Switch Gear & Protection Practical Notes_Electrical Engineering GTU.pdf

PRACTICAL – 1
FLOATING NEUTRAL
NAME : Gajjar Parth Sunilkumar
Enrollment : 180420109009
Submitted To :
1. Prof. Krishna Vakharia
2. Prof. Sharad Patel
Date Of Submission : 9th
July, 2021

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
• Aim : To study the effect of floating neutral on 3-Ø voltages.
• Theory :
The process of connecting the metallic frame (i.e., non-current carrying part) of
electrical equipment or some electrical part of the system (e.g., neutral point in a
star- connected system, one conductor of the secondary of a transformer etc.) to
earth (i.e., soil) is called grounding or earthing.
It is strange but true that grounding of electrical systems is less understood aspect
of power system. Nevertheless, it is a very important subject. If grounding is done
systematically in the line of the power system, we can effectively prevent accidents
and damage to the equipment of the power system and at the same time continuity
of supply can be maintained.
Grounding or earthing may be classified as: (i) Equipment grounding (ii) System
grounding. Equipment grounding deals with earthing the non-current-carrying metal
parts of the electrical equipment. On the other hand, system grounding means
earthing some part of the electrical system.
e.g., earthing of neutral point of star-connected system in generating stations and
sub- stations.
When the star point of star-connected load is isolated from neutral point of the
supply system, potential of the star point is subjected to variations according to
unbalance of load.
Under certain condition of loading, a considerable difference of potential may exist
between the star point of the load and neutral point. Such isolated star point is called
a “Floating star point”. All star connected loads supplied from poly-phase systems
without a neutral wire have floating star point.
Also as balancing of load is not in the control of the electricity supply boards (since
it is dependent on use of power by consumer), unbalancing is bound to occur; i.e.,
3-Ø currents IR + IY +IB. this unbalancing of the star point but also the variations of
not only potential of the star point but also the variation of the voltage of the 3 lines
with respect to star point of load.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
These voltages may assume values much higher or much lower than the rated
voltage atconsumer’s premise. The consumer getting supplies from less loaded
phase get higher voltage than the rated value. This can damage equipment’s like
T.V., Lamps etc.
But consumers getting supply from more loaded phase get lower voltage then the
rated one. This cause more current to be drawn from supply if IM is connected. This
high current can damage the induction motor.
To avoid such problem, it is common practice to connect the star point of load with
neutral point of system. Such a system is called 3-Ø, 4-wire system & is
extensively used for distribution of power.
In case of the floating star point VN = VR + VY + VB, where V n is voltage of displaced
star point with respect to neutral, & VR , VY , VB are phase voltage for such a 3-
Ø, 4 wire system, the unequal phase voltages can be plotted in an equilateral triangle
of the sides.Equal to rated line voltage.
For star point connected with neutral (3-Ø, 4-wire system), the 3 unbalanced
currentcan be added graphically to obtain magnitude & phase angle IN.
• Procedure :
PART-1 :- NEUTRAL CONNECTED WITH STAR POINT (3-Ø, 4-WIRE SYSTEM)
− Connect circuit. As shown in figure.
− Keep one way switch “ON”.
− Switch on the mains.
− Adjust some balance load with the help of rheostats.
− Note down 3-line current IR, IY & IB and 3-Ø voltage VR, VY &
VB usingvoltmeter selector switch.
− Observe that the voltage of the star point with respect to neutral of
system is zeroi.e., VN = 0
− Observe that neutral current IN = 0
− Adjust the condition of unbalanced load using variable rheostat.
− Note down IR, IY & IB and VR, VY & VB.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
− Note that VN = 0 but in = IR + IY + IB.
− A very important observation can be mode here that the phase voltages
are same and are equal to rated phase to neutral voltage even though
the load is not balanced. This is a very important advantage of 3-Ø, 4-
wire system. Obviously,requirement of phase to neutral voltage being
constant at consumer’s premises is satisfied in such a system of
distribution. Hence, such a system of distribution is always preferred at
the cost of 4th
wire connected right from distribution transformer to the
consumer’s premises.
− Draw vector diagram & prove the result. IN = IR + IY + IB graphically.
PART-2 :- Y-POINT ISOLATED FROM NEUTRAL OF SYSTEM (3-Ø, 4-WIRE SYSTEM)
− Make one-way switch off.
− Adjust balanced load.
− Note down values of IR, IY, IB & VR, VY, VB.
− Observe that the voltage of the star point with respect to neutral (VN) is
zero. The whole distribution system behaves in a similar manner to a 3-
Ø, 4-wire system with balance load.
− For unbalanced loads, however, voltage V n, of star point with respect
to neutral will assume some value & amp; also, VR, VY & VB are not
same. The constancy of voltage is not maintained. This is the
disadvantage of 3-Ø, 4-wire system, though thecost of one wire is saved.
− Vector diagram can be drawn for 3-line voltages from an equilateral
triangle. The arccan be drawn with vertices of this triangle as the centre
& amp; 3 observed phase voltages as radii, respectively. The 3 arcs will
intersect at a single point at which is floated neutral VN, then is the
voltage found graphically which is equal to length of line drawn from
neutral of balanced load condition to the new floated neutral.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
• MATLAB Circuit Diagram :
powergui
Scope
R1 10
Display 1
I_Y
Discrete
Fourier 1
R2
10
I_B R3
Display 2
R-ph Y-ph B-ph
10
Display 3
100
Display 4
Discrete
Fourier 4
100
Display 5
100
Constant
Display 6
Discrete
Fourier 6
Neutral Current
I_N
Ideal Switch
Neutral Current
Scope
Continuous

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
MATLAB Waveforms :
1. Balance Load with Neutral Open

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
3-Ph
Current
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
I_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
I_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
I_B
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-4
-2
0
2
4
x 10
-14
I_N
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
3-Ph
Voltage
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
V_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
V_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
Time (s)
V_B
2. Balance Load with Neutral Closed

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
0
10
3-Ph
Current
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
0
10
I_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
0
10
I_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
I_B
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
0
100
3-Ph
Voltage
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-50
0
50
V_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
0
100
V_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
0
100
Time (s)
V_B
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-4
-2
0
2
4
x 10
-4
I_N
3. Unbalanced Load with Neutral Open

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-20
-10
0
10
20
3-Ph
Current
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-20
-10
0
10
20
I_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
-5
0
5
10
I_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-5
0
5
I_B
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
3-Ph
Voltage
Waveform
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
V_R
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
V_Y
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-100
-50
0
50
100
Time (s)
V_B
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-10
0
10
I_N
4. Unbalanced Load with Neutral Closed

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
• Observation Table
1. Balanced Load with Neutral Open
2. Balanced Load with Neutral Closed
PHASE
Per Phase
Voltage
Per Phase Current
Resistance of
Load
R 100 10 10
Y 100 10 10
B 100 10 10
N -3.812*10^-18
3. Unbalanced Load with Neutral Open
PHASE
Per Phase
Voltage
Per Phase Current
Resistance of
Load
R 68.64 13.73 5
Y 113.5 11.335 10
B 125 8.332 15
N -7.837*10^-5
PHASE
Per Phase
Voltage
Per Phase Current
Resistance of
Load
R 100 10 10
Y 100 10 10
B 100 10 10
N 1.243*10^-18

