6)Breaking capacity –it is the current in (r.

ms) that a circuit

TOPIC 6 breaker is capable of breaking at given recovery voltage
under specific condition e,g p.f, rate of rise of re-striking
SWITCH GEARS
voltage.
A switch gear refers to apparatus for switching, controlling and 7)Making capacity – Peak value of current (including d.c
protecting electrical circuits and equipment current) during the first cycle of current wave after the first
closure of circuit breaker.
ESSENTIAL PROPERTIES/FEATURES OF SWITCH GEARS 8)Short – time rating – Period for which circuit breaker is
able to carry a fault current while remaining close.
a)Reliability
b)Quick Operation Construction of a circuit breaker
c)Provision for manual operation -It consists of two contacts
d)Ability to discriminate between a faulty section and healthy section of
-a fixed contact and a moving contact.
power lines  A handle is attached at the end of the moving contact.
e)Provision for instrument required for specific area of applications.  It can be operated manually or automatically.
 The automatic operation needs a separate mechanism
Examples of Switch Gears which consists of a trip coil.
1)Switches – advice used to open or close electrical circuit
 The trip coil is energized by secondary of current
ii)Fuses – a short piece of wire which melts when excess current flows
transformer.
in the circuit
 The terminals of the circuit breaker are brought to the
iii)Circuit Breakers- Is an equipment which may close or open acircuit
supply.
under certain conditions i.e fault, full-load or no-load
iv)Relays – a device which detects the faults and sends an information
to circuit breaker for circuit interruption.

CIRCUIT BREAKERS

What is a circuit breakers (CB)

A Circuit Breaker (CB) is basically a piece of equipment or a switch
which can make or break circuit in order to protect the circuit from
excess current..
-It consist of a fixed and moving contact.
-Interrupting capacity of the circuit breaker is therefore expressed in
MVA Basic principle of operation of a circuit breaker
•
Purpose of circuit breakers (switchgear) -Under normal working conditions, the e.m.f produced in
the secondary winding of the transformer is insufficient to
The main purpose of a circuit breaker is to:
energize the trip coil completely for its operation. Thus the
• Switch load currents
• Break normal and fault currents contacts remain in closed position carrying the normal
• Carry fault current without blowing itself open (or up!) i.e. no working current.
distortion due to magnetic forces under fault conditions
-Under abnormal or faulty conditions, high current in the
primary winding of the current transformer induces
TERMS USED IN CIRCUIT BREAKERS
sufficiently high e.m.f in the secondary winding so that the trip
1)Arc voltage – voltage that appears across the contact of the circuit
breaker during the arcing period coil is energized.
2)Re-striking voltage – Transient Voltage that appears across the Arcing Time
contact of a near current zero during arcing period.
This is the time interval from the contact separation to the
3)Recovery voltage –The Normal frequency (50HZ) that appears
extinction of arc
across the contact of the circuit breakers after final arc extinction.
4)Current chopping – is the phenomenon of current interruption
before natural current. Formation of an Arc
As the chops occurs at current I, the energy stored in the inductance is  Under faulty conditions heavy current flows through the
𝐿𝑖 2 contacts of the circuit breaker before they are opened.
this energy will be transferred to the capacitance C charging the  As soon as the contacts start separating, the area of
2
latter to a prospective voltage given by contact decreases which will increase the current density
𝑳𝒊𝟐 𝑪𝒆𝟐 and consequently rise in the temperature.
. =
𝟐 𝟐
 The medium between the contacts of circuit breaker may
𝑳
Or e = 𝒊√ volts be air or oil.
𝑪
 The heat, which is produced, is sufficient to ionize air or
5)Rate of re-striking voltage – is the rate of increase of re-striking
oil which will act as conductor.
voltage

38 | P a g e COMPILED BY: ODIPO 0729692775

 Thus an arc is struck between the contacts.
The medium could be :
NOTE: The arcing time depends on arcing resistance.
• Oil
Factors affecting Arcing Period • Air
• SF6.
a)Degree of ionization:- If there is less number of ionized particles • Vacuum
between the contacts then the arc resistance increases.
1)Oil Circuit Breakers
b)Length of arc: The arc resistance is a function of length of arc
which is nothing but separation between the contacts. More the length
more is the arc resistance.

