Power System Analysis

ELE401

Consider the voltage source VS with a

source (series) impedance of ZS as

shown in Fig.(a). Using Norton’s

theorem this circuit can be replaced by a

current source IS with a parallel

admittance of YS as shown in Fig.(b).

The relations between the original

system and the Norton equivalent are:

2

Consider the 4-bus power system shown in Fig. This contains two

generators G1 and G2 that are connected through transformers T1 and T2

to buses 1 and 2.

Denote the synchronous reactances of G1 and G2 by XG1 and XG2

respectively and the leakage reactances of T1 and T2 by XT1 and XT2

respectively. 3

Zij , i=1,…,4 and j=1,…,4

denote the line impedance

between buses i and j.

Z11=j(XG1+XT1).

Z22= j(XG2+XT2).

Then the system impedance diagram is as shown in Fig.

4

In this diagram Yij = 1/Zij ,

i =1,…,4 and j=1,…,4.

The voltage sources EG1 and

EG2 are converted into the

equivalent current sources I1

and I2 respectively using the

Norton’s theorem.

The impedance diagram is converted into an equivalent admittance

diagram shown in Fig.

5

We would like to determine the voltage-current relationships of the

network. 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:

In this example:

I3 =0 and I4 =0

6

The voltage-current relationships of the network can also be written as

follows:

In this example:

I3 =0 and I4 =0

7

Zbus is called Bus Impedance Matrix and Ybus is called Bus

Admittance Matrix.

It is too difficult to write Zbus of n-bus power system using the

impedance diagram, but there is a simple way to write Ybus of

power system using the admittance diagram.

To find Zbus of n-bus power system you should:

1) Draw admittance diagram of power system

2) Calculate Ybus of power system

3) Calculate the inverse of Ybus : (Zbus =Ybus -1

)

8

In general the format of the Ybus matrix for an n-bus power system

is as follows:

The main diagonal (leading

diagonal) of matrix includes:

Where Y’kk is the sum of all

admittances connected to bus k.

The values of –Y’kj in the other arrays are :

-1× (sum of all admittances between bus k and bus j)

9

Example 1 :Consider the impedance diagram of Fig. in which the

system parameters are given in per unit by:

Z11 = Z22 = j0.25 pu.

Z12 = j0.2 pu.

Z13 = j0.25 pu.

Z23 = Z34 = j0.4 pu

Z24 = j0.5pu.

Find the matrices Ybus and Zbus for this power system.

10

Solution : The system admittance can then be written in per unit as:

Y11 = Y22 = (j0.25) -1 pu = -j4 pu.

Y12 = (j0.2) -1 pu = -j5 pu.

Y13 = ( j0.25) -1 pu = -j4 pu.

Y23 = Y34 = (j0.4) -1 pu = -j2.5 pu.

Y24 = (j0.5) -1 pu = -j2 pu.

11

Now we can form the admittance matrix as follow:

12

Consequently the bus impedance matrix can be computed by:

Zbus =Ybus -1

It can be seen

that like the Ybus

matrix the Zbus

matrix is also

symmetric.

13

Example 2: In Example 1 , assume that the voltages EG1 and EG2

are given by , find the voltage

of each bus in perunit.

14

Solution:

The current sources I1 and I2 are given by:

15

Zbus is already calculated

in Example 1.

16

Example 3: Consider the impedance diagram of Fig. , assume

that , find the voltage of each bus in perunit.

17

Zg=Rg+jXg=0.05+j0.2 pu

ZT=RT+jXT=0.01+j0.07 pu

Z11=Zg+ZT=0.06+j0.27 pu

ZTL=RTL+jXTL=1.5+j2.5 pu

Z12 = Z21= ZTL =1.5+j2.5 pu

ZLoad=RLoad+jXLoad=1.2+j2 pu

Z22=ZLoad=1.2+j2.0 pu

Solution:

18

Z11=0.06+j0.27 pu Y11 = (0.06+j0.27)-1 = 0.784 - j3.53 pu

Z12 = Z21= 1.5+j2.5 pu Y12=Y21 =(1.5+j2.5)-1= 0.176 - j0.294 pu

Z22=1.2+j2.0 pu Y22 =(1.2+j2.0)-1 = 0.22 - j0.368 pu

Admittance diagram of power system

1

19

pu

1

20

-2.096-j2.046

21

pu

22

1

23