SYMMETRICAL COMPONENTS

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Chapter 3 SYMMETRICAL COMPONENTS

• Introduction
• Fortescues Theorem
• The a Operator
• Matrix Relations
• Power in terms of Sym Components
• Examples

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Introduction

• SLD, Reactance Diagrams assumes Symmetry
• Faults can cause unbalanced conditions
• Methods to address unsymmetrical systems
• Symmetrical components method

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• Fortescues Theorem
• One set of n related phasors can be transformed
  into n set of bal. phasors
• The n phasors of each set are balanced
• Applicable to any unbalanced polyphase system
• For a 3-ph system 3 unbalanced phasors can be
  resolved into 3 balanced systems of 3 phasors
  each

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• For 3-ph System, the balanced sets
• of component are called as
• Positive-sequence components Equal in
  magnitude 1200 phase displacement same sequence
  as original phasors
• Negative-sequence components Equal in
  magnitude 1200 phase displacement opposite
  sequence to original phasors
• Zero-sequence components
• Equal in magnitude 00 phase displacement

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• Let Va, Vb, Vc be the Phase voltages
• According to Fortescue, these can be transformed
  into
• Positive-seq. voltages Va1, Vb1, Vc1
• Negative-seq. voltages Va2, Vb2, Vc2
• zero-sequence voltages Va0, Vb0, Vc0
• Thus, Va Va1 Va2 Va0
• Vb Vb1 Vb2 Vb0
• Vc Vc1 Vc2 Vc0

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Vc1
Va1
Va2
Va0
Va1
Vb1
Vc
Va
Vb
Vb2
Va2
Vc2
Va0 Vb0 Vc0
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The a operator a 1lt1200 -0.5 j 0.866
a I rotates I by 1200 a2 1lt2400 -0.5 j
0.866 a3 1lt3600 1lt00 1 j 0 1 a
a2 0
a
-a2
1
-1
-a
a2
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Matrix Relations Vb1 a2Va1 Vc1 a Va1 Vb2
a Va2 Vc2 a2 Va2 Vb0 Va0 Vc0
Va0 Thus, Va Va0 Va1 Va2 Vb
Va0 a2Va1 a Va2 Vc Va0 a Va1
a2Va2
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Matrix Relations Let And Inverse of A is

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Matrix Relations Similarly currents or
any other variables can be obtained using their
symmetrical components
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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)
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Matrix Relations
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• Numerical Example
• The line currents in a 3-ph 4 wire system are Ia
  100lt300 Ib 50lt3000 Ic 30lt1800. Find the
  symmetrical components and the neutral current.
• Solution
• Ia0 1/3(Ia Ib Ic) 27.29 lt 4.70 A
• Ia1 1/3(Ia a Ib a2Ic) 57.98 lt 43.30 A
• Ia2 1/3(Ia a2 Ib a Ic) 18.96 lt 24.90 A
• In Ia Ib Ic 3 Ia0 81.87 lt4.70 A

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Numerical Example 2. The sequence component
voltages of phase voltages of a 3-ph system are
Va0 100 lt00 V Va1 223.6 lt -26.60 V Va2
100 lt1800 V. Determine the phase voltages.
Solution Va Va0 Va1 Va2 223.6
lt-26.60 V Vb Va0 a2Va1 a Va2 213 lt
-99.90 V Vc Va0 a Va1 a2 Va2 338.6 lt
66.20 V
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Numerical Example 3. The two seq. components and
the corresponding phase voltage of a 3-ph system
are Va0 1lt-600 V Va12lt00 V Va 3 lt00
V. Determine the other phase voltages. Solution
Va Va0 Va1 Va2 ? Va2 Va Va0
Va1 1 lt600 V Vb Va0 a2Va1 a Va2 3
lt -1200 V Vc Va0 a Va1 a2 Va2 0 V

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Numerical Example 4. Determine the sequence
components if Ia 10lt600 A Ib 10lt-600 A
Ic 10 lt1800 A. Solution Ia0 1/3 (Ia
Ib Ic) 0 A Ia1 1/3 (Ia a Ib
a2Ic) 10lt600 A Ia2 1/3 (Ia a2 Ib a
Ic) 0 A Thus, If the phasors are balanced,
Two Sequence components will be zero.
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Ia
Ia
Ib
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Numerical Example 5. Determine the sequence
components if Va 100 lt300 V Vb 100
lt1500 V and Vc 100 lt-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) 100lt300 V
Observation If the phasors are balanced, Two
sequence components will be zero.
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Numerical Example 6. The line b of a 3-ph line
feeding a balanced Y-load with neutral grounded
is open resulting in line currents Ia 10lt00 A
Ic 10lt1200 A. Determine the sequence current
components. Solution Ib 0 A.
Ia0 1/3(Ia Ib Ic) 3.33lt600 A Ia1
1/3(Ia a Ib a2Ic) 6.66lt00 A Ia2
1/3(Ia a2 Ib a Ic) 3.33lt-600 A
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Numerical Example 7. One conductor of a 3-ph line
feeding a balanced delta-load is open. Assuming
that line c is open, if current in line a is
10lt00 A , determine the sequence components of
the line currents. Solution Ic 0 A
Ia 10lt00 A. ? Ib 10lt1200 A Ia0
1/3(Ia Ib Ic) 0 A Ia1 1/3(Ia a Ib
a2Ic) 5.78lt-300 A Ia2 1/3(Ia a2
Ib a Ic) 5.78lt 300 A Note The zero-seq.
components of line currents of a delta
load (3-ph 3-wire) system are zero.
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Power Calculation S P jQ Va Ia Vb
Ib Vc Ic VTI where, VT Va Vb
Vc IT Ia Ib Ic
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Using Sym. Comps. for voltages and currents, S
VpT Ip A VsTA Is VsT AT
A Is 3 VsTIs 3Va0 Ia0 3Va1 Ia1
3Va2 Ia2 3-phase power sum of sym.
Components power. Thus the power in the
unsymmetrical 3-ph system is the sum of powers in
the equivalent symmetrical components. Note S
Va0 Ia0 Va1 Ia1 Va2 Ia2 if V and I
are in per unit values
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Numerical Example 8. The sequence voltages and
the corresponding sequence currents in per unit
of a 3-phase system on a 80MVA base are Va11
Va2(0.4j).3) Va0(0.1-j0.1)) and
Ia1(0.8j0.6) Ia1(0.2-j0.1) Ia0(0.05j0.08)
respectively. Find the 3-phase MVA, MW, and MVAR
of the system. Solution S 3Va0 Ia0 3Va1
Ia1 3Va2 Ia2 if values are in volts and amps
S Va0 Ia0 Va1 Ia1 Va2 Ia2 if
values are in per units Hence S
(1.15lt33.970)pu ( 76.3j51) MVA 92lt33.970
MVA P 76.3 MW and Q 51 MVAR.
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• Thank you
• Dr G.K. Purushothama
• e-mail gkuttama_at_rediffmail.com
• gkuttama_at_gmail.com
• Phone [phone]

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