MAHARASHTRA STATE BOARD OF TECHNICAL...

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

Subject Code : 12143 (SAP)

Important Instructions to examiners:

1) The answers should be examined by key words and not as word

scheme.

2) The model answer and the answer written by candidate may vary but the examiner may should assess

the understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more importance

applicable for subject English and Communication Skills).

4) While assessing figures, examiner may give credit for principal

The figures drawn by candidate and model answer may vary. The examiner may give credit for any

equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant value

may vary and there may be some difference in the candidate’s answers and model answer.

6) In case of some questions credit may be given by judgement on part of examiner of relevant answer

based on candidate’s understanding.

7) For programming language papers, credit may be given to any other program based o on equivalent

concept.


(Autonomous)

(ISO/IEC-27001-2005 Certified)

Summer – 2014 Examinations

Subject Code : 12143 (SAP) Model Answer

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model answer


anding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more importance

applicable for subject English and Communication Skills).

4) While assessing figures, examiner may give credit for principal components indicated in the figure.


5) Credits may be given step wise for numerical problems. In some cases, the assumed constant value



based on candidate’s understanding.

pers, credit may be given to any other program based o on equivalent


Page No : 1 of 24

word as given in the model answer


3) The language errors such as grammatical, spelling errors should not be given more importance (Not

components indicated in the figure.


5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values



pers, credit may be given to any other program based o on equivalent



1 a) Attempt any three:

1 a) i) State importance of following switchgear elements of protective elements.

1) Fuse

2) Isolator

3) Circuit breaker

4) Relay

Ans:

Fuse: provides short circuit protection.

Isolator: Theses are used in addition with circuit breakers and are provided on each side

of every C.B. to provide isolation and enable maintenance,

isolate a power system element such as transformer etc

only after operating the CB.

Circuit breaker: It performs interruption function. Its function is to make or break the

circuit manually or remotely under normal condition and automatically under fault

condition, used to switch and bring the power system components in circuit, protect the

related sections from any abnormal conditions of operation such as overload, over voltage,

overcurrent, earth faults, phase to phase faults, under frequency etc.

Relay: used to give suitable signals for certain

overcurrent, faults, switching etc. to circuit breaker mechanism for operation.

1 a) ii) Explain with neat sketch the action of arc runners and arc splitters in Air

Ans:

Arc runners:

Fall in the high resistance method of arc extinction:

- Arc path resistance is increased to reduce the current to low values while

interrupting the arc. Arc resistance = v

The arc resistance is increased by lengthening of the

1 – initial position of arc, 2 – final position of arc, 3

4 - Electromagnetic force, 5 –


(Autonomous)



Subject Code : 12143 (SAP) Model Answer Page No :

State importance of following switchgear elements of protective elements.

circuit protection.

Theses are used in addition with circuit breakers and are provided on each side

of every C.B. to provide isolation and enable maintenance, it’s a no load switch, used to

isolate a power system element such as transformer etc. via circuit breaker and operated

It performs interruption function. Its function is to make or break the


switch and bring the power system components in circuit, protect the


overcurrent, earth faults, phase to phase faults, under frequency etc.

table signals for certain conditions such as low/no voltage,


Explain with neat sketch the action of arc runners and arc splitters in Air CB.

Fall in the high resistance method of arc extinction:

Arc path resistance is increased to reduce the current to low values while

interrupting the arc. Arc resistance = varc/iarc.

increased by lengthening of the arc over runners:

final position of arc, 3 – arc runner,

– force due to field.


Page No : 2 of 24

12

Theses are used in addition with circuit breakers and are provided on each side

it’s a no load switch, used to

. via circuit breaker and operated

It performs interruption function. Its function is to make or break the


switch and bring the power system components in circuit, protect the


conditions such as low/no voltage,


1 mark

each = 4

marks

CB.

Arc path resistance is increased to reduce the current to low values while

1 mark

diagram

and 1

mark

descriptio

n = 2

marks



Arc runners are horn like blades of conducting materials connected to arcing contacts

of CB with their tips radiating upward in ‘V’ shape. The arc on getting struck is blown

over the arc runners upwards by electromagnetic force which results in increase in arc

resistance due to which the arc is extinguished.

Arc splitters:

1 – arc runner, 2- arc splitters,

5 – blow out coil, 6 – arc origin.

Plates of resin bonded fibre glass (arc splitters) placed in path of arc (perpendicular to arc

length). Arc is pulled over the plates by electromagnetic force leading to its elonga

splitting, cooling and finally extinction.

1 a) iii) List the various steps for calculating the actual relay operating ti

Ans:

Expected points:

The following quantities need to be known to calculate the actual relay operating time

1) Time versus Plug Setting Multiplier relationship (graphical curve) (time/PSM

graph)

2) Current Setting (CS) of relay.

3) Time setting on relay.

4) Fault current (IF)

5) CT ratio (K).

The procedure to calculate the actual operating time of relay is as given below:

a) Transform the fault current to the relay coil current by the CT ratio (I

b) Express the relay current I’

I’R = IR /CS = IF/[CS * K] which is the plug setting multiplier PSM to be done.

c) Referring the time/PSM graph note the time ‘t

PSM (I’R).

d) Then the actual time of relay operation T

1 a) iv) List out all important faults which occur on alternator.


(Autonomous)





radiating upward in ‘V’ shape. The arc on getting struck is blown


resistance due to which the arc is extinguished.

arc splitters, 3 – elongated arc, 4 – travelling arc,

arc origin.


Arc is pulled over the plates by electromagnetic force leading to its elonga

splitting, cooling and finally extinction.

List the various steps for calculating the actual relay operating time.

The following quantities need to be known to calculate the actual relay operating time

Time versus Plug Setting Multiplier relationship (graphical curve) (time/PSM

Current Setting (CS) of relay.


Transform the fault current to the relay coil current by the CT ratio (I

Express the relay current I’R as a multiple of current setting (CS) as

/[CS * K] which is the plug setting multiplier PSM to be done.

Referring the time/PSM graph note the time ‘tO’ of operation for the calculated

Then the actual time of relay operation TO = tO x (Time setting on relay

List out all important faults which occur on alternator.