c)Cross-section of arc: If the area of cross-section of the arc is less

then arc resistance is large

Arc Extinction
It is essential that arc should be extinguished as early as possible.
There are two methods of extinguishing the arc in circuit breakers
 High resistance method It uses oil as arc extinction agent, the contacts opens under oil
 Low resistance or current zero method and an arc is struck between them. The heat of the arc
evaporates the surrounding oil and dissociates it into a
a)High Resistance Method substantial volume of gases such as hydrogen at high
This method is employed in only d.c circuit pressure.
In this method, the arc resistance is increased with time. The hydrogen gas occupies a large volume. The oil therefore
 This will reduce the current to such a value, which will be is pushed away from the arc and expanding hydrogen gas
bubble surrounds the arc region and adjacent portions of the
insufficient to maintain the arc thus the current is interrupted, and
contact.
the arc is extinguished.
Achieved by two processes.
– The resistance of the arc may be increased by: 1)First -The hydrogen gas has high conductivity and cools the arc.
– lengthening the arc 2)second-the gas sets up the turbulence in the oil and forces it into
– cooling the arc the spaces between the contacts thus eliminating the arcing product
– reducing the cross-section of the arc from the arc path.
– splitting the arc.
b).Low Resistance (current zero )Method Properties of oil that makes it good insulation for circuit
breaker.
-This method is employed for arc extinction in ac. Circuits and circuit i)Has good dielectric strength
breakers. ii)Good absorbent of arc
ii)excellent cooling properties
 In this method arc resistance is kept low until current zero where
extinction of arc takes place naturally and is prevented from Advantages of Oil cct Disadvantages of Oil cct
restriping. Breaker breaker
 The current becomes zero two tires in a cycle. So at each current Excellent cooling properties It is inflammable and there
zero point the arc vanishes for small instant and again it appears. as it discomposes oil is a risk of a fire.
 But in auxiliary circuit breakers the arc is interrupted at a current It acts as an insulator and It may form an explosive
zero point. The space between the contacts is ionized and this permits smaller clearance mixture with air
produces high voltage across the contacts thus the dielectric between live conductors and
strength must be very high to extinction of the arc. earthed components High Maintenance cost due
to replacement of oil
TYPES OF CIRCUIT BREAKERS
Circuit breakers are classified as:
i)Single break Maintenance e of Oil Circuit Breakers
i)Double break. i)Check the current carrying parts and arcing contacts. If the
burning is severe, the contacts should be replaced.
a)Single break type ,only the bus-bar end is isolated ii)Check the dielectric strength of the oil. If the oil is badly
b)Double break type, both bus-bar (source) and cable (load) ends are discolored, it should be changed or reconditioned. The oil in
broken. good condition should withstand 30 kV for one minute in a
However, the double break is the most common and accepted type in standard oil testing cup with 4 mm gap between electrodes.
modern installation iii)Check the insulation for possible damage. Clean the
surface and remove carbon deposits with a strong and dry
NB:Circuit breakers may be further classified depending on the fabric.
medium in which the breaker opens and closes. iv)Check the oil level.
iv)Check closing and tripping mechanism.

39 | P a g e COMPILED BY: ODIPO 0729692775

2) Air Circuit Breaker 3)Sulphur Hexafluoride (SF6) Circuit Breaker
-Uses atmospheric pressure air as an arc-extinguishing medium.
-the contacts are opened in a flow of air blast established by opening
the blast valve. The air blast cools the arc and sweeps away the arcing
products to the atmosphere. This rapidly increases the dielectrics
strength of the medium between the contacts and prevents from re-
establishing the arc. Hence the arc is extinguished and current flow is
interrupted. The principle of high resistance interruption is employed.
They are of two types:
a)axial blast circuit breaker
b)Cross –blast air circuit breaker
 The inert SF6 gas is used for arc extinction. It has the strong
a)Axial blast air circuit breaker – when the fault occurs, the tripping tendency of absorbing free electrons. .
impulse cause opening of the circuit breaker reservoir to the arcing  The contacts of the breaker arc opened in a high-pressure
chamber flow of SF6 gas and an arc is struck between them, the
conducting free electrons are rapidly captured by the gas to
form relatively immobile negative ions. This loss of
conducting electrons in arc quickly build up enough
insulation to extinguish the arc.They are effective for high
power and voltage. Mostly suitable where explosive hazards
exists, non-inflammable SF6 hence no risk of fire.
 It is thermally stable, non- poisonous, odorless gas, good
dielectric strength and arc extinguishing properties.

