Module 4 Analysis of Faulted Power System

Module 4

Short Circuit analysis

4.1 ZBUS formation without mutual coupling between ele-ments

For a network with ‘m’ buses and a reference bus, one can write a relation between bus currentsand bus voltages as

[IBUS] = [YBUS] [VBUS] (4.1)

Where,

IBUS is (m × 1) bus current injection vector

VBUS is (m × 1) bus voltage vector

YBUS is (m ×m) bus admittence matrixequation (4.1) can also be written as

[VBUS] = [ZBUS] [IBUS] (4.2)

Where,

ZBUS is m ×m bus impedance matrix and is given by,

[ZBUS] = [YBUS]−1

From equation (4.2) for the ith bus one can write

Vi = Zi1I1 + Zi2I2 +⋯ZiiIi +⋯ + ZimIm (4.3)

From equation (4.3), Zij can be written as

Zij =Vi

Ij∣Ik = 0; ∀ k = 1, 2, ⋯m, ≠ j

(4.4)

99

Zii =Vi

Ii∣Ik = 0; ∀ k = 1, 2, ⋯m, ≠ i

(4.5)

Following points should be noted for the ZBUS matrix

• Zij is the off-diagonal element of ZBUS matrix and is called the ‘open-circuit transferimpedance’ between ith and jth bus.

• Zii is the diagonal element of ZBUS matrix and is called the ‘open-circuit driving pointimpedance’ of ith bus.

• If the YBUS matrix is symmetrical, then the matrix ZBUS is also symmetrical i.e. Zik = Zki.

• Since in a power network each bus is connected to very few other buses,the YBUS matrix ofthe network has large number of zero elements and is therefore, sparse in nature. The ZBUS,matrix on the other hand, is invariably a full matrix.

The ZBUS matrix of a network can be found out by inverting the YBUS matrix of the network.This is not an efficient method as every time there is a modification in the network, the YBUS matrixis modified and inversion has to be done again to obtain the modified the ZBUS matrix.

A step-by-step ZBUS building algorithm overcomes these problems. It avoids the inversionprocess and network modifications are easily incorporated in the existing ZBUS.

Few terms need to be defined before the step by step process can be explained. These are :

• Graph : The graph of a network describes the geometrical structure of the network showingthe interconnections of network elements.

• Tree : A tree of a graph is a connected sub graph that connects all the nodes without forminga closed path or a loop. A graph can have a number of distinct trees.

• Branches : The elements of a tree are called branches. The number of branches ‘b’ of a treewith ‘n’ nodes, including reference, is given by

b = n − 1 (4.6)

• Links : The elements of a graph not included in the tree of the graph are called links. Eachlink is associated with a loop. If ‘e’ is the number of elements in a graph, then the number oflinks ‘`’ is given by

` = e − b = e − n + 1 (4.7)

The above definitions are explained with the help of illustrations as shown below :Fig. 4.1 is a single line diagram of a power system. It has 4 buses, bus(1) to bus(4) and six

elements element e1 to element e6 . In this figure, bus(0) is taken as the reference bus.Fig. 4.2 shows the graph of the network depicting the interconnection of the elements and the

reference node.

100

Figure 4.1: Single Line Diagram of a Power System

Figure 4.2: A graph of the Power system of Fig. 4.1

A tree of the graph of Fig. 4.2 is shown in Fig. 4.3. The branches and the links have been shownwith solid lines and dotted lines respectively.

Following points should be noted from Fig. 4.3 :

• The total number of nodes (including reference node) is 5 (i.e. n = 5)

• The number of branches is b = n − 1 = 5 − 1 = 4 . As can be as in Fig. 4.3 where

101

Figure 4.3: A tree of the graph of Fig. 4.2

e1, e2, e5, e6, are such a set of branches that form a tree of the graph.

• The total number of elements in the graph is e = 6 .

• The number of links is ` = e − n + 1 = 6 − 5 + 1 = 2. The two links in the graph are e3 ande4 shown with dotted lines in Fig. 4.3.

Figure 4.4: Partial network with ‘m’ buses

The bus impedance matrix is built up starting with a branch connected to the reference andsubsequently the elements are added one by one till all the nodes and elements are considered. Let

102

us assume that the ZBUS matrix for a partial network with ‘m’ buses and a reference bus ‘0’, asshown in Fig. 4.4, exists.

The bus voltages and bus currents for the partial network satisfy the relation

[VmBUS] = [Zm

BUS] [ImBUS] (4.8)

Where,Vm

BUS is m × 1 bus voltage vector

ImBUS is m × 1 bus current injection vector

ZmBUS is m ×m bus impedence matrix of the partial network

To build ZBUS, one element at a time is added to the partial network, till all the elements areadded to the network. The added element may be a branch or a link and hence the four possibleelement additions to a partial network are:

a. Addition of a branch between a new node and the reference

b. Addition of a branch between a new node and an existing node

c. Addition of a link between an existing node and the reference

d. Addition of a link between two existing nodes

Let us now discuss these four cases one-by-one in detail.

4.1.1 Addition of a branch between a new node and the reference node(case 1):

Fig. 4.5 shows the addition of a branch between a new node ‘q’ and the reference ‘0’.The addition ofa new node to the partial network increases the size of ZBUS to (m+1)×(m+1) with the additionof a new row and a new column corresponding to the new node ‘q’, Let the impedance of this branchbe zq0. The new network equation can be written as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

V2

⋮Vp⋮Vm⋯Vq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1p ⋯ Z1m ⋮ Z1q

Z21 Z22 ⋯ Z2p ⋯ Z2m ⋮ Z2q

⋮ ⋮Zp1 Zp2 ⋯ Zpp ⋯ Zpm ⋮ Zpq⋮

Zm1 Zm2 ⋯ Zmp ⋯ Zmm ⋮ Zmq⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯Zq1 Zq2 ⋯ Zqp ⋯ Zqm ⋮ Zqq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

I2

⋮Ip⋮Im⋯Iq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.9)

103

Figure 4.5: Addition of a branch between a new node and the reference

The addition of branch does not change the elements of the original matrix ZBUS. Only the elementsof the added new row and column corresponding to qth bus need to be calculated. Further,since thepower system elements are linear and bilateral, Zqi = Ziq, ∀ i = 1, 2, ⋯m.

Now since,

Zqq =Vq

Iq∣Ik = 0; ∀ k = 1, 2, ⋯m

a current source of Iq = 1 p.u is connected to the qth bus, with all the others buses open, and thevoltage of qth bus (Vq) is computed, as shown in Fig. 4.6.

From Fig. 4.6 one gets Vq = zq0Iq, and thus with Iq = 1 p.u.

Zqq =Vq

Iq∣Ik = 0; ∀ k = 1, 2, ⋯m,

= zqo

For finding out Zqi, a current source Ii = 1 p.u. is connected between ith bus and the referencebus with all other buses open circuited as shown in Fig. 4.7.

From Fig. 4.7 , Vq = 0, and hence with Ii = 1 p.u.

Zqi =Vq

Ii∣Ik = 0; ∀ k = 1, 2, ⋯m, ≠ i

= 0

This implies that all the off-diagonal elements Zq1, Zq2, ⋯ Zmq and Z1q, Z2q, ⋯ Zqm are equalto zero.

Hence, the modified ZBus matrix after addition of an element between the new bus ‘q’ and the

104

Figure 4.6: Calculation of Zqq for Case 1

reference bus ‘0′ is given as,

ZBUS =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1p ⋯ Z1m 0Z21 Z22 ⋯ Z2p ⋯ Z2m 0⋮Zp1 Zp2 ⋯ Zpp ⋯ Zpm 0⋮

Zm1 Zm2 ⋯ Zmp ⋯ Zmm 00 0 ⋯ 0 ⋯ 0 zqo

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.10)

4.1.2 Addition of a branch between a new node and an existing node(Case 2):

Let a branch with impedance zpq be connected between an existing node ‘p’ and a new node ‘q’ asshown in Fig. 4.8. In this case also, the size of ZBus matrix increases by one to (m+1)×(m+1) dueto the addition of a new node ’q’ to the network. The modified network equations can be written

105

Figure 4.7: Calculation of Zqi for case 1

as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

V2

⋮Vp⋮Vm⋯Vq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1p ⋯ Z1m ⋮ Z1q

Z21 Z22 ⋯ Z2p ⋯ Z2m ⋮ Z2q

⋮ ⋮Zp1 Zp2 ⋯ Zpp ⋯ Zpm ⋮ Zpq⋮ ⋮

Zm1 Zm2 ⋯ Zmp ⋯ Zmm ⋮ Zmq⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯Zq1 Zq2 ⋯ Zqp ⋯ Zqm ⋮ Zqq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

I2

⋮Ip⋮Im⋯Iq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.11)

Even after the addition of branch p-q, the original matrix ZmBus remains unchanged. Only the

additional elements corresponding to the qth row and column need to be calculated.

For calculating Zqq one can write

Zqq =Vq

Iq∣Ik = 0; ∀ k = 1, 2, ⋯m

To evaluate Zqq current source of Iq = 1 p.u is connected to the qth bus, with all the others busesopen circuited, and the voltage of qth bus Vq is computed, as shown in Fig. 4.9. From Fig. 4.9 with

106

Figure 4.8: Addition of a branch between an existing node ‘p’ and a new node ‘q’

Iq = 1 p.u. and Ik = 0, ∀ k = 1, 2, ⋯, m one can write,

V1 = Z1qIq = Z1q

V2 = Z2qIq = Z2q

⋮Vp = ZpqIq = Zpq

⋮Vm = ZmqIq = ZmqVq = ZqqIq = Zqq

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4.12)

From the Fig. 4.10, the voltages Vp and Vq can be related as

Vq = Vp − vpq = ZpqIq − zpq ipq = Zpq + zpq (4.13)

Because, from the Fig. 4.10 , ipq = −Iq = −1 pu and from equation (4.12) Vp = Zpq and Vq = Zqq.Thus,

Zqq = Zpq + zpq (4.14)

For calculating Zqi one can write

Zqi =Vq

Ii∣Ik = 0; ∀ k = 1, 2, ⋯m, ≠ i

Hence, to compute Zqi a current source of I1 = 1 p.u is connected to the ith bus, with all the others

107

Figure 4.9: Calculation of Zqq

Figure 4.10: Relation between Vp and Vq

buses open circuited, and the bus voltage Vi is computed for all the buses, as shown in Fig. 4.11.From equation (4.11) one gets

V1 = Z1iIi = Z1i

V2 = Z2iIi = Z2i

⋮Vp = ZpiIi = Zpi

⋮Vm = ZmiIi = ZmiVq = ZqiIi = Zqi

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

108

Figure 4.11: Calculation of Zqi for case 2

From Fig. 4.11, Vq = Vp as the current in the branch p − q is zero. Hence, from the above equationsone gets

Zqi = Zpi; ∀ i = 1, 2, ⋯m (4.15)

Hence, the modified ZBus matrix after addition of an element between an existing bus ‘p’ thenew bus ‘q’ is given as,

ZBUS =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1p ⋯ Z1m Z1p

Z21 Z22 ⋯ Z2p ⋯ Z2m Z2p

⋮Zp1 Zp2 ⋯ Zpp ⋯ Zpm Zpp⋮

Zm1 Zm2 ⋯ Zmp ⋯ Zmm ZmpZp1 Zp2 ⋯ Zpp ⋯ Zpm Zpq + zqp

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.16)

So far in this lecture, we have considered the cases of addition of branches only. In the next lecturewe will consider the case of addition of links.

109

4.1.3 Addition of a link between an existing node and the referencenode (Case 3):

When an element is connected between an existing node and the reference, it creates a loop andthus, the addition of this element is equivalent to the addition of a link. This will not generate anynew node and the size of modified ZBus matrix remains unchanged. However, all the elements aremodified and need to be recalculated. Let the added element,with an impedance of zqo, be connectedbetween an existing node ‘q’ and the reference node ‘0’ as shown in the Fig. 4.12. I` is the current

Figure 4.12: Addition of a link between an existing node ‘q’ and the reference

through the link as shown in the Fig. 4.12. This current modifies the current injected into qth busfrom Iq to Iq − I`. The modified network equations can be written as,

V1 = Z11I1 + Z12I2 +⋯ + Z1q(Iq − I`) +⋯ + Z1mIm⋮ ⋮Vq = Zq1I1 + Zq2I2 +⋯ + Zqq(Iq − I`) +⋯ + ZqmIm⋮ ⋮Vm = Zm1I1 + Zm2I2 +⋯ + Zmq(Iq − I`) +⋯ + ZmmIm

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4.17)

AlsoVq = zqoI` (4.18)

Substuting Vq from the equation (4.17) into the equation (4.18) one can write

Zq1I1 + Zq2I2 +⋯ + Zqq(Iq − I`) +⋯ + ZqmIm = zqoI`

110

or0 = −Zq1I1 − Zq2I2 −⋯ − ZqqIq +⋯ − ZqmIm + (Zqq + zqo)I` (4.19)

Equations equation (4.17) and equation (4.19) together form the set of (m + 1) simultaneousnetwork equations which can be expressed in matrix form as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

V2

⋮Vq⋮Vm⋯0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1q ⋯ Z1m ⋮ −Z1q

Z21 Z22 ⋯ Z2q ⋯ Z2m ⋮ −Z2q

⋮ ⋮Zq1 Zq2 ⋯ Zqq ⋯ Zqm ⋮ −Zqq⋮ ⋮

Zm1 Zm2 ⋯ Zmq ⋯ Zmm ⋮ −Zmq⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯−Zq1 −Zq2 ⋯ −Zqq ⋯ −Zqm ⋮ Zqq + zqo

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

I2

⋮Iq⋮Im⋯I`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.20)

The link current I` has to be eliminated and hence, the last row and column of modified ZBus matrixhave to be eliminated. The partitioned matrix relation of equation (4.20) can be written in compactform as: ⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

[VmBus]

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [

(m) (1)

(m) ZmBus [∆Z]

(1) [∆Z]T Z``]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

ImBus

I`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.21)

where,[∆Z] = [−Z1q − Z2q⋯− Zqq⋯− Zmm]

T

From equation (4.20) one can write

0 = [∆Z]T[ImBus] + Z``I`

or,

I` = −[∆Z]T[Im

Bus]Z``

(4.22)

From equation (4.21) one can also write,

[VmBus] = [Zm

Bus][IBus] + [∆Z]I` (4.23)

Substituting I` from equation (4.22) into equation (4.23) one obtains,

[VmBus] = [[Zm

Bus] −[∆Z][∆Z]T

Z``] [Im

Bus] (4.24)

111

Hence,[Vm

Bus] = [ZBus][ImBus] (4.25)

where

[ZBus] = [[ZmBus] −

[∆Z][∆Z]T

Z``] (4.26)

It is worth observing that the [ZBus] matrix is an m ×m matrix i.e. the size of the [ZBus] matrixdoes not increase when a link is added to the partial network of ‘m’ buses as no new node is created.

4.1.4 Addition of a link between two existing nodes (Case 4):

Let an element with impedance zpq be connected between two existing nodes ‘p’ and ‘q’. This is anaddition of a link as it forms a loop encompassing nodes ‘p’ and ‘q’ as shown in Fig. 4.13. Let I`

Figure 4.13: Addition of a link between two existing nodes ‘p’ and ‘q’

be the current through the link as shown in the Fig. 4.13. This link current changes the injectedcurrent at pth node from Ip to (Ip − I`) , while the injected current at node qth is modified from Iqto (Iq + I`). The modified network equations can be written as:

112

V1 = Z11I1 + Z12I2 +⋯ + Z1p(Ip − Il) +⋯ + Z1q(Iq + I`) +⋯ + Z1mIm⋮

Vp = Zp1I1 + Zp2I2 +⋯ + Zpp(Ip − Il) +⋯ + Zpq(Iq + I`) +⋯ + ZpmIm⋮

Vq = Zq1I1 + Zq2I2 +⋯ + Zqp(Ip − Il) +⋯ + Zqq(Iq + I`) +⋯ + ZqmIm⋮

Vm = Zm1I1 + Zm2I2 +⋯ + Zmp(Ip − Il) +⋯ + Zmq(Iq + I`) +⋯ + ZmmIm

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4.27)

Also from Fig. 4.13, the relation between Vp and Vq in terms of I` and zpq can be written as

Vp − Vq = zpqI` (4.28)

or0 = −Vp + Vq + zpqI` (4.29)

Substituting Vp and Vq from the equation (4.27) into the equation (4.29) in the following relationis obtained:

0 =(Zq1 − Zp1)I1 + (Zq2 − Zp2)I2 +⋯ + (Zqp − Zpp)Ip +⋯(Zqq −Zpq)Iq +⋯ + (Zqm − Zpm)Im+ (Zpp + Zqq − 2Zpq + zpq)I`

(4.30)

Equations (4.27) and (4.30) form a set of (m + 1) simultaneous equation which can be written inmatrix form as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

V2

⋮Vq⋮Vm⋯0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1q ⋯ Z1m ⋮ Z1`

Z21 Z22 ⋯ Z2q ⋯ Z2m ⋮ Z2`

⋮ ⋮Zq1 Zq2 ⋯ Zqq ⋯ Zqm ⋮ Zq`⋮ ⋮

Zm1 Zm2 ⋯ Zmq ⋯ Zmm ⋮ Zm`⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯Z`1 Z`2 ⋯ Z`q ⋯ Z`m ⋮ Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

I2

⋮Iq⋮Im⋯I`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.31)

Where,Z1` = Z`1 = (Z1q − Z1p) ; Z2` = Z`2 = (Z2q − Z2p)Zq` = Z`q = (Zqq − Zqp) ; Zm` = Z`m = (Zmq − Zmp)

also

and

113

Z`` = Zqq + Zpp − 2Zpq + zpq.For eliminating the link current I` equation (4.31) can be written in the compact form as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

VmBus

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= [

(m) (1)

(m) ZmBus [∆Z]

(1) [∆Z]T Z``]

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

ImBus

I`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.32)

where,[∆Z] = [(Z1q − Z1p)⋯(Zpq − Zpp)⋯(Zqq − Zqp)⋯(Zmq − Zmp)]

T

Now using equation (4.26), the [ZBus] matrix, after the elemination of the link current I`, can bedetermined. It is worth noting that the [ZBus] is still (m ×m) in size as no new node has beencreated.

Summarizing the step-by-step procedure for building the [ZBus] as follows:Step 1: Draw the graph of the network and select a tree of the graph. Identify the branches and

the links of the graph. A tree of a graph with branches and links is shown in Fig. 4.14.

