This document discusses symmetrical faults in power systems. It begins by defining a symmetrical or three-phase fault as one where all three conductors are shorted simultaneously, resulting in equal fault currents with 120 degree displacement. Methods for calculating symmetrical fault current are presented, including using percentage reactance and a common base kVA. Several examples demonstrate calculating fault current and kVA at different points in sample systems. The importance of determining fault levels is discussed to select properly rated protective devices and switchgear.

1. Lecture 2
Faults in Power
System (Symmetrical
Faults)
EEE436 Switchgear and Protection

2. EEE436 Short Circuit
• The most common type of fault
occurs in power system is short
circuit fault.
• Whenever a fault occurs on a
network such that a large current
flows in one or more phases, a short
circuit is said to have occurred.
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3. EEE436 Causes of Short Circuit
(i) Internal effects :
• Breakdown of equipment or transmission lines
• From deterioration of insulation in a generator,
transformer etc. (ageing of insulation, inadequate
design or improper installation)
(ii) External effects:
• Insulation failure due to lightning surges
• Overloading of equipment causing excessive heating
• Mechanical damage by public etc.
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4. EEE436 Effects of Short Circuit
(i) The heavy current due to short-circuit causes
excessive heating which may result in fire or
explosion.
For example: a spark on a transmission line not cleared
quickly will burn the conductor severely causing it to
break, resulting in a long-time interruption of the line.
(ii) The low voltage created by the fault has a very
harmful effect on the service rendered by the power
system.
For example: If the voltage remains low for even a few
seconds, the consumers’ motors may be shut down and
generators on the power system may become unstable.
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5. 3-phase system can be classified into two main categories:
(i) Symmetrical faults
(ii) Unsymmetrical faults
(i) Symmetrical faults
That fault which gives rise to symmetrical (same
amount of) fault currents (i.e. equal fault currents
with 120o displacement) is called a symmetrical fault.
The most common example of symmetrical fault is
when all the three conductors of a 3-phase line are
brought together simultaneously into a short-circuit
condition.
Faults in a power system
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6. • Referring to the Figure fault currents
IR, IY and IB will be equal in
magnitude with 120-degree
displacement among them.
• Because of balanced nature of fault,
only one phase(line-to-neutral basis)
need be considered in calculations
since condition in the other two phases
will also be similar.
Symmetrical Faults on 3-Phase
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Figure 2.1: Schematic
diagram of Symmetrical
Fault occurrence

7. (ii) Unsymmetrical faults
Those faults which give rise to unsymmetrical currents (i.e. unequal line
currents with unequal displacement) are called unsymmetrical faults. 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
Faults in a power system
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Figure 2.2: Schematic diagram of UnSymmetrical Fault occurrence

8. EEE436 Importance of Short Circuit Current Calculation (symmetrical fault current)
(i) A short-circuit on the power system is cleared by a circuit breaker or
a fuse. It is necessary, therefore, to know the maximum possible
values of short-circuit current so that switchgear of suitable
rating may be installed to interrupt them.
(ii) The magnitude of short-circuit current determines the setting and
sometimes the types and location of protective system.
(iii) The calculation of short-circuit currents enables us to make proper
selection of the associated apparatus (e.g. bus-bars, current
transformers etc.) so that they can withstand the forces that arise
due to the occurrence of short circuits.
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9. Calculation of Symmetrical Fault Current
The reactance of generators, transformers, reactors
etc. is usually expressed in percentage reactance
(instead of ohmic reactance) to permit rapid short
circuit calculations
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G=2500 KVA, 10% T=500 KVA, 5%
Figure 2.3: Schematic diagram of Symmetrical calculation

10. • It is the percentage of the total phase voltage drop
when full load current flows through it.
Percentage Reactance
….Equation (1)
……Equation (3)
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11. Percentage Reactance
……Equation (3)
….Equation (2)
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Here, KV= line voltage not phase voltage

12. • Generally, the various equipments used in the power
system have different kVA ratings.
• Therefore, it is necessary to find the percentage
reactance of all the elements on a common kVA
rating. This common kVA rating is known as base
kVA. The value of this base kVA can be considered as
any number as :
(i) equal to that of the largest plant
(ii) equal to the total plant capacity
(iii) any arbitrary value
Base KVA
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13. When a base KVA has been considered, the % reactance
can be found by using the following relation
% reactance at base kVA =
Base kVA
Rated kVA
× % reactance at
rated kVA
Percentage Reactance & Base KVA
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G=2500 KVA, 10% T=500 KVA, 5%
For example,
The Base KVA = 2500KVA
So, the Percentage Reactance of G=10%
the Percentage Reactance of T= 25%
[
5𝑋2500
500
= 25%] Figure 2.5: Symmetrical fault calculation Method

14. Line Current Calculation from Base KVA
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Based on this the line current can be calculated as:
=
𝑩𝒂𝒔𝒆 𝑲𝑽𝑨
𝟑𝑽𝑳−𝑳,
[In 3 Phase System,
Real Power, P= 3𝑉𝑙𝐼𝑙𝑐𝑜𝑠∅
Apparent Power, S= 3𝑉𝑙𝐼𝑙
As, 𝑐𝑜𝑠∅=
𝑃
𝑆
So,
𝐼𝑙 =
𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
3𝑉𝑙
]

15. If X is the only reactance element in the circuit, then short-
circuit current is given by :
𝐼𝑠𝑐 =
𝑉
𝑋
=
𝑉
%𝑋𝑉
100𝐼
= I ×
100
%𝑋
{ Replacing the value of X =
%𝑋 𝑉
100𝐼
in the above equation}
Example:
If the percentage reactance of an element is 20% and the full-
load current of the circuitry is 50 A, then short-circuit current
will be 50 × 100/20 = 250 A when only that element is in the
circuit.
Short Circuit Current Calculation
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16. We know, % reactance at base
kVA =
Base kVA
Rated kVA
× % reactance
at rated kVA
We know, full load current
I=
𝐵𝑎𝑠𝑒 𝐾𝑉𝐴
3𝑉𝐿−𝐿,
Symmetric Fault Calculation
We know, Short Circuit current
𝐼𝑠𝑐 = I ×
100
%𝑋
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17. From the above illustration, it is clear that whatever may be the value
of base kVA, short-circuit current is the same.
Does Different Base KVA values differs the value of short circuit current?
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18. Reactor Control of Short-Circuit Current(?)
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Find Your Answer from the Book
HAPPY HOUR!