Advantages of SF6 Disadvantages of SF6

Have short arcing time due to SF6 breakers are costly due to
superior quenching property the high cost of SF6
b)Cross-blast air breaker – air blast is directed at aright angle to the Can interrupt much larger Need for additional equipment
arc. It lengthens the arc for extinction current since,SF6 gas has to be
Noise-less operation since no reconditioned after every
exhaust to atmosphere as in air operation of the breaker,
blast additional
No Moisture problem
SF6 is non-inflammable hence
no risk of fire
No carbon deposits and hence
insulation problems eliminated
Low maintenance cost
Applicable in Hazardous
areas

Advantages of Air-blast Disadvantages of Air-blast

cct breaker circuit breaker
No risk of fire Has inferior arc extinction FUSES
properties A fuse is a short piece of wire of low melting point that
Reduced size of the Need for the maintenance of melts in case of excess current in the circuit and hence
device the compressor protecting the circuit equipment.
No arcing products and Sensitive to variation in the
expense of regular oil rate of re-striking voltage Desirable characteristics of a fuse
replacement is avoided i)low melting point e.g., tin, lead.
Arcing time is small ii)high conductivity e.g., silver, copper.
hence no burning of the iii)free from deterioration due to oxidation e.g., silver.
contacts iv)low cost e.g., lead, tin, copper.
Suitable where frequent
operation is required

40 | P a g e COMPILED BY: ODIPO 0729692775

Important terms used in Fuses of fault current are to be interrupted. It consists of a base and
a fuse carrier
When a fault occurs, the fuse element is blown out and the
circuit is interrupted. The fuse carrier is taken out and the
blown out fuse element is replaced by the new one. The fuse
carrier is then re- inserted in the base to restore the supply.
Advantage Disadvantage
The detach- able fuse Possibility of renewable
carrier permits the by the fuse wire of wrong
replacement of fuse size and material
element without any Has low breaking
danger of coming in capacity and hence
contact with live parts. cannot be used in high
fault level
Protective capacity of
such a fuse is uncertain
subjected to deterioration
The following terms are much used in the analysis of fuses : due to oxidation through
a)Current rating of fuse element. It is the current which the fuse the continuous heat- ing
element can normally carry without overheating or melting. up of the element.
b)Fusing current. It is the minimum current at which the fuse element
melts and thus disconnects the circuit protected by it. Obviously, its 2)HIGH RUPTURING CAPACITY FUSE (HRC)
value will be more than the current rating of the fuse element. The filling powder is either chalk or plaster of paris or Quartz
The fusing current depends upon the various factors such as : which acts as an arc quenching or coolling agent.
i)material of fuse element
ii)length – the smaller the length, the greater the current because a
short fuse can easily conduct away all the heat
ii)diameter
iv)size and location of terminals
v)previous history
vi)type of enclosure used

