Network Calculations

ADVANCED POWER SYSTEM

2-NETWORK CALCULATIONS

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 Kirchhoff’s 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 points’ 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.

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

Xc XT3

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.

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 Kirchhoff’s current Law, we have

At node 1

𝐼1 = 𝑉1𝑌𝑎 + 𝑉1 − 𝑉3 𝑌𝑓 + 𝑉1 − 𝑉4 𝑌𝑑

At node 2 𝐼2 = 𝑉2𝑌𝑏 + 𝑉2 − 𝑉3 𝑌𝑔 + 𝑉2 − 𝑉4 𝑌𝑕

At node 3 𝐼3 = 𝑉3𝑌𝑐 + 𝑉3 − 𝑉1 𝑌𝑓 + 𝑉3 − 𝑉2 𝑌𝑔 + 𝑉3 − 𝑉4 𝑌𝑒

At node 4 0 = 𝑉4 − 𝑉1 𝑌𝑑 + 𝑉4 − 𝑉2 𝑌𝑕 + 𝑉4 − 𝑉3 𝑌𝑒

Rearranging these equations will yield to; At node 1

𝐼1 = 𝑉1𝑌𝑎 + 𝑉1 − 𝑉3 𝑌𝑓 + 𝑉1 − 𝑉4 𝑌𝑑

𝑰𝟏 = 𝑽𝟏 𝒀𝒂 + 𝒀𝒇 + 𝒀𝒅 − 𝑽𝟑𝒀𝒇 − 𝑽𝟒𝒀𝒅

At node 2

𝐼2 = 𝑉2𝑌𝑏 + 𝑉2 − 𝑉3 𝑌𝑔 + 𝑉2 − 𝑉4 𝑌𝑕

𝑰𝟐 = 𝑽𝟐 𝒀𝒃 + 𝒀𝒈 + 𝒀𝒉 − 𝑽𝟑𝒀𝒈 + −𝑽𝟒𝒀𝒉

At node 3

𝐼3 = 𝑉3𝑌𝑐 + 𝑉3 − 𝑉1 𝑌𝑓 + 𝑉3 − 𝑉2 𝑌𝑔 + 𝑉3 − 𝑉4 𝑌𝑒

𝑰𝟑 = −𝑽𝟏𝒀𝒇−𝑽𝟐𝒀𝒈 + 𝑽𝟑 𝒀𝒈 + 𝒀𝒄 + 𝒀𝒇 + 𝒀𝒆 − 𝑽𝟒𝒀𝒆

At node 4

0 = 𝑉4 − 𝑉1 𝑌𝑑 + 𝑉4 − 𝑉2 𝑌𝑕 + 𝑉4 − 𝑉3 𝑌𝑒

𝟎 = −𝑽𝟏𝒀𝒅 − 𝑽𝟐𝒀𝒉 − 𝑽𝟑𝒀𝒆 + 𝑽𝟒 𝒀𝒅 + 𝒀𝒉 + 𝒀𝒆 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.

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,

𝑰𝟏𝑰𝟐𝑰𝟑𝑰𝟒

𝒀𝟏𝟏 𝒀𝟏𝟐

𝒀𝟐𝟏 𝒀𝟐𝟐

𝒀𝟏𝟑 𝒀𝟏𝟒

𝒀𝟐𝟑 𝒀𝟐𝟒

𝒀𝟑𝟏 𝒀𝟑𝟐

𝒀𝟒𝟏 𝒀𝟒𝟐

𝒀𝟑𝟑 𝒀𝟑𝟒

𝒀𝟒𝟑 𝒀𝟒𝟒

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

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.

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

𝑰𝟏𝑰𝟐𝑰𝟑

𝒀𝟏𝟏 𝒀𝟏𝟐 𝒀𝟏𝟑

𝒀𝟐𝟏 𝒀𝟐𝟐 𝒀𝟐𝟑

𝒀𝟑𝟏 𝒀𝟑𝟐 𝒀𝟑𝟑

𝑽𝟏

𝑽𝟐

𝑽𝟑

And expanding this to equations, we have

𝑰𝟏 = 𝒀𝟏𝟏𝑽𝟏 + 𝒀𝟏𝟐𝑽𝟐 + 𝒀𝟏𝟑𝑽𝟑 𝑰𝟐 = 𝒀𝟐𝟏𝑽𝟏 + 𝒀𝟐𝟐𝑽𝟐 + 𝒀𝟐𝟑𝑽𝟑 𝑰𝟑 = 𝒁𝟑𝟏𝑽𝟏 + 𝒁𝟑𝟐𝑽𝟐 + 𝒁𝟑𝟑𝑽𝟑

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

𝒀𝟐𝟐 = 𝑰𝟐𝑽𝟐

𝑽𝟏=𝑽𝟑=𝟎

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.

To find the mutual admittance, at node 1 and node 3 the equation obtained are

𝒀𝟏𝟐 = 𝑰𝟏𝑽𝟐

𝑽𝟏=𝑽𝟑=𝟎

𝒀𝟑𝟐 = 𝑰𝟑𝑽𝟐

𝑽𝟏=𝑽𝟑=𝟎

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.

If we invert Ybus to Zbus, by definition, we will have

𝑍𝑏𝑢𝑠 = 𝑌𝑏𝑢𝑠−1

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

𝑽 = 𝒁𝒃𝒖𝒔𝑰 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

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝒁𝟏𝟏 𝒁𝟏𝟐 𝒁𝟏𝟑

𝒁𝟐𝟏 𝒁𝟐𝟐 𝒁𝟐𝟑

𝒁𝟑𝟏 𝒁𝟑𝟐 𝒁𝟑𝟑

𝑰𝟏𝑰𝟐𝑰𝟑

Converting the matrix form to equation form, we obtain

𝑽𝟏 = 𝒁𝟏𝟏𝑰𝟏 + 𝒁𝟏𝟐𝑰𝟐 + 𝒁𝟏𝟑𝑰𝟑 𝑽𝟐 = 𝒁𝟐𝟏𝑰𝟏 + 𝒁𝟐𝟐𝑰𝟐 + 𝒁𝟐𝟑𝑰𝟑 𝑽𝟑 = 𝒁𝟑𝟏𝑰𝟏 + 𝒁𝟑𝟐𝑰𝟐 + 𝒁𝟑𝟑𝑰𝟑

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,

𝒁𝟐𝟐 = 𝑽𝟐

𝑰𝟐 𝑰𝟏=𝑰𝟑=𝟎

To enable us to measure some transfer impedances, current sources I1 and I3 should be open circuited, so we can have

𝒁𝟏𝟐 = 𝑽𝟏

𝑰𝟐 𝑰𝟏=𝑰𝟑=𝟎

𝒁𝟑𝟐 = 𝑽𝟑

𝑰𝟐 𝑰𝟏=𝑰𝟑=𝟎

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.

THEVENIN’S 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=ΔI, the new bus voltages are given by the superposition equation,

where ΔV represents the changes in the bus voltages from their original values

Figure (a) shows network with bus k and reference node extracted voltages ΔVn, at bus n is caused by current ΔIk 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 I° and voltages V0 are zero. Then, into bus k a current of Δk amp (or Δk 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 ΔVk to ΔVn, are give n by

with the only nonzero entry in the current vector equal to ΔIk 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 ΔIk. Adding these voltage changes to the original voltages at the buses according to equation above yields to bus k.

𝑽𝒌 = 𝑽𝒌𝟎+𝒁𝒌𝒌∆𝑰𝒌

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

𝒁𝒕𝒉 = 𝒁𝒌𝒌 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 ΔIj at bus j and ΔIk at bus k.

