Symmetrical Faults

Dr. Habib-ur Rehman

Electrical Engineering Department
American University of Sharjah

21:16

Chapter # 7
Symmetrical Faults
Week # 11,
11 Lecture 31
Holiday

Chapter # 7
Symmetrical Faults
Week # 11,
11 Lecture 32

Balanced and Unbalanced Faults

The steady-State operating mode of electric power system is
a balanced three phase alternating current (ac).

21:16

Due to sudden external or internal changes in the system,

the steady-state conditions is disrupted.

21:16

A Fault:

When the insulation of the system fails

or a conducting object comes in touch with a
live ppoint,, a short circuit or a fault occurs.

There are two Classes of Faults:

1. Balanced Fault (Symmetrical Fault)
A Fault
au involving
vo v g aall thee three
ee pphases.
ases.
2. Unbalanced Faults (Unsymmetrical Faults)
A fault involving only one or two phases.
21:16

Three Phases Fault (Balanced Fault)

IFa

IFb
Transformer

IFc
21:16

Unbalanced Faults
There are three types of Unbalanced Faults:
1. Single Line to Ground (SLG fault)
2. Line to Line Fault (LL fault)
3 Double
3.
D bl Li
Line tto G
Ground
d (LLG ffault)
lt)

21:16

Single Line to Ground Fault

IF

IF

IF

Transformer

21:16

IF

Line to Line Fault

IF
Transformer

21:16

IF

IF

Double line to ground Fault

IF
IFb
Transformer

21:16

IF

IFc
IF

The majority of the faults are unsymmetrical.

Fault calculations involve finding the voltage
and current distribution throughout the system
during fault condition.
Why do we need that?
Too adjus
adjust and
a d set
se thee protective
p o ec ve devices
dev ces so
we can detect any fault and isolate the faulty
portion of the system.
To protect the human being and the equipment
during the abnormal operating conditions.
21:16

MVA Level
When fault occurs at a ppoint in a power
p
system,
y
the corresponding MVA of the fault is referred
to as the fault level at that point.
Different elements in an electric system are
designed and selected based on the short circuit
MVA level.
21:16

RL Circuit Analysis

To understand fault analysis we need to review the behavior of an RL

circuit.
circuit
Apply KVL

Ldi (t )
Ri(t ) 2V sin(t )
dt
Before the switch is closed obviously i(t) = 0. When the switch is closed at t=0 the
currentt will
ill have
h
two
t components:
t
1) a steady-state value
2) a transient (dc offset) value

i (t ) iac (t ) idc (t )
21:16

2V
sin(t ) sin( )e t / T
Z

RL Circuit Analysis, contd

iac (t )

2V
sin(t )
Z

idc (t )

2V
sin( )e t / T
Z

Where:
Z R 2 (L) 2

tan 1

L
R

tan 1

X
R

L X

R R

The total current is called the asymmetrical

y
fault current and is plotted
p
as
follows:

21:16

Chapter # 7
Symmetrical Faults
Week # 11,
11 Lecture 33
Visit to Power House

Chapter # 7
Symmetrical Faults
Week # 12,
12 Lecture 34

RL Circuit Analysis, contd

iac (t )

22V
V
sin(t )
Z

idc (t )

2V
sin( )e t / T
Z

The RMS AC fault current is Iac = V/Z

The magnitude of the dc offset depends on .
So it will vary from 0 when = to 2Iac when
= ( /2)

Since we are interested in the largest fault current, we chose

= ( - /2) then,
then

i (t ) 2 I ac sin(t / 2) e t / T

21:16

Where Iac = V/Z

RL Circuit Analysis, contd

Since i(t)
( ) is not strictly
y pperiodic,, its RMS value is not strictly
y defined.
However, we can treat the exponential term as a constant, we can stretch the
RMS concept to calculate the RMS asymmetrical fault current as follows:
I RMS (t )

I ac 2 I dc (t )2

I ac 2

2 I ac e t / T

I RMS (t ) I ac 1 2e 2t / T

It is convenient to use T = X/(2fR) and t=/f where is time in cycles and:

I RMS (t ) k ( ) I ac

Where

k ( ) 1 2e 4 /( X / R )

Form these equations we can say that, the rms asymmetrical fault current equals
the rms ac fault current times an asymmetry factor, k(). IRMS() decreases
from 3Iac when =0 to Iac when is large.
21:16Also, the higher X to R rations (X/R) give higher value of IRMS().

