ = ๐ ๐ We will divide I matrix by Y matrix which will be,
๐ผ
๐ = ๐
or ๐ผ ๐ โ1 = ๐
To get [Y]-1, remember the inverse matrix formula,
๐โ๐ =๐
๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐ป
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To start with, a. Form the cofactor matrix of matrix Y,
=
+
โ๐๐. ๐ ๐๐. ๐ ๐๐
๐๐. ๐ โ๐๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ โ
๐ ๐๐. ๐ ๐๐
๐๐ โ๐๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐
โ
๐ ๐๐ ๐๐
๐๐. ๐ โ๐๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ +
โ๐๐. ๐ ๐๐ ๐๐
๐๐ โ๐๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐
+
๐ โ๐๐. ๐ ๐๐
๐๐ ๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ โ
๐ โ๐๐. ๐ ๐๐. ๐
๐๐ ๐๐. ๐ โ๐๐๐. ๐
๐๐ ๐๐ ๐๐
โ
โ๐๐. ๐ ๐ ๐๐
๐๐ ๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ +
โ๐๐. ๐ ๐ ๐๐
๐๐ ๐๐. ๐ โ๐๐๐. ๐
๐๐ ๐๐ ๐๐
+ ๐ ๐๐ ๐๐
โ๐๐. ๐ ๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ _
โ๐๐. ๐ ๐๐ ๐๐
๐ ๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐
โ ๐ ๐๐ ๐๐
โ๐๐. ๐ ๐๐. ๐ ๐๐
๐๐. ๐ โ๐๐๐. ๐ ๐๐ +
โ๐๐. ๐ ๐๐ ๐๐
๐ ๐๐. ๐ ๐๐
๐๐ โ๐๐๐. ๐ ๐๐
+ โ๐๐. ๐ ๐ ๐๐
๐ โ๐๐. ๐ ๐๐
๐๐ ๐๐ โ๐๐๐ โ
โ๐๐. ๐ ๐ ๐๐
๐ โ๐๐. ๐ ๐๐. ๐
๐๐ ๐๐ ๐๐
โ โ๐๐. ๐ ๐ ๐๐
๐ ๐๐. ๐ ๐๐
๐๐ ๐๐. ๐ ๐๐ +
โ๐๐. ๐ ๐ ๐๐
๐ โ๐๐. ๐ ๐๐. ๐
๐๐ ๐๐. ๐ โ๐๐๐. ๐
=
๐1059.62 ๐822.5๐822.5 ๐1081.22
๐892.1 ๐919.3๐870.5 ๐915.7
๐892.1 ๐870.5๐919.3 ๐915.7
๐1011.62 ๐939.22๐939.22 ๐1050.452
b. Form the transpose matrix of the cofactor matrix of matrix Y that is
๐ ๐ =
๐1059.62 ๐822.5๐822.5 ๐1081.22
๐892.1 ๐919.3๐870.5 ๐915.7
๐892.1 ๐870.5๐919.3 ๐915.7
๐1011.62 ๐939.22๐939.22 ๐1050.452
c. Evaluate the determinant of matrix Y
๐ ๐๐ ๐ =
โ๐๐. ๐ ๐
๐ โ๐๐. ๐
๐๐ ๐๐
๐๐. ๐ ๐๐
๐๐ ๐๐. ๐
๐๐ ๐๐
โ๐๐๐. ๐ ๐๐
๐๐ โ๐๐๐
๐ ๐๐ ๐ = ๐๐๐๐.๐๐๐ * It can be solved by different methods discussed in advanced math like modification, chioโs and pivotal method.
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d. Divide each element of the transpose cofactor of Y by the value of
determinant of Y.
๐โ๐ =๐
๐ ๐๐ ๐ ๐๐๐๐๐๐๐๐ ๐๐ ๐ ๐ป
=1
2219.376
๐1059.62 ๐822.5
๐822.5 ๐1081.22
๐892.1 ๐919.3
๐870.5 ๐915.7
๐892.1 ๐870.5
๐919.3 ๐915.7
๐1011.62 ๐939.22
๐939.22 ๐1050.452
๐โ๐ =
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
2. After finding the inverse of the admittance matrix or what we may call now
bus impedance matrix Zbus, multiply it by current matrix, thus
๐0.4774 ๐0.3706๐0.3706 ๐0.4872
๐0.4020 ๐0.4142๐0.3922 ๐0.4126
๐0.4020 ๐0.3922๐0.4142 ๐0.4126
๐0.4558 ๐0.4232๐0.4232 ๐0.4733
0 โ ๐1.2โ0.72 โ ๐0.96
0 โ ๐1.20
=
๐1
๐2
๐3
๐4
3. Performing the indicated matrix multiplication will yield to,
๐.๐๐๐ โ ๐๐. ๐๐๐๐๐. ๐๐๐๐ โ ๐๐.๐๐๐๐๐. ๐๐๐๐ โ ๐๐.๐๐๐๐๐. ๐๐๐๐ โ ๐๐.๐๐๐๐
=
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
And so, the node voltages are
๐ฝ๐ = ๐. ๐๐๐ โ ๐๐. ๐๐๐๐ = ๐. ๐๐๐โ โ ๐๐.๐๐ ๐๐ ๐ฝ๐ = ๐. ๐๐๐๐ โ ๐๐.๐๐๐๐ = ๐. ๐๐๐โ โ ๐๐. ๐๐ ๐๐ ๐ฝ๐ = ๐. ๐๐๐๐ โ ๐๐.๐๐๐๐ = ๐. ๐๐๐โ โ ๐๐. ๐๐ ๐๐ ๐ฝ๐ = ๐. ๐๐๐๐ โ ๐๐.๐๐๐๐ = ๐. ๐๐๐โ โ ๐๐. ๐๐ ๐๐
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2. If a capacitor having a reactance of 5.0 per unit is connected to node 4 of the
circuit in problem 1. Find the current drawn by the capacitor and the resulting voltages at nodes 1, 2, 3 and 4 if Ic is of negative value providing Ea, Eb and Ec remains the same. Reqโd:
