Power System Lab Manual Original and Modified

Power System Lab Manual

INDEX

List of Experiments

1. To the study time vs.voltage characteristcs of over voltage induction relay

2. To the study time vs. current characteristics of over current induction relay

3. To the study time vs. differential current characteristics of differential relay

4. To study the time vs. current characteristics of earth fault relay

5. To study the time vs. temperature of thermal relay characteristics

6. To study and verify the operation of buchholz relay.

7. To study and performance of the long T/M line under no load test and under light load

(R,L,C) condition and Ferranti effect.

8. To find the location of a fault in a UG Cable using Murray loop test method

9. Study the single line to ground fault.

10. To study the three phase faults as practical application in transmission line.

EXPERIMENT NO. 1

OVER VOLTAGE RELAY

AIM:

1. Study the operating current & de-operating voltage of disc. 2. Study the characteristics between voltage and time.

MATERIAL REQUIRED: 1. Voltmeter (0-300 V) 96 x 96 mm 2. Auto Transformer 0-270V 3. Indicating Light 4. Over Voltage Relay (Electromechanical Type) 5. Timer with Start & Stop facility 6. Push Button for Timer START & STOP 7. Rotary Switch 8. MCB/Isolator 9. Insulating terminals

INTRODUCTION

There are several instruments in power system which may burn at over voltage like motor, transformer, capacitors etc. To protect these instruments from over voltage, the over voltage relay is necessary. The several type of over voltage relays is available. One of the type from these relays is studying here. It is electromechanical type. It is very much rough & tough in use, its failure almost negligible. As the accuracy required to the electric instrument, the sufficient accuracy is available in this type of relay. THEORY

Here the relay have an operating coil which core developed a rotating flux in disk due to this the disk rotates at certain voltage. Before the certain voltage a helical spring makes force in opposite direction so the disk is not in the condition to rotate.

A socket is given to different setting of tripping voltage. Time setting multiplier is given to setting the tripping time. A flag relay is given to extending DC voltage to the circuit breaker. The resistances and diode is providing to operating of flag relay.

We apply the tripping voltage to pin No 9 and 10 to this relay as given in Fig –‘1’. The pin No. 9 will extend to a transformer, disk operating coil and through moving this contact to a bridge. Pin No 10 will extend to the common pole of plug setting socket. On the plug setting socket all the primary tapings are connected and secondary will fed to operating disk coil and bridge rectifier. The output of bridge is filtered by capacitor C1 and resistance R1. The filtered DC voltage is fed to flag relay through resistance R2. Resistance R2 and R3 is meant for increasing the operating voltage capacity of flag relay. Resistance R4 and R5 is meant for increasing the operating voltage capacity of disk operating coil.

PROCEDURE: Study.1. Study the operating current & de-operating current of disc. (i) Keep the voltage source at minimum. (ii) The amp adj / relay test rotary switch is kept at VOLT ADJ. (iii) Switch ON the test set. (iv) Push the ON button. (v) Increase the voltage slowly and pay attention at disc of relay. (vi) At certain voltage, it just moves in forward direction, this voltage is operating

voltage and note it. (vii) Now decrease the voltage through voltage source and pay hard attention at disc. (viii) The disc will stop at certain voltage and moves in reverse direction just after

reducing the voltage. This current is de-operating voltage and note its value.

Observation Table:

2. Study the characteristics between PSM & time (i) We have to study the graph between PSM and time at TMS-1 etc. (ii) Keep the voltage source at minimum. (iii) The volt adj / relay test rotary switch is kept at VOLT ADJ. (iv) Switch ON the MCB/Isolator. (v) Push the ON button.

(vi) Set the voltage at which the time has to observe and note down this voltage in table. (vii) Push the OFF button. (viii) The volt adj / relay test rotary switch is kept at Relay Test. (ix) Push the ON button. (x) The relay will trip the system then note down the time in given table. (xi) Repeat the above procedure to observe the time for different PSM and note these

PRECAUTIONS 1. The relay should be used very carefully and fast. 2. Suppose the 1 amp relay can withstand upto 1 amp continuously and if more than 1

amp the relay should trip the system within few seconds

Result:

EXPERIMENT NO.2

I.D.M.T. OVER CURRENT RELAY

AIM: 1. Study the operating current & de-operating current of disc. 2. Study the time current characteristics at various multiples of plug setting current.