Denoting the changes in the bus voltages resulting from the combination of these two current injections by ΔV1 to ΔVn, we obtain

in which the right-hand vector is numerically equal to the product of ΔIj at bus j and ΔIk at bus k of Zbus. Adding these voltages changes to original bus voltages, we obtain at buses j and k

𝑽𝒋 = 𝑽𝒋𝟎 + 𝒁𝒋𝒋∆𝑰𝒋 + 𝒁𝒋𝒌∆𝑰𝒌

𝑽𝒌 = 𝑽𝒌𝟎+𝒁𝒌𝒋∆𝑰𝒋 + 𝒁𝒌𝒌∆𝑰𝒌

Adding and subtracting 𝒁𝒌𝒋∆𝑰𝒋 and likewise 𝒁𝒋𝒌∆𝑰𝒌 in the equation above will give,

𝑽𝒋 = 𝑽𝒋𝟎 + 𝒁𝒋𝒋 − 𝒁𝒋𝒌 ∆𝑰𝒋 + 𝒁𝒋𝒌 ∆𝑰𝒋 + ∆𝑰𝒌

𝑽𝒌 = 𝑽𝒌𝟎 + 𝒁𝒌𝒋 ∆𝑰𝒋 + ∆𝑰𝒌 + 𝒁𝒌𝒌 − 𝒁𝒌𝒋 ∆𝑰𝒌

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

𝑽𝒌𝟎 − 𝑽𝒋

𝟎, and the impedance encountered by the short-circuit Isc from bus k to bus j is

evidently the Thevenin’s impedance

𝒁𝒕𝒉,𝒌𝒋 = 𝒁𝒋𝒋 + 𝒁𝒌𝒌 − 𝟐𝒁𝒋𝒌

This result is readily confirmed by substituting Isc = ΔIj = -ΔIk in above equations and by

setting the difference 𝑽𝒌𝟎 − 𝑽𝒋

𝟎 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 𝒁𝒋𝒋 = 𝒁𝒋𝒋 − 𝒁𝒋𝒌 + 𝒁𝒋𝒌 and the open-circuit voltage 𝑽𝒋𝟎.

From bus k to the reference node we have the Thevenin impedance 𝒁𝒌𝒌 =

𝒁𝒌𝒌 − 𝒁𝒋𝒌 + 𝒁𝒌𝒋, and the open- circuit 𝑽𝒌𝟎; and between buses j and k , the Thevenin

equation of 𝒁𝒕𝒉,𝒌𝒋 = 𝒁𝒋𝒋 + 𝒁𝒌𝒌 − 𝟐𝒁𝒋𝒌 and open voltage of 𝑽𝒌𝟎 − 𝑽𝒋

𝟎 is evident. Finally

when the branch impedance Zb is connected between buses j and k, the resulting current Ib is given by

𝑰𝒃 =𝑽𝒌

𝟎 − 𝑽𝒋𝟎

𝒁𝒕𝒉,𝒌𝒋 + 𝒁𝒃=

𝑽𝒌 − 𝑽𝒋

𝒁𝒃

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.

j1.15 j0.1Ea

j1.15 j0.1

Eb j1.15 j0.1

j0.125

Req’d: a.) node equations in matrix b.) Find the bus voltages, V1, V2, V3 and V4

-j 4.0

-j 2.5

-j 5.0

-j 8.0

-j 5.0

Sol’n:

a.) node equations in matrix

𝒀𝟏𝟏 𝒀𝟏𝟐

𝒀𝟐𝟏 𝒀𝟐𝟐

𝒀𝟏𝟑 𝒀𝟏𝟒

𝒀𝟐𝟑 𝒀𝟐𝟒

𝒀𝟑𝟏 𝒀𝟑𝟐

𝒀𝟒𝟏 𝒀𝟒𝟐

𝒀𝟑𝟑 𝒀𝟑𝟒

𝒀𝟒𝟑 𝒀𝟒𝟒

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

1. Draw the equivalent reactance diagram converted to current sources and admittances,

2. Solve for the current sources,

𝐈𝟏 = 𝐈𝟑 =1.5∠0

j 1.25= 𝟏. 𝟐∠ − 𝟗𝟎 = 𝟎 − 𝐣𝟏.𝟐𝟎 𝐩𝐮

𝐈𝟐 =1.5∠ − 36.87

j 1.25= 𝟏. 𝟐∠ − 𝟏𝟐𝟔. 𝟖𝟕 = −𝟎. 𝟕𝟐 − 𝐣𝟎. 𝟗𝟔 𝐩𝐮

3. Solve for the self admittances, Y11, Y22, Y33 and Y44. Thus,

𝒀𝟏𝟏 = 𝒀𝒂 + 𝒀𝒇 + 𝒀𝒅

= −𝑗5 − 𝑗4 − 𝑗0.8 𝒀𝟏𝟏 = −𝒋𝟗. 𝟖

𝒀𝟐𝟐 = 𝒀𝒃 + 𝒀𝒈 + 𝒀𝒉

= −𝑗0.8 − 𝑗2.5 − 𝑗5 𝒀𝟐𝟐 = −𝒋𝟖. 𝟑

𝒀𝟑𝟑 = 𝒀𝒄 + 𝒀𝒆 + 𝒀𝒇 + 𝒀𝒈

= −𝑗0.8 − 𝑗8 − 𝑗4 − 𝑗2.5 𝒀𝟑𝟑 = −𝒋𝟏𝟓.𝟑

𝒀𝟒𝟒 = 𝒀𝒅 + 𝒀𝒆 + 𝒀𝒉

= −𝑗5 − 𝑗5 − 𝑗0.8 𝒀𝟒𝟒 = −𝒋𝟏𝟖

4. Solve for the mutual admittances,

𝒀𝟏𝟐 = 𝒀𝟐𝟏 = 𝟎 𝒀𝟏𝟑 = 𝒀𝟑𝟏 = +𝒋𝟒 𝒀𝟏𝟒 = 𝒀𝟒𝟏 = +𝒋𝟓

𝒀𝟐𝟑 = 𝒀𝟑𝟐 = +𝒋𝟐. 𝟓 𝒀𝟐𝟒 = 𝒀𝟒𝟐 = +𝒋𝟓 𝒀𝟑𝟒 = 𝒀𝟒𝟑 = +𝒋𝟖

Hence, the node equations in matrix form is,

𝟎 − 𝒋𝟏.𝟐−𝟎. 𝟕𝟐 − 𝒋𝟎. 𝟗𝟔

𝟎 − 𝒋𝟏.𝟐𝟎

−𝒋𝟗.𝟖 𝟎𝟎 −𝒋𝟖.𝟑

𝒋𝟒 𝒋𝟓𝒋𝟐.𝟓 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓𝒋𝟓 𝒋𝟓

−𝒋𝟏𝟓.𝟑 𝒋𝟖𝒋𝟖 −𝒋𝟏𝟖

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

b.) Find the bus voltages, V1, V2, V3 and V4

1. Get the inverse of the bus admittance matrix, that is From,

𝐼 = 𝑌 𝑉 We will divide I matrix by Y matrix which will be,

𝑌 = 𝑉

or 𝐼 𝑌 −1 = 𝑉

To get [Y]-1, remember the inverse matrix formula,

𝒀−𝟏 =𝟏

𝒅𝒆𝒕 𝒀 𝒄𝒐𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒀 𝑻

To start with, a. Form the cofactor matrix of matrix Y,

−𝒋𝟖. 𝟑 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑 𝒋𝟖

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖 −

𝟎 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟒 −𝒋𝟏𝟓. 𝟑 𝒋𝟖

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖

𝟎 𝒋𝟒 𝒋𝟓

𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑 𝒋𝟖

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖 +

−𝒋𝟗. 𝟖 𝒋𝟒 𝒋𝟓

𝒋𝟒 −𝒋𝟏𝟓. 𝟑 𝒋𝟖

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖

𝟎 −𝒋𝟖. 𝟑 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓 𝒋𝟖

𝒋𝟓 𝒋𝟓 −𝒋𝟏𝟖 −

𝟎 −𝒋𝟖. 𝟑 𝒋𝟐. 𝟓

𝒋𝟒 𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑

𝒋𝟓 𝒋𝟓 𝒋𝟖

−𝒋𝟗. 𝟖 𝟎 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓 𝒋𝟖

𝒋𝟓 𝒋𝟓 −𝒋𝟏𝟖 +

−𝒋𝟗. 𝟖 𝟎 𝒋𝟒

𝒋𝟒 𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑

𝒋𝟓 𝒋𝟓 𝒋𝟖

+ 𝟎 𝒋𝟒 𝒋𝟓

−𝒋𝟖. 𝟑 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖 _

−𝒋𝟗. 𝟖 𝒋𝟒 𝒋𝟓

𝟎 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟓 𝒋𝟖 −𝒋𝟏𝟖

− 𝟎 𝒋𝟒 𝒋𝟓

−𝒋𝟖. 𝟑 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑 𝒋𝟖 +

−𝒋𝟗. 𝟖 𝒋𝟒 𝒋𝟓

𝟎 𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟒 −𝒋𝟏𝟓. 𝟑 𝒋𝟖

+ −𝒋𝟗. 𝟖 𝟎 𝒋𝟓

𝟎 −𝒋𝟖. 𝟑 𝒋𝟓

𝒋𝟓 𝒋𝟓 −𝒋𝟏𝟖 −

−𝒋𝟗. 𝟖 𝟎 𝒋𝟒

𝟎 −𝒋𝟖. 𝟑 𝒋𝟐. 𝟓

𝒋𝟓 𝒋𝟓 𝒋𝟖

− −𝒋𝟗. 𝟖 𝟎 𝒋𝟓

𝟎 𝒋𝟖. 𝟑 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓 𝒋𝟖 +

−𝒋𝟗. 𝟖 𝟎 𝒋𝟒

𝟎 −𝒋𝟖. 𝟑 𝒋𝟐. 𝟓

𝒋𝟒 𝒋𝟐. 𝟓 −𝒋𝟏𝟓. 𝟑

𝑗1059.62 𝑗822.5𝑗822.5 𝑗1081.22

𝑗892.1 𝑗919.3𝑗870.5 𝑗915.7

𝑗892.1 𝑗870.5𝑗919.3 𝑗915.7

𝑗1011.62 𝑗939.22𝑗939.22 𝑗1050.452

b. Form the transpose matrix of the cofactor matrix of matrix Y that is

𝑌 𝑇 =

𝑗1059.62 𝑗822.5𝑗822.5 𝑗1081.22

𝑗892.1 𝑗919.3𝑗870.5 𝑗915.7

𝑗892.1 𝑗870.5𝑗919.3 𝑗915.7

𝑗1011.62 𝑗939.22𝑗939.22 𝑗1050.452

c. Evaluate the determinant of matrix Y

𝒅𝒆𝒕 𝒀 =

−𝒋𝟗. 𝟖 𝟎

𝟎 −𝒋𝟖. 𝟑

𝒋𝟒 𝒋𝟓

𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓

𝒋𝟓 𝒋𝟓

−𝒋𝟏𝟓. 𝟑 𝒋𝟖

𝒋𝟖 −𝒋𝟏𝟖

𝒅𝒆𝒕 𝒀 = 𝟐𝟐𝟏𝟗.𝟑𝟕𝟔 * It can be solved by different methods discussed in advanced math like modification, chio’s and pivotal method.

d. Divide each element of the transpose cofactor of Y by the value of

determinant of Y.

𝒀−𝟏 =𝟏

𝒅𝒆𝒕 𝒀 𝒄𝒐𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒀 𝑻

2219.376

𝑗1059.62 𝑗822.5

𝑗822.5 𝑗1081.22

𝑗892.1 𝑗919.3

𝑗870.5 𝑗915.7

𝑗892.1 𝑗870.5

𝑗919.3 𝑗915.7

𝑗1011.62 𝑗939.22

𝑗939.22 𝑗1050.452

𝒀−𝟏 =

𝒋𝟎. 𝟒𝟕𝟕𝟒 𝒋𝟎. 𝟑𝟕𝟎𝟔

𝒋𝟎. 𝟑𝟕𝟎𝟔 𝒋𝟎. 𝟒𝟖𝟕𝟐

𝒋𝟎. 𝟒𝟎𝟐𝟎 𝒋𝟎. 𝟒𝟏𝟒𝟐

𝒋𝟎. 𝟑𝟗𝟐𝟐 𝒋𝟎. 𝟒𝟏𝟐𝟔

𝒋𝟎. 𝟒𝟎𝟐𝟎 𝒋𝟎. 𝟑𝟗𝟐𝟐

𝒋𝟎. 𝟒𝟏𝟒𝟐 𝒋𝟎. 𝟒𝟏𝟐𝟔

𝒋𝟎. 𝟒𝟓𝟓𝟖 𝒋𝟎. 𝟒𝟐𝟑𝟐

𝒋𝟎. 𝟒𝟐𝟑𝟐 𝒋𝟎. 𝟒𝟕𝟑𝟑

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

𝑗0.4774 𝑗0.3706𝑗0.3706 𝑗0.4872

𝑗0.4020 𝑗0.4142𝑗0.3922 𝑗0.4126

𝑗0.4020 𝑗0.3922𝑗0.4142 𝑗0.4126

𝑗0.4558 𝑗0.4232𝑗0.4232 𝑗0.4733

0 − 𝑗1.2−0.72 − 𝑗0.96

0 − 𝑗1.20

3. Performing the indicated matrix multiplication will yield to,

𝟏.𝟒𝟏𝟏 − 𝒋𝟎. 𝟐𝟔𝟔𝟖𝟏. 𝟑𝟖𝟑𝟎 − 𝒋𝟎.𝟑𝟓𝟎𝟖𝟏. 𝟒𝟎𝟓𝟗 − 𝒋𝟎.𝟐𝟖𝟐𝟒𝟏. 𝟒𝟎𝟎𝟗 − 𝒋𝟎.𝟐𝟗𝟕𝟏

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

And so, the node voltages are

𝑽𝟏 = 𝟏. 𝟒𝟏𝟏 − 𝒋𝟎. 𝟐𝟔𝟔𝟖 = 𝟏. 𝟒𝟑𝟔∠ − 𝟏𝟎.𝟕𝟏 𝒑𝒖 𝑽𝟐 = 𝟏. 𝟑𝟖𝟑𝟎 − 𝒋𝟎.𝟑𝟓𝟎𝟖 = 𝟏. 𝟒𝟐𝟕∠ − 𝟏𝟒. 𝟐𝟒 𝒑𝒖 𝑽𝟑 = 𝟏. 𝟒𝟎𝟓𝟗 − 𝒋𝟎.𝟐𝟖𝟐𝟒 = 𝟏. 𝟒𝟑𝟒∠ − 𝟏𝟏. 𝟑𝟔 𝒑𝒖 𝑽𝟒 = 𝟏. 𝟒𝟎𝟎𝟗 − 𝒋𝟎.𝟐𝟗𝟕𝟏 = 𝟏. 𝟒𝟑𝟐∠ − 𝟏𝟏. 𝟗𝟕 𝒑𝒖

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. Req’d:

1.) Ic 2.) Find the bus voltages, V1, V2, V3 and V4 if IC is negative of the value

obtained in the solution Illustration:

j1.15 j0.1Ea

j1.15 j0.1

Eb j1.15 j0.1

j0.125

Sol’n:

1. To solve for IC, a.) Create the Thevenin’s equivalent circuit for node 4. Making sources short

circuited except node 4. Thus,

j0.125

It becomes

V4 = Vth

Z44 = Zth

XC = ZC

b.) To solve for Thevenin’s equivalent of the circuit behind node 4, the thevenin’s

voltage at node 4 before the capacitor is connected has an emf of

𝑉4 = 𝑉𝑇𝑕 = 1.4009 − 𝑗0.2971 = 1.432∠ − 11.97 𝑝𝑢 c.) To find the Thevenin’s 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,

𝑉4 = 𝑍41𝐼1 + 𝑍42𝐼2 + 𝑍43𝐼3 + 𝑍44𝐼4

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,

𝑍𝑇𝑕 = 𝑍44 = 𝑗0.4733

d.) Thus the current drawn by the capacitor is,

𝑉𝑇𝑕 = 𝑍𝑇𝑕 + 𝑍𝐶 𝐼𝐶

𝐼𝐶 = 𝑉𝑇𝑕

𝑍𝑇𝑕 + 𝑍𝐶

= 1.432∠ − 11.97 𝑝𝑢

𝑗0.4733 − 𝑗5

𝑰𝑪 = 𝑰𝟒 = 𝟎. 𝟑𝟏𝟔∠𝟕𝟖.𝟎𝟑 𝒑𝒖

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

𝑉1 = 𝐼4𝑍14 = −0.316∠78.03 𝑗0.4142 = 0.1309∠− 11.97° 𝑉2 = 𝐼4𝑍24 = −0.316∠78.03 𝑗0.4126 = 0.1304∠− 11.97° 𝑉3 = 𝐼4𝑍34 = −0.316∠78.03 𝑗0.4232 = 0.1337∠− 11.97° 𝑉4 = 𝐼4𝑍44 = −0.316∠78.03 𝑗0.4733 = 0.1496∠ − 11.97°

By superposition, the resulting voltages determined by adding the

voltages caused by the injected current with emfs shorted to the node voltages are;

𝑽𝟏𝒏𝒆𝒘 = 𝑉1 + 𝑉1𝑜𝑙𝑑 = 0.1309∠ − 11.97° + 1.436∠ − 10.71 = 𝟏. 𝟓𝟔𝟕∠ − 𝟏𝟎. 𝟖𝟏 𝒑𝒖 𝑽𝟐𝒏𝒆𝒘 = 𝑉2 + 𝑉2𝑜𝑙𝑑 = 0.1304∠ − 11.97° + 1.427∠ − 14.24 = 𝟏.𝟓𝟓𝟕∠ − 𝟏𝟒. 𝟒𝟏𝒑𝒖 𝑽𝟑𝒏𝒆𝒘 = 𝑉3 + 𝑉3𝑜𝑙𝑑 = 0.1337∠ − 11.97° + 1.434∠ − 11.36 = 𝟏. 𝟓𝟔𝟖∠ − 𝟏𝟏. 𝟒𝟏 𝒑𝒖 𝑽𝟒𝒏𝒆𝒘 = 𝑉4 + 𝑉4𝑜𝑙𝑑 = 0.1496∠ − 11.97° + 1.432∠ − 11.97 = 𝟏. 𝟓𝟖𝟐∠ − 𝟏𝟏. 𝟗𝟕𝒑𝒖

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.