Figure 4.14: Tree of a graph

Step 2: Select a branch connected to the reference node to initiate the [ZBus] matrix buildingprocess. From Fig. 4.14, it is evident that the first branch selected could be either 1 or 3 or 4 asthese are the only branches connected to the reference node. Let the branch 1 be selected as thestarting branch and zpo be the impedance of the branch then

Z(m)Bus = [(1)

(1) zp0 ]

Step 3: Pick up another element from the graph. It should either be connected to an existing nodeor the reference node. Never select an element connected to two new nodes as it will be isolated fromthe existing partial network and this will result in the elements of [ZBus] matrix becoming infinite.For instance, with reference to Fig. 4.14, in the next step of [ZBus] matrix building process, if

114

element 5 is next added to the partial network as shown in Fig. 4.15, the resultant network isdisjointed. This is an incorrect choice. The proper choice could be any one of the elements 2 or 3 or4 or 6.

Figure 4.15: Selecting a wrong element in the step-by-step process

If the bus impedance matrix of a partial network with m-nodes, [ZmBus], is known, then depending

on whether the added (m + 1)th element is a branch or a link, the following steps are to be followedto obtain the new [ZBus] matrix:

(a) If the added element is a branch between a new node ‘q’ and the reference node with an impedancezqo, then the size of the new [ZBus] matrix will increase by one and the new matrix is given as :

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (⋯) (m) (q)

(1) Z11 ⋯ Z1m 0⋮ ⋮ ⋮ ⋮ ⋮(m) Zm1 ⋯ Zmm 0(q) 0 0 0 zq0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(b) If the added element is a branch between an existing node ‘p’ and a new node ‘q’ with animpedance zpq, then a new row and column corresponding to the new node ‘q’ is added to theexisting [Zm

Bus] matrix. The new matrix is calculated as follows:

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) ⋯ (p) ⋯ (m) (q)

(1) Z11 ⋯ Z1p ⋯ Z1m Z1p

⋮ ⋮ ⋮ ⋮ ⋮(p) Zp1 ⋯ Zpp ⋯ Zpm Zpp⋮ ⋮ ⋮ ⋮ ⋮(m) Zm1 ⋯ Zmp ⋯ Zmm Zmp

(q) Zp1 ⋯ Zpp ⋯ Zpm Zpp + zpq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

115

(c) If the added element is a link between an existing node ‘q’ and the reference node with animpedance zqo, then no new node is added to the network. A two-step procedure has to befollowed to find the new bus impedance matrix.

In the first step, a column and a row will be temporarily added to existing [ZmBus] matrix as:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) ⋯ (q) ⋯ (m) (`)

(1) Z11 ⋯ Z1q ⋯ Z1m −Z1q

⋮ ⋮ ⋮ ⋮ ⋮(q) Zq1 ⋯ Zqq ⋯ Zqm −Zqq⋮ ⋮ ⋮ ⋮ ⋮(m) Zm1 ⋯ Zmq ⋯ Zmm −Zmq(`) −Zq1 ⋯ −Zqq ⋯ −Zqm Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦where,

Z`` = Zqq + zq0

The additional row and column have to be deleted so that [ZBus] matrix is (m ×m) in size.The elimination process is carried out using


[∆Z][∆Z]T

Z``]]

where,[∆Z] = [−Z1q − Z2q⋯− Zqq⋯− Zmq]

T

(d) If the added element is a link between two existing nodes ‘p’ and ‘q’ with an impedance zpq,then again the two step procedure as outlined in (step c) is to be followed. The temporaryimpedance matrix Z(temp)

Bus is calculated as:

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (p) (q) (m) (`)

(1) Z11 ⋯ Z1p Z1q ⋯ Z1m (Z1q − Z1p)⋮

(p) Zp1 ⋯ Zpp Zpq ⋯ Zpm (Zpq − Zpp)⋮

(q) Zq1 ⋯ Zqp Zqq ⋯ Zqm (Zqq − Zqp)

(m) Zm1 ⋯ Zmp Zmq ⋯ Zmm (Zmq − Zmp)(`) (Zq1 − Zp1) ⋯ (Zqp − Zpp) (Zqq − Zpq) ⋯ (Zqm − Zpm) Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where,Z`` = Zpp + Zqq − 2Zpq + zpq

116

Next eliminate the added row and column ‘`’ using the expression:


[∆Z][∆Z]T

Z``]

where,[∆Z] = [(Z1q − Z1p)⋯(Zpq − Zpp)⋯(Zqq − Zqp)⋯(Zmq − Zmp)]

T

Step 4: Repeat Step 3 till all the elements are considered.

In the next lecture, we will be looking at an example of [ZBus] matrix building algorithm.

117

4.2 Example of [ZBus] matrix building algorithm

The single line diagram of a power system is shown in the Fig. 4.16. The line impedances in pu arealso given. The step-by-step procedure for [ZBus] matrix formulation is explained as given below:

Figure 4.16: Single Line Diagram of the Power System for the example

Preliminary Step: The graph of the network and a tree is shown in Fig. 4.17. Elements 1,2,4and 5 are the tree branches while 3, 6 and 7 are the links.

Figure 4.17: Graph and a tree of the network of Fig. 4.17

Step 1: The step-by-step [ZBus] matrix building algorithm starts with element 1, which isa tree branch connected between nodes 1 and the reference node 0 and has an impedance ofz10 = j0.10 pu. This is shown in the accompanying figure ,Fig. 4.18.

118

Figure 4.18: Partial network of Step 1

The resulting [ZBus] matrix is

ZBus = [(1)

(1) z10 ] = [(1)

(1) j0.10 ]

Step 2: Next, the element 2 connected between node 2 (q = 2) and the reference node ‘0’ isselected. This element has an impedance of z20 = j0.10 p.u. As this is the addition of a tree branchit will add a new node ‘2’ to the existing [ZBus] matrix. This addition is illustrated in Fig. 4.19.


The new bus impedance matrix is given by :

ZBus = [

(1) (2)

(1) j0.10 0(2) 0 z20

] = [

(1) (2)

(1) j0.1 0(2) 0 j0.10

]

Step 3: Element 3 connected between existing nodes, node 1 (p = 1) and node 2 (q = 2),having an impedance of z12 = j0.20 p.u. is added to the partial network, as shown in Fig. 4.20.

Since this is an addition of a link to the network a two step procedure is to be followed. In the

119


first step a new row and column is added to the matrix as given below :

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (`)

(1) j0.10 0.0 (Z12 − Z11)(2) 0.0 j0.10 (Z22 − Z21)(`) (Z21 − Z11) (Z22 − Z12) Z``

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (`)

(1) j0.10 0.0 −j0.10(2) 0.0 j0.10 j0.10(`) −j0.10 j0.10 j0.40

⎤⎥⎥⎥⎥⎥⎥⎦

where,Z`` = Z11 + Z22 − 2Z12 + z20 = j0.10 + j0.10 − 0.0 + j0.20 = j0.40 p.u.

Next this new row and column is eliminated to restore the size of [ZBus] matrix as given below:

[ZBus] = [j0.10 0.00.0 j0.10

] −[−j0.10j0.10

] [−j0.10 j0.10]

j0.40

Hence, the impedance matrix after the addition of element 3 is found out to be :

[ZBus] = [

(1) (2)

(1) j0.075 j0.025(2) j0.025 j0.075

]

Step 4: The element 4 , which is added next, is connected between an existing node, node 2(p = 2) and a new node, node 3 (q = 3). The impedance of this element is z23 = j0.30 p.u. andit is a tree branch hence, a new node, node 3 is added to the partial network. This addition, shownin Fig. 4.21, thus increases the size of [ZBus] to (3 × 3).

120


The new impedance matrix can be calculated as:

ZBus =⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3)

(1) j0.075 j0.025 Z12

(2) j0.025 j0.0.075 Z22

(3) Z21 Z22 Z22 + z23

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3)

(1) j0.075 j0.025 j0.025(2) j0.025 j0.0.075 j0.075(3) j0.025 j0.075 j0.375

⎤⎥⎥⎥⎥⎥⎥⎦

Step 5: Element 5 is added next to the existing partial network. This is a tree branch connectedbetween an existing node, node 3 (p = 3) and a new node, node 4 (q = 4). This is illustratedin Fig. 4.22.

Since a new node is added to the partial network, the size of [ZBus] increases to (4 × 4). Theimpedance of the new element is z34 = j0.15 p.u. The new bus impedance matrix is :

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4)

(1) j0.075 j0.025 j0.025 Z31

(2) j0.025 j0.075 j0.075 Z32

(3) j0.025 j0.075 j0.375 Z33

(4) Z13 Z23 Z33 Z33 + z34

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4)

(1) j0.075 j0.025 j0.025 j0.025(2) j0.025 j0.075 j0.075 j0.075(3) j0.025 j0.075 j0.375 j0.375(4) j0.025 j0.075 j0.375 j0.525

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Step 6: Next,the element 6 connected between two existing nodes node 1 (p = 1) and node4 (q = 4) is added to the network, as shown in the Fig. 4.23. The impedance of this elementis z23 = j0.25 p.u. As this is a link addition, the two step procedure is used. The bus impedance

121



matrix is modified by adding a new row and column as given below:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)

(1) j0.075 j0.025 j0.025 j0.025 (Z14 − Z11)(2) j0.025 j0.075 j0.075 j0.075 (Z24 − Z21)(3) j0.025 j0.075 j0.375 j0.375 (Z34 − Z31)(4) j0.025 j0.075 j0.375 j0.525 (Z44 − Z41)(`) (Z41 − Z11) (Z42 − Z12) (Z43 − Z13) (Z44 − Z14) Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

122

Substituting the values of appropriate [ZBus] matrix elements in the last row and column theintermediate impedance matrix is:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)

(1) j0.075 j0.025 j0.025 j0.025 −j0.05(2) j0.025 j0.075 j0.075 j0.075 j0.05(3) j0.25 j0.075 j0.375 j0.375 j0.35(4) j0.025 j0.075 j0.375 j0.525 j0.50(`) −j0.05 0.05 j0.35 j0.50 j0.80

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where,

Z`` = Z44 + Z11 − 2Z14 + z14 = j0.075 + j0.525 − 2 × j0.025 + j0.25 = j0.80 p.u.

The additional row and column ‘`’ are to be eliminated to restore the impedance matrix size to(m ×m), and the [ZBus] matrix after the addition of element 6 is calculated as:

[ZBus] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

j0.075 j0.025 j0.025 j0.025j0.025 j0.075 j0.075 j0.075j0.25 j0.075 j0.375 j0.375j0.025 j0.075 j0.375 j0.525

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−j0.05j0.50j0.35j0.50

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

[−j0.05 j0.05 j0.35 j0.50]

j0.80

Hence,

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4)


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Step 7: Finally the element 7 connected between two existing nodes node 2 (p = 2) and node 4(q = 4) is added to the partial network of step 6. The impedance of this element is is z23 = j0.40pu. This is also a link addition, as shown in Fig. 4.24 and hence the two step precedure will befollowed to obtain the [ZBus] matrix. In the first step the Z(temp)

Bus is calculated after a row and a

123


column are added to the exiting ZBus as follows:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Substituing the values of the elements of impedance matrix one gets:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)

(1) j0.0719 j0.0281 j0.0469 j0.0563 j0.281(2) j0.02810 j0.0719 j0.0531 j0.0437 −j0.281(3) j0.0469 j0.0531 j0.2219 j0.1562 j1031(4) j0.0563 j0.0437 j0.1562 j0.2125 j0.1688(`) j0.281 −j0.281 j0.1031 j0.1688 j0.5969

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where,

Z`` = Z22 + Z44 − 2Z24 + z24 = j0.0719 + j0.2125 − 2 × j0.0563 + j0.40 = j0.5969 p.u.

The additional row and column ‘`’ are to be eliminated to restore the impedance matrix size to

124

(m ×m), and [ZBus] after the addition of element 7 is calculated as:

[ZBus] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

j0.0281−j0.0281j0.1031j0.1688

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

[j0.0281 −j0.0281 j0.1031 j0.1688]

j0.5969

Hence,

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4)


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

As can be seen that the final [ZBus] matrix is a (4 × 4) matrix, as the network has 4 nodes anda reference node. As there are 7 elements is the network, 7 steps are required for the formation of[ZBus] matrix.

4.2.1 Modifications in the existing [ZBus] :

If in an existing network, for which the [ZBus] matrix is known, some modification such as lineremoval or line impedance alteration is carried out then the [ZBus] matrix can be easily modifiedwithout any need of reconstructing the matrix from scratch.

As an example, let the ZBus matrix be the final bus impedance matrix given for the network ofFig. 4.16. Next, let the element 7 connecting nodes 2 and 4 be removed from the network and itis required to find the modified ZBus.

Removal of element 7 is equivalent to setting its impedance z24 to infinite. This can be obtainedby connecting a fictitious element zadd24 in parallel to the existing element zorg24 such that the resultantimpedance zresult24 is infinite i.e.

1zresult24

= 1zorg24

+ 1zadd24

= 1∞ = 0

orzadd24 = −zorg24 = −j0.40 p.u.

Hence, by adding an element zadd24 = −j0.4 p.u. in parallel to zorg24 the removal of line between nodes2 and 4 can be simulated. The new added fictitious element is a link addition between the two nodes,p = 2 and q = 4 and is shown in Fig. 4.25 . Hence, this will require a two-step procedure. Theaddition of the fictitious element 8 , which is a link, will introduce a temporary row and column.

125

Figure 4.25: Adding a link to simulate the removal of element 7

The Z(temp)Bus is given as:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Substituting the appropriate values one gets:

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4) (`)

(1) j0.0705 j0.0295 j0.0420 j0.0483 j0.0188(2) j0.0295 j0.0705 j0.0580 j0.0517 −j0.0188(3) j0.0420 j0.0580 j0.2041 j0.1271 j0.0691(4) j0.0483 j0.0517 j0.1271 j0.1648 j0.1131(`) j0.0188 −j0.0188 j0.0691 j0.1131 −j0.2681

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

where,

Z`` = Z22 + Z44 − 2Z24 + zadd24 = j0.0705 + j0.1648 − 2 × j0.0483 + (−j0.40) = −j0.2681 p.u.

126

The additional row and column is eliminated in the following step:

[ZBus] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

j0.0188−j0.0188j0.0691j0.1131

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

[j0.0188 −j0.0188 j0.0691 j0.1131]

−j0.2681

Thus, the final impedance matrix after the removal of element 7 is :

ZBus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (4)


⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The obtained ZBus matrix is identical to the ZBus matrix obtained in step 6 of the previousexample, which is the impdance matrix of the network before the addition of element 7.

So far we have considered the ZBus matrix building algorithm without any presence of mutuallycoupled elements. In the next lecture, we will take into account the presence of mutually coupledelements while forming the ZBus matrix.

127

4.3 ZBUS formation considering mutual coupling betweenelements

Assume that the bus impedance matrix [ZmBUS] is known for a partial network of ‘m’ nodes and

a reference node. The bus voltage and bus current relation for the partial network, shown in Fig.4.26, can be expressed as:

Figure 4.26: Partial Network with m-buses

[VBUS] = [ZmBUS] [IBUS] (4.33)

In equation (4.33),

VBUS is (m × 1) bus voltage vector

IBUS is (m × 1) bus current vector

ZmBUS is (m ×m) bus impedance matrix

The new added element ‘p-q’ may be a branch or may be a link as discussed in the previousalgorithm.

128

4.3.1 Addition of a branch to this partial network:

The performance equation of the network with an added branch ‘p-q’ is

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

V2

⋮Vp⋮VmVq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 Z12 ⋯ Z1p ⋯ Z1m Z1q

Z21 Z22 ⋯ Z2p ⋯ Z2m Z2q

⋮Zp1 Zp2 ⋯ Zpp ⋯ Zpm Zpq⋮

Zm1 Zm2 ⋯ Zmp ⋯ Zmm ZmqZq1 Zq2 ⋯ Zqp ⋯ Zqm Zqq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

I2

⋮Ip⋮ImIq

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.34)

The network is assumed to contain bilateral passive elements and hence, Zqi = Ziq for i =1,2,⋯m, i ≠ q. The added branch ‘p-q’ is assumed to be mutually coupled with one or moreelements of the partial network.

To determine element Zqi, inject a current at ith node and calculate the voltage at qth node withrespect to reference, as shown in Fig. 4.27.

Figure 4.27: Calculation of Zqi for the addition of branch

Calculation of Zqi

As all other bus currents are zero, bus voltages can be written as,

129

V1 = Z1iIiV2 = Z2iIi

⋮Vp = ZpiIi

⋮Vm = ZmiIiVq = ZqiIi

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4.35)

Also from Fig. 4.27, Vp and Vq can be related as,

Vq = Vp − vpq (4.36)

Where vpq is the voltage across the added element ‘p − q’. Also, the currents in the elements ofthe network can be related to the voltages across the elements as,

[ipqiρσ

] = [ ypq,pq ypq,ρσ

yρσ,pq yρσ,ρσ] [ vpq

vρσ] (4.37)

Where,

ipq = the current through the added element ‘p − q’.iρσ = (m × 1) current vector of the elements of the partial network.vρσ = (m × 1) voltage vector of the elements of the partial network.ypq,pq = Self-admittance of the added element.ypq,ρσ = (m × 1) vector of mutual admittances between the added element ‘p − q’ and the

elements ‘ρ − σ’ of the partial network.yρσ,ρσ = (m ×m) primitive admittance matrix of the partial network.yρσ,pq = [ypq,ρσ]

T

The diagonal elements of primitive impedance matrix [z] are the self impedance of the individualelements while the off-diagonal elements are mutual impedances between the elements. The inverseof primitive impedance matrix is the primitive admittance matrix [y]. This can be explained withthe help of an illustrative example.

A single line diagram of a power system is shown in Fig. 4.28. The self-impedances of lines arewritten by the side of the line and are in p.u.. The two lines between nodes 1 and 3 are mutuallycoupled with a mutual impedance of j0.10. The primitive impedance matrix for the network can bewritten as,

[z] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

j0.60 0 0 0 00 j0.50 0 0 00 0 j0.50 0 00 0 0 j0.25 j1.00 0 0 j1.0 j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

130

Figure 4.28: Sample Power System

The inverse of [z] is [y], the primitive admittance matrix.