MATERIALS REQUIREDS: 1. Voltmeter (0-300 V) Digital 2. Ammeter (0-10 A) Digital 3. Loading C.T. 4. Auto Transformer 0-270V 5. Indicating Light 6. I.D.M.T. Relay Type CDG 7. Timer with Start & Stop facility 8. Push Button for Timer START & STOP 9. Rotary Switch 10. DP Switch 11. Insulating terminals

Theory:

There are several over current protection such as fuse, thermal relay & IDMT Relay. IDMT

(Inverse Definite Minimum Time) Relay is a high accuracy over current relay. If we does not want

to flow the current in lines more than 1 Amp, we will set the tripping current in our relay 1 Amp.

As the current will become 1.10 or 1.20, the relay disc will start forward and trip the breaker after

certain time. It is widely used to prevent over current on transmission lines, power transformers etc,

because the error & tripping time of the relay is tolerable by the lines and transformer.

As the requirement of system is that the faulted line should be open instantaneously. If the

faulted line breaker fails to open the faulted line, the next supply breaker have to be open to for

making dead the faulty line. The next breaker may be at higher voltage line or the same voltage.

The next breaker should open only after the first breaker failure. So we will allow approx 0.4 sec

time to operate first breaker. If first breaker dose not become open within 0.4 sec than it will be

assume failure and the next breaker will become functional. These time and current distinguish is

made by IDMT relay. 1

The time multiplier setting for an inverse time relay is defined as :-

T

TMS =

Tm

Where T = the required time of operation.

Tm = The time obtained from the relay characteristics curve at TMS = 1.0 and using PSM

equivalent to max fault current.

Current setting is adjusted by means of a tapped plug bridge hence known as PSM.

Fault Current

PSM =

Plug Setting

The operating time of an over current relay tends to become asymptotic to a definite

minimum value with increase in the value of current. This is inherent in electromagnetic relays due

to saturation of magnetic circuit. So, by varying the saturation, different characteristics are

obtained. If the core is made to saturate at large stage, it is called IDMT relay. The time

characteristics in inverse over the same range and then after saturation occurs and definite min.

time is reached. At low value of operating current the shape of current is determined by effect of

restraining force of control spring while at higher values the effect of saturation predominates.

Different time multiplier setting (TMS) are obtained by varying the travel of disc or cup

required to close the contracts. Higher is the TMS, greater is the spring restraining force. In order to

make the relay operate at constant value of minimum trip current for any TMS greater holes are cut

in the disc.

TYPE CDG RELAY

Type CDG Relay consists of an induction disc unit with an operation indicator and in some

cases, an instantaneous high set unit all assembled on a standard frames.

The disc shaft carried a silver rod moving contact which completes the aux-unit circuit

through the fixed contact. Permanent magnet is used to control the disc speed.

The setting is adjusted by the movement of the back stop which is controlled by rotating a

knurled moulded disc at the base of the graduated time multiplier.

2

The self powered relay has an auxiliary unit which is powered by a secondary winding on

the electro-magnet through a bridge rectifier. When the current through the relay exceeds its

setting, the disc unit operated and closes its contacts to complete the path for secondary winding.

The auxiliary unit connected across the rectifier now picks up and one of its own N/O contacts

reinforces the disc contact. The other contacts of this auxiliary unit are available for tripping and

alarm purposes.

Relay modes fitted with instantaneous element are also available.

The instantaneous high set unit, type CAG13/CAG17 attracted armature relay, if powered,

is fitted at the top left hand corner of the relay, with an operation indicator reset by a separate push

rod protruding through the relay case. The pick up current of CAG 13 unit is adjusted by means of

means of knurled knob which rotates a calibrated scale. The pick up current of CAG 17 unit is

adjusted by rotating calibrated white dial fitted above the CAG 17 unit. A separate publication

covers the operation and commissioning instruction for these relays.

For exact connections, references should be made to the panel schematics and relay wiring

diagram. The self powered auxiliary lelement is fitted with indicator. The single disc contacts is

rated to make and carry for 0.5 Sec. 2500 VA with maximum of 10 Amps and 660 Volts AC or

DC. The two pairs of electrically separate, self or and rest contacts provoded in the auxiliary unit

are rated to make and carry of 0.5 Sec. 7500 VA with maximum of 30 amps and 660 volts AC or

DC.

SELECTION OF TIME SETTING

For selective operation when there are a no of relays connected in series, the relay farthest

from the source should be set to operate in the min possible time. For succeeding relays towards

source a time delay step is given. For over current relays (inverse time) the time setting should be

done at the max fault current. If the relay has proper selectivity at max. fault current it will

automatically have a higher selectivity at the min fault current as the curve is more inverse on lower

current region.