RL Circuit Analysis, contd

The above series R-L short circuit currents are summarized in the table below:

21:16

10

Example 7.1:
A bolted short circuit occurs in the series RL circuit of the figure below. V
=20 kV, X = 8 ohms, and R=0.8 ohm with maximum DC offset. The
circuit breaker opens 3 cycles after fault inception. Determine:
a) The RMS AC fault current.
b) The RMS momentary current at = 0.5 cycles.
c) The RMS asymmetrical fault current that the breaker interrupts.

21:16

Example 1-solution
a) rms ac fault current

I ac

20 103

82 0.82

2.488 kA

b) rms ac momentary current at = 0.5 .

( X / R) 8 / 0.8 10

K (0.5 cycle) 1 2e 4 ( 0.5) /10 1.438

I momentryy K (0.5 cycle) I ac 1.4382.488 3.576 kA

c) The RMS asymmetrical fault current that the breaker interrupts at 3 cycles

K (3 cycle) 1 2e 4 (3) /10 1.023

21:16

11

I momentry K (3 cycle) I ac 1.0232.488 2.544 kA

Chapter # 7
Symmetrical Faults
Week # 12,
12 Lecture 35

Generator Modeling During Faults

During a fault the only devices that can contribute fault current are those
with
ith energy storage
t
like
lik inductors.
i d t
Thus the models of generators (and other rotating machines) are very
important since they contribute the bulk of the fault current.
Generators can be approximated as a constant voltage behind a timevarying reactance.
One way to investigate a three phase short circuit at the terminals of
synchronous machine is to perform a test on an actual machine.

Ea'
21:16

12

Generator Modeling During Faults

(Reference
article
5.10
Chapman) When a fault occurs
on a synchronous generator, the
resulting current flow in the
phases of generator can appear
as shown in the figure.
The current in each phase can be
represented
t d as a dc
d transient
t
i t
component added on the top of a
symmetrical ac component.

21:16

Generator Modeling During Faults

21:16

13

Before the fault, only ac voltages and currents were present within the
generator,
t while
hil after
ft the
th fault
f lt both
b th ac andd dc
d currents
t are present.
t Where
Wh
did the dc current come from?
The synchronous generator is basically inductive it is modeled by
internal generated voltage in series with the synchronous reactance. Also,
recall that a current can not change instantaneously in an inductor.
When a fault occurs, the ac component of current jumps to a very large
value, but the total current cannot change at that instant. The dc
component of current is just large enough that the sum of ac and dc
components just after the fault equals ac current flowing just before the
fault.
Since the instantaneous values of current at the moment of fault is
different in each phase, the magnitude of dc component of current will be
different in each phase.

Generator Modeling During Faults

Figure shows an oscillogram of the ac fault current in one phase of an the

machine.
hi
Th dc
The
d offset
ff t has
h been
b
removedd from
f
th oscillograms.
the
ill
Th
The
amplitude of the sinusoidal waveform decreases from a high initial value to
a lower steady state value.
The ac symmetrical component
of the current can be divided into
roughly three periods.
During the first cycle or so after
th fault
the
f lt occurs, the
th ac currentt is
i
very large and falls very rapidly.
This period of time is called
subtransient period and is
modeled
by
direct
axis
subtransient reactance Xd//.

21:16

Generator Modeling During Faults

21:16

14

After the subtransient period is over, the current continues to fall at a

slower
l
rate,
t until
til att last
l t it reaches
h a steady
t d state
t t value.
l The
Th period
i d off time
ti
during which it falls at a slower rate is called transient period and is
modeled by direct axis transient reactance Xd/.
The time in which it reaches steady state condition is known as steady
state period and is modeled by direct axis synchronous reactance Xd.