[y] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−j1.67 0 0 0 00 −j2.0 0 0 00 0 −j2.0 0 00 0 0 −j5.0 j2.50 0 0 j2.5 −j6.25

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The current ipq in the added branch ‘p − q’ equal to zero as node ‘q’ is open.

ipq = 0 (4.38)

The voltage vpq, however, is not zero as the added branch is mutually coupled to one or more elementsof the partial network. Thus, the voltage across other elements of the network can be expressed as,

vρσ = Vρ − Vσ (4.39)

where Vρ and Vσ are the voltages of the nodes of the partial network. With ipq = 0 from equation(4.37) one can write,

ypq,pqvpq + ypq,ρσvρσ = 0

Hence,vpq = −

ypq,ρσvρσypq,pq

(4.40)

131

Substituting vρσ from equation (4.39) and vpq equation (4.36) in equation (4.40) one gets,

Vq = Vp +ypq,ρσ(Vρ − Vσ)

ypq,pq

Substitution of Ii = 1 pu in equation (4.35) results in Vp, Vq, Vρ and Vσ being replaced by theircorresponding impedances and hence,

Zqi = Zpi +ypq,ρσ(Zρi − Zσi)

ypq,pq(4.41)

∀i = 1,2,⋯,m, i ≠ q

For calculating the self impedance Zqq, a current Iq = 1 p.u. is injected into qth node with allother currents equal to zero as shown in Fig. 4.29. Then the voltages of the nodes are calculatedfrom equation (4.35) , as

V1 = Z1qIqV2 = Z2qIq

⋮Vp = ZpqIq

⋮Vm = ZmqIqVq = ZqqIq

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

(4.42)

With Iq = 1 p.u., Zqq Can be calculated directly by calculating Vq.

also,Vq = Vp − vpq (4.43)

andipq = −Iq = −1 (4.44)

Hence, from equation (4.37) one gets

ipq = −1 = ypq,pqvpq + ypq,ρσvρσ (4.45)

And thus vpq can be written as,

vpq = −1 + ypq,ρσvρσ

ypq,pq(4.46)

Substituting vpq and vρσ, the above equation can be rewritten as,

Vq = Vp +1 + ypq,ρσ(Vρ − Vσ)

ypq,pq(4.47)

132

Figure 4.29: Calculation of Zqq for the addition of a branch

With Iq = 1 p.u., from equation (4.35) Vp, Vq, Vρ and Vσ can be replaced by respective transferimpedances,

Zqq = Zpq +1 + ypq,ρσ(Zρq − Zσq)

ypq,pq(4.48)

4.3.2 Addition of a link to this partial network:

If the added element p− q is a link, then a fictitious node ` is created by connecting an ideal voltagein series with the added element, as shown in Fig. 4.30.

The value of the source voltage is selected such that the current I` through the added link iszero. If e` is the voltage of node ` with respect to node q and I` is the current injected into node `from node q. The performance equation of the partial network with the added link p − ` and idealseries voltage source e` is,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1

⋮Vp⋮Vme`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 . . . Z1p . . . Z1m Z1`

⋮Zp1 . . . Zpp . . . Zpm Zp`⋮

Zm1 . . . Zmp . . . Zmm Zm`Z`1 . . . Z`p . . . Z`m Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

I1

⋮Ip⋮ImI`

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.49)

133

Figure 4.30: Calculation of Zì for the addition of a link

Here, Zli represents the transfer impedance relating the current, Ii, injected into the ith busand the voltage of the added source e`, connected between nodes ` and q . They are conceptuallydifferent from the elements Zij of ZBUS matrix which relate the current injected into the jth busand the voltage of the ith bus with repect to the reference.

Ase` = V` − Vq,

The elements Zì, 1 = 1,2,⋯m, i ≠ `, of the added row and column, can be calculated byinjecting a current Ii into the ith node and determining the voltage of `th node with repect to qth

node.Hence,

Zì =e`

Ii∣ , Ik = 0, k = 1,2,⋯m, k ≠ `

Also,e` = Vp − Vq − vp` (4.50)

The current ip` through the link can be written as,

ip` = yp`,p`vp` + yp`,pσvρσ (4.51)

Since the current through the link is zero, ip` = ipq = 0Hence,

vp` = −yp`,ρσvρσyp`,p`

(4.52)

134

Since the voltage source is ideal source, one can write,

yp`,ρσ = ypq,ρσ

andyp`,p` = ypq,pq (4.53)

So,

vp` = −ypq,ρσvρσypq,pq

= − ypq,ρσ(Vρ − Vσ)ypq,pq

(4.54)

With Ii = 1 p.u., substituting Vp, Vq, Vp and Vσ from equation (4.35) and vp` from equation(4.52) in equation (4.50) one gets

Z`i = Zpi − Zqi +ypq,ρσ(Zpi − Zσi)

ypq,pq(4.55)

∀i = 1,2,⋯m, i ≠ `

To calculate Z``, a current is injected at the ‘`th’ node with respect to node ‘q’, as shown in Fig.4.31. As all other node currents are zero, the node voltages can be written as,

Figure 4.31: Calculation of Z`` for the addition of a link

Vk = Zk`I`, ∀k = 1,2,⋯m

135

e` = Z``I` (4.56)

With I` = 1 p.u., Z`` can be directly computed by calculating e`. The current in the element p− ` is

ip` = −I` = 1 p.u. (4.57)

From equation (4.37) one getsip` = yp`,p`vp` + yp`,pσvpσ = −1

Further as yp`,p` = ypq,pq and yp`,pσ = ypq,pσ, hence vp` can be expressed as

vp` = −1 + ypq,pσvpσ

ypq,pq(4.58)

Substituting vp` from equation (4.50) , one can write

e` = Vp − Vq +1 + ypq,pσ(Vp − Vσ)

ypq,pq

With I` = 1 p.u., substituting Vp, Vq, Vρ, Vσ and e` from equation (4.56), Z`` is obtained as

Z`` = Zpi − Zqi +1 + ypq,ρσ(Zp` − Zσ`)

ypq,pq(4.59)

∀i = 1,2,⋯m, i ≠ `

In the case of link addition the additional row and column corresponding to fictitious node ` are tobe eliminated.

For this the fictitious series voltage source e` is short circuited. From equation (4.49) the busvoltages can be written in compact from as

[VmBus] = [Zm

Bus][ImBus] + [∆Z][I`] (4.60)

Where,

[∆Z] = [Z1` Z2` ⋯ Zm`]Tis an (m×1) vector comprising of the entries of the column added

to the ZmBus matrix

[ImBus], [Vm

Bus] = (m × 1) bus current and voltage vectors respectively, of the partial network beforethe addition of element p − `.[Zm

Bus] = (m ×m) [ZBus] matrix of the partial network before the addition of element p − `.I` = current injected in the link. Also,

[e`] = [∆Z]T [IBus] + Z``I` = 0 (4.61)

136

On substituting I` from equation (4.61) into equation (4.60) [VBus] can be written as,

[VBus] = [ZmBus] − [∆Z.∆ZT

Z``][IBus] (4.62)

Hence, the final [ZBus] which is (m ×m) in size can be written as

[ZFinalBus ] = [Zm

Bus] − [∆Z.∆ZT

Z``] (4.63)

An example illustrating the [ZBus] building procedure will be discussed in the next lecture.

137

4.4 Example of [ZBus] matrix formulation in the presence ofmutual impedances

Consider the network shown in Fig. 4.32.

Figure 4.32: The power system for [ZBus] example

A tree for the network is shown in Fig. 4.32. The system data is given in Table 4.1.

Figure 4.33: Tree of the network

Table 4.1: System data

Element no.Self Mutual

Bus code Impedance Bus code Impedancep - q zpq,pq (p.u.) r - s zpq,rs (p.u.)

1 0 - 1(1) j0.4 0 - 1(2) j0.22 0 - 1(2) j0.53 0 - 2 j0.5 0 - 1(1) j0.14 2 - 3 j0.45 1 - 3 j0.6

138

Step 1: The algorithm starts building the [ZBus] matrix element by element. To initiate theprocess, start with element 1 connected between nodes p = 0 and q = 1, shown in Fig. 4.34. The[ZBus] matrix of the partial network is given as,

Figure 4.34: Partial network in Step 1

[ZBus] = [(1)

(1) j0.4 ]

Step 2: Next add element 2 connected between p = 0 and q = 1 which is mutually coupled to theexisting element 1, connected between ρ = 0 and σ = 1. This new element is a link as it does notcreate a new node, the partial network for this step is shown in Fig. 4.35. The augmented [Z(temp)

Bus ]matrix after the addition of this element, is given by


Z(temp)Bus = [

(1) (`)

(1) j0.4 Z1`

(`) Z`1 Z``]

Z`1 = Z01 − Z11 +y0−1(2),0−1(1)(Z01 − Z11)

y0−1(2),0−1(2)

Z`` = Z0` − Z1` +1 + y0−1(2),0−1(1)(Z0` − Z1`)

y0−1(2),0−1(2)

where, Z01 and Z0` are the elements of [ZBus] matrix associated with the reference node.The primitive impedance matrix [z] for the partial network is

[z] = [

0−1(1) 0−1(2)

0−1(1) j0.4 j0.20−1(2) j0.2 j0.5

]

The primitive admittance matrix [y] for the partial network in nothing but the inverse of primitive

139

impedance matrix [z] and is given by

[y] = [z]−1 = [

0−1(1) 0−1(2)

0−1(1) −j3.125 j1.250−1(2) j1.25 −j2.5 ]

With Z01 = 0, since 0 is the reference node, Z`1 is evaluated as

Z`1 = −j0.4 +j1.25(−j0.4)

−j2.5 = −j0.2 = Z1`

Also as Z0` = 0, since 0 is the reference node, and hence, Z`` is calculated as

Z`` = j0.2 +1 + j1.25(j0.2)

−j2.5 = j0.50

The augmented Z(temp)bus matrix is given as

Z(temp)Bus = [

(1) (`)

(1) j0.4 −j0.2(`) −j0.2 j0.5

]

The row and column corresponding to the `th row and column corresponding to a link addi-tion,(shown in red in the above matrix), need to be eliminated as the link addition does not createa new node. The [ZBus] matrix, after the addition of second element to the partial network, iscalculated using the following expression

[ZBus] = [ZBus] −Z1`Z`1

Z``

= j0.4 − (−j0.2)(−j0.2)j0.5

ZBus = [(1)

(1) j0.32 ]

Note that the size of ZBus matrix is still (1×1) as no new node has been added to the partial networkas yet.Step 3: Next add element 3, which is connected between the nodes p = 0 and q = 2. This is abranch addition as a new node, node 2 is created. This element is mutually coupled to the existingelement 1. Hence, the primitive [z] matrix of the partial network, shown in Fig. 4.36, is

140


[z] =⎡⎢⎢⎢⎢⎢⎢⎣

0−1(1) 0−1(2) 0−2

0−1(1) j0.4 j0.2 j0.10−1(2) j0.2 j0.5 0

0−2 j0.1 0 j0.5

⎤⎥⎥⎥⎥⎥⎥⎦

The primitive [y] matrix is calculated as [z]−1 and is equal to

[y] =⎡⎢⎢⎢⎢⎢⎢⎣

0−1(1) 0−1(2) 0−2

0−1(1) j3.333 j1.333 j0.6670−1(2) j1.333 −j2.533 −j0.2667

0−2 j0.667 −j0.2667 j2.133

⎤⎥⎥⎥⎥⎥⎥⎦

The modified [ZBus] matrix is expressed as

[ZBus] = [

(1) (2)

(1) j0.32 Z12

(2) Z21 Z22]

For this element p = 0 and q = 2 and the set of elements [ρσ] mutually coupled to this elementis [0 − 1(1) 0 − 1(2)]

Z21 = Z01 +[y0−2,0−1(1) y0−2,0−1(2)] [

Z02 − Z11

Z01 − Z11]

y0−2,0−2

Z01 and Z02 are the transfer impedances associated with the reference node and are equal to zero.

Z21 =[j0.667 −j0.2667] ∗ [−j0.32

−j0.32]

j2.133 = j0.06

141

Hence, Z12 = Z21 = j0.06

Z22 = Z02 +1 + [y0−2,0−1(1) y0−2,0−1(2)] [

Z02 − Z12

Z01 − Z12]

y0−2,0−2

Z21 =1 + [j0.667 −j0.2667] ∗ [−j0.32

−j0.32]

j2.133 = j0.48

The modified [ZBus] matrix is

[ZBus] = [

(1) (2)

(1) j0.32 j0.06(2) j0.06 j0.48

]

Step 4: On adding element 4 between p = 2 and q = 3, a new node, node 3 is created. Hence, thisis a branch addition and is shown in Fig. 4.37. The modified [ZBus] matrix can be written as


ZBus =⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3)

(1) j0.32 j0.06 Z13

(2) j0.06 j0.48 Z23

(3) Z31 Z32 Z33

⎤⎥⎥⎥⎥⎥⎥⎦

As this element is not mutually coupled to other elements the elements of vector ypq,ρσ are zero.Hence, the new elements of [ZBus] matrix can be calculated, using the expression given in (4.41),as :

Off-diagonal elements

Zqi = Zpi ∀ i = 1,2,3 i ≠ q

142

Z31 = Z21 = j0.06

Z32 = Z22 = j0.48

Z13 = Z31 = j0.06

Z23 = Z32 = j0.48

Diagonal elementUsing the expression of (4.48) with no mutual coupling, the diagonal element can be written as:

Zqq = Zpq + zpq,pq

hence,Z33 = Z23 + z23,23 = j0.48 + j0.4 = j0.88

ZBus =⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3)

(1) j0.32 j0.06 j0.06(2) j0.06 j0.48 j0.48(3) j0.06 j0.48 j0.88

⎤⎥⎥⎥⎥⎥⎥⎦Step 5: Finally add element 5 between nodes p = 1 and q = 3. This is an addition of a link hence atemporary row and column are added. Fig. 4.38 showns the final network after the addition of thiselement. The modified Z(temp)

Bus matrix can be written as

Figure 4.38: The complete network after the addition of link in step 5

Z(temp)Bus =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (`)

(1) j0.32 j0.06 j0.06 Z1`

(2) j0.06 j0.48 j0.48 Z2`

(3) j0.06 j0.48 j0.88 Z3`

(`) Z`1 Z`2 Z`3 Z``

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

143

Since this element is not mutually coupled to other elements, the new elements of [Z(temp)Bus ]

matrix can be calculated, using the expression of (4.55), as :Off-diagonal elements

Z`i = Zpi − Zqi ∀ i = 1,2,3

Z1` = Z11 − Z13 = j0.32 − j0.06 = j0.26 = Z`1

Z2` = Z21 − Z23 = j0.06 − j0.48 = −j0.42 = Z`2

Z3` = Z31 − Z33 = j0.06 − j0.88 = −j0.82 = Z`3

Diagonal elementFor calculating the diagonal element, the expression given in (4.59) is used. Hence,

Z`` = Zp` − Zq` + zpq,pq

Z`` = Z1` − Z3` + z13,13 = j0.26 + j0.82 + j0.6 = j1.68

Hence, the temporary [Z(temp)Bus ] matrix can be written as

[Z(temp)Bus ] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3) (`)

(1) j0.32 j0.06 j0.06 j0.26(2) j0.06 j0.48 j0.06 −j0.42(3) j0.06 j0.48 j0.88 −j0.82(`) j0.26 −j0.42 −j0.82 j1.68

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The `th row and `th column are to be eleminated to restore the size of ZBus to 3×3. The eliminationis done using the relation

ZBus = Z(temp)Bus − ∆Z ∗∆ZT

Z``

∆ZT = [j0.26 −j0.42 −j0.82]

ZBus =⎡⎢⎢⎢⎢⎢⎢⎣

j0.32 j0.06 j0.06j0.06 j0.48 j0.48j0.06 j0.48 j0.88

⎤⎥⎥⎥⎥⎥⎥⎦−

⎡⎢⎢⎢⎢⎢⎢⎣

j0.23−j0.42−j0.82

⎤⎥⎥⎥⎥⎥⎥⎦∗ [−j0.26 −j0.42 −j0.82]

j1.68

144

Hence, the final matrix [ZBus] is

[ZBus] =⎡⎢⎢⎢⎢⎢⎢⎣

(1) (2) (3)

(1) j0.2798 j0.1250 j0.1869(2) j0.1250 j0.3750 j0.2750(3) j0.1869 j0.2750 j0.4798

⎤⎥⎥⎥⎥⎥⎥⎦

After this discussion of formulation of [ZBus] matrix, we are now ready to discuss fault analysis,which we will start from the next lecture.

145

4.5 Fault Analysis:

Under normal conditions, a power system operates under balanced conditions with all equipmentscarrying normal load currents and the bus voltages within the prescribed limits. This condition canbe disrupted due to a fault in the system. A fault in a circuit is a failure that interferes with thenormal flow of current. A short circuit fault occurs when the insulation of the system fails resultingin low impedance path either between phases or phase(s) to ground. This causes excessively highcurrents to flow in the circuit, requiring the operation of protective equipments to prevent damageto equipment. The short circuit faults can be classified as:

• Symmetrical faults

• Unsymmetrical faults

4.6 Symmetrical faults:

A three phase symmetrical fault is caused by application of three equal fault impedances Zf to thethree phases, as shown in Fig. 4.39. If Zf = 0 the fault is called a solid or a bolted fault. These faultscan be of two types: (a) line to line to line to ground fault (LLLG fault) or (b) line to line to linefault (LLL fault). Since the three phases are equally affected, the system remains balanced. Thatis why, this fault is called a symmetrical or a balanced fault and the fault analysis is done on perphase basis. The behaviour of LLLG fault and LLL fault is identical due to the balanced natureof the fault. This is a very severe fault that can occur in a system and if Zf = 0, this is usually themost severe fault that can occur in a system. Fortunately, such faults occur infrequently and onlyabout 5% of the system faults are three phase faults.

Figure 4.39: Symmetrical Fault

146

4.7 Unsymmetrical faults:

Faults in which the balanced state of the network is disturbed are called unsymmetrical or unbalancedfaults. The most common type of unbalanced fault in a system is a single line to ground fault (LGfault). Almost 60 to 75% of faults in a system are LG faults. The other types of unbalanced faultsare line to line faults (LL faults) and double line to ground faults (LLG faults). About 15 to 25%faults are LLG faults and 5 to 15% are LL faults. These faults are shown in Fig. 4.40.

