Download pdf - UNSYMMETRICAL-FAULTS- - University of · PDF fileMostof-the-faults-thatoccur-on-power-systems-are-unsymmetrical-faults ... applied-to-the-analysis-of-unsymmetrical-faults.- ... of-the-original-balanced-system

UNSYMMETRICAL FAULTS

updated 11/11/13

Unsymmetrical Faults (c) 2013 H. Zmuda 1 11/11/13

11/11/13 Unsymmetrical Faults (c) 2013 H. Zmuda 2

Introductory Comments Most of the faults that occur on power systems are unsymmetrical faults, which may consist of unsymmetrical short circuits, unsymmetrical faults through impedances, or open conductors. Unsymmetrical faults occur as single line-‐to-‐ground faults, line-‐to-‐line faults, or double line-‐to-‐ground faults. The path of the fault current from line to line or line to ground may or may not contain impedance. One or two open conductors result in unsymmetrical faults, through either the breaking of one or two conductors or the acHon of fuses and other devices that may not open the three phases simultaneously.


Introductory Comments Since any unsymmetrical fault causes unbalanced currents to flow in the system, the method of symmetrical components is very used in the analysis to determine the currents and voltages in all parts of the system aLer the occurrence of the fault. We will consider faults on a power system by applying Thévenin's theorem, which allows us to find the current in the fault by replacing the enHre system by a single generator and series impedance, and we will show how the bus impedance matrix is applied to the analysis of unsymmetrical faults.


Unsymmetrical Faults In the derivaHon of equaHons for the symmetrical components of currents and voltages in a general network the currents flowing out of the original balanced system from phases a , b, and c at the fault point will be designated as Ia, lb, and lc, respecHvely. We can visualize these currents by as follows: This shows the three lines a, b, and c of the three-‐phase system at the part of the network where the fault occurs. The flow of current from each line into the fault is indicated by arrows shown beside hypotheHcal stubs connected to each line at the fault locaHon.

I fa

I fb

I fc

a

b

c


Unsymmetrical Faults Appropriate connecHons of the stubs represent the various types of fault. For instance, direct connecHon of stubs b and c produces a line-‐to-‐line fault through zero impedance. The current in stub a is then zero, and lb equals -‐ lc. The line-‐to-‐ground voltages at any bus j of the system during the fault will be designated Vja, Vjb and Vjc and we shall conHnue to use superscripts 1, 2, and 0, respecHvely, to denote posiHve-‐, negaHve-‐, and zero-‐sequence quanHHes. Thus, for example, V(1)

ja, V(2)jb and V(0)

jc will denote, respecHvely, the posiHve-‐, negaHve-‐, and zero-‐sequence components of the line-‐to-‐ground voltage Vja at bus j during the fault.


Unsymmetrical Faults The line-‐to-‐neutral voltage of phase a at the fault point before the fault occurs will be designated simply by Vf, which is a posiHve-‐sequence voltage since the system is balanced. We considered the prefault voltage Vf previously when calculaHng the currents in a power system with a symmetrical three-‐phase fault applied.


Unsymmetrical Faults Consider a single-‐line diagram of a power system containing two synchronous machines. This simple system is sufficiently general that the equaHons derived are applicable to any balanced system regardless of the complexity. The point where a fault is assumed to occur is marked P, and in this example it is called bus k on the single-‐line diagram and in the sequence networks.

Single line diagram of a balanced three-‐phase system

P

k

Unsymmetrical Faults

Thévenin Equivalent of PosiKve-‐Sequence Network PosiKve-‐Sequence Network


P

k

Reference

P k

+

Vf

−

I fa

1( )

+−

Vf

P I fa

1( )

k

Zkk1( )

+

Vka1( )

−


Thévenin Equivalent of NegaKve-‐Sequence Network NegaKve-‐Sequence Network


P

k

Reference

P k

I fa

2( )P

I fa2( )

k

Zkk2( )

+

Vka2( )

−


Thévenin Equivalent of Zero-‐Sequence Network Zero-‐Sequence Network


P

k

Reference

P k

I fa

0( )P

I fa0( )

k

Zkk0( )

+

Vka0( )

−


Unsymmetrical Faults Machines are represented by their subtransient internal voltages in series with their subtransient reactances when subtransient fault condiHons are being studied. Previously we used the bus impedance matrix composed of posiHve-‐sequence impedances to determine currents and voltages upon the occurrence of a symmetrical three-‐phase fault. The method can be easily extended to apply to unsymmetrical faults by realizing that the negaHve-‐ and zero-‐sequence networks also can be represented by bus impedance matrices. The bus impedance matrix will now be wriben symbolically for the posiHve-‐, negaHve-‐, and zero-‐sequence networks in the following form…



Ζbus0,1,2( ) =

Z110,1,2( ) Z12

0,1,2( ) Z1k0,1,2( ) Z1N

0,1,2( )

Z210,1,2( ) Z22

0,1,2( ) Z2k0,1,2( ) Z2 N

0,1,2( )

Zk10,1,2( ) Zk 2

0,1,2( ) Zkk0,1,2( ) ZkN

0,1,2( )

ZN10,1,2( ) ZN 2

0,1,2( ) ZNk0,1,2( ) ZNN

0,1,2( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥


Unsymmetrical Faults The Thévenin equivalent circuit between the fault point P and the reference node in each sequence network can be used for the analysis. As before, the voltage source in the posiHve-‐sequence network and its Thévenin equivalent circuit is Vf, the prefault voltage to neutral at the fault point P, which happens to be bus k in this illustraHon. The Thévenin impedance measured between point P and the reference node of the posiHve-‐sequence network is Z(1)kk, and its value depends on the values of the reactances used in the network. Recall that subtransient reactances of generators and 1.5 Hmes the subtransient reactances (or else the transient reactances) of synchronous motors are the values used in calculaHng the symmetrical current to be interrupted.


Unsymmetrical Faults There are no negaHve-‐ or zero-‐sequence currents flowing before the fault occurs, and the prefault voltages are zero at all buses of the negaHve-‐ and zero-‐sequence networks. Therefore, the prefault voltage between point P and the reference node is zero in the negaHve-‐ and zero-‐sequence networks and no electromoHve forces (emfs) appear in their Thévenin equivalents. The negaHve-‐ and zero-‐sequence impedances between point P at bus k and the reference node in the respecHve networks are represented by the the impedances Z(2)kk and Z(0)kk, the Diagonal elements of Z(2)bus and Z(0)bus, respecHvely.


Unsymmetrical Faults Since Ifa is the current flowing from the system into the fault, its symmetrical components flow out of the respecHve sequence networks and their equivalent circuits at point P, as shown. Thus, the currents –I(1)fa, –I(2)g and –I(0)fc represent injected currents into the faulted bus k of the posiHve-‐, negaHve-‐, and zero-‐sequence networks due to the fault. These current injecHons cause voltage changes at the buses of the posiHve-‐, negaHve-‐, and zero-‐sequence networks, which can be calculated from the bus impedance matrices in the manner similare to what we have done before.


Unsymmetrical Faults For instance, due to the injecHon –I(1)fa into bus k, the voltage changes in the posiHve-‐sequence network of the N-‐bus system are given in general terms by:

ΔV1a1( )

ΔV2a1( )

ΔVka1( )

ΔVNa1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

Z111( ) Z12

1( ) Z1k1( ) Z1N

1( )

Z211( ) Z22

1( ) Z2k1( ) Z2 N

1( )

Zk11( ) Zk 2

1( ) Zkk1( ) ZkN

1( )

ZN11( ) ZN 2

1( ) ZNk1( ) ZNN

1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

00

− I fa1( )

0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

−Z1k1( )I fa

1( )

−Z2k1( )I fa

1( )

−Zkk1( )I fa

1( )

−ZNk1( ) I fa

1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥


Unsymmetrical Faults Once again, it is industry pracHce to regard all prefault currents as being zero and to designate the voltage Vf as the posiHve-‐sequence voltage at all buses of the system before the fault occurs. Using superposiHon, the total posiHve-‐sequence voltage of phase a at each bus during the fault is:

V1a1( )

V2a1( )

Vka1( )

VNa1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

Vf

Vf

Vf

Vf

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

+

ΔV1a1( )

ΔV2a1( )

ΔVka1( )

ΔVNa1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

Vf − Z1k1( )I fa

1( )

Vf − Z2k1( )I fa

1( )

Vf − Zkk1( )I fa

1( )

Vf − ZNk1( ) I fa

1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥


Unsymmetrical Faults This is the same equaKon as found for symmetrical faults, the only difference being the added superscripts and subscripts denoKng the posiKve-‐sequence components of the phase a quanKKes.