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
4. Unbalanced Load with Neutral Closed
PHASE
Per Phase
Voltage
Per Phase Current
Resistance of
Load
R 99.99 20 5
Y 100 10 10
B 100 6.667 15
N -2.886
• Question - Answers
Why distribution side transformer connection is of delta-star? Why primary side
delta and secondary side star only? Describe in detail.
Ans : Main reason for delta star connection in distribution transformer is:
- Cross section area on Primary side is less due to delta connection.
- Generally, on power distribution side three phase four wire system is more
preferable as most domestic loads are unbalanced.
- Star connection on secondary provides a Neutral connection at the Star point
which gives unbalanced current a path through it, the phase/line voltage
applied to appliances remains as per requirement, and hence the better
performance is obtained.
- Neutral dose not shift and remains at ground potential so better voltage
regulation is obtained even in fault condition, on any phase, and complete
black out can be avoided.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
- Easy relaying of ground protection
- Protective relaying is easier on a delta-wye transformer because
ground faults on the secondary side are isolated from the primary,
making coordination much easier. If there is upstream relaying on a
delta-wye transformer, any zero-sequence current can be assumed to
be from a primary ground fault, allowing very sensitive ground fault
protection. Actually, ground fault protection is one of the primary
advantages of delta-wye units.
- Suppression of 3rd & multiples of 3rd order harmonics:
- The magnetizing current must contain odd harmonics for the induced
voltages to be sinusoidal and the third harmonic is the dominant
harmonic component.
- In a three-phase system the third harmonic currents of all three phases
are in phase with each other because they are zero-sequence currents
and they are able to circulate around the path formed by the delta
connected winding.
- Doesn’t allow Zero sequence unbalance voltage and current from LV to
HV.
- Any unbalance in the zero-sequence voltage or current in LV secondary
side doesn’t reflect on the HV Primary side.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
Draw phasor diagram of balance and unbalance load condition of 3-
phase system with equations.
The same current flows through phase winding as well as in the line
conductor as it is connected in series with the phase winding.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
Phase current and line current will be
We can therefore, remove the neutral conductor without affecting the
voltage or current in the circuit. This is only applicable for perfectly
balanced system i.e., system with identical load on each phase. And we
can call such system as BALANCED SYSTEM.
Properties of Balanced System :
⁻ Waveform is perfectly sinusoidal i.e. in-terms of magnitude
and phase shift of 120°
⁻ Current flowing through each phase is identical.
⁻ No current flows through the neutral.
⁻ Power loss is very low or not present.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
Properties of Unbalanced System :
⁻ Waveforms are disturbed in terms of magnitude & phase angle.
⁻ Current flowing through phases is not same.
⁻ Neutral in needed.
⁻ Power losses are more.
What is Neutral, Grounding and Earthing?
Ans : Neutral ;
⁻ Neutral wire carries the circuit back to the original power source. More
specifically, neutral wire brings the circuit to a ground or busbar usually
connected at the electrical panel.
⁻ This point is earthed. Ideally, in an AC system, neutral and earth should
be at the same potential.
⁻ Neutral Wire is always charged.
Grounding ;
⁻ In grounding, the current carrying parts are directly connected to the
ground. The grounding provides the return path for the leakage current
and hence protect the power system equipment from damage.
⁻ When the fault occurs in the equipment, the current in all the three
phases of the equipment become unbalance. The grounding discharges
the fault current to the ground and hence makes the system balance.
⁻ The grounding provides the great safety to the equipment and improves
the service reliability.
Earthing ;
⁻ Earthing means connecting the dead part (i.e. which doesn’t carry
current under normal condition) to the earth. Eg: Frames, Enclosures,
supports of Equipment’s.
⁻ When the fault occurs in the system, then the potential of the non-
current part of the equipment raises, and when any human or stray
animal touch the body of the equipment, then they may get shocked.
⁻ The earthing discharges the leakage current to the earth and hence avoid
the personnel from the electric shock. It also protects the equipment
from lightning strokes and provides the discharge path for the surge
arrester, gap and other devices.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
⁻ The earthing is achieved by connecting the parts of the installation to the
earth by using the earth conductor or earth electrode in intimate contact
with the soil placed with some distance below the ground level.
⁻ For grounding, the black colour wire is used, and for earthing the green
colour, the wire is used.
What is MCB & ELCB ?
Ans : MCB – Miniature Circuit Breaker
⁻ Miniature Circuit Breaker
⁻ It is an electrical safety device that serves to disconnect electricity when
over-current occurs, which may be caused by lightning stroke or short
circuit.
⁻ MCB trips only when there is surge that exceeds the limit of it.
⁻ MCB can’t save us when electrocuted. To get protection from
Electrocution we need to install ELCB.
ELCB – Earth Leakage Circuit Breaker
⁻ Earth leakage circuit breaker.
⁻ It is an electrical safety device for protection against Electric leakage.
⁻ It can also protect in case of short circuit and there is large surge.
⁻ As it provides human safety as well as protection from various
disturbances, it is better to use instead of MCB.
State difference between Fuse & Circuit Breaker
BASIS FUSE CIRCUIT BREAKER
Working Principle
Fuse works on the electrical
and thermal properties of
the conducting materials.
Circuit breaker works on the
Electromagnetism and
switching principle.
Reusability
Fuses can be used only
once.
Circuit breakers can be used
a number of times.
Status indication
It does not give any
indication.
It gives an indication of the
status
Auxiliary contact
No auxiliary contact is
required.
They are available with
auxiliary contact.
Switching Action
Fuse cannot be used as as
an ON/OFF switch.
The Circuit breaker is used
as an ON/OFF switches.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
Characteristic Curve
The Characteristic curve
shifts because of the ageing
effect.
The characteristic curve
does not shift.
Protection
The Fuse provides
protection against only
power overloads
Circuit breaker provides
protection against power
overloads and short circuits.
Function
It provides both detection
and interruption process.
Circuit breaker performs
only interruption. Faults are
detected by relay system.
Breaking capacity
Breaking capacity of the
fuse is low as compared to
the circuit breaker.
Breaking capacity is high.
Operating time
Operating time of fuse is
very less (0.002 seconds)
Operating time is
comparatively more than
that of the fuse. (0.02 – 0.05
seconds)
Mode of operation
Completely automatically. Manually as well as
automatically operated.
Cost
Cost of fuse is low. Cost of circuit breaker is
high.
Difference between balanced and unbalanced loads.
Ans : Balanced Load
⁻ Balanced load in 3 phase system is a condition where all three phases
(lines) carry same magnitude of current, with evenly spaced phase
difference. If the load is star connected, with neutral as return path,
this neutral will carry no current. This is so since vector sum of all
three phase currents meeting at common neutral point is zero i.e. IR
+IY +IB - IN =0
Unbalanced Loads
⁻ Unbalanced load makes the lines / phases to carry different current
magnitudes, and sum total of these at neutral point is not zero. Load
in each phase is different, carrying its own current. Neutral in this
case carries the net unbalanced current.
⁻ If there are reactive impedance components in load, there will be
phase difference in line currents from respective voltages, and even if
magnitude becomes same, the phase difference will result in
unbalanced current in neutral.

PRACTICAL – 1
FLOATING NEUTRAL
SG & P Practical - 1 180420109009
• Conclusion :
⁻ For balanced load with Neutral switch open and Neutral switch closed, the phase
voltages as well as load currents are identical so there is negligible current (Almost
0) passes through neutral i.e. if loads are balanced then floating neutral doesn’t
affect much.
⁻ For unbalanced load with Neutral switch closed, Voltage is same in all the phase
which is the advantage of using 3 phase, 4 wire system. In this case Some current
will flow through Neutral wire due to load imbalance.
⁻ For unbalanced load with Neutral switch open i.e. floating neutral, Phase voltage
is not constant due to unbalanced load, which is worst case for consumers as
Voltage fluctuation may cause damage to their loads and may also create faults &
hazards.
⁻ From this Practical we can conclude that Floating Neutral in an unbalanced system
can cause a lot of issues, so we should use 3-phase, 4 wire system with proper
neutral connection in which even unbalanced loads doesn’t affect the Phase
voltages and lead to a stable system operation.

PRACTICAL – 2
CT POLARITY
Submitted To :
July, 2021

PRACTICAL – 2
CT POLARITY
SG & P Practical - 2 180420109009
• Aim : To check the polarity of Current Transformer.
• Apparatus : Galvanometer, 9V battery, Current Transformer
• Procedure :
- CT polarity check is necessary to see the relative polarity of the primary and
secondary terminals of current transformer when terminals are not marked or to
establish the correction of the marking if already marked.
- A simple test circuit for this purpose is shown in the figure. The primary is
connected across a low voltage battery through push button switch.
- When press push button switch, galvanometer deflect in positive direction,
polarity of current transformer is correct.
• Steps :
- Remove all external connections before doing test.
- Connect analog galvanometer to secondary core winding (+ve to S1 and -ve to S2).
- Apply dc voltage (through cell) in impulse fashion (apply and remove) to primary
conductor (+ve to P1 and -ve to P2).
- Observe the deflection in meter.
- If the deflection is in clock wise then its polarity is correct otherwise polarity is
wrong.

PRACTICAL – 2
CT POLARITY
SG & P Practical - 2 180420109009
CT Terminal Images CT Rating

PRACTICAL – 2
CT POLARITY
SG & P Practical - 2 180420109009
• Question & Answers :
Q1. What is the importance of doing polarity test ?
- The importance of the polarity test is to make sure that all single-pole devices like
switches, circuit breakers, and fuses are allied only in the phase conductor.
- Also, it is necessary to see the relative polarity of the primary and secondary
terminals of current transformer when terminals are not marked or to establish the
correction of the marking if already marked.
- And polarity is very sensitive in case of differential and restricted earth fault
protection, so it is necessary to identify the polarity of Current Transformer.
Q2. Why galvanometer is used for test, not ammeter ?
Ammeter has accuracy more than the Galvanometer but we cannot see the
direction of the current in the ammeter. Galvanometer shows both, the direction
and the magnitude of the current. So, in our practical, as the direction of the flow
of current is also important to know we use galvanometer to measure current.
Q3. Why we apply DC voltage is in impulse fashion, not continuously ?
Pure DC voltage would generate a constant flux and we would need a time
varying flux for the development of the Transformer Action. To create a
momentary flux, we apply Impulsating DC voltage.
Q4. Why AC supply is not given for testing purpose ?
AC voltage cannot be applied because it changes its direction nearly constantly
so it would not help in detect the polarity. Pointer in galvanometer will
constantly oscillate and thus we would not be able to identify the polarity.

PRACTICAL – 2
CT POLARITY
SG & P Practical - 2 180420109009
Q5. How to identify the primary and secondary winding of CT without giving any
supply ?
The primary and secondary winding of Current Transformer can visually be
identified :
H1,H2 :- Indicates Primary Winding current where,
• H1 – Primary current, Line facing direction
• H2 – Primary current, Load facing direction
X1,X2 – Indicates Secondary Winding current
• Conclusion :
In this practical, we observed that the CT Polarity test is used to prove that the
predicted direction of secondary CT current(leaving) is correct for a given direction
of the Primary CT current (entering). Taking care to observe proper polarity is
important when installing and connecting current transformer to power metering
Protective relays. So, to reduce the risks of confusing polarity leading to mixing up
connections significantly, we perform CT polarity test before initial use.

