lecture 1. balanced three-phase systems from network to single-phase equivalent circuit power plant...
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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
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)
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.
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.