Advanced Power System

Network Admittance and Impedance Matrices

Overview

As we have seen in Chapter 1 that a power system network can be converted into an equivalent impedance diagram. This diagram forms the basis of power flow (or load flow) studies and short circuit analysis. In this chapter we shall discuss the formation of bus admittance matrix (also known as Y_bus matrix) and bus impedance matrix (also known as Z_bus matrix). These two matrices are related by

(3.1)

We shall discuss the formation of the Y_bus matrix first. This will be followed by the discussion of the formation of the Z_bus matrix.

Section I: Formation of Bus Admittance Matrix

Formation of Bus Admittance Matrix

Consider the voltage source V_S with a source (series) impedance of Z_S as shown in Fig. 3.1 (a). Using Norton's theorem this circuit can be replaced by a current source I_S with a parallel admittance of Y_S as shown in Fig. 3.1 (b). The relations between the original system and the Norton equivalent are

We shall use this Norton's theorem for the formulation of the Y_bus matrix.

Fig. 3.1 (a) Voltage source with a source impedance and (b) its Norton equivalent.

For the time being we shall assume the short line approximation for the formulation of the bus admittance matrix. We shall thereafter relax this assumption and use the π -representation of the network for power flow studies. Consider the 4-bus power system shown in Fig. 3.2. This contains two generators G₁ and G₂ that are connected through transformers T₁ and T₂ to buses 1 and 2. Let us denote the synchronous reactances of G₁ and G₂ by X_G1 and X_G2 respectively and the leakage reactances of T₁ and T₂ by X_T1 and X_T2 respectively. Let Z_ij, i = 1, ..., 4 and j = 1, ... , 4 denote the line impedance between buses i and j .

Fig. 3.2 Single-line diagram of a simple power network.

Then the system impedance diagram is as shown in Fig. 3.3 where Z₁₁ = j ( X_G1 + X_T1 ) and Z₂₂ = j ( X_G2 + X_T2).

Fig. 3.3 Impedance diagram of the power network of Fig. 3.2.

In this figure the nodes with the node voltages of V₁ to V₄ indicate the buses 1 to 4 respectively. Bus 0 indicates the reference node that is usually the neutral of the Y-connected system. The impedance diagram is converted into an equivalent admittance diagram shown in Fig. 3.4. In this diagram Y_ij = 1/ Z_ij, i = 1,..., 4 and j = 1, ... , 4. The voltage sources E_G1 and E_G2 are converted into the equivalent current sources I₁ and I₂ respectively using the Norton's theorem discussed before.

Fig. 3.4 Equivalent admittance diagram of the impedance of Fig. 3.3.

We would like to determine the voltage-current relationships of the network shown in Fig. 3.4. It is to be noted that this relation can be written in terms of the node (bus) voltages V₁ to V₄ and injected currents I₁ and I₂ as follows

(3.3)

or,

(3.4)

It can be easily seen that we get (3.1) from (3.3) and (3.4).

Consider node (bus) 1 that is connected to the nodes 2 and 3. Then applying KCL at this node we get

(3.5)

In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations

	(3.6)
	(3.7)
	(3.8)

Combining (3.5) to (3.8) we get

(3.9)

Comparing (3.9) with (3.3) we can write

(3.10)

In general the format of the Y_bus matrix for an n -bus power system is as follows

(3.11)

where

(3.12)

It is to be noted that Y_bus is a symmetric matrix in which the sum of all the elements of the k th column is Y_kk .

Example 3.1

Consider the impedance diagram of Fig. 3.2 in which the system parameters are given in per unit by

Z₁₁ = Z₂₂ = j0.25, Z₁₂ = j0.2, Z₁₃ = j0.25, Z₂₃ = Z₃₄ = j0.4 and Z₂₄ = j0.5

The system admittance can then be written in per unit as

Y₁₁ = Y₂₂ = - j4, Y₁₂ = - j5, Y₁₃ = - j4, Y₂₃ = Y₃₄ = - j2.5 and Y₂₄ = - j2

The Y_bus is then given from (3.10) as

per unit

Consequently the bus impedance matrix is given by

per unit

It can be seen that like the Y_bus matrix the Z_bus matrix is also symmetric.