MATRIX PARTTITIONING: A use method of matrix manipulation is called parttitioning. Consider the 3x3 matrix multiplied by 3x1, 𝑪 = 𝑨𝑩 . . .eq.1

𝐶 =

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

𝑏11

𝑏21

𝑏31

The matrix A will be partitioned into four submatrices by the horizontal and vertical dashed lines. The matrix may be written,

𝑨 =

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

= 𝑫 𝑬𝑭 𝑮

where:

𝑫 = 𝑎11 𝑎12

𝑎21 𝑎22 𝑬 =

𝑎13

𝑎23

𝑭 = 𝑎31 𝑎32 𝑮 = 𝑎33

Also the matrix B will be parttitioned,

𝑩 =

𝑏11

𝑏21

𝑏31

= 𝑯𝑱

where:

𝑯 = 𝑏11

𝑏21 𝑱 = 𝑏31

Replacing the eq.1 by letter symbols, we have

𝑪 = 𝐴𝐵 = 𝐷 𝐸𝐹 𝐺

𝐻𝐽 =

𝑫𝑯 + 𝑬𝑱𝑭𝑯 + 𝑮𝑱

= 𝑴𝑵

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 𝑵 = 𝑭𝑯 + 𝑮𝑱

= 𝑎31 𝑎32 𝑏11

𝑏21 + 𝑎33𝑏31

Hence, 𝑵 = 𝒂𝟑𝟏𝒃𝟏𝟏 + 𝒂𝟑𝟐𝒃𝟐𝟏 + 𝒂𝟑𝟑𝒃𝟑𝟏 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

𝑰 = 𝒀𝒃𝒖𝒔 𝑽 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

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

𝑰𝑨𝑰𝑿

= 𝑲 𝑳𝑳𝑻 𝑴

𝑽𝑨

𝑽𝑿

where: 𝑰𝑿 = 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.

𝑽𝑿 = submatrix composed of the voltages at the node at which Ix enters

𝑲 = is composed self- and mutual- admittances which are those identified only with nodes to be retained.

𝑴 = 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.

𝑳 𝑎𝑛𝑑 𝑳𝑻 = 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 𝐼𝐴 = 𝐾𝑉𝐴 + 𝐿𝑉𝑋 eq.1

and 𝐼𝑋 = 𝐿𝑇𝑉𝐴 + 𝑀𝑉𝑋 eq.2

Since all elements of Ix = 0, then

𝐼𝑋 = 𝐿𝑇𝑉𝐴 + 𝑀𝑉𝑋 0 = 𝐿𝑇𝑉𝐴 + 𝑀𝑉𝑋

−𝐿𝑇𝑉𝐴 = 𝑀𝑉𝑋

making, −𝑀−1𝐿𝑇𝑉𝐴 = 𝑉𝑋 - substitute to eq.1

will give,

𝐼𝐴 = 𝐾𝑉𝐴 + 𝐿𝑉𝑋 𝐼𝐴 = 𝐾𝑉𝐴 + 𝐿 −𝑀−1𝐿𝑇𝑉𝐴

Factor out VA, then

𝐼𝐴 = 𝑉𝐴 𝐾 + 𝐿 −𝑀−1𝐿𝑇

𝐼𝐴𝑉𝐴

= 𝐾 + 𝐿 −𝑀−1𝐿𝑇

Therefore,

𝒀𝑨 = 𝑲 − 𝑳𝑴−𝟏𝑳𝑻 B.) GAUSSIAN ELIMINATION (KRON REDUCTION) In simple circuits, node elimination can be accomplished by Y-Δ

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:

1. If we have the below matrix,

𝒀𝟏𝟏 𝒀𝟏𝟐

𝒀𝟐𝟏 𝒀𝟐𝟐

𝒀𝟏𝟑 𝒀𝟏𝟒

𝒀𝟐𝟑 𝒀𝟐𝟒

𝒀𝟑𝟏 𝒀𝟑𝟐

𝒀𝟒𝟏 𝒀𝟒𝟐

𝒀𝟑𝟑 𝒀𝟑𝟒

𝒀𝟒𝟑 𝒀𝟒𝟒

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

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,

𝒀𝒃𝒖𝒔 = 𝒀𝟏𝟏 𝒀𝟏𝟐 𝒀𝟏𝟑

𝒀𝟐𝟏 𝒀𝟐𝟐 𝒀𝟐𝟑

𝒀𝟑𝟏 𝒀𝟑𝟐 𝒀𝟑𝟑

𝒀𝟒𝟒 𝒀𝟏𝟒

𝒀𝟐𝟒

𝒀𝟑𝟒

𝒀𝟒𝟏 𝒀𝟒𝟐 𝒀𝟒𝟑

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,

𝒀𝒌𝒋 𝒏𝒆𝒘 = 𝒀𝒌𝒋 𝒐𝒓𝒊𝒈 −𝒀𝒌𝒏𝒀𝒏𝒋

𝒀𝟒𝟒

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,

𝒀𝟑𝟏 𝒀𝟑𝟐

𝒀𝟒𝟏 𝒀𝟒𝟐

𝑽𝟑

𝑽𝟒 =

𝑰𝟑𝑰𝟒

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.

Req’d:

a) Ybus by matrix partition b) Ybus by gaussian method

c) equivalent circuit with 3 and 4 eliminated

d) complex power and e) V1

Sol’n: The bus admittance matrix of the circuit partitioned for elimination of nodes 3 and 4 is