Figure 4.40: Unsymmetrical Fault

Majority of the faults occur on transmission lines as they are exposed to external elements.Lightening strokes may cause line insulators to flashover, high velocity winds may cause tower failure,ice loading and wind may result in mechanical failure of line or insulator and tree branches may causeshort circuit. Much less common are the faults on cables, circuit breakers, generators, motors andtransformers.

Fault analysis is necessary for selecting proper circuit breaker rating and for relay settings and co-ordination.The symmetrical faults are analysed on per phase basis while the unsymmetrical faultsare analyzed using symmetrical components. Further, the ZBUS matrix is very usefull for shortcircuit studies .

4.8 Symmetrical or Balanced three phase fault analysis:

In this type of faults all three phases are simultaneously short circuited. Since the network remainsbalanced, it is analyzed on per phase basis. The other two phases carry identical currents but with aphase shift of 120○. A fault in the network is simulated by connecting impedances in the network atthe fault location. The faulted network is then solved using Thevenin’s equivalent network as seenfrom the fault point. The bus impedance matrix is convenient to use for fault studies as its diagonalelements are the Thevenin’s impedance of the network as seen from different buses. Prior to theoccurrence of fault, the system is assumed to be in a balanced steady state and hence per phasenetwork model is used. The generators are represented by a constant voltage source behind a suitablereactance which may be sub-transient, transient or normal d-axis reactance. The transmission linesare represented by their π-models with all impedances referred to a common base. A typical bus

147

Figure 4.41: Fault at kth bus of a power system

of an n- bus power system network is shown in Fig. 4.41. Further, a balanced three phase fault,through a fault impedance Zf is assumed to occur at kth bus as shown in the figure. A pre-faultload flow provides the information about the pre-fault bus voltage.

Let [VBUS(0)] be the prefault bus voltage vector =[V1(0) . . . Vk(0) . . . Vn(0)]T

p.u. The faultat kth bus through an impedance Zf will cause a change in the voltage of all the buses [∆VBUS]due to the flow of heavy currents through the transmission lines. This change can be calculated byapplying a voltage Vk(0) at kth bus and short circuiting all other voltage sources. The sources andloads are replaced by their equivalent impedances. This is shown in Fig. 4.42. In Fig. 4.42, Zi

Figure 4.42: Network representation for calculating [∆VBUS]

and Zk are the equivalent load impedances as bus i and k respectively, zik is the impedance of line

148

between ith and kth buses. xdi is the appropriate generator reactance, Zf is the fault impedance,Ik(F ) is the fault current and Vk(0) is the prefault voltage at kth bus. From the superpositiontheorem, the bus voltages due to a fault can be obtained as the sum of prefault bus voltages and thechange in bus voltages due to fault,i.e.,

[VBUS(F)] = [VBUS(0)] + [∆VBUS] (4.64)

where,[VBUS(F)] = Vector of bus voltages during fault =[V1(F ) . . . Vi(F ) . . . Vn(F )]

T

[VBUS(0)] = Vector of pre-fault bus voltages =[V1(0) . . . Vi(0) . . . Vn(0)]T

[∆VBUS] = Vector of change in bus voltages due to fault= [∆V1 . . .∆Vk . . .∆Vn]T

Also the bus injected current [IBUS] can be expressed as,

[IBUS] = [YBUS] [VBUS] (4.65)

where, [VBUS] is the bus voltage vector and [YBUS] is the bus admittance matrix.

With all the bus currents, except of the faulted bus k, equal to zero, the node equation for thenetwork of Fig. 4.42 can be written as

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0⋮

−Ik(F )⋮0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Y11 ⋯ Y1k ⋯ Y1n

⋮Yk1 ⋯ Ykk ⋯ Ykn⋮Yn1 ⋯ Ynk ⋯ Ynn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆V1

⋮∆Vk⋮

∆Vn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.66)

As the fault current Ik(F ) is leaving the bus it is taken as a negative current entering the bus.Hence,

[IBUS(F)] = [YBUS] [∆VBUS] (4.67)

[∆VBUS] can be calculated as:

[∆VBUS] = [YBUS]−1 [IBUS(F)] = [ZBUS] [IBUS(F)] (4.68)

where, [ZBUS] is the bus impedance matrix = [YBUS]−1.

Substituting the expression of [∆VBUS] from equation (4.68) in equation (4.64) one can write,

[VBUS(F)] = [VBUS(0)] + [ZBUS(F)] [IBus(F)] (4.69)

Expanding the above equation one can write,

149

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1(F )⋮

Vk(F )⋮

Vn(F )

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

V1(0)⋮

Vk(0)⋮

Vn(0)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

+

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z11 ⋯ Z1k ⋯ Z1n

⋮Zk1 ⋯ Zkk ⋯ Zkn⋮Zn1 ⋯ Znk ⋯ Znn

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(0)⋮

−Ik(F )⋮0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.70)

The bus voltage of kth bus can be expressed as:

Vk(F ) = Vk(0) − ZkkIk(F ) (4.71)

Also from Fig. 4.41Vk(F ) = ZF Ik(F ) (4.72)

For a bolted fault Zf = 0 and hence, Vk(F ) = 0. Thus the fault current Ik(F ) for bolted fault canbe expressed using equation (4.71) as,

Ik(F ) = Vk(0)Zkk

(4.73)

For faulty with non-zero fault impedance Zf , the fault current can be calculated as:

Ik(F ) = Vk(0)Zkk + Zf

(4.74)

The quantity Zkk in equation (4.73) and equation (4.74) is the Thevenin’s impedance or open-circuit impedance of the network as seen from the faulted bus k. From equation (4.70), the busvoltage after fault for the unfaulted or healthy buses can be written as:

Vi(F ) = Vi(0) − ZikIk(F ) ∀i = 1,2,⋯n, i ≠ k (4.75)

Substituting Ik(F ) from equation (4.73) , Vi(F ) can be expressed as:

Vi(F ) = Vi(0) −Zik

Zkk + ZfVk(0) (4.76)

The fault current Iij(F ) flowing in the line connecting ith and jth bus can be calculated as

Iij(F ) = Vi(F ) − Vj(F )zij

(4.77)

where zij is the impedance of line connecting buses i and j.

150

4.9 Unsymmetrical or Unbalanced fault analysis:

For the analysis of unsymmetrical or unbalanced faults, symmetrical component method is used.The use of symmetrical components simplifies the analysis procedure of unbalanced system and alsohelps in improving the understanding of the system behavior during fault conditions.

A review of symmetrical components is presented next.

4.9.1 Symmetrical components:

Any unbalanced set of three phase voltage or current phasors can be replaced by three balanced setsof three phase voltage or current phasors. These three balanced set of voltage or current phasorsare called symmetrical components of voltages or currents. Let Ia, Ib, and Ic be an arbitrary set ofthree current phasors representing phase currents. Then using symmetrical components they can beexpressed as:

⎡⎢⎢⎢⎢⎢⎢⎣

IaIbIc

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

Ia0

Ib0Ic0

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

Ia1

Ib1Ic1

⎤⎥⎥⎥⎥⎥⎥⎦+⎡⎢⎢⎢⎢⎢⎢⎣

Ia2

Ib2Ic2

⎤⎥⎥⎥⎥⎥⎥⎦(4.78)

Or,[I]

abc= [I0] + [I1] + [I2]

where,[I]

abc= [Ia Ib Ic]

Tis the arbitrary set of three current phasors of phase currents.

[I0] = [Ia0 Ib0 Ic0]Tis the set of zero sequence components.The magnitudes of the three zero

sequence components are equal i.e.∣Ia0∣ = ∣Ib0∣ = ∣Ic0∣ and they are co-phasors.

[I1] = [Ia1 Ib1 Ic1]T

is the set of positive sequence components, with Ia1 = ∣Ia1∣∠0○, Ib1 =∣Ib1∣∠− 120○, and Ic1 = ∣Ic1∣∠120○, with ∣Ia1∣ = ∣Ib1∣ = ∣Ic1∣.

[I2] = [Ia2 Ib2 Ic2]T

is the set of negative sequence components, with Ia2 = ∣Ia2∣∠0○, Ib2 =∣Ib2∣ ∠120○, and Ic2 = ∣Ic2∣∠− 120○,with ∣Ia2∣ = ∣Ib2∣ = ∣Ic2∣

The graphical representation of the sequence components is shown in Fig. 4.43.Let an operator ‘a’ be defined such that a = ∠120○ . Any phasor multiplied by ‘a’ undergoes a

counter clockwise rotation of 120○ without any change in the magnitude. Further,

a = 1∠120○

a2 = 1∠240○

a3 = 1∠360○

also 1 + a + a2 = 0Ia1 = Ia1∠θ1

where, ∠θ1 is the angle of phase ‘a’ positive sequence current.

151

Figure 4.43: Representation of Symmetrical Components

Ib1 = a2Ia1

Ic1 = aIa1

The phase sequence of the positive component set is ‘abc’.

Similarly the negative sequence set can be written as:

Ia2 = Ia2∠θ2

where, ∠θ2 is the angle of phase ‘a’ negative sequence current.

Ib2 = aIa2

Ic2 = a2Ia2

152

The phase sequence of the negative component set is ‘acb’.The zero-sequence component set can be written as:

Ia0 = Ia0∠θ0 = Ib0 = Ic0

where, ∠θ0 is the angle of phase ‘a’ zero sequence current.Hence, equation (4.78) can be simplified as:

⎡⎢⎢⎢⎢⎢⎢⎣

IaIbIc

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Ia0

Ia1

Ia2

⎤⎥⎥⎥⎥⎥⎥⎦(4.79)

It can also expressed in a compact form as:

[I]abc

= [A] [I]012

(4.80)

where, [I]abc

= set of phase quantities = [Ia Ib Ic]T

[I]012

= set of sequence quantities = [Ia0 Ia1 Ia2]T

A =⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦is the symmetrical component transformation matrix.

The symmetrical components [I]012

can be written in terms of phase quantities [I]abc

as:

[I]012

= [A]−1 [I]abc

(4.81)

where, A−1 = 13

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a a2

1 a2 a

⎤⎥⎥⎥⎥⎥⎥⎦thus,

Ia0 =13[Ia + Ib + Ic]

Ia1 =13[Ia + aIb + a2Ic] (4.82)

Ia2 =13[Ia + a2Ib + aIc]

To summarize:

• For voltage:

[V]abc

= [A] [V]012

(4.83)

153

[V]012

= [A]−1 [V]abc

(4.84)

where, [V]abc is the set of phase voltages, and [V]012 is the set of sequence voltages.

• For current:

[I]abc

= [A] [I]012

(4.85)

[I]012

= [A]−1 [I]abc

(4.86)

where, [I]abc is the set of phase voltages, and [I]012 is the set of sequence voltages.Before starting unbalanced fault analysis, it is necessary to learn about the sequence networks of

different power system components, which we will discuss in the next lecture.

154

4.9.2 Sequence Networks of a loaded Synchronous Generator:

A three-pahse synchronous generator, having a synchronous impedance of Zs per phase, with itsneutral grounded through a impedance Zn is shown in Fig.4.44. The generator is supplying abalanced three phase load. The generator voltages Ea, Eb and Ec are balanced and hence treated

Figure 4.44: Three phase synchronous generator supplying a balanced load

as positive sequence set of voltage phasors and can be expressed as:

[E]abc

=⎡⎢⎢⎢⎢⎢⎢⎣

1a2

a

⎤⎥⎥⎥⎥⎥⎥⎦[Ea] (4.87)

As the generator is supplying a three-phase balanced load, the following KVL equations can bewritten for each phase :

Va = Ea − ZsIa − ZnIn

Vb = Eb − ZsIb − ZnIn (4.88)

Vc = Ec − ZsIc − ZnIn

155

Substituting the neutral current In = Ia + Ib + Ic in equation (4.88), and writing the resulting equa-tion in matrix form, we get:

⎡⎢⎢⎢⎢⎢⎢⎣

VaVbVc

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

Ea

Eb

Ec

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

Zs + Zn Zn ZnZn Zs + Zn ZnZn Zn Zs + Zn

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

IaIbIc

⎤⎥⎥⎥⎥⎥⎥⎦(4.89)

The above matrix equation can be expressed in a compact form as:

[V]abc

= [E]abc

− [Z]abc

[I]abc

(4.90)

where,

[V]abc = [Va Vb Vc]Tis the vector of terminal phase voltages.

[I]abc = [Ia Ib Ic]Tis the vector of terminal phase currents.

[Z]abc is the impedance matrix which can be easily identified from equation (4.89).Replacing the phase quantities of equation (4.90) by corresponding sequence quantities, using

the transformation equation (4.83) and equation (4.85) one can write:

[A] [V]012

= [A] [E]012

− [Z]abc

[A] [I]012

(4.91)

Premultiplying bothsides of the equation (4.91) by [A]−1and after simplifications one gets:

[V]012

= [E]012

− [Z]012

[I]012

(4.92)

where,[Z]012

is Generator Sequence Impedance Matrix and is defined as:

[Z]012

= [A]−1 [Z]abc

[A] =⎡⎢⎢⎢⎢⎢⎢⎣

Zs + 3Zn 0 00 Zs 00 0 Zs

⎤⎥⎥⎥⎥⎥⎥⎦

[E]012

is the generated sequence voltage vector and is defined as

⎡⎢⎢⎢⎢⎢⎢⎣

0Ea

0

⎤⎥⎥⎥⎥⎥⎥⎦since the generated voltages

are always balanced and contain only the positive sequence component.Substituting [E]

012and [Z]

012in equation (4.92) we get:

⎡⎢⎢⎢⎢⎢⎢⎣

Va0

Va1

Va2

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0Ea

0

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

Z0 0 00 Z1 00 0 Z2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Ia0

Ia1

Ia2

⎤⎥⎥⎥⎥⎥⎥⎦(4.93)

where, Z1 = Zs is the positive sequence generator impedance, Z2 = Zs is the negative sequence gener-

156

ator impedance and Z0 = Zs + 3Zn is the zero sequence generator impedance. Expanding the aboveequation, one can write seperate equation for each of the sequence components as :

Va0 = −Z0Ia0

Va1 = Ea − Z1Ia1 (4.94)

Va2 = −Z2Ia2

From equation (4.94), it is evident that the three sequence components are independent of eachother.The current of a particular sequence produces a voltage drop of that sequence only, hencethe three sequences are decoupled from each other. The three sequence networks of a synchronousgenerator are shown in Fig. 4.45.

Figure 4.45: The sequence networks of a synchronous generator

4.9.3 Sequence networks of a transmission line:

For a static device such as a transmission line, the phase sequence of voltages and currents have noeffect on the impedance offered by the line as both positive and negative phase sequences encounteridentical line geometry. Hence, the positive and negative sequence impedances offered by a line areidentical i.e. Z1 = Z2.

The zero sequence currents, however, are in phase and flow through the conductors and returnthrough grounded neutral and/or ground wires. As a result, the ground or ground wire are to beincluded in the path of the zero sequence currents. The zero sequence impedance Z0 is, therefore,different from Z1 and Z2 due to the inclusion of the ground return path. Z0 is usually more thanthree times of Z1 or Z2. The three sequence networks of the transmission lines are shown in Fig.4.46.

4.9.4 Sequence networks of a tranformer:

For short circuit studies, the shunt magnetizing branch of transformer is neglected as the currentthrough it is negiligible as compared to short circuit current. The transformer is, therefore, mod-elled with an equivalent series leakage impedance. Since the transformer is also a static device like a

157

Figure 4.46: The sequence networks of a transmission line

transmission line, the series leakage impedance will not change if the phase sequence of applied volt-age is reversed. Therefore, the positive and negative sequence impedances offered by a transformerare equal. The zero sequece current flows through a transformer if paths for it to flow exist on theprimary as well the secondary sides. For such transformers the zero sequence impedance is equal tothe leakage impedance, as a consequence Z0 = Z1 = Z2. The positive and negative sequence networksof a transformer are identical to the positive and negative sequence networks of a transmission lineas shown in Fig. 4.46 (a) and (b). The sequence networks for zero sequence depends on the windingconnections and whether or not the neutrals are grounded. To derive these circuits for differenttransformer connections, one has to keep in mind that an open circuit will exist on the primary(secondary) side if there is no ground return for primary (secondary) currents or if there is no corre-sponding path for secondary(primary) zero-sequence currents. The different three-phase transformerconnections and their equivalent zero-sequence networks are discussed next. It is assumed that theneutrals, if grounded, are solidly grounded.

Figure 4.47: The zero-sequence equivalent circuit of a Star-Star transformer with both neutralsgrounded

(a) Star-Star connections with both neutrals grounded: Since both the neutrals are grounded,thephasor sum of three unbalanced phase currents is equal to three times the zero sequence currentIa0 (equation (4.82)). Hence, the zero sequence currents can flow in the primary and secondary

158

windings and the transhomer, therefore, can be represented by the equivalent zero-sequenceleakage impedance. The equivalent circuit is shown in Fig. 4.47.

(b) Star-Star connections with only one neutral grounded: When the neutral of only one winding isgrounded, the phase currents of the ungrounded winding must add up to zero. This implies thatthe zero sequence currents can not exist in the ungrounded winding and hence the zero sequencecurrents can not exist even in the transformer side with neutral grounded. The transformer inthis case, is represented as an open circuit between primary and secondary windings and theequivalent circuit is shown in Fig. 4.48.

Figure 4.48: The zero-sequence equivalent circuit of a Star-Star transformer with only one neutralgrounded

(c) Star-Star connections with only no neutral grounded: In this case also the phasor sum of thephase curents of both the windings is zero and hence the zero sequence currents can not exist onany winding in this case also. The zero sequence equivalent network is represented as an opencircuit between the two windings and is shown in Fig. 4.49.

Figure 4.49: The zero-sequence equivalent circuit of a Star-Star transformer with both neutralsungrounded

159

(d) Star-delta connections with neutral grounded: A zero sequence current on the grounded star-winding will cause a circulating zero-sequence current in the closed delta-winding. However,the zero-sequence current on the delta-winding can not exist on line side of the winding andis confined only to the closed delta-winding. As a result, an open circuit exists between thestar and the delta sides. But, as the zero-sequence currents can exist on the line-side of thegrounded star winding, the zero-sequence leakage impedance of the transformer is connected toground on the star side of the transformer and an open circuit exists between the two windings.The equivalent circuit for this connection is shown in Fig. 4.50.