Page No : 3 of 24


radiating upward in ‘V’ shape. The arc on getting struck is blown



Arc is pulled over the plates by electromagnetic force leading to its elongation,

1 mark

diagram

and 1

mark

descriptio

n = 2

marks

The following quantities need to be known to calculate the actual relay operating time;

Time versus Plug Setting Multiplier relationship (graphical curve) (time/PSM


Transform the fault current to the relay coil current by the CT ratio (IR = IF/K).

g (CS) as

/[CS * K] which is the plug setting multiplier PSM to be done.

’ of operation for the calculated

(Time setting on relay).

All items

1 mark

1 mark

1 mark

½ mark

½ mark



Ans:

Under frequency

Rotor earth fault.

Over voltages

Thermal over loading

Alternator bus bar faults as short circuit.

Stator winding earth fault.

Stator winding turn short fault

Stator winding phase to phase fault

Field system open circuit

1 b) Attempt any one:

1 b) i) Explain with neat diagram construction and working of vacuum CB

applications of VCB.

Ans:

Vacuum circuit breaker:

Operation: (minimum points)

On operation of the breaker the moving contact separates from the fixed contact resulting

in arching between them. The arc consists of metal ions of surface of

arc gets extinguished quickly and vacuum has good recovery of dielectric strength. The

arc extinction occurs at a small vacuum gap of about 0.6 to 0.7 cm.

Applications: used in conditions where maintenance free circuit breakers are re

as

1)remote areas,

2)employed for outdoor applications from 22 kV to 66kV

3) preferred for indoor switchgear up to 36 kV, 750 MVA.

1 b) ii) Draw block diagram of Static over


(Autonomous)




as short circuit.

Stator winding turn short fault

Stator winding phase to phase fault

Explain with neat diagram construction and working of vacuum CB. State any two


in arching between them. The arc consists of metal ions of surface of the contacts. The


arc extinction occurs at a small vacuum gap of about 0.6 to 0.7 cm.

Applications: used in conditions where maintenance free circuit breakers are re

2)employed for outdoor applications from 22 kV to 66kV

3) preferred for indoor switchgear up to 36 kV, 750 MVA.

Draw block diagram of Static over current relay and explain its working.


Page No : 4 of 24

½ mark

each any

8 faults =

4 marks

6 marks

. State any two


the contacts. The


Applications: used in conditions where maintenance free circuit breakers are required such

2 marks

2 marks

Any two

=

2 marks



Ans:

Fig. illustrates simplified functions blocks in a single actuating quantity static over current

relay. The secondary current of line C.T. is generally not suitable for static relay

operations. It is higher. The line

(such as 4000/5), the intermediate CT (Auxiliary CT) reduces the current further to 1 amp.

So that it is suitable for static relay circuits. The alternating current derived from auxiliary

CT may contain harmonics, hence filters and spike suppressors are provided in the input

stage of static relay.

The current rectified in full wave rectifier and is smoothened in smoothing circuits

comprising resistors and capacitors. The smoothing circuit e

output waveform of the rectifier. The output of rectifier is supplied to level detector.

In single actuating quantity relay, the comparators are generally not necessary.

2 Attempt any four:

2 a) Definitions related to CB:

i) Rated current: RMS value of

overheating or exceeding the permissible temperature limits.

ii) Breaking capacity: rated brea

iii) Making capacity: it’s the maximum RMS current (ac and dc components) against

which the breaker is capable of closing and immediately opening without welding

or damage of contacts.

iv) Short time rating: is the highest current including the dc component that the


(Autonomous)






operations. It is higher. The line CT may be selected with a suitable higher current ratio



tain harmonics, hence filters and spike suppressors are provided in the input


comprising resistors and capacitors. The smoothing circuit eliminates the ripple in the



RMS value of current (Amperes) it can carry continuously without

or exceeding the permissible temperature limits.

rated breaking capacity is = √3 Vrated Irated breaking current capacity.

Making capacity: it’s the maximum RMS current (ac and dc components) against


is the highest current including the dc component that the


Page No : 5 of 24



CT may be selected with a suitable higher current ratio



tain harmonics, hence filters and spike suppressors are provided in the input


liminates the ripple in the



Descripti

on 2

marks

Diagram

4 marks

16 marks

it can carry continuously without

rated breaking current capacity.

Making capacity: it’s the maximum RMS current (ac and dc components) against


is the highest current including the dc component that the

1 mark

each = 4

marks



breaker shall be required to carry without damage for the specified time interval.

2 b) Explain working of ELCB:

Ans:

ELCB :-Works on principle of relaying when t

value. ELCB is used for protection against electric leakage in the circuit of 50 Hz or 60 Hz

, rated voltage single phase 240 V, 3 ph. 4 kv. Rated current up to 60 Amp.

earth fault occurs, the ELCB cuts

protect the personnel.

Under normal conditions (IL –

phase and neutral senses the differential current under earth fault and actuates

operate (open). The difference current I

earth. If this value exceeds a preset value then the CB opens. Normally it is around 35 mA

for tripping in domestic installations with tripping time bei

2 c) With help of neat diagram explain working of pantograph isolator in brief.

Ans:

Working: the isolator is made of light tubes of aluminum magnesium alloy that has

mechanical strength, good elasticity and high conductivity. Upper pantograph bars are

fitted with silver plated contact strips. In closed condition the upper two arms close on the

overhead busbar giving a grip.

upper busbar to lower one.

While opening the rotary insulator is rotated about its axis. The pantograph arms collapse

in vertical plane and vertical isolation is achieved between line terminal and the

pantograph upper terminal. By rotating th

pantograph are brought nearer while closing it again.

Normally these isolators are used at UHV substations.


(Autonomous)





Works on principle of relaying when the current in the earth path exceeds a set

ELCB is used for protection against electric leakage in the circuit of 50 Hz or 60 Hz

, rated voltage single phase 240 V, 3 ph. 4 kv. Rated current up to 60 Amp.

earth fault occurs, the ELCB cuts off the power within the time of 0.1 sec. automatically to

– IN) = If is very low or nearly zero. The CT surrounding the

phase and neutral senses the differential current under earth fault and actuates

operate (open). The difference current If through fault path resistance Re is the leakage to


for tripping in domestic installations with tripping time being as low as 25msec.

With help of neat diagram explain working of pantograph isolator in brief.