Generator Modeling During Faults

Where Xd//< Xd/< Xd. The subscript d refers to the direct axis. There are
similar
i il quadrature
d t
axis
i reactances
t
Xq//<Xq/< Xq. However,
H
th quadrature
the
d t
axis reactances do not significantly affect the short circuit current.
The instantaneous ac fault current using direct axis reactances can be
written as:

1
//
1
1
1 t / Td/
1

e
iac (t ) 2 E g // / e t / Td /

Sin(t )....(1)
X
X
X
X
X
2
d
d
d
d
d

21:16

Where Eg is the rms line-to-neutral prefault terminal voltage of the

unloaded synchronous machine. Armature resistance is neglected.
At t=0, when the fault occurs, the current is called the rms subtransient
fault current, I//, the rms value of iac is:
E
I ac (0) g// I //
Xd

Generator Modeling During Faults

21:16

15

The duration of I// is determined by Td//, called the direct axis shortcircuit subtransient time constant.
At a later time, when t is large compared to Td// but small compared to the
direct axis short-circuit transient time constant Td/ , the first exponential
term in Eq. (1) has decayed almost to zero, but the second exponential
has not yet decayed significantly.
The rms ac fault current then equals the rms transient fault current, given
by.
Eg
I/ /
Xd
When t is much larger than Td/, the rms ac fault current approaches its
steady state value given by
I ac ()

Eg
Xd

Generator Modeling During Faults

Since the three phase no load voltages are displaced by 120o from each
other, the three phase fault currents are also displaced 120o apart from
each other.
In addition to the ac fault current, each phase has different dc offset. The
maximum dc offset in any one phase, which occurs when =0 is
idc max (t )

2Eg
X d//

e t / TA 2 I // e t / TA

Where TA is called armature time constant. Note that the magnitude of

the maximum dc offset depends upon only on the rms subtransient fault
current I//.

21:16

Generator Modeling, contd

21:16

16

Synchronous machine short-circuit currents are summarized in the table

below.
below
Machine reactances and time constants are provided by the manufacturers
which can be obtained through the tests as well.

Generator Modeling, contd

21:16

Chapter # 7
Symmetrical Faults
Week # 12,
12 Lecture 36

17

Generator Short Circuit Example

Example
p 7.2:
A 500 MVA, 20 kV, 60 Hz, 3 synchronous generator with reactances of
Xd//=0.15, Xd/=0.24, Xd =1.1 per unit and time constants Td//=0.035,
Td/=2.0, TA=0.20 sec. is connected to a circuit breaker. The generator is
operating at 5% above rated voltage and at no load when a bolted three
phase short circuit occurs on the load side of breaker. The breaker
interrupts the fault at 3 cycles after the fault inception. Determine:
a) The subtransient fault current in per unit and in KA rms.
b) Maximum dc offset as a function of time; and
c) rms symmetrical fault current, which the circuit breaker interrupts
assuming maximum dc offset.

21:16

Generator S.C. Example, cont'd

Substituting in the values
1 t 2.0
1 1

1.1 0.24 1.1 e

I ac (t ) 1.05

1 1 e t 0.035

0.15 0.24

I ac (0) 1.05
Ibase

0 15
0.15

7 p.u.

500 106
14,433 A I ac (0) 101,000 A
3 20 103

IDC (0) 101 kA 2 e

21:16

18

0.2

143 k A I RMS (0) 175 kA

Generator S.C. Example, cont'd

Evaluating at t = 0.05 seconds for breaker opening
1 0.05 2.0
1 1

1.1 0.24 1.1 e

I ac (0.05) 1.05

1 1 e 0.05 0.035

0.15 0.24

I ac (0
(0.05)
05) 70.8
70 8 kA
I DC (0.05) 143 e

0.05

0.2

kA 111 k A

I RMS (0.05) 70.82 1112 132 kA

21:16

Chapter # 7
Symmetrical Faults
Week # 13,
13 Lecture 37

19

Thevenin Equivalent Circuit

According to Thevenin
Theveninss Theorem
Theorem, any linear network
containing any number of voltage sources and impedances can
be replaced by:
A single voltage source (EMF) known as Thevenins voltage.
This voltage (EMF) is equal to open circuit voltage as seen
from the terminals under consideration.
A single impedance known as Thevenins impedance. This
impedance is equal to the total impedance under open circuit
condition as seen from the terminals under consideration and
after ignoring the sources.
21:16