SELECTION OF CURRENT SETTING

It is necessary to calculate the max fault current which can occur at each relay position. On

a radial system the lowest setting must be at the farthest end, the setting being in creased for the

subsequent relay towards the source. As per Indian standards the operating value should not exceed

130% of the setting.

3

Minimum Sht. Ckt Current

I setting =

1.3

Procedure:

1.Study the operating current & de-operating current of disc.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Keep the current source at minimum.

The amp adj / relay test rotary switch is kept at AMP ADJ.

Switch ON the test set.

Increase the current source slowly and pay attention at disc of relay.

At certain current, it just moves in forward direction, this current is operating

current and note the current.

Now decrease the current through current source and pay hard attention at disc.

The disc will stop at certain current and moves in reverse direction just after

reducing the current. This current is de-operating current and note its value.

Observations:

S NO.

1.

2.

3.

4.

5.

6.

7.

PLUG SETTING

1 A

OPERATING

CURRENT

1.06 A

DE-OPERATING

CURRENT

0.90 A

5

3. Study the time current characteristics at various multiples of plug setting current.

(i)

(ii)

(iii)

(iv)

(v)

We have to study the graph between PSM and time at TMS-1 etc.

Keep the current source at minimum.

The amp adj / relay test rotary switch is kept at AMP ADJ.

Switch ON the MCB/Isolator.

Increase the fault current upto required PSM (refer table). it is quit possible that

while adjusting the fault current the FLAG of the Relay might trip for that you have

to RESET the FLAG by moving the marked shaft UPWARD denoted by (RELAY

FLAG RESET) for resetting the FLAG the Toggle switch must be brought in OFF

position and the marked shaft move UPWARD.

(vi) Now the desire Fault Current is SET and relay FLAG RESET - Only when the disk

has move fully anti clockwise. Now move the Toggle Switch in Relay Test and press

the green push button and timer counting will START and counting will STOP once

the relay is operated. Note down the time in seconds.

(vii) Now for various T.M.S. (Time Multiplier Setting) and P.S.M. (Plug Setting

Multiplier), the time taken by the relay to operate at various fault current may be

noted down.

(viii) Now plot the graph between time take for the relay to operate Vs Plug Setting

Multiplier at various T.M.S.

Observations: S

NO.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

PLUG

SETTING (PS)

1 A

PSM

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

10.0

FAULT CURRENT

= PS x PSM

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

10

TRIPPING

TIME

6

4. Study to find out the plug setting and TMS (Time Multiplier Setting).

(i) When we want to use this relay, we know only the fault current and the delay time,

the system can withstand the fault.

(ii) The fault current is given 1000 amp, the delay time is 0.9 secs, and the CT used

100/1. Normal running current less than 100 amps and our system can tolerate upto

100 amps.

(iii) Plug setting = Tolerable current / Primary current of CT.

= 100 / 100 = 1 Amp

(iv) PSM = Fault Current in secondary of CT / PS

= 10 Amp / 1 = 10 Amp

(v) From the already discussed graph at PSM 10, TMS at 1, the delay time is shown 3.5

sec approximate.

(vi)

(vii)

TMS = T / Tm

= 0.9 / 3.5 = 0.25

Verify the practically delay time by setting the TMS and PS as calculated.

5.

(viii) Practically delay time = 0.8686

Study of front plate numerical values.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

At front plate plug setting (PS) is given to set the relay tripping current.

Some numerical values given between plug setting multiplier (PSM) and time.

We will calculate the PSM from the formula as discussed.

The seconds shows the delay in tripping at given PSM.

Auxiliary volt is used in some relays to operating the flag relay.

The amp shows approx tolerance of relay i.e. 1 amp relay should not in use more

than 1 amp.

7

RESULT:

MERITS OF THE RELAY

1.

2.

3.

The relay is electromechanical type can tolerate rough-tough use.

High accuracy in respect of thermal relays.

Normally no false tripping .

DE-MERITS OF THE RELAY

1.

2.

The relay is electromechanical type so it should be used in dust free environment.

The frequency variations can change the parameters.

PRECAUTIONS

1. The relay should be used very carefully and fast, beyond its rating.

2. Suppose the 1 amp relay can withstand upto 1 amp continuously and if more than 1

amp the relay should trip the system within few seconds.

EXPERIMENT NO. 3

PERCENTAGE DIFFERENTIAL RELAY STATIC TYPE

AIM:

Study of operation of relay on different operating currents.

