The Impedance Model and Network Calculation

• The Bus Admittance and Impedance Matrices

• Thevenin’s Theorem and Zbus

• Modification of An Existing Zbus

• Direct Determination of Zbus

The Bus Admittance and Impedance Matrixes 1

Properties:

1. It is symmetrical.

2. It is a highly dense matrix .

3. Zbus matrix is very useful in the fault analysis.

1bus bus

Z Y

1. The impedance elements of Zbus on the principal diagonal are called driving-point impedances of the buses

2. Off-diagonal elements are called the transfer impedances of the buses

The Bus Admittance and Impedance Matrixes 2

Analyze the elements of Y bus:

A three bus example:

+

-

1

3

1I

2I

3I2V

02

112 31

| VVV

IY

02

222 31

| VVV

IY

0ijY If i and j are not directly connected

2

busI Y V

2 21 1 22 2 23 3I Y V Y V Y V

If V1 and V3 are reduced to zero by shorting buses 1 and 2 to the reference node and V2

is applied at bus 2…

1 11 1 12 2 13 3I Y V Y V Y V

The Bus Admittance and Impedance Matrixes 3

Analyze the elements of Z bus

02

222 31

| III

VZ

02

332 31

| III

VZ

02

112 31

| III

VZ

Zij means bus i is isolated from bus j

+

-

1 2

3

3I 2V1V

3V

+

-

+

-

1I 2I

busV Z I

1 11 1 32 2 33 3V Z I Z I Z I

2 21 1 22 2 23 3V Z I Z I Z I

3 31 1 32 2 33 3V Z I Z I Z I

Thevenin’s Theorem and Zbus 1

From the definition for Thevenin’s equivalent impedance, at bus 2

22Z

iiZ

jiijZ

It equals to in Z bus

----- driving -point impedance (Thevenin’s

equivalent , impedance at bus i)

------ transfer impedance

+

-

1 2

3

3I 2V1V

3V

+

-

+

-

2I1I

1 3

20

2

|ii th I I

VZ Z

I

Figure 8.3

3V

2V

1V

nV

kV

kIReference

OriginalNetwork

busZ

3

2

1

0

n

k

(a)

kI

Originalnetwork

busZ+

-

okV

th kkZ Z

Reference0

k

kV

+

-

(b)

Figure 8.3

th kkZ Z

Thevenin impedance between two buses j and k of the network.

Figure 8.4

3V

2V

1VjV jI

Reference

OriginalNetwork

busZ

3

2

1

0

k

j

(a)

kV kI

(b)

Figure 8.4

jI

Reference

OriginalNetwork

busZ

0

k

j

kI+

+

0kV0jV

kk kjZ Z

jj jkZ Zjk kjZ Z

k

j

k

jbZscI scI

(c) (d), 2th jk jj kk jkZ Z Z Z

sc j kI I I 0 0

,

k j k jb

th jk b b

V V V VI

Z Z Z

Thevenin’s Theorem and Z bus 4

Example 8.1

A capacitor having a reactance of 5.0 per unit is connected between the reference node and bus of a given circuit . The original emfs and the corresponding external current injection at buses and are same as in those examples. Find the current drawn by the capacitor .

Solution :

43

4

4

0

-j1.43028135738 110 74660. .

+

-

V4

5.0j

Z4

0

j0.69890

0.94866 20.7466

+

-

V4

44Z

capI 5.0j

(a)

Figure 8.5

capI

07466.20

40V

4V

4V

(b)

0.94866 20.74660.22056 69.2534

5.0 0.69890capI puj j

Thevenin’s Theorem and Z bus 6

Example 8.2

If an additional current equal to is injected into

the network at bus of Example 7.6, find the resulting voltage at bus , , , and .