Figure 4.50: The zero-sequence equivalent circuit of a Star-Delta transformer with Star-side neutralgrounded

(e) Star-delta connections with ungrounded neutral grounded: Since the neutral is isolated, no zero-sequence current can exist in the star side of the transformer and as a consequence zero-sequencecurrents can not exist in the delta side. The transformer is, therefore, represented as an open-circuit and the equivalent circuit is shown in Fig. 4.51.

Figure 4.51: The zero-sequence equivalent circuit of a Star-Delta trnasformer with Star-side neutralungrounded

(f) Delta-delta connections with ungrounded neutral grounded: In this case, the zero-sequence cur-rents can only circulate within the closed delta windings and can not exit on line sides of both

160

the windings. Hence, an open circuit exists between the two windings as far as zero-sequence cur-rents are concerned. To permit the circulating zero-sequence current to exist, the zero-sequenceleakage impedance is represented as a closed path with the ground. The equivalent circuit isshown in Fig. 4.52.

Figure 4.52: The zero-sequence equivalent circuit of a Delta-Delta transformer

Figure 4.53: The zero-sequence equivalent circuit of a Star-Delta transformer with neutral groundedthrough impedance

Point to remember: If the neutral of a transformer is grounded through a grounding impedanceZn, as shown in Fig. 4.53, then, the total zero-sequence equivalent impdance to be used in theequivalent circuit is

Z0total = Z0 + 3Zn (4.95)

This is due to the fact that the neutral current is 3 times the zero-sequence current per phase.Next, the concepts of unsymmetrical fault analysis are developed with help of Thevenin’s equiv-

alent circuit of sequence networks and symmetrical components in the next lecture.

161

4.9.5 Single line to ground (LG) fault analysis :

An unloaded balanced three-phase synchronous generator with neutral grounded through an impedanceZn is shown in Fig. 4.54. Suppose a single line to ground fault (LG) occurs on phase ‘a’ thoughan impedance Zf .

Figure 4.54: LG fault on phase ‘a’ of an unloaded generator

Since the generator is unloaded, the following terminal conditions exist at the fault point:

Va = Zf Ia

Ib = 0 (4.96)

Ic = 0

Substituting Ib = Ic = 0 in equation (4.86), the symmetrical components of currents can becalculated as: ⎡⎢⎢⎢⎢⎢⎢⎣

Ia0

Ia1

Ia2

⎤⎥⎥⎥⎥⎥⎥⎦= 1

3

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a a2

1 a2 a

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Ia00

⎤⎥⎥⎥⎥⎥⎥⎦(4.97)

162

Solving the above equation, the values of the symmetrical components of fault current Ia are:

Ia0 = Ia1 = Ia2 =13 Ia (4.98)

The voltage of phase a can be expressed in terms of symmetrical components from equation (4.83),as

Va = Va0 + Va1 + Va2 (4.99)

Substituing in the equation the values of Va0, Va1 and Va2 from equation (4.94) into equation (4.99),Va can be written as (with Ia0 = Ia1 = Ia2 from equation (4.98)):

Va = Ea − (Z0 + Z1 + Z2)Ia0 (4.100)

From equations (4.96) and (4.98), Va = Zf Ia = 3Zf Ia0. Hence, equation (4.100) can be expressed as:

3Zf Ia0 = Ea − (Z0 + Z1 + Z2)Ia0

or,

Ia0 =Ea

Z0 + Z1 + Z2 + 3Zf(4.101)

The fault current, therefore, is:

If = Ia = 3Ia0 =3Ea

Z0 + Z1 + Z2 + 3Zf(4.102)

From equations (4.98) and (4.101), it be easily interpreted that the three sequence networks areconnected in series as shown in Fig. 4.55.

Figure 4.55: Connection of sequence networks for LG fault

Note that, for solidly grounded generator, Zn = 0 and for bolted fault Zf = 0.Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s equiv-

163

alent circuit (as seen from the fault point) is obtained, individually for the three sequence networks.For the positive sequence network Vth, the open circuit pre-fault voltage at the fault point, and Z1th,the positive sequence Thevenin’s equivalent impedance as seen from the fault point are determined.For negative and zero sequence networks, only the Thevenin’s equivalent impedances Z2th and Z0th,respectively are calculated. The three Thevenin’s equivalent networks are then connected in series.

4.9.6 Line to Line (LL) fault analysis :

Fig. 4.56 shows a line to line fault (LL) between phases ‘b’ and ‘c’ through an impedance Zf , onan unloaded three phase generator. The terminal conditions at the fault point are:

Figure 4.56: LL fault between phases ‘b’ and ‘c’ of an unloaded generator

Vb − Vc = Zf Ib

Ib + Ic = 0 (4.103)

164

Ia = 0

Substituting Ia = 0 and Ib = −Ic in equation (4.86), the symmetrical components of cuurents can becalculated as: ⎡⎢⎢⎢⎢⎢⎢⎣

Ia0

Ia1

Ia2

⎤⎥⎥⎥⎥⎥⎥⎦= 1

3

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a a2

1 a2 a

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0Ib−Ib

⎤⎥⎥⎥⎥⎥⎥⎦(4.104)

Solving the above equation, the values of the symmetrical components of the current Ia are:

Ia0 = 0

Ia1 =13(a − a2)Ib (4.105)

Ia2 =13(a

2 − a)Ib = −Ia1

From equation (4.83), we have

Vb − Vc = (a2 − a)(Va1 − Va2) = Zf Ib (4.106)

Substituting Va1 and Va2 from equation (4.94) and noting that Ia1 = −Ia2, one can write:

(a2 − a) [Ea − (Z1 + Z2)Ia1] = Zf Ib (4.107)

Also from equation (4.105),

Ib =3Ia1

(a − a2) (4.108)

Substituting this value of Ib in equation (4.107), we get:

[Ea − (Z1 + Z2)Ia1] =3Zf Ia1

(a − a2)(a2 − a)

Since, (a − a2)(a2 − a) = 3, the above expression can be simplified and written as:

Ia1 =Ea

(Z1 + Z2 + Zf)(4.109)

The phase currents during fault can be calculated as:

⎡⎢⎢⎢⎢⎢⎢⎣

IaIbIc

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0Ia1

−Ia1

⎤⎥⎥⎥⎥⎥⎥⎦(4.110)

Solving for the phase currents, the expressions for Ib and Ic can be written as:

165

Ib = −Ic = (a2 − a)Ia1 (4.111)

Substituting Ib from equation (4.111) in equation (4.106) one gets:

(Va1 − Va2) = Zf Ia1

The equivalent circuit of the fault in terms of the sequence networks is shown in Fig. 4.57. Thecircuit has been drawn on the basis of equation (4.105) and the above equation. It shows that thepositive sequence and negative sequence networks are connected in phase opposition bridged by thefault impedance Zf . Also, since Ia0 = 0, the zero sequence network is open circuited and hence isnot shown in the diagram.

Figure 4.57: Connection of sequence networks for an LL fault between phases ‘b’ and ‘c’ of anunloaded generator

Extending the above concept to LL fault calculations in a power system, it can be concluded that,the Thevenin’s equivalent positive and negative sequence networks, as seen from the fault point, canbe connected in phase opposition through the fault impdedance for calculating fault current.

4.9.7 Double Line to ground (LLG) fault analysis :

Fig. 4.58 shows a double line to ground (LLG) fault on phases ‘b’ and ‘c’ through an impedanceZf on an unloaded three phase generator. The terminal conditions at the fault point are:

Vb = Vc = Zf If = Zf(Ib + Ic)

Ia = Ia1 + Ia2 + Ia0 = 0 (4.112)

From equation (4.83), Vb and Vc can be written as:

Vb = Va0 + a2Va1 + aVa2

Vc = Va0 + aVa1 + a2Va2 (4.113)

166

Figure 4.58: LLG fault between phases ‘b’ and ‘c’ of an unloaded generator

Since Vb = Vc, from equation (4.113), one can write

Va1 = Va2 (4.114)

Substituting Ib and Ic in terms of their sequence components from equation (4.85), voltage of phase’b’ can be expressed as:

Vb = Zf(Ia0 + a2Ia1 + aIa2 + Ia0 + aIa1 + a2Ia2)

167

Vb = Zf(Ia0 + a2Ia1 + aIa2 + Ia0 + aIa1 + a2Ia2)= Zf(2Ia0 + (a2 + a)(Ia1 + Ia2))= Zf(2Ia0 − (Ia1 + Ia2))

Since 1 + a + a2 = 0 and Ia = Ia1 + Ia2 + Ia0 = 0, hence

Vb = 3Zf Ia0 (4.115)

Further substituting Vb from equation (4.115) and the condition of equation (4.114) in equation(4.113), we get:

3Zf Ia0 = Va0 + (a2 + a)Va1

= Va0 − Va1 (4.116)

Substituting Va0 and Va1 from equation (4.94) in equation (4.116), the zero sequence component of

Figure 4.59: Connection of sequence networks for an LLG fault between phases ‘b’ and ‘c’ of anunloaded generator

current Ia0 is given by:

Ia0 = −(Ea − Z1Ia1)(Z0 + 3Zf)

(4.117)

For calculating the negative sequence component of current, Ia2, substitute Va1 and Va2 from equation(4.94) in equation (4.114). The expression for Ia2 is:

Ia2 = −(Ea − Z1Ia1)

Z2(4.118)

Finally, by substituting Ia0 and Ia2 from equations (4.117) and (4.118) in equation (4.112), the value

168

of the positive sequence component of current Ia1 is found out as:

Ia1 =Ea

Z1 +Z2(Z0 + 3Zf)

(Z0 + Z2 + 3Zf)

(4.119)

Since Vb = Zf If , from equation (4.115) one can conclude that

If = 3Ia0 (4.120)

The equivalent circuit for the fault in terms of the sequence networks is shown in Fig. 4.59. Thecircuit shown in Fig. 4.59 is based on equations (4.114) and (4.116). For LLG fault calculations ina power system, the Thevenin’s equivalent of the three sequence networks, as seen from the faultpoint, are found out. The positive and negative sequence equivalents are connected in parallel andthe combination is then connected to the zero sequence network through 3Zf . In the next lecture,we will study the procedure of unbalanced fault analysis using ZBUS matrix.

169

4.10 Unbalanced fault analysis using ZBUS matrix:

In the previous section, it is observed that, for fault calculations the Thevenin’s equivalent networks,at the fault point, are needed for the three sequence networks. Since the three sequence networks areindependent, the ZBUS matrices of these sequence networks can be found seperately. The diagonalelements of the three sequence ZBUS matrices infact, are the Thevenin’s equivalent impedancesof the sequence networks as seen from the various buses. Let, the three sequence bus impedancematrices for zero sequence, positive sequence and negative sequence networks be represented as[Z(0)BUS], [Z

(1)BUS] and [Z(2)BUS] respectively. If the fault is at the kth bus, then Z(0)kk , Z(1)kk and Z(2)kk

of the sequence bus impedance matrices are the zero, positive and negative Thevenin’s equivalentimpedances, respectively, as seen from the faulted bus. Hence, the first step in the fault analysisusing ZBUS matrix is the determination of the three sequence networks and subsequently, findingthe bus impedance matrix for each sequence network.

To illustrate this step, consider the single line diagram of the power system shown in Fig. 4.60.

Figure 4.60: Single line diagram of Power System

The positive sequence equivalent network for the system is shown in Fig. 4.61. In this figure allthe elements of the system have been represented by their positive sequence equivalents.

Similarly by representing all elements by their negative sequence impedances, the negative se-quence equivalent network can be obtained. The negative sequence network is shown in Fig. 4.62.For the zero sequence equivalent network, the generator neutral connections and transformer con-nections have to be considered. The zero sequence equivalent network is shown in Fig. 4.63. Inthe next step, the [ZBus] matrix for the three sequence networks is found using [ZBus] buildingalgorithm. Once [Z(0)Bus], [Z

(1)Bus] and [Z(2)Bus] matrices are known, the following procedure is followed

for the fault analysis of the given network.

170

Figure 4.61: The positive sequence equivalent network

Figure 4.62: The negative sequence equivalent network

(a) LG fault: Let the fault be on phase ‘a’ of bus ‘k’ with a fault impedance Zf as shown in Fig.4.64.

From equations (4.98) and (4.101), it can be seen that the three equivalent sequence networksare in series for calculating the sequence components of the fault currents. Hence, generalizingequation (4.101) for fault at kth bus, the expression for sequence component of fault current canbe written as:

I(0)k (F ) = I(1)k (F ) = I(2)k (F ) = Vk(0)Z(1)(kk) + Z

(2)(kk) + Z

(0)(kk) + 3Zf

(4.121)

• Z(0)kk , Z(1)kk and Z(2)kk are the kth diagonal elements of [Z(0)Bus], [Z

(1)Bus] and [Z(2)Bus] matrices

respectively.

• Vk(0) is the prefault voltage of kth bus, usually taken as 1∠0○ pu.

171

Figure 4.63: The zero sequence equivalent network

Figure 4.64: LG fault on phase ‘a’ of kth bus

The fault current is given by:

[I(abc)k (F)] = [A] [I(012)

k (F)] (4.122)

(b) LL fault: Let the fault be between phases phase ‘b’ and phase ‘c’ of bus ‘k’ through animpedance Zf as shown in Fig. 4.65. From equation (4.109) and Fig. 4.57 it is observed that thepositive sequence and negative sequence equivalent networks are connected in phase opposition.Thus, the expression of equation (4.109) for the sequence components of fault current at bus kcan be generalized as:

I(0)k (F ) = 0

172

Figure 4.65: LL fault between phase ‘b’ and phase ‘c’ of kth bus

and

I(1)k (F ) = Vk(0)Z(0)kk + Z(1)kk + Z(2)kk

= −I(2)k (F ) (4.123)

The phase components of fault current is the calculated from equation (4.122)

Ik(F ) = I(b)k (F ) = −I(c)k (F ) (4.124)

(c) LLG fault: Fig. 4.66 shows an LLG fault involving phases phase ‘b’ and phase ‘c’ of bus ‘k’through an impedance Zf . Referring to equation (4.119) and Fig.4.66, the generalized expression

Figure 4.66: LLG fault involving phase ‘b’ and phase ‘c’ of kth bus

for sequence components of fault current at bus k can be written as

I(1)k (F ) = Vk(0)

Z(1)kk + Z(2)kk (Z(0)kk + 3Zf)

Z(2)kk + Z(0)kk + 3Zf

173

I(2)k (F ) = − Vk(0) − Z(1)kk I

(1)k (F )

Z(2)kk

(4.125)

I(0)k (F ) = − Vk(0) − Z(1)kk I

(1)k (F )

Z(0)kk + 3Zf

The phase currents can be obtained from equation (4.122), the fault current is then calculatedas

Ik(F ) = I(b)k (F ) + I(c)k (F ) (4.126)

4.10.1 Calculation of Bus voltages and Line currents during fault:

To calculate the voltages of buses during fault equation (4.94) can be generalized as:

V (0)i (F ) = −Z(0)ik I(0)k (F )

V (1)i (F ) = V (1)i (0) − Z(1)ik I(1)k (F ) (4.127)

V (2)i (F ) = −Z(2)ik I(2)k (F )

The pre fault voltage V 1i (0)is usually set as 1.0 ∠00 pu.

The bus phase voltage during fault is calculated from the following relation.

V (abc)i (F ) = [A] [V (012)i (F )] (4.128)

where [A] is the symmetrical component transformation matrix.

To calculate the symmetrical components of line currents in the line from bus i to bus j thefollowing relation is used:

I(0)ij (F ) =V (0)i (F ) − V (0)j (F )

z(0)ij

I(1)ij (F ) =V (1)i (F ) − V (1)j (F )

z(1)ij(4.129)

I(2)ij (F ) =V (2)i (F ) − V (2)j (F )

z(2)ij

where z(0)ij , z(1)ij and z(2)ij are the zero, positive and negative sequence impedance respectively of

the line between bus i and bus j. The phase currents for the line can be calculated from thesymmetrical components using the relation:

174

[Iabcij (F )] = [A][I012ij (F )] (4.130)

The process of fault analysis of a power system network is illustrated in the next lecture withthe help of an example.

175

4.11 Example of fault calculation for three phase and LGfaults in power system network

A single line diagram of a power system is shown in Fig. 4.67 and the system data is as follows:-

• Generators G1 and G2 : X1 = X2 = 0.2 pu, X0 = 0.05 pu

• Transformers T1 and T2 : X1 = X2 = X0 = X` = 0.05 pu

• Transmission Lines L1, L2 and L3 : X1 = X2 = 0.1 pu, X0 = 0.3 pu

Figure 4.67: Single line diagram of the power System of the example

Prefault voltage for all buses is taken as Vi(0) = 1.0∠00 pu ∀ i = 1,2,3.

We wish to carry out the complete short-circuit analysis of the system for:

(a) three phase bolted fault at bus 5

(b) LG fault with Zf = 0.1 pu at bus 5

(c) LL fault with Zf = 0.1 pu at bus 5

(d) LLG fault with Zf = 0.0 pu at bus 5

Solution:

(a) Three phase fault at bus 5

For the three phase bolted fault, only positive sequence network and the positive sequence busimpedance matrix [Z(1)Bus] is required. The positive sequence network for the power system ofFig. 4.67 is shown in Fig. 4.68. In this diagram all the elements have been replaced by their perunit positive sequence impedances.