Find operational characteristics of the relay.

MATERIAL REQUIREDS: 1. Relay Single Pole Version 1 A (Numerical Type) 'AREVA' make MBCH-12

2. Timer

3. Auto Transformer 0-270V, 10 A

4. Ammeter (10 A, AC) - 2 Nos.

5. Neon lamp 1A, AC, 230 V

6. Rheostat 5 A, 45 Ohms - 1 Nos.

7. Rheostat 10 A, 20 Ohms - 1 Nos.

8. Isolation Transformer

9. Auxiliary DC Supply Unit with Transformer

THEORY

It is a very important protection of the transformer. It is based on the ratio of

H.T. current and L.T. current should be constant. Consider the Fig No '1', here we

considering the single pole of 132/33 KV Transformer. It's H.T. current and L.T. current

ratio will be 1:4. If the CT of H.T. side is considered 100/1 Amp, so the CT of L.T. side will

be 400/1 Amp. The secondary current of L.T. side CT and H.T. side CT will always equal in

normal condition. Both the secondary of CTs will enter in Numerical type % Differential

Relay. The secondary of CT connection is make in such a way that the CT current will flow

only through coil circuit and no extra current is to flow from Differential coil. As soon as the

fault occurs in transformer, the H.T. current will high. The ratio of H.T. current and

L.T. current will change. The secondary of H.T. side CT current will become high with

respect to secondary of L.T. side CT current. So the difference of current will flow through

differential winding. The secondary of differential winding transformer will go to an

electronic circuit that will operate a tripping relay to trip the breaker of main transformer. The

through windings are used to restraining the differential relay. It will more clearly by drawing

the curve between through current and differential current.

PRINCIPAL OF OPERATION

Considering the drawing given below, that two types of proportionate DC is available.

One is proportional to through current and another is proportional to differential current.

2

The through current always restraining the tripping and differential current always operating

the tripping relay. With a very simple way, we are understanding the circuit that in an

amplifier we applies differential current DC to the base of transistor to operate the tripping

coil and through current DC applied to the emitter of transistor to restrain/block the tripping.

By the resistance we set the blocking and tripping of relay as per our requirement.

PROCEDURE:

Experiment No 1.

Study of operation of relay on different operating currents.

1. Turn the current source to minimum.

2. Connect the Rheostat 5 Amp, 45 Ohms to differential point.

3. Select the moving part of Rheostat at center.

4. Select the current by DIP switch at the relay.

5. Switch ON the MCB.

6. Push the green start button.

7. Increase the differential current gradually by the current source.

8. Observe that the differential current will increase the set value by DIP switch,

9. The relay will issue the tripping command and trip LED will become glow.

10. We can reset the trip relay by resetting button.

Experiment No 2.

Find operational characteristics of the relay

1. Turn the current source to minimum.

2. Connect the Rheostat 5 Amp, 45 Ohms to differential points.

3. Connect the Rheostat 10 Amp, 20 Ohms to through points.

4. Select the moving part of Rheostats 5 Amp, 45 Ohms at maximum ohms.

5. Select the moving part of Rheostats 10 Amp, 20 Ohms at center.

6. Select the current by DIP switch at the relay.

7. Switch ON the MCB.

8. Push the green start button.

9. Increase the through current gradually as given in table by the current source.

10. Increase the differential current gradually upto the tripping of the relay and record in the table.

11. Repeat the same process to get the differential values at different through current.

Observation Table:

S.No Restraining Current (A) Operating Current (A)

Precautions: 1. The relay is designed to handle 1 Amp current. The study of relay more than 1 Amp, required fast

operation.

2. Do not touch any electronic part with hand, otherwise the relay may damaged by static current.

RESULT:

EXPERIMENT NO. 4

EARTH FAULT RELAY

AIM: Find operational characteristics of the relay for time and current setting.

MATERIAL REQUIREDS: 1. Voltmeter (0-300 V)

2. Ammeter (0-10 A)

3. Loading C.T.

4. Auto Transformer 0-270V

5. Neon lamp 230 V

6. Earth Fault Relay Electromechanical type CDG

7. Timer with Start & Stop facility

8. Push Button for Timer START & STOP

9. DP Switch

THEORY: This manual covers the commissioning and maintenance instruction for non-

directional inverse time over current relays belonging to the CDG 11 family, which are self

powered. There is no need for any separate auxiliary DC or AC supply for these type of

relays. These relays are available in the standard current setting ranges of 50-200% for

present case, 20-90% and 10-40% of 1 A or 5 A. Triple Pole version designate as CDG 31

consists of identical 3 units.