1 2 3

4

0.22056 69.2534

0.73128 0.69140 0.61323 0.63677

0.69140 0.71966 0.60822 0.64178

0.61323 0.60822 0.69890 0.55110

0.63677 0.64178 0.55110 0.69890

bus

j j j j

j j j j

j j j j

j j j j

Z

1 14 0.22056 69.2534 0.63677 0.14045 20.7466capV I Z j

2 24 0.22056 69.2534 0.64178 0.14155 20.7466capV I Z j

3 34 0.22056 69.2534 0.55110 0.12155 20.7466capV I Z j

4 44 0.22056 69.2534 0.69890 0.15415 20.7466capV I Z j

Thevenin’s Theorem and Z bus 7

1

2

3

4

0.96903 18.4189 0.14045 20.7466

0.96734 18.6028 0.14155 20.7466

0.99964 15.3718 0.12155 20.7466

0.94866 20.7466 0

1.10938 18.7135

1.10880 1

.15415 20.

8.8764

1.12071 15.9539

746

V

V

V

V

1.1026 81 20.7466

Modification of An Existing Zbus 1

How an existing Z bus may be modified to add new buses or to connect new lines to established

buses?

Case 1:

Adding from a new bus to the reference node

The addition of the new bus connected to the reference node through without a connection to any of the buses of the original network cannot alter the original bus voltages when a current Ip is injected at the new bus.

bZ p

bZ

Modification of An Existing Zbus 2

The voltage at the new bus is equal to .

p

N

bp

N

I

I

I

I

ZV

V

V

V

2

1

0

02

01

000

0

0

0

.........

origZ

...pV bpZI

Increase the rank of the Z bus

)(newbusZ

Modification of An Existing Zbus 3

Case 2 : Adding from a new bus to an existing bus p

bZ

kI

pI

k

0

p

Reference

Original network with bus and the reference node extracted

k

Go next slide

bZ k

The current Ip flow into the network at bus will increase the original voltage by the voltage :

k0

kV kkpZI

Modification of An Existing Zbus 4

kkpkk ZIVV 0

pV kV bpZI

origZ bV

will be larger than the new by the voltage :

bpkkpkp ZIZIVV 0

Substituting for kV

kV 0

The new now which must be added to in order to find is

Modification of An Existing Zbus 5

Since must be a square matrix a round the principal diagonal , a new column which is the transpose of the new row must be added .

p

N

bkkkNkk

Nk

k

k

p

N

I

I

I

I

ZZZZZ

Z

Z

Z

V

V

V

V

2

1

21

2

1

2

1

origZ

...

... ...

...

p

p

)(newbusZ

busZ

Increase the rank of the Z bus

Modification of An Existing Zbus 6

b

innhorighinewhi ZZ

ZZZZ

kk

)1()1()()(

Case 3 : Adding exiting bus to the reference mode

1 . Add a new bus connected through to bus

2 . Short circuit bus to the reference mode by letting equal zero to yield

the same matrix equation as in case 2 expect that is zero .

3 . Eliminate the (N+1) row and (N+1) column .Each element in the new matrix :

pV

bZ

hiZ

p

p

k

k

bZ

pV

Modification of An Existing Zbus 7

Case 4 : Adding between two exiting buses and

Extract these buses from the original network . bZ kj

bZkI

iI

0 Reference

Original network with bus and the reference node extracted

bI

bi II

bk II

j

k

kj

Modification of An Existing Zbus 8

The change in voltage at each bus cased by the injection at bus

and at bus is given by

The matrix equation

bI

aIbhkhih IZZV )(

j

k

k

b

N

k

j

bb

N

k

i

I

I

I

I

I

Z

V

V

V

V 11

0 (row j - row k) of

( col . j - col . k) oforigZ

origZ

origZ

Modification of An Existing Zbus 9

The coefficient of in the last row of previous matrix is denoted by

The new row is column minus column of with in the (N+1) row .

The new row is the transpose of the new column .

