SYMMETRICAL COMPONENTS
In 1918 one of the most powerful tools for
dealing with unbalanced polyphase circuit was
discussed by C.L. Fortescue at a meeting of the
America Institute of Electrical Engineers. Since the
time the method of symmetrical components has
become of great importance and has been the
subject of many articles and experimental
investigations.
Unsymmetrical
faults
on
transmission systems, which may consist of short
circuits, impedance between lines, impedance from
one or two lines to ground, or open conductors, are
studied by the method of symmetrical components.
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
Fortescue’s work proves that an
unbalanced system of n related phasors can be
resolved into n systems of balanced phasors
called the symmetrical components of the
original phasors. The n phasors of each set of
components are equal in length, and the angles
between adjacent phasors of the set are equal.
Although the method is applicable to any
unbalanced polyphase system, we shall confine
our discussion to three-phase systems.
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
It is customary, when solving a problem by
symmetrical components, to designate the three phases
of the system as A, B, and C in such a manner that the
phase sequence of the voltages and currents in the
system is ABC. Thus the phase sequence of the positivesequence components of the unbalanced phasors is ABC,
and the phase sequence of the negative-sequence
components is ACB. If the original phasors are voltages,
they may be designated 𝑉𝑎 , 𝑉𝑏 & 𝑉𝑐 . The three sets of
symmetrical components are designated by the
additional subscript 1 for the positive-sequence
components, 2 for the negative-sequence components,
and 0 for the zero-sequence components.
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
The positive-sequence components of
𝑉𝑎 , 𝑉𝑏 & 𝑉𝑐 𝑎𝑟𝑒 𝑉𝑎1 , 𝑉𝑏1 & 𝑉𝑐1 .
Similarly, the
negative-sequence components 𝑉𝑎2 , 𝑉𝑏2 & 𝑉𝑐2 ,
and the zero-sequence components are
𝑉𝑎0 , 𝑉𝑏0 & 𝑉𝑐0 . Figure 1 shows three such sets of
symmetrical components. Phasor representing
currents will be designated by I with subscripts
as for voltages.
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
Since each of the original unbalanced
phasors is the sum of its components, the
original phasors expressed interms of their
components are
𝑉𝑎 = 𝑉𝑎1 + 𝑉𝑎2 + 𝑉𝑎0
𝑉𝑏 = 𝑉𝑏1 + 𝑉𝑏2 + 𝑉𝑏0
𝑉𝑐 = 𝑉𝑐1 + 𝑉𝑐2 + 𝑉𝑐0
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
Figure 1 Three sets of balanced phasors which are the symmetrical
components of three unbalanced phasors.
SYNTHESIS OF UNSYMMETRICAL PHASORS FROM
THEIR SYMMETRICAL COMPONENTS
•
Figure 2 Graphical addition of the components shown in Fig. 1 to obtain three unbalanced phasors.
OPERATORS
Because of the phase displacement of
symmetrical components of the voltages and
currents in the three phase system, it is convenient
to have a shorthand method of indicating the
rotation of a phasor through 120 degree. The result
of the multiplication of two complex numbers is the
product of their magnitudes and the sum of their
angles. If the complex number expressing a phasor
is multiplied by a complex number of unit
magnitude and angle 𝜃 , the resulting complex
number represents a phasor equal to the original
phasor displaced by the angle 𝜃.
OPERATORS
The letter a is commonly used to
designated the operator that causes a rotation
of 120 degree in the counterclockwise direction.
Such an operator is a complex number of unit
magnitude with an angle 120 degree and is
defined by
𝑎 = 1∠120° = 1𝑒 𝑗2𝜋/3 = −0.5 + 𝑗0.866
OPERATORS
Figure 3 Phasor diagram of the various power of the operator a.
OPERATORS
If the operator a is applied to a phasor twice
is succession, the phasor is rotated through 240
degree. Three successive application of a rotate the
phasor through 360 degree. Thus,
𝑎2 = 1∠240° = −0.5 − 𝑗0.866
and
𝑎3 = 1∠360° = 1∠0° = 1
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
We have seen (Fig. 2) the synthesis of
three unsymmetrical phasors from three sets of
symmetrical phasors. The synthesis was made in
accordance with the following equations:
𝑉𝑎 = 𝑉𝑎1 + 𝑉𝑎2 + 𝑉𝑎0
𝑉𝑏 = 𝑉𝑏1 + 𝑉𝑏2 + 𝑉𝑏0
𝑉𝑐 = 𝑉𝑐1 + 𝑉𝑐2 + 𝑉𝑐0
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Now let us examine these same equations to determine how
to resolve three unsymmetrical phasors into their symmetrical
components.
