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

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[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]

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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 ).

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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.

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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

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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

Z(2)th,ij − z(2)ij

(4.148)

Z(0)kk′ =−(z(0)ij )2

Z(0)th,ij − z(0)ij

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

V (1)th,kk′ =(−z(1)ij )2

Z(1)th,ij − z(1)ij

(V (1)i − V (1)j ) (4.149)

substitutingZ(1)kk′

z(1)ij= −

z(1)ij

Z(1)th,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.

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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)

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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 :

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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