This document discusses unsymmetrical faults in power systems. It defines unsymmetrical faults as faults that result in unequal line currents and phase displacements. There are three types of unsymmetrical faults: single line-to-ground faults, line-to-line faults, and double line-to-ground faults. Single line-to-ground faults are the most common, accounting for 75-80% of faults, occurring when a conductor contacts ground. Line-to-line faults occur when conductors contact each other. Double line-to-ground faults involve two lines contacting each other and ground. Symmetrical component analysis is used to analyze unsymmetrical faults by resolving currents into positive, negative, and zero sequence components.

1. Switchgear and Protection
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Lecture- 3
Unsymmetrical Faults in
Power System
EEE436

2. • (ii) Unsymmetrical faults
• Those faults which give rise to unsymmetrical currents (i.e. unequal line
currents with unequal displacement) are called unsymmetrical faults.
• On the occurrence of an unsymmetrical fault, the currents in the three
lines become unequal and so is the phase displacement among them.
• The unsymmetrical faults may take one of the following forms:
(a) Single line-to-ground fault
(b) Line-to-line fault
(c) Double line-to-ground fault
Unsymmetrical Faults in Power System
Navila Rahman Nadi
EEE436

3. • A short circuit between any of the conductors and
earth is called a Single Line-to Ground/ Phase to
ground fault.
• Generally, a single line-to-ground fault on a
transmission line occurs when one conductor
drops to the ground or comes in contact with
the neutral conductor.
• Other reason is-due to the failure of the
insulation between a phase conductor and the
earth.
• Such types of failures may occur in power system
due to many reasons like high-speed wind,
falling off a tree, lightning, etc.
• A single line-to-ground (LG) fault is one of the
most common faults and experiences show that
75-80% of the faults that occur in power system
are of this type.
Single Line-to-Ground Faults
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4. • A line-to-line fault occurs when a live conductor gets in
contact with another live conductor.
• Heavy winds are the major cause for this fault during
which swinging of overhead conductors may touch together.
• These are less severe faults and its occurrence range varies
between 15-20%
Line-to-line Faults
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EEE436

5. • In double line to ground faults, two lines come into
the contact with each other as well as with ground.
• These are severe faults, and the occurrence of these
faults is about 10% when compared with total system
faults.
• This can be a result of a tree falling on two of the
power lines
Double Line-to-Ground Faults
Navila Rahman Nadi
EEE436

6. q Symmetrical components method
• In this method, any unbalanced system of 3-phase currents (or voltages)
are regarded as being composed of three separate sets of balanced
vectors*. These vectors are called as three symmetrical
components:
• Positive Sequence:
A balanced system of 3-phase currents having positive† (or normal) phase as
the original sequence.
• Negative sequence:
A balanced system with the opposite phase sequence as the original
sequence.
• Zero Sequence:
Three phasors that are equal in magnitude and phase.
*Here, A balanced system of 3-phase currents implies that three currents
are equal in magnitude having 120º displacement from each other.
Unsymmetrical Fault Calculation
Navila Rahman Nadi
EEE436

7. Suppose an unsymmetrical fault occurs on
a 3-phase system having phase sequence
RYB.
According to symmetrical components theory,
the resulting unbalanced currents IR , IY and IB
can be resolved into :
(i) A balanced system of 3-phase currents, IR1 ,
IY1 and IB1 having positive phase sequence (i.e.
RYB) as original sequence-shown in below
figure. These are the positive phase sequence
components.
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Unsymmetrical Fault Calculation Illustration
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Fig 3.5 Original sequence
Fig 3.6: Positive Phase Sequence Components

8. ii) A balanced system of 3-phase currents
IR2 , IY2 and IB2 having negative phase
sequence (i.e. RBY) as shown in Figure
below. These are the negative phase sequence
components.
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Unsymmetrical Fault Calculation
EEE436
Fig 3.7: Negative Phase Sequence Components

9. iii) A system of three currents equal in
magnitude and having zero phase
displacement. These are called zero phase
sequence components.
Unsymmetrical Fault Calculation
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Fig 3.8: Zero Phase Sequence Components
Fig 3.5 Original sequence

10. Navila Rahman Nadi
Unsymmetrical Fault Calculation Sequences
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Fig 3.5 Original sequence