Working: the isolator is made of light tubes of aluminum magnesium alloy that has



overhead busbar giving a grip. Conducting pantograph bars carry the current from the



pantograph upper terminal. By rotating the rotary insulator column the linkage of the

pantograph are brought nearer while closing it again.

Normally these isolators are used at UHV substations.


Page No : 6 of 24


he current in the earth path exceeds a set

ELCB is used for protection against electric leakage in the circuit of 50 Hz or 60 Hz

, rated voltage single phase 240 V, 3 ph. 4 kv. Rated current up to 60 Amp. When the

ime of 0.1 sec. automatically to

is very low or nearly zero. The CT surrounding the

phase and neutral senses the differential current under earth fault and actuates the CB to

is the leakage to


25msec.

Diagram

or

equivalen

t 3 marks

Descripti

on 1 mark

Working: the isolator is made of light tubes of aluminum magnesium alloy that has good



rent from the



e rotary insulator column the linkage of the

Diagram

2 marks

Descripti

on 2

marks



2 d) Explain with neat sketch the w

Ans:

HRC fuse : -

HRC fuse mainly consists of heat resisting ceramic body. The current carrying element

is compactly surrounded by the filling powder. Filling material acts as an arc quenching

and cooling medium when the fuse e

under abnormal conditions. Under no

below its melting point. Therefore, it carries the normal current without overheating.

When a fault occurs, the current increases and the heat produced is sufficient to

melt these elements. Fuse element melt before the fault current rea

value. Vaporization of metallic

results in the formation of high resista


(Autonomous)




Explain with neat sketch the working of HRC fuse.



and cooling medium when the fuse element blows off due to excessive heat g

ormal conditions. Under normal condition, the fuse element is at a temperature

. Therefore, it carries the normal current without overheating.

rs, the current increases and the heat produced is sufficient to

melt these elements. Fuse element melt before the fault current reaches its first peak

metallic silver element chemically reacts with filling powder and

e formation of high resistance substance and helps in quenching


Page No : 7 of 24



ws off due to excessive heat generated

n, the fuse element is at a temperature

. Therefore, it carries the normal current without overheating.

rs, the current increases and the heat produced is sufficient to

s its first peak

silver element chemically reacts with filling powder and

nce substance and helps in quenching the arc.

Diagram

2 marks

Descripti

on 2

marks



2 e) State any four frequently occurring faults in 3 phase IM and suggest forms of protection

for them.

Ans:

Faults / abnormal conditions due to external conditions:

1) Unbalanced supply voltages:

three line voltages are summed to give a resultant voltage or output to trip the

motor control circuit or negative phase sequence currents

unbalance supply to give the trip signal.

2) Under voltages: protected by the hold on coil which releases/opens the main closed

power contacts if the voltage falls below a set threshold.

3) Reverse phase sequence starting:

supply the motor control circuit does not energize till pr

restored thus motor is protected.

4) Single phasing: protected by single phasing preventer which prevents energizing

the motor circuit till all the phases are healthy restored.

Faults / abnormal conditions due to internal

1) Stalling/Bearing failures:

2) Earth faults: it is detected by measuring element whose pickup value is set

potentiometer or similar device.

3) Overloads: overload relays (normally bimetallic thermal relays) are used

proper fuses may be used for small motors.

4) Stator over heating: for large motors embedded semiconductor resistance

temperature detectors are

5) Blocked rotor / prolonged starting:

bimetallic thermal relay.

1) Heavy Short circuited winding or severe earth fault: detected / protected by fuse of

proper rating.

2 f) Draw a neat labeled diagram of single phasing preventer used for motor protection.

Ans:

Single phasing preventor:


(Autonomous)




State any four frequently occurring faults in 3 phase IM and suggest forms of protection

conditions due to external conditions:

Unbalanced supply voltages: protected by voltage summation scheme wherein the


motor control circuit or negative phase sequence currents are sensed in the case of

unbalance supply to give the trip signal.

protected by the hold on coil which releases/opens the main closed

power contacts if the voltage falls below a set threshold.

Reverse phase sequence starting: protection: in the case of reverse phase sequence

supply the motor control circuit does not energize till proper phase sequence is

restored thus motor is protected.

protected by single phasing preventer which prevents energizing

l all the phases are healthy restored.

Faults / abnormal conditions due to internal faults/conditions:

Bearing failures: stalling protection by bimetallic thermal relay.

it is detected by measuring element whose pickup value is set

potentiometer or similar device.

overload relays (normally bimetallic thermal relays) are used

proper fuses may be used for small motors.

for large motors embedded semiconductor resistance

temperature detectors are used to sense and trip the supply.

Blocked rotor / prolonged starting: similar to stalling condition, protection by

bimetallic thermal relay.

Heavy Short circuited winding or severe earth fault: detected / protected by fuse of

Draw a neat labeled diagram of single phasing preventer used for motor protection.


Page No : 8 of 24

State any four frequently occurring faults in 3 phase IM and suggest forms of protection

protected by voltage summation scheme wherein the


are sensed in the case of

protected by the hold on coil which releases/opens the main closed

in the case of reverse phase sequence

oper phase sequence is

protected by single phasing preventer which prevents energizing

stalling protection by bimetallic thermal relay.

it is detected by measuring element whose pickup value is set on

overload relays (normally bimetallic thermal relays) are used, also

for large motors embedded semiconductor resistance

similar to stalling condition, protection by

Heavy Short circuited winding or severe earth fault: detected / protected by fuse of

1 mark

each any

four = 4

marks

(1/2 mark

for fault

and ½

mark for

protection

other

valid

points

may be

considere

d if

relevant.

Draw a neat labeled diagram of single phasing preventer used for motor protection.

Labeled

diagram 4

marks

Partially

labeled 2

to 3

marks

Unlabeled

1 mark



Connected in secondaries of line CTs. Mainly contain negative sequence filter whose

output is fed to level detector which further sends tripping command to starter or CB. Thus

it protects the motor from damage.

3 Attempt any four

3 a) State fundamental requirements of protective relaying.