Type CDG 11 is an inverse time over current relay with a definite minimum time. It is

used for protection against phase and earth faults. The 50 Hz version has a time current

characteristics which confirms to IS 3231 (1965).

The time multiplier setting for an inverse time relay is defined as :-

T

TMS =

Tm

Where T = the required time of operation.

Tm = The time obtained from the relay characteristics curve at TMS = 1.0 and

using PSM equivalent to max fault current.

PROCEDURE: 1. Switch ON the MCB, the Voltmeter will show the line voltage.

2. Initially Rotary Switch should be in Amp Adjust Position.

3. Now to set the desired fault current we will be using current source. For that

switch ON the Rotary switch and move the current source till the desired fault

current is indicated on the Ammeter, it is quit possible that while adjusting the

fault current the FLAG of the Relay might trip for that you have to RESET the

FLAG by moving the marked shaft UPWARD denoted by (RELAY FLAG

RESET) for resetting the FLAG the Toggle switch must be brought in OFF

position and the marked shaft move UPWARD.

4. Now the desire Fault Current is SET and relay FLAG RESET - Only when the

disk has move fully anti clockwise. Now move the Rotary Switch in Relay Test

and press the green push button and timer counting will START and counting will

STOP once the relay is operated. Note down the time in seconds.

5. Now for various T.M.S. (Time Multiplier Setting) and P.S.M. (Plug Setting

Multiplier), the time taken by the relay to operate at various fault current may be

noted down.

6. Now plot the graph between time take for the relay to operate Vs Plug Setting

Multiplier at various T.M.S.

OBSERVATION:

(i) TMS = 1.00, 0.9, 0.7, 0.5 PS = 1.0 Amp

Current Time PSM

RESULT:

EXPERIMENT NO.5

THERMAL RELAY

EXPERIMENT NO.5

THERMAL RELAY

AIM: 1. Find operational characteristics of the relay

2. Determine time current characteristics of given fuse.

MATERIAL REQUIREDS: 1. Auto Transformer

2. Thermal Relay

3. Fuse 3A

4. Ammeter 10 Amp

5. Loading C.T.

6. Timer

THEORY: Thermal relays are current sensitive devices. They operate on thermal effect of

electric current. This relay senser current by temperature rise, which is produced by the

higher current. These relays can also respond to understand 3 currents, which cause rise in

temperature due ti their negative sequence component.

Fuse:

Fuse is a current interrupting device which opens the circuit in which it is inserted,

when the current in circuit exceeds a certain value. The part of the fuse which is designed to

melt, when the fuse melts it is called fuse link

PROCEDURE:

For Thermal relay:

1. Switch ON the DP Isolator Switch.

2. Select both the rotary switches to Thermal relay side.

3. Now put the current source at minimum setting.

4. Push the Green Push Button marked on right hand side of the panel.

5. Now set desired current by current source and note the corresponding tripping time of the relay.

For Fuse:

1. Switch ON the DP Isolator Switch.

2. Select both the rotary switches to Fuse Test Side.

3. Now put the current source at minimum setting.

4. Push the Green Push Button marked on right hand side of the panel.

5. Now set desired current-by-current source and note the corresponding tripping time of the relay.

6. Press the push button for fuse test till the fuse block and timer stops thus giving the blowing time of the

fuse.

RESULT:

EXPERIMENT NO. 6

BUCHHOLZ'S RELAY AIM:

To study the Buchholz's Relay

MATERIAL REQUIREDS:

1. Experimental setup consisting of Buchholz Relay,

2. Transformer Tank,

3. Air Cylinder,

4. Conservator,

5. Breather and Air Pump THEORY:

Buchholz's Relay is a gas actuated relay used for protecting oil immersed

transformer against all types of internal faults and marks use of the fact that faults

decompose oil thus generating gases.

Fig-'1' shows such as a relay. It is connected by means of pipes between the

top of the transformer tank and the conservator as shown in Fig-'2' and is, therefore,

under normal condition of operation, full of oil.

It consists of a cast housing containing a hinged hollow float, inside of which

is mercury switch, the float being located in the upper part of the housing, and a

hinged flat value to which is attached a similar mercury switch, located directly in the

path of the oil between the bank and the conservator.