c)Fuse factor - is the ration of minimum fusing current to the current

rating of the fuse element. And its value is always more than 1. I.e the
When a fault occurs, the current increases and the fuse
greater is the difficulty in avoiding deterioration due to
element melts before the fault current reaches its first peak.
overheating and oxidation at rated carrying current The heat produced in the process vaporizes the silver element.
𝑀𝑖𝑛 𝐹𝑢𝑠𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
Fuse factor = The Chemical reaction between the silver vapors and the
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑟𝑎𝑡𝑖𝑛𝑔 𝑂𝑓 𝐹𝑢𝑠𝑒
d)Prospective Current- It is the r.m.s. value of the first loop of the fault filling powder results in the formation of high resistance
current obtained if the fuse is replaced by an ordinary conductor of element, which helps in arc extinction.
negligible resistance.
e)Cut-off current. It is the maximum value of fault current actually Advantages of HRC Disadvantages of HRC
reached before the fuse melts. Fuse Fuse
f)Pre-arcing time. It is the time between the commencement of fault and Capable of clearing high Have to be replaced after
the instant when cut off occurs. and low faults each operation
g)Total operating time. It is the sum of pre-arcing and arcing times. Do not deteriorate with Heat produced by the arc
It may be noted that operating time of a fuse is generally quite low (say age may affects associated
0·002 sec.) as compared to a circuit breaker (say 0·2 sec or so). This is Have High speed of switches.
an added advantage of a fuse over a circuit breaker. Operation
Provides High reliable
TYPES OF FUSES operations
In general, fuses may be classified into : Requires no Maintenance
i)Low voltages fuses Cheaper than any other
ii) High voltage fuses Circuit interruption
Device
a)Low voltages fuses Permits Consistency of
Low voltage fuses can be subdivided into two classes viz., Performance
1)Semi-enclosed re-wireable fuse –
2)High rupturing capacity (H.R.C.) cartridge fuse.
H.R.C. fuse with tripping device. Sometime, H.R.C. cartridge
fuse is provided with a tripping device. When the fuse blows
1)SEMI-ENCLOSED RE-WIREABLE FUSE –
out under fault conditions, the tripping device causes the
Rewireable fuse (also known as kit-kat type) is used where low values
circuit breaker to operate.

41 | P a g e COMPILED BY: ODIPO 0729692775

 melts, the spring retracts part of it through a baffle (or liquid
director) and draws it well into the liquid. The small quantity
of gas generated at the point of fusion forces some part of
liquid into the passage through baffle and there it effectively
extinguishes the arc.

When a fault occurs, the silver fuse elements are the first to
be blown out and then current is transferred to the tungsten
wire. The weak link in series with the tungsten wire gets fused
and causes the chemical charge to be detonated. This forces the
plunger outward to operate the circuit breaker. The travel of the
plunger is so set that it is not ejected from the fuse body under
fault conditions
Advantages. H.R.C. fuse with a tripping device has
the following advantages over a H.R.C. fuse
without tripping device :
i)Can prevent ‘single phasing ‘ in case of fault in a three phase system
by the plunger enabling tripping of all the three phases iii)Metal clad fuses. Metal clad oil-immersed fuses have
ii)Prevents short circuit in been developed with the object of providing a substitute for
ii)The fuse-tripped circuit breaker is generally capable of dealing with the oil circuit breaker. Such fuses can be used for very high
fairly small fault cur rents itself. This avoids the necessity for replacing voltage circuits and operate most satisfactorily under short-
the fuse except after highest currents for which it is intended. circuit conditions approaching their rated capacity

Current Capacity of Fuse element

b)High Voltage Fuses Ordinary fuse Law
They include The current carrying capacity of a fuse element mainly
i) Cartridge type depends on the metal use and the cross-sectional area but if
ii) Liquid type affected also by the length, the state of surface and the
surrounding of the fuse. When the fuse element attains steady
a) Cartridge type - This is similar in general construction to the low temperature.
voltage cartridge type except that special design features are Heat produced per sec =Heat lost per second by Modes of
incorporated. High voltage cartridge fuses are used upto 33 kV with heat
2
breaking capacity of about 8700 A at that voltage. Rating of the order of 𝐼 𝑅 = Constant x Effective surface area
𝑙
200 A at 6·6 kV and 11 kV and 50 A at 33 kV are also available 𝐼 2 (𝜌 ) = constant x d x l
𝑎
2 𝜌𝑙
𝐼 𝜋 = constant x d x l
[ ]𝑑 2
4
2 3
𝐼 ∝𝑑