First, we note that the number of unknown quantities can be
reduced by expressing each component of 𝑉𝑏 and 𝑉𝑐 as the product of
some function of the operator a and a component of 𝑉𝑎 . Reference to
Fig. 11.1 verifies the following relations:
𝑉𝑏1 = 𝑎2 𝑉𝑎1
𝑉𝑏2 = 𝑎𝑉𝑎2
𝑉𝑏0 = 𝑉𝑎0
𝑉𝑐1 = 𝑎𝑉𝑎1
𝑉𝑐2 = 𝑎2 𝑉𝑎2
𝑉𝑐0 = 𝑉𝑎0
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
𝑉𝑎 = 𝑉𝑎1 + 𝑉𝑎2 + 𝑉𝑎0
𝑉𝑏 = 𝑎2 𝑉𝑎1 + 𝑎𝑉𝑎2 + 𝑉𝑎0
𝑉𝑐 = 𝑎𝑉𝑎1 + 𝑎2 𝑉𝑎2 + 𝑉𝑎0
In Matrix form
𝑉𝑎
1 1
𝑉𝑏 = 1 𝑎2
𝑉𝑐
1 𝑎
1
𝑎
𝑎2
𝑉𝑎0
𝑉𝑎1
𝑉𝑎2
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
𝑉𝑎
Let 𝑉𝑝= 𝑉𝑏
𝑉𝑐
1 1 1
𝐴 = 1 𝑎2 𝑎
1 𝑎 𝑎2
𝑉𝑎0
𝑉𝑠 = 𝑉𝑎1
𝑉𝑎2
The matric equation will become
𝑉𝑝 = 𝐴𝑉𝑠
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Inverse of A=A-1
−1
𝐴
=
1
3
1 1
1 𝑎
1 𝑎2
1
𝑎2
𝑎
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Pre-multiplying the inverse in the matric
equation to yields
𝐴−1 𝑉𝑝 = 𝐴−1 𝐴𝑉𝑠
𝑉𝑎0
1 1
𝑉𝑎1 = 1
3
𝑉𝑎2
1
1
𝑎
𝑎2
1
𝑎2
𝑎
𝑉𝑎0
𝑉𝑎1
𝑉𝑎2
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
1
𝑉𝑎0 = 𝑉𝑎 + 𝑉𝑏 + 𝑉𝑐
3
1
𝑉𝑎1 = 𝑉𝑎 + 𝑎𝑉𝑏 + 𝑎2 𝑉𝑐
3
1
𝑉𝑎2 = 𝑉𝑐 + 𝑎2 𝑉𝑏 + 𝑎𝑉𝑐
3
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
The preceding equations could have been written for any set of related
phasors, and we might have written them for currents instead of for voltages. They
may be solved either analytically or graphically. Because some of the preceding
equations are so fundamental, they are summarized for currents:
𝐼𝑎 = 𝐼𝑎1 + 𝐼𝑎2 + 𝐼𝑎0
𝐼𝑏 = 𝑎2 𝐼𝑎1 + 𝑎𝐼𝑎2 + 𝐼𝑎0
𝐼𝑐 = 𝑎𝐼𝑎1 + 𝑎2 𝐼𝑎2 + 𝐼𝑎0
1
𝐼𝑎0 = 3 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐
1
𝐼𝑎1 = 3 𝐼𝑎 + 𝑎𝐼𝑏 + 𝑎2 𝐼𝑐
1
𝐼𝑎2 = 3 𝐼𝑐 + 𝑎2 𝐼𝑏 + 𝑎𝐼𝑐
In a three-phase Y-connected system, the neutral current In is the sum of the line
currents:
𝐼𝑛 = 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Problem:
Calculate the sequence components of the
following balanced line-to-neutral voltages with
ABC sequence:
𝑉𝑎𝑛
277∠0𝑜
𝑉𝑝 = 𝑉𝑏𝑛 = 277∠ −120𝑜 𝑣𝑜𝑙𝑡𝑠
𝑉𝑐𝑛
277∠120𝑜
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Solution:
𝑉0 =
1
3
277∠0𝑜 + 277∠ −120𝑜 +277∠120𝑜 = 0
𝑉1 =
1
3
277∠0𝑜 + 𝑎277∠ −120𝑜 +𝑎2277∠120𝑜
= 277∠0 𝑣𝑜𝑙𝑡𝑠 = 𝑉𝑎𝑛
𝑉2 =
1
3
277∠0𝑜 + 𝑎2277∠ −120𝑜 +𝑎277∠120𝑜 = 0
This example illustrates the fact that balanced three-phase systems
with ABC sequence (or positive sequence) have no zero-sequence or
negative-sequence components. For this example, the positive-sequence
voltage V1 equals Van, and the zero-sequence and negative-sequence voltages
are both zero.