Let us now assume that the voltages E_G1 and E_G2 are given by

The current sources I₁ and I₂ are then given by

We then get the node voltages from (3.4) as

per unit

Solving the above equation we get the node voltages as

per unit

Node Elimination by Matrix Partitioning

Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave. Let (3.3) be written as

(3.13)

In the above equation I_A is a vector containing the currents that are injected, I_x is a null vector and the Y_bus matrix is portioned with the matrices K , L and M . Note that the Y_bus matrix contains both L and L^T due to its symmetric nature.

We get the following two sets of equations from (3.13)

	(3.14)
	(3.15)

Substituting (3.15) in (3.14) we get

(3.16)

Therefore we obtain the following reduced bus admittance matrix

Example 3.2

Let us consider the system of Example 3.1. Since there is no current injection in either bus 3 or bus 4, from the Y_bus computed we can write

We then have

per unit

Substituting I₁ = 4Ð-60° per unit and I₂ = 4Ð-90° per unit we shall get the same values of V₁ and V₂ as given in Example 3.1.

Inspecting the reduced Y_bus matrix we can state that the admittance between buses 1 and 2 is - j 6.8978. Therefore the self admittance (the admittance that is connected in shunt) of the buses 1 and 2 is - j4 per unit (= - j 10.8978 + j 6.8978). The reduced admittance diagram obtained by eliminating nodes 3 and 4 is shown in Fig. 3.5. It is to be noted that the impedance between buses 1 and 2 is the Thevenin impedance between these two buses. The value of this impedance is 1/( - j 6.8978) = j 0.145 per unit.

Fig. 3.5 Reduced admittance diagram after the elimination of buses 3 and 4.

Node Elimination by Kron Reduction

Consider an equation of the form

(3.18)

where A is an ( n X n ) real or complex valued matrix, x and b are vectors in either Rⁿ or Cⁿ . assume that the b vector has a zero element in the n th row such that (3.18) is given as

(3.19)

We can then eliminate the k^th row and k^th column to obtain a reduced ( n - 1) number of equations of the form

(3.20)

The elimination is performed using the following elementary operations

(3.21)

Example 3.3

Let us consider the same system of Example 3.1. We would like to eliminate the last two rows and columns. Let us first eliminate the last row and last column. Some of the values are given below

In a similar way we can calculate the other elements. Finally eliminating the last row and last column, as all these elements are zero, we get the new Y_bus matrix as

Further reducing the last row and the last column of the above matrix using (3.21), we obtain the reduced Y_bus matrix given in Example 3.2.

Inclusion of Line Charging Capacitors

So far we have assumed that the transmission lines are modeled with lumped series impedances without the shunt capacitances. However in practice, the Y_bus matrix contains the shunt admittances for load flow analysis in which the transmission lines are represented by its π -equivalent. Note that whether the line is assumed to be of medium length or long length is irrelevant as we have seen in Chapter 2 how both of them can be represented in a p -equivalent.

Consider now the power system of Fig. 3.2. Let us assume that all the lines are represented in an equivalent- π with the shunt admittance between the line i and j being denoted by Y_chij . Then the equivalent admittance at the two end of this line will be Y_ch_ij/2. For example the shunt capacitance at the two ends of the line joining buses 1 and 3 will be Y_ch₁₃/2. We can then modify the admittance diagram Fig. 3.4 as shown in Fig. 3.6. The Y_bus matrix of (3.10) is then modified as

(3.22)

where

(3.23)

Fig. 3.6 Admittance diagram of the power system Fig. 3.2 with line charging capacitors.

Section II: Elements Of The Bus Impedance And Admittance Matrices

Equation (3.1) indicates that the bus impedance and admittance matrices are inverses of each other. Also since Y_bus is a symmetric matrix, Z_bus is also a symmetric matrix. Consider a 4-bus system for which the voltage-current relations are given in terms of the Y_bus matrix as

(3.24)

We can then write

(3.25)

This implies that Y₁₁ is the admittance measured at bus-1 when buses 2, 3 and 4 are short circuited. The admittance Y₁₁ is defined as the self admittance at bus-1. In a similar way the self admittances of buses 2, 3 and 4 can also be defined that are the diagonal elements of the Y_bus matrix. The off diagonal elements are denoted as the mutual admittances . For example the mutual admittance between buses 1 and 2 is defined as

(3.26)

The mutual admittance Y₁₂ is obtained as the ratio of the current injected in bus-1 to the voltage of bus-2 when buses 1, 3 and 4 are short circuited. This is obtained by applying a voltage at bus-2 while shorting the other three buses.