Ans:

Fundamental requirements of protective relaying:

i) Selectivity: - It is the ability of protective system to select correctly that part of

system in trouble and disconnect the faulty part without disturbing the rest

system.

ii) Speed: The relay system should disconnect the faulty section as fast as possible to

prevent the electrical apparatus from damage and for system stability.

iii) Sensitivity: - It is the ability of the relay system to operate with low value of

actuating quantity.

iv) Reliability: - It is the ability of the relay system to operate under predetermined

conditions.

v) Simplicity: - The relay system should be simple so that it can be easily maintained.

vi) Economy: - The most important factor in the choice of p

is the economic aspect. The protective gear should not cost more than 5% of the

total cost of equipment to be protected.

3 b) Draw a neat labeled diagram of induction type directional over

Ans:


(Autonomous)




econdaries of line CTs. Mainly contain negative sequence filter whose


it protects the motor from damage.

requirements of protective relaying.

Fundamental requirements of protective relaying:

It is the ability of protective system to select correctly that part of

system in trouble and disconnect the faulty part without disturbing the rest

Speed: The relay system should disconnect the faulty section as fast as possible to


It is the ability of the relay system to operate with low value of

It is the ability of the relay system to operate under predetermined

The relay system should be simple so that it can be easily maintained.

The most important factor in the choice of particular protection scheme


total cost of equipment to be protected.

d diagram of induction type directional over-current relay.


Page No : 9 of 24

econdaries of line CTs. Mainly contain negative sequence filter whose


16 marks

It is the ability of protective system to select correctly that part of

system in trouble and disconnect the faulty part without disturbing the rest of the

Speed: The relay system should disconnect the faulty section as fast as possible to


It is the ability of the relay system to operate with low value of

It is the ability of the relay system to operate under predetermined

The relay system should be simple so that it can be easily maintained.

articular protection scheme


2 points

1 mark,

4 points

2 marks,

5 points

3 marks,

6 points

4 marks.

current relay.



3 c) Explain with neat sketch operation of attracted armature relay.

Ans: Attracted Armature Relay :

It consist of laminated electromagnet carrying a coil and a pivoted laminated armature.

The armature is balanced by a counter weight and carries a pair of spring contact fingers at

its free and under normal operating conditions the current through the relay coil is such

that counter weight holds the armature in the position shown. When a short circuit occurs,

the current through the relay coil increases sufficiently and the relay armature is attracted

upwards. This completes the trip circuit which results in the opening of th


(Autonomous)




Explain with neat sketch operation of attracted armature relay.

Attracted Armature Relay : -

d electromagnet carrying a coil and a pivoted laminated armature.



ounter weight holds the armature in the position shown. When a short circuit occurs,


upwards. This completes the trip circuit which results in the opening of the circuit


Page No : 10 of 24

Labeled 4

marks

Unlabeled

2 marks

Incomplet

e 1 mark.

d electromagnet carrying a coil and a pivoted laminated armature.



ounter weight holds the armature in the position shown. When a short circuit occurs,


e circuit breaker

Any type

attracted

armature

diagram 2

marks

2 marks



3 d) Define following terms related to CT.

i) CT burden: it is the load imposed on the secondary of the

measured in VA (product of volts and amps).

ii) Ratio error: Ratio error : It is the ratio of error in the secondary side current w.r.t.

primary side given by [(N.I

currents, N = current ratio}

iii) Accuracy class: The class assigned to the C.T. with the

and phase angle error.

iv) Phase Angle error: It is the phase angle between primary current vector and the

reversed secondary current vector.

3 e) State functions of substation earthing system. Also s

earthing system.

Ans:

Functions:

i) Provide sufficiently low earth electrode resistance to ensure reliable operation of the

switchgear under fault/abnormal conditions.

ii) Provide means to carry and dissipate electric current i

abnormal conditions of working.

iii) Minimize arcing grounds.

iv) Provide degree of safety to personnel working/walking around the earthed

equipment of the substation.

v) Carry the maximum earth fault currents to avoid overheating, mechanica

unduly drying of soil around substation.

vi) Provide protection such that non

sheaths, armouring etc. do not suffer heavy

vii) Avoid dangerous potential areas in substation vicinity.

Parts of substation earthing systems:

i) Earth mats (mesh of steel strips or rods).

ii) Copper strips. Bars.

iii) Overhead earth/shield

iv) Earthing transformers.

v) Earthing switches.

3 f) Give any four differences between equipment earthing and neutral earthing.

Ans:

Sr. no Equipment earthing

1


(Autonomous)




Define following terms related to CT.

is the load imposed on the secondary of the CT at rated current and is

measured in VA (product of volts and amps).

Ratio error: Ratio error : It is the ratio of error in the secondary side current w.r.t.

given by [(N.IS – IP)/IP] x100 %. {IP, IS are primary and secondary

currents, N = current ratio}

Accuracy class: The class assigned to the C.T. with the specified limits of ratio error

Phase Angle error: It is the phase angle between primary current vector and the

reversed secondary current vector.

State functions of substation earthing system. Also state different parts of substation

Provide sufficiently low earth electrode resistance to ensure reliable operation of the

switchgear under fault/abnormal conditions.

Provide means to carry and dissipate electric current into earth under normal and

abnormal conditions of working.

Minimize arcing grounds.

Provide degree of safety to personnel working/walking around the earthed

equipment of the substation.

maximum earth fault currents to avoid overheating, mechanica

unduly drying of soil around substation.

Provide protection such that non-current carrying parts of electrical equipment as

sheaths, armouring etc. do not suffer heavy fault currents.

Avoid dangerous potential areas in substation vicinity.

ts of substation earthing systems:

Earth mats (mesh of steel strips or rods).

Copper strips. Bars.

Overhead earth/shield wires.

Earthing transformers.

Give any four differences between equipment earthing and neutral earthing.

Equipment earthing Neutral earthing


Page No : 11 of 24

at rated current and is

Ratio error: Ratio error : It is the ratio of error in the secondary side current w.r.t.

are primary and secondary

specified limits of ratio error

Phase Angle error: It is the phase angle between primary current vector and the

1 mark

each = 4

marks

tate different parts of substation

Provide sufficiently low earth electrode resistance to ensure reliable operation of the

nto earth under normal and

Provide degree of safety to personnel working/walking around the earthed

maximum earth fault currents to avoid overheating, mechanical damage or

current carrying parts of electrical equipment as

½ mark

each any

four = 2

marks.

Any two

parts 1

mark

each = 2

marks.