Using Tevenins
Theorem,
The circuit can
be simplified
as shown

The Calculation of the

Fault Currents can be
easily done Using
Tevenins Theorem
21:16

20

Symmetrical Faults (Balanced Faults)

A Fault involving all the three
phases.

Three Phase
Short Circuit

It is an important type of fault

and
can
be
easily
calculated.
The circuit breaker rated MVA
breaking capacity is selected
b d on the
based
th three
th
phase
h
short
h t
circuit MVA.

Transformer

21:16

Symmetrical Faults (Balanced Faults)

The following assumptions are made in three phase fault
calculation :
1) The EMF of all generators are 1 per unit making zero angle.
angle
o
V 10 This means that the system voltage is at its nominal
value and the system is operating at no-load conditions at the
time of fault. Therefore, All the generators can be replaced by a
single generator since all EMFs are equal and are in phase.
2) The shunt elements in the transformer model that account for
magnetizing
i i current andd core losses
l
are neglected.
l
d
3) The shunt capacitances of the transmission Lines are neglected.
4) The system resistances are neglected and only the inductive
reactance of different elements are taken into account.
21:16

21

Symmetrical

Three Phase Fault

Calculation

Fault Calculation

1 Simple Circuits

Load Ignored
Use Thevenin's Equivalent

Three Phase
Short Circuit

2 Simple Circuits
Load not Ignored
Use Thevenin's Equivalent
Find Pre-fault voltage
Transformer

3 Large Circuits

Construct
The Bus Impedance Matrix
21:16

1. Simple Circuits and Load is ignored

The Calculations for the three phase fault are easy because the
circuit is completely symmetrical and calculations can be done
for only one phase.

T1

G1
3 phase
fault e

G2
2

21:16

22

T2

Steps For Calculating Symmetrical Faults:

j 0.15

j 0.10

1.

Draw a single line diagram for the

system.

2.

Select a common base and find out the

per unit reactances of all generators,
generators
transformers, transmission lines, etc.

4.

From the single line diagram of the system draw a single line reactance
diagram showing one phase and neutral. Indicate all the reactances, etc. on
the single line reactance diagram.

5.

Reduce the single line reactance diagram by using series parallel and DeltaWye transformations keeping the identity of the fault point intact. Find the
total reactance of the system as seen from the fault point (Using Thevenin
Theveninss
Theorem).

5.

Find the fault current and the fault MVA in per unit. Convert the per unit
values to actual values.

10

10

j 0.20 j 0.16

j 0.23

j 0.23

6.

Retrace the steps to calculate the voltages and the currents throughout
different parts of the power system.
21:16

Symmetrical Fault Calculation, Load Ignored

Example 3:
A three phase fault occurs in the system as shown in the Figure.
Find the total fault current, the fault level and fault current
supplied by each generator.
G1

100 MVA,
11 kV
15% Reactance

T1

G2
50 MVA,
11 kV
10% Reactance
21:16

23

100 MVA, 11/132 kV

10% Reactance

T2
2

50 MVA, 11/132 kV
8% Reactance

X=0.2 ohm/phase/km

200 km

3 phase
fault e

Example 3 -Solution:
Step 1: Draw a single line diagram for the system.
G1

100 MVA, 11/132 kV

10% Reactance

100 MVA,
11 kV
15% Reactance

T1

G2

X=0.2 ohm/phase/km

200 km

T2

50 MVA,
11 kV
10% Reactance

3 phase
ffault e

50 MVA
MVA, 11/132 kV
8% Reactance

The single line diagram for the system is given in the

example as shown.