The [Z(1)Bus] matrix for the network of the Fig. 4.68 is given below:

176

Figure 4.68: Positive sequence equivalent network of Fig. 4.67

[Z(1)Bus] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 3 4 5

1 j0.1294 j0.0706 j0.1118 j0.0882 j0.102 j0.0706 j0.1294 j0.0882 j0.1118 j0.103 j0.1118 j0.0882 j0.1397 j0.1103 j0.12504 j0.0882 j0.1118 j0.1103 j0.1397 j0.12505 j0.10 j0.10 j0.1250 j0.1250 j0.1750

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

pu

The sequence component of three phase fault current at bus 5 are given as, from equation (4.73):

[I(012)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

01Z1

550

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

01

j0.17500

⎤⎥⎥⎥⎥⎥⎥⎥⎦

=⎡⎢⎢⎢⎢⎢⎢⎣

0−j5.7143

0

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of the fault current are calculated using equation (4.122):

[I(abc)5 (F)] = [A][I(012)

5 (F)]

[I(abc)fault ] = [I(abc)

5 (F)] =⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0−j5.7143

0

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

5.7143∠− 900

5.7143∠1500

5.7143∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus voltages during fault

Bus 1:

177

V (1)1 (F ) = V (1)1 (0) − Z(1)15 I(1)5 (F )

= 1.0 − j0.10 ∗ (−j5.7143)= 0.42857∠00pu

Since this is a balanced fault, V (a)1 (F ) = V (1)1 (F )

[V(abc)1 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.42857∠00

0.42857∠− 1200

0.42857∠1200

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 2:

V (1)2 (F ) = V (1)2 (0) − Z(1)25 I(1)5 (F )

= 1.0 − j0.10 ∗ (−j5.7143)= 0.42857∠00pu

Since this is a balanced fault V (a)2 (F ) = V (1)2 (F )

[V(abc)2 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.42857∠00

0.42857∠− 1200

0.42857∠1200

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 3:

V (1)3 (F ) = V (1)3 (0) − Z(1)35 I(1)5 (F )

= 1.0 − j0.125 ∗ (−j5.7143)= 0.28571∠00pu


[V(abc)3 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.28571∠00

0.28571∠− 1200

0.28571∠1200

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 4:

178

V (1)4 (F ) = V (1)4 (0) − Z(1)45 I(1)5 (F )

= 1.0 − j0.125 ∗ (−j5.7143)= 0.28571∠00pu


[V(abc)4 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.28571∠00

0.28571∠− 1200

0.28571∠1200

⎤⎥⎥⎥⎥⎥⎥⎦pu

The bus voltage of bus 5 under faulted condition is V(abc)5 (F) =

⎡⎢⎢⎢⎢⎢⎢⎣

000

⎤⎥⎥⎥⎥⎥⎥⎦pu because the fault

impedance is zero.

Line Currents during fault

For line L1 from bus 3 to bus 4 the positive sequence component for line current (I(1)34 (F )) iscalculated as:

I(1)34 (F ) = V(1)

3 (F ) − V (1)4 (F )Z(1)34

= 0.28571 − 0.28571j0.1 = 0

Hence, the phase components of line current are

I(abc)34 (F) =

⎡⎢⎢⎢⎢⎢⎢⎣

000

⎤⎥⎥⎥⎥⎥⎥⎦pu


I(1)35 (F ) = V(1)

3 (F ) − V (1)5 (F )Z(1)35

= 0.28571 − 0.0j0.1 = 2.8571∠− 900pu


[I(abc)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

179


I(1)45 (F ) = V(1)

4 (F ) − V (1)5 (F )Z(1)45

= 0.28571 − 0.0j0.1 = 2.8571∠− 900pu


[I(abc)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer Currents during fault

For transformer T1 between bus 1 and bus 3 the positive sequence component fault current(I(1)13 (F )) is calculated as:

I(1)13 (F ) = V(1)

1 (F ) − V (1)3 (F )z(1)T1

= 0.42857 − 0.28571j0.05 = 2.8571∠− 900pu

The phase components of the transformer T1 current are:

[I(abc)31 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

For transformer T2 between bus 2 and bus 4 the positive sequence component fault current(I(1)24 (F )) is calculated as:

I(1)24 (F ) = V(1)

2 (F ) − V (1)4 (F )z(1)T2

= 0.42857 − 0.28571j0.05 = 2.8571∠− 900pu

The phase components of the transformer T2 current are:

[I(abc)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

Generator Currents during fault

For generator G1 connected at bus 1 the positive sequence component fault current (I(1)G1(F ))

is calculated as:

180

I(1)G1(F ) = Ea − V (1)1 (F )

z(1)G2

= 1.0 − 0.42857j0.2 = 2.8571∠− 900pu

The phase components of the generator G1 current are:

[I(abc)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

For Generator G2 connected at bus 2 the positive sequence component fault current (I(1)G2(F ))

is calculated as:

Figure 4.69: Flow of fault current in the network

I(1)G2(F ) = Ea − V (1)2 (F )

z(1)G2

= 1.0 − 0.42857j0.2 = 2.8571∠− 900pu

The phase components of the generator G2 current can be calculated as:

[I(abc)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

2.8571∠− 900

2.8571∠1500

2.8571∠300

⎤⎥⎥⎥⎥⎥⎥⎦pu

The flow of fault current in the system is shown in the single line diagram of Fig. 4.69.

(b) Single line to ground fault at bus 5

181

In this case all the sequence networks are required. The positive sequence network is same asthe one shown in the Fig. 4.68 and [Z(1)Bus] is identical to the matrix used in three phase faultanalysis.

The negative sequence equivalent network for this network is as shown in Fig. 4.70. The network

Figure 4.70: Negative sequence equivalent network

is identical to the network of Fig. 4.68 except for the voltage sources. Hence, [Z(2)Bus] = [Z(1)Bus].The zero sequence equivalent network is drawn next considering the transformer connectionsand grounding as well as generator grounding. The equivalent zero sequence networks is shownin Fig. 4.71.

Figure 4.71: Zero sequence equivalent network

An explanation of the equivalent circuit will be in order. Generators G1 and G2 have their neu-trals grounded, so their zero sequence impedances are connected to the reference. TransformerT1 has both the windings connected in star, with both neutrals solidly grounded. As a result,the zero sequence impedance of the transformer is directly connected between buses 1 and 2.

182

Transformer T2 has both the winding connected in delta, hence, no connection exists betweenthe primary and secondary sides for zero sequence currents to flow. To represent circulatingzero sequence currents in the delta connected transformer winding, it is represented as a shortcircuited winding.

[Z(0)Bus], the zero sequence bus impedance matrix is then calculated using the step-by-step ZBus

building algorithm. The zero sequence bus impedance matrix is given as:

[Z(0)Bus] =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 3 4 5

1 j0.05 0.0 j0.05 j0.05 j0.052 0.0 j0.05 0.0 0.0 0.03 j0.05 0.0 j0.10 j0.10 j0.104 j0.05 0.0 j0.10 j0.30 j0.205 j0.05 0.0 j0.10 j0.20 j0.30

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

pu

Fault current:

The sequence component of the fault current at bus 5 are given as, from equation (4.121):

I(0)5 (F ) = I(1)5 (F ) = I(2)5 (F ) = Vk(0)Z(1)55 + Z(2)55 + Z(0)55

= 1.0j0.175 + j0.175 + j0.30 = −j1.538 pu

[I(abc)fault ] = [I(abc)

5 (F)] =⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a

1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦

[I(abc)fault ] =

⎡⎢⎢⎢⎢⎢⎢⎣

4.6154∠− 900

00

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus voltages:

The bus voltage in sequence components, during fault, are calculated using equation (4.127)written in compact form as:

⎡⎢⎢⎢⎢⎢⎢⎣

V (0)i (F )V (1)i (F )V (2)i (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0Vi0

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

Z(0)ik 0 00 Z(1)ik 00 0 Z(2)ik

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)k (F )I(1)k (F )I(2)k (F )

⎤⎥⎥⎥⎥⎥⎥⎦(4.131)

where, k represents the faulted bus number and Z(0)ik , Z(1)ik and Z(2)ik are the elements of the

respective sequence bus impedance matrices.

I(0)k , I(1)k and I(2)k represent the sequence components of fault current at kth bus.

Vi(0) is the pre fault bus voltage of ith bus.

Bus 1: The sequence voltages are:

183

⎡⎢⎢⎢⎢⎢⎢⎣

V (0)1 (F )V (1)1 (F )V (2)1 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.05 0 00 j0.10 00 0 j0.10

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦Or,

[V(012)1 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−0.07690.8462−0.1538

⎤⎥⎥⎥⎥⎥⎥⎦pu

The bus voltage in the phase form is calculated using equation (4.128).

[V(abc)1 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.6154∠00

0.9638∠− 116.040

0.9638∠− 116.040

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)2 (F )V (1)2 (F )V (2)2 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

0.0 0 00 j0.10 00 0 j0.10

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[V(012)2 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.00.8462−0.1538

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)2 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.6923∠00

0.9326∠− 111.790

0.9326∠− 111.790

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)3 (F )V (1)3 (F )V (2)3 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.10 0 00 j0.125 00 0 j0.125

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[V(012)3 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−0.15380.8077−0.1923

⎤⎥⎥⎥⎥⎥⎥⎦pu

184


[V(abc)3 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.4615∠00

0.9813∠− 118.050

0.9813∠− 118.050

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)4 (F )V (1)4 (F )V (2)4 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.20 0 00 j0.125 00 0 j0.125

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[V(012)4 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−0.30760.8077−0.1923

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)4 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.3077∠00

1.0624∠− 125.400

1.0624∠− 125.400

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)5 (F )V (1)5 (F )V (2)5 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.30 0 00 j0.175 00 0 j0.175

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j1.538−j1.538

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[V(012)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−0.46150.7308−0.2692

⎤⎥⎥⎥⎥⎥⎥⎦pu

The bus voltage in the phase form is calculated using equation (4.128)

[V(abc)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.0∠00

1.087∠− 128.640

1.087∠− 128.640

⎤⎥⎥⎥⎥⎥⎥⎦pu

Observe that the phase voltage of the faulted phase ’a’ is zero due to a zero impedance fault.

Line Currents

185

The sequence components of line currents during fault are calculated using equation (4.129),written here in compact form as

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)ij (F )I(1)ij (F )I(2)ij (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1z(0)ij

0 0

0 1z(1)ij

0

0 0 1z(2)ij

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

V (0)i (F ) − V (0)j (F )V (1)i (F ) − V (1)j (F )V (2)i (F ) − V (2)j (F )

⎤⎥⎥⎥⎥⎥⎥⎦(4.132)

In equation (4.132), the line is between ith and jth buses.

z(0)ij , z(1)ij , z

(2)ij represent the respective sequence impedances of the line iÐ→ j

V (0)i (F ), V (1)i (F ), V (2)i (F ), V (0)j (F ), V (1)j (F ), V (2)j (F ) are the sequence components of volt-ages of ith and jth buses respectively during fault.

Line 1: The sequence components of line current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)34 (F )I(1)34 (F )I(2)34 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−0.1538 − (−0.3076)0.8077 − 0.8077

−0.1923 − (−0.1923)

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[I(012)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−j0.512800

⎤⎥⎥⎥⎥⎥⎥⎦pu

The line current in phase form is calculated as:

[I(abc)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.5128∠− 900

0.5128∠− 900

0.5128∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

I(0)35 (F )I(1)35 (F )I(2)35 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−0.1538 − (−0.4615)0.8077 − 0.7308

−0.1923 − (−0.2692)

⎤⎥⎥⎥⎥⎥⎥⎦

186

Or,

[I(012)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.0256−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.5641∠− 900

0.2564∠− 900

0.2564∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

I(0)45 (F )I(1)45 (F )I(2)45 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−0.3077 − (−0.4615)0.8077 − 0.7308

−0.1923 − (−0.2692)

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[I(012)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−j0.5128−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

2.0513∠− 900

0.2564∠− 900

0.2564∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer Currents

Transformer T1 ∶ The sequence components of line current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)13 (F )I(1)13 (F )I(2)13 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.05 0

0 0 1.0j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

−0.0769 − (−0.1538)0.8462 − 0.8077

−0.1538 − (−0.1923)

⎤⎥⎥⎥⎥⎥⎥⎦

Or,

[I(012)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu

187


[I(abc)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

3.0769∠− 900

0.7692∠− 900

0.7692∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer T2 ∶ The sequence components of line current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)24 (F )I(1)24 (F )I(2)24 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0∞ 0 0

0 1.0j0.05 0

0 0 1.0j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0 − (−0.3076)0.8462 − 0.8077

−0.1538 − (−0.1923)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.538∠− 900

0.7692∠− 900

0.7692∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu

Generator Currents

The sequence components of generator currents during fault are calculated using the expression

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)Gi (F )I(1)Gi (F )I(2)Gi (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1z(0)gi

0 0

0 1z(1)gi

0

0 0 1z(2)gi

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

E(0)Gi (F ) − V (0)ti (F )E(1)Gi (F ) − V (1)ti (F )E(2)Gi (F ) − V (2)ti (F )

⎤⎥⎥⎥⎥⎥⎥⎦(4.133)

where,

E(0)Gi (F ), E(1)Gi (F ), E(2)Gi (F ) the zero, positive and negative sequence generated voltages respec-tively of ith generator.V (0)ti (F ), V (1)ti (F ), V (2)ti (F ) are the zero, positive and negative sequence terminal voltages re-spectively of ith generator after fault.z(0)gi (F ), z(1)gi (F ) and z(2)gi (F ) are the sequence impedances of the ith generator.

Generator 1 : The sequence components of generator 1 current are

188

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G1(F )

I(1)G1(F )

I(2)G1(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.20 0

0 0 1.0j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0 − (−0.0769)1 − 0.8462

0 − (−0.1538)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

−j1.538−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components generator current are calculated as:

[I(abc)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

3.0769∠− 900

0.7692∠− 900

0.7692∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu

Generator 2 : The sequence components of Generator 1 current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G2(F )

I(1)G2(F )

I(2)G2(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.20 0

0 0 1.0j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0 − 01 − 0.8462

0 − (−0.1538)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0−j0.7692−j0.7692

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components generator current are calculated as:

[I(abc)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

1.538∠− 900

0.7692∠− 900

0.7692∠− 900

⎤⎥⎥⎥⎥⎥⎥⎦pu

The flow of sequence currents in the sequence networks is shown next in the Fig. 4.72. FromFig. 4.72 the following points are worth observing:

• Both generators contribute equal amount of positive and negative sequence currents as thenetwork is symmetrical as seen from the fault point.

• Since the positive and negative sequence fault voltages are equal for buses 3 and 4, thepositive and negative sequence currents through line L1 between buses 3 and 4 are zero.

189

Figure 4.72: Flow of sequence currents for LG fault at bus 5

• The zero sequence circuit of generator G2 is open circuited due to ∆ − ∆ transformer T2

as a result, G2 does not contribute any zero sequence current to the fault. Generator G1

has to provide the entire zero sequence current.

190

• Moreover, the zero sequence network is not symmetrical, hence, zero sequence voltages ofbuses 3 and 4 are not equal and as a result a zero sequence current flows through line L1.

In the next lecture, we will look into the examples of short circuit fault calculation for LL andLLG faults.

191

4.12 Example of fault calculation for LL and LLG faultin power system network

(b) Double line (LL) fault between phases ‘b’ and ‘c’ at bus 5

In this case only positive and negative sequence networks are required. Hence [Z(1)Bus] and [Z(2)Bus]as calculated previously will be used. Let the fault impedance be Zf = j0.1 pu.

Fault current calculations: The sequence components of fault current are calculated using equa-tion (4.123).

I(1)5 (F ) = 0

As zero sequence current can not flow without a ground path.

I(1)5 (F ) = V (0)5

Z(1)55 + Z(2)55 + Zf= 1.0j0.175 + j0.175 + j0.1 = −j2.2222 pu

I(2)5 (F ) = −I(1)5 (F ) = j2.222 pu

Hence

[I(012)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j2.222j2.222

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of the fault current are

[I(abc)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−3.8493.849

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus voltage calculations:

Bus 1: The sequence components of bus 1 voltage are calculated as

192

Zero SequenceV (0)1 (F ) = 0.0 pu

Positive Sequence

V (1)1 (F ) = V (1)1 (0) − Z(1)15 I(1)5 (F ) = 1.0 − j0.10 ∗ (−j2.222)

V (1)1 (F ) = 0.7778 pu

Negative Sequence

V (2)1 (F ) = −Z(2)15 I(2)5 (F ) = −j0.10 ∗ (j2.222)

V (2)1 (F ) = 0.2222 pu

Hence, the sequence components of bus 1 voltage are:

[V0121 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.77780.2222

⎤⎥⎥⎥⎥⎥⎥⎦pu

bus 1 voltage in phase form

[V(abc)(1) (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.00.6939∠− 1360

0.6939∠1360

⎤⎥⎥⎥⎥⎥⎥⎦pu



Positive Sequence

V (1)2 (F ) = V (1)2 (0) − Z(1)25 I(1)5 (F ) = 1.0 − j0.10 ∗ (−j2.222)

V (1)2 (F ) = 0.7778 pu

Negative SequenceV (2)2 (F ) = −Z(2)25 I

(2)5 (F ) = −j0.10 ∗ (j2.222)

193

V (2)2 (F ) = 0.2222 pu


[V0122 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.77780.2222

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)(2) (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.00.6939∠− 1360

0.6939∠1360

⎤⎥⎥⎥⎥⎥⎥⎦pu



Positive Sequence

V (1)3 (F ) = V (1)3 (0) − Z(1)35 I(1)5 (F ) = 1.0 − j0.125 ∗ (−j2.222)

V (1)3 (F ) = 0.7222 pu

Negative Sequence

V (2)3 (F ) = −Z(2)35 I(2)5 (F ) = −j0.125 ∗ (j2.222)

V (2)3 (F ) = 0.2778 pu


[V0123 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.72220.2778

⎤⎥⎥⎥⎥⎥⎥⎦pu


194

[V(abc)(3) (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.00.6310∠− 142.410

0.6310∠142.410

⎤⎥⎥⎥⎥⎥⎥⎦pu



Positive Sequence

V (1)4 (F ) = V (1)4 (0) − Z(1)45 I(1)5 (F ) = 1.0 − j0.125 ∗ (−j2.222)

V (1)4 (F ) = 0.7222 pu

Negative Sequence

V (2)4 (F ) = −Z(2)45 I(2)5 (F ) = −j0.125 ∗ (j2.222)

V (2)4 (F ) = 0.2778 pu


[V0124 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.72220.2778

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)(4) (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.00.6310∠− 142.410

0.6310∠142.410

⎤⎥⎥⎥⎥⎥⎥⎦pu



Positive Sequence

V (1)5 (F ) = V (1)5 (0) − Z(1)55 I(1)5 (F ) = 1.0 − j0.175 ∗ (−j2.222)

195

V (1)5 (F ) = 0.6111 pu

Negative Sequence

V (2)5 (F ) = −Z(2)55 I(2)5 (F ) = −j0.175 ∗ (j2.222)

V (2)5 (F ) = 0.3889 pu


[V0125 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.61110.3889

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)(5) (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.00.5358∠− 158.950

0.5358∠158.950

⎤⎥⎥⎥⎥⎥⎥⎦pu

Line current calculations:


⎡⎢⎢⎢⎢⎢⎢⎣

I(0)34 (F )I(1)34 (F )I(2)34 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

00.7222 − 0.72220.2778 − 0.2778

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

000

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

000

⎤⎥⎥⎥⎥⎥⎥⎦pu


196

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)35 (F )I(1)35 (F )I(2)35 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

00.7222 − 0.61110.2778 − 0.3889

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu


⎡⎢⎢⎢⎢⎢⎢⎣

I(0)45 (F )I(1)45 (F )I(2)45 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.3 0 0

0 1.0j0.10 0

0 0 1.0j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

00.7222 − 0.61110.2778 − 0.3889

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer current calculations:

Transformer 1: The sequence components of transformer 1 current are:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)13 (F )I(1)13 (F )I(2)13 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.05 0

0 0 1.0j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0(0.7778 − 0.7222)(0.2222 − 0.2778)

⎤⎥⎥⎥⎥⎥⎥⎦

197

[I(012)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu

The transformer current in phase form is calculated as:

[I(abc)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer 2: The sequence components of transformer 2 current are:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)24 (F )I(1)24 (F )I(2)24 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.05 0

0 0 1.0j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0(0.7778 − 0.7222)(0.2222 − 0.2778)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu

The transformer current in phase form is calculated as:

[I(abc)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu

Generator current calculations:

Generator 1: The sequence components of generator 1 current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G1(F )

I(1)G1(F )

I(2)G1(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.20 0

0 0 1.0j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0(1 − 0.7778)(0 − 0.2222)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu

198

The generator current in phase form is calculated as:

[I(abc)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu

Generator 2: The sequence components of generator 2 current are

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G2(F )

I(1)G2(F )

I(2)G2(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0j0.05 0 0

0 1.0j0.20 0

0 0 1.0j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

0(1 − 0.7778)(0 − 0.2222)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.1111j1.1111

⎤⎥⎥⎥⎥⎥⎥⎦pu

The generator current in phase form is calculated as:

[I(abc)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0−1.92451.9245

⎤⎥⎥⎥⎥⎥⎥⎦pu

From Fig. 5.54 it can be seen that:

V (1)5 (F ) = V (2)5 (F ) + Zf I(1)5 (F ) = 0.3889 + j0.1 ∗ −j2.222

V (1)5 (F ) = 0.6111pu

This value is same as the one calculated earlier.