Give any four differences between equipment earthing and neutral earthing.

Any four

1 mark

each = 4

marks



2

Connection of

carrying metallic part

electrical equipment

called as equipment earthing.

3

It is provided for protection of

human being from electric

shocks.

4 It has nothing to do with

stability

5

Equipment earthing is provided

through Pipe earthing, Plate

earthing.

6

It does not provide any means

for protection system against

earth fault

4 a) Attempt any three of the following:

4 a) i) Define short circuit. Discuss possible causes of

Ans:

Definition: whenever the charged (live) conductor/line/lines of an electric supply path

make contact with a line/conductor/path of different potential to circulate heavy

heavy currents due to the unintended

occur between phases, phase/s and neutral or phase/s and earth.

Possible causes of short circuit in power station:

1) Failure of insulation between lines or lines and earth

(equipment body included) due to following reasons:

i) Deteriorated insulation due to aging.

ii) Voltage surge due to external causes as lightning, switching of external

devices.

iii) Voltage surge due to internal causes as switching of devices.

iv) Bad very humid atmos

2) Generator bus terminals insulation

3) Transformer bus terminals

4) Sustained equipment overloading leading to overheating and finally failure o

insulation.

5) Improper maintenance of power station equipment.

4 a) ii)

Explain Ring system and Tie bar system of bus reactors.

Ans:

Reactors are suitable inductive coils placed in the power

limit the short circuit fault currents between the points they are connected.

Ring system:


(Autonomous)




Connection of the non-current

carrying metallic parts of the

electrical equipment to earth is

s equipment earthing.

Connection of the neutral

of three phase transformer,

generators, motors etc. to earth

is neutral earthing.

It is provided for protection of

human being from electric

It is provided for eliminating

arching ground and over voltage

surge.

It has nothing to do with Stability of the system is

increased.

Equipment earthing is provided

through Pipe earthing, Plate

Neutral earthing is provided

through solid earthing, r

earthing and reactance earthing.

It does not provide any means

for protection system against It provides suitable means for

earth fault protection system.

Attempt any three of the following:

Define short circuit. Discuss possible causes of short circuit in the power station.


make contact with a line/conductor/path of different potential to circulate heavy

due to the unintended connection it is said to be a short circuit. Short may

occur between phases, phase/s and neutral or phase/s and earth.

Possible causes of short circuit in power station:

Failure of insulation between lines or lines and earth in the power station

nt body included) due to following reasons:

Deteriorated insulation due to aging.

Voltage surge due to external causes as lightning, switching of external

Voltage surge due to internal causes as switching of devices.

Bad very humid atmosphere with dirty conditions.

Generator bus terminals insulation failure (over heating, bad maintenance etc.)

Transformer bus terminals, internal phase short circuits due to overheating etc.

Sustained equipment overloading leading to overheating and finally failure o

Improper maintenance of power station equipment.

Explain Ring system and Tie bar system of bus reactors.

Reactors are suitable inductive coils placed in the power systems at proper locations to

limit the short circuit fault currents between the points they are connected.


Page No : 12 of 24

neutral point

transformer,

etc. to earth

It is provided for eliminating

over voltage

Stability of the system is

Neutral earthing is provided

resistance

earthing.

It provides suitable means for

earth fault protection system.

12

short circuit in the power station.


make contact with a line/conductor/path of different potential to circulate heavy to very

it is said to be a short circuit. Short may

in the power station

Voltage surge due to external causes as lightning, switching of external

Voltage surge due to internal causes as switching of devices.

, bad maintenance etc.)

, internal phase short circuits due to overheating etc.

Sustained equipment overloading leading to overheating and finally failure of

2 marks

Any four

valid

points ½

mark

each = 2

marks

systems at proper locations to



Each feeder is mostly fed by a single generator. The reactors have high reactances such

that the generator busbars are nearly

little voltage across them. Under fault on one of the feeders it is mainly fed by the

respective generator. The other generators feed the faulty feeder through the reactors

which limit the fault current.

Tie-bar system:

Here the reactors are placed in the path as shown where the faulty feeder is fed directly by

its generator and through the tie bar by the other generators. The reactors between the tie

bar and generator limit the fault current better than the

that the number of units in tie bar can be i

switchgear.

4 a) iii) Let common base be 3.3 kV, 1.5 MVA.

The % reactance of 1.5 MVA alternator remains altered at X

voltage and MVA are altered.

while that of 1 MVA alternator will be

XB = old reactance x (new base MVA/old base MVA)

= 10 (1.5/1) = 15 %.

These two alternators in paralle


(Autonomous)





that the generator busbars are nearly isolated from each other and the reactors have very





bar and generator limit the fault current better than the ring system. Another advantage is

that the number of units in tie bar can be increased without the difficulty of replacement of

Let common base be 3.3 kV, 1.5 MVA.

tance of 1.5 MVA alternator remains altered at XA = 18 % as the base values of

while that of 1 MVA alternator will be

= old reactance x (new base MVA/old base MVA)

These two alternators in parallel feed the fault at their common busbar.


Page No : 13 of 24


ted from each other and the reactors have very





ring system. Another advantage is

ncreased without the difficulty of replacement of

1 mark

1 mark

1 mark

1 mark

= 18 % as the base values of

Diagram

not

compulso

ry

1 mark



The equivalent reactance of the fault path (alternators circuit) feeding the fault in parallel

will be

XEQ = XAXB/( XA + XB )

= 18 x 15/(18 + 15)

= 8.18 %.

Short circuit MVA = [base MVA x 100]/(% reactance of fault path)

= 1.5 x 100/X

= 1.5 x 100/ 8.18

= 18.33 MVA.


(Autonomous)





cuit MVA = [base MVA x 100]/(% reactance of fault path)

= 1.5 x 100/XEQ.

= 1.5 x 100/ 8.18


Page No : 14 of 24


1 mark

1 mark

1 mark



4 a) iv) Explain with neat diagram the operation of Buchholz relay:

Terminals ‘A’ : are for alarm and Terminals ‘T’: are for

Operation:

Alarm is obtained due to slow deterioration of insulation when gas accumulates at top

below cock and tilts the hollow float about its hinge making the mercury operate.