Unsymmetrical Faults The negaHve-‐ and zero-‐sequence voltage changes due to the fault at bus k of the N-‐bus system are similarly wriben with the superscripts changed accordingly. Because the prefault voltages are zero in the negaHve-‐ and zero-‐sequence networks, the voltage changes express the total negaHve-‐ and zero-‐sequence voltages during the fault, namely,

V1a2( )

V2a2( )

Vka2( )

VNa2( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

−Z1k2( )I fa

2( )

−Z2k2( )I fa

2( )

−Zkk2( )I fa

2( )

−ZNk2( )I fa

2( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

,

V1a0( )

V2a0( )

Vka0( )

VNa0( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

−Z1k0( )I fa

0( )

−Z2k0( )I fa

0( )

−Zkk0( )I fa

0( )

−ZNk0( )I fa

0( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥


Unsymmetrical Faults When the fault is at bus k, note that only the entries in columns k of Z(2)bus and Z(0)bus are involved in the calculaHons of negaHve-‐ and zero-‐sequence voltages. Thus, knowing the symmetrical components I(0)fa, I(1)fa and I(2)fa of the fault currents at bus k, we can determine the sequence voltages at any bus j of the system from the jth rows of

V1a1( )

V2a1( )

Vka1( )

VNa1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

Vf − Z1k1( )I fa

1( )

Vf − Z2k1( )I fa

1( )

Vf − Zkk1( )I fa

1( )

Vf − ZNk1( ) I fa

1( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

,

V1a2( )

V2a2( )

Vka2( )

VNa2( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

−Z1k2( )I fa

2( )

−Z2k2( )I fa

2( )

−Zkk2( )I fa

2( )

−ZNk2( )I fa

2( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

,

V1a0( )

V2a0( )

Vka0( )

VNa0( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

−Z1k0( )I fa

0( )

−Z2k0( )I fa

0( )

−Zkk0( )I fa

0( )

−ZNk0( )I fa

0( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥


Unsymmetrical Faults That is, during the fault at bus k the voltages at any bus j are: If the prefault voltage at bus CD is not Vf, then simply replace Vf in by the actual value of the prefault (posiHve-‐sequence) voltage at that bus. Since Vf is by definiHon the actual prefault voltage at the faulted bus k, we always have at that bus: and these are the terminal voltage equaHons for the Thévenin equivalents of the sequence networks previously shown.

Vja

0( ) = −Z jk0( )I fa

0( ) , Vja1( ) =Vf − Z jk

1( )I fa1( ) , Vja

2( ) = −Z jk2( )I fa

2( )

Vka

0( ) = −Zkk0( )I fa

0( ) , Vka1( ) =Vf − Zkk

1( )I fa1( ) , Vka

2( ) = −Zkk2( )I fa

2( )


Unsymmetrical Faults It is important to remember that the currents I(0)fa, I(1)fa and I(2)fa are symmetrical-‐component currents in the stubs hypotheHcally abached to the system at the fault point. These currents take on values determined by the parHcular type of fault being studied, and once they have been calculated, they can be regarded as negaHve injecHons into the corresponding sequence networks.


Unsymmetrical Faults If the system has Δ-‐Y transformers, some of the sequence voltages may have to be shiLed in phase angle before being combined with other components to form the new bus voltages of the faulted system. There are no phase shiLs involved in when the voltage Vf at the fault point is chosen as reference, which is customary.

Vka

0( ) = −Zkk0( )I fa

0( ) , Vka1( ) =Vf − Zkk

1( )I fa1( ) , Vka

2( ) = −Zkk2( )I fa

2( )

Unsymmetrical Faults In a system with Δ-‐Y transformers the open circuits encountered in the zero-‐sequence network requires some care in the Zbus building algorithm. Consider, for instance, the solidly grounded Δ-‐Y transformer connected between buses m and n as shown along with their posiHve and zero-‐sequence circuits:

Δ-‐Y transformer with PosiKve-‐sequence Zero-‐sequence leakage impedance Z circuit circuit

The negaHve-‐sequence circuit is the same as the posiHve-‐sequence circuit.


n m n m

Reference

Z

n m Reference

Z


Unsymmetrical Faults It is straighlorward using the circuit representaHons shown to generate the bus impedance matrices Zbus (0,1,2). This will be done subsequently. Suppose, however, that we wish to represent removal of the transformer connecHons from bus n in a computer algorithm which cannot make use of circuit (schemaHc) representaHons. We can easily undo the connecHons to bus n in the posiHve-‐ and negaHve-‐sequence networks by applying the building algorithm the Zbus (1,2) matrices in the usual manner, i.e., by adding the negaHve of the leakage impedance Z between buses m and m in the posiHve-‐ and negaHve-‐sequence networks. (Next slide)


Unsymmetrical Faults PosiKve-‐sequence circuit:

n m Reference

Z

-‐ Z

n m Reference

è


Unsymmetrical Faults This strategy does not apply to the zero-‐sequence matrix Zbus (0) if it has been formed directly from the schemaHc representaHon shown. Adding -‐ Z between buses m and m does not remove the zero-‐sequence connecHon from bus n.

n m Reference

Z

-‐ Z


Unsymmetrical Faults To permit uniform procedures for all sequence networks, one strategy is to include an internal node p, as shown below. Note that the leakage impedance is now subdivided into two parts between node p and the other nodes as shown. ConnecHng –Z/2 between buses n and p in each of the sequence circuits will open the transformer connecHons to bus n.

n m Reference

Z

n m Reference

Z

n m Reference

Z/2

n m Reference

Z/2

Z/2 p Z/2 p

è

è


Unsymmetrical Faults ConnecHng –Z/2 between buses n and p in each of the sequence circuits will open the transformer connecHons to bus n.

n m Reference

Z/2

n m Reference

Z/2

Z/2

p

Z/2

p

-‐ Z/2 -‐ Z/2

n m Reference

Z/2

n m Reference

Z/2 p

p

è

è


Unsymmetrical Faults The faults to be discussed in succeeding secHons may involve impedance Zf between lines and from one or two lines to ground. When Zf = 0, we have a direct short circuit, which is called a bolted fault . Although such direct short circuits result in the highest value of fault current and are therefore the most conservaHve values to use when determining the effects of anHcipated faults, the fault impedance is seldom zero.


Unsymmetrical Faults Most faults are the result of insulator flashovers, where the impedance between the line and ground depends on the resistance of the arc, of the tower itself, and of the tower fooHng if ground wires are not used. Tower-‐fooHng resistances form the major part of the resistance between line and ground and depend on the soil condiHons. The resistance of dry earth is 10 to 100 Hmes the resistance of swampy ground.


Unsymmetrical Faults ConnecHons of the hypotheHcal stubs for faults through impedance Zf are as follows:

Three-‐Phase Fault

I fa

I fb

I fc

a

b

c

Z f

Z f

Z f



Single Line-‐to-‐Ground Fault

I fa

I fb

I fc

a

b

c

Z f



Line-‐to-‐Line Fault

I fa

I fb

I fc

a

b

c

Z f



Double Line-‐to-‐Ground Fault

I fa

I fb

I fc

a

b

c

Z f


Unsymmetrical Faults Other types of faults:

Open-‐Conductor Faults

Ia

Ib

Ic

a

b

c


Unsymmetrical Faults Other types of faults:

Open-‐Conductor Faults

Ia

Ib

Ic

a

b

c


Unsymmetrical Faults A balanced system remains symmetrical aLer the occurrence of a three-‐phase fault having the same impedance between each line and a common point. Only posiHve-‐sequence currents flow. With the fault impedance Zf equal in all phases, as in a three-‐phase fault, we simply add impedance Zf to the usual (posiHve-‐sequence) Thévenin equivalent circuit of the system at the fault bus k and calculate the fault current from the equaHon: For each of the other types of faults, formal derivaHons of the equaHons for the symmetrical-‐component currents follow. In each case the fault point P is designated as bus k.

I fa

1( ) =Vf

Zkk1( ) + Z f


Example Two synchronous machines are connected through three-‐phase transformers to the transmission line as shown. The raHngs and reactances of the machines and transformers are:

Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5%

Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8%

On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15% and X0 = 50%. Determine Zbus for the sequence networks.

T1 T2 1 2 3 4 Machine 1 Machine 2


Example PosiHve-‐ and negaHve-‐sequence circuit: The raHngs and reactances of the machines and transformers are: Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20%

X0 = 4%, Xn = 5% Transformers T1, T2: 100 MVA, 20Δ/345Y kV, X = 8% (split as 4% + 4%) Transmission line reactances are X1 = X2 = 15% and X0 = 50%.