Example
A fuse wire of circular cross-section has a radius of
0.8mm. The wire blows off at a current of 8A. calculate
the radius of the wire that will blow off at a current of 1A
ii)Liquid type -These fuses are filled with carbon tetrachloride and have Sln
the widest range of application to HV systems. They may be used for 𝐼 2 ∝ 𝐼 3
circuits up to about 100 A rated current on systems up to 132 kV and 𝐼2 2 𝑟2 3
may have breaking capacities of the order of 6100 A. . It consists of a [𝐼1] = [𝑟1]
2
glass tube filled with carbon tetrachloride solution and sealed at both 𝑟2 3 𝐼2 2 𝐼2 3
ends with brass caps. The fuse wire is sealed at one end of the tube and 𝑟1 = √ [ ] =[ ]
𝐼1 𝐼1
the other end of the wire is held by a strong phosphor bronze spiral 2 2
𝐼2 3 1 3
spring fixed at the other end of the glass tube. When the current 𝑟2 = 𝑟1 × [ ] = 0.8 x[ ] = 0.2mm
𝐼1 8
exceeds the prescribed limit, the fuse wire is blown out. As the fuse

42 | P a g e COMPILED BY: ODIPO 0729692775

 𝑉𝐿
But 𝐸𝑚𝑎𝑥 = √2 𝑉𝑝 and 𝑉𝑃 =
√3
Differences between a fuse and circuit Breaker 𝑉𝐿 11𝑘𝑉
𝐸𝑚𝑎𝑥 = √2 𝑥 = √2 𝑥 = 8.981kV
√3 √3
No Particular Fuse Circuit Hence,
Breaker Peak re-striking voltage = 2𝐸𝑚𝑎𝑥 = 2x 8.981 =17.96 kV
1 Function Performs both detection Performs
& interruption functions interruption ii)Frequency of oscillation
functions 1 1
only. 𝑓𝑟 = = −11
= 12621.808Hz
2𝜋√𝐿𝐶 2𝜋√ 0.0159 𝑥 10
Detection of iii)Average rate of rise of re-Striking Voltage up to the
fault is done fuse peak.
by relay It occurs at a given time t
2 Operation Inherently completely Requires 1
automatic elaborate t= = 𝜋√𝐿𝐶
2𝑓𝑟
equipment i.e 1 1
relay for
t= =t= = 3.9614 x 10−5 Sec
2𝑓𝑟 2 𝑥 12628𝐻𝑧
automatic Average rate of rise of re-Striking Voltage
action 𝑅𝑒−𝑠𝑡𝑟𝑖𝑐𝑘𝑖𝑛𝑔 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
=
3 Breaking Small Very large 𝑡𝑖𝑚𝑒 𝑢𝑝 𝑡𝑜 𝑓𝑖𝑟𝑠𝑡 𝑝𝑒𝑎𝑘
17.96𝑘𝑉
capacity =
Operating time Very small Comparatively 36.9𝜇𝑠
4
0.002 sec large (0.1 to = 453 x103 kV/sec
0.2 sec
5 Replacement Requires replacement No SYMETRICAL FAULT CALCULATION
after every operation replacement a)Symmetrical faults. That fault which gives rise to equal
after operation faults currents with 120o displacement . The most common
example of symmetrical fault is when all the three conductors
of a 3-phase line are brought together simultaneously into a
Example short-circuit condition.
A circuit breaker is rated as 1500A, 1000MVA ,33KV , 3second, 3 –
phase oil circuit breaker, Find: b)Unsymmetrical faults. Those faults which give rise to
a)Rated Normal current unequal line currents with unequal displacement.
b)Breaking Capacity
c)Rated Symetrical breaking current Steps for Symmetrical Fault Calculations
d)Rated making current It has already been discussed that 3-phase short-circuit faults
e)Short –time rating result in symmetrical fault currents i.e. fault currents in the
f)Rated service Voltage. three phases are equal in magnitude but displaced 120 o
Sln electrical from one another. Therefore, problems involving
a) Rated normal current = 1500A such faults can be solved by considering one phase only as the
b) Breaking Capacity = 1000MVA same conditions prevail in the other two phases. The
Breaking Capacity procedure for the solution of such faults involves the
c) Rated symmetrical braking current =
√3 x Voltage following steps :
1000 𝑥106
= = 1.7496 A (rms) i)Draw a single line diagram of the complete network
√3 𝑥 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
indicating the rating, voltage and percentage reactance of
d) Rated making current = 2.55 x 17496 = 44614.8A each element of the network.
e) Short –time rating = 17496 A for 3 seconds ii)Choose a numerically convenient value of base kVA and
f) Rated service Voltage. = 33kV r.m.s convert all percentage reactance to this base value.
iii)Corresponding to the single line diagram of the network,
draw the reactance diagram showing one phase of the system
Example and the neutral. Indicate the % reactance on the base kVA in
A 50Hz, 11kV, 3phase alternator with earthed neutral has the reactance diagram. The transformer in the system should
reactance of 50Ω per phase and is connected to a bus-bar through a be represented by a reactance in series.
circuit breaker. The distributed capacitance up to circuit breaker iv)Find the total % reactance of the network upto the point
between the phase and neutral is 0.01p.Farad Determine: of fault. Let it be X%.
v)Find the full-load current corresponding to the selected
i)Peak re-striking Voltage base kVA and the normal system voltage at the fault point.
ii)Frequency of oscillation Let it be I.
iv)Then various short-circuit calculations are :
iii) Average rate of rise of re-Striking Voltage up to the fuse peak.
Sln 𝟏𝟎𝟎
i)Peak re-striking Voltage Short-circuit current , 𝑰𝑺𝑪 = 𝑰 ( )
%𝑿
𝑋𝐿 = 2𝜋𝑓𝐿 Hence: Short- circuit KVA = Base kVA x
𝟏𝟎𝟎
𝑋𝐿 5 %𝑿
L= = = 0.0159Henry
2𝜋𝑓 2𝑥3.142 𝑥 50
Peak re-striking voltage = 2𝐸𝑚𝑎𝑥