19. Examples of Symmetric Fault Calculation
Example 01:
The below figure shows the single line diagram of a 3-phase system. The percentage
reactance of each alternator is given. Find the short-circuit current that will occur at
the point of F.
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REACTANCE DIAGRAM IS A MUST
Considering base KVA as 35,000KVA

20. Navila Rahman Nadi

21. • The product of normal system voltage and short-circuit current at the
point of fault expressed in kVA is known as short-circuit kVA
• Let V = normal phase voltage in volts
I = full-load current in amperes at base kVA
%X = percentage reactance of the system on base kVA up to
the fault point
• So, Short-circuit kVA for 3-phase circuit:
Short Circuit KVA
As, 𝐼𝑆𝐶 = I ×
100
%𝑋
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22. Examples of Symmetric Fault Calculation
Example 02:
An 11 kV generating station has four identical 3-phase alternators A,
B, C and D, each of 10 MVA capacity and 12% reactance. There are
two sections of bus-bar, P and Q linked by a reactor rated at 10 MVA
with 24% reactance. Alternators A and B are connected to bus-bar
section P and alternators C and D to bus-bar section Q. From each
section, load is taken through a number of 6 MVA, 11 kV/66 kV step-
up transformers, each having a reactance of 3%. Find the short circuit
current at the transformer end.
.
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23. Navila Rahman Nadi
Let 10 MVA is the base
MVA,

24. Example 03:
3-phase transmission line operating at 10 kV and having a resistance of 1Ω and
reactance of 4 Ω is connected to the generating station bus-bars through 5 MVA
step-up transformer having a reactance of 5%. The bus-bars are supplied by a 10
MVA alternator having 10% reactance.
Calculate the short-circuit kVA fed to symmetrical fault between phases if it
occurs
(i) at the load end of transmission line
(ii) at the high voltage terminals of the transformer
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Examples of Symmetric Fault Calculation

25. Let 10,000 KVA is the base KVA
The line impedance is given in ohms. So by using equation

27. Example 4:
The plant capacity of a 3-phase generating station consists of two 10,000 kVA
generators of reactance 12% each and one 5000 kVA generator of reactance 18%. The
generators are connected to the station bus-bars from which load is taken through three
5000 kVA step-up transformers each having a reactance of 5%. Determine the maximum
fault MVA which the circuit breakers on (i) low voltage side and (ii) high voltage side
may have to deal with.
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Examples of Symmetric Fault Calculation

28. Examples of Symmetric Fault Calculation
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29. EEE436
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Examples of Symmetric Fault Calculation
Short circuit current=
𝐹𝑎𝑢𝑙𝑡 𝐾𝑉𝐴
3×𝑙𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
Short circuit current= 𝐼𝑙 ×
100
%𝑋

30. Example 5:
A 3-phase, 20 MVA, alternator is connected 10 kV line and has internal
reactance of 5% and negligible resistance. Find the External reactance
per phase to be connected in series with the alternator so that steady
current on short-circuit does not exceed 8 times the full load current.
Examples of Symmetric Fault Calculation
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Single Line Diagram
As, Phase voltage× 3 = 𝐿𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 ,
Short Circuit current
𝐼𝑠𝑐 = I ×
100
%𝑋

31. Example 5:
A 3-phase, 20 MVA, alternator is connected 10 kV line and has internal
reactance of 5% and negligible resistance. Find the External reactance
per phase to be connected in series with the alternator so that steady
current on short-circuit does not exceed 8 times the full load current.
Examples of Symmetric Fault Calculation
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Single Line Diagram

32. EEE436
Assignment 1:
A small generating station has two alternators of 3000 kVA and 4500 kVA and
percentage reactance's of 7% and 8% respectively. The feeders are provided with the
circuit breakers which have a rupturing capacity of 150 MVA. There is an extended
system added to the bus bar by a supply from the grid via a transformer of 7500 kVA
which has 7·5% reactance.
Find the reactance (ohmic) of the reactor connected in the bus-bar section after the
two alternators to prevent the circuit breakers being overloaded, if a symmetrical
short-circuit occurs on an outgoing feeder connected right after the B alternator.
Assume the bus voltage = 3300 V
Examples of Symmetric Fault Calculation
Short Circuit KVA= 150MVA

33. Assignment 2:
The bus-bars of a power station are in two sections P and Q separated by a
reactor. Connected in section P are two 15 MVA generators of 12 % reactance each and to
Q one 8 MVA generator of 10% reactance. The reactor is rated at 10 MVA and 15%
reactance. Feeders are connected to bus-bar P through transformers, each rated at 5 MVA
and 4% reactance. Determine the maximum short-circuit MVA with which circuit breakers
on the outgoing side of the transformers have to deal.
Examples of Symmetric Fault Calculation
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[Wrong reactance diagram with
calculation has been done in the
book]

34. References & Helpful Site
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https://www.electrical4u.com/
V. K. Mehta, “Principles of power system”, Chapter 17.
Practice the examples (except 17.8, 17.9 and 17.10)