1.) Ic 2.) Find the bus voltages, V1, V2, V3 and V4 if IC is negative of the value
obtained in the solution Illustration:
4
1
3
2
j1.15 j0.1Ea
0
j1.15 j0.1
Eb j1.15 j0.1
Ec
j0.25
j0.4
j0.2
j0.125
j0.2
Solโn:
1. To solve for IC, a.) Create the Theveninโs equivalent circuit for node 4. Making sources short
circuited except node 4. Thus,
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From,
4
1
3
2
Ea
0
Eb
Ec
j0.25
j0.4
j0.2
j0.125
j0.2
It becomes
V4 = Vth
Z44 = Zth
XC = ZC
b.) To solve for Theveninโs equivalent of the circuit behind node 4, the theveninโs
voltage at node 4 before the capacitor is connected has an emf of
๐4 = ๐๐๐ = 1.4009 โ ๐0.2971 = 1.432โ โ 11.97 ๐๐ข c.) To find the Theveninโs impedance, the emfs are short-circuited or current
sources are open circuited, and the impedance between node 4 and the reference node must be determined. Hence,
๐4 = ๐41๐ผ1 + ๐42๐ผ2 + ๐43๐ผ3 + ๐44๐ผ4
With emfs short-circuited (or with emfs and their series impedances replaced by
the equivalent current sources and shunt admittances with the current source open) no current is entering the circuit from the sources at nodes 1, 2, and 3. The ratio of a voltage applied at node 4 to current in the network is Z44, and this impedance is known since Zbus was calculated. Then,
๐๐๐ = ๐44 = ๐0.4733
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d.) Thus the current drawn by the capacitor is,
๐๐๐ = ๐๐๐ + ๐๐ถ ๐ผ๐ถ
๐ผ๐ถ = ๐๐๐
๐๐๐ + ๐๐ถ
= 1.432โ โ 11.97 ๐๐ข
๐0.4733 โ ๐5
๐ฐ๐ช = ๐ฐ๐ = ๐. ๐๐๐โ ๐๐.๐๐ ๐๐
2.) Find the bus voltages, V1, V2, V3 and V4
With original emfs short-circuited, the voltages at the nodes due only to the injected current will be calculated by making use of the bus impedance network. The required impedances are in column 4 of Zbus. The voltages with all emfs shorted are
๐1 = ๐ผ4๐14 = โ0.316โ 78.03 ๐0.4142 = 0.1309โ โ 11.97ยฐ ๐2 = ๐ผ4๐24 = โ0.316โ 78.03 ๐0.4126 = 0.1304โ โ 11.97ยฐ ๐3 = ๐ผ4๐34 = โ0.316โ 78.03 ๐0.4232 = 0.1337โ โ 11.97ยฐ ๐4 = ๐ผ4๐44 = โ0.316โ 78.03 ๐0.4733 = 0.1496โ โ 11.97ยฐ
By superposition, the resulting voltages determined by adding the
voltages caused by the injected current with emfs shorted to the node voltages are;
๐ฝ๐๐๐๐ = ๐1 + ๐1๐๐๐ = 0.1309โ โ 11.97ยฐ + 1.436โ โ 10.71 = ๐. ๐๐๐โ โ ๐๐. ๐๐ ๐๐ ๐ฝ๐๐๐๐ = ๐2 + ๐2๐๐๐ = 0.1304โ โ 11.97ยฐ + 1.427โ โ 14.24 = ๐.๐๐๐โ โ ๐๐. ๐๐๐๐ ๐ฝ๐๐๐๐ = ๐3 + ๐3๐๐๐ = 0.1337โ โ 11.97ยฐ + 1.434โ โ 11.36 = ๐. ๐๐๐โ โ ๐๐. ๐๐ ๐๐ ๐ฝ๐๐๐๐ = ๐4 + ๐4๐๐๐ = 0.1496โ โ 11.97ยฐ + 1.432โ โ 11.97 = ๐. ๐๐๐โ โ ๐๐. ๐๐๐๐
Since the changes in voltages due to the injected current are all the same angle and this angle differs little from the angles of the originalvoltages, an approximation will give satisfactory answers. The change in voltage magnitude at a bus is about equal to the product of the magnitude of the per unit current and the magnitude of the appropriate impedances.These values are added to the original voltage magnitude will give the magnitudes of the new voltages very closely.This approximation is valid because the network is purely reactive, but it provides a good estimate where the reactance is considerably larger than the resistance, as usually the case.