The voltage-current relation can be written in terms of the Z_bus matrix as

(3.27)

The driving point impedance at bus-1 is then defined as

(3.28)

i.e., the driving point impedance is obtained by injecting a current at bus-1 while keeping buses 2, 3 and 4 open-circuited. Comparing (3.26) and (3.28) we can conclude that Z₁₁ is not the reciprocal of Y₁₁ . The transfer impedance between buses 1 and 2 can be obtained by injecting a current at bus-2 while open-circuiting buses 1, 3 and 4 as

(3.29)

It can also be seen that Z₁₂ is not the reciprocal of Y₁₂ .

Section II: Modification of Bus Impedance Matrix

Modification of Bus Impedance Matrix

Equation (3.1) gives the relation between the bus impedance and admittance matrices. However it may be possible that the topology of the power system changes by the inclusion of a new bus or line. In that case it is not necessary to recompute the Y_bus matrix again for the formation of Z_bus matrix. We shall discuss four possible cases by which an existing bus impedance matrix can be modified.

Let us assume that an n -bus power system exists in which the voltage-current relations are given in terms of the bus impedance matrix as

(3.30)

The aim is to modify the matrix Z_orig when a new bus or line is connected to the power system.

Adding a New Bus to the Reference Bus

It is assumed that a new bus p ( p > n ) is added to the reference bus through an impedance Z_p . The schematic diagram for this case is shown in Fig. 3.7. Since this bus is only connected to the reference bus, the voltage-current relations the new system are

(3.31)

Fig. 3.7 A new bus is added to the reference bus.

Adding a New Bus to an Existing Bus through an Impedance

This is the case when a bus, which has not been a part of the original network, is added to an existing bus through a transmission line with an impedance of Z_b . Let us assume that p ( p > n ) is the new bus that is connected to bus k ( k < n ) through Z_b. Then the schematic diagram of the circuit is as shown in Fig. 3.8. Note from this figure that the current I_p flowing from bus p will alter the voltage of the bus k . We shall then have

(3.32)

In a similar way the current I_p will also alter the voltages of all the other buses as

(3.33)

Furthermore the voltage of the bus p is given by

(3.34)

Therefore the new voltage current relations are

(3.35)

It can be noticed that the new Z_bus matrix is also symmetric.

Fig. 3.8 A new bus is added to an existing bus through an impedance.

Adding an Impedance to the Reference Bus from an Existing Bus

To accomplish this we first assume that an impedance Z_b is added from a new bus p to an existing bus k . This can be accomplished using the method discussed in Section 3.3.2. Then to add this bus k to the reference bus through Z_b, we set the voltage V_p of the new bus to zero. However now we have an ( n + 1) X ( n + 1) Z bus matrix instead of an n X n matrix. We can then remove the last row and last column of the new Z_bus matrix using the Kron's reduction given in (3.21).

Adding an Impedance between two Existing Buses

Let us assume that we add an impedance Z_b between two existing buses k and j as shown in Fig. 3.9. Therefore the current injected into the network from the bus k side will be I_k- I_b instead of I_k. Similarly the current injected into the network from the bus j side will be I_j + I_b instead of I_j. Consequently the voltage of the i^th bus will be

(3.36)

Similarly we have

(3.37)

and

(3.38)

Fig. 3.9 An impedance is added between two existing buses.

We shall now have to eliminate I_b from the above equations. To do that we note from Fig. 3.9 that

(3.39)

Substituting (3.37) and (3.38) in (3.39) we get

(3.40)

We can then write the voltage current relations as

(3.41)

where

(3.42)

We can now eliminate the last row and last column using the Kron's reduction given in (3.21).

Direct Determination of Z_bus Matrix

We shall now use the methods given in Sections 3.3.1 to 3.3.4 for the direct determination of the Z_bus matrix without forming the Y_bus matrix first. To accomplish this we shall consider the system of Fig. 3.2 and shall use the system data given in Example 3.1. Note that for the construction of the Z bus matrix we first eliminate all the voltage sources from the system.