PRACTICAL – 3
CT & PT
Submitted To :
October, 2021

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
• Aim : To study Current Transformer and Potential Transformer.
CURRENT TRANSFORMER
Define : A current transformer is a device that is used for the transformation of current
from a higher value into a proportionate current to a lower value. It transforms the high
voltage current into the low voltage current due to which the heavy current flows through
the transmission lines is safely monitored by the ammeter.
The current transformer is used with the AC instrument, meters or control apparatus
where the current to be measured is of such magnitude that the meter or instrument coil
cannot conveniently be made of sufficient current carrying capacity. The current
transformer is shown in the figure below.
The primary and secondary current of the current transformers are proportional to
each other. The current transformer is used for measuring the high voltage current
because of the difficulty of inadequate insulation in the meter itself. The current
transformer is used in meters for measuring the current up to 100 amperes.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Construction of CT :
The core of the current transformer is built up with lamination of silicon steel. The
primary windings of the current transformers carry the current which is to be measured,
and it is connected to the main circuit. The secondary windings of the transformer carry
the current proportional to the current to be measured, and it is connected to the
current windings of the meters or the instruments.
The primary and the secondary windings are insulated from the cores and each other.
The primary winding is a single turn winding (also called a bar primary) and carries the
full load current. The secondary winding of the transformers has a large number of turns.
The ratio of the primary current and the secondary current is known as a current
transformer ratio of the circuit. The current ratio of the transformer is usually high. The
secondary current ratings are of the order of 5A, 1A and 0.1A. The current primary
ratings vary from 10A to 3000A or more.
The working principle of the current transformer is slightly different from the power
transformer. In a current transformer, the load’s impedance or burden on the
secondary has slightly differed from the power transformers. Thus, the current
transformer operates on secondary circuit conditions.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Burden of CT :
The burden of a current transformer is the value of the load connected across the
secondary transformer. It is expressed as the output in volt-amperes (VA). The rated
burden is the value of the burden on the nameplate of the CT. The rated burden is the
product of the voltage and current on the secondary when the CT supplies the
instrument or relay with its maximum rated value of current.
Effect of Open Secondary Windings of a CT :
Under normal operating conditions the secondary winding of a CT is connected to its
burden, and it is always closed. When the current flows through the primary windings,
it always flows through secondary windings and amperes turns of each winding are
subsequently equal and opposite.
The secondary turns will be 1% and 2% less than the primary turns and the difference
being used in the magnetising core. Thus, if the secondary winding is opened and the
current flows through the primary windings, then there will be no demagnetizing flux
due to the secondary current.
Due to the absence of the counter ampere turns of the secondary, the unopposed
primary MMF will set up an abnormally high flux in the core. This flux will produce core
loss with subsequent heating, and a high voltage will be induced across the secondary
terminal.
This voltage caused the breakdown of the insulation and also the loss of accuracy in
the future may occur because the excessive MMF leaves the residual magnetism in the
core. Thus, the secondary of the CT may never be open when the primary is carrying
the current.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Phasor Diagram of Current Transformer
The phasor diagram of the current transformer is shown in the figure below. The main
flux is taken as a reference. The primary and secondary induced voltages are lagging
behind the main flux by 90º. The magnitude of the primary and secondary voltages
depends on the number of turns on the windings. The excitation current induces by the
components of magnetising and working current.
where, Is – secondary current
Es – secondary induced voltage
Ip -primary current
Ep – primary induced voltage
Kt – turn ratio, number of secondary turn/number of primary turn
I0 – excitation current
Im – magnetising current
Iw – working component
The secondary current lags behinds the secondary induced voltage by an angle θ. The
secondary current relocates to the primary side by reversing the secondary current and
multiply by the turn ratio. The current flows through the primary is the sum of the
exciting current I0 and the product of the turn ratio and secondary current Kt*Is.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Errors of Current Transformer :
The current transformer has two errors – ratio error and a phase angle error.
Current Ratio Errors – The current transformer is mainly due to the energy component
of excitation current and is given as
Ratio Error =
Kt Is − Ip
Ip
Where Ip is the primary current. Kt is the turn ratio and is the secondary current.
Phase Angle Error – In an ideal current transformer the vector angle between the
primary and reversed secondary current is zero. But in an actual current transformer,
there is a phase difference between the primary and the secondary current because
the primary current has also supplied the component of exciting current. Thus, the
difference between the two phases is termed as a phase angle error.
POTENTIAL TRANSFORMER
Define : The potential transformer may be defined as an instrument transformer
used for the transformation of voltage from a higher value to the lower value. This
transformer steps down the voltage to a safe limit value which can be easily measured
by the ordinary low voltage instrument like a voltmeter, wattmeter and watt-hour
meters, etc.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Construction of Potential Transformer :
The potential transformer is made with high-quality core operating at low flux density
so that the magnetising current is small. The terminal of the transformer should be
designed so that the variation of the voltage ratio with load is minimum and the phase
shift between the input and output voltage is also minimum.
The primary winding has a large number of turns, and the secondary winding has a
much small number of turns. For reducing the leakage reactance, the co-axial winding
is used in the potential transformer. The insulation cost is also reduced by dividing the
primary winding into the sections which reduced the insulation between the layers.
The potential transformer is connected in parallel with the circuit. The primary
windings of the potential transformer are directly connected to the power circuit
whose voltage is to be measured. The secondary terminals of the potential transformer
are connected to the measuring instrument like the voltmeter, wattmeter, etc. The
secondary windings of the potential transformer are magnetically coupled through the
magnetic circuit of the primary windings.
The primary terminal of the transformer is rated for 400V to several thousand volts,
and the secondary terminal is always rated for 400V. The ratio of the primary voltage
to the secondary voltage is termed as transformation ratio or turn ratio.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Burden of PT :
The burden is the total external volt-amp load on the secondary at rated secondary
voltage. The rated burden of a PT is a VA burden which must not be exceeded if the
transformer is to operate with its rated accuracy. The rated burden is indicated on the
name plate.
The limiting or maximum burden is the greatest VA load at which the potential
transformer will operate continuously without overheating its windings beyond the
permissible limits. This burden is several times greater than the rated burden.
Phasor Diagram of a Potential Transformer :
The phasor diagram of the potential transformer is shown in the figure below.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Where, Is – secondary current
Es – secondary induced emf
Vs – secondary terminal voltage
Rs – secondary winding resistance
Xs – secondary winding reactance
Ip – Primary current
Ep – primarily induced emf
Vp – primary terminal voltage
Rp – primary winding resistance
Xp – primary winding reactance
Kt – turn ratio
Io – excitation current
Im – magnetising component of Io
Iw – core loss component of Io
Φm – main flux
Β- phase angle error
The main flux is taken as a reference. In instrument transformer, the primary current
is the vector sum of the excitation current Io and the current equal to the reversal
secondary current Is multiplied by the ratio of 1/kt. The Vp is the voltage applied to the
primary terminal of the potential transformer.
The voltage drops due to resistance and reactance of primary winding due to primary
current is given by IpXp and IpRp. When the voltage drop subtracts from the primary
voltage of the potential transformer, the primarily induced emf will appear across the
terminals.
This primary emf of the transformer will transform into secondary winding by mutual
induction and converted into secondary induced emf Es. This emf will drop by the
secondary winding resistance and reactance, and the resultant voltage will appear
across the secondary terminal voltage, and it is denoted by Vs.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Errors of Potential Transformer :
In an ideal potential transformer, the primary and the secondary voltage is exactly
proportional to the primary voltage and exactly in phase opposition. But this cannot be
achieved practically due to the primary and secondary voltage drops. Thus, both the
primary and secondary voltage is introduced in the system.
Voltage Ratio Error – The voltage ratio error is expressed in regarding measured voltage,
and it is given by the formula as shown below.
Ratio Error =
Kn Vs − Vp
Vp
Where Kn is the nominal ratio, i.e., the ratio of the rated primary voltage and the rated
secondary voltage.
Phase Angle Error – The phase angle error is the error between the secondary terminal
voltage which is exactly in phase opposition with the primary terminal voltage.
The increases in the number of instruments in the relay connected to the secondary of the
potential transformer will increase the errors in the potential transformers.

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Difference Between CT & PT
COMPARISION CURRENT TRANSFORMER POTENTIAL TRANSFORMER
Definition
Transform the current form high
value to the low value.
Transform the voltage from high
value to the low value.
Circuit Symbol
Core
Usually built up with lamination
of silicon steel.
It is made up of with high quality
steel operating at low flux
densities.
Primary
Winding
It carries the current which is to
be measured.
It carries the current which is to
be measured.
Secondary
winding
It is Connected to the current
winding of the instrument.
It is connected to the meter or
instrument.
Connection
Connected in series with the
instrument.
Connected in parallel with the
instrument.
Primary Circuit Has a small number of turns. Has a large number of turns.
Secondary
Circuit
Has a large number of turns and
cannot be open Ckt.
Has a small number of turns and
can be open Ckt.
Range 5 A or 1 A 110 V
Burden
Does not depend on secondary
burden.
Depends on the secondary
burden.
Types
Two types (wound and close
core)
Two types (Electromagnetic and
Capacitor voltage)
Application
Measuring current and power,
monitoring the power grid
operation, for operating
protective relay.
Measurement power source,
operating protective relay

PRACTICAL – 3
CT & PT
SG & P Practical - 3 180420109009
Difference between Measuring CT and Protective CT
MEASURING C.T PROTECTING C.T
It is operated in the ankle point region.
It is operated over range between ankle
point and knee point.
It is accurate up to approximately 1.2
times its rated current.
It is transform primary current on to the
secondary side about 20 times is full load
current.
It saturates for current about 120% of its
rated value.
It can handle very large fault currents.
For some volt – ampere rating and
material.
It is required large cross sectional area of
core.
Small size. Large size.
Core material used in nickel – iron alloys
having low exciting ampere – turns per
unit length and low flux density.
Core material used in grain oriented steel
having very high saturation levels.