1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

(1)

(2) (3) (4) (5)


Example Add branch (1): Add branch (2):

1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

(1)

(2) (3) (4)

(5)

Z11,2( ) = j0.20⎡⎣ ⎤⎦

Z2

1,2( ) =j0.20 j0.20j0.20 j0.20+ j0.08

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

j0.20 j0.20j0.20 j0.28

⎡

⎣⎢⎢

⎤

⎦⎥⎥


Example Add branch (3):

1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

(1)

(2) (3) (4)

(5)

Z31,2( ) =

j0.20 j0.20 j0.20j0.20 j0.28 j0.28j0.20 j0.28 j0.28+ j0.15

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

j0.20 j0.20 j0.20j0.20 j0.28 j0.28j0.20 j0.28 j0.43

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥



1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

(1)

(2) (3) (4)

(5)

Z41,2( ) =

j0.20 j0.20 j0.20 j0.20j0.20 j0.28 j0.28 j0.28j0.20 j0.28 j0.43 j0.43j0.20 j0.28 j0.43 j0.43+ j0.08

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=

j0.20 j0.20 j0.20 j0.20j0.20 j0.28 j0.28 j0.28j0.20 j0.28 j0.43 j0.43j0.20 j0.28 j0.43 j0.51

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥



1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

(1)

(2) (3) (4)

(5)

Z41,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Zij

new = Zij −ZikZkj

Zkk + Zb


Example


Example

Zbus1,2( ) :

Example Zero-‐sequence circuit: The raHngs and reactances of the machines and transformers are:

Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5% Transformers T1 & T2: 100 MVA, 20Δ/345Y kV, X = 8%

Transmission line reactances are X1 = X2 = 15% and X0 = 50%.


1 2 3 4

j0.04

j0.04

Reference

j0.04 j0.5 j0.08 j0.04

3Xn = j0.15

(1) (2) (3) (4) (5) (6)

3Xn = j0.15

See Slide 199 of Symmetrical Components

Transformer Node Bus


See Slide 196 of Symmetrical Components


Example

1 2 3 4

j0.08

Reference

j0.5 j0.08

j0.19 j0.19(1) (2)

(3) (4)

(5)


Example Add branch (1): Add branch (3): Add branch (2):

1 2 3 4

j0.08

Reference

j0.5 j0.08

j0.19 j0.19(1) (2)

(3) (4)

(5)

Z10( ) = j0.19⎡⎣ ⎤⎦

Z2

0( ) =j0.19 0

0 j0.08

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Z30( ) =

j0.19 0 00 j0.08 j0.080 j0.08 j0.08+ j0.5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥



1 2 3 4

j0.08

Reference

j0.5 j0.08

j0.19 j0.19(1) (2)

(3) (4)

(5)

Zbus0( ) =

j0.19 0 0 00 j0.08 j0.08 00 j0.08 j0.58 00 0 0 j0.19

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥


Example We’ll use these in subsequent examples.

Zbus0( ) =

j0.19 0 0 00 j0.08 j0.08 00 j0.08 j0.58 00 0 0 j0.19

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥


Single Line-‐to-‐Ground Faults The single line-‐to-‐ground fault, the most common type, is caused by lightning or by conductors making contact with grounded structures. A single line-‐to-‐ground fault on phase a through impedance Zf is shown below: The relaHons to be developed will apply only when the fault is on phase a, but any phase can be designated as phase a. The condiHons at the fault bus k are expressed by the following equaHons:

I fa

I fb

I fc

a

b

c

Z f

k

I fb = I fc = 0, Vka = Z f I fa


Single Line-‐to-‐Ground Faults The symmetrical components of the current are, with : But recall from Slide 21: Thus:

I fb = I fc = 0

Vka

0( ) = −Zkk0( )I fa

0( ) , Vka1( ) =Vf − Zkk

1( )I fa1( ) , Vka

2( ) = −Zkk2( )I fa

2( )

Vka

0( ) = −Zkk0( )I fa

0( ) , Vka1( ) =Vf − Zkk

1( )I fa0( ) , Vka

2( ) = −Zkk2( )I fa

0( )

I fa0( )

I fa1( )

I fa2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

I fa

00

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥⇒ I fa

0( ) = I fa1( ) = I fa

2( ) =I fa

3


Single Line-‐to-‐Ground Faults Summing, and noHng that Solve for This has the circuit representaHon shown on the next slide.

Vka =Vka0( ) +Vka

1( ) +Vka2( )

=Vf − Zkk0( ) + Zkk

1( ) + Zkk2( )( ) I fa

0( ) = 3Z f I fa0( )

Vka = 3Z f I fa

0( )

I fa

0( )

I fa

0( ) = I fa1( ) = I fa

2( ) =Vf

Zkk0( ) + Zkk

1( ) + Zkk2( ) + 3Z f


Single Line-‐to-‐Ground Faults

V f

+− Zkk

1( )

Zkk2( )

Zkk0( )

3Z f

I fa

1( )

I fa

2( )

I fa

0( )

I fa

0( ) = I fa1( ) = I fa

2( )

k

k

k

+−

Vka1( )

+−

Vka0( )

+−

Vka2( )


Single Line-‐to-‐Ground Faults The last result are the fault current equaHons parHcular to the single line-‐to-‐ground fault through impedance Zf and they are used with the symmetrical-‐component relaHons to determine all the voltages and currents at the fault point P. If the Thévenin equivalent circuits of the three sequence networks of the system are connected in series, as shown on the previous slide, we see that the resulHng currents and voltages saHsfy the above equaHons – for the Thévenin impedances looking into the three sequence networks at fault bus k are then in series with the fault impedance 3Zf and the prefault voltage source Vf.


Single Line-‐to-‐Ground Faults With the equivalent circuits so connected, the voltage across each sequence network is the corresponding symmetrical component of the voltage Vka at the fault bus k, and the current injected into each sequence network at bus k is the negaHve of the corresponding sequence current in the fault. The series connecHon of the Thévenin equivalents of the sequence networks, as shown on Slide 55 is a convenient means of remembering the equaHons for the soluHon of the single line-‐to-‐ground fault, for all the necessary equaHons for the fault point can be determined from the sequence-‐network connecHon.


Single Line-‐to-‐Ground Faults Once the sequence components of the fault currents are known, the components of voltages at all other buses of the system can be determined from the bus impedance matrices of the sequence networks according to Slide 21, namely:

Vja0( ) = −Z jk

0( )I fa0( )

Vja1( ) =Vf − Z jk

1( )I fa1( )

Vja2( ) = −Z jk

2( )I fa2( )


EXAMPLE – Single Line-‐to-‐Ground Faults Two synchronous machines are connected through three-‐phase transformers to the transmission line as shown. The raHngs and reactances of the machines and transformers are:

Machine 1 & 2: 100MVA, 20 kV, Xd” = X1 = X2 = 20% X0 = 4%, Xn = 5%

Transformers T1 & T2: 100 MVA, 20Y/345Y kV, X = 8%



EXAMPLE – Single Line-‐to-‐Ground Faults Both transformers are solidly grounded on on two sides. On a chosen base of 100 MVA, 345 kV in the transmission line circuit the line reactances are X1 = X2 = 15 % and X0 = 50 %. The system is operaHng at nominal voltage without prefault currents when a bolted (Zf = 0) single line-‐to -‐ground fault occurs on phase A at bus 3. Using the bus impedance matrix for each of the three sequence networks, determine the subtransient current to ground at the fault, the line-‐to-‐ground voltages at the terminals of machine 2, and the subtransient current out of phase c of machine 2.


