2. Signal Processing, CE00039-2
Contents
• Introduction
• Symmetrical Faults on 3-Phase System
• Percentage Reactance and Base KVA
• Steps for Symmetrical Fault Calculations
• Unsymmetrical Faults on 3-Phase system
• Significance of Positive, Negative and Zero sequence
components
• Average three phase power in terms of symmetrical
components
• Sequence impedances
• Single Line-to-Ground Fault
• Line-to-Line Faults
• Double Line-to-Ground Faults
3. Signal Processing, CE00039-2
Introduction
Short-Circuit
“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”.
Causes of short-circuit
(i) Internal effects are caused by breakdown of
equipment or transmission lines, from deterioration
of insulation in a generator, transformer etc. Such
troubles may be due to ageing of insulation,
inadequate design or improper installation.
(ii) External effects causing short circuit include
insulation failure due to lightning surges,
overloading of equipment causing excessive
heating; mechanical damage by public etc.
4. Signal Processing, CE00039-2
Effects of short-circuit
• When a short-circuit occurs, the current in the system increases to
an abnormally high value while the system voltage decreases to a
low value.
• Short-circuit causes excessive heating which may result in fire or
explosion. Sometimes short-circuit takes the form of an arc and
causes considerable damage to the system
• Low voltage created by the fault has a very harmful effect on the
service rendered by the power system. 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.
5. Signal Processing, CE00039-2
Short-Circuit Current Calculations
The calculations of the short-circuit currents are important for the
following reasons:
(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 magnitude of short-circuit current determines the size of the
protective reactors which must be inserted in the system so that the
circuit breaker is able to withstand the fault current.
(iv) 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.
6. Signal Processing, CE00039-2
Faults in a Power System
• Symmetrical faults : That fault which gives rise to symmetrical fault
currents (i.e. equal faults currents with 120o displacement) is called
a symmetrical fault.
Example: when all the three conductors of a 3-phase line are
brought together simultaneously into a short-circuit condition.
• Unsymmetrical faults: Those faults which give rise to
unsymmetrical currents (i.e. unequal line currents with unequal
displacement) are called unsymmetrical faults.
• Single line-to-ground fault
• Line-to-line fault
• Double line-to-ground fault
Common: a short-circuit from one line to ground
7. Signal Processing, CE00039-2
Symmetrical Faults on 3-Phase System
“That fault on the power system
which gives rise to symmetrical
fault currents (i.e. equal fault
currents in the lines with 120o
displacement) is called a
symmetrical fault”
• symmetrical fault rarely occurs
in practice
• The symmetrical fault is the
most severe and imposes
more heavy duty on the circuit
breaker
8. Signal Processing, CE00039-2
Limitation of Fault Current
• When a short circuit occurs at
any point in a system, the
short-circuit current is limited
by the impedance of the
system up to the point of fault
• The knowledge of the
impedances of various
equipment and circuits in the
line of the system is very
important for the determination
of short-circuit currents
10. Signal Processing, CE00039-2
Conclusion
• For instance, if the percentage reactance of an element is 20% and
the full-load current is 50 A, then short-circuit current will be 50 ×
100/20 = 250 A when only that element is in the circuit.
• It may be worthwhile to mention here the advantage of using
percentage reactance instead of ohmic reactance in short-circuit
calculations. Percentage reactance values remain unchanged as
they are referred through transformers, unlike ohmic reactances
which become multiplied or divided by the square of transformation
ratio. This makes the procedure simple and permits quick
calculations.
16. Signal Processing, CE00039-2
Reactor Control of Short-Circuit Currents
• With the fast expanding power system, the fault level (i.e. the power
available to flow into a fault) is also rising
• The circuit breakers connected in the power system must be
capable of dealing with maximum possible short-circuit currents that
can occur at their points of connection
• Generally, the reactance of the system under fault conditions is low
and fault currents may rise to a dangerously high value
• If no steps are taken to limit the value of these short-circuit currents,
not only will the duty required of circuit breakers be excessively
heavy, but also damage to lines and other equipment will almost
certainly occur
• Solution: additional reactance's known as reactors are connected in
series with the system at suitable points
• A reactor is a coil of number of turns designed to have a large
inductance as compared to its ohmic resistance
19. Signal Processing, CE00039-2
Example 1
• Fig. shows the single line diagram of a
3-phase system. The percentage
reactance of each alternator is based on
its own capacity. Find the short-circuit
current that will flow into a complete 3-
phase short-circuit at F
21. Signal Processing, CE00039-2
Example 2
• A 3-phase, 20 MVA, 10 kV alternator 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.