PRACTICAL – 4
RELAY TESTING BY
SECONDARY
INJECTION METHOD
Submitted To :
Date Of Performance : 23rd
July, 2021

PRACTICAL – 4
RELAY TESTING BY SECONDARY INJECTION
METHOD
SG & P Practical - 4 180420109009
AIM : To study relay testing by secondary injection method and calculate error of over
current relay.
APPARATUS : Secondary injection testing kit, over current relay, CDG 11 AREVA relay.
OPERATING PROCEDURE :
- Connect the circuit as per the diagram.
- Set desire value of plug setting and time setting of IDMT non directional relay.
- Calculate the value of injected current to relay (by using formula of PSM, plug
setting and CT ratio).
- Inject current in to relay by secondary injection kit.
- Observe tripping of relay and measure time which is displayed by timer of
secondary injection kit.
- Calculate time of relay operation by using relay characteristics for same plug
setting and time setting.
- Calculate percentage error of relay by formula.
- Similarly find out of percentage error of relay for different values of plug setting
CIRCUIT DIAGRAM :

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
Relay Testing Panel Zoomed view of IDMT Relay
Relay Testing Kit Over Current Relay (Inverse)
Relay Testing Panel
Connection

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
OBSERVATION TABLE :
Sr.
No.
Plug
Setting
TMS
Value of
injected
current
(If)
(A)
Pick-up
Current
(Ip)
(A)
PSM
Actual
Operating T
ime
(sec.)
Calculated
Time
(sec.)
Error
(%)
1. 0.5 1 3 0.5 6 3.9 3.85 1.28
2. 0.5 0.5 4 0.5 8 1.67 1.66 0.6
3. 0.5 0.2 5 0.5 10 0.6 0.6 0
4. 1 1 3 1 3 6.5 6.28 3.38
5. 1 0.5 4 1 4 2.5 2.49 0.4
6. 1 0.2 5 1 5 0.9 0.89 1.11
7. 1.25 1 3 1.25 2.4 8.5 7.25 14.70
8. 1.25 0.5 4 1.25 3.2 3 2.875 4.16
9. 1.25 0.2 5 1.25 4 1 1 0
Definite Time Over Current
Relay
Relay Testing Panel
Connection with Kit

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
CALCULATION (Analytical Method) :
a. For Reading – 1
b. For Reading – 2
c. For Reading – 3
Pick-up current = 0.5 × 1 = 0.5 A
PSM =
Injected current (If)
Pick−up current (Ip)
=
3
0.5
= 6
Operating Time of relay =
3 × TMS
log10 PSM
=
3 × 1
log10 6
= 3.85 sec
Error =
(3.9 − 3.85)
3.9
× 100 = 1.28%
PSM =
=
4
0.5
= 8
3 × TMS
log10 PSM
=
3 × 0.5
log10 8
= 1.66 sec
Error =
(1.67 − 1.66)
1.67
× 100 = 0.6%
PSM =
=
5
0.5
= 10
3 × TMS
log10 PSM
=
3 × 0.2
log10 10
= 0.6 sec
Error =
(0.6 − 0.6)
0.6
× 100 = 0%

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
d. For Reading – 4
e. For Reading – 5
CALCULATION (Graphical Method) :
Here, the manufacturer of this relay has provided us the characteristic of Operating Time
V/s PSM for TMS = 1.
So, we can calculate the operating time for different TMS values by multiplying the
given TMS with the operating time obtained for TMS=1.
∴ Operating Time (For new TMS) = (Value of New TMS) × (Operating time for TMS=1)
Pick-up current = 1 × 1 = 1 A
PSM =
=
3
1
= 3
3 × TMS
log10 PSM
=
3 × 1
log10 3
= 6.28 sec
Error =
(6.5 − 6.28)
6.5
× 100 = 3.38%
PSM =
=
4
1
= 4
3 × TMS
log10 PSM
=
3 × 0.5
log10 4
= 2.49 sec
Error =
(2.5 − 2.49)
2.5
× 100 = 0.4%

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
f. For Reading – 6
g. For Reading – 7
PSM =
=
5
1
= 5
TMS = 0.2
From operating time of relay V/s PSM graph:
Operating Time (For new TMS) = (Value of New TMS) × (Operating time for TMS=1)
∴ Operating Time = 0.2 × 4.45 = 0.89 sec
Error =
(0.9 − 0.89)
0.9
× 100 = 1.11%
PSM =
=
3
1.25
= 2.4
TMS = 1
∴ Operating Time = 1 × 7.25 = 7.25 sec
Error =
(8.5 − 7.25)
8.5
× 100 = 14.7%

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
h. For Reading – 8
i. For Reading – 9
PSM =
=
4
1.25
= 3.2
TMS = 0.5
∴ Operating Time = 0.5 × 5.75 = 2.875 sec
Error =
(3 − 2.875)
3
× 100 = 4.16%
PSM =
=
5
1.25
= 4
TMS = 0.2
∴ Operating Time = 0.2 × 5 = 1 sec
Error =
(1 − 1)
1
× 100 = 0%

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
CONCLUSION :
From this experiment we concluded that, for the same value of the Plug setting, the value
of injected current and PSM is increased, when the TMS and operating time of relay is
reduced.
QUESTION & ANSWERS
Q.1 Explain primary and secondary injection method. Give Application of both methods.
Primary Injection Method
- Primary injection verifying a protection scheme testing is essential in
commissioning and verifying a protection scheme since the secondary injection
test does not check all the components in the system and the condition of the
overall protection installation unlike primary injection testing, the secondary
injection tests do not simulate the real operating conditions, it would not reveal a
defective CT, whether that CT is the correct ratio or polarity, or whether the
secondary wiring is correct and serviceable.
- The primary injection testing is the only way to prove correct installation and
operation of the whole protection chain, as involves testing the entire circuit
- Current transformer primary and secondary windings, relays, trip and alarm
circuits, circuit breakers and all wiring are checked.
- Primary injection testing is often the last test performed in the commissioning and
maintenance process, or after any modification, as a functional test of the whole
system, not excluding the also needed secondary injection tests that must be
performed first.

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
Secondary Injection Method
- Secondary injection tests are performed to verify the correct operation of the
protection system with regard to everything that is connected to the secondary
of the transformer.
- The secondary protection system may include protective relays, auxiliary relay,
protection circuits, communications and control systems, meeting devices low –
voltage devices, alarm, etc.
- Accordingly, the secondary injection test covers not only the testing of protective
relays and their tripping settings, but also the checking of all associated circuits
and devices involved in the proper performance of the secondary protection
system.
- The secondary injection test involves disconnection of the device and/or circuits
its normal CT/PT connection and it might include, among others, the following
tests.
Q.2 Give definitions of TMS; Plug setting, PSM, Pickup current, Injected current.
TMS : The adjustment of travelling distance of an electromechanical relay is known as
time setting. These adjustments commonly known as time multiplier of relay.
The time setting dial is calibrated from 0 to 1 in steps of 0.1 / 0.05.
Plug Setting : This is given in terms of either ampere or percentage of relay – rated
current plug setting is the threshold above which the relay will start operating.

PRACTICAL – 4
METHOD
SG & P Practical - 4 180420109009
Relay rated current = C.T. secondary
Pick up current (Primary) = plug setting × CT (primary) Ratings
Pick up current (Secondary) = plug setting × CT (Secondary) Ratings
PSM – Plug setting multiplier of relay is ratio of fault current to pick-up current of the
relay.
PSM =
Fault current in relay coil (If)
=
Fault current in relay coil (If)
Plug setting ×Relay rated current
• Pick-up current – The minimum value at which relay starts operating is called the
pickup current. The relay does not operate and the circuit breaker is not operated
when the value of the relay current is less than the pickup value. Relay operates when
the value of the relay current becomes equal to or exceeds the pick-up value, the
circuit of the trip coil closes and the circuit breaker trips.
• Injected current – In general, injection current is the fault current that induces a
voltage on the internal structures higher than the normal operating voltage of the
device which can degrade the lifetime of the relay and other internal circuitry.