EXAMPLE – Single Line-‐to-‐Ground Faults The system is the same as on Slide 39, except that the transformers are now Y-‐Y connected. Therefore, we can conHnue to use Zbus (1,2) from Slide 51, however, because the transformers are solidly grounded on both sides, the zero-‐sequence network is fully connected, as shown below, and has the bus impedance matrix

1 2 3 4

j0.15

j0.04 j0.15

j0.15

j0.08 j0.08 j0.04

Reference


EXAMPLE – Single Line-‐to-‐Ground Faults The bus impedance matrix is:

1 2 3 4

j0.15

j0.04 j0.15

j0.15

j0.08 j0.08 j0.04

Reference

Zbus0( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥


EXAMPLE – Single Line-‐to-‐Ground Faults Since the line-‐to-‐ground fault is at bus 3:

The total current in the fault is:

V f = 1.0 Z33

1( )

Z332( )

Z330( )

3Z f = 0

I fA

1( )

I fA

2( )

I fA

0( )

I fA

3

k

k

k

+−

V3a1( )

+−

V3a0( )

+−

V3a2( )

±

Zbus0( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

I fA0( ) = I fA

1( ) = I fA2( ) =

Vf

Zkk0( ) + Zkk

1( ) + Zkk2( ) + 3Z f

= 1j0.1999+ j1.696+ j1.696+ 0

= − j1.8549

⇒ I fA = 3I fA0( ) = − j5.5648


EXAMPLE – Single Line-‐to-‐Ground Faults Since the base current in the high-‐voltage transmission line is:

amps Then: amps The phase-‐a sequence voltages at bus 4, the terminals of machine 2, are (from Slide 21 with k = 3 and j = 4):

per unit

Ibase =

100,0003 × 345

= 167.35

I fA = − j5.5648×167.35= 931∠270°

V4a0( ) = −Z43

0( )I fA0( ) = − j0.1407( ) − j1.8549( ) = −0.261

V4a1( ) =Vf − Z43

1( )I fA1( ) = 1− j0.1211( ) − j1.8549( ) = −0.7754

V4a2( ) = −Z43

2( )I fA2( ) = − j0.1211( ) − j1.8549( ) = −0.2246


EXAMPLE – Single Line-‐to-‐Ground Faults Note that the subscripts A and a denote voltages in the high-‐voltage and low-‐voltage circuits, respecHvely, of the Y-‐Y connected transformer. No phase shiL is involved. From the symmetrical components we can calculate the a-‐b-‐c line-‐to-‐ground voltages at bus 4 as:

per unit

V4a

V4b

V4c

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 13

1 1 11 a2 a1 a a2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

V4a0( )

V4a1( )

V4a2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a2 a1 a a2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−0.261−0.7754−0.2246

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=0.2898

−0.5346− j0.8660−0.5346− j0.8660

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥


EXAMPLE – Single Line-‐to-‐Ground Faults To express the line-‐to-‐ground voltages of machine 2 (in kV) mulHply by 20/31/2:

kV

V4a

V4b

V4c

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 203

0.2898−0.5346− j0.8660−0.5346− j0.8660

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

3.3461.0187∠−121.8°1.0187∠−121.8°

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥


EXAMPLE – Single Line-‐to-‐Ground Faults To determine phase-‐c current out of machine 2 we must first calculate the symmetrical components of the phase-‐a current in the branches represenHng the machine in the sequence networks. From the zero-‐sequence circuit, the zero-‐sequence current out of the machine is:

per unit

1 2 3 4

j0.15

j0.04 j0.15

j0.15

j0.08 j0.08 j0.04

Reference

+

V4a0( )

−

Ia0( )

Ia

0( ) = −V4a

0( )

jX0

= 0.261j0.04

= − j6.525


EXAMPLE – Single Line-‐to-‐Ground Faults Similarly for the posiHve-‐ and negaHve-‐sequence currents:

per unit Ia

1( ) =Vf −V4a

1( )

j ′′X= 1− 0.7754

j0.2= − j1.123

1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.15 j0.20

j0.04 j0.04 j0.04

Reference

+

V4a1,2( )

−

Ia1,2( )

Ia

2( ) = −V4a

1( )

jX2

= 0.2246j0.2

= − j1.123


EXAMPLE – Single Line-‐to-‐Ground Faults The phase-‐c current in machine 2 is

per unit Since the base current in the machine circuit is:

amps

amps

Ic = Ia0( ) + aIa

1( ) + a2Ia2( )

= − j6.525+ a − j1.123( ) + a2 − j1.123( )= − j5.402

Ibase =

100,0003 × 20

= 2886.751

Ic = 15,994


Line-‐to-‐Line Faults To represent a line-‐to-‐line fault through impedance Zf the hypotheHcal stubs on the three lines at the fault are connected as shown. Bus k is again the fault point P, and without any loss of generality, the line-‐to-‐line fault is regarded as being on phases b and c. Clearly:

I fa

I fb

I fc

a

b

c

Z f

k

k

k

I fa = 0

I fb = − I fc

Vkb −Vkc = Z f I fb


Line-‐to-‐Line Faults SubsHtuHng: The voltages throughout the zero-‐sequence network must be zero since there are no zero-‐sequence sources, and because , current is not being injected into that network due to the fault. Hence, line-‐to-‐line fault calculaHons do not involve the zero-‐sequence network, which remains the same as before the fault -‐ a dead network.

I fa0( )

I fa1( )

I fa2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0I fb

− I fc

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⇒ I fa0( ) = 0

⇒ I fa1( ) = − I fa

2( )

I fa

0( ) = 0


Line-‐to-‐Line Faults To saHsfy the requirement that we connect the Thévenin equivalents of the posiHve-‐ and negaHve-‐sequence networks in parallel, as shown:

I fa

1( ) = − I fa2( )

V f

+−

Zkk1( )

Zkk2( )

Z f

I fa

1( ) I fa

2( )

k

+

Vka1( )

−

k

+

Vka2( )

−Reference


Line-‐to-‐Line Faults To show that this connecHon of the networks also saHsfies the voltage equaHon: we now expand each side of that equaHon separately as follows: and:

Vkb −Vkc = Z f I fb

Vkb −Vkc = Vkb1( ) +Vkb

2( )( )− Vkc1( ) +Vkc

2( )( )= Vkb

1( ) −Vkc1( )( ) + Vkb

2( ) −Vkc2( )( )

= a2 − a( )Vka1( ) + a − a2( )Vka

2( )

= a2 − a( ) Vka1( ) −Vka

2( )( )

Z f I fb = Z f I fb

1( ) + I fb2( )( ) = Z f a2I fa

1( ) + aI fa2( )( )


Line-‐to-‐Line Faults Equate both terms and set: But this is precisely the voltage drop across Zf in the figure on Slide 66.

Vkb −Vkc = a2 − a( ) Vka1( ) −Vka

2( )( ) = Z f I fb = Z f a2I fa1( ) + aI fa

2( )( )

⇒ a2 − a( ) Vka1( ) −Vka

2( )( ) = Z f a2 − a( ) I fa1( )

⇒Vka1( ) −Vka

2( ) = Z f I fa1( )

I fa

1( ) = − I fa2( )


Line-‐to-‐Line Faults Thus, all the fault condiHons are saHsfied by connecHng the posiHve-‐ and negaHve-‐sequence networks in parallel through impedance Zf as was shown. The zero-‐sequence network is inacHve and does not enter into the line-‐to-‐line calculaHons. The equaHon for the posiHve-‐sequence current in the fault can be determined directly from the circuit as:


Example – Line-‐to-‐Line Faults The same system as before is operaHng at nominal system voltage without prefault currents when a bolted line-‐to-‐line fault occurs at bus 3. Determine the currents in the fault, the line-‐to-‐line voltage at the fault bus, and the line-‐to-‐line voltages at the terminals of machine 2.


Example – Line-‐to-‐Line Faults Note that we will not need the zero-‐sequence bus impedance matrix since the fault is line-‐to-‐line (see Slide 71). The Thévenin equivalent circuit for the posiHve-‐ and negaHve-‐ sequence is: From the circuit:

per unit

V f

Z331( )

Z332( )

Z f = 0

I fA

1( ) I fA

2( )

3

+

V3A1( )

−

3

+

V3A2( )

−Reference

±

Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

I fA

1( ) = − I fA2( ) =

Vf

Z331( ) + Z33

2( ) =1.0

2× j0.1696= − j2.9481


Example – Line-‐to-‐Line Faults Since (Uppercase since the fault is in the high-‐voltage transmission line) Then and since Then Also per unit and per unit MulHply these by Ibase = 167.35 A to get the actual currents.

I fA = I fA

1( ) + I fA2( )

I fA

0( ) = 0

I fA

1( ) = − I fA2( )

I fA = 0

I fB = a2I fA

1( ) + aI fA2( ) = −5.1061

I fC = − I fB = 5.1061


Example – Line-‐to-‐Line Faults The symmetrical components of phase-‐A voltage to ground at bus 3 are:

per unit

V3A0( ) = 0

V3A1( ) =V3A

2( ) = 1− Zkk1( )I fA

1( ) = 1− j0.1696( ) − j2.9481( )= 0.5+ j0


Example – Line-‐to-‐Line Faults The line-‐to-‐ground voltages at fault bus 3 are: all per unit

V3A =V3A0( ) +V3A

1( ) +V3A2( ) = 0+ 0.5+ 0.5= 1.0

V3B =V3B0( ) + a2V3B

1( ) + 2V3B2( ) = 0+ a2 × 0.5+ a × 0.5= −0.5

V3C =V3B = −0.5


Example – Line-‐to-‐Line Faults The line-‐to-‐line voltages at fault bus 3 are: all per unit. MulHply these by 345/31/2 to obtain the actual voltage.