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MATRIX PARTTITIONING: A use method of matrix manipulation is called parttitioning. Consider the 3x3 matrix multiplied by 3x1, ๐ช = ๐จ๐ฉ . . .eq.1
๐ถ =
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
๐11
๐21
๐31
The matrix A will be partitioned into four submatrices by the horizontal and vertical dashed lines. The matrix may be written,
๐จ =
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
= ๐ซ ๐ฌ๐ญ ๐ฎ
where:
๐ซ = ๐11 ๐12
๐21 ๐22 ๐ฌ =
๐13
๐23
๐ญ = ๐31 ๐32 ๐ฎ = ๐33
Also the matrix B will be parttitioned,
๐ฉ =
๐11
๐21
๐31
= ๐ฏ๐ฑ
where:
๐ฏ = ๐11
๐21 ๐ฑ = ๐31
Replacing the eq.1 by letter symbols, we have
๐ช = ๐ด๐ต = ๐ท ๐ธ๐น ๐บ
๐ป๐ฝ =
๐ซ๐ฏ + ๐ฌ๐ฑ๐ญ๐ฏ + ๐ฎ๐ฑ
= ๐ด๐ต
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The product is finally determined where C is composed of submatrices M and N. If we wish to find only the submatrix N, we have ๐ต = ๐ญ๐ฏ + ๐ฎ๐ฑ
= ๐31 ๐32 ๐11
๐21 + ๐33๐31
Hence, ๐ต = ๐๐๐๐๐๐ + ๐๐๐๐๐๐ + ๐๐๐๐๐๐ NODE ELIMINATION There are many ways to remove the need for matrix inversion when solving nodal equations of large scale power system. At the same time, it is also apparent that elimination of variables is identical to network reduction since it leads to a sequence of reduced-order network equivalents by node elimination at each step. This may be important in analyzing a large interconnected power system if there is special interest in the voltages at only some of the buses of the overall system. For instance, electric utility company with interconnection to other companies may wish to continue its study of voltage levels to those substations within its own service territory. By selective numbering of the buses, different methods are used to reduce the Ybus equation of the overall systems to a set which contains only those bus voltages of special interest. The coefficient matrix in the reduced order set of equations then represents the Ybus for an equivalent network containing only those buses which are to be retained. All other buses are eliminated in the mathemarical sense that their bus voltages and current injections do not appear explicitly. Such reduction in size of the equation set leads to efficiency of computation and helps to focus more directly on that portion of the overall network which is the primary interest. Below are following methods used to reduce node or eliminate nodes: A.) By MATRIX PARTITIONING (MATRIX-ALGEBRA) The standard node equations in matrix notation are expressed as
๐ฐ = ๐๐๐๐ ๐ฝ where V and I are column matrices and Ybus is a symmetrical square matrix. The
column matrices must be so arranged that elements associated with nodes to be
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eliminated are in the lower rows of the matrices. Elements of the square admittance matrix are located correspondingly. If the standard node equations in matrix notation will be expanded to its general form, we will have
๐ฐ๐จ๐ฐ๐ฟ
= ๐ฒ ๐ณ๐ณ๐ป ๐ด
๐ฝ๐จ
๐ฝ๐ฟ
where: ๐ฐ๐ฟ = submatrix composed of the currents entering the nodes to be
eliminated. Every element in it is zero, for the nodes could not be eliminated otherwise.
๐ฝ๐ฟ = submatrix composed of the voltages at the node at which Ix enters
๐ฒ = is composed self- and mutual- admittances which are those identified only with nodes to be retained.
๐ด = is composed self- and mutual- admittances which are those identified only with nodes to be eliminated. It is a square matrix whose order is equal to the number of nodes to be eliminated.
๐ณ ๐๐๐ ๐ณ๐ป = are composed of only those mutual admittances common to a node to be retained and to be eliminated.
If we will observe the above formula, we can see that column matrices
are partitioned so that the elements associated with nodes to be eliminated are separated from the other elements. The admittance matrix is partitioned so that elements identified only with nodes to be eliminated are separated from the other elements by horizontal and vertical lines.
If we will perform the multiplication indicated in the general formula, it
will give ๐ผ๐ด = ๐พ๐๐ด + ๐ฟ๐๐ eq.1
and ๐ผ๐ = ๐ฟ๐๐๐ด + ๐๐๐ eq.2
Since all elements of Ix = 0, then
๐ผ๐ = ๐ฟ๐๐๐ด + ๐๐๐ 0 = ๐ฟ๐๐๐ด + ๐๐๐
โ๐ฟ๐๐๐ด = ๐๐๐
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making, โ๐โ1๐ฟ๐๐๐ด = ๐๐ - substitute to eq.1
will give,
๐ผ๐ด = ๐พ๐๐ด + ๐ฟ๐๐ ๐ผ๐ด = ๐พ๐๐ด + ๐ฟ โ๐โ1๐ฟ๐๐๐ด
Factor out VA, then
๐ผ๐ด = ๐๐ด ๐พ + ๐ฟ โ๐โ1๐ฟ๐
๐ผ๐ด๐๐ด
= ๐พ + ๐ฟ โ๐โ1๐ฟ๐
Therefore,
๐๐จ = ๐ฒ โ ๐ณ๐ดโ๐๐ณ๐ป B.) GAUSSIAN ELIMINATION (KRON REDUCTION) In simple circuits, node elimination can be accomplished by Y-ฮ
transformations and by working with series and parallel combinations of impedances. The matrix partitioning method is a general method which is thereby more suitable for computer solutions. However, for the elimination of large number of nodes, matrix M whose inverse must be found will be large.
Inverting matrix is avoided by eliminating one node at a time. This
process is what we call gaussian method. Gaussian method is very simple. One voltage bus (also called node) at a time is sequentially removed from the original system of N equations in N unknown. The node to be eliminated must b the highest numbered node and renumbering may be required. The matrix M becomes a single element and M-1 is the reciprocal of the element. The steps of this method are as follows:
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1. If we have the below matrix,
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
=
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
the variable V1 does not explicitly appear in the resultant (N-1)x(N-1)
system, which fully represents the original network if the actual value of the voltage V1 at bus 1 is not of direct interest. Making
become,
According to equation,
๐๐๐๐ = ๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
โ1
๐๐๐ ๐๐๐
๐๐๐
๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
and when indicated manipulation of matrices is accomplished, the
element in row k and column j of the resultiing (N-1)x(N-1) matrix will be,
๐๐๐ ๐๐๐ = ๐๐๐ ๐๐๐๐ โ๐๐๐๐๐๐
๐๐๐
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4
1
3
2
!1
0
!2
!3
Ya
Yb
Yc
Yf
Yd
Yg
Ye
Yh
The procedure is to multiply the element to be modified in the last element of the same column and row and then its product must be divided by the cornermost element Y44 (in other books is Ynn) in the matrix to be modified. Finally, subtract the answer to the original element being modified.
