Lecture 1

Balanced Three-phase SystemsFrom Network to Single-phase Equivalent Circuit

Power plant Transformer Switchingstation

Line Load

Three-phase system

Single-line diagram

Equivalent single-phase circuit

Balanced Three-phase Systems•A three-phase system can be analyzed by means of a single-phase equivalent when:

– The source voltages are balanced or symmetrical

– The electrical parameters of the system are symmetrical

– The loads are balanced

•Three-phase quantities can be determined from single phase voltages and currents when symmetry is assumed between phases.

Balanced and Unbalanced faults

•Balanced Cases– three-phase fault – (symmetrical) load flow

•Unbalanced Cases– Single line to ground fault– Line to line fault– Double line to ground fault– (unsymmetrical load flow)

Analyzing unbalanced system using Fortescue’s Theorem

–Unbalanced faults in power systems require a phase by phase solution method or other techniques. –One of the most useful techniques to

deal with unbalanced networks is the “symmetrical component” method, developed in 1918 by C.L. Fortescue.

Symmetrical Components

•Reasons for use of symmetrical components:–Unbalanced systems are difficult to handle

• >-several independent balanced systems are easier to handle than one unbalanced system.

–Transformation of one unbalanced 3-phase system into 3 balanced 3-phase systems.

• >-for each system only one phase has to be considered

Analyzing unbalanced system using Fortescue’s Theorem

Any three unbalanced set of voltages or currents can be resolved into three balanced systems of voltages or currents, referred to as the system symmetrical components, defined as follows:

Positive Sequence components: three phasors of equal magnitude displaced 120 degrees from each other following the positive sequence

Negative Sequence components: three phasors of equal magnitude displaced 120 degrees of each other following the negative sequence

Zero Sequence components: three parallel phasors having equal magnitude and angle

For a 3-ph system: 3 unbalanced phasors can be resolved into 3 balanced systems of 3 phasors each

Let Va, Vb, Vc be the Phase voltages According to Fortescue, these can be

transformed into1. Positive-seq. voltages: Va1, Vb1, Vc1

2. Negative-seq. voltages: Va2, Vb2, Vc2

3. zero-sequence voltages: Va0, Vb0, Vc0

Thus, Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0

Va1

Vb1

Vc1

Va2

Vc2

Vb2

Va1

Va2Va0

VaVc

Vb

Va0 Vb0 Vc0

The ‘a’ operatora = 1<1200 = -0.5 + j

0.866 a I rotates I by 1200

a2 = 1<2400 = -0.5 – j 0.866

a3 = 1<3600 = 1<00 = 1 + j 0

1 + a + a2 = 0

1

a2

a

-a

-a2

-1

From figure previous figures

Vb1 = a2Va1 Vc1 = a Va1

Vb2 = a Va2 Vc2 = a2 Va2

Vb0 = Va0 Vc0 = Va0

sub. In Eq. (Slide 8) we get:Thus, Va = Va0 + Va1 + Va2

Vb = Va0 + a2Va1 + a Va2

Vc = Va0 + a Va1 + a2Va2

Matrix RelationsLet

And Inverse of A is

a a02

b a12

c a2

v v 1 1 1

Vp = v ; Vs = v ; A = 1 a a

v v 1 a a

-1 213

2

1 1 1

A = 1 a a

1 a a

Matrix Relations

Similarly currents can be obtained using their symmetrical components

2

1

0

2

2

1

1

111

a

a

a

c

b

a

v

v

v

aa

aa

v

v

v

Matrix Relations Vp = A Vs; Vs = A-1Vp

Va0 = 1/3 (Va + Vb + Vc)

Va1 = 1/3 (Va + aVb + a2Vc)

Va2 = 1/3 (Va + a2Vb + aVc)

Matrix Relations

c

b

a

a

a

a

v

v

v

aa

aa

v

v

v

2

231

2

1

0

1

1

111

Numerical Example1. The line currents in a 3-ph 4 –wire

system are Ia = 100<300 ; Ib = 50<3000 ; Ic = 30<1800. Find the symmetrical components and the neutral current.

Solution :

Ia0 = 1/3(Ia + Ib + Ic) = 27.29 < 4.70 AIa1 = 1/3(Ia + a Ib + a2Ic) = 57.98 < 43.30 AIa2 = 1/3(Ia + a2 Ib + a Ic) = 18.96 < 24.90 AIn = Ia + Ib + Ic = 3 Ia0 = 81.87 <4.70 A

Numerical Example2. The sequence component voltages of

phase voltages of a 3-ph system are: Va0 = 100 <00 V; Va1 = 223.6 < -26.60 V ; Va2 = 100 <1800 V. Determine the phase voltages.

Solution:

Va = Va0 + Va1 + Va2 = 223.6 <-26.60 V

Vb = Va0 + a2Va1 + a Va2 = 213 < -99.90 V

Vc = Va0 + a Va1 + a2 Va2 = 338.6 < 66.20 V

Numerical Example3. The two seq. components and the

corresponding phase voltage of a 3-ph system are Va0 =1<-600 V; Va1=2<00 V ; & Va = 3 <00 V.

Determine the other phase voltages.Solution:

Va = Va0 + Va1 + Va2

Va2 = Va – Va0 – Va1 = 1 <600 V

Vb = Va0 + a2Va1 + a Va2 = 3 < -1200 V

Vc = Va0 + a Va1 + a2 Va2 = 0 V

Numerical Example4. Determine the sequence components if Ia =10<600 A; Ib = 10<-600 A ; & Ic = 10

<1800 A.

Solution:

Ia0 = 1/3 (Ia + Ib + Ic) = 0 A

Ia1 = 1/3 (Ia + a Ib + a2Ic) = 10<600 A

Ia2 = 1/3 (Ia + a2 Ib + a Ic) = 0 A

Thus, If the phasors are balanced, Two Sequence components will be zero.

Ia

Ib

Ia

Numerical Example5. Determine the sequence components

if Va = 100 <300 V; Vb = 100 <1500

V ; and Vc = 100 <-900 V.

Solution:

Va0 = 1/3(Va + Vb + Vc) = 0 V

Va1 = 1/3(Va + a Vb + a2Vc) = 0 V

Va2 = 1/3(Va + a2 Vb + a Vc) = 100<300 V Observation: If the phasors are balanced, Two sequence components will be zero.

Three phase power in symmetrical components

S = VpT Ip* = [A Vs]T[A Is]*

= VsT AT A* Is* = 3 Vs

TIs*

= 3Va0 Ia0* + 3Va1 Ia1* + 3Va2 Ia2*note that AT = A