V3,AB =V3A −V3B = 1+ j0( )− −0.5+ j0( ) = 1.5

V3,BC =V3B −V3C = −0.5+ j0( )− −0.5+ j0( ) = 0

V3,CA =V3C −V3A = −0.5+ j0( )− 1+ j0( ) = −1.5


Example – Line-‐to-‐Line Faults For the moment, let us avoid phase shiLs due to the Δ-‐Y transformer connected to machine 2 and calculate the sequence voltages of phase A at bus 4 using the the results from Slide 21 with k = 3 and j = 4: per unit

Vja0( ) = −Z jk

0( )I fa0( ) , Vja

1( ) =Vf − Z jk1( )I fa

1( ) , Vja2( ) = −Z jk

2( )I fa2( )

V4a0( ) = −Z43

0( )I fa0( ) = 0

V4a1( ) =Vf − Z jk

1( )I fa1( ) = 1− j0.1211( ) − j2.9481( ) = 0.643

V4a2( ) = −Z43

2( )I fa2( ) = − j0.1211( ) − j2.9481( ) = 0.357


Example – Line-‐to-‐Line Faults To account for phase shiLs in stepping down from the high-‐voltage transmission line to the low-‐voltage terminals of machine 2, we must retard the posiHve-‐sequence voltage and advance the negaHve-‐sequence voltage by 30°. At machine 2 terminals, indicated by lowercase a, the voltages are:

V4a0( ) = −Z43

0( )I fa0( ) = 0

V4a1( ) =V4a

1( )∠− 30° = 0.643∠− 30° = 0.5569− j0.3215

V4a2( ) =V4a

2( )∠30° = 0.357∠30° = 0.3092+ j0.1785

V4a =V4a0( ) +V4a

1( ) +V4a2( ) = 0.8861− j0.1430 = 0.8778∠− 9.4°


Example – Line-‐to-‐Line Faults Phase-‐b voltages at the terminals of machine 2 are:

V4b0( ) =V4a

0( ) = 0

V4b1( ) = a2V4a

1( ) = 1∠240°× 0.643∠− 30° = −0.5569− j0.3215

V4b2( ) = aV4a

2( ) = 1∠120°× 0.357∠30° = −0.3092+ j0.1785

V4b =V4b0( ) +V4b

1( ) +V4b2( ) = −0.8861− j0.1430 = 0.8778∠−170.6°


Example – Line-‐to-‐Line Faults Phase-‐c voltages at the terminals of machine 2 are:

V4c0( ) =V4a

0( ) = 0

V4c1( ) = aV4a

1( ) = 1∠120°× 0.643∠− 30° = 0.643∠90°

V4c2( ) = a2V4a

2( ) = 1∠240°× 0.357∠30° = 0.357∠− 90°

V4c =V4c0( ) +V4c

1( ) +V4c2( ) = j0.286


Example – Line-‐to-‐Line Faults Line-‐to-‐line voltages at the terminals of machine 2 are:

per unit For the voltage, mulHply by 20 kV/31/2

kV

V4,ab0( ) =V4a

0( ) −V4b0( ) = 1.7322

V4,bc0( ) =V4b

0( ) −V4c0( ) = −0.8661− j0.429

V4,ca0( ) =V4c

0( ) −V4a0( ) = −0.8661+ j0.429

V4,ab0( ) = 1.7322× 20

3= 20

V4,bc0( ) = 11.2∠−153.65°

V4,ca0( ) = 11.2∠153.65°


Double Line-‐to-‐Ground Faults Again it is clear, with fault taken on phases b and c, that the relaHons at fault bus k are:

I fa

I fb

I fc

a

b

c

Z f

I fa = 0

Vkb =Vkc = Z f I fb + I fc( )

k

k

k


Double Line-‐to-‐Ground Faults Since Thus SubsHtuHng into:

I fa = 0⇒ I fa

0( ) = 13

I fa + I fb + I fc( ) = 13

I fb + I fc( ) Vkb =Vkc = Z f I fb + I fc( ) = 3Z f I fa

0( )

Vka0( )

Vka1( )

Vka2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Vka

Vkb

Vkc

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Vka

Vkb

Vkb

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥


Double Line-‐to-‐Ground Faults Expanding the second and third equaHons:

Vka0( )

Vka1( )

Vka2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Vka

Vkb

Vkb

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Vka1( ) = 1

3a + a2( )Vkb

Vka2( ) = 1

3a2 + a( )Vkb

⇒Vka1( ) =Vka

2( )


Double Line-‐to-‐Ground Faults From the first equaHon: recall:

Vka0( )

Vka1( )

Vka2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Vka

Vkb

Vkb

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

3Vka0( ) =Vka + 2Vkb = Vka

0( ) +Vka1( ) +Vka

2( )( ) + 2 3Z f I fa0( )( )

Vkb =Vkc = Z f I fb + I fc( ) 3I fa

0( ) = I fb + I fc


Double Line-‐to-‐Ground Faults Collect zero-‐sequence terms and recall that Now solve: Thus: and since Ifa = 0 These last two results characterize the double line-‐to-‐ground fault.

Vka1( ) =Vka

2( )

3Vka

0( ) =Vka0( ) + 6Z f I fa

0( ) + 2Vka1( )

Vka

1( ) =Vka0( ) − 3Z f I fa

0( )

Vka

1( ) =Vka2( ) =Vka

0( ) − 3Z f I fa0( )

I fa

0( ) + I fa1( ) + I fa

2( ) = 0


Double Line-‐to-‐Ground Faults These can be realized by puxng all three sequence networks in parallel as follows: Clearly:

Zkk2( )

Zkk0( )

3Z f

I fa

2( ) I fa

0( )k k

+−

Vka0( )

+−

Vka2( )

V f

+− Zkk

1( ) I fa

1( )k

+−

Vka1( )

I fa1( ) =

Vf

Zkk1( ) + Zkk

0( ) + 3Z f( ) Zkk2( ) =

Vf

Zkk1( ) +

Zkk0( ) + 3Z f( )Zkk

2( )

Zkk0( ) + Zkk

2( ) + 3Z f


Double Line-‐to-‐Ground Faults By current divider:

Zkk2( )

Zkk0( )

3Z f

I fa

2( ) I fa

0( )k k

+−

Vka0( )

+−

Vka2( )

V f

+− Zkk

1( ) I fa

1( )k

+−

Vka1( )

I fa

2( ) = − I fa1( ) Zkk

0( ) + 3Z f

Zkk0( ) + Zkk

2( ) + 3Z f

, I fa0( ) = − I fa

1( ) Zkk2( )

Zkk0( ) + Zkk

2( ) + 3Z f


Double Line-‐to-‐Ground Faults For a bolted fault Zf is set equal to 0. When Zf = ∞, the zero-‐sequence circuit becomes an open circuit, no zero-‐sequence current an flow, and the equaHons revert back to those for the line-‐to-‐line fault. Again we observe that the sequence currents, once calculated, can be treated as negaHve injecHons into the sequence networks at the fault bus k and the sequence voltage changes at all buses of the system can then be calculated from the bus impedance matrices, as we have done in all along.


Example Double Line-‐to-‐Ground Faults Find the subtransient currents and the line-‐to-‐line voltages at the fault under subtransient condiHons when a double line-‐to-‐ground fault with Zf = 0 occurs at the terminals of machine 2 in the system below. Assume that the system in unloaded and operaHng at rated voltage when the fault occurs.



Example Double Line-‐to-‐Ground Faults Model (the fault is a bus k = 4):

Zkk2( )

Zkk0( )

3Z f

I fa

2( ) I fa

0( )k k

+−

Vka0( )

+−

Vka2( )

V f

+− Zkk

1( ) I fa

1( )k

+−

Vka1( )

Zbus0( ) =

j0.19 0 0 00 j0.08 j0.08 00 j0.08 j0.58 00 0 0 j0.19

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

, Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥


Example Double Line-‐to-‐Ground Faults Model (the fault is a bus k = 4):

per unit

j0.1437 j0.19 I fa

2( ) I fa

0( )4 4

+−

V4a0( )

+−

V4a2( )

1.0 +

−

I fa

1( )4

+−

V4a1( ) j0.1437

I fa1( ) =

Vf

Zkk1( ) +

Zkk0( ) + 3Z f( )Zkk

2( )

Zkk0( ) + Zkk

2( ) + 3Z f

= 1.0

Z441( ) +

Z440( )Z44

2( )

Z440( ) + Z44

2( )

= 1.0

j0.1437 +j0.19( ) j0.1437( )j0.19+ j0.1437

= − j4.4342


Example Double Line-‐to-‐Ground Faults The sequence voltages at the fault are:

per unit

j0.1437 j0.19 I fa

2( ) I fa

0( )4 4

+−

V4a0( )