When an incipient fault (such as insulation fault between the turns, breakdown

or core insulation, core heating, bad switch contact, fault joints) occurs in the

transformer bubbles of gas are evolved by the heat generated and rise upto the tank

surface and thence into housing of the devices where their through passage is

prevented by the flap valve. The bubble pass to the top of the housing, causing the oil

level to fall, where upon the hollow float tilts and when the mercury switch contacts

are short-circuited by the mercury, an alarm device is operated. If however a serious

fault such as internal short-circuited between phases or turns, earth fault puncture of

bushing insulator inside tank occurs the volume of gas generated is considerable

which in moving through the relay causes the gas surge flap value to be deflected,

thereby closing the mercury contact switch and energising the trip coils of the circuit

breakers.

The connecting pipe between the tank and conservator should be as straight

possible and should slope upwards towards the conservator at a small angle from the

horizontal.

PROCEDURE:

1. Connect the pump to air cylinder and close the outlet valve outlet valve of air cylinder.

2. Pressurize the air cylinder upto the pressure of the 30 lbs.

3. Release the air in the transformer tank by opening outlet valve of air cylinder.

4. NC contact of relay will change to open position and NO contact will

change to close position and will comeback to open position.

5. Change in contact position is indicated by glowing of lamp on the panel board.

6. Release the air from the relay by slowly opening the air release valve of the relay.

7. Properly close the air release valve.

RESULT:

EXPERIMENT NO.7

PERFORMANCE OF LONG TRANSMISSION LINE UNDER NO & LIGHT LOAD

AIM: To study the performance of a long transmission line under no load and under light load conditions.

MATERIAL REQUIREDS: 1. Transmission line load model of Pie network

2. Two Digital voltmeter and Ttwo digital ammeter

3. Resistive, Inductive, Capacitive loa

4. LPF wattmeter

5. Patch cards

THEORY: Ferranti Effect:

A simple explanation of Ferranti Effect can be given by approximating the distributed parameters of the line

by lumped impedance as shown in Fig -'1'.

Line representation (Lumped) under No Load Condition Fig 1

Phasor Diagram for Line representation (Lumped) under No Load Condition Fig 2

Since usually the capacitor reactance of the line is quite large as compared to the

inductive reactance under no load or lightly loaded conditions the line current is of

leading P.F. The phasor diagram is given as fig -'2' for thus operating conditions.

The charging current produces drop in the reactance if the line which is in phase

opposition to the receiving end voltage and hence the sending end voltage becomes smaller

than the receiving end voltage.

For finding out inductance KVAR required to obtain zero% voltage regulation

variable shunt inductor is connected at the receiving end and its value is changed till we get

voltage at receiving end equal to that of voltage at the sending end.

PROCEDURE: 1. Apply the voltage (200 V max.) to the sending end and connect power factor

meter. Also connect 1 ammeter and voltmeter to each end (receiving and

sending).

2. Connect the load comprising of R, L and C at the receiving end and note down

the value of receiving end voltage.

3. Now remove the load from the receiving end and note down the voltage on

receiving end. This voltage at the receiving end is quite large as compared to

sending end voltage.

OBSERVATION:

Load Vs Is Vr Ir Resistive Inductive Capacitance No Load

RESULT:

EXPERIMENT NO. 8

FAULT LOCATION IN CABLE BY MURRAY LOOP TEST AIM: To locate fault in a cable by Murray Loop Test.

MATERIAL REQUIREDS:

1. Rheostat 1.1 A, 800 Ohms - 2 Nos 2. Digital Multimeter

3. Galvanometer - 1 No.

4. Measuring Tape (5M) - 1 No

5. 3Core Cable (25M) - 1 No

6. DC Power Source - 1 No

THEORY: MURRAY LOOP TEST

This test is carried out for locating a ground or a short circuit fault, provided

that a cable runs along with the grounded cable or with two cables (or with two cores

of a multi-core cable) which are short circuited. The advantage of loop test is that the

resistance of the fault does not affect the results obtained. Provided this resistance is

not very high. Otherwise it may adversely affect the sensitivity.

The connection diagram for conducting Murray Loop Test is shown in Fig '1'

.

Connection diagram for locating short circuit fault by Murray Loop Test – Fig – 2

Connections of Fig -'1' are used for locating a ground fault and those of

Fig -'2' for a short circuit fault. As is evident from these figures, both circuits are essential

1. Take a multicore cable (say 3 core) of known length (say 25M).

2.Measure the resistance of each core.

3. Make connections as shown in Fig - '1'. Short circuit the two cores of

the cable at the other end. Adjust P and Q such that balance is obtained.