Step-1 : Start with bus-1. Assume that no other buses or lines exist in the system. We add this bus to the reference bus with the impedance of j 0.25 per unit. Then the Z_bus matrix is

(3.43)

Step-2 : We now add bus-2 to the reference bus using (3.31). The system impedance diagram is shown in Fig. 3.10. We then can modify (3.43) as

(3.44)

Fig. 3.10 Network of step-2.

Step-3 : We now add an impedance of j 0.2 per unit between buses 1 and 2 as shown in Fig. 3.11. The interim Z bus matrix is then obtained by applying (3.41) on (3.44) as

Eliminating the last row and last column using the Kron's reduction of (3.31) we get

(3.45)

Step-4 : We now add bus-3 to bus-1 through an impedance of j 0.25 per unit as shown in Fig. 3.12. The application of (3.35) on (3.45) will then result in the following matrix

Fig. 3.11 Network of step-3.

(3.46)

Fig. 3.12 Network of step-4.

Step-5 : Connect buses 2 and 3 through an impedance of j 0.4 per unit as shown in Fig. 3.13. The interim Z_bus matrix is then formed from (3.41) and (3.46) as

Fig. 3.13 Network of step-5.

Using the Kron's reduction we get the following matrix

(3.47)

Step-6 : We now add a new bus-4 to bus-2 through an impedance of j 0.5 as shown in Fig. 3.14. Then the application of (3.35) on (3.47) results in the following matrix

(3.48)

Fig. 3.14 Network of step-6.

Step-7 : Finally we add buses 3 and 4 through an impedance of j 0.4 to obtain the network of Fig. 3.3 minus the voltage sources. The application of (3.41) on (3.48) results in the interim Z_bus matrix of

Eliminating the 5^th row and column through Kron's reduction we get the final Z_bus as

(3.49)

The Z_bus matrix given in (3.49) is the as that given in Example 3.1 which is obtained by inverting the Y_bus matrix.

Section IV: Thevenin Impedance And Z_bus Matrix

To establish relationships between the elements of the Z_bus matrix and Thevenin equivalent, let us consider the following example.

Example 3.4

Consider the two bus power system shown in Fig. 3.15. It can be seen that the open-circuit voltages of buses a and b are V_a and V_b respectively. From (3.11) we can write the Y_bus matrix of the system as

Fig. 3.15 Two-bus power system of Example 3.4.

The determinant of the above matrix is

Therefore the Z_bus matrix is

Solving the last two equations we get

(3.50)

Now consider the system of Fig. 3.15. The Thevenin impedance of looking into the system at bus- a is the parallel combination of Z_aa and Z_ab + Z_bb, i.e.,

(3.51)

Similarly the Thevenin impedance obtained by looking into the system at bus- b is the parallel combination of Z_bb and Z_aa + Z_ab, i.e.,

(3.52)

Hence the driving point impedances of the two buses are their Thevenin impedances.

Let us now consider the Thevenin impedance while looking at the system between the buses a and b . From Fig. 3.15 it is evident that this Thevenin impedance is the parallel combination of Z_ab and Z_aa + Z_bb, i.e.,

With the values given in (3.50) we can write

Comparing the last two equations we can write

(3.53)

As we have seen in the above example in the relation V = Z_bus I , the node or bus voltages V_i, i = 1, ... , n are the open circuit voltages. Let us assume that the currents injected in buses 1, ... , k - 1 and k + 1, ... , n are zero when a short circuit occurs at bus k . Then Thevenin impedance at bus k is

(3.54)

From (3.51), (3.52) and (3.54) we can surmise that the driving point impedance at each bus is the Thevenin impedance.

Let us now find the Thevenin impedance between two buses j and k of a power system. Let the open circuit voltages be defined by the voltage vector V° and corresponding currents be defined by I° such that

(3.55)

Now suppose the currents are changed by ΔI such that the voltages are changed by ΔV . Then

(3.56)

Comparing (3.55) and (3.56) we can write

(3.57)

Let us now assume that additional currents ΔI_k and ΔI_k are injected at the buses k and j respectively while the currents injected at the other buses remain the same. Then from (3.57) we can write

(3.58)

We can therefore write the following two equations form (3.58)

The above two equations can be rewritten as

	(3.59)
	(3.60)

Since Z_jk = Z_kj the network can be drawn as shown in Fig. 3.16. By inspection we can see that the open circuit voltage between the buses k and j is