11. The current in any phase is equal to the vector sum of
positive, negative and zero phase sequence currents in that
phase.
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Unsymmetrical Fault Calculation
EEE436

12. • As the symmetrical component theory involves the
concept of 120o displacement in the positive
sequence set and negative sequence set, therefore, it is
desirable to evolve some operator which should cause
120o rotation.
• For this purpose, operator ‘a’ (symbols h or λ are
sometimes used instead of ‘a’) is used.
• The operator ‘a’ is one, which when multiplied to a
vector, it rotates the vector through 120o in the
anticlockwise direction.
Operator “a”
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13. Navila Rahman Nadi
Symmetrical Components in terms of operator “a”
…..(v)
…..(vi)
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14. • Consider a vector I represented by OA
as shown in the figure.
• If this vector is multiplied by operator
‘a’, the vector is rotated through 120o
in the anticlockwise direction and
assumes the position OB.
• ∴ a I = I ∠120o
= I (cos 120o+ j sin 120o)
= I (− 0·5 + j 0·866)
∴ a = − 0·5 + j 0·866 ... (i)
Operator “a”
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EEE436

15. • If the vector assuming position OB is
multiplied by operator ‘a’, the vector is further
rotated through 120o in the anticlockwise
direction and assumes the position OC.
∴ a2I = I ∠2400
= I (cos 2400 + j sin 2400)
= I (− 0·5 − j 0·866)
∴ a2 = − 0·5 − j 0·866 ... (ii)
• This is the same as turning the vector through
120o in clockwise direction.
∴ a2 I = I ∠− 120o
Similarly, a3I = I ∠3600
= I (cos 3600 + j sin 3600)
∴ a3 = 1 ... (iii)
Operator “a”
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EEE436

16. ∴ a = − 0·5 + j 0·866 ... (i)
∴ a2 = − 0·5 − j 0·866 ... (ii)
∴ a3 = 1 ... (iii)
(i) Adding exp. (i) and (ii), we get,
a + a2 = (− 0·5 + j 0·866) + (− 0·5 − j0.866)
= −1
∴ 1 + a + a2 = 0
Properties of Operator “a”
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(ii) Subtracting exp. (ii) from exp. (i), we get,
a − a2 = (− 0·5 + j 0·866) − (− 0·5 − j 0·866)
= j 1·732
∴ a − a2 = j 3

17. Navila Rahman Nadi
Symmetrical Components in terms of operator “a”
…..(iv)
…..(v)
…..(vi)
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18. Symmetrical Components in terms of Phase currents
…..(iv)
…..(v)
…..(vi)
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19. Symmetrical Components in terms of Phase Currents
…..(iv)
…..(v)
…..(vi)
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20. Symmetrical Components in terms of Phase Currents
…..(iv)
…..(v)
…..(vi)
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21. (i) The currents I1 , I2 and I0 are the symmetrical components of R-phase.
Because of the symmetry of each set, the symmetrical components of
yellow and blue phases can be easily known.
(i) Although the treatment has been made considering currents, the method
applies equally to voltages. Thus the symmetrical voltage components
of R-phase in terms of phase voltages shall be :
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Symmetrical Components in terms of Phase Currents
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22. • Prove that:
(1)
!"#!
#"#! = −𝑎 2
!"#
!$#! = 1 − 𝑎%
Assignments
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23. • (i) A balanced 3-phase system consists of positive sequence
components only; the negative and zero sequence components
being zero.
• (ii) The presence of negative or zero sequence currents in a 3-
phase system introduces asymmetry and is indicative of an
abnormal condition of the circuit in which these components
are found.
• (iii) The vector sum of the positive and negative sequence
currents of an unbalanced 3-phase system is zero.
• The resultant solely consists of three zero sequence currents
i.e.
Some Important Facts
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24. Example 01
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In a 3-phase system, the currents in R, Y and B lines; under abnormal conditions of
loading are as under :
Calculate the positive, negative and zero sequence currents in the R-line
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25. Example 02
The currents in a 3-phase unbalanced system are :
Calculate the zero, positive and negative sequence currents of the RYB phase.
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26. Example 02
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27. Example 03

31. Example 04