21:16

Step 2: Select a common base and find the per unit reactances of all
generators, transformers, etc.
Select the common base as:
100 MVA (100,000 kVA)
11 kV ffor T
Transformer
f
llow voltage
lt
side
id (LV)
132 kV for Transformer high voltage side (HV)

G1

100 MVA, 11/132 kV

10% Reactance

V Base 132kV

100 MVA,
11 kV
15% Reactance

T1

X=0.2 ohm/phase/km

V Base 11kV

G2
50 MVA,
11 kV
10% Reactance
21:16

24

T2

200 km

3 phase
fault e

50 MVA, 11/132 kV
8% Reactance

G1

The per unit reactance

XG1 j 0.15

G2

The per unit reactance

XG 2 j 0.1 *

T1

The per unit reactance

XT1 j 0.1

T2

The p
per unit reactance

XT2 j 0.08 *

100
j 0.2
50

100
j 0.16
50

100 MVA

LINE
( j 0.2 * 200 )
j 0.23
TL The per unit reactance X LINE Z
(132kV ) 2
Base

G1

100 MVA, 11/132 kV

10% Reactance

V Base 132kV

100 MVA,
11 kV
15% Reactance

T1

X=0.2 ohm/phase/km

V Base 11kV

G2

200 km

T2

50 MVA,
11 kV
10% Reactance

21:16

3 phase
fault e

50 MVA, 11/132 kV
8% Reactance

From the single line diagram of the system draw a single line reactance diagram
showing one phase and neutral. Indicate all the reactances, etc. on the single line
reactance diagram.

T1

G1
3 phase
fault e

G2
2

T2

j 0.15

10

j 0.10

10
21:16

25

j 0.20 j 0.16

j 0.23

j 0.23

Find the total impedance (reactance) of the system as seen from the
fault side.
j 0 . 25

1 0

1 0

j 0 . 115

j 0 . 36

j 0 . 25

j 0 . 115

j 0 . 36

X Total
X Total

21:16

j 0.25 * j 0.36
j 0.115
j 0.25 j 0.36
j 0.2625 pu

j 0.2625

10

IF

10
j 3.8095 pu
j 0.2625

| I base | L

( I Base )132kV side

| S Base
ase |3

3 | V Base | LL

100 * 1000

437.4 A

3 * 132

( I F ) Actual ( I F ) pu * I Base
( I F ) Actual j 3.8095 * 437.4 1666.27 90 o A
21:16

26

V Base 11kV

100 MVA, 11/132 kV

10% Reactance

G1

100 MVA,
11 kV
15% Reactance

V Base 132kV

T1

X=0.2 ohm/phase/km

IF

( I F )G 1
G2

200 km

T2

50 MVA,
11 kV
10% Reactance

3 phase
fault e

50 MVA, 11/132 kV
8% Reactance

( I F )G 2

The Fault Level is:

( MVA) Level 3.8095 pu
( MVA) Base 100

21:16

( MVA) Actual Level 3.8095 * 100 380.95

( I F )G 1

V Base 11kV

j 0.15

10

V Base 132kV

j 0.10

10

j 0.20 j 0.16

j 0.23

( I F )Total

j 0.23

( I F )G 2

At 11 kV Side:
( I Base )11kV side

100 * 1000

5248.8 A

3 * 11

The Fault Current at 11 kV side supplied by the two generators is:

( I F ) Actual j 3.8095 * 5248.8 19995 90 o A
21:16

27

( I F )G 1

10

( I F )G 1
G1

( I F )Total

j0.25
T1

j0.36
G

G2

10

T2

( I F )G 2
( I F )G 2

( I F )G1 (19995 90)

j 0.36
11800.3 90 A
j 0.36 j 0.25

( I F )G 2 ( I F )T ,11kV ( I F )G1
( I F )G 2 8194.7 90 A
21:16

Chapter # 7
Symmetrical Faults
Week # 13,
13 Lecture 38

28

j0.115

Symmetrical Fault Calculation, Load Ignored

Example 4: The single-line diagram
of a power system is shown in the
Figure. The transient reactance of
each part of the system is as shown
and expressed in pu on a common 100
MVA base.
Assuming that all generators are
working on the rated voltages, when a
th
three-phase
h
f lt with
fault
ith impedance
i
d
off
j0.16 pu occurs at bus 5. Find: The
fault currents.