+−

V4a2( )

1.0 +

−

I fa

1( )4

+−

V4a1( ) j0.1437

V4a1( ) =V4a

2( ) =V4a0( ) =Vf − Z44

1( )I fa1( )

= 1.0− j j0.1437( ) − j4.4342( ) = 0.3628


Example Double Line-‐to-‐Ground Faults Now that we have

per unit

j0.1437 j0.19 I fa

2( ) I fa

0( )4 4

+−

V4a0( )

+−

V4a2( )

1.0 +

−

I fa

1( )4

+−

V4a1( ) j0.1437

I fa2( ) = − I fa

1( ) Zkk0( ) + 3Z f

Zkk0( ) + Zkk

2( ) + 3Z f

= − I fa1( ) Zkk

0( )

Zkk0( ) + Zkk

2( ) = − I fa1( ) Z44

0( )

Z440( ) + Z44

2( )

= j4.4342 j0.19j0.19+ j0.1437

= j2.5247

I fa

1( )


Example Double Line-‐to-‐Ground Faults Now that we have

per unit

j0.1437 j0.19 I fa

2( ) I fa

0( )4 4

+−

V4a0( )

+−

V4a2( )

1.0 +

−

I fa

1( )4

+−

V4a1( ) j0.1437

I fa0( ) = − I fa

1( ) Zkk2( )

Zkk0( ) + Zkk

2( ) + 3Z f

= − I fa1( ) Z44

2( )

Z440( ) + Z44

2( )

= j4.4342 j0.1437j0.19+ j0.1437

= j1.9095

I fa

1( )


Example Double Line-‐to-‐Ground Faults The currents out of the system at the fault point are: The current If into ground is:

all in per unit

I fa = I fa0( ) + I fa

1( ) + I fa2( ) = j1.9095− j4.4342+ j2.5247 = 0

I fb = I fa0( ) + a2I fa

1( ) + aI fa2( ) = −6.0266+ j2.8642

I fc = I fa0( ) + aI fa

1( ) + a2I fa2( ) = 6.0266+ j2.8642

I f = I fa + I fb + I fc = 2I fa

0( ) = j5.7285


Example Double Line-‐to-‐Ground Faults CalculaHng the a-‐b-‐c voltages at the fault bus:

V4a =V4a0( ) +V4a

1( ) +V4a2( ) = 1.0884

V4b =V4c = 0

V4,ab =V4a −V4b = 1.0884

V4,bc =V4b −V4c = 0

V4,ca =V4c −V4a = −1.0884


Example Double Line-‐to-‐Ground Faults Recall the base current in the circuit of machine 2 is:

amps and the base line-‐to-‐neutral voltage in machine 2 is:

kV

Ibase =

100×103

3 × 20= 2887

Vbase =

203


Phase Shias The previous two examples show that phase shiLs due to Δ-‐Y transformers do not enter into the calculaHons of sequence currents and voltages in that part of the system where the fault occurs provided Vf at the fault point is chosen as the reference voltage for the calculaHons. However, for those parts of the system which are separated by Δ-‐Y transformers from the fault point, the sequence currents, and voltages calculated by bus impedance matrix must be shiLed in phase before being combined to form the actual voltages. This is because the bus impedance matrices of the sequence networks are formed without consideraHon of phase shiLs, and so they consist of per-‐unit impedances referred to the part of the network which includes the fault point.


Example Phase Shias Solve for the subtransient voltages to ground at bus 2, the end of the transmission line remote from the double line-‐to-‐ground fault for the same system.


Zbus0( ) =

j0.19 0 0 00 j0.08 j0.08 00 j0.08 j0.58 00 0 0 j0.19

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

, Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥


Example Phase Shias We just solved for the values of the fault-‐current components. NeglecHng the phase shiL of the Δ-‐Y transformer for the moment we have, from Slide 21: At bus 2:

Vja

0( ) = −Z jk0( )I fa

0( ) , Vja1( ) =Vf − Z jk

1( )I fa1( ) , Vja

2( ) = −Z jk2( )I fa

2( )

V2a0( ) = −Z24

0( )I fa0( ) = − 0( ) j1.9095( ) = 0

V2a1( ) =Vf − Z24

1( )I fa1( ) = 1.0− j0.0789( ) − j4.4342( ) = 0.6501

V2a2( ) = −Z24

2( )I fa2( ) = − j0.0789( ) j2.5247( ) = 0.1992


Example Phase Shias AccounHng for phase shiL in stepping up to the transmission-‐line circuit from the fault at bus 4 we have: Now the required voltages can be calculated: all per unit

V2 A0( ) = 0

V2 A1( ) =V2 A

1( )∠30° = 0.6501∠30°

V2 A2( ) =V2 A

2( )∠− 30° = 0.1992∠− 30°

V4 A =V4 A0( ) +V4 A

1( ) +V4 A2( ) = 0.7355+ j0.2255

V4 B =V4 B0( ) + a2V4 B

1( ) + aV4 B2( ) = −0.1275 j0.5535

V4C =V4C0( ) + aV4C

1( ) + a2V4C2( ) = −0.5656+ j0.1274


Example Phase Shias The per-‐unit values can be converted to volts by mulHplying by the line-‐to-‐neutral base voltage of:

kV of the transmission line.

Vbase =

3453


Open-‐Conductor Faults When one phase of a balanced three-‐phase system opens, an unbalance is created and asymmetrical currents flow. A similar type of unbalance occurs when any two of the three phases are opened while the third phase remains closed. These unbalanced condiHons are caused, for example, when one-‐ or two-‐phase conductors of a transmission line are physical broken by accident or storm. In other circuits, due to current overload, fuses or other switching devices may operate in one or two conductors and fail to operate in other conductors. Such open-‐conductor faults can be analyzed by means of the bus impedance matrices of the sequence networks, as we now show.


Open-‐Conductor Faults Consider a secHon of a three-‐phase circuit in which the line currents in the respecHve phases are la, Ib, and Ic, with posiHve direcHon from bus m to bus n, with phase a open between points p and p’:

Ia

Ib

Ic

a

b

c

m

m

m

n

n

n

p p’


Open-‐Conductor Faults Also consider the case where phases b and c are open between points p and p’:

Ia

Ib

Ic

a

b

c

m

m

m

n

n

n

p p’


Open-‐Conductor Faults The same open-‐conductor fault condiHons will result is all three phases are first opened between points p and p’ and short circuits are then applied in those phases which are shown to be closed in the preceding figures. The ensuing development follows this reasoning. Opening the three phases is the same a removing the line m-‐n altogether then adding appropriate impedances from buses m and n to the points p and p’.


Open-‐Conductor Faults If line m-‐n has the sequence impedances Z0, Z1, and Z2, simulate the opening of the three phases by adding the negaHve impedances – Z0, – Z1, and – Z2 between buses m and n in the corresponding Thévenin equivalents of the three sequence networks of the intact system.


Open-‐Conductor Faults For example, consider the connecHon of – Z1 to the posiHve-‐sequence Thévenin equivalent between buses m and n: Voltages Vm and Vn are the normal (posiHve-‐sequence) voltages of phase a at buses m and n before the open-‐conductor faults occur.

Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( ) = Zmm1( ) + Znn

1( ) − 2Zmn1( )


Open-‐Conductor Faults The posiHve-‐sequence impedances kZ1 and (1 – k)Z1, 0<k<1, is added to represent the fracHonal lengths of the broken line m-‐n from bus m to point p and bus n to point p’.

Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( ) = Zmm1( ) + Znn

1( ) − 2Zmn1( )

kZ1

1− k( )Z1

p

p’

+

Va1( )

−

Z p ′p

1( )


Open-‐Conductor Faults Simplify…

Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( )

kZ1

1− k( )Z1

p

p’

+

Va1( )

−

Add the do-‐nothing source



Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( )

kZ1

1− k( )Z1

p

p’

+

Va1( )

−

Combine these



Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( )

Z1 p

p’

+

Va1( )

−

Perform a source conversion



Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( )

p

p’

Va1( )

Z1

Open circuit

Z1


Open-‐Conductor Faults Final result (posiHve-‐sequence equivalent circuit):

Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( )

Zth,mn

1( )

Va1( )

Z1


Open-‐Conductor Faults The above consideraHons for the posiHve-‐sequence network also apply directly to the negaHve-‐ and zero-‐sequence networks, but we must remember that the laber networks do not contain any internal sources of their own.