Trip signal is obtained due to sudden flow of oil (due to a sudden major violent fault in the

oil tank) from tank to conservator that tilts the flap valve to operate the mercury switch.

OR

The relay in located in the path of the oil from transformer tank to conservator.

As seen from diagram the upper mercury switch operates the alarm circuit due to

tilting of the float by accumulation of gas evolved slowly in the transformer

minor faults which may develop into major ones if the alarm is not investigated.

Further lower mercury switch operates the trip circuit to switch off the circuit

breaker related to the transformer when there is a sudden flow of oil fr

tank to the conservator. Such flow occurs when there is series fault in the transformer tank.

Here the float (lower) is placed in such a manner that it senses the sudden violent

movement of oil from transformer tank to conservator

4 b) Attempt any one of following:

4 b) i) Explain with neat labeled diagram the working of the SF

CB.

Ans:


(Autonomous)




Explain with neat diagram the operation of Buchholz relay:

are for alarm and Terminals ‘T’: are for tripping.


d tilts the hollow float about its hinge making the mercury operate.





tilting of the float by accumulation of gas evolved slowly in the transformer



breaker related to the transformer when there is a sudden flow of oil from the transformer



movement of oil from transformer tank to conservator

Attempt any one of following:

Explain with neat labeled diagram the working of the SF6 CB. State two advantages of SF


Page No : 15 of 24


d tilts the hollow float about its hinge making the mercury operate.





tilting of the float by accumulation of gas evolved slowly in the transformer tank due to



om the transformer



Labeled

Diagram

2 marks,

unlabeled

1 mark

1 mark

1 mark

OR

1 mark

1 mark

06

CB. State two advantages of SF6



Single pressure puffer type br

Gas is compressed by the moving cylinder and is released through the nozzle to extinguish

the arc formed.

The moving cylinder (1) connected to moving contact(2) against the fixed piston (5). Due

to relative motion between

released through nozzle for arc extinction. This happens by puffing action. At current zero

the diameter becomes too small and arc gets extinguished.

Double pressure type breaker:

Here the gas is made to flow from

The flowing gas covers the arc. In the divergent section the speed of the gas is very high

and carries away most of the heat from the periphery of arc that results in the reduction of

diameter of arc which becomes nearly zero at current zero leading to arc being

extinguished. Finally the gas enters the contact space increasing its the dielectric strength.


(Autonomous)




Single pressure puffer type breaker:



to relative motion between (1) and (5) the gas gets compressed in enclosure (6) and is


the diameter becomes too small and arc gets extinguished.

OR

Double pressure type breaker:

Here the gas is made to flow from area P1 to P2 through a convergent-divergent



ich becomes nearly zero at current zero leading to arc being

as enters the contact space increasing its the dielectric strength.


Page No : 16 of 24



compressed in enclosure (6) and is


divergent nozzle.



ich becomes nearly zero at current zero leading to arc being

as enters the contact space increasing its the dielectric strength.

Any one

of given

two:

Diagram

2 marks,

descriptio

n 2 marks



Advantages of SF6 breaker:

1. Higher speed interruption

2. Long arc length not needed for arc interruptio

3. Lower arcing time.

4. No frequent contact maintenance.

5. Cleaner operation.

6. Cleaner maintenance as it is less bulky and cleaner than oil

4 b) ii) Distance Relay –

Principle: The action of relay depends u

where the relay is installed and the point of fault.

of voltage and current in the fault path.

certain value. Higher current indic

Types --

1) Impedance relay

2) Mho relay

3) Reactance relay

Definite distance relay:

From diagram the restraining force F

connected) is overcome by force F

FI > FV. but these are proportional to respective electric quantities. F

k2V2. From this when (V/IF

on the distance of fault on line from relay as Z

of line and L = distance at which fault has occurred)

5 Attempt any four

5 a) Describe with neat diagram the balanced earth fault protection for alternator.

Ans:

Under normal working the phase currents summed up by three line CTs is equal to that of

the CT in neutral line earthed and hence no operation of breaker occurs. For fault at F

outside protected zone again the summation of the three CTs is equal to the neutral CT

current and the balance is maintained with no operation of the relay. But for fault in

protected zone the summation current of the three line CTs differs and is not balanced by


(Autonomous)




Advantages of SF6 breaker:

Higher speed interruption

Long arc length not needed for arc interruption.

No frequent contact maintenance.

maintenance as it is less bulky and cleaner than oil

The action of relay depends upon the distance (impedance) between the point

where the relay is installed and the point of fault. The operation is dependent on the ratio

in the fault path. The relay operates when this ratio falls below a

certain value. Higher current indicates faulty condition for a particular distance.

iagram the restraining force FV due to voltage electromagnet (PT: constant voltage

connected) is overcome by force FI due to current electromagnet (operating force) that is

. but these are proportional to respective electric quantities. FI = k

F ) < √(k1/k2) or ZF < √(k1/k2) (fault path impedance depends

on the distance of fault on line from relay as ZF = z L, where z = impedance per unit length

of line and L = distance at which fault has occurred)

Describe with neat diagram the balanced earth fault protection for alternator.


in neutral line earthed and hence no operation of breaker occurs. For fault at F


current and the balance is maintained with no operation of the relay. But for fault in



Page No : 17 of 24

Any two

advantage

1 mark

each = 2

marks

) between the point

The operation is dependent on the ratio

The relay operates when this ratio falls below a

ates faulty condition for a particular distance.

due to voltage electromagnet (PT: constant voltage

due to current electromagnet (operating force) that is

= k1IF2 and FV =

) (fault path impedance depends

= z L, where z = impedance per unit length

2 marks.

2 marks

2 marks

16 marks

Describe with neat diagram the balanced earth fault protection for alternator.


in neutral line earthed and hence no operation of breaker occurs. For fault at F2


current and the balance is maintained with no operation of the relay. But for fault in the


Descripti

on 2

marks



the earthed neutral line CT current and the unbalance current in the relay operates it to trip

the breaker related and protect form the earth fault.

5 b) State and explain in brief the inter

Ans:

Under normal working conditions the two currents in the stator winding sections S

are identical and by virtue of the cross connected CT secondaries the relay current is zero,

hence no relay operation. But when one of the windings is faulty (inter turn fault) its

current differs and hence the two CT secondary currents are different, due which the

difference current is diverted through the relay coil to operate it leading to isolation of the

alternator from the power system.