2. If knowledge of V2 is also not of prime interest, we can interpret the resultant as (N-2) x (N-2) system as replacing the actual network by an (N-2) bus equivalent with buses 1 and 2 removed and so on. Thus,
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐ฝ๐
๐ฝ๐ =
๐ฐ๐๐ฐ๐
SAMPLE PROBLEM:
1.) If the generator and transformer at bus 3 are removed from the circuit of the figure below, eliminate nodes 3 and 4 by the matrix-algebra and gaussian method just described and find the equivalent circuit with these nodes eliminated. Find also the complex power transferred into or out of the network at nodes 1 and 2. Also find the voltage at node 1. The emfs are Ea=1.5 cis 0, Eb = 1.5 cis -36.87, and Ec = 1.5 cis 0, all in per unit.
Reqโd:
a) Ybus by matrix partition b) Ybus by gaussian method
c) equivalent circuit with 3 and 4 eliminated
d) complex power and e) V1
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Solโn: The bus admittance matrix of the circuit partitioned for elimination of nodes 3 and 4 is
4
1
3
2
-j0.8
!1
0
-j 4.0
-j 2.5
-j 5.0
-j0.8
!2
-j 8.0
-j 5.0
Ya
Yb
Yf
Yd
Yg
Ye
Yh
๐๐๐๐ =
โ๐๐. ๐ ๐
๐ โ๐๐. ๐
๐๐ ๐๐
๐๐. ๐ ๐๐
๐๐ ๐๐. ๐
๐๐ ๐๐
โ๐๐๐. ๐ ๐๐
๐๐ โ๐๐๐
a) Ybus by matrix partition
๐๐๐๐ = ๐ฒ ๐ณ๐ณ๐ป ๐ด
= ๐ฒโ๐ณ๐ดโ๐๐ณ๐ป
=
โ๐๐. ๐ ๐
๐ โ๐๐. ๐
๐๐ ๐๐
๐๐. ๐ ๐๐
๐๐ ๐๐. ๐
๐๐ ๐๐
โ๐๐๐. ๐ ๐๐
๐๐ โ๐๐๐
The inverse of the submatrix in the lower right position is,
๐โ1 =1
๐๐๐ก ๐ ๐๐๐๐๐๐ก๐๐ ๐ ๐
=1
โ197 โ๐18 โ๐8
โ๐8 โ๐14.5 ๐
=1
โ197 โ๐18 โ๐8
โ๐8 โ๐14.5
๐ดโ๐ = ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
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Then,
๐๐๐๐ = ๐ฒโ๐ณ๐ดโ๐๐ณ๐ป
= โ๐9.8 0
0 โ๐8.3 โ
๐4 ๐5
๐2.5 ๐5
๐0.0914 ๐0.0406
๐0.0406 ๐0.0736
๐4 ๐2.5
๐5 ๐5
= โ๐9.8 0
0 โ๐8.3 โ โ
๐4.9264 ๐4.0736
๐4.0736 ๐3.4264
๐๐๐๐ = โ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ โ๐๐. ๐๐๐๐
b) Ybus by gaussian method First, we remove node 4 then node 3. Thus,
๐๐๐ข๐ =
โ๐9.8 0
0 โ๐8.3
๐4 ๐5
๐2.5 ๐5
๐4 ๐2.5
๐5 ๐5
โ๐14.5 ๐8
๐8 โ๐18
To get the (N-1)x(N-1) resultant, use
๐๐๐ ๐๐๐ = ๐๐๐ ๐๐๐๐ โ๐๐๐๐๐๐
๐๐๐
and so,
๐๐๐ ๐๐๐ = โ๐9.8 โ๐5 ๐5
โ๐18= โ๐๐.๐๐๐
๐๐๐ ๐๐๐ = ๐๐๐ ๐๐๐ = 0 โ๐5 ๐5
โ๐18= ๐๐. ๐๐๐๐
๐๐๐ ๐๐๐ = ๐๐๐ ๐๐๐ = ๐2.5 โ๐5 ๐8
โ๐18= ๐๐. ๐๐๐๐
๐๐๐ ๐๐๐ = โ๐8.3 โ๐5 ๐5
โ๐18= โ๐๐.๐๐๐๐
๐๐๐ ๐๐๐ = ๐๐๐ ๐๐๐ = ๐2.5 โ๐5 ๐8
โ๐18= โ๐๐. ๐๐๐๐
๐๐๐ ๐๐๐ = โ๐14.5 โ๐5 ๐5
โ๐18= โ๐๐๐. ๐๐๐๐
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After determining the new value of elements, Reducing the above matrix to node 3,
๐๐๐ข๐ =
โ๐8.411 ๐1.3889 ๐6.2222๐1.3889 โ๐6.9111 โ๐4.7222๐6.2222 โ๐4.7222 โ๐10.9444
Having,
๐๐๐ ๐๐๐๐ = ๐๐๐ ๐๐๐๐ = โ๐8.411 โ๐6.2222 ๐6.2222
โ๐10.9444= โ๐๐.๐๐๐๐
๐๐๐ ๐๐๐๐ = ๐๐๐ ๐๐๐๐ = ๐1.3889 โโ๐4.7222 ๐6.2222
โ๐10.9444= โ๐๐.๐๐๐๐
Therefore,
๐๐๐๐ = โ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
๐๐. ๐๐๐๐ โ๐๐. ๐๐๐๐
c) equivalent circuit with 3 and 4 eliminated
To get the equivalent circuit, we can examine from the Ybus matrix that the admittance between the bus 1 and 2 is โj4.0736. Making,
1
2
!1
0
!2
Ya
Yb
Y12=-j4.0376
But we still donโt know the value of admittance between bus 1 and reference, Ya. To get it, we consider the value of the self admittance Y11 formula, which is
๐๐๐ = ๐๐ + ๐๐๐ + ๐๐๐
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But since bus 4 has been eliminated then Y14=0, yet we have
๐๐๐ = ๐๐ + ๐๐๐ โ๐4.8736 = ๐๐ + โ j4.0736
๐๐ = โ๐๐.๐
๐๐๐ = ๐๐ + ๐๐๐ โ๐4.8736 = ๐๐ + โ j4.0736
๐๐ = โ๐๐. ๐ The resulting circuit now is,
1
2
I1
0
I2
Ya=-j0.8
Yb=-j0.8
Y12=-j4.0376