The flow of sequence currents in the sequence networks is shown next in the Fig.4.73.

(c) Double line to ground fault (LLG) fault involving phases ‘b’ and ‘c’ at bus 5

In this case all the three sequence networks are required. Hence, [Z(0)Bus],[Z(1)Bus],and [Z(2)Bus] as

calculated earlier will be used. It is assumed that the fault impedance Zf = 0.

Fault current calculations:The sequence components of fault current are calculated using equation (4.125) as follows:

Positive Sequence Current

199

Figure 4.73: Flow of sequence currents for LL fault at bus 5

I(1)5 (F ) = 1.0

Z(1)55 + Z(2)55 (Z(0)55 + 3Zf)

Z(2)55 + Z(0)55 + 3Zf

I(1)5 (F ) = 1.0

j0.175 + j0.175(j0.175 + 3 ∗ 0)j0.175 + j0.3 + 3 ∗ 0

I(1)5 (F ) = −j3.5023 pu

The negative and zero sequence currents are calculated using current division as:

Negative Sequence Current

I(2)5 (F ) = − Z(0)55

Z(0)55 + Z(2)55

∗ I(1)5 (F )

= − j0.3(j0.3 + j0.175) ∗ (−j3.5023)

I(2)5 (F ) = j2.212 pu

Zero Sequence Current

200

I(0)5 (F ) = − Z(2)55

Z(0)55 + Z(2)55

∗ I(1)5 (F )

= − j0.175(j0.3 + j0.175) ∗ (−j3.5023)

I(0)5 (F ) = j1.2903 pu

Hence, the fault current in sequence components is :

[I(012)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of the fault current are :

[I(abc)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

05.3137∠158.640

5.3137∠21.360

⎤⎥⎥⎥⎥⎥⎥⎦pu

The total fault current

I(F ) = I(b)5 (F ) + I(c)5 (F )= 5.3137∠158.640 + 5.3137∠21.360

I(F ) = 3.871∠900 pu

Bus Voltage Calculations:

Bus 1:

The sequence components of bus 1 voltage are calculated as

⎡⎢⎢⎢⎢⎢⎢⎣

V (0)1 (F )V (1)1 (F )V (2)1 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.05 0 00 j0.10 00 0 j0.10

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦

201

[V(012)1 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.06450.64970.2212

⎤⎥⎥⎥⎥⎥⎥⎦pu

The bus voltage in the phase form is calculated as:

[V(abc)1 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.9355∠00

0.5248∠− 134.990

0.5248∠134.990

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 2:


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)2 (F )V (1)2 (F )V (2)2 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

0 0 00 j0.10 00 0 j0.10

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦

[V(012)2 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

00.64970.2212

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)2 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.8710∠00

0.5722∠− 139.560

0.5722∠139.560

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 3:


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)3 (F )V (1)3 (F )V (2)3 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.10 0 00 j0.125 00 0 j0.125

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦

[V(012)3 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.12900.56220.2765

⎤⎥⎥⎥⎥⎥⎥⎦pu


202

[V(abc)3 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.9766∠00

0.3815∠− 139.560

0.3815∠139.560

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 4:


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)4 (F )V (1)4 (F )V (2)4 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.20 0 00 j0.125 00 0 j0.125

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦

[V(012)4 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.25810.56220.2765

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)4 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.0968∠00

0.2954∠− 123.100

0.2954∠123.100

⎤⎥⎥⎥⎥⎥⎥⎦pu

Bus 5:


⎡⎢⎢⎢⎢⎢⎢⎣

V (0)5 (F )V (1)5 (F )V (2)5 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

01.00

⎤⎥⎥⎥⎥⎥⎥⎦−⎡⎢⎢⎢⎢⎢⎢⎣

j0.30 0 00 j0.175 00 0 j0.175

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j3.503j2.212

⎤⎥⎥⎥⎥⎥⎥⎦

[V(012)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.38710.38710.3871

⎤⎥⎥⎥⎥⎥⎥⎦pu


[V(abc)5 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

1.1613∠00

00

⎤⎥⎥⎥⎥⎥⎥⎦pu

Line Current Calculations:

Line L1 ∶

203

The sequence components of Line 1 current are calculated as

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)34 (F )I(1)34 (F )I(2)34 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.30 0 0

0 1j0.10 0

0 0 1j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.1290 − 0.2581)(0.5622 − 0.5622)(0.2765 − 0.2765)

⎤⎥⎥⎥⎥⎥⎥⎦

Hence,

[I(012)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j0.430100

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of line 1 fault current are

[I(abc)34 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j0.4301j0.4301j0.4301

⎤⎥⎥⎥⎥⎥⎥⎦pu

Line L2 ∶The sequence components of Line 2 current are calculated as

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)35 (F )I(1)35 (F )I(2)35 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.30 0 0

0 1j0.10 0

0 0 1j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.1290 − 0.3871)(0.5622 − 0.3871)(0.2765 − 0.3871)

⎤⎥⎥⎥⎥⎥⎥⎦

Hence,

[I(012)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j0.8602−j1.7511j1.1060

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)35 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.2151∠900

2.7425∠154.250

0.2954∠25.550

⎤⎥⎥⎥⎥⎥⎥⎦pu

Line L3 ∶

204

The sequence components of Line 3 current are calculated as

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)45 (F )I(1)45 (F )I(2)45 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.30 0 0

0 1j0.10 0

0 0 1j0.10

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.2581 − 0.3871)(0.5622 − 0.3871)(0.2765 − 0.3871)

⎤⎥⎥⎥⎥⎥⎥⎦

Hence,

[I(012)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j0.4301−j1.7511j1.1060

⎤⎥⎥⎥⎥⎥⎥⎦pu


[I(abc)45 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.2151∠− 900

2.5863∠163.080

2.5863∠16.920

⎤⎥⎥⎥⎥⎥⎥⎦pu

Transformer Current Calculations:

Transformer T1 ∶

The sequence components of Transformer 1 current are calculated as:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)13 (F )I(1)13 (F )I(2)13 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.05 0 0

0 1j0.05 0

0 0 1j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.0645 − 0.1290)(0.6497 − 0.5622)(0.2212 − 0.2765)

⎤⎥⎥⎥⎥⎥⎥⎦

Hence,

[I(012)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j1.750j1.106

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of Transformer 1 fault current are

[I(abc)13 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.6452∠900

2.9536∠146.900

2.9536∠33.100

⎤⎥⎥⎥⎥⎥⎥⎦pu

205

Transformer T2 ∶The sequence components of Transformer 2 current are calculated as

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)24 (F )I(1)24 (F )I(2)24 (F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1∞ 0 0

0 1j0.05 0

0 0 1j0.05

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.0 − 0.2581)(0.6497 − 0.5622)(0.2212 − 0.2765)

⎤⎥⎥⎥⎥⎥⎥⎦

Hence,

[I(012)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0−j1.750j1.106

⎤⎥⎥⎥⎥⎥⎥⎦pu

The phase components of Transformer 2 fault current are

[I(abc)24 (F)] =

⎡⎢⎢⎢⎢⎢⎢⎣

0.6452∠− 900

2.4953∠172.570

2.4953∠7.430

⎤⎥⎥⎥⎥⎥⎥⎦pu

Figure 4.74: Flow of sequence currents for LLG fault at bus 5

Generator Current Calculations:

Line G1 ∶ The sequence components of Generator 1 current are calculated as:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G1(F )

I(1)G1(F )

I(2)G1(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.05 0 0

0 1j0.20 0

0 0 1j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.0 − 0.0645)(1.0 − 0.6497)(0.0 − 0.2212)

⎤⎥⎥⎥⎥⎥⎥⎦

206

[I(012)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

j1.290−j1.750j1.106

⎤⎥⎥⎥⎥⎥⎥⎦pu

[I(abc)G1

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0.6452∠00

2.9536∠146.900

2.9536∠33.100

⎤⎥⎥⎥⎥⎥⎥⎦pu

Line G2 ∶ The sequence components of Generator 2 current are calculated as:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)G2(F )

I(1)G2(F )

I(2)G2(F )

⎤⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1j0.05 0 0

0 1j0.20 0

0 0 1j0.20

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.0 − 0.0)(1.0 − 0.6497)(0.0 − 0.2212)

⎤⎥⎥⎥⎥⎥⎥⎦

[I(012)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0.0−j1.750j1.106

⎤⎥⎥⎥⎥⎥⎥⎦pu

[I(abc)G2

(F)] =⎡⎢⎢⎢⎢⎢⎢⎣

0.6452∠00

2.4953∠172.570

2.4953∠7.430

⎤⎥⎥⎥⎥⎥⎥⎦pu

The flow of sequence currents in the sequence networks is shown in the Fig.4.74.

We will be discussing open-conductor faults in the next lecture.

207

4.13 Open Conductor Faults:

When one or two phases of a balanced three-phase line opens it creates an unbalance in thesystem and results in the flow of unbalanced currents. Such conditions occur in the system whenone or two conductors of a trnansmission line are broken due to storm or if fuses, isolators orcircuit breakers operate only on one or two phases leaving others connected. Such open conductorfaults can also be analysed with the help of [ZBus] matrices of sequence networks.

Figure 4.75: Open Conductor faults on a section of three phase system

In Fig. 4.75, a section of a three phase system between buses i and j is shown. Fig. 4.75 (a)shows one conductor open while Fig. 4.75 (b) shows two conductors open between points k andk′. The positive direction of currents Ia, Ib and Ic are shown in the figure. For the analysis ofsuch faults, the Thevenin’s impedance between two buses i and j is required and the relationshipbetween the elements of [ZBus] and Thevenin’s impedances at each bus of the network needsto be established.

Let [V0] be the vector of open-circuit bus voltages corresponding to the initial (pre-fault) valueof bus current vector [I0] injected in a network with bus impedance matrix [ZBus]. We canthen write

208

[V0] = [ZBus][I0] (4.134)

If the bus currents are changed to a new value, [I0 +∆I], the new bus voltage [V] can beexpressed as:

[V] = [ZBus] [I0 +∆I]= [ZBus][I0] + [ZBus][∆I] (4.135)

= [V0] + [∆V]

where,[∆V] represents the change in the values of the original bus voltage [V0].

Fig. 4.76 represents a power system with buses i and j taken out along with the reference node.The circuit is not energised so that [V0] and [I0] are zero. Currents [∆Ii] and [∆Ij] areinjected into the ith and jth buses respectively, through current sources connected between thenode and the reference node.

Figure 4.76: Change in bus voltage [∆V] due to current [∆Ii] and [∆Ij]

209

The changes in bus voltage [∆V] can be calculated from equation (4.135) as

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆V1

⋮∆Vi∆Vj⋮

∆VN

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 i j N

1 Z11 ⋯ Z1i Z1j ⋯ Z1N

⋮ ⋮ ⋮ ⋮i Zi1 ⋯ Zii Zij ⋯ ZiNj Zj1 ⋯ Zji Zjj ⋯ ZjN

⋮ ⋮ ⋮ ⋮N ZN1 ⋯ ZNi ZNj ⋯ ZNN

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆I1

⋮∆Ii∆Ij⋮

∆IN

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

∆V1

⋮∆Vi∆Vj⋮

∆VN

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

Z1i∆Ii + Z1j∆Ij⋮

Zii∆Ii + Zij∆IjZji∆Ii + Zjj∆Ij

⋮ZNi∆Ii + ZNj∆Ij

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(4.136)

The modified voltage at ith bus can be written as :

Vi = V 0i +∆Vi = V 0

i + Zii∆Ii + Zij∆Ij (4.137)

adding and subtracting Zij∆Ii in equation (4.137), one obtains

Vi = V 0i + (Zii − Zij)∆Ii + Zij(∆Ii +∆Ij) (4.138)

Similarly the modified voltage at jth bus can be written as

Vj = V 0j + ∆Vj = V 0

j + Zji∆Ii + Zjj∆Ij (4.139)

adding and subtracting Zji∆Ij in equation (4.139) , one obtains

Vj = V 0j + (Zjj − Zji)∆Ij + Zji(∆Ii +∆Ij) (4.140)

Since the network is symmetricalZji = Zij (4.141)

Thus the equations (4.138) and (4.140) can be represented by an equivalent circuit shown inFig. 4.77, which is also the Thevenin’s Equivalent circuit of the network as seen from the ith

and jth buses.

From the figure it can be observed that the Thevenin’s open circuit voltage between ith and jth

buses is (V 0i − V 0

j ).

210

Figure 4.77: The Thevenin’s Equivalent of the original network

To calculate the open circuit impedance between ith and jth buses, the initial voltages V 0i and

V 0j are set equal to zero and an ideal current source I is connected between the two busses.

Next, the resulting voltages Vi and Vj are calculated. Note that ∆Ii = I and ∆Ij = −I .

Vi = (Zii − Zij)I (4.142)

Vj = (Zjj − Zji)(−I) (4.143)

Next, calculate the voltage difference ∆Vij between ith and jth buses as:

∆Vij = Vi − Vj = (Zii + Zjj − 2Zij)I (4.144)

Hence,

∆ZThevenin,ij =∆VijI

= (Zii + Zjj − 2Zij) (4.145)

Once the Thevenin’s equivalent is established, the analysis of open-conductor faults can proceedfurther. The opening of all the three phases is equivalent to the removal of the line i → j totallyfrom the network. If z(0)ij , z(1)ij and z(2)ij are the the three sequence impedance of the line i →j, then the removal of this line from the network can be simulated by adding −z(0)ij , −z(1)ij and−z(2)ij to the corresponding Thevenin’s equivalent network of the three sequence networks of theoriginal network as seen from ith and jth buses.

Let x represents the fractional length of the broken line i → j from ith bus to the break point‘k’, where 0 ≤ x ≤ 1.

211

The positive sequence impedance of the conductor segment between the ith bus and the point ofbreak k is xz(1)ij , and the positive sequence impedance of the remaining conductor from point k tojth bus is (1−x)z(1)ij . These two impedances are then added to represent the broken conductor.This is illustrated in Fig. 4.78

Figure 4.78: Positive sequence equivalent network with line open between buses k and k’

If V (a)kk′ , V(b)kk′ and V (c)kk′ represent the phase component of voltage drops between points k and

k’, then V (0)kk′ , V(1)kk′ and V (2)kk′ represent the sequence components of the voltage drops between

points k and k’. These sequence voltages have different values depending on the type of openconductor fault.

To further simplify the circuit, the voltage V (1)kk′ and the total series impedance [xz(1)ij + (1 −

x)z(1)ij ] = z(1)ij is replaced by a current sourceV (1)kk′

z(1)ijand a parallel impedance z(1)ij as shown in

Fig. 4.79.

Further, the parallel combination of z(1)ij and −z(1)ij is∞ and hence, is replaced by an open circuit.The final simplified positive sequence impedance equivalent circuit is shown in Fig. 4.80

Similarly, the negative sequence and zero sequence equivalent networks are shown in Fig.4.81 (a)and (b) repectively. These equivalent networks are identical to the positive sequence equivalentnetwork but do not contain any internal voltage sources.

The equivalent currentsV (1)kk′

z(1)ij,V (2)kk′

z(2)ijand

V (0)kk′

z(0)ijare due to open conductor fault between k and

k’. If no conductor is open then the sequence voltages are all zero and the current sources arenot present in the equivalent circuit. Further, the current sources can be regarded as currentinjections into the buses i and j of the original sequence networks. All through the calculation

212

Figure 4.79: Thevenin’s Equivalent with transformed current source

Figure 4.80: Final positive sequence Thevenin’s Equivalent circuit repesenting the opening of line i→ j between buses k and k’

process, the bus impedance matrices [Z(0)Bus], [Z(1)Bus] and [Z(2)Bus] of the original network areused.