PRACTICAL – 5
SIMULATE FAULT IN RADIAL
FEEDER & RING MAIN
PROTECTION
Submitted To :
Date Of Performance : 16th
July, 2021

PRACTICAL – 5
SIMULATE FAULT IN RADIAL FEEDER &
RING MAIN PROTECTION
SG & P Practical - 5 180420109009
• Aim : To study and simulate fault in radial feeder and ring main protection scheme.
• Apparatus :
- Radial feeder and Ring main protection panel main components.
- Directional Over current relay: 4 Nos.
- Non-Directional over current relay: 2 Nos.
- Selector Switches, push button switches, rheostats, contactors: 12 Nos.
• Operating Procedure :
- Connect the circuit as per diagram.
- Select the Position 1 in TPDT Switch for radial protection.
RADIAL PROTECTION SYSTEM
a. Switch on the MCB-1(Main MCB).
b. Press all Green push button of main contactors.
c. Set the TMS & plug Setting of all Relays (1, 3 &5) as per study calculation.
d. To create a line to ground fault through rheostat, set the value of rheostat in such
e. a way that value of fault current is greater or equal to pick up current of Relays.
f. Fault in any section is created using switch (F1, F3, F5).
- The rheostats are connected between 1 & 2, 3 & 4, and 5 & 6 all represents feeder
resistance (set the resistance value of all rheostats connected between terminals 1
& 2, 3 & 4, and 5 & 6 in such a way that Fault current is higher than the pickup
current).
- Simulate the fault current in section 3 by switching on switch F5. (Before switch on
ensure that rheostats set for calculated value of resistance). Maximum fault
current should not me more than 10 Amperes.
- Similarly create fault in section 2 and 1 and measure time of operation.

PRACTICAL – 5
SG & P Practical - 5 180420109009
RING MAIN PROTECTION SYSTEM
- Set Selector Switch position 2 IN TPDT Switch and also switch on MCB 2 for ring
main protection scheme.
- Set plug setting and time setting of relay 1, relay 2, relay 3, relay 4, relay 5, relay
6 as per study calculation.
- Press all on push button of main contactors.
- Simulate the fault current in section 1 by switching on switch F1. (Before switch
on ensure that rheostats set for calculated value of resistance). Maximum fault
current should not me more than 10Amp.
- Observe the tripping of relay 1 and relay 2(From Relay 1 and Relay 2, which relay
would operate depends upon the TMS & plug setting of relay). Supply would
continue via MCB 2 from Right side to Left side.
Radial Transmission System
Ring Main Transmission System

PRACTICAL – 5
SG & P Practical - 5 180420109009
Power Circuit for Radial & Ring Main Protection Scheme
Control Circuit for Radial & Ring Main Protection Scheme

PRACTICAL – 5
SG & P Practical - 5 180420109009
CIRCUIT CONNECTION :

PRACTICAL – 5
SG & P Practical - 5 180420109009
OBSERVATION TABLE FOR RADIAL SYSTEM :
Assume, CT rating: 10/5; Assume PS = 0.5; Assume TMS of R5 = 0.1 and then calculate
time of R5. From this Calculate TMS of R3 and R1 by considering 0.5 second margin
between each relay. Pick up current = PS × CT rating.
Sr.
No.
Relay
Fault
created
at
Section
PS TMS PSM
Relay Tripping time during
fault %
Error
Fault
current
Value
Observed Calculated
1. R1 1 (F1) 0.5 0.3 4 - 1.5 Sec. - 10 A
2. R3 2 (F3) 0.5 0.2 4 - 1.0 Sec. - 10 A
3. R5 3 (F5) 0.5 0.1 4 - 0.498 Sec. - 10 A
OBSERVATION TABLE FOR RING MAIN SYSTEM :
Assume, CT rating: 10/5 ; PS = 0.5 ; Fault current 10 A ; TMS of R2 & R5 is 0.1
Sr. No. Relay PS TMS PSM
1. R1 0.5 0.3 4
2. R2 0.5 0.1 4
3. R3 0.5 0.2 4
4. R4 0.5 0.2 4
5. R5 0.5 0.1 4
6. R6 0.5 0.3 4
Sr.
No.
Fault
created at
Section
Relay operate during
fault
Relay Tripping time during
fault % Error
Fault
current
Value
Observed Calculated
1. 1 (F1)
R1 - 1.5 Sec. -
10 A
R2 - 0.498 Sec. -
2. 2 (F3)
R3 - 1.0 Sec. -
10 A
R4 - 1.0 Sec. -
3. 3 (F5)
R5 - 0.498 Sec. -
10 A
R6 - 1.5 Sec. -

PRACTICAL – 5
SG & P Practical - 5 180420109009
Pick-up current = PS × C.T rating
∴ Pick-up current = 0.5 × 5 = 2.5 A
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
TMS = 0.1
Operating Time of Relay =
3 ×TMS
log10 PSM
∴ Operating Time of R5 =
3 ×0.1
log10 4
= 0.498 Sec.
CALCULATION FOR RADIAL SYSTEM :
Assume, CT rating: 10/5; Assume PS = 0.5; Assume TMS of R5 = 0.1 and then calculate
time of R5. From this Calculate TMS of R3 and R1 by considering 0.5 second margin
between each relay.
1. For Relay R5
2. For Relay R3
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
Here, it is given that the time margin between two relays is of 0.5 Sec.
So, operating time of R3 will be 1.0 Sec.
3 ×TMS
log10 PSM
∴ 1 =
3 ×TMS
log10 4
∴ TMS = 0.2

PRACTICAL – 5
SG & P Practical - 5 180420109009
3. For Relay R1
CALCULATION FOR RADIAL MAIN SYSTEM :
Assume, CT rating: 10/5; PS = 0.5; Fault current 10 A; TMS of R2 and R5 is 0.1 Then find
for other relays.
1. For Relays R5 & R2
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
So, operating time of R1 will be 1.5 Sec.
3 ×TMS
log10 PSM
∴ 1.5 =
3 ×TMS
log10 4
∴ TMS = 0.3
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
TMS = 0.1
3 ×TMS
log10 PSM
∴ Operating Time of R5 and R2 =
3 ×0.1
log10 4
= 0.498 Sec.

PRACTICAL – 5
SG & P Practical - 5 180420109009
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
So, operating time of R3 and R4 will be 1.0 Sec.
3 ×TMS
log10 PSM
∴ 2 =
3 ×TMS
log10 4
∴ TMS = 0.2
PSM =
Fault Current
Pick−up Current
=
10
2.5
= 4
So, operating time of R1 and R6 will be 1.5 Sec.
3 ×TMS
log10 PSM
∴ 1.5 =
3 ×TMS
log10 4
∴ TMS = 0.3

PRACTICAL – 5
SG & P Practical - 5 180420109009
CONCLUSION : In this experiment, we observed the operation of the protective
relays during the fault conditions in their respective protected zones for both Radial
and Ring Main system.
Radial System : The consumers are dependent only on single feeder and single
distributor. Therefore, during any fault condition on the feeder or distributor side, the
power supply will cut off to the consumers who are at the other side of the fault, away
from the sub-station. Due to this limitation, this system is generally used for short
distances only.
Ring Main System : This system is more reliable as each distributor is fed via two
feeders. In the event of fault on any section of the feeder, the continuity of supply is
maintained. Suppose, if fault occurs at any point of section of the feeder, then that
section of the feeder can be isolated for repairs and the same time continuity of supply
is maintained to all the consumers via the second feeder.

PRACTICAL – 5
SG & P Practical - 5 180420109009
Q.1 Give comparisons between Radial and Ring Main System.
RADIAL SYSTEM RING MAIN SYSTEM
In this system, separate feeders radiate
from a single sub-station and feed the
distributors at one end only.
In this system, the primaries of distribution
transformers would form a loop and returns
to sub-station.
The consumers are dependent on a
single feeder and single distributor.
The system is very reliable as each
distributor is fed via two feeders.
In case of fault, on the feeder or
distributor cuts off supply to the
consumers who are on the side of the
fault away from the sub-station.
In the event of fault on any section of the
feeder, the continuity of supply is
maintained.
The consumers at the distant end of the
distributor would be subjected to serious
voltage fluctuations when the load on
the distributor changes.
There are less voltage fluctuations at
consumer’s terminals.

PRACTICAL – 6
SIMULATE FAULT
DIFFERENTIAL RELAY
Submitted To :
July, 2021

PRACTICAL – 6
SIMULATE FAULT DIFFERENTIAL RELAY
SG & P Practical - 6 180420109009
• Aim : To study and simulate external and internal fault of differential protection
scheme for 3-phase Transformer.
• Apparatus :
- 3 Phase transformer (415/220 Volt, 6kVA)
- AREVA DDT 32 relay, push button switches, Lamp indicators, Current Transformer
total Nos.-6, switches, rheostats.
• Operating Procedure :
- Connect the circuit as per the diagram.
- Switch on supply, press ON push button (transformer primary winding) and
also press ON push button (load side).
- Switch on switch of load trolley. (Ensure current in three phases should be
balance).
- Observe the differential relay would not give tripping in such normal loading
condition.
- Switch on two switches S4 and S5 and press ON push button for external fault
(before switch on ensure that rheostats position set for maximum resistance).
- Measure the value of fault current, and observe differential relay operation.
- Increase the fault current by decreasing resistance of rheostat, and measure
the fault current when differential relay give tripping.
- Switch on two switches S1 and S2 and press ON push button for internal
fault(before switch on ensure that rheostats position set for maximum
resistance).
- Observe the tripping of differential relay in the case of internal fault.