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Problem:
A wye-connected load has balanced currents
with ACB sequence given by
𝐼𝑎
10∠0𝑜
𝐼𝑝 = 𝐼𝑏 = 10∠120𝑜 𝐴𝑚𝑝𝑟𝑒𝑠
𝐼𝑐
10∠ − 120𝑜
Calculate the sequence currents.
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Solution:
1
𝐼0 = 10∠0𝑜 + 10∠120𝑜 + 10∠ − 120𝑜 = 0
3
1
𝐼1 = 10∠0𝑜 + 𝑎10∠120𝑜 + 𝑎210∠ − 120𝑜
3
=0
1
3
10∠0𝑜 + 𝑎210∠120𝑜 + 𝑎10∠ − 120𝑜
= 10∠0 = 𝐼𝑎
This example illustrates the fact that balanced threephase systems with ACB sequence (or negative sequence) have
no zero-sequence or positive-sequence components. For this
example the negative-sequence current I2 equals Ia, and the
zero-sequence and positive-sequence currents are both zero.
𝐼2 =
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Problem:
A three-phase line feeding a balanced-Y load has
one of its phases (phase B) open. The load neutral is
grounded, and the unbalanced line currents are
𝐼𝑎
10∠0𝑜
𝐼𝑝 = 𝐼𝑏 =
𝐴𝑚𝑝𝑟𝑒𝑠
0
𝐼𝑐
10∠120𝑜
Calculate the sequence currents and the neutral
current.
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
THE SYMMETRICAL COMPONENTS OF
UNSYMMETRICAL PHASORS
Solution:
𝐼0 =
1
3
10∠0𝑜 + 0 + 10∠120𝑜 = 3.333∠60
1
𝐼1 = 10∠0𝑜 + 0 + 𝑎210∠120𝑜 = 6.667∠0
3
1
𝐼2 = 10∠0𝑜 + 0 + 𝑎10∠120𝑜 = 3.333∠60
3
𝐼𝑛 = 10∠0𝑜 + 0 + 10∠120𝑜 = 10 ∠60A=3𝐼0
This example illustrates the fact that unbalanced threephase systems may have nonzero values for all sequence
components. Also, the neutral current equals three times
the zero-sequence current.
Definition of Sequence Networks
Definition of Sequence Networks
Definition of Sequence Networks
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Sequence Network Models of Power System Components
Simplified Derivation of Transformer ZeroSequence Circuit Modeling
Simplified Derivation of Transformer ZeroSequence Circuit Modeling
Simplified Derivation of Transformer ZeroSequence Circuit Modeling
Simplified Derivation of Transformer ZeroSequence Circuit Modeling
Simplified Derivation of Transformer ZeroSequence Circuit Modeling
Transformers
(Zero-Sequence Circuit Model)
Transformers
(Zero-Sequence Circuit Model)
Transformers
(Zero-Sequence Circuit Model)
Transformers
(Zero-Sequence Circuit Model)
Transmission Lines
(Positive Sequence Network)
Transmission Lines
(Negative Sequence Network)
The same model as the positive-sequence
network is used for transmission lines in as
much as the positive-sequence and negativesequence impedances of transmission lines are
the same
Transmission Lines
(Zero Sequence Network)
The zero-sequence network model for a
transmission line is the same as that of the
positive- and negative-sequence networks. The
sequence impedance of the model is of course
the zero-sequence impedance of the line. This is
normally higher than the positive- and negativesequence impedances because of the influence
of the earth’s resistivity and the ground wire/s.