(Autonomous)





the breaker related and protect form the earth fault.

State and explain in brief the inter-turn fault protection in alternator.

Under normal working conditions the two currents in the stator winding sections S

y virtue of the cross connected CT secondaries the relay current is zero,



e current is diverted through the relay coil to operate it leading to isolation of the

alternator from the power system.


Page No : 18 of 24


Diagram

2 marks

Under normal working conditions the two currents in the stator winding sections S1 and S2

y virtue of the cross connected CT secondaries the relay current is zero,



e current is diverted through the relay coil to operate it leading to isolation of the

Diagram

2 marks

Unlabeled

1 mark

Descripti

on 2

marks



5 c) Explain with neat diagram the reverse power protection in

Ans:

Reverse power protection of alternator:

Operating characteristics of reverse power relay:

On failure of input to the prime mover the alternator works as sy

load (3 to 12 % of rated) and p

protect the alternator from such reverse power flow

relay that senses direction of power flow in alternator is employed. The CTs for this a

placed either at neutral end or busbar end of alternator. An intentional time lag is used to

avoid operation during temporary system disturbances when power may flow in the

reverse. When the current in the CT reverses the aluminum disc rotates to trip t


(Autonomous)




Explain with neat diagram the reverse power protection in alternator.

Reverse power protection of alternator: A Directional power relay is used for the same.

Operating characteristics of reverse power relay:

On failure of input to the prime mover the alternator works as synchronous motor with low

% of rated) and power factor of operation depends on the excitation. To

or from such reverse power flow a single element directional

relay that senses direction of power flow in alternator is employed. The CTs for this a



When the current in the CT reverses the aluminum disc rotates to trip t


Page No : 19 of 24

A Directional power relay is used for the same.

nchronous motor with low

the excitation. To

directional power

relay that senses direction of power flow in alternator is employed. The CTs for this are



When the current in the CT reverses the aluminum disc rotates to trip the circuit.

1½ mark

½ mark

2 marks



5 d) Draw circuit diagram of biased differential protection of

Ans:

5 e) State common faults in power transformer suggesting protection for each fault.

Ans:

No. Faults

1 Earth faults (winding to earth)

2 Through faults

3 Lightning surge (over voltage surges)

(external cause)

4 Over loads (not a fault hence not compulsory)

5

Incipient faults in oil tank (phase to phase

windings short or arcing, phase to core/body

short or arcing) that lead to gradual gas

formation and release.


(Autonomous)




Draw circuit diagram of biased differential protection of D/Y power transformer.

mon faults in power transformer suggesting protection for each fault.

Faults Suggested protection

Earth faults (winding to earth) Earth fault relay.

Differential protection.

Graded time lag over current

relay

HRC fuses.

ghtning surge (over voltage surges)

Lightning Arrester at

incoming lines.

Surge arresters.

RC surge suppressors.

Horn gap arresters.

Over loads (not a fault hence not compulsory)

Over load relays (over

current)

Thermal over load relays.

Temperature indicator relays

with sound facility (alarm).

Incipient faults in oil tank (phase to phase

windings short or arcing, phase to core/body

short or arcing) that lead to gradual gas

formation and release.

Buchholz relay (gas actuated

alarm)


Page No : 20 of 24

/Y power transformer.

Labeled

diagram 4

marks,

Partially

labeled 3

marks,

Unlabeled

2 marks

mon faults in power transformer suggesting protection for each fault.

Suggested protection

Earth fault relay.

Differential protection.

Graded time lag over current

Lightning Arrester at

RC surge suppressors.

Horn gap arresters.

Over load relays (over

Thermal over load relays.

Temperature indicator relays

with sound facility (alarm).

Buchholz relay (gas actuated

½ mark

each

fault, ½

mark

protection

, any four

sets = 4

marks



6

Sudden fault (mild explosion) in oil tank due

to faults between phase windings or to

earth/core/body leading to sudden flow of oil

towards conservator.

7 Sudden violent explosion in oil tank.

8 Magnetic core saturation

5 f) 66 kV/11 kV, star delta transformer,

Assume 250 A is flowing in lines on LT side.

(V1 & V2 are primary (HT) and secondary

and secondary (LT) line current

Primary kVA = secondary kVA,

√3 V1I1 = √3 V2I2

Hence I1 = I2(V2/V1) = 250 x 11/66 =

on HT.

On delta side (LT) the CT secondaries are star connected. Their secondary current is 5 A.

in pilot wires between HT and LT CTs.

On HT side (star connection of power transformer)

connected have currents = (5/√

Hence CT ratio on HT side is I

6 Attempt any four

6 a) State abnormalities and faults on busbars.

Ans:

Abnormalities and faults on busbars are:

1) Weakening and failure of insulation due to aging and

2) CB failure.

3) Earth fault due to failure of support insulator.

4) Flashover due to sustained excessive over vol

5) Errors/mistakes in operation and mainten


(Autonomous)




Sudden fault (mild explosion) in oil tank due

to faults between phase windings or to

earth/core/body leading to sudden flow of oil

Buchholz relay trip.

PRV at tank top

Sudden violent explosion in oil tank.

PRV (pressure release val

at top of tank that bursts out

to release the pressure by

spilling oil out.

Percentage differential

protection

Magnetic core saturation

Over voltage protection

Over fluxing protection

66 kV/11 kV, star delta transformer,

0 A is flowing in lines on LT side.

and secondary (LT) line voltages, I1 & I2 are primary

currents).

Primary kVA = secondary kVA,

) = 250 x 11/66 = (125/3) A = 41.66 A, which is primary CT current

the CT secondaries are star connected. Their secondary current is 5 A.

in pilot wires between HT and LT CTs.

star connection of power transformer) CT secondaries which are delta

√3) A in them.

Hence CT ratio on HT side is I1 : (5/√3) = (125/3) : (5/√3) = (25/√3) :1.

State abnormalities and faults on busbars.

s and faults on busbars are:

and failure of insulation due to aging and deterioration.

fault due to failure of support insulator.

due to sustained excessive over voltages.

mistakes in operation and maintenance of switchgear.


Page No : 21 of 24

Buchholz relay trip.