Note P, Q and calculate distance of fault x. Take three-four observations

and take the mean of calculated value of length of the fault from each set

of readings. This length should be equal to the distance of fault from the

lower end of resistance Q.

4.Note down the actual distance of fault by measuring the actual distance

of fault and calculate the % error.

5.Make connections as shown in Fig -'2'. Short circuit any two cores of

cable to create short circuit fault. Adjust P and Q such that balance is

obtained. Note P and Q in the observation table. Calculate x with.

6.Take three-four observations and find average of x. calculate the

distance of short circuit fault from the measuring end of the cable.

PROCEDURE

7.Note down the actual distance of fault by measuring of actual fault

distance and calculate the %age error.

Result:

OBSERVATION

Total length of the cable = 25 Meter

r = resistance per meters = 0.9/25 = 0.036

Localization of Earth Fault :-

S P Q x

No. (Ohm) (Ohm) = 2l [Q/(P+Q)]

1.

2.

3.

699

609

799

172.7

148.3

194

9.91

9.79

9.77

Distance of fault

from measuring

end = x (m)

(Average of x1,

x2, x3 )

9.82

Actual %

location Error

of fault

10.00 1.8

Localization of Short Circuit Fault :-

S P Q X

No. (Ohm) (Ohm) = 2l [P/(P+Q)]

1.

2.

3.

161

185

165

670

761

678

10.32

9.78

9.77

Distance of fault

from measuring

end = x (m)

(Average of x1,

x2, x3 )

9.96

Actual %

location Error

of fault

10.00 0.4

EXPERIMENT NO. 9

SINGLE LINE TO GROUND FAULT

AIM: To Study Single Line to Ground Fault as practical application in transmission lines.

MATERIAL REQUIREDS:

Theory:

1. Instantaneous Earth Fault Relay Type CAG

2. 'Areva' make Electro Mechanical Type

3. Digital AC Voltmeter 0-300 V,

4. Push Button

5. Rotary Switch

6. Indicating Light

7 .Fuse

8. DP Switch

9. Insulating Terminals

10. Transformers

11. Line Impedances

12.1 Ph Variac with loading CT

The Single line to Ground fault comes under Shunt type of fault (Short Circuit) this

fault occur due to one broken conductor on transmission line. The shunt faults are

characterise by increase in voltage & frequency and fall in current in the faulted phase.

The need for line fault study is because these fault mostly occur in transmission lines.

These are broadly classified as 3 Phase fault & 1 Ph to ground fault. Fault point which may

occur in transmission line should be isolated from rest supply system after the fault occurs.

The rest transmission line should remain normal as usual without hindrance.

Here to study the above fault in the lab we will be creating mock faults at various

points and first calculating the theoretical values of current at the point where the fault occurs

and then practically verifying the calculated values.

For this purpose we have considered a single phase system (generally used in railway

Traction). Operating voltage for MAIN TRANSFORMER referred to as (T) is 230/125 V and

the transmission line impedance to be 2+2 Ohms each and the secondary transformer referred

to as (t) as 125/74 V. Also we have provided an Instantaneous Earth Fault Relay (Electro

Mechanical Type) showing the tripping of line at the time when fault occurs.

WORKING OF RELAY

1. Whenever breakdown or fault occurs on a transmission line some kind of protection is

essential in order to isolate the faulty point from the rest of the transmission line. The

Instantaneous Earth Fault Relay is one of the protective relays.

2. This relay is generally used in the distributing lines such as 33 KV, 11 KV and 0.4

KV etc.

3. As soon as a fault occurs on any line tripping command issues from this relay to

breakers, 24 V DC is extended through this Relay to tripping coils of the 33 KV or 11 KV

Breakers. In the Breakers of 0.4 KV, AC/DC 220V/110V is given as per design and the line

is again charged. In our lab experiment we are using a Contactor as a Breaker. This tripping

Relay shorts the coils of an other Auxiliary relay thereby coil denergizes and the supply to

contactor coil is disconnected. Locking and also over current is recovered the Instantaneous

Earth Fault Relay becomes normal and Auxiliary Relay whose coil was short circuited by the

Instantaneous Earth fault Relay is again operated, therefore our system again comes in the

normal manually start position. If it happens practically on site, the patrolling is making in

Fault area, fault locates and removes.

WORKING OF PANEL

At first we have to see that any fault has not been framed on the panel. If it is so then

we have to make it free from any such fault by disconnecting all the leads.

1. VOLTAGE RATIO TEST

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Turn the knob of the variac to bring it to zero indication.