33. Signal Processing, CE00039-2
Example 1
In a 3-phase, 4-wire system, the currents in R, Y and B
lines under abnormal conditions of loading are as under:
IR = 100 ∠30º A ; IY = 50 ∠300º A ; IB = 30 ∠180º A
Calculate the positive, negative and zero sequence
currents in the R-line and return current in the neutral
wire.
35. Signal Processing, CE00039-2
Example 2
The currents in a 3-phase unbalanced system are :
IR = (12 + j 6) A ; IY = (12 − j 12) A ; IB = (−15 + j 10) A
The phase sequence is RYB. Calculate the zero, positive
and negative sequence components of the currents.
38. Signal Processing, CE00039-2
Example 3
The sequence voltages in the red phase are as under :
ER0 = 100 V ; ER1 = (200 − j 100) V ; ER2 = − 100 V
Find the phase voltages ER ,EY and EB .
40. Signal Processing, CE00039-2
Example 4
A balanced star connected load takes 90 A from a
balanced 3-phase, 4-wire supply. If the fuses in the Y
and B phases are removed, find the symmetrical
components of the line currents
(i) before the fuses are removed (ii) after the fuses are
removed
48. Signal Processing, CE00039-2
Sequence Impedances
Each element of power system will offer impedance to
different phase sequence components of current which
may not be the same. Therefore, in unsymmetrical fault
calculations, each piece of equipment will have three
values of impedance—one corresponding to each
sequence current viz.
(i) Positive sequence impedance (Z1)
(ii) Negative sequence impedance (Z2)
(iii) Zero sequence impedance (Z0)
50. Signal Processing, CE00039-2
Analysis of Unsymmetrical Faults
In the analysis of unsymmetrical faults, the following
assumptions will be made :
(i) The generated e.m.f. system is of positive sequence
only.
(ii) No current flows in the network other than due to
fault i.e. load currents are neglected.
(iii) The impedance of the fault is zero.
(iv) Phase R shall be taken as the reference phase.
In each case of unsymmetrical fault, e.m.fs’ per phase
are denoted by ER, EY and EB and the terminal p.d. per
phase by VR, VY and VB.
62. Signal Processing, CE00039-2
Observations on Faults
• Positive sequence currents are present in all type of
faults
• Negative sequence currents are present in all
unsymmetrical faults
• Zero sequence currents are present when the neutral of
the system is grounded and faults also involves the
ground and magnitude of the neutral current is equal to
3IR0
63. Signal Processing, CE00039-2
Example 1
A 3-phase, 10 MVA, 11 kV generator with a solidly earthed neutral
point supplies a feeder. The relevant impedances of the generator
and feeder in ohms are as under :
Generator feeder
Positive sequence impedance j 1·2 j 1·0
Negative sequence impedance j 0·9 j 1·0
Zero sequence impedance j 0·4 j 3·0
If a fault from one phase to earth occurs on the far end of the feeder,
calculate
(i) the magnitude of fault current
(ii) line to neutral voltage at the generator terminal
66. Signal Processing, CE00039-2
Example 2
A 3-phase, 11 kV, 25 MVA generator with X0 = 0·05 p.u.,
X1 = 0·2 p.u. and X2= 0·2 p.u. is grounded through a
reactance of 0·3 Ω. Calculate the fault current for a
single line to ground fault.
69. Signal Processing, CE00039-2
Example 3 Two 11 KV, 12 MVA, 3f, star connected
generators operate in parallel. The positive, negative and zero
sequence reactance's of each being j0.09, j0.05 and j0.04 pu
respectively. A single line to ground fault occurs at the terminals of
one of the generators.
Estimate (i) the fault current (ii) current in grounding resistor (iii)
voltage across grounding resistor.
72. Signal Processing, CE00039-2
Example 4
• A 50 MVA, 11 kV three-phase alternator was subjected to different
types of faults. The fault currents are as under :
3-phase fault = 2000 A ; Line-to-Line fault = 2600 A ; Line-to-ground
fault = 4200 A
The generator neutral is solidly grounded. Find the values of the
three sequence reactance's of the alternator. Ignore resistances.