Classification of Power System
Short Circuits
Shunt Faults
Single line-to-ground faults
Double line-to-ground faults
Line-to-line faults
Three-phase faults
Classification of Power System
Short Circuits
Combination of Shunt and Series Faults
Single line-to-ground and one-line open
Double line-to-ground and one-line open
faults
Line-to-line and one-line open faults
Three-phase and one-line open faults
Classification of Power System
Short Circuits
Combination of Shunt and Series Faults
Single line-to-ground and two-line open
faults
Double line-to-ground and two-line open
faults
Line-to-line and two-line open faults
Three-phase and two-line open faults
Balanced Faults
(Symmetrical or Three-Phase Faults)
Balanced Faults
(Symmetrical or Three-Phase Faults)
Unbalanced Faults
(Single Line-to-Ground Faults)
Unbalanced Faults
(Single Line-to-Ground Faults)
Unbalanced Faults
(Line-to-Line Faults)
Unbalanced Faults
(Line-to-Line Faults)
Unbalanced Faults
(Line-to-Line Faults)
Unbalanced Faults
(Double-to-line Ground Fault)
Unbalanced Faults
(Double-to-line Ground Fault)
Unbalanced Faults
(Double-to-line Ground Fault)
Unbalanced Faults
(Double-to-line Ground Fault)
Unbalanced Faults
(Double-to-line Ground Fault)
Assumptions Made to Simplify
Fault Calculations
1. Pre-fault load currents are neglected.
2. Pre-fault voltages are assumed equal to 1.0 per
unit.
3. Resistances are neglected (only for 115kV & up).
4. Mutual impedances, when not appreciable are
neglected.
5. Off-nominal transformer taps are equal to 1.0 per
unit.
6. Positive- and negative-sequence impedances are
equal.
Outline of Procedures for
Short Circuit Calculations
1. Setup the network impedances expressed in
per unit on a common MVA base in the form of
a single-line diagram
2. Determine the single equivalent (Thevenin’s)
impedance of each sequence network.
3. Determine the distribution factor giving the
current in the individual branches for unit total
sequence current.
Outline of Procedures for
Short Circuit Calculations
4. Interconnect the three sequence networks for
the type of fault under considerations and
calculate the sequence currents at the fault
point.
5. Determine the sequence current distribution
by the application of the distribution factors to
the sequence currents at the fault point.
6 Synthesize the phase currents from the
sequence currents.
Outline of Procedures for
Short Circuit Calculations
7. Determine the sequence voltages throughout
the networks from the sequence current
distribution and branch impedances.
8. Synthesize the phase voltages from the
sequence voltage components.
9. Convert the pre unit currents and voltages to
actual physical units.
FAULT CALCULATIONS
Problem:
A 50 MVA, 11 kV, synchronous generator has a substransient
reactance of 20 %. The generator supplies two motors over a
transmission line with transformers at both ends as shown in
the figure. The motors have rated inputs of 30 and 15 MVA,
both 10 kV, with 25% subtransient reactance. The three-phase
transformers are both rated 60 MVA, 10.6/121 kV, with
leakage reactance of 10% each. Assume zero-sequence
reactances for the generator and motors of 6 % each. Current
limiting reactors of 2.5 ohms each are connected in the
neutral of the generator and motor No. 2. The zero sequence
reactance of the transmission line is 300 ohms. The series
reactance of the lines is 100 ohms. Draw the positive, negative
and zero sequence networks.
FAULT CALCULATIONS
Problem:
Assume that the negative sequence reactance of
each machine is equal to its subtransient reactance
FAULT CALCULATIONS
Solution:
Assume base power=50 MVA, Base voltage 11 kV
FAULT CALCULATIONS
FAULT CALCULATIONS
Positive sequence network
FAULT CALCULATIONS
Negative sequence network
FAULT CALCULATIONS
Zero sequence network
FAULT CALCULATIONS
Assignment: (To be passed next meeting)
1.The phase voltages across a certain load are
given as:
𝑉𝑎 = 176 − 𝑗132 𝑣𝑜𝑙𝑡𝑠
𝑉𝑏 = −128 − 𝑗96 𝑣𝑜𝑙𝑡𝑠
𝑉𝑐 = −160 + 𝑗100 𝑣𝑜𝑙𝑡𝑠
Compute positive, negative and zero sequence
component of voltage.
FAULT CALCULATIONS
Assignment: (To be passed next meeting)
2. A balance delta-connected load is connected
to a three-phase system and supplied to it’s a
current 0f 15 ∠0 amperes. If the fuse in one of
the lines melts, compute the symmetrical
components of the line currents.
FAULT CALCULATIONS
Assignment: (To be passed next meeting)
FAULT CALCULATIONS
Assignment: (To be passed next meeting)
3. Draw the positive, negative and zero
sequence networks of the power system shown
below.
FAULT CALCULATIONS
Assignment: (To be passed next meeting)
4. Figure shown below is a sample power system
network. Draw the zero sequence network. Data
is given below.
FAULT CALCULATIONS
Assignment: (To be passed next meeting)