PRV (pressure release valve)

at top of tank that bursts out

to release the pressure by

Percentage differential

Over voltage protection

Over fluxing protection

are primary (HT)

, which is primary CT current

the CT secondaries are star connected. Their secondary current is 5 A.

CT secondaries which are delta

1 mark

1 mark

1 mark

1 mark

16 marks

½ mark

each any

eight = 4

marks



6) Mechanical damage due to improper working of personnel or earth quake.

7) Accidents due to other bodies falling

8) Flashover due to heavily polluted insulator.

9) Loosened fittings due to

6 b) Describe fault bus protection with diagram for busbars.

Ans:

Fault bus protection for bus bars:

- Each phase bus is surrounded by an earthed metal mesh or structure

fault bus)

- Hence the busbar fault w

- The earth path current is sensed by the CT and if it is above the relay setting value

the relay trips the concerned circuit breaker indicating bus bar fault.

6 c) Explain with neat diagram pilot wire protection used for transmission lines.

Ans:


(Autonomous)




damage due to improper working of personnel or earth quake.

due to other bodies falling across on them.

due to heavily polluted insulator.

ned fittings due to vibration etc.

Describe fault bus protection with diagram for busbars.

Fault bus protection for bus bars:

Each phase bus is surrounded by an earthed metal mesh or structure

Hence the busbar fault will never be inter-phase one.

The earth path current is sensed by the CT and if it is above the relay setting value


ram pilot wire protection used for transmission lines.


Page No : 22 of 24

damage due to improper working of personnel or earth quake.

Each phase bus is surrounded by an earthed metal mesh or structure – (called as

The earth path current is sensed by the CT and if it is above the relay setting value


Diagram:

fully

labeled 2

marks,

partially

labeled 1

mark.

2 marks

ram pilot wire protection used for transmission lines.

Complete

Diagram

3 marks,

Partially

labeled 2

marks,

unlabeled

1 mark.



Current entering one end of lines is equal to that leaving from other end under healthy

condition. Hence CT secondar

hence no relay current. But if there is fault in between then the two side CTs have different

induced voltages resulting in current through relay coil

6 d) State various causes of over voltages

Ans:

Causes of over voltages:

Causes external to power system.

Lightning strokes during strong or rainy weather. These can be so high that the magnitudes

of voltage waves created can be is the range of 800 kV to 1500kV.

Causes internal to power system.

• Switching surges occur when switching operations of circuit breakers are carried out

especially under abnormal conditions.

• Switching an unloaded line sets

doubling effect to produce an instantaneous max voltage of 2

rms (line voltage) value. Similar effect is obtained when an unloaded line is switched

off.

• Resonance in power systems.

• Interrupting currents before their natural zero.

• Arcing grounds.

• Insulation failures.

6 e) Explain with neat sketch the working of Thyrit

Ans:

Normal system voltage is insuffici

When a high over voltage surge reaches it, breakdown of the series gap occurs and the

surge current is conducted to earth through the

resistance under these conditions (

After the surge is conducted to earth through low resistance path the non

offer high resistance to low currents and thus normal condition is restored.


(Autonomous)




entering one end of lines is equal to that leaving from other end under healthy

ries have equal and opposite voltages at two ends

if there is fault in between then the two side CTs have different

induced voltages resulting in current through relay coils and operation to trip.

State various causes of over voltages

Causes external to power system.


of voltage waves created can be is the range of 800 kV to 1500kV.

Causes internal to power system.

Switching surges occur when switching operations of circuit breakers are carried out

especially under abnormal conditions.

Switching an unloaded line sets up travelling waves that on reflection create a

doubling effect to produce an instantaneous max voltage of 2 2 E where ‘E’ is the


in power systems.

Interrupting currents before their natural zero.

with neat sketch the working of Thyrite LA.

is insufficient to cause electric breakdown of the air gap assembly

a high over voltage surge reaches it, breakdown of the series gap occurs and the

ducted to earth through the non-linear resistors which have low

resistance under these conditions (these have inverse resistance / current characteristics

the surge is conducted to earth through low resistance path the non-linear

w currents and thus normal condition is restored.


Page No : 23 of 24

entering one end of lines is equal to that leaving from other end under healthy

ies have equal and opposite voltages at two ends of line and

if there is fault in between then the two side CTs have different

operation to trip.

1 mark


Switching surges occur when switching operations of circuit breakers are carried out

up travelling waves that on reflection create a

E where ‘E’ is the


1 mark

Any three

1 mark

each = 3

marks

breakdown of the air gap assembly.

a high over voltage surge reaches it, breakdown of the series gap occurs and the

linear resistors which have low

characteristics).

linear resistors

2 marks

Diagram

labeled 3

marks,

partially

labeled 2

marks,

unlabeled

1 mark.



6 f) Define insulation coordination.

power system with LA.

Ans:

Insulation coordination is the co

with the characteristics of protective devices such that the insulation of the whole power

system is protected from excessive over voltages.

Curve A is volt time curve of protective device and curve B is that of equipment protected.

Insulation coordination is needed to protect the e

Especially the costly transformers, generators etc from the harmful over voltage that occur

in the system. The Basic Impulse (Insulation) levels of the various devices and equipment

are graded selectively to provide protection to the costly equipment.

equipment it will be having its insulation level more on

A: protection level voltage characteristic of LA

B: impulse insulation level of

Between B & C Isolators and

C: impulse insulation level of

Between C & D, CBs

D: impulse insulation level of transformer


(Autonomous)




e insulation coordination. Also show the coordination of various equipme

Insulation coordination is the co-relation of the insulation of electric equipment and lines


is protected from excessive over voltages.


Insulation coordination is needed to protect the equipment in the power systems.

ly transformers, generators etc from the harmful over voltage that occur


are graded selectively to provide protection to the costly equipment. Costlier

equipment it will be having its insulation level more on the higher side.

voltage characteristic of LA

line insulator.

B & C Isolators and normal low cost switching devices.

of CTs, PTs etc.

of transformer (costliest equipment).


Page No : 24 of 24

of various equipments in

relation of the insulation of electric equipment and lines



quipment in the power systems.

ly transformers, generators etc from the harmful over voltage that occur


ostlier the

1 mark

1 mark

1 mark

Diagram

1 mark

Date post:	18-Mar-2018
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