43 | P a g e COMPILED BY: ODIPO 0729692775

Example
The figure below shows the single line diagram of a 3phase system. EXAMPLE 2:
The percentage reactance of each alternator is based on its own A 3 – phase, 20MVA , 10kV alternator has internal
capacity. Find the short –circuit current that will flow into a reactance of 5% and negligible resistance . Find the
complete 3 –phase short at F. 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.

Sln
20 𝑥 106
Full load current I = = 1154.7 𝐴
√3 𝑥 10 𝑥 103
10 𝑥 103 10 000
Voltage per phase , V = = 𝑉
√3 √3
As the short circuit –current is to be 8 times the full load
current
Thus: Total % reactance required
𝐹𝑢𝑙𝑙−𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
Sln = 𝑥 100
𝑆ℎ𝑜𝑟𝑡−𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡
Method 1 1
= 𝑥 100 = 12.5 %
Let the base kVA be 35000 kVA 8

%Reactance of alternator A at the base kVA is

35000 External % reactance required
% 𝑋𝐴 = 𝑥 30 = 70% = 12.5 – 5 = 7.5%
15000
Let XΩ = be the per phase external reactance required
% Reactance of alternator B at the base kVA is 𝐼𝑋
35000
Now , percentage reactance = 𝑥 100
𝑉
% 𝑋𝐵 = 𝑥 50 = 87.5% 1154.7 𝑋
20000 7.5 = 10 000 x 100
Line current corresponding to 35000 at kV is √3
35000 𝑥 103 7.5 𝑥 10000
𝐼 = = 1684 𝐴 X= = 𝟎. 𝟑𝟕𝟓 Ω
√3 𝑥 12 𝑥 103 √3 𝑥 100 𝑥 1154.7

Total % reactance from generator neutral up to fault point is Example

% X = 𝑋𝐴 //𝑋𝐵 The section bus-bars A and B are linked by bus bar
𝑋 𝑋 70 𝑥 87.5 reactor rated at 5000 kVA with 10% reactance. On the
= 𝐴 𝐵 = = 38.89 %
𝑋𝐴 +𝑋𝐵 70+87.5 Bus bar A, There are two generators each of 10000 kVA
𝟏𝟎𝟎 𝟏𝟎𝟎
Hence, 𝑰𝑺𝑪 = 𝑰 ( ) = 1684 x ( ) = 𝟒𝟑𝟑𝟎 𝑨 with 10% reactance and on B two generators each of 8000
%𝑿 𝟑𝟖.𝟖𝟗
kVA with 12% reactance. Find the steady MVA fed into a
Alternative Method 2 dead short circuit between all phases on B with bus – bar
The problem can also be solved by solved by component short- reactor in the circuit
circuit current method. Each alternator supplies short circuit
current to the fault. The total current fed to the fault is the sum
of the two.