To get the value the Power from reference to bus 1 and reference to bus 2, S1 and S2, we redraw the circuit and solve for the V and the current. For current, transform the above ckt to voltage-impedance ckt. Making this,
1 2
I1= 0-j1.2
0
I2= -0.72- j0.96Ya=-j0.8 Yb=-j0.8
Y12=-j4.0376
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to
1 2
0
E2=1.5 -36.87
Za=j1.25 Zb=j1.25Z12=-j0.2455
E1 =1.5 0I
But,
๐ผ =๐ธ1 โ ๐ธ2
๐๐ + ๐12 + ๐๐
=1.5โ 0 โ 1.5โ โ 36.87
๐1.25 + ๐0.2455 + ๐1.25
๐ฐ = ๐. ๐๐๐๐โ โ ๐๐. ๐๐ ๐๐
Hence; ๐1 = ๐ผโ๐ธ1
= 0.3455โ 18.44 1.5โ 0 ๐บ๐ = ๐. ๐๐๐ + ๐๐. ๐๐๐ ๐๐
๐2 = ๐ผโ๐ธ2 = 0.3455โ 18.44 1.5โ โ 36.87
๐บ๐ = ๐. ๐๐๐ โ ๐๐. ๐๐๐ ๐๐
The reactive power in the circuit equals to, ๐ = ๐ผ2๐๐
= 0.3455 2 ๐1.25 + ๐0.2455 + ๐1.25 ๐ธ = ๐.๐๐๐ ๐๐ The voltage at node 1 will be, ๐1 = ๐ธ1 โ ๐ผ๐๐
= 1.5โ 0 โ 0.3455โ โ 18.44 ๐1.25 ๐ฝ๐ = ๐. ๐๐๐ โ ๐๐. ๐๐๐ ๐๐
ADVANCED POWER SYSTEM
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
MODIFICATION OF EXISTING BUS IMPEDANCE
Since Zbus is such an important tool in power-system analysis, we shall examine how an existing Zbus may be modified to add new buses or connected new lines to established buses. Of course we create new Ybus and invert it, but direct methods are very much simpler than a matrix inversion even for small number of buses. Also when we know how to modify Zbus we can see how to build it directly.
We recognize several types of modifications involving the addition of a branch
having impedance Zb to a network whose original bus Zbus is known and is identified as Zorig, an nxn matrix.
In our analysis, existing buses will be identified by numbers or letters and letter k will designate a new bus to be added to the network to convert Zorig to an (n+1) x (n+1) matrix. Case I: Adding Zb from a new bus k to the reference bus.
Zb is added from a new bus to the reference bus r (i.e. a new branch is added and the dimension of Zbus goes up by one). The addition of the new bus k connected to the reference bus through Zb without a connection to any buses of the original network cannot alter the original bus voltages when a current Ik is injected at the new bus. Thus,
The new Zbus will be,
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
Making,
๐ฝ๐
๐ฝ๐
โฎ๐ฝ๐
๐ฝ๐
=
๐๐๐๐๐
๐๐โฎ๐
๐ ๐ โฏ ๐ ๐๐
๐ฐ๐๐ฐ๐โฎ๐ฐ๐๐ฐ๐
Note; Vk = ZbIk ; Zb = Zkk ; Zki = Zik = 0 Case II: Adding Zb from a new bus k to an existing bus j.
Zb is added from a new bus k to old bus j. The addition of the new bus k through Zb to an existing bus j with Ik injected at bus k will cause the current entering the original network at bus j to become the sum of Ij which is injected at bus j plus the current Ik coming through Zb. Thus,
The new bus becomes,
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
Making,
๐ฝ๐
๐ฝ๐
โฎ๐ฝ๐
๐ฝ๐
=
๐๐๐๐๐
๐๐๐
๐๐๐
โฎ๐๐๐
๐๐๐ ๐๐๐ โฏ ๐๐๐ ๐๐๐ + ๐๐
๐ฐ๐๐ฐ๐โฎ๐ฐ๐๐ฐ๐
where: ๐๐ = ๐ผ1๐๐1 + ๐๐2๐ผ2 + โฏ + ๐๐๐ ๐ผ๐ + ๐๐๐ + ๐๐ ๐ผ๐
= ๐๐ ๐๐๐๐ + ๐๐๐ + ๐๐ ๐ผ๐
= ๐๐ ๐๐๐๐ + ๐๐๐ ๐ผ๐ + ๐ผ๐๐๐
= ๐๐ ๐๐๐ค + ๐ผ๐๐๐
๐ฝ๐ = ๐ฝ๐ + ๐ฐ๐๐๐
Case III: Adding Zb from an existing bus j to the reference bus r. To see how to alter Z orig by connecting an imedance Zb from an existing bus j to the reference bus, we shall add a new bus k through the Zb to the bus j. Thus,
Then we short circuit bus k to the reference by letting Vk equal to zero to yield the same matrix equation as in case II except that Vk is zero. So for modification, we proceed to create new row and column exactly the same as in case 2 but we then eliminate the (n+1) x (n+1) column, which is possible because of the zero in column matrix of voltages.
