Unsymmetrical Faults

Method of Symmetrical

Components

� This method is based on the fact that a set of three-phase unbalanced phasors can be resolved into three sets of symmetrical components, which are termed the positive sequence, negative-sequence, and zero-sequence components.

� a. positive phase sequence system – a

balanced three-phase system in the normal

sequence

� b. negative phase sequence system - a

balanced three-phase system in the

reversed sequence

� c. zero phase sequence system – three

phasors equal in magnitude and direction

revolving in the positive phase

Prepared by: Mr. John Mateo

Prepared by: Mr. John Mateo

In particular, we haveV a= V a1 + V a2 + V a3

V b= V b1 + V b2 + V b3

V c = V c1 + V c2 + V c3

We now introduce an operator a that causes a counterclcokwiserotation of 120° (just as the j operator produces a 90°rotation), such that

1 + a + a 2 = 0

Prepared by: Mr. John Mateo

� V b1 = a 2V a1

� V c1 = aV a1

� V b2 = aVa2

� V c2 = a2V a2

� V a0 = V b0 = V co

in terms of components of phase a,V a = V a0 + V a1 + Va2

Vb= V a0 + a2 a1 1 + aVa2

Vc = V a0 + aV a1 + a2Va2

components of a given sequence in terms of any chosen component.

Prepared by: Mr. John Mateo

Solving for the sequence components of aV a0 = 1/3 (V a + V b + Vc) V a1 = 1/3 (V a + aVb + a2Vc) V a 2= 1/3 (V a + a2Vb + aVc)

A quantity (current, voltage, impedance, power) that is given in terms of symmetrical components is sometimes called the sequence quantity, as in “sequence current.”

SEQUENCE NETWORKS OF

GENERATORS

(1) Since the generator is designed to supply balanced three phase voltages, the generator voltages are of positive sequence only and hence only appear in the positive sequence network

(2) Since the neutral impedance connected to the generator neutral carries three zero sequence current, the neutral generator impedance is represented by three times its value in the zero sequence network

Transformer phase

representation

� Zero-sequence

currents free to

flow in both primary

and secondary

circuits

Wye-wye grounded

� No path for zero-

sequence currents

in primary circuit

Delta-wye

� Single phase

currents can

circulate in the

delta but not

outside it

Delta - wye

� No flow of zero-

sequence currents

possible

Delta-delta

� No flow of zero-

sequence currents

possible

� Tertiary windings

provides path for

zero-sequence

currents

Types of Faults

� Single line-to-earth fault

� Io=1/3(Ia +Ib + Ic)

� I1 = 1/3(Ia +a2Ic + aIb)

� I2 = 1/3(Ia +a2Ib + aIc)

� Io=I1=I2 = Ia/3

� Ib = Ic = 0

� Va = E - I1Z1 – I2Z2 -1oZo = 0

� E = I1Z1 – I2Z2 -1oZo

� If = 3E / (Z1 + Z2 + Zo)

Sequence inteconnection

Line –to-line fault

� Io=0

� I1 = 1/3 Ib(a - a2)

� I2 = 1/3 Ib(a2 - a)

� I1= E/ (Z1 + Z2)

Sequence inteconnection

Line –to-line-to-earth fault

� Ia = 0

� I1 = E / (Z1 + Z2Zo/(Z2 +Zo)

� I2 = - I1 (Zo /(Zo+ Z2 ))

� I0 = - I1 (Z2/(Zo+ Z2 ))

� If = 3E / (Z1 + 3Zg +Z2 + Zo)

� The phase currents in a wye-connected, unbalanced load are Ia = (44 – j33), Ib = -(32 + j24), and Ic = -(40 + j25)A. Determine the sequence currents.

Prepared by: Mr. John Mateo

� I a0 = 1/3 [(44 – j33) – (32 + j24) + (-40 +

j25)]

= -9.33 – j10.67 = 14.17∠-131.2°A

� I a1 = 1/3 [(44 – j33) – (-0.5 + j0.866)(32 +

j24) – (0.5 – j0.866)(-40 + j25)]

= 40.81 – j8.77 = 41.7<-12.1°A

� I a2 = 1/3 [(44 – j33) + (0.5 – j0.866)(32 +

j24) + (-0.5 + j0.866)(-40 + j25)]

= 12.52 – j13.48 = 18.37 ∠ -47°A

Prepared by: Mr. John Mateo

Example

A synchronous machine A generating 1 p.u. voltage is connected through star-star transformer, reactance 0.12 p.u. to two lines in parallel. The other ends of the lines are connected through star-star transformer of reactance 0.1 p.u. to a second machine B, also generating 1 p.u. voltage. For both transformers, X1 = X2 = Xo.

� The relevant per unit reactances of the plant, all referred to the sme base are as follows:

� For each line, X1=X2=0.3; X0=0.7

� The star points of machine A and B are solidly earthed.

X1 X2 X0

Machine A 0.3 0.2 0.05

Machine B 0.25 0.15 0.03

Positive- sequence

Negative sequence

Zero-sequence

Example

� A 75 MVA 115/33 kV power transformer at the P.T.T. Central Substation in Rayong, Thailand, developed a line-to-ground short-circuit fault at a high voltage winding terminal.

� The percentage impedance of the transformer is 12% at 75 MVA, and the transformer connection is delta on the HV windings and star on the LV windings. If the supply impedance of 115 kV power supply is 3% based on 75 MVA, draw the sequence impedance networks for the ground fault, and calculate the ground fault current. The 115 kV network can be considered as a star connection with centre point solidly earthed.

� For a single line-to-ground fault on the “A” phase, the positive, negative and zero sequence networks are to be connected in series in order to calculate the sequence short-circuit current,

� Total sequence impedances = 3 x 3% = 9%

If = 1/9% = 11.11 p.u.

� Short-circuit current on “A” phase

= 3 x 11.11 p.u.

= 33.33 x 376.5

= 12,549 Ampere