G1

j0.2 G2

j0.1

Bus 2

Bus 1

j0.1

j0.2

Bus 3

Bus 4

j0.8

j0.4

j0.4

Bus 5
21:16

j0.1

j0.2

Bus 1

Bus 2

j 0.1

Bus 3

G1

j0.2

j0.8

Bus 4

j0.2 G2

j0.1

Bus 2

Bus 1

j0.4

j0.4
j0.1
Bus 3

j0.2

j0.4
j0.2
Bus 3

Bus 4

j0.8

j0.16

j0.8

Bus 4

Bus 5

j0.4

j0.4
j0.4

j0.4

j0.16
21:16

29

Bus 5

Bus 5

j0.4

j0.2

I FG1

I FG1

Bus 4

Bus 3

IF

I FG2
j0.6

j0.4
j0.2

j0.2

j0.34

I FG2

Bus 5

j0.16

j0.1

j0.16

IF

j0.1

Bus 5

IF

j0.16

Bus 5

IF

Bus 5

j0.16

j0.34

ZTH j0.34 j0.16 j0.5

( I F )Bus5

10
j 2.0 pu
j0.5

( I F )G1 I F
( I F )G2 I F

21:16

Chapter # 7
Symmetrical Faults
Week # 13,
13 Lecture 38

30

j0.6
j1.2 pu
j0.6 j0.4
j0.4
j0.8 pu
j0.6 j0.4

Circuit Breaker and Fuse Selection

A circuit breaker is a mechanical switch capable of
interrupting fault current and re-closing.
re closing
When the circuit contact separate while carrying current, an
arc forms. The breaker is designed to extinguish the arc by
elongating and cooling it.
The fact that AC arc current naturally passes through zero
twice during its 60 Hz cycle aids the arc extinction process.
Circuit breakers are classified as power circuit breakers when
they are intended for service in ac circuits above 1500 V,
V as
low voltage circuit breakers in ac circuits up to 1500 v.
There are different types of circuit breaker depending on the
medium: Air, Oil, SF6 or Vacuum in which arc is
elongated. Also, the arc can be elongated either by a magnetic
force or by a blast of air.

21:16

Circuit Breaker and Fuse Selection

21:16

31

Some circuit breakers are equipped with a high speed

automatic reclosing capability. Since most faults are
temporary and self clearing, reclosing is based on the
idea that if a circuit is deenergized for a short time, it is
likely that whatever caused the fault has disintegrated
and the ionized arc in the fault has dissipated.
When reclosing the breakers are employed in EHV
systems, standard practice is to reclose it only once,
approximately 10 to 15 cycles (depending on operating
voltage)
lt ) after
ft the
th breaker
b k interrupts
i t
t the
th fault.
f lt If the
th fault
f lt
persists and EHV breaker recloses into it, the breaker
interrupts the fault current and then locks out requires
operator resetting.
Multiple shot-reclosing in EHV systems is not a
standard practice because transient stability may be
compromised.

Circuit Breaker and Fuse Selection

However for distribution systems

y
((2.4 kV to 46 kV)) where
customer outages are of concern, standard reclosures are
equipped with two or more reclosures.
For low-voltage applications, molded case circuit breakers
with dual trip capability are available. There is magnetic
instantaneous trip for large fault currents above a specified
threshold, and a thermal trip with time delay for smaller fault
currents.
M d
Modern
circuit
i it breakers
b k
are based
b d on symmetrical
ti l
interrupting current. It is usually necessary to calculate only
symmetrical fault current and then select a breaker with a
symmetrical interrupting capability equal to or above the
calculated current.

21:16

Circuit Breaker and Fuse Selection

21:16

32

When selecting circuit breakers for generators,

generators two cycle
breakers arc employed in practice, and the subtransient fault
current is calculated; therefore subtransient machine
reactances Xd// are used in fault calculations.
For synchronous motors, subtransient reactances Xd// or
transient reactances Xd/ are used, depending on breaker
speed.
Also, induction motors can momentarily contribute to fault
current. Large induction motors are usually modeled as
sources in series with Xd// or Xd/, depending on breaker
speed. Smaller induction motors (below 50 hp) are often
neglected entirely.