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Thus,
๐ฝ๐
๐ฝ๐
โฎ๐ฝ๐
๐
=
๐๐๐๐๐
๐๐๐
๐๐๐
โฎ๐๐๐
๐๐๐ ๐๐๐ โฏ ๐๐๐ ๐๐๐ + ๐๐
๐ฐ๐๐ฐ๐โฎ๐ฐ๐๐ฐ๐
Eliminate Ik in the set of equations contained in the matrix operation
0 = ๐ผ1๐๐1 + ๐๐2๐ผ2 + โฏ + ๐๐๐ ๐ผ๐ + ๐๐๐ + ๐๐ ๐ผ๐
โ ๐๐๐ + ๐๐ ๐ผ๐ = ๐ผ1๐๐1 + ๐๐2๐ผ2 + โฏ + ๐๐๐ ๐ผ๐
๐ฐ๐ =โ๐
๐๐๐+๐๐ ๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐ โฆโฆ eq.1
Now,
๐ฝ๐ = ๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐ + ๐๐๐๐ฐ๐ โฆ. eq.2
Substituting eq.2 to eq.1, we have
๐ฝ๐
๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐ + ๐๐๐=
โ๐
๐๐๐ + ๐๐ ๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐
Equation can be written in the matrix form,
๐๐๐๐ (๐๐๐) = ๐๐๐๐ (๐๐๐๐) โ๐
๐๐๐ + ๐๐
๐๐๐
โฎ๐๐๐
๐๐๐ โฏ ๐๐๐
Or simply,
๐๐๐ (๐๐๐) = ๐๐๐ (๐๐๐๐) โ๐๐ (๐+๐)๐(๐+๐)๐
๐๐๐ + ๐๐
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
Case IV: Adding Zb between two existing bus j and i
To add a branch impedance Zb between already established buses i and j, we examine the figure below which shows these buses extracted from the original network. Thus,
The current Ik is shown flowing through Zb from bus j to bus i. We now write some equations for node voltages.
๐ฝ๐ = ๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐ ๐ฐ๐ + ๐ฐ๐ + ๐๐๐ ๐ฐ๐ โ ๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐
Similar equations follow for other buses. The voltages of the buses i and j are, however, constrained by the equation ๐ฝ๐ = ๐ฝ๐ + ๐ฐ๐๐๐
๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐ ๐ฐ๐ + ๐ฐ๐ + ๐๐๐ ๐ฐ๐ โ ๐ฐ๐ + โฏ + ๐๐๐๐ฐ๐= ๐ฐ๐๐๐๐ + ๐๐๐๐ฐ๐ + โฏ + ๐๐๐ ๐ฐ๐ + ๐ฐ๐ + ๐๐๐ ๐ฐ๐ โ ๐ฐ๐ + โฏ
+ ๐๐๐๐ฐ๐ + ๐ฐ๐๐๐
Rearranging, we have
0 = ๐๐๐ โ ๐๐๐ ๐ฐ๐ + โฏ + ๐๐๐ โ ๐๐๐ ๐ฐ๐ + ๐๐๐ โ ๐๐๐ ๐ฐ๐ + โฏ +
๐๐๐ โ ๐๐๐ ๐ฐ๐ + ๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐ โ ๐๐๐ ๐ฐ๐
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Collecting equations similar to case 2 and case 3, we can write
๐ฝ๐
๐ฝ๐
โฎ๐ฝ๐
๐
=
๐๐๐๐๐
๐๐๐ โ ๐๐๐
๐๐๐ โ ๐๐๐
โฎ๐๐๐ โ ๐๐๐
๐๐๐ โ ๐๐๐ ๐๐๐ โ ๐๐๐ โฏ ๐๐๐ โ ๐๐๐ ๐๐๐
๐ฐ๐๐ฐ๐โฎ๐ฐ๐๐ฐ๐
Where:
๐๐๐ = ๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐๐
Eliminate Ik in the set of equations contained in the matrix operation
๐๐๐๐ (๐๐๐) = ๐๐๐๐ (๐๐๐๐) โ๐
๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐๐
๐๐๐ โ ๐๐๐
โฎ๐๐๐ โ ๐๐๐
๐๐๐ โ ๐๐๐ โฏ ๐๐๐ โ ๐๐๐
Or simply,
๐๐๐ (๐๐๐) = ๐๐๐ (๐๐๐๐) โ๐๐ (๐+๐)๐(๐+๐)๐
๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐๐
With the use of four relationships, bus impedance matrix can be built by a step-by-step procedure (bringing in one branch at a time). This procedure being a mechanical one can be easily computerized. When network is undergoes changes, the modification procedures can be employed to revise the bus impedance matrix of the network.
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
SAMPLE PROBLEM:
1. Modify the bus matrix of the example 2 (page 52) to account for the connection of the capacitor having a reactance of 5.0 per unit between bus 4 and the reference bus. Then find V4 using the impedance of the new matrix and the current source. Compare this value of V4 with that found on example 2.
Reqโd: V4 Solโn: We use and recognize that Zorig is the 4x4 matrix. That k = 4 and that Zb = -j5.0 pu to find.
Before we use the case 2 modification, we first recognize the original Zbus of
the example before the capacitor is being installed from bus 4 to the reference.