PRACTICAL – 6
SG & P Practical - 6 180420109009
• Connection Steps :
1. Source Side
- Make connection of R1, Y1, B1 of panel to the transformer R1, Y1, B1.
- Make connection of R2, Y2, B2 of panel to the transformer of R2, Y2, B2.
- On panel, connect terminal 3 with 4 and C with D.
- Connect terminal 1 of panel to fixed terminal of Rheostat. Connect Rheostat
variable terminal with ammeter (0-10A AC). Connect other terminal of
ammeter with 2 number terminal of panel.
2. Load Side
- Connect load side R,Y,B terminal of panel with ammeter and load trolley (1-
phase)
- Connect R terminal of panel with 0-10A ammeter. Connect ammeter other
terminal with P terminal of load trolley. Same steps follow for Y and B
terminal. Connect N terminal of all load trolley. (Y-connected load).

PRACTICAL – 6
SG & P Practical - 6 180420109009
• Circuit Connection :

PRACTICAL – 6
SG & P Practical - 6 180420109009
• Performance Steps :
1. For Internal Fault
- Make connection with 1 and 2 and disconnect terminal A and B of panel.
- Set 1 A current in all ammeters.
- Make load on, S1 and S2 on
- Change Rheostat value from max to min and observe reading of fault
current at the instant of tripping.
- Reset the flag.
2. For External Fault
- Make connection with A and B and disconnect terminal 1 and 2 of panel.
- Set 1 A current in all ammeters.
- Make load on, S4 and S5 on
- Change Rheostat value from max to min and observe reading of fault
current at the instant of tripping.
- Reset the flag.
• Observation Table :
For Internal
fault
Load Current
IL
Fault current
Ifmin
1. 3.08 3.15
For External
fault
Load Current
IL
Fault current
Ifmax
1. 2.04 3.81

PRACTICAL – 6
SG & P Practical - 6 180420109009

PRACTICAL – 6
SG & P Practical - 6 180420109009
• Conclusion :
From this Experiment, we have simulated external & internal fault of differential
protection scheme of 3 phases Transformer.
We have created fault in the system to check the operation of the relay. The relay
will not trip if there is external fault. After certain rage of value in external fault
the relay will still operate in internal circuit.
We have measured the fault current and calculate ICT ratio and percentage bias
ratio.

PRACTICAL – 7
INDUCTION MOTOR
PROTECTION SCHEME
Submitted To :
August, 2021

PRACTICAL – 7
INDUCTION MOTOR PROTECTION SCHEME
SG & P Practical - 7 180420109009
• Aim : To study Induction motor protection scheme and simulate phase to phase
and earth fault protection scheme of three phase induction motor.
• Apparatus :
- Protection panel main components.
- L&T make MPR 300 relay and L & T make MC 31A relay, contactors, push
button switches, current transformer 10/5 ratio: 3 Nos, Lamp indicator, MCB
- 3-Ø induction motor-alternator set.
• Procedure :
- Connect the circuit as per circuit diagram.
- Switch on the supply, press ON push button.
- Motor starter current can be controlled by auto transformer starter, (Initially
output voltage of auto transformer starter set for 40% of input voltage, after
motor starting it will be reset for 100% of input voltage)
- To simulate phase to phase fault, switch on switches of two phases (R & Y).
Before switch on ensure that rheostats position ser for maximum resistance.
Procedure for simulate phase to phase fault between other phases
- Connect rheostats between Y and B phases and by switching on switches of
phase Y and Phase B.
- Connect rheostats between R and B phases and by switching on switches of
phase R and Phase B.
- Measure the value of fault current and observe different relay operation.
- To simulate phase to ground fault switch on switches of R phase and ground
switch. Before switch on ensure that rheostats position ser for maximum
resistance.
- Observe the tripping of relays in the case of phase-to-phase fault and earth
(ground) fault.

PRACTICAL – 7
SG & P Practical - 7 180420109009
⚫ Power and Control Circuit :

PRACTICAL – 7
SG & P Practical - 7 180420109009
⚫ Observation Table :
1. Under Healthy Condition ,
Sr. No. Phase Current Amperes
1. IR 0.91
2. IY 0.95
3. IB 0.92
2. Under Line – to – Line Fault Condition ,
Time Lag
(Sec)
1. IR 0.886
1.47
2. IY 0.915
3. IB 0.959
Panel of 3-Ø induction motor protection scheme

PRACTICAL – 7
SG & P Practical - 7 180420109009
3. Under Line – to – Ground Fault ,
• Conclusion :
From this experiment we conclude that, 3 phase induction motor connected with
three phase supply through starter, MPR 300 relay and MC-31A relay. When line to
ground and line to line fault occurs then relay should be operated at selected
current value and protect to motor from fault.
Fault Current
If (A)
Time Lag
(Sec)
1. IR 0.943
2.868 0.92
2. IY 0.947
3. IB 0.923

PRACTICAL – 8
CT ERRORS
Submitted To :
August, 2021

PRACTICAL – 8
CT ERRORS
SG & P Practical - 8 180420109009
• Aim : To find out the C.T Errors.
• Apparatus : CT testing kit, Standard CT, Test CT, Burdens.
• Theory : The current transformer has two errors – ratio error and a phase angle
error.
Current Ratio Errors : The current transformer is mainly due to the energy component
of excitation current and is given as
Ratio Error =
n Is − Ip
Ip
Where Ip is the primary current. n is the turn ratio and is the secondary current.
Phase Angle Error – In an ideal current transformer the vector angle between the
primary and reversed secondary current is zero. But in an actual current transformer,
there is a phase difference between the primary and the secondary current because
the primary current has also supplied the component of exciting current. Thus, the
difference between the two phases is termed as a phase angle error.

PRACTICAL – 8
CT ERRORS
SG & P Practical - 8 180420109009
• Procedure :
- Connect the Secondary of the "Standard C.T." And "Test C.T." At the terminals
marked "STD" respectively.
- Connect the burden on the Secondary of the Standard C.T. On which
characteristics of the same are known at the terminals "BURDEN" normally these
are shorted.
- Connect the burden at which the "X" C.T is to be tested at the terminals burden.
- Connect the 3-pin plug top to the mains (230V ±10% 50 Hz).
- ON the power-ON (1) switch. The % current meter will display OO.
- Connect the primaries of Standard Transformer and X Transformer of same
current ratio in series to inject the same, current primary from the current source.
- Increase the current up to 10%. In case reverse polarity is found, interchange the
connection of any one of the C.T.
- The result will be obtained.
- Adjust various percentage of Primary current and obtain the reading.
⚫ Images :

PRACTICAL – 8
CT ERRORS
SG & P Practical - 8 180420109009
• Observation Table :
Sr.
No.
Primary Current
(Amp.)
Ratio Error of
Standard C.T
Ratio Error of Test
C.T
Phase angle error
of Test C.T
1. 2.7 5.3 5.2 -193.5
2. 5.05 10.0 9.8 -143.0
3. 10.39 20.6 20.7 -55.5
• Conclusion : In this experiment, we observed and found out the C.T errors i.e.,
Current ratio error and the phase angle error.

PRACTICAL – 9
ANALYSIS OF SG
FAILURE
Submitted To :
Date Of Performance : 3rd
September, 2021

PRACTICAL – 9
ANALYSIS OF SG FAILURE
SG & P Practical - 9 180420109009
• Aim : Analysis of top 5 switchgear failure, causes and how to avoid them.
As per electrical power system is growing in size and complexity in all sectors such
as Generation, Transmission, Distribution, and load system.
Types of faults like Short Circuit Conditions in the power system network result in
severe economic losses and reduced the reliability of the electrical system.
An electrical fault is an abnormal condition, caused by equipment failures such as
transformers and rotating machines, human errors, and environmental conditions.
These faults cause interruption to electric flows, equipment damages and even
cause the death of humans, bird, and animals.
There may be lot of probabilities of faults to appear in the power system network,
including lighting, wind, tree falling on lines, apparatus failure, etc.
The fault inception also involves in insulation failures and conducting path failures
which results short circuit and open circuit of conductors.

PRACTICAL – 9
SG & P Practical - 9 180420109009
1. Loose Connections :
Loose and faulty connections cause an increase of resistance at that localized point.
The increased resistance causes increased heat in accordance with Ohm’s law, P= I2
R
The increase in heat will escalate until complete thermal failure of the connection
occurs or the nearby insulation fails resulting in a fault. One major insurance carrier
estimate that approximately 25 percent of all electrical failures occur due to loose
connections.
Remedies – The solution to avoiding these types of failures is to per-form regular
infrared inspections of all switchgear. Infrared viewing ports should be installed and
medium-voltage switchgear should have ports that also pass ultraviolet light so that
corona cameras can be used to inspect for corona and surface partial discharge
activity. Figure shows a thermal image of a loose switchgear connection that could lead
to future failure if not repaired
Thermal Image of loose switchgear
connection
Failure due to loose connection

PRACTICAL – 9
SG & P Practical - 9 180420109009
2. Insulation Breakdown :
Medium-voltage insulation systems are much more complex due to the greater
voltage stresses that exist. Areas within the switchgear that are overstressed will
initially fail over a small portion of the insulation but will then escalate over time until
complete failure occurs. The most likely areas for these problems to occur are:
- Medium voltage jumper cable
- Switchgear bus support barrier
- Cable terminations
Remedies – Keeping low-voltage insulation failures in check mostly involves keeping
the insulation dry and clean and ensuring clearances are adequate.
The solution to preventing medium-voltage switchgear failures begins with utilizing the
hand-held partial discharge detector equipped with an ultrasonic sensor to detect
surface insulation defects and a transient earth voltage sensor to detect internal
insulation defects.
Unshielded Medium-
Voltage Jumper Cable
Termination Failure
Switchgear Bus
Support Barrier

PRACTICAL – 9
SG & P Practical - 9 180420109009
3. Moisture :
Water intrusion or immersion due to natural disasters or accidents can create instant
short circuits, long term insulation damage, and long term metallic component
corrosion, among other complications. Medium-voltage switchgear that is exposed to
high humidity conditions will absorb moisture, and voltage stresses will attack the
hydrophobic insulation surfaces which were designed to inhibit moisture absorption.
Remedies – For medium-voltage switchgear, using the partial dis-charge detector
described above will prevent long term moisture-related insulation faults while the
infrared camera can detect abnormal heating of corroded connections.
The upper left photo shows good
insulation with hydro-phobic
surface qualities that progressively
worsen until the insulation
becomes hydrophilic in the lower
right photo due to voltage stresses.