-j 4.0

-j 2.5

-j 5.0

-j 8.0

-j 5.0

𝒀𝒃𝒖𝒔 =

−𝒋𝟗. 𝟖 𝟎

𝟎 −𝒋𝟖. 𝟑

𝒋𝟒 𝒋𝟓

𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓

𝒋𝟓 𝒋𝟓

−𝒋𝟏𝟒. 𝟓 𝒋𝟖

𝒋𝟖 −𝒋𝟏𝟖

a) Ybus by matrix partition

𝒀𝒃𝒖𝒔 = 𝑲 𝑳𝑳𝑻 𝑴

= 𝑲−𝑳𝑴−𝟏𝑳𝑻

−𝒋𝟗. 𝟖 𝟎

𝟎 −𝒋𝟖. 𝟑

𝒋𝟒 𝒋𝟓

𝒋𝟐. 𝟓 𝒋𝟓

𝒋𝟒 𝒋𝟐. 𝟓

𝒋𝟓 𝒋𝟓

−𝒋𝟏𝟒. 𝟓 𝒋𝟖

𝒋𝟖 −𝒋𝟏𝟖

The inverse of the submatrix in the lower right position is,

𝑀−1 =1

𝑑𝑒𝑡 𝑀 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟 𝑀 𝑇

−197 −𝑗18 −𝑗8

−𝑗8 −𝑗14.5 𝑇

−197 −𝑗18 −𝑗8

−𝑗8 −𝑗14.5

𝑴−𝟏 = 𝒋𝟎. 𝟎𝟗𝟏𝟒 𝒋𝟎. 𝟎𝟒𝟎𝟔

𝒋𝟎. 𝟎𝟒𝟎𝟔 𝒋𝟎. 𝟎𝟕𝟑𝟔

𝒀𝒃𝒖𝒔 = 𝑲−𝑳𝑴−𝟏𝑳𝑻

= −𝑗9.8 0

0 −𝑗8.3 −

𝑗4 𝑗5

𝑗2.5 𝑗5

𝑗0.0914 𝑗0.0406

𝑗0.0406 𝑗0.0736

𝑗4 𝑗2.5

𝑗5 𝑗5

= −𝑗9.8 0

0 −𝑗8.3 − −

𝑗4.9264 𝑗4.0736

𝑗4.0736 𝑗3.4264

𝒀𝒃𝒖𝒔 = −𝒋𝟒. 𝟖𝟕𝟑𝟔 𝒋𝟒. 𝟎𝟕𝟑𝟔

𝒋𝟒. 𝟎𝟕𝟑𝟔 −𝒋𝟒. 𝟖𝟕𝟑𝟔

b) Ybus by gaussian method First, we remove node 4 then node 3. Thus,

𝑌𝑏𝑢𝑠 =

−𝑗9.8 0

0 −𝑗8.3

𝑗4 𝑗5

𝑗2.5 𝑗5

𝑗4 𝑗2.5

𝑗5 𝑗5

−𝑗14.5 𝑗8

𝑗8 −𝑗18

To get the (N-1)x(N-1) resultant, use

𝒀𝒌𝒋 𝒏𝒆𝒘 = 𝒀𝒌𝒋 𝒐𝒓𝒊𝒈 −𝒀𝒌𝒏𝒀𝒏𝒋

𝒀𝟒𝟒

and so,

𝒀𝟏𝟏 𝒏𝒆𝒘 = −𝑗9.8 −𝑗5 𝑗5

−𝑗18= −𝒋𝟖.𝟒𝟏𝟏

𝒀𝟏𝟐 𝒏𝒆𝒘 = 𝒀𝟐𝟏 𝒏𝒆𝒘 = 0 −𝑗5 𝑗5

−𝑗18= 𝒋𝟏. 𝟑𝟖𝟖𝟗

𝒀𝟏𝟑 𝒏𝒆𝒘 = 𝒀𝟑𝟏 𝒏𝒆𝒘 = 𝑗2.5 −𝑗5 𝑗8

−𝑗18= 𝒋𝟔. 𝟐𝟐𝟐𝟐

𝒀𝟐𝟐 𝒏𝒆𝒘 = −𝑗8.3 −𝑗5 𝑗5

−𝑗18= −𝒋𝟔.𝟗𝟏𝟏𝟏

𝒀𝟐𝟑 𝒏𝒆𝒘 = 𝒀𝟑𝟐 𝒏𝒆𝒘 = 𝑗2.5 −𝑗5 𝑗8

−𝑗18= −𝒋𝟒. 𝟕𝟐𝟐𝟐

𝒀𝟑𝟑 𝒏𝒆𝒘 = −𝑗14.5 −𝑗5 𝑗5

−𝑗18= −𝒋𝟏𝟎. 𝟗𝟒𝟒𝟒

After determining the new value of elements, Reducing the above matrix to node 3,

𝑌𝑏𝑢𝑠 =

−𝑗8.411 𝑗1.3889 𝑗6.2222𝑗1.3889 −𝑗6.9111 −𝑗4.7222𝑗6.2222 −𝑗4.7222 −𝑗10.9444

Having,

𝒀𝟏𝟏 𝒏𝒆𝒘𝟐 = 𝒀𝟐𝟐 𝒏𝒆𝒘𝟐 = −𝑗8.411 −𝑗6.2222 𝑗6.2222

−𝑗10.9444= −𝒋𝟒.𝟖𝟕𝟑𝟔

𝒀𝟏𝟐 𝒏𝒆𝒘𝟐 = 𝒀𝟐𝟏 𝒏𝒆𝒘𝟐 = 𝑗1.3889 −−𝑗4.7222 𝑗6.2222

−𝑗10.9444= −𝒋𝟒.𝟎𝟕𝟑𝟔

Therefore,

𝒀𝒃𝒖𝒔 = −𝒋𝟒. 𝟖𝟕𝟑𝟔 𝒋𝟒. 𝟎𝟕𝟑𝟔

𝒋𝟒. 𝟎𝟕𝟑𝟔 −𝒋𝟒. 𝟖𝟕𝟑𝟔

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 –j4.0736. Making,

Y12=-j4.0376

But we still don’t 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

𝒀𝟏𝟏 = 𝒀𝒂 + 𝒀𝟏𝟐 + 𝒀𝟏𝟒

But since bus 4 has been eliminated then Y14=0, yet we have

𝒀𝟏𝟏 = 𝒀𝒂 + 𝒀𝟏𝟐 −𝑗4.8736 = 𝑌𝑎 + – j4.0736

𝒀𝒂 = −𝒋𝟎.𝟖

𝒀𝟐𝟐 = 𝒀𝒃 + 𝒀𝟐𝟏 −𝑗4.8736 = 𝑌𝑎 + – j4.0736

𝒀𝒃 = −𝒋𝟎. 𝟖 The resulting circuit now is,

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,

I1= 0-j1.2

I2= -0.72- j0.96Ya=-j0.8 Yb=-j0.8

Y12=-j4.0376

E2=1.5 -36.87

Za=j1.25 Zb=j1.25Z12=-j0.2455

E1 =1.5 0I

𝐼 =𝐸1 − 𝐸2

𝑍𝑎 + 𝑍12 + 𝑍𝑏

=1.5∠0 − 1.5∠ − 36.87

𝑗1.25 + 𝑗0.2455 + 𝑗1.25

𝑰 = 𝟎. 𝟑𝟒𝟓𝟓∠ − 𝟏𝟖. 𝟒𝟒 𝒑𝒖

Hence; 𝑆1 = 𝐼∗𝐸1

= 0.3455∠18.44 1.5∠0 𝑺𝟏 = 𝟎. 𝟒𝟗𝟐 + 𝒋𝟎. 𝟏𝟔𝟒 𝒑𝒖

𝑆2 = 𝐼∗𝐸2 = 0.3455∠18.44 1.5∠ − 36.87

𝑺𝟐 = 𝟎. 𝟒𝟗𝟐 − 𝒋𝟎. 𝟏𝟔𝟒 𝒑𝒖

The reactive power in the circuit equals to, 𝑄 = 𝐼2𝑍𝑇

= 0.3455 2 𝑗1.25 + 𝑗0.2455 + 𝑗1.25 𝑸 = 𝟎.𝟑𝟐𝟕 𝒑𝒖 The voltage at node 1 will be, 𝑉1 = 𝐸1 − 𝐼𝑍𝑎

= 1.5∠0 − 0.3455∠ − 18.44 𝑗1.25 𝑽𝟏 = 𝟏. 𝟑𝟔𝟑 − 𝒋𝟎. 𝟒𝟏𝟎 𝒑𝒖

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,

Making,

𝑽𝟏

𝑽𝟐

⋮𝑽𝒏

𝑽𝒌

𝒁𝒐𝒓𝒊𝒈

𝟎𝟎⋮𝟎

𝟎 𝟎 ⋯ 𝟎 𝒁𝒃

𝑰𝟏𝑰𝟐⋮𝑰𝒏𝑰𝒌

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,

Making,

𝑽𝟏

𝑽𝟐

⋮𝑽𝒏

𝑽𝒌

𝒁𝟏𝒋

𝒁𝟐𝒋

⋮𝒁𝒏𝒋

𝒁𝒋𝟏 𝒁𝒋𝟐 ⋯ 𝒁𝒋𝒏 𝒁𝒋𝒋 + 𝒁𝒃

where: 𝑉𝑘 = 𝐼1𝑍𝑗1 + 𝑍𝑗2𝐼2 + ⋯ + 𝑍𝑗𝑛 𝐼𝑛 + 𝑍𝑗𝑗 + 𝑍𝑏 𝐼𝑘

= 𝑉𝑘 𝑜𝑟𝑖𝑔 + 𝑍𝑗𝑗 + 𝑍𝑏 𝐼𝑘

= 𝑉𝑘 𝑜𝑟𝑖𝑔 + 𝑍𝑗𝑗 𝐼𝑘 + 𝐼𝑘𝑍𝑏

= 𝑉𝑘 𝑛𝑒𝑤 + 𝐼𝑘𝑍𝑏

𝑽𝒌 = 𝑽𝒋 + 𝑰𝒌𝒁𝒃

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.