๐๐๐ข๐ ๐๐๐๐ =
๐0.4774 ๐0.3706๐0.3706 ๐0.4872
๐0.4020 ๐0.4142๐0.3922 ๐0.4126
๐0.4020 ๐0.3922๐0.4142 ๐0.4126
๐0.4558 ๐0.4232๐0.4232 ๐0.4733
Then apply case 2, we have
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐
=
๐0.4774 ๐0.3706
๐0.3706 ๐0.4872
๐0.4020 ๐0.4142
๐0.3922 ๐0.4126๐0.4020 ๐0.3922
๐0.4142 ๐0.4126
๐0.4558 ๐0.4232
๐0.4232 ๐0.4733
๐๐๐
๐๐๐
โฎ๐๐๐
๐๐๐ ๐๐๐ โฏ ๐๐๐ ๐๐๐ + ๐๐
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
Then,
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐
=
๐0.4774 ๐0.3706
๐0.3706 ๐0.4872
๐0.4020 ๐0.4142
๐0.3922 ๐4126๐4020 ๐0.3922
๐0.4142 ๐0.4126
๐0.4558 ๐0.4232
๐0.4232 ๐0.4733
๐๐. ๐๐๐๐
๐๐. ๐๐๐๐๐๐. ๐๐๐๐
๐๐. ๐๐๐๐๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ + (โ๐๐)
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
ADVANCED POWER SYSTEM
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
So;
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐ฝ๐
๐
=
๐0.4774 ๐0.3706
๐0.3706 ๐0.4872
๐0.4020 ๐0.4142
๐0.3922 ๐0.4126๐0.4020 ๐0.3922
๐0.4142 ๐0.4126
๐0.4558 ๐0.4232
๐0.4232 ๐0.4733
๐๐. ๐๐๐๐
๐๐. ๐๐๐๐๐๐. ๐๐๐๐
๐๐. ๐๐๐๐๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ โ๐๐. ๐๐๐๐
๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐๐ฐ๐
The terms in the fifth row and column were obtained by repeating the forh row and column of Zorig and noting that
๐๐๐ = ๐๐๐ + ๐๐ = ๐๐๐ + ๐๐ = ๐๐. ๐๐๐๐ + โ๐๐ = โ๐๐. ๐๐๐๐
Then eliminating the fifth row and column, using the formula,
๐๐๐ (๐๐๐) = ๐๐๐ (๐๐๐๐) โ๐๐ (๐+๐)๐(๐+๐)๐
๐๐๐ + ๐๐
we obtain for Zbus new,
๐๐๐ (๐๐๐) = ๐0.4774 โ๐0.4142 ๐0.4142
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.3706 โ๐0.4142 ๐0.4126
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4020 โ๐0.4142 ๐0.4232
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4142 โ๐0.4142 ๐0.4733
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.3706 โ๐0.4142 ๐0.4142
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4872 โ๐0.4142 ๐0.4126
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.3922 โ๐0.4142 ๐0.4232
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4126 โ๐0.4142 ๐0.4733
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ ๐๐๐ = ๐0.4020 โ๐0.4142 ๐0.4142
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.3922 โ๐0.4142 ๐0.4126
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4558 โ๐0.4142 ๐0.4232
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4232 โ๐0.4142 ๐0.4733
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ ๐๐๐ = ๐0.4142 โ๐0.4142 ๐0.4142
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4126 โ๐0.4142 ๐0.4126
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4232 โ๐0.4142 ๐0.4232
โ๐4.5267= ๐๐. ๐๐๐๐
๐๐๐ (๐๐๐) = ๐0.4733 โ๐0.4142 ๐0.4733
โ๐4.5267= ๐๐. ๐๐๐๐
Then weโll have,
๐๐๐๐ ๐๐๐ =
๐๐. ๐๐๐๐ ๐๐.๐๐๐๐๐๐. ๐๐๐๐ ๐๐.๐๐๐๐
๐๐. ๐๐๐๐ ๐๐.๐๐๐๐๐๐. ๐๐๐๐ ๐๐.๐๐๐๐
๐๐. ๐๐๐๐ ๐๐.๐๐๐๐๐๐. ๐๐๐๐ ๐๐.๐๐๐๐
๐๐. ๐๐๐๐ ๐๐.๐๐๐๐๐๐. ๐๐๐๐ ๐๐.๐๐๐๐
ADVANCED POWER SYSTEM
2-NETWORK CALCULATIONS
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Prepared by: Joana Joy N. Abonalla and Pal Maleter Domingo
The column matrix of currents by which the new Zbus is multiplied to obtain the new bus voltages is the same as that of the previous example. ๐ฝ ๐ (๐๐๐) = ๐41๐ผ1 + ๐42๐ผ2 + ๐43๐ผ3
= ๐0.4575 โ๐1.2 + ๐0.4575 0.72 โ ๐0.96 + ๐0.4674 โ๐1.2 Hence;
๐ฝ ๐ (๐๐๐) = ๐. ๐๐๐โ โ ๐๐. ๐๐๐๐
DIRECT DETERMINATION OF A BUS IMPEDANCE MATRIX We have seen how to determine Zbus by first finding Ybus and inverting it. However, formulation of Zbus directly is straightforward process on the computer and simpler than inverting Ybus for a large network. To begin, we have a list of impedances showing the buses to which they are connected. We start by writing the equation for one bus connected through impedance Za to the reference bus as
๐ฝ ๐ = ๐๐๐ฐ ๐ And this can be considered as a matrix equation where each of the three matrices has one row and one column. Now, we might add new bus connected to the first bus or to the reference bus. For instance, if the second bus is connected to the reference but through Zb, we have the matrix equation
๐ฝ ๐
๐ฝ ๐ =
๐๐ ๐๐ ๐๐
๐ฐ ๐
๐ฐ ๐
And we proceed to modify our matrix by odding other buses following the procedures described. Usually, the buses of the network must be renumbered to agree with the order in which they are to be added to Zbus as it is build up.
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EXAMPLE: 1. Determine the Zbus for the network shown below where impedances are shown
in per unit. Preserve all three nodes.
32
j 0.3
j 1.5
j 0.15
1
j 1.2
0
j 0.2
Solโn:
1. We start by establishing bus 1 with the impedance to the reference bus and write
๐ฝ ๐ = ๐๐๐๐ฐ ๐ = ๐๐. ๐ ๐ฐ ๐
Then we have 1x1 bus impedance matrix,
๐๐๐๐ = ๐๐. ๐ 2. To establish bus 2 with its impedance to bus 1, we follow
๐๐๐๐ = ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ + ๐๐๐ =
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ =
๐๐. ๐ ๐๐.๐ ๐๐. ๐ ๐๐. ๐ + ๐๐. ๐
๐๐๐๐ = ๐๐.๐ ๐๐.๐ ๐๐.๐ ๐๐. ๐
Note: Elements Z01 in the new row and column is repitition of the elements of row 1 and column 1 being modified.