The current injections at the buses i and j can be tabulated as:

213

Figure 4.81: Final (a) negative sequence (b) zero sequence Thevenin’s Equivalent circuit repesentingthe opening of line i → j between k and k’

Positive Sequence Negative Sequence Zero sequence

at ith busV (1)kk′

z(1)ij

V (2)kk′

z(2)ij

V (0)kk′

z(0)ij

at jth bus − V(1)kk′

z(1)ij− V

(2)kk′

z(2)ij− V

(0)kk′

z(0)ij

The sequence voltage drops ∆V (0)n , ∆V (1)n and ∆V (2)n at any bus ‘n’ due to the current injectionsat the buses ‘i’ and ‘j’ can be calculates from equation (4.136) as:

∆V (0)n =(Z(0)ni − Z

(0)nj )V

(0)kk′

z(0)ij

∆V (1)n =(Z(1)ni − Z

(1)nj )V

(1)kk′

z(1)ij(4.146)

∆V (2)n =(Z(2)ni − Z

(2)nj )V

(2)kk′

z(2)ij

Next, the Thevenin’s equivalent impedances for each sequence network, as seen from the buseskand k’, are calculated as follows:

From Fig.4.78, the positive sequence equivalent impdeance Z(1)kk′ is found out as :

Z(1)kk′ = xz(1)ij +

Z(1)th,ij(−z(1)ij )

Z(1)th,ij + (−z(1)ij )+ (1 − x)z(1)ij

214

Z(1)kk′ =−(z(1)ij )2

Z(1)th,ij − z(1)ij

(4.147)

Similarly from Fig. 4.81 (a) and (b), the negative sequence and zero sequence Thevenin’s equiv-alent impedances can be expressed as:

Z(2)kk′ =−(z(2)ij )2


(4.148)

Z(0)kk′ =−(z(0)ij )2


The open-circuit voltage from point k to k’ can be calculated as:

V (1)th,kk′ =(−z(1)ij )2


(V (1)i − V (1)j ) (4.149)

substitutingZ(1)kk′

z(1)ij= −

z(1)ij


from equation (4.147) in equation (4.149), the final value of

V (1)th,kk′ is obtained as:

V (1)th,kk′ =Z(1)kk′

z(1)ij(V (1)i − V (1)j ) (4.150)

Also prior to the occurance of open-conductor fault on any conductor, the current I(1)ij flowingin phase a is the positive sequence component and is given by the relation:

I(1)ij =(V (1)i − V (1)j )

z(1)ij(4.151)

Substituting equation (4.151) in equation (4.150), one gets

V (1)th,kk′ = Z(1)kk′ I

(1)ij (4.152)

The Thevenin’s equivalent network as seen from points k and k’ for the three sequence networksare shown in Fig. 4.82

We are now ready to discuss the two possible cases of open-circuit fault i.e. (a) open phase open(b) two phases open. We will be discussing them in the next lecture.

215

Figure 4.82: Thevenin’s Equivalent networks as seen from k and k’

4.13.1 One Phase open:

Consider that phase a conductor is open as shown in Fig. 4.75(a), hence phase a current Ia = 0.As a result:

I(1)a + I(2)a + I(0)a = 0 (4.153)

where, I(1)a , I(2)a and I(0)a are the symmetrical components of phase a current. Since phases b andc are closed , the voltage drops .

Vkk′,b = 0 Vkk′,c = 0 (4.154)

216

The symmetrical components of voltage drops across the fault point can be calculated as :

⎡⎢⎢⎢⎢⎢⎢⎣

V (0)a

V (1)a

V (2)a

⎤⎥⎥⎥⎥⎥⎥⎦= 1

3

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a a2

1 a2 a

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Vkk′,a00

⎤⎥⎥⎥⎥⎥⎥⎦= 1

3

⎡⎢⎢⎢⎢⎢⎢⎣

Vkk′,aVkk′,aVkk′,a

⎤⎥⎥⎥⎥⎥⎥⎦(4.155)

Hence,

V (0)a = V (1)a = V (2)a = 13 Vkk

′,a (4.156)

It implies that open conductor in phase a causes equal voltages to appear across points k andk’ of each sequence network. Hence, the three equivalent sequence networks can be connected inparallel across points k and k’ as shown in Fig. 4.83.

Figure 4.83: Connection of Equivalent sequence networks to represent open phase a between k andk’

The current I(1)a is given as :

I(1)a = IijZ(1)kk′

Z(1)kk′ +Z(2)kk′ Z

(0)kk′

Z(2)kk′ + Z(0)kk′

simplifying

I(1)a = IijZ(1)kk′ [Z

(2)kk′ + Z

(0)kk′ ]

Z(0)kk′ Z(1)kk′ + Z

(1)kk′ Z

(2)kk′ + Z

(2)kk′ Z

(0)kk′

(4.157)

The sequence voltage drops V (1)kk′ , V(2)kk′ and V (0)kk′ can be calculated with reference to Fig.4.83 as:

V (1)kk′ = I(1)a

Z(2)kk′ Z(0)kk′

Z(2)kk′ + Z(0)kk′

Substituting I(1)a and simplifying we get:

217

V (1)kk′ = V(2)kk′ = V

(0)kk′ = Iij

Z(1)kk′ Z(2)kk′ Z

(0)kk′

Z(0)kk′ Z(1)kk′ + Z

(1)kk′ Z

(2)kk′ + Z

(2)kk′ Z

(0)kk′

(4.158)

• Z(1)kk′ , Z(2)kk′ , and Z

(0)kk′ are obtained from the impedance parameters of the sequence networks

[equation (4.147) and equation (4.148)].

• Iij is the pre-fault current or load current in phase a of the line i → j

Next, the equivalent injected currentsV (1)kk′

z(1)ij,V (2)kk′

z(2)ijand

V (0)kk′

z(0)ijare calculated.

Further, ∆V (0)i ,∆V (1)i and ∆V (2)i representing the changes in the symmetrical components ofbus voltage are calculated using equation (4.146).

Finally, the bus voltages after fault are calculated using superposition principle as :

V (1)i (F ) = V (1)i (0) +∆V (1)i

V (2)i (F ) = ∆V (1)i (4.159)

V (0)i (F ) = ∆V (2)i

4.13.2 Two Phases open:

When two phases b and c are open then,

V (1)kk′,a = V (0)a + V (1)a + V (2)a = 0Ib = 0 (4.160)

Ic = 0

The sequence components of line current are:

⎡⎢⎢⎢⎢⎢⎢⎣

I(0)a

I(1)a

I(2)a

⎤⎥⎥⎥⎥⎥⎥⎦= 1

3

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a a2

1 a2 a

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Ia00

⎤⎥⎥⎥⎥⎥⎥⎦Simplifying one gets the condition:

I(0)a = I(1)a = I(2)a = 13 Ia (4.161)

equation (4.161) indicates that the three equivalent sequence networks are in series and to ensureV (0)a + V (1)a + V (2)a = 0 the circuit should be closed.The interconnection of the sequence networks isshown in Fig.4.84.

From the equivalent circuit of Fig.4.84, the sequence currents can be calculated as :

218

Figure 4.84: Connection of Equivalent sequence networks to represent open phases b and c betweenk and k’

I(1)a = I(2)a = I(0)a = IijZ(1)kk′

Z(0)kk′ + Z(1)kk′ + Z

(2)kk′

(4.162)

Iij is the pre-fault current in phase a.The sequence voltage can be calculated as:

V (1)kk′ = I(1)a (Z(0)kk′ + Z(2)kk′ ) = Iij

Z(1)kk′ (Z(0)kk′ + Z

(2)kk′ )

Z(0)kk′ + Z(1)kk′ + Z

(2)kk′

V (2)kk′ = −I(2)a Z(2)kk′ = −IijZ(1)kk′ Z

(2)kk′

Z(0)kk′ + Z(1)kk′ + Z

(2)kk′

(4.163)

V (0)kk′ = −I(0)a Z(0)kk′ = −IijZ(1)kk′ Z

(0)kk′

Z(0)kk′ + Z(1)kk′ + Z

(2)kk′

Remaining calculations are similar to those performed for single conductor open case. In thenext lecture we will be looking at an example of open conductor fault analysis.

219

4.14 Example of calculations for open conductor fault inpower system:

The power system described in Fig. 4.67 is considered again.The prefault bus voltages for buses 3, 4 and 5 are :

⎡⎢⎢⎢⎢⎢⎢⎣

V3(0)V4(0)V5(0)

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0.9165∠− 8.7580

0.9152∠− 10.0980

0.8858∠− 12.9610

⎤⎥⎥⎥⎥⎥⎥⎦pu

ONE CONDUCTOR OPENLet one conductor of line 7 between buses 4 and 5 be open. The pre-fault current in phase a of

the faulted line is I45 = (0.3821 − j0.3779) pu. The sequence impedance matrices [Z(0)Bus], [Z(1)Bus]

and [Z(2)Bus] are same as used in the example for calculation of short circuit faults on the powersystem of Fig. 4.67.

The Thevenin’s impedance of the network as seen from buses 4 and 5 is calculated using equation(4.145) with i=4 and i=5 as:

Z(1)th,45 = Z(1)44 + Z(1)55 − 2Z(1)45 = j0.1397 + j0.1750 − 2 ∗ 0.1250 = j0.0647 pu

Z(2)th,45 = Z(1)th,45 = j0.0647 pu

Z(0)th,45 = Z(0)44 + Z(0)55 − 2Z(0)45 = j0.30 + j0.30 − 2 ∗ 0.20 = j0.20 pu

Next,Z(1)kk′ , Z(2)kk′andZ

(0)kk′ are calculated from equation (4.147) and equation (4.148).

Z(1)kk′ =(−z(1)45 )2

Z(1)th,45 − z(1)45

= −(j0.1)2

j0.0647 − j0.10 = j0.2833 pu

Z(2)kk′ = Z(1)kk′ = j0.2833 pu

Z(0)kk′ =(−z(0)45 )2

Z(0)th,45 − z(0)45

= −(j0.30)2

j0.20 − j0.30 = j0.90 pu

The current I(1)a is calculated using equation (4.157)

I(1)a = IijZ(1)kk′ [Z

(2)kk′ + Z

(0)kk′ ]

Z(0)kk′ Z(1)kk′ + Z

(1)kk′ Z

(2)kk′ + Z

(2)kk′ Z

(0)kk′

I(1)a = (0.3821 − j0.3779) j0.2833(j0.2833 + j0.3)j0.9 ∗ j0.2833 + j0.2833 ∗ j0.2833 + j0.2833 ∗ j0.9 = (0.2170 − j0.2146) pu

220

The sequence voltage drops V (1)kk′ , V(2)kk′ and V (0)kk′ are then calculated using the equation:

V (1)kk′ = I(1)a

Z(2)kk′ Z(0)kk′

Z(2)kk′ + Z(0)kk′

V (1)kk′ = (0.2170 − j0.2146)j0.2833 ∗ j0.9j0.2833 + j0.9 = (0.0463 + j0.0468) pu

V (2)kk′ = V(0)kk′ = V

(1)kk′ = (0.0463 + j0.0468) pu

As a check calculate

I(2)a = − V(2)kk′

Z(2)kk′

= −0.0463 + j0.0468j0.2833 = (−0.1651 + j0.1633) pu

I(0)a = − V(0)kk′

Z(0)kk′

= −0.0463 + j0.0468j0.90 = (−0.052 + j0.0514) pu

Ia = I(0)a + I(1)a + I(2)a = (0.2170 − j0.2146) + (−0.1651 + j0.1633) + (−0.052 + j0.0514)= 0 Q.E.D.

Then we calculate the changes in bus voltages using equation (4.146).

Bus 3

∆V (0)3 = (Z(0)34 − Z(0)35 )V (0)kk′

z(0)45

= j0.10 − j0.10j0.30 ∗ (0.0463 + j0.0468) = 0 pu

∆V (1)3 = (Z(1)34 − Z(1)35 )V (1)kk′

z(1)45

= j0.1103 − j0.1250j0.1 ∗ (0.0463 + j0.0468) = (−0.0068 − j0.0069) pu

∆V (2)3 = ∆V (1)3 = (−0.0068 − j0.0069) pu

Hence,

∆V3 = ∆V (0)3 +∆V (1)3 +∆V (2)3 = (−0.0136 − j0.0138) pu

Bus 4

∆V (0)4 = (Z(0)44 − Z(0)45 )V (0)kk′

z(0)45

= j0.30 − j0.20j0.3 ∗ (0.0463 + j0.0468) = (0.0154 + j0.0156) pu

221

∆V (1)4 = (Z(1)44 − Z(1)45 )V (1)kk′

z(1)45

= j0.01397 − j0.125j0.10 ∗ (0.0463 + j0.0468) = (0.0068 + j0.0069) pu

∆V (2)4 = ∆V (1)4 = (0.0068 + j0.0069) pu

∆V4 = ∆V (0)4 +∆V (1)4 +∆V (2)4 = (0.0290 − j0.0293) pu

Bus 5

∆V (0)5 = (Z(0)54 − Z(0)55 )V (0)kk′

z(0)45

= j0.20 − j0.30j0.3 ∗ (0.0463 + j0.0468) = (−0.0154 − j0.0156) pu

∆V (1)5 = (Z(1)54 − Z(1)45 )V (1)kk′

z(1)45

= j0.1250 − j0.1750j0.3 ∗ (0.0463 + j0.0468) = (−0.0231 − j0.0234) pu

∆V (2)5 = ∆V (1)5 = (−0.0231 − j0.0234) pu

∆V5 = ∆V (0)5 +∆V (1)5 +∆V (2)5 = (−0.0617 − j0.0624) pu

The bus voltages during fault are

V3(F ) = V 03 +∆V3 = (0.9058 − j0.1395) + (−0.0136 − j0.138) = (0.8922 − j0.1533) pu

V3(F ) = 0.9053∠− 9.750 pu

V4(F ) = V 04 +∆V4 = (0.9010 − j0.1605) + (0.0290 + j0.0293) = (0.9300 − j0.1311) pu

V4(F ) = 0.9392∠− 8.020 pu

V5(F ) = V 05 +∆V5 = (0.8632 − j0.1987) + (−0.0617 − j0.0624) = (0.8016 − j0.2610) pu

222

V5(F ) = 0.8430∠− 18.040 pu

The phase components of current in line 4-5 are:

⎡⎢⎢⎢⎢⎢⎢⎣

I(a)45

I(b)45

I(c)45

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(−0.052 + j0.0514)(0.2170 − j0.2146)(−0.1651 + j0.1633)

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

00.4782∠− 147.840

0.4782∠58.560

⎤⎥⎥⎥⎥⎥⎥⎦pu

TWO CONDUCTORS OPENThe sequence currents can be calculated with the help of equation (4.162) as:

I(1)a = IijZ(1)kk′

Z(0)kk′ + Z(1)kk′ + Z

(2)kk′

= (0.3821 − j0.3779) ∗ j0.2833j0.9 + j0.2833 + j0.2833

I(1)a = I(2)a = I(0)a = (0.0738 − j0.073) pu

The sequence voltages of Thevenin’s equivalent sequence network is calculated using equation(4.163)

V (1)kk′ = I(1)a (Z(0)kk′ + Z(2)kk′ ) = (0.0738 − j0.073) ∗ (j0.9 + j0.2833) = (0.0864 + j0.0873) pu

V (2)kk′ = −I(2)a Z(2)kk′ = −(0.0738 − j0.073)(j0.2833) = −(0.0207 + j0.0209) pu

V (0)kk′ = −I(0)a Z(0)kk′ = −(0.0738 − j0.073)(j0.9) = −(0.0657 + j0.0664) pu

The changes in bus voltages are calculated using equation (4.146).Bus 3

∆V (0)3 = (Z(0)34 − Z(0)35 )V (0)kk′

z(0)45

= j0.1 − j0.10j0.3 ∗ (−0.0657 − j0.0664) = 0 pu

∆V (1)3 = (Z(1)34 − Z(1)35 )V (1)kk′

z(1)45

= j0.1103 − j0.1250j0.1 ∗ (0.0864 + j0.0873) = (−0.0127 − j0.0128) pu

∆V (2)3 = (Z(2)34 − Z(2)35 )V (2)kk′

z(2)45

= j0.1103 − j0.1250j0.1 ∗ (−0.0207 − j0.0208) = (0.0030 + j0.0030) pu

∆V3 = ∆V (0)3 +∆V (1)3 +∆V (2)3 = (−0.0097 − j0.0098) pu

223

Hence, the voltage of bus 3 during fault is:

V3(F ) = V 03 +∆V3 = (0.9058 − j0.1395) + (−0.0097 − j0.0098) = (0.8962 − j0.1493) pu

V3(F ) = 0.9085∠− 9.460 pu

Bus 4

∆V (0)4 = (Z(0)44 − Z(0)45 )V (0)kk′

z(0)45

= j0.30 − j0.20j0.30 ∗ (−0.0657 − j0.0664) = (−0.0219 − j0.0221) pu

∆V (1)4 = (Z(1)44 − Z(1)45 )V (1)kk′

z(1)45

= j0.1397 − j0.1250j0.1 ∗ (0.0864 + j0.0873) = (0.0127 + j0.0128) pu

∆V (2)4 = (Z(2)44 − Z(2)45 )V (1)kk′

z(2)45

= j0.1397 − j0.1250j0.1 ∗ (0.0207 − j0.0209) = (−0.0030 − j0.0031) pu

∆V4 = ∆V (0)4 +∆V (1)4 +∆V (2)4 = −(0.0122 + j0.0124) pu


V4(F ) = V 04 +∆V4 = (0.9010 − j0.1605) + (−0.0122 − j0.0124) = (0.8888 − j0.1728) pu

V4(F ) = 0.9054∠− 11.00 pu

Bus 5

∆V (0)5 = (Z(0)54 − Z(0)55 )V (0)kk′

z(0)45

= j0.20 − j0.30j0.30 ∗ (−0.0657 − j0.0664) = (0.0219 + j0.0221) pu

∆V (1)5 = (Z(1)54 − Z(1)55 )V (1)kk′

z(1)45

= j0.1250 − j0.1750j0.10 ∗ (0.0864 + j0.0873) = (−0.0432 − j0.0437) pu

224

∆V (2)5 = (Z(2)54 − Z(2)55 )V (2)kk′

z(2)45

= j0.1250 − j0.1750j0.10 ∗ (−0.0207 − j0.0209) = (0.0103 + j0.0105) pu

∆V5 = ∆V (0)5 +∆V (1)5 +∆V (2)5 = −(0.0111 + j0.01111) pu


V5(F ) = V 05 +∆V5 = (0.8632 − j0.1987) + (−0.0111 − j0.01111) = (0.8523 − j0.2097) pu

V5(F ) = 0.8777∠− 13.830 pu

The phase components of current in line 4-5 are:

⎡⎢⎢⎢⎢⎢⎢⎣

I(a)45

I(b)45

I(c)45

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

1 1 11 a2 a1 a a2

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

(0.0738 − j0.073)(0.0738 − j0.073)(0.0738 − j0.073)

⎤⎥⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎢⎣

0.3114∠− 44.690

00

⎤⎥⎥⎥⎥⎥⎥⎦pu

225

Module 4 Analysis of Faulted Power System

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Transcript of Module 4 Analysis of Faulted Power System