𝑽𝟏

𝑽𝟐

⋮𝑽𝒏

𝒁𝟏𝒋

𝒁𝟐𝒋

⋮𝒁𝒏𝒋

Eliminate Ik in the set of equations contained in the matrix operation

0 = 𝐼1𝑍𝑗1 + 𝑍𝑗2𝐼2 + ⋯ + 𝑍𝑗𝑛 𝐼𝑛 + 𝑍𝑗𝑗 + 𝑍𝑏 𝐼𝑘

− 𝑍𝑗𝑗 + 𝑍𝑏 𝐼𝑘 = 𝐼1𝑍𝑗1 + 𝑍𝑗2𝐼2 + ⋯ + 𝑍𝑗𝑛 𝐼𝑛

𝑰𝒌 =−𝟏

𝒁𝒋𝒋+𝒁𝒃 𝑰𝟏𝒁𝒋𝟏 + 𝒁𝒋𝟐𝑰𝟐 + ⋯ + 𝒁𝒋𝒏𝑰𝒏 …… eq.1

𝑽𝒊 = 𝑰𝟏𝒁𝒊𝟏 + 𝒁𝒊𝟐𝑰𝟐 + ⋯ + 𝒁𝒊𝒏𝑰𝒏 + 𝒁𝒊𝒋𝑰𝒌 …. eq.2

Substituting eq.2 to eq.1, we have

𝑽𝒊

𝑰𝟏𝒁𝒊𝟏 + 𝒁𝒊𝟐𝑰𝟐 + ⋯ + 𝒁𝒊𝒏𝑰𝒏 + 𝒁𝒊𝒋=

−𝟏

𝒁𝒋𝒋 + 𝒁𝒃 𝑰𝟏𝒁𝒋𝟏 + 𝒁𝒋𝟐𝑰𝟐 + ⋯ + 𝒁𝒋𝒏𝑰𝒏

Equation can be written in the matrix form,

𝒁𝒃𝒖𝒔 (𝒏𝒆𝒘) = 𝒁𝒃𝒖𝒔 (𝒐𝒓𝒊𝒈) −𝟏

𝒁𝒋𝒋 + 𝒁𝒃

𝒁𝟏𝒋

⋮𝒁𝒏𝒋

𝒁𝒋𝟏 ⋯ 𝒁𝒋𝒏

Or simply,

𝒁𝒉𝒊 (𝒏𝒆𝒘) = 𝒁𝒉𝒊 (𝒐𝒓𝒊𝒈) −𝒁𝒉 (𝒏+𝟏)𝒁(𝒏+𝟏)𝒊

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.

𝑽𝒊 = 𝑰𝟏𝒁𝒊𝟏 + 𝒁𝒊𝟐𝑰𝟐 + ⋯ + 𝒁𝟏𝒊 𝑰𝒊 + 𝑰𝒌 + 𝒁𝒊𝒋 𝑰𝒋 − 𝑰𝒌 + ⋯ + 𝒁𝒊𝒏𝑰𝒏

Similar equations follow for other buses. The voltages of the buses i and j are, however, constrained by the equation 𝑽𝒋 = 𝑽𝒊 + 𝑰𝒌𝒁𝒃

𝑰𝟏𝒁𝒋𝟏 + 𝒁𝒋𝟐𝑰𝟐 + ⋯ + 𝒁𝒋𝒊 𝑰𝒊 + 𝑰𝒌 + 𝒁𝒋𝒋 𝑰𝒋 − 𝑰𝒌 + ⋯ + 𝒁𝒋𝒏𝑰𝒏= 𝑰𝟏𝒁𝒊𝟏 + 𝒁𝒊𝟐𝑰𝟐 + ⋯ + 𝒁𝟏𝒊 𝑰𝒊 + 𝑰𝒌 + 𝒁𝒊𝒋 𝑰𝒋 − 𝑰𝒌 + ⋯

+ 𝒁𝒊𝒏𝑰𝒏 + 𝑰𝒌𝒁𝒃

Rearranging, we have

0 = 𝒁𝒊𝟏 − 𝒁𝒋𝟏 𝑰𝟏 + ⋯ + 𝒁𝒊𝒊 − 𝒁𝒋𝒊 𝑰𝒊 + 𝒁𝒊𝒋 − 𝒁𝒋𝒋 𝑰𝒋 + ⋯ +

𝒁𝒊𝒏 − 𝒁𝒋𝒏 𝑰𝒏 + 𝒁𝒃 + 𝒁𝒊𝒊 + 𝒁𝒋𝒋 − 𝒁𝒊𝒋 − 𝒁𝒋𝒊 𝑰𝒌

Collecting equations similar to case 2 and case 3, we can write

𝑽𝟏

𝑽𝟐

⋮𝑽𝒏

𝒁𝟏𝒊 − 𝒁𝟏𝒋

𝒁𝒊𝒊 − 𝒁𝒊𝒋

⋮𝒁𝒏𝒊 − 𝒁𝒏𝒋

𝒁𝒊𝟏 − 𝒁𝒋𝟏 𝒁𝒊𝒊 − 𝒁𝒋𝒊 ⋯ 𝒁𝒊𝒏 − 𝒁𝒋𝒏 𝒁𝒃𝒃

Where:

𝒁𝒃𝒃 = 𝒁𝒃 + 𝒁𝒊𝒊 + 𝒁𝒋𝒋 − 𝟐𝒁𝒊𝒋

Eliminate Ik in the set of equations contained in the matrix operation

𝒁𝒃𝒖𝒔 (𝒏𝒆𝒘) = 𝒁𝒃𝒖𝒔 (𝒐𝒓𝒊𝒈) −𝟏

𝒁𝒃 + 𝒁𝒊𝒊 + 𝒁𝒋𝒋 − 𝟐𝒁𝒊𝒋

𝒁𝟏𝒊 − 𝒁𝟏𝒋 ⋯ 𝒁𝒏𝒊 − 𝒁𝒏𝒋

Or simply,

𝒁𝒃 + 𝒁𝒊𝒊 + 𝒁𝒋𝒋 − 𝟐𝒁𝒊𝒋

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.

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.

Req’d: V4 Sol’n: 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.

𝑍𝑏𝑢𝑠 𝑜𝑟𝑖𝑔 =

𝑗0.4774 𝑗0.3706𝑗0.3706 𝑗0.4872

𝑗0.4020 𝑗0.4142𝑗0.3922 𝑗0.4126

𝑗0.4020 𝑗0.3922𝑗0.4142 𝑗0.4126

𝑗0.4558 𝑗0.4232𝑗0.4232 𝑗0.4733

Then apply case 2, we have

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

𝑗0.4774 𝑗0.3706

𝑗0.3706 𝑗0.4872

𝑗0.4020 𝑗0.4142

𝑗0.3922 𝑗0.4126𝑗0.4020 𝑗0.3922

𝑗0.4142 𝑗0.4126

𝑗0.4558 𝑗0.4232

𝑗0.4232 𝑗0.4733

𝒁𝟏𝒋

𝒁𝟐𝒋

⋮𝒁𝒏𝒋

𝑰𝟏𝑰𝟐𝑰𝟑𝑰𝟒𝑰𝒌

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

𝑗0.4774 𝑗0.3706

𝑗0.3706 𝑗0.4872

𝑗0.4020 𝑗0.4142

𝑗0.3922 𝑗4126𝑗4020 𝑗0.3922

𝑗0.4142 𝑗0.4126

𝑗0.4558 𝑗0.4232

𝑗0.4232 𝑗0.4733

𝒋𝟎. 𝟒𝟏𝟒𝟐

𝒋𝟎. 𝟒𝟏𝟐𝟔𝒋𝟎. 𝟒𝟐𝟑𝟐

𝒋𝟎. 𝟒𝟕𝟑𝟑𝒋𝟎. 𝟒𝟏𝟒𝟐 𝒋𝟎. 𝟒𝟏𝟐𝟔 𝒋𝟎. 𝟒𝟐𝟑𝟐 𝒋𝟎. 𝟒𝟕𝟑𝟑 𝒋𝟎. 𝟒𝟕𝟑𝟑 + (−𝒋𝟓)