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3. To establish bus 3 with its impedance to bus 1, we follow
๐๐๐๐ =
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ + ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐ + ๐๐๐
=
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐
=
๐๐.๐ ๐๐. ๐ ๐๐. ๐๐๐.๐ ๐๐. ๐ ๐๐. ๐๐๐.๐ ๐๐. ๐ ๐๐. ๐ + ๐๐.๐
๐๐๐๐ =
๐๐.๐ ๐๐.๐ ๐๐.๐๐๐.๐ ๐๐.๐ ๐๐.๐๐๐.๐ ๐๐.๐ ๐๐.๐
Note: The other elements in new row and column are repitition of the elements of row 1 and column 1 of the matrix being modified since the new node is being connected to bus 1.
4. If we now decide to add Z30 = Zb =j1.5 from node 3 to the reference bus, we
follow case2 to connect new bus 4 through Z30, thus
๐๐๐๐ =
๐๐๐ ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ + ๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐ + ๐๐๐
๐๐๐
๐๐๐
๐๐๐ + ๐๐๐
๐๐๐ ๐๐๐ ๐๐๐ + ๐๐๐ ๐๐๐ + ๐๐๐ + ๐๐๐
=
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐ + ๐๐๐
=
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
๐๐๐ ๐๐๐
=
๐๐.๐ ๐๐.๐ ๐๐.๐๐๐.๐ ๐๐.๐ ๐๐.๐๐๐.๐ ๐๐.๐ ๐๐.๐
๐๐. ๐๐๐. ๐๐๐. ๐
๐๐.๐ ๐๐.๐ ๐๐.๐ ๐๐. ๐ + ๐๐.๐
๐๐๐๐ =
๐๐. ๐ ๐๐. ๐ ๐๐. ๐๐๐. ๐ ๐๐. ๐ ๐๐. ๐๐๐. ๐ ๐๐. ๐ ๐๐. ๐
๐๐.๐๐๐.๐๐๐.๐
๐๐. ๐ ๐๐. ๐ ๐๐. ๐ ๐๐
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Note: The other elements in new row and column are repitition of the elements of row 3 and column 3 of the matrix being modified since bus 3 is the one which is connecting tothe reference bus through Z30.
5. We now eliminate row 4 and column 4. Some of the elements of the new matrix
from equation
๐๐๐ (๐๐๐) = ๐๐๐ (๐๐๐๐) โ๐๐ (๐+๐)๐(๐+๐)๐
๐๐๐ + ๐๐
are,
๐๐๐ (๐๐๐) = ๐1.2 โ๐1.2 ๐1.2
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = ๐1.2 โ๐1.2 ๐1.2
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = ๐1.2 โ๐1.2 ๐1.5
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = ๐1.2 โ๐1.2 ๐1.2
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = ๐1.4 โ๐1.2 ๐1.2
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = ๐1.2 โ๐1.2 ๐1.5
๐3= ๐๐. ๐๐
๐๐๐ ๐๐๐ = 1.2 โ๐1.2 ๐1.5
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = 1.2 โ๐1.2 ๐1.5
๐3= ๐๐. ๐๐
๐๐๐ (๐๐๐) = 1.5 โ๐1.5 ๐1.5
๐3= ๐๐. ๐๐
When all the elements are determined, we have
๐๐๐๐ ๐๐๐ =
๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐
6. Finally, we add impedance Z32 = Zb = j0.15 between buses 2 and 3. If we let i and j in case 4 equal to 2 and 3, respectively, we obtain the elements for row 4 and column 4.
๐ฝ๐
๐ฝ๐
โฎ๐ฝ๐
๐
=
๐๐๐๐๐
๐๐๐ โ ๐๐๐
๐๐๐ โ ๐๐๐
โฎ๐๐๐ โ ๐๐๐
๐๐๐ โ ๐๐๐ ๐๐๐ โ ๐๐๐ โฏ ๐๐๐ โ ๐๐๐ ๐๐๐
๐ฐ๐๐ฐ๐โฎ๐ฐ๐๐ฐ๐
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where:
๐๐๐ = ๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐๐
which is ๐๐๐ = ๐๐๐ + ๐๐๐ + ๐๐๐ โ ๐๐๐๐
= ๐0.15 + ๐0.92 + ๐0.75 โ 2 ๐0.60
๐๐๐ = ๐ ๐. ๐๐
And,
๐๐๐ = ๐๐๐ โ ๐๐๐ = ๐๐. ๐๐ โ ๐๐. ๐๐ = ๐๐. ๐๐ ๐๐๐ = ๐๐๐ โ ๐๐๐ = ๐๐. ๐๐ โ ๐๐. ๐๐ = ๐๐. ๐๐
๐๐๐ = ๐๐๐ โ ๐๐๐ = ๐๐. ๐๐ โ ๐๐. ๐๐ = โ๐๐. ๐๐ So we write,
๐๐๐๐ =
๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐๐๐. ๐๐ ๐๐. ๐๐ ๐๐.๐๐
๐๐. ๐๐๐. ๐๐โ๐๐.๐๐
๐๐. ๐ ๐๐. ๐๐ โ๐๐.๐๐ ๐๐. ๐๐
7. Repeating procedure 5. Eliminating row 4 and column 4 of 4x4 to get 3x3 matrix. We find,
๐๐๐๐ ๐๐๐ =
๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐ ๐๐. ๐๐๐๐
which is the bus impedance matrix to be determined.