PRACTICAL – 9
SG & P Practical - 9 180420109009
4. Breaker Racking:
Racking in a closed-circuit breaker onto an energized bus can quickly cause severe
personal injury or death and immediate severe equipment damage. Additionally, the
breaker may not always line up properly or may encounter other difficulties as it is
being racked, and these problems can cause a sudden severe fault. Unfortunately, the
traditional act of breaker racking requires personnel to manually perform this task
directly in harm’s way.
Remedies – The solution to this problem is to always make sure that mechanical and
electrical interlocks are functional and all breaker and cell components are properly
inspected and serviced. To ensure personnel safety, strong consideration should be
given to employment of a remote circuit breaker racking device such as shown in
figure.
Severe Breaker Damage Caused by Improper Rack-In Operation

PRACTICAL – 9
SG & P Practical - 9 180420109009
5. Faulty ground fault protection:
Unlike the items above, a defective ground fault protective device will not create a fault
itself. However, it will not offer protection from an arcing ground fault which is a
common failure mode in solidly-grounded switchgear. These types of faults are very
destructive but do not draw high enough currents to trip breakers or cause fuses to
open and can persist until catastrophic failure of the switchgear occurs.
Remedies – The solution to this problem requires an outage and manually testing the
ground fault protection system by current injection. Just as important is to pay close
attention to ensure that the equipment is properly installed. Sensor polarities must be
tested when applicable and the neutral ground connection must be located in the
correct position so that the sensors will detect fault currents properly.
Severe Damage from Arcing Ground Fault

PRACTICAL – 10
ANALYSIS IN 3 Φ
SYSTEM
Submitted To :
October, 2021

PRACTICAL – 10
ANALYSIS IN 3 Φ SYSTEM
SG & P Practical - 10 180420109009
• Aim : Fault analysis in 3 phase system using MATLAB Simulink.
Symmetrical & Asymmetrical Fault :
MATLAB Model :
Asymmetrical Fault between Phase A & C :

PRACTICAL – 10
SG & P Practical - 10 180420109009
Waveforms :
Symmetrical Fault :

PRACTICAL – 10
SG & P Practical - 10 180420109009
Waveforms :
Fault is symmetric, so the voltage in all 3 phase is 0 during fault
condition.
Fault Analysis using OC relay :

PRACTICAL – 10
SG & P Practical - 10 180420109009
Fault Analysis Vabc

PRACTICAL – 10
SG & P Practical - 10 180420109009
Fault Analysis Iabc :

PRACTICAL – 10
SG & P Practical - 10 180420109009
Conclusion :
In this practical, we simulated Symmetrical and asymmetrical fault in 3 phase
system using Matlab where we observed that during Symmetrical faults, Current
flows through all the 3 phases while voltage is 0 during fault.
For asymmetrical faults, healthy phase has no current flowing through it and
Voltage is also normal during fault.

PRACTICAL – 11
DIFFERENT CIRCUIT
BREAKERS
Submitted To :
Date Of Performance :

PRACTICAL – 11
DIFFERENT CIRCUIT BREAKERS
SG & P Practical - 11 180420109009
• Aim : To study Air, Vacuum and SF6 Circuit Breakers.
AIR CIRCUIT BREAKER
In air break circuit breaker, the arc is initiated and extinguish in substantially static air in
which the arc moves. Such breakers are used for low voltages, generally up to 15KV and
rupturing capacities of 500MVA. Air circuit breaker has several advantages over the oil, as
an arc quenching medium.
These are,
- Elimination of risk and maintenance associated with the use of oil.
- The absence of mechanical stress that is set up by gas pressure and oil movement.
- Elimination of the cost of regular oil replacement that arises due to deterioration
of oil with the successive breaking operation.
In the air break, circuit breaker the contact separation and arc extinction take place in air
at atmospheric pressure. In air break circuit breaker high resistance principle is employed.
In this circuit breaker arc is expanded by the mean of arc runners, arc chutes, and arc
resistance is increased by splitting, cooling and lengthening.
The arc resistance is increased to such an extent that the voltage drop across the arc
becomes more than the system voltage, and the arc gets extinguished at the current zero
of AC wave.
Air break circuit breakers are employed in DC circuits and Ac circuits up to 12,000 voltages.
Such breakers are usually of indoor type and installed on vertical panels or indoor draw
out switch gear. AC circuit breakers are widely employed indoor medium voltage and low
voltage switchgear.

PRACTICAL – 11
SG & P Practical - 11 180420109009
1. Plain Break Type Air Break Circuit Breaker
It is the simplest one in which contacts
are made in the shape of two horns. The
air initially strikes across the shortest
distance between the horns and is
driven steadily upwards by the
convection currents caused by heating
of air during arcing and the interaction
of the magnetic and the electric fields. The arc extends from one tip to the other
when the horns are fully separated resulting in lengthening and cooling arc.
The relative slowness of the process and the possibility of arc spreading of
adjacent metal works limits the application of about 500V and too low power
circuits.
2. Magnetic Blow-Out Type Air Break Circuit Breaker
Some air circuit breakers are used in the circuits
having voltage up to 11 KV, the arc extinction is
accomplished using magnetic field provided by the
current in blowout coils connected in series with the
circuit being interrupted. Such coils are called blow
out the coil. The magnetic field itself does not
extinguish the arc. It simply moves the arc into chutes
where the arc is lengthened, cooled and
extinguished. The arc shields prevent arc spreading
to an adjacent network.
It is important to connect the coils at correct polarity so that the arc is directed
upwards. As the breaking action becomes more effective with large currents, this
principle has resulted in increasing the rupturing capacities of such breakers to
higher values. Arc chute is an efficient device for arc extinction in air and performs
the following three interrelated functions.
- It confines the arc within a restricted space.
- It provides magnetic control over the arc movement so as to make arc extinction
within the devices.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Air Chute Air Break Circuit Breaker :
The normal arrangement of air-chute air break circuit breaker employed for low and
medium voltage circuits is shown in the figure below. There are two sets of contacts called
the main contacts and arcing or auxiliary contacts. Main contacts are usually of copper and
conduct the current in the closed position of the breakers. They have low contact
resistance and are silver plated.
The arcing contacts are hard, heat resistant and usually of copper alloy. Arcing contacts are
used to relieve the main contacts from damage due to arcing. The arcing contacts are easily
renewable when required. The auxiliary and arcing contacts close before and open after
the main contacts during the operation.
Here the blowouts consist of a steel insert in the arcing chutes. These are so arranged that
the magnetic field induced in them by the current in the arc moves it upwards still faster.
The steel plates divide the arc into a number of arcs in the series.
The distribution of voltage along the arc length is not linear, but it is accompanied by a
rather large anode and cathode drops. In case the total sum of anode and drops of all the
short arcs in series exceeds the system voltage, conditions for the quick extinction of the
arc are automatically established.
When the contact has come in contact with the relatively cool surfaces of the steel plants
gets rapidly and effectively cooled. The movement of the arc may be naturally or aided by
a magnetic blowout. Thus, the arc is extinguished by lightening and increasing the power
loss of the arc.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Working Principle Air Break Circuit Breaker :
When the fault occurs, the main contacts are separate first, and the current is shifted to
the arcing contacts. Now the arcing contacts are separate, and the arc is drawn between
them. This arc is forced upwards by the electromagnetic forces and thermal action. The arc
ends travel along the arc runner. The arc moves upward and is split by the arc splitter
plates. The arc is extinguished by lengthening, cooling, splitting, etc.
Applications of Air Break Circuit Breaker :
Air break circuit breaker is suitable for the control of power station auxiliaries and
industrial plants. They do not require any additional equipment such as compressors, etc.
They are mainly used in a place where there are possibilities of fire or explosion hazards.
Air break principle of lengthening of the arc, arc runners magnetic blow-up is employed for
DC circuit breakers up to 15 KV.
Drawback of Air Break Circuit Breaker :
A drawback of arc chute principle is its inefficiency at low currents where the
electromagnetic fields are weak. The chute itself is not necessarily less efficient in its
lengthening and deionizing action than at high currents, but the arc movement into the
chute tends to become slower, and high-speed interruption is not necessarily obtained.