Open-‐Conductor Faults NegaHve-‐sequence equivalent circuit:

Zmm2( ) − Zmn

2( )

Reference

m

n

0

Zmn2( ) = Znm

2( )

Znn2( ) − Znm

2( ) −Z2

Zth,mn

2( ) = Zmm2( ) + Znn

2( ) − 2Zmn2( )

kZ2

1− k( )Z2

p

p’

+

Va2( )

−

Z p ′p

2( )


Open-‐Conductor Faults NegaHve-‐sequence equivalent circuit simplified:

Zmm2( ) − Zmn

2( )

Reference

m

n

0

Zmn2( ) = Znm

2( )

Znn2( ) − Znm

2( )

Zth,mn

2( )

Va2( )

Z2


Open-‐Conductor Faults Zero-‐sequence equivalent circuit:

Zmm0( ) − Zmn

0( )

Reference

m

n

0

Zmn0( ) = Znm

0( )

Znn0( ) − Znm

0( ) −Z0

Zth,mn

0( ) = Zmm0( ) + Znn

0( ) − 2Zmn0( )

kZ0

1− k( )Z0

p

p’

+

Va0( )

−

Z p ′p

0( )


Open-‐Conductor Faults Zero-‐sequence equivalent circuit simplified:

Zmm0( ) − Zmn

0( )

Reference

m

n

0

Zmn0( ) = Znm

0( )

Znn0( ) − Znm

0( )

Zth,mn

0( )

Va0( )

Z0


Open-‐Conductor Faults Let the voltage denote the phase-‐a posiHve-‐sequence component of the voltage drops Vpp’a, Vpp’b, and Vpp’c from p to p' in the phase conductors. We will soon see that and the corresponding negaHve-‐ and zero-‐sequence components and , take on different values depending on which one of the open-‐conductor fauIts is being considered.

Va1( )

Va1( )

Va2( )

Va0( )


Open-‐Conductor Faults In drawing the sequence equivalent circuits it is understood that the currents sources owe their origin to the open-‐conductor fault between points p and p' in the system. If there is no open conductor, the voltages: are all zero and the current sources disappear.

Va0( ) , Va

1( ) , Va2( )


Open-‐Conductor Faults Note that each of the sequence currents sources: can be regarded in turn as a pair of injecHons into buses m and n of the corresponding sequence network of the intact system. Hence, we can use the bus impedance sequence matrices of the normal configuraHon of the system to determine the voltage changes due to the open-‐conductor faults.

Va0( )

Z0

,Va

1( )

Z1

,Va

2( )

Z2


Open-‐Conductor Faults First we must find expressions for the symmetrical components of Va (i.e., of the voltage drops across the fault points p and p’ for each type of fault, (one or two open lines). These voltage drops can be regarded as giving rise to the following sets of injecHon currents into the sequence networks of the normal system configuraHon:

POSITIVE NEGATIVE ZERO SEQUENCE SEQUENCE SEQUENCE

At bus m: At bus n:

Va1( ) Z1 Va

2( ) Z2 Va0( ) Z0

−Va1( ) Z1 −Va

2( ) Z2 −Va0( ) Z0


Open-‐Conductor Faults Recall Slide 87 from Network CalculaKons:

ΔV1

ΔVj

ΔVk

ΔVN

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

Z1 jΔI j + Z1kΔIk

Z jjΔI j + Z jkΔIk

ZkjΔI j + ZkkΔIk

ZNjΔI j + ZNkΔIk

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥


Open-‐Conductor Faults The changes in the symmetrical components of the phase-‐a voltage of each bus i is::

Zero-‐sequence:

PosiHve-‐sequence:

NegaHve-‐sequence:

ΔVi

0( ) =Zim

0( ) − Zin0( )

Z0

Va0( )

ΔVi

1( ) =Zim

1( ) − Zin1( )

Z1

Va1( )

ΔVi

2( ) =Zim

2( ) − Zin2( )

Z2

Va2( )


Open-‐Conductor Faults Before developing the equaHons for the sequence components of the voltage for each type of open-‐ conductor fault, let us derive expressions for the Thévenin equivalent impedances of the sequence networks as seen from fault points p and p’.


Open-‐Conductor Faults Looking into the posiHve-‐sequence network between p and p’, we see the impedance:

Zmm1( ) − Zmn

1( )

Reference

m

n

0

Zmn1( ) = Znm

1( )

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( ) = Zmm1( ) + Znn

1( ) − 2Zmn1( )

kZ1

1− k( )Z1

p

p’

+

Va1( )

−

Z p ′p

1( )

Z p ′p

1( ) = kZ1 + Zth,mn1( ) −Z1( ) + 1− k( )Z1 = Z1 −

Z1Zth,mn1( )

Zth,mn1( ) − Z1

=−Z1

2

Zth,mn1( ) − Z1


Open-‐Conductor Faults The open-‐circuit voltage from p to p’ is: But: Thus:

Vm

Zmm1( ) − Zmn

1( )

+−

Reference

m

n

0

−+

Zmn1( ) = Znm

1( )

Vn

Znn1( ) − Znm

1( ) −Z1

Zth,mn

1( ) = Zmm1( ) + Znn

1( ) − 2Zmn1( )

kZ1

1− k( )Z1

p

p’

+

Va1( )

−

Z p ′p

1( )

Vp ′p

1( ) = −Z1

Vm −Vn

Zth,mn1( ) − Z1

Z p ′p

1( ) =−Z1

2

Zth,mn1( ) − Z1

Vp ′p1( ) =

Z p ′p1( )

Z1

Vm −Vn( )


Open-‐Conductor Faults Before any conductor opens, the current Imn in phase a of the line m-‐n is posiHve sequence and is given by: Thus: Where: Similarly: and: Now we can find:

Vp ′p

1( ) = Z p ′p1( ) Imn

Imn =

Vm −Vn

Z1

Z p ′p

1( ) =−Z1

2

Zth,mn1( ) − Z1

Z p ′p

2( ) =−Z2

2

Zth,mn2( ) − Z2

Z p ′p0( ) =

−Z02

Zth,mn0( ) − Z0

Va0( ) , Va

1( ) , Va2( )


Open-‐Conductor Faults Thévenin equivalents:

PosiHve-‐Sequence NegaHve-‐Sequence Zero-‐Sequence:

+

ImnZ p ′p1( )

−

p

p’

+

Va1( )

−

Z p ′p

1( )

p

p’

+

Va2( )

−

Z p ′p

2( )

p

p’

+

Va0( )

−

Z p ′p

0( )

Ia0( )

Ia2( )

Ia1( )


Open-‐Conductor Faults – One Open Conductor

Ia

Ib

Ic

a

b

c

m

m

m

n

n

n

p p’

Ia = 0⇒ Ia0( ) + Ia

1( ) + Ia2( ) = 0

Vp ′p ,b = 0

Vp ′p ,c = 0


Open-‐Conductor Faults – One Open Conductor The open conductor in phase a causes equal voltage drops to appear from p to p' in each of the sequence networks. We can saHsfy this requirement by connecHng the Thévenin equivalents of the sequence networks in parallel at the points p and p’.

Va0( )

Va1( )

Va2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Vp ′p ,a

00

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 1

3

Vp ′p ,a

Vp ′p ,a

Vp ′p ,a

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⇒Va0( ) =Va

1( ) =Va2( ) =

Vp ′p ,a

3


Open-‐Conductor Faults – One Open Conductor From this circuit the expression for the posiHve-‐sequence current is found as:

Z p ′p

2( ) Z p ′p

0( ) Ia

2( ) Ia

0( ) p p

+−

Va0( )

+−

Va2( )

+

ImnZ p ′p1( )

−

Z p ′p

1( ) Ia

1( )p

+−

Va1( )

p’ p’ p’

Ia1( ) =

ImnZ p ′p1( )

Z p ′p1( ) + Z p ′p

0( ) Z p ′p2( ) =

ImnZ p ′p1( )

Z p ′p1( ) +

Z p ′p0( )Z p ′p

2( )

Z p ′p0( ) + Z p ′p

2( )

= Imn

Z p ′p1( ) Z p ′p

0( ) + Z p ′p2( )( )

Z p ′p0( )Z p ′p

1( ) + Z p ′p1( ) Z p ′p

2( ) + Z p ′p0( )Z p ′p

2( )


Open-‐Conductor Faults – One Open Conductor The sequence voltage drops are:

Z p ′p

2( ) Z p ′p

0( ) Ia

2( ) Ia

0( ) p p

+−

Va0( )

+−

Va2( )

+

ImnZ p ′p1( )

−

Z p ′p

1( ) Ia

1( )p

+−

Va1( )

p’ p’ p’

Va

0( ) =Va1( ) =Va

2( ) = Ia1( ) Z p ′p

0( )Z p ′p2( )

Z p ′p0( ) + Z p ′p

2( ) = Imn

Z p ′p0( )Z p ′p

1( ) Z p ′p2( )

Z p ′p0( )Z p ′p

1( ) + Z p ′p1( ) Z p ′p

2( ) + Z p ′p0( )Z p ′p

2( )

These terms are known from the impedance parameters of the sequence networks and the prefault current in phase a of the line m-‐n.


