Download For a 3-ph system

Lecture 1
Balanced Three-phase Systems
From Network to Single-phase Equivalent Circuit
Power plant Transformer
Switching
station
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 into
1. 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
Vc1
Va1
Va1
Vb1
Vc
Va2
Vc2
Va0 Vb0 Vc0
Va0
Va
Vb
Vb2
Va2
The ‘a’ operator
a = 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
a
-a2
1
-1
a2
-a
From figure previous figures
Vb1 = a2Va1
Vb2 = a Va2
Vb0 = Va0
Vc1 = a Va1
Vc2 = a2 Va2
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 Relations
Let
 va 
Vp =  vb  ;
 vc 
 va0 
Vs =  va1  ;
 va2 
And Inverse of A is
1 1
A = 1 a2
1 a
1
a 
a2 
1 1 1 
A-1 = 31 1 a a2 
1 a2 a 
Matrix Relations
va 
1 1
v   1 a 2
 b

vc 
1 a
1

a
2
a 
va 0 
v 
 a1 
va 2 
Similarly currents can be obtained using their
symmetrical components
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
va 0 
v  
 a1 
va 2 
1 1
1
a
3 1
2
1 a
1
2
a 
a 
va 
v 
 b
vc 
Numerical Example
1.
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
Ia1
Ia2
In
= 1/3(Ia + Ib + Ic)
= 27.29 < 4.70 A
= 1/3(Ia + a Ib + a2Ic) = 57.98 < 43.30 A
= 1/3(Ia + a2 Ib + a Ic) = 18.96 < 24.90 A
= Ia + Ib + Ic = 3 Ia0 = 81.87 <4.70 A
Numerical Example
2. 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
3.
Numerical Example
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 Example
4. Determine the sequence components if
Ia =10<600 A; Ib = 10<-600 A ; & Ic = 10 <1800 A.
Solution:
Ia0 = 1/3 (Ia + Ib + Ic)
=0A
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
Ia
Ib
Numerical Example
5. 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)
=0V
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 VsTIs*
= 3Va0 Ia0* + 3Va1 Ia1* + 3Va2 Ia2*
note that AT = A