PRACTICAL – 11
SG & P Practical - 11 180420109009
VACCUM CIRCUIT BREAKER
A breaker which used vacuum as an arc extinction medium is called a vacuum circuit
breaker. In this circuit breaker, the fixed and moving contact is enclosed in a permanently
sealed vacuum interrupter. The arc is extinct as the contacts are separated in high vacuum.
It is mainly used for medium voltage ranging from 11 KV to 33 KV.
Vacuum circuit breaker has a high insulating medium for arc extinction as compared to the
other circuit breaker. The pressure inside the vacuum interrupter is approximately 10-4
torrent and at this pressure, very few molecules are present in the interrupter.
The vacuum circuit breaker has mainly two phenomenal properties.
- High insulating strength: In comparison to various other insulating media used
in circuit breaker vacuum is a superior dielectric medium. It is better than all
other media except air and SF6, which are employed at high pressure.
- When an arc is opened by moving apart the contacts in a vacuum, an
interruption occurs at the first current zero. With the arc interruption, their
dielectric strength increases up to a rate of thousands time as compared to other
breakers.
The above two properties make the breakers more efficient, less bulky and cheaper in cost.
Their service life is also much greater than any other circuit breaker, and almost no
maintenance is required.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Construction of VCB :
It is very simple in construction as compared to any other circuit breaker. Their
construction is mainly divided into three parts, i.e., fixed contacts, moving contact and arc
shield which is placed inside the arc interrupting chamber.
The outer envelope of vacuum circuit breaker is made up of glass because the glass
envelope help in the examination of the breaker from outside after the operation. If the
glass becomes milky from its original finish of silvery mirror, then it indicates that the
breaker is losing vacuum.
The fixed and moving contacts of the breaker are placed inside the arc shield. The pressure
in a vacuum interrupter at the time of sealing off is kept at about 10^(-6)torr. The moving
contacts of the circuit breaker are move through a distance of 5 to 10 mm depending upon
the operating voltage.
The metallic bellows made of stainless steel is used to move the moving contacts. The
design of the metallic bellows is very important because the life of the vacuum circuit
breaker depends on the ability of the component to perform repeated operations
satisfactorily.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Working Vacuum Circuit Breaker :
When the fault occurs in the system, the contacts of the breaker are moved apart and
hence the arc is developed between them. When the current carrying contacts are pulled
apart, the temperature of their connecting parts is very high due to which ionization
occurs. Due to the ionization, the contact space is filled with vapour of positive ions which
is discharged from the contact material.
The density of vapour depends on the current in the arcing. Due to the decreasing mode
of current wave their rate of release of vapour fall and after the current zero, the medium
regains its dielectric strength provided vapour density around the contacts reduced.
Hence, the arc does not restrike again because the metal vapour is quickly removed from
the contact zone.
Current Chopping in Vacuum Circuit Breaker :
Current chopping in vacuum circuit breaker depends on the vapour pressure and the
electron emission properties of the contact material. The chopping level is also influenced
by the thermal conductivity–lower the thermal conductivity, lower is the chopping level.
It is possible to reduce the current level at which chopping occurs by selecting a contact
material which gives out sufficient metal vapour to allow the current to come to a very low
value or zero value, but this is rarely done as it affects the dielectric strength adversely.
Vacuum Arc recovery of Vacuum Circuit Breaker :
High vacuum possesses extremely high dielectric strength. At zero current the arc is
extinguished very quickly, and the dielectric strength is established very quickly. This return
of dielectric strength is because of the vaporized metal which is localized between the
contacts diffuses rapidly due to the absence of gas molecules. After arc interruption, the
recovery strength during the first few microseconds is 1 kV/µs second for an arc current of
100A.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Property of contact material :
The contact material of the vacuum circuit breaker should have the following property.
The material should have high electrical conductivity so as to pass normal load currents
without overheating.
- The contact material should have low resistance and high density.
- The material should possess high thermal conductivity so as to dissipate rapidly
the large heat generated during arcing.
- The material should have a high arc withstand ability and low current chopping
level.
Advantages of Vacuum Circuit Breaker :
- Vacuum circuit breaker does not require any additional filling of oil or gas. They do
not need periodic refilling.
- Rapid recovery of high dielectric strength on current interruptions that only a half
cycle or less arcing occurs after proper contact separation.
- Breaker unit is compact and self-contained. It can be installed in any required
orientation.
- Because of the above reasons together with the economic advantage offered,
vacuum circuit breaker has high acceptance.
Disadvantage of Vacuum Circuit Breaker :
- Requirements of high technology for production of vacuum interrupters.
- It needs additional surge suppressors for the interruption of low magnetizing
currents in a certain range.
- Loss of vacuum due to transit damage or failure makes the entire interrupter useless,
and it cannot be repaired on site.

PRACTICAL – 11
SG & P Practical - 11 180420109009
SULPHUR HEXAFLUORIDE (SF6) CIRCUIT BREAKER :
A circuit breaker in which SF6 under pressure gas is used to extinguish the arc is called
SF6 circuit breaker. SF6 (sulphur hexafluoride) gas has excellent dielectric, arc quenching,
chemical and other physical properties which have proved its superiority over other arc
quenching mediums such as oil or air. The SF6 circuit breaker is mainly divided into three
types.
- Non-puffer piston circuit breaker
- Single- puffer piston circuit breaker.
- Double-puffer piston circuit breaker.
The circuit breaker which used air and oil as an insulating medium, their arc extinguishing
force builds up was relatively slow after the movement of contact separation. In the case
of high voltage circuit breakers quick arc extinction properties are used which require less
time for quick recovery, voltage builds up. SF6 circuit breakers have good properties in
these regards compared to oil or air circuit breakers. So in high voltage up to 760 kV, SF6
circuit breakers is used.
Properties of Sulphur hexafluoride :
- It is colourless, odourless, non-toxic, and non-inflammable gas.
- SF6 gas is extremely stable and inert, and its density is five times that of air.
- It has high thermal conductivity better than that of air and assists in better cooling
current carrying parts.
- SF6 gas is strongly electronegative, which means the free electrons are easily removed
from discharge by the formation of negative ions.
- It has a unique property of fast recombination after the source energising spark is
removed. It is 100 times more effective as compared to arc quenching medium.
- Its dielectric strength is 2.5 times than that of air and 30% less than that of the dielectric
oil. At high pressure the dielectric strength of the gas increases.
- Moisture is very harmful to SF6 circuit breaker. Due to a combination of humidity and
SF6 gas, hydrogen fluoride is formed (when the arc is interrupted) which can attack the
parts of the circuit breakers.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Construction of SF6 Circuit Breakers :
SF6 circuit breakers mainly consist of two parts, namely (a) the interrupter unit and (b)
the gas system.
Interrupter Unit – This unit consists of moving and fixed contacts comprising a set of
current-carrying parts and an arcing probe. It is connected to the SF6 gas reservoir. This
unit consists slide vents in the moving contacts which permit the high-pressure gas into
the main tank.
Gas System – The closed-circuit gas system is employed in SF6 circuit breakers. The SF6
gas is costly, so it is reclaimed after each operation. This unit consists low and high-
pressure chambers with a low-pressure alarm along with warning switches. When the
pressure of the gas is very low due to which the dielectric strength of gases decreases
and an arc quenching ability of the breakers is endangered, then this system gives the
warning alarm.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Working Principle of SF6 Circuit Breaker :
- In the normal operating conditions, the contacts of the breaker are closed. When
the fault occurs in the system, the contacts are pulled apart, and an arc is struck
between them. The displacement of the moving contacts is synchronised with the
valve which enters the high-pressure SF6 gas in the arc interrupting chamber at a
pressure of about 16kg/cm^2.
- The SF6 gas absorbs the free electrons in the arc path and forms ions which do not
act as a charge carrier. These ions increase the dielectric strength of the gas and
hence the arc is extinguished. This process reduces the pressure of the SF6 gas up to
3kg/cm^2 thus; it is stored in the low-pressure reservoir. This low-pressure gas is
pulled back to the high-pressure reservoir for re-use.
- Now a day puffer piston pressure is used for generating arc quenching pressure
during an opening operation by mean of a piston attached to the moving contacts.
Advantage of SF6 circuit breaker :
- SF6 gas has excellent insulating, arc extinguishing and many other properties which
are the greatest advantages of SF6 circuit breakers.
- The gas is non-inflammable and chemically stable. Their decomposition products
are non-explosive and hence there is no risk of fire or explosion.
- Electric clearance is very much reduced because of the high dielectric strength of
SF6.
- Its performance is not affected due to variations in atmospheric condition.
- It gives noiseless operation, and there is no over voltage problem because the arc
is extinguished at natural current zero.
- There is no reduction in dielectric strength because no carbon particles are formed
during arcing.
- It requires less maintenance and no costly compressed air system is required.
- SF6 performs various duties like clearing short-line faults, switching, opening
unloaded transmission lines, and transformer reactor, etc. without any problem.

PRACTICAL – 11
SG & P Practical - 11 180420109009
Disadvantages of SF6 circuit breakers :
- SF6 gas is suffocating to some extent. In the case of leakage in the breaker tank, the
SF6 gas being heavier than air and hence SF6 are settled in the surroundings and
lead to the suffocation of the operating personnel.
- The entrance of moisture in the SF6 breaker tank is very harmful to the breaker, and
it causes several failures.
- The internal parts need cleaning during periodic maintenance under clean and dry
environment.
- The special facility requires for transportation and maintenance of quality of gas.

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