Open-‐Conductor Faults – One Open Conductor The currents: for injecHon into the sequence networks can now be determined.

Va0( )

Z0

,Va

1( )

Z1

,Va

2( )

Z2

Zmm0,1,2( ) − Zmn

0,1,2( )

Reference

m

n

0

Zmn0,1,2( ) = Znm

0,1,2( )

Znn0,1,2( ) − Znm

0,1,2( )

Zth,mn

0,1,2( )

Va0,1,2( )

Z0,1,2


Open-‐Conductor Faults – Two Open Conductors Clearly:

Ia

Ib

Ic

a

b

c

m

m

m

n

n

n

p p’

Ib = 0

Ic = 0

Vp ′p ,a =Va

0( ) +Va1( ) +Va

2( ) = 0


Open-‐Conductor Faults – Two Open Conductors Resolving the current into its symmetrical components: Along with: These can be saHsfied by connecHng the Thévenin equivalents of the sequence networks in series between the points p and p’.

Ia0( )

Ia1( )

Ia2( )

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= 13

1 1 11 a a2

1 a2 a

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Ia

00

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥= 1

3

Ia

Ia

Ia

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⇒ Ia0( ) = Ia

1( ) = Ia2( ) =

Ia

3

Vp ′p ,a =Va

0( ) +Va1( ) +Va

2( ) = 0


Open-‐Conductor Faults – Two Open Conductors

Z p ′p

1( )

Z p ′p

2( )

Z p ′p

0( )

Ia1( )

Ia2( )

Ia0( )

Ia0( ) = Ia

1( ) = Ia2( )p

p

p

+−

Va1( )

+−

Va0( )

+−

Va2( )

p’

p’

p’

+

ImnZ p ′p1( )

−



Z p ′p

1( )

Z p ′p

2( )

Z p ′p

0( )

Ia1( )

Ia2( )

Ia0( )

Ia0( ) = Ia

1( ) = Ia2( )p

p

p

+−

Va1( )

+−

Va0( )

+−

Va2( )

p’

p’

p’

+

ImnZ p ′p1( )

−

Ia0( ) = Ia

1( ) = Ia2( )

=ImnZ p ′p

1( )

Z p ′p0( ) + Z p ′p

1( ) + Z p ′p2( )



Z p ′p

1( )

Z p ′p

2( )

Z p ′p

0( )

Ia1( )

Ia2( )

Ia0( )

p

p

p

+−

Va1( )

+−

Va0( )

+−

Va2( )

p’

p’

p’

+

ImnZ p ′p1( )

−

Va1( ) = ImnZ p ′p

1( ) − Z p ′p1( ) Ia

1( )

= ImnZ p ′p1( ) − Z p ′p

1( ) ImnZ p ′p1( )

Z p ′p0( ) + Z p ′p

1( ) + Z p ′p2( )

= ImnZ p ′p1( ) Z p ′p

0( ) + Z p ′p2( )

Z p ′p0( ) + Z p ′p

1( ) + Z p ′p2( )

Va

2( ) = −Z p ′p2( ) Ia

2( ) = −ImnZ p ′p

1( ) Z p ′p2( )

Z p ′p0( ) + Z p ′p

1( ) + Z p ′p2( )

Va

0( ) = −Z p ′p0( ) Ia

0( ) = −ImnZ p ′p

0( )Z p ′p1( )

Z p ′p0( ) + Z p ′p

1( ) + Z p ′p2( )


Open-‐Conductor Faults – Two Open Conductors In each of these equaHons the right-‐hand side quanHHes are all known before the fault occurs. The net effect of the open conductors on the posiHve-‐sequence network is to increase the transfer impedance across the line in which the open-‐conductor fault occurs. For one open conductor this increase in impedance equals the parallel combinaHon of the negaHve-‐ and zero-‐sequence networks between points p and p’. For two open conductors the increase in impedance equals the series combinaHon of the negaHve-‐and zero-‐sequence networks between points p and p'.


Open-‐Conductor Faults – EXAMPLE For the system of Slide 39, consider that Machine 2 is a motor drawing a load equivalent to 50 MVA at 0.8 power-‐factor lagging and nominal system voltage of 345 kV at bus 3. Determine the change in voltage at bus 3 when the transmission line undergoes a.  a one-‐open-‐conductor fault and b.  a two-‐open-‐conductor fault along its span between buses 2 and 3.

Choose a base of 100 MVA, 345 kV in the transmission line.



Open-‐Conductor Faults – EXAMPLE Prefault current (per unit) in line 2-‐3:

S = 50Sbase

= 50100

= 0.5

S =V3I23* ⇒ I23 =

P − jQV3

* = SP − jQ( ) S

1.0

= 0.50.8− j0.61.0

= 0.4− j0.3


Open-‐Conductor Faults – EXAMPLE Recall: Also recall: From Slide 108:

Zbus0( ) =

j0.19 0 0 00 j0.08 j0.08 00 j0.08 j0.58 00 0 0 j0.19

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

, Zbus1,2( ) =


⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Z1 = Z2 = j0.15, Z0 = j0.5

Z p ′p

1( ) =−Z1

2

Zth,mn1( ) − Z1

Z p ′p0( ) =

−Z02

Zth,mn0( ) − Z0

Zth,mn

1( ) = Zmm1( ) + Znn

1( ) − 2Zmn1( )

Z p ′p1( ) = Z p ′p

2( ) =−Z1

2

Z221( ) + Z33

1( ) − 2Z231( ) − Z1

=− j0.15( )2

j0.1696+ j0.1696− 2 j0.1104( )− j0.15= j0.712

Z p ′p0( ) =

−Z02

Z220( ) + Z33

0( ) − 2Z230( ) − Z0

=− j0.5( )2

j0.08+ j0.58− 2 j0.08( )− j0.5= ∞


Open-‐Conductor Faults – EXAMPLE Thus, if the line from bus 2 to bus 3 is opened, then an infinite impedance is seen looking into the zero-‐sequence network between points p and p' of the opening. The zero-‐sequence circuit confirms this fact since bus 3 would be isolated from the reference by opening the connecHon between bus 2 and bus 3. Slide 47:

1 2 3 4

j0.04

j0.04

Reference

j0.04 j0.5 j0.08 j0.04

3Xn = j0.15

(1) (2) (3) (4) (5) (6)

3Xn = j0.15




Open-‐Conductor Faults – EXAMPLE One open conductor: From Slide 118, From Slide 104:

Va0( ) =Va

1( ) =Va2( ) = Imn

Z p ′p0( )Z p ′p

1( ) Z p ′p2( )

Z p ′p0( )Z p ′p

1( ) + Z p ′p1( ) Z p ′p

2( ) + Z p ′p0( )Z p ′p

2( ) →Z p ′p

0( )→∞Imn

Z p ′p1( ) Z p ′p

2( )

Z p ′p1( ) + Z p ′p

2( )

= 0.4− j0.3( ) j0.712( )2

2 j0.712( ) = 0.1068+ j1424

ΔVi

0( ) =Zim

0( ) − Zin0( )

Z0

Va0( ) , ΔVi

1( ) = ΔVi2( ) =

Zim1( ) − Zin

1( )

Z1

Va1( )


Open-‐Conductor Faults – EXAMPLE One open conductor:

ΔV30( ) =

Z320( ) − Z33

0( )

Z0

Va0( ) , ΔV3

1( ) = ΔV32( ) =

Z321( ) − Z33

1( )

Z1

Va1( )

ΔV30( ) = j0.08− j0.58

j0.50.1068+ j1424( ) = −0.1068− j0.1424

ΔV31( ) = ΔV3

2( ) = j0.1104− j0.1696j0.15

0.1068+ j1424( ) = −0.0422− j0.0562

ΔV3 = ΔV30( ) + ΔV3

1( ) + ΔV32( ) = −0.1912− j0.24548

V3new =V3 + ΔV3 = 1.0− 0.1912− j0.24548 = 0.8088− j0.2548


Open-‐Conductor Faults – EXAMPLE Two open conductors: InserHng the infinite impedance of the zero-‐sequence network in series between points p and p' of the posiHve-‐sequence network causes an open circuit in the laber . No power transfer can occur in the system, since power cannot be transferred by only one phase conductor of the transmission line in this case since the zero-‐sequence network offers no return path for current. 1 2 3 4

+−

Vf V f

+−

j0.20 j0.04 j0.20

j0.04 j0.04 j0.04

Reference

Open (1)

(2) (3) (4) (5)