Full-load current delivered by alternator A,

𝑅𝑎𝑡𝑒𝑑 𝑘𝑉𝐴 𝑜𝑓 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑜𝑟 𝐴
𝐼𝐴 =
√3 𝑥 𝐵𝑢𝑠−𝑏𝑎𝑟 𝑣𝑜𝑙𝑡𝑎𝑔𝑒

Sln
15000 𝑥 103
= = 721.7 𝐴 Let 10000 kVA be the base kVA
√3 𝑥 12 𝑥 103
Short circuit current fed to fault by alternator A % reactance of generator 1 and 2 on the base kvA
10000
𝐼𝑆𝐴 = 𝐼𝐴 𝑥
100 = 10 x = 10%
10000
% 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴
100 % reactance of generator 3 and 4 on the base kvA
= 721.7 x ( = 2405.5 A 10000
30 = 12 x = 15%
8000
𝐹𝑢𝑙𝑙 load line current delivered by alternator B
20000 𝑥 103
% reactance of bus bar reactor on the base kvA
𝐼𝐵 = = 962.28 𝐴 = 10 x
10000
= 20%
√3 𝑥 12 𝑥 103
5000
Short circuit current fed to fault by alternator B
100
𝐼𝑆𝐵 = 𝐼𝐵 𝑥
% 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐵
100
𝐼𝑆𝐵 = 962.28 𝑥 = 1924.5𝐴
50
Total short –current fed to fault
𝐼𝑆𝐶 = 𝐼𝑆𝐴 +𝐼𝑆𝐵 = 2405.5 + 1924.5 = 𝟒𝟑𝟑𝟎 𝑨

NOTE: In Complex cases:, The first method has advantage

over the 2nd method.

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 Total % reactance from generator neutral up to fault
When fault occurs on section B (point F in Fig. 17.14), the reactance point F1
diagram at the selected base kVA will be as shown in Fig. 17.15 (i). This = % 𝑋𝐴 + % 𝑋𝑇 = 10 + 10 = 20%
𝟏𝟎𝟎
series parallel circuit is further reduced to Fig. 17.15 (ii). Referring to Short – circuit kVA = 10000 x = 50000 kVA
𝟐𝟎
Fig. 17.15 (ii), it is clear that reactance from generator neutral upto
the fault point F is (5% + 20%) in parallel with 7·5% i.e.

Total % reactance from generator neutral up to fault point F

= (5% + 20% ) // 7.5 % (They are in parallel)
25 𝑥 7,5
= = 5.77%
25+7.5
10000 𝑥 100
Fault kVA = = 1.73310
5.77
Full MVA = 173.31

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

Sln
% reactance of alternator on base kVA
10000
% 𝑋𝐴 = 3 𝑥 10 = 10%
10 𝑥 10

% reactance of transformer on base kVA

10000
% 𝑋𝐴 = 𝑥 5 = 10%
5 𝑥 103

The line impedance is given in Ohms. It can be converted into

percentage impedance
(𝑘𝑉𝐴) 𝑥 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 Ω
% 𝑋𝐿 = 2
10 𝑥 (𝑘𝑉)
10000 𝑥 4
% 𝑋𝐿 = = 40%
10 𝑥 (10)2
% resistance of transmission line
10000 𝑥 1
% 𝑅𝐿 = 2 = 10%
10 𝑥 (10)

i)The reactance diagram of the network on the selected base

kVA is shown . The fault at the end of a transmission line (poit F2)
Total % reactance = % 𝑋𝐴 + % 𝑋𝑇 + % 𝑋𝐿
= 10 + 10 + 40 = 60%
% resistance = 10%
% Impedence from generator neutral up to fault point F2.
= √(60)2 + (10)2 = 60.83%
100
Short-circuit kVA = 10000 x = 16440𝑘𝑉𝐴
60.83
ii)For a fault at high voltage terminals of the transformer (Point F1),

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