21:16

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33

Circuit Breaker and Fuse Selection

Voltage ratings:

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Rated maximum voltage: The maximum RMS line-line

operating voltage. The breaker should be used in systems
with an operating voltage less than or equal to this rating.
Rated low frequency withstand voltage: The maximum 60
Hz line to line voltage that the circuit breaker can withstand
without insulation damage.
Rated Impulse
Imp lse withstand
ithstand voltage:
oltage: The maximum
ma im m crest
voltage of a voltage pulse with standard rise and delay
times that the breaker insulation can withstand.
Rated voltage range factor K: The range of voltage for
which the symmetrical interrupting capability time the
operating voltage is constant.

Circuit Breaker and Fuse Selection

Current ratings:

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34

Rated continuous current: The maximum RMS current

that the breaker can carry continuously while it is in the
closed position without overheating.
Rated short-circuit current: The maximum RMS
symmetrical current that the breaker can safely interrupt at
rated maximum voltage.
voltage
Rated momentary current: The maximum rms
asymmetrical current that the breaker can withstand while
in the closed position without damage. Rated momentary
current of standard breakers is 1.6 time the symmetrical
interrupting capability.

Circuit Breaker and Fuse Selection

Current ratings:

Rated interrupting time: The time in cycles on a 60 Hz

basis from the instant the trip coil is energized to the
instant the fault current is cleared.
Rated interrupting MVA: For a three phase circuit
breaker, this is 3 times rated maximum voltage in kV
times the rated short-circuit current in kA.
kA It is more
common to work with the current and voltage ratings than
with the MVA ratings.

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Circuit Breaker and Fuse Selection

For example, the symmetrical interrupting capability of the 69-kV class
b k is
breaker
i shown
h
in
i the
th following
f ll i graph.
h
The symmetrical interrupting capability increases from its rated short-current
of 19 KA at rated maximum voltage of 72.5 kV up to 23 kA at 60 KV.
Breaker of 115 KV class and higher have a voltage range factor of k=1.0.
Imax=KI=(1.21)(19)=23 kA at an
operating voltage
Vmin=Vmax/K=72.5/1.21= 60 kV.
Below Vmin the symmetrical
interrupting capability remains at
Imax=23 kA. At operating voltages
between Vmin and Vmax, the
symmetrical interrupting capability is:
I Vmax 1378
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kA
V
V

35

Fuses
Fuses are one of the simplest over
overcurrent devices.

The fuse consists of a metal link

encapsulated in a tube and packed in
filler material.

At normal operation, the current will

flow between the terminal through the
link.

If the current increased more than

certain level, the link will melt and a
gap is formed and an electric arc is
established.

The arc will further burn the link till

the resistance of the arc reaches a
high value that the arc cant sustained
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and an open circuit will be formed.

Fuses

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36

Fuses

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Voltage rating: This RMS voltage determines the ability of a

fuse to suppress the internal arc that occurs after the fuse link
melts. A blown fuse should be able to withstand its voltage
rating.
Continuous current rating: The fuse should carry this RMS
current indefinitely without melting.
Interrupting
p g current rating:
g This is the largest
g current that the
fuse can safely interrupt.
Time response: The melting and clearing time of a fuse
depends on the magnitude of the over-current or fault current.
This time is specified by the time-current curve as shown in
the following slide.

Fuses
This is the time-current characteristics curve
of a 15.5 kV, 100 A current limiting fuse.
For example, the current will melt after 2
seconds and clear after 5 seconds for a 500 A.
For 5000 A, the fuse will melt in less than
0.01 second and clears within 0.015 second.
Fuses are inexpensive
inexpensive, fast operating
operating, easily
coordinated and they do not require protective
relays or instrument transformers.
On the hand, the melted fuse must be
replaced manually.
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37