Eliminating the (N+1) row and (N+1) column each element in the

new matrix is

bI

jorigZk bbZ

)(newhiZ

bjkkk

iNNhhinewhi ZZZZ

ZZZZ

jj

2)1()1(

)(

bjkkkjjbjkthbb ZZZZZZZ 2,

Direct Determination of Zbus 1

1 . Start by writing the equation for one bus connected through a branch impedance

to the reference as

11 IZV a2 . Add a new bus connected to the first bus or to the reference node . For instance, if

the second bus is connected to the reference node through , the matrix equations :

2

1

2

1

0

0

I

I

Z

Z

V

V

b

a

3 . Add other buses and branch following the procedures described in

previous section .

aZ

bZ1

2

2

1

1

1

Direct Determination of Zbus 2

Example 8.4

Determine the Zbus

125.0j

2.0j

25.0j

25.1j

4.0j

25.1j

0

1 2 3

4

(1)

(2) (3)

(4)

(5)

(6)

Direct Determination of Zbus 3

11 25.1 IjV

1

1

25.11, jZbus

1

1

The establish bus with its impedance to bus , Follow CASE TWO :

5.125.1

25.125.12, jj

jjZbus

1

1

2

2

2 1

The term j1.50 above is the sum of j1.25 and j0.25 .

25.1j

25.0j

1 2

CASE ONE

25.1j

1

Direct Determination of Zbus 4

Bus with the impedance connecting it to bus is established by writing

90.150.125.1

50.150.125.1

25.125.125.1

3,

jjj

jjj

jjj

Zbus

1

2

2

2

1

3

3

The term j1.90 above is the sum of Z22 of the matrix being modified and the impedance Zb of the branch being connected to bus from bus .2 3

3

25.1j

25.0j

1 2

4.0j

3

CASE TWO

Direct Determination of Zbus 5

To add the impedance from bus to the reference node, follow

case three to connect a new bus through and obtain the impedance matrix

25.1jZb

15.390.150.125.1

90.190.150.125.1

50.150.150.125.1

25.125.125.125.1

4,

jjjj

jjjj

jjjj

jjjj

Zbus

1 2 3

3

2

1

p

p

3

pbZ

Where j3.15 above is the sum of .

bZZ 33

CASE THREE

25.1j

25.0j1

2

4.0j3

25.1j

Direct Determination of Zbus 6

After the row p and column p are eliminated , the new matrix is

75397.059524.049603.0

59524.078571.065476.0

49603.065476.075397.0

5,

jjj

jjj

jjj

Zbus

1 2

2

3

3

1

No change in rank

Direct Determination of Zbus 7

Some of the elements of the new matrix are calculates by :

75397.015.3

)25.1)(25.1(25.1)(11 j

j

jjjZ new

78571.015.3

)50.1)(50.1(50.1)(22 j

j

jjjZ new

15.3

)90.1)(50.1(50.1)(32)(23 j

jjjZZ newnew

59524.0j

Direct Determination of Zbus 8

To add the impedance from bus to establish bus , follow the

CASE TWO

95397.075397.059524.049603.0

75397.075397.059524.049603.0

59524.059524.078571.065476.0

49603.049603.065476.075397.0

6,

jjjj

jjjj

jjjj

jjjj

Zbus

1

1

2

2

3

3 4

4

The new diagonal element is the sum of Z33 of the previous matrix and Zb = j0.20

3 420.0jZb

CASE TWO

25.1j

25.0j1

2

4.0j

3

25.1j

4

Direct Determination of Zbus 9

Finally , to add impedance between buses and ,

follow the CASE FOUR

125.0jZb 2 4

15873.049603.065476.0141215 jjjZZZ

19047.059524.0.078571.0.0242225 jjjZZZ 15873.0.075397.059524.0343235 jjjZZZ

35873.095397.059524.0444245 jjjZZZ

bZZZZZ 24442255 2 125.0)}59524.0(2)95397.078571.0{( jj 67421.0j

CASE FOUR

25.1j

25.0j1

2

4.0j

3

25.1j

4

Direct Determination of Zbus 10

The 5*5 matrix is

67421.035873.015873.019147.015873.0

35873.0

15873.0

19047.0

15873.0

jjjjj

j

j

j

j6,busZ

q

q

Direct Determination of Zbus 11

After the row q and column q are eliminated, the bus impedance matrix to be determined is:

76310.066951.069659.058049.0

66951.071660.064008.053340.0

69659.064008.073109.060992.0

58044.053340.060992.071660.0

jjjj

jjjj

jjjj

jjjj

Zbus

1 2

1

2

3

3

4

4

No change in rank