𝑽𝟏

𝑽𝟐

𝑽𝟑

𝑽𝟒

𝑗0.4774 𝑗0.3706

𝑗0.3706 𝑗0.4872

𝑗0.4020 𝑗0.4142

𝑗0.3922 𝑗0.4126𝑗0.4020 𝑗0.3922

𝑗0.4142 𝑗0.4126

𝑗0.4558 𝑗0.4232

𝑗0.4232 𝑗0.4733

𝒋𝟎. 𝟒𝟏𝟒𝟐

𝒋𝟎. 𝟒𝟏𝟐𝟔𝒋𝟎. 𝟒𝟐𝟑𝟐

𝒋𝟎. 𝟒𝟕𝟑𝟑𝒋𝟎. 𝟒𝟏𝟒𝟐 𝒋𝟎. 𝟒𝟏𝟐𝟔 𝒋𝟎. 𝟒𝟐𝟑𝟐 𝒋𝟎. 𝟒𝟕𝟑𝟑 −𝒋𝟒. 𝟓𝟐𝟔𝟕

The terms in the fifth row and column were obtained by repeating the forh row and column of Zorig and noting that

𝒁𝟓𝟓 = 𝒁𝒋𝒋 + 𝒁𝒃 = 𝒁𝟒𝟒 + 𝒁𝒃 = 𝒋𝟎. 𝟒𝟕𝟑𝟑 + −𝒋𝟓 = −𝒋𝟒. 𝟓𝟐𝟔𝟕

Then eliminating the fifth row and column, using the formula,

we obtain for Zbus new,

𝒁𝟏𝟏 (𝒏𝒆𝒘) = 𝑗0.4774 −𝑗0.4142 𝑗0.4142

−𝑗4.5267= 𝒋𝟎. 𝟓𝟏𝟓𝟑

𝒁𝟏𝟐 (𝒏𝒆𝒘) = 𝑗0.3706 −𝑗0.4142 𝑗0.4126

−𝑗4.5267= 𝒋𝟎. 𝟒𝟎𝟖𝟒

𝒁𝟏𝟑 (𝒏𝒆𝒘) = 𝑗0.4020 −𝑗0.4142 𝑗0.4232

−𝑗4.5267= 𝒋𝟎. 𝟒𝟒𝟎𝟕

𝒁𝟏𝟒 (𝒏𝒆𝒘) = 𝑗0.4142 −𝑗0.4142 𝑗0.4733

−𝑗4.5267= 𝒋𝟎. 𝟒𝟓𝟕𝟓

𝒁𝟐𝟏 (𝒏𝒆𝒘) = 𝑗0.3706 −𝑗0.4142 𝑗0.4142

−𝑗4.5267= 𝒋𝟎. 𝟓𝟏𝟓𝟑

𝒁𝟐𝟐 (𝒏𝒆𝒘) = 𝑗0.4872 −𝑗0.4142 𝑗0.4126

−𝑗4.5267= 𝒋𝟎. 𝟓𝟐𝟒𝟖

𝒁𝟐𝟑 (𝒏𝒆𝒘) = 𝑗0.3922 −𝑗0.4142 𝑗0.4232

−𝑗4.5267= 𝒋𝟎. 𝟒𝟑𝟎𝟖

𝒁𝟐𝟒 (𝒏𝒆𝒘) = 𝑗0.4126 −𝑗0.4142 𝑗0.4733

−𝑗4.5267= 𝒋𝟎. 𝟒𝟓𝟓𝟕

𝒁𝟑𝟏 𝒏𝒆𝒘 = 𝑗0.4020 −𝑗0.4142 𝑗0.4142

−𝑗4.5267= 𝒋𝟎. 𝟒𝟒𝟎𝟕

𝒁𝟑𝟐 (𝒏𝒆𝒘) = 𝑗0.3922 −𝑗0.4142 𝑗0.4126

−𝑗4.5267= 𝒋𝟎. 𝟒𝟑𝟎𝟖

𝒁𝟑𝟑 (𝒏𝒆𝒘) = 𝑗0.4558 −𝑗0.4142 𝑗0.4232

−𝑗4.5267= 𝒋𝟎. 𝟒𝟗𝟓𝟒

𝒁𝟑𝟒 (𝒏𝒆𝒘) = 𝑗0.4232 −𝑗0.4142 𝑗0.4733

−𝑗4.5267= 𝒋𝟎. 𝟒𝟔𝟕𝟒

𝒁𝟒𝟏 𝒏𝒆𝒘 = 𝑗0.4142 −𝑗0.4142 𝑗0.4142

−𝑗4.5267= 𝒋𝟎. 𝟒𝟓𝟕𝟓

𝒁𝟒𝟐 (𝒏𝒆𝒘) = 𝑗0.4126 −𝑗0.4142 𝑗0.4126

−𝑗4.5267= 𝒋𝟎. 𝟒𝟓𝟓𝟕

𝒁𝟒𝟑 (𝒏𝒆𝒘) = 𝑗0.4232 −𝑗0.4142 𝑗0.4232

−𝑗4.5267= 𝒋𝟎. 𝟒𝟔𝟕𝟒

𝒁𝟒𝟒 (𝒏𝒆𝒘) = 𝑗0.4733 −𝑗0.4142 𝑗0.4733

−𝑗4.5267= 𝒋𝟎. 𝟓𝟐𝟐𝟖

Then we’ll have,

𝒁𝒃𝒖𝒔 𝒏𝒆𝒘 =

𝒋𝟎. 𝟓𝟏𝟓𝟑 𝒋𝟎.𝟒𝟎𝟖𝟒𝒋𝟎. 𝟒𝟎𝟖𝟒 𝒋𝟎.𝟓𝟐𝟒𝟖

𝒋𝟎. 𝟒𝟒𝟎𝟕 𝒋𝟎.𝟒𝟓𝟕𝟓𝒋𝟎. 𝟒𝟑𝟎𝟖 𝒋𝟎.𝟒𝟓𝟓𝟕

𝒋𝟎. 𝟒𝟒𝟎𝟕 𝒋𝟎.𝟒𝟑𝟎𝟖𝒋𝟎. 𝟒𝟓𝟕𝟓 𝒋𝟎.𝟒𝟓𝟓𝟕

𝒋𝟎. 𝟒𝟗𝟓𝟒 𝒋𝟎.𝟒𝟔𝟕𝟒𝒋𝟎. 𝟒𝟔𝟕𝟒 𝒋𝟎.𝟓𝟐𝟐𝟖

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. 𝑽 𝟒 (𝒏𝒆𝒘) = 𝑍41𝐼1 + 𝑍42𝐼2 + 𝑍43𝐼3

= 𝑗0.4575 −𝑗1.2 + 𝑗0.4575 0.72 − 𝑗0.96 + 𝑗0.4674 −𝑗1.2 Hence;

𝑽 𝟒 (𝒏𝒆𝒘) = 𝟏. 𝟓𝟖𝟐∠ − 𝟏𝟏. 𝟗𝟕𝒑𝒖

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

𝑽 𝟏 = 𝒁𝒂𝑰 𝟏 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

𝑽 𝟏

𝑽 𝟐 =

𝒁𝒂 𝟎𝟎 𝒁𝒃

𝑰 𝟏

𝑰 𝟐

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.

EXAMPLE: 1. Determine the Zbus for the network shown below where impedances are shown

in per unit. Preserve all three nodes.

j 0.15

Sol’n:

1. We start by establishing bus 1 with the impedance to the reference bus and write

𝑽 𝟏 = 𝒁𝟎𝟏𝑰 𝟏 = 𝒋𝟏. 𝟐 𝑰 𝟏

Then we have 1x1 bus impedance matrix,

𝒁𝒃𝒖𝒔 = 𝒋𝟏. 𝟐 2. To establish bus 2 with its impedance to bus 1, we follow

𝒁𝒃𝒖𝒔 = 𝒛𝟎𝟏 𝒛𝟎𝟏

𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟐 =

𝒁𝟏𝟏 𝒁𝟏𝟐

𝒁𝟏𝟐 𝒁𝟐𝟐 =

𝒋𝟏. 𝟐 𝒋𝟏.𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟐 + 𝒋𝟎. 𝟐

𝒁𝒃𝒖𝒔 = 𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏. 𝟒

Note: Elements Z01 in the new row and column is repitition of the elements of row 1 and column 1 being modified.

3. To establish bus 3 with its impedance to bus 1, we follow

𝒁𝒃𝒖𝒔 =

𝒛𝟎𝟏 𝒛𝟎𝟏 𝒛𝟎𝟏

𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟐 𝒛𝟎𝟏

𝒛𝟎𝟏 𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟑

𝒁𝟏𝟏 𝒁𝟏𝟐 𝒁𝟏𝟑

𝒁𝟐𝟏 𝒁𝟐𝟐 𝒁𝟐𝟑

𝒁𝟑𝟏 𝒁𝟑𝟐 𝒁𝟑𝟑

𝒋𝟏.𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟐𝒋𝟏.𝟐 𝒋𝟏. 𝟒 𝒋𝟏. 𝟐𝒋𝟏.𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟐 + 𝒋𝟎.𝟑

𝒁𝒃𝒖𝒔 =

𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟐𝒋𝟏.𝟐 𝒋𝟏.𝟒 𝒋𝟏.𝟐𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟓

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

𝒁𝒃𝒖𝒔 =

𝒛𝟎𝟏 𝒛𝟎𝟏 𝒛𝟎𝟏

𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟐 𝒛𝟎𝟏

𝒛𝟎𝟏 𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟑

𝒛𝟎𝟏

𝒛𝟎𝟏 + 𝒛𝟏𝟑

𝒛𝟎𝟏 𝒛𝟎𝟏 𝒛𝟎𝟏 + 𝒛𝟏𝟑 𝒛𝟎𝟏 + 𝒛𝟏𝟑 + 𝒛𝟑𝟎

𝒁𝟏𝟏 𝒁𝟏𝟐

𝒁𝟐𝟏 𝒁𝟐𝟐

𝒁𝟏𝟑 𝒁𝟏𝟒

𝒁𝟐𝟑 𝒁𝟐𝟒

𝒁𝟑𝟏 𝒁𝟑𝟐

𝒁𝟒𝟏 𝒁𝟒𝟐

𝒁𝟑𝟑 𝒁𝟑𝟒

𝒁𝟒𝟑 𝒁𝟑𝟑 + 𝒛𝟑𝟎

𝒁𝟏𝟏 𝒁𝟏𝟐

𝒁𝟐𝟏 𝒁𝟐𝟐

𝒁𝟏𝟑 𝒁𝟏𝟒

𝒁𝟐𝟑 𝒁𝟐𝟒

𝒁𝟑𝟏 𝒁𝟑𝟐

𝒁𝟒𝟏 𝒁𝟒𝟐

𝒁𝟑𝟑 𝒁𝟑𝟒

𝒁𝟒𝟑 𝒁𝟒𝟒

𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟐𝒋𝟏.𝟐 𝒋𝟏.𝟒 𝒋𝟏.𝟐𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟓

𝒋𝟏. 𝟐𝒋𝟏. 𝟐𝒋𝟏. 𝟓

𝒋𝟏.𝟐 𝒋𝟏.𝟐 𝒋𝟏.𝟓 𝒋𝟏. 𝟓 + 𝒋𝟏.𝟓

𝒁𝒃𝒖𝒔 =

𝒋𝟏. 𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟐𝒋𝟏. 𝟐 𝒋𝟏. 𝟒 𝒋𝟏. 𝟐𝒋𝟏. 𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟓

𝒋𝟏.𝟐𝒋𝟏.𝟐𝒋𝟏.𝟓

𝒋𝟏. 𝟐 𝒋𝟏. 𝟐 𝒋𝟏. 𝟓 𝒋𝟑

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

𝒁𝟏𝟏 (𝒏𝒆𝒘) = 𝑗1.2 −𝑗1.2 𝑗1.2

𝑗3= 𝒋𝟎. 𝟕𝟐

𝒁𝟏𝟐 (𝒏𝒆𝒘) = 𝑗1.2 −𝑗1.2 𝑗1.2

𝑗3= 𝒋𝟎. 𝟕𝟐

𝒁𝟏𝟑 (𝒏𝒆𝒘) = 𝑗1.2 −𝑗1.2 𝑗1.5

𝑗3= 𝒋𝟎. 𝟔𝟎

𝒁𝟐𝟏 (𝒏𝒆𝒘) = 𝑗1.2 −𝑗1.2 𝑗1.2

𝑗3= 𝒋𝟎. 𝟕𝟐

𝒁𝟐𝟐 (𝒏𝒆𝒘) = 𝑗1.4 −𝑗1.2 𝑗1.2

𝑗3= 𝒋𝟎. 𝟗𝟐

𝒁𝟐𝟑 (𝒏𝒆𝒘) = 𝑗1.2 −𝑗1.2 𝑗1.5

𝑗3= 𝒋𝟎. 𝟔𝟎

𝒁𝟑𝟏 𝒏𝒆𝒘 = 1.2 −𝑗1.2 𝑗1.5

𝑗3= 𝒋𝟎. 𝟔𝟎

𝒁𝟑𝟐 (𝒏𝒆𝒘) = 1.2 −𝑗1.2 𝑗1.5

𝑗3= 𝒋𝟎. 𝟔𝟎

𝒁𝟑𝟑 (𝒏𝒆𝒘) = 1.5 −𝑗1.5 𝑗1.5

𝑗3= 𝒋𝟎. 𝟕𝟓

When all the elements are determined, we have

𝒋𝟎. 𝟕𝟐 𝒋𝟎. 𝟕𝟐 𝒋𝟎.𝟔𝟎𝒋𝟎. 𝟕𝟐 𝒋𝟎. 𝟗𝟐 𝒋𝟎.𝟔𝟎𝒋𝟎. 𝟔𝟎 𝒋𝟎. 𝟔𝟎 𝒋𝟎.𝟕𝟓

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.

𝑽𝟏

𝑽𝟐

⋮𝑽𝒏

𝒁𝒊𝒊 − 𝒁𝒊𝒋

𝒁𝒊𝟏 − 𝒁𝒋𝟏 𝒁𝒊𝒊 − 𝒁𝒋𝒊 ⋯ 𝒁𝒊𝒏 − 𝒁𝒋𝒏 𝒁𝒃𝒃

where:

𝒁𝒃𝒃 = 𝒁𝒃 + 𝒁𝒊𝒊 + 𝒁𝒋𝒋 − 𝟐𝒁𝒊𝒋

which is 𝒁𝟒𝟒 = 𝒛𝟐𝟑 + 𝒁𝟐𝟐 + 𝒁𝟑𝟑 − 𝟐𝒁𝟐𝟑

= 𝑗0.15 + 𝑗0.92 + 𝑗0.75 − 2 𝑗0.60

𝒁𝟒𝟒 = 𝒋 𝟎. 𝟔𝟐

𝒁𝟏𝟒 = 𝒁𝟏𝟐 − 𝒁𝟏𝟑 = 𝒋𝟎. 𝟕𝟐 − 𝒋𝟎. 𝟔𝟎 = 𝒋𝟎. 𝟏𝟐 𝒁𝟐𝟒 = 𝒁𝟐𝟐 − 𝒁𝟐𝟑 = 𝒋𝟎. 𝟗𝟐 − 𝒋𝟎. 𝟔𝟎 = 𝒋𝟎. 𝟑𝟐

𝒁𝟑𝟒 = 𝒁𝟑𝟐 − 𝒁𝟑𝟑 = 𝒋𝟎. 𝟔𝟎 − 𝒋𝟎. 𝟕𝟓 = −𝒋𝟎. 𝟏𝟓 So we write,

𝒁𝒃𝒖𝒔 =

𝒋𝟎. 𝟕𝟐 𝒋𝟎. 𝟕𝟐 𝒋𝟎.𝟔𝟎𝒋𝟎. 𝟕𝟐 𝒋𝟎. 𝟗𝟐 𝒋𝟎.𝟔𝟎𝒋𝟎. 𝟔𝟎 𝒋𝟎. 𝟔𝟎 𝒋𝟎.𝟕𝟓

𝒋𝟏. 𝟐𝒋𝟎. 𝟑𝟐−𝒋𝟎.𝟏𝟓

𝒋𝟏. 𝟐 𝒋𝟎. 𝟑𝟐 −𝒋𝟎.𝟏𝟓 𝒋𝟎. 𝟔𝟐

7. Repeating procedure 5. Eliminating row 4 and column 4 of 4x4 to get 3x3 matrix. We find,

𝒋𝟎. 𝟔𝟗𝟔𝟖 𝒋𝟎. 𝟔𝟓𝟖𝟏 𝒋𝟎. 𝟔𝟐𝟗𝟎𝒋𝟎. 𝟔𝟓𝟖𝟏 𝒋𝟎. 𝟕𝟓𝟒𝟖 𝒋𝟎. 𝟔𝟕𝟕𝟒𝒋𝟎. 𝟔𝟐𝟗𝟎 𝒋𝟎. 𝟔𝟕𝟕𝟒 𝒋𝟎. 𝟕𝟏𝟑𝟕

which is the bus impedance matrix to be determined.

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