Variac O/P is connected to the Transformer (T) I/P.

Connect the transformer (T) Input & Output to the Voltmeter.

Press the 'ON' push button.

Increase the Variac voltage upto 230 volts & the secondary voltage should be

125 volts.

Feed the Output from Transformer (T) to the Input of Transformer (t).

The Output of Transformer (t) shall be 74 volts.

st

Voltage Ratio Test 1 Transformer (T)

Input - 230 Volts

Output - 125 Volts

-

-

115 x 2 = 230

62 x 2 = 124

Voltage Ratio Test 2 nd Transformer (t)

Input - 125 Volts

Output - 74 Volts

-

-

62 x 2 = 124

36 x 2 = 72

2. PERCENTAGE IMPEDANCE OF TRANSFORMER (%Z)

Formula =

Full load short ckt voltage

Applied Voltage

x 100%

(a) for Transformer (T)

(i) Full Load current of Transformer (T) 124 Volts side

1000

= = 8.06 A

124

Zero the Variac knob.

Connect the Variac Output to 230 Volts side of Transformer (T).

Connect the Voltmeter to Variac output.

Short the 125 Volts side.

Connect the Tong tester at 125 volts side lead.

Increase the voltage by Variac gradually so that the current in tong tester

reaches 8.00 A.

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii) Read the voltage at output of Variac it should be 14.0.

11.62

% impedance of Transformer (T) = x 100

230

= 5.05 %

(b) for Transformer (t)

(i) Full Load current of transformer (t) 74 Volts side

500

= = 6.94 A

72



Connect the transformer (T) 125 V side to 125 V side of transformer (t).

Connect the voltmeter to 125 volts of transformer (T).



(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii) Increase the voltage by Variac gradually so that the current in tong tester

reaches 6.76 A.

(ix) Read the voltage at output of Transformer (T) it should be 12 Volts.

% impedance of Transformer (t) =

14.14

124

= 11.40 %

x 100

% IMPEDANCE OF TRANSFORMER 'Z' ON 100 MVA BASE

Formula : %Z on base MVA =

% Z x 100

MVA (Transformer Capacity)

For Transformer (T) %Z on base MVA

5.05 x 100

=

0.001

For Transformer (t) %Z on base MVA

11.40 x 100

=

0.0005

= 505217.39 %

= 2280645.16 %

LINE IMPEDANCE ON 100 MVA BASE

Line Impedance x 100 x 100

Line Impedance On 100 MVA Base = 2

(KV)

Line Impedance = 2 + 2 = 4

KV = 0.125

4 x 100 x 100

= = 2560000% 2

(.125)

Framing of Network for faults

Red Jumpers are to be connected as network design or samples as given below.

1.

2.

3.

SAMPLE - I

Only one transformer & Line.

One transformer line & other transformer.

Line & other transformer etc.

As a sample-I we are considering the given design & fault at point F. Considering zero line

impedance.

Equivalent N/W design on 100 MVA Base

I1 I2 I3

505217.39 2280645.16

Fault Impedance = 609000 + 1920000 = 2785862.55 %

100 x 100

Formula : Fault MVA =

Fault MVA =

Fault Impedance

100 x 100

2560000

= 0.003589552

Formula I =

Fault MVA x 1000

Volts in KVA

0.003589552 x 1000

= 49.85 Amp

.072

I3 =

I2 =

I1 =

0.003589552 x 1000

.124

0.003589552 x 1000

.230

= 28.94 Amp

= 15.60 Amp

Practical Value

(i)

(ii)

(iii)


Design the N/W as per design on panel.

Apply the voltage 23 Volts at port 230 Volts side of Transformer (T).

The measured current values

are at 23V :-

I3

I

2

I

1

= 5.03 A

= 3.10 A

= 1.78 A

So for 23 x 10 = 230 V

I3

I2

I1

= 5.03 x 10 = 50.3 A

= 3.10 x 10 = 31.0 A

= 1.78 x 10 = 17.8 A

Theoretical Value :-

I3

I2

I1

= 49.85 A

= 28.94 A

= 15.60 A

Hence the practical value & theoretical values are nearly same and little difference due to

lead impedance only.

SAMPLE - II

As a sample II we are considering the given design fault at point F & line impedance 4 Ohms.


I1 I2 I2 I3

505217.39 2280645.16 2560000

Fault Impedance = 609000 + 1920000 + 2560000 = 5345862.55

100 x 100

Fault MVA = = 0.001870605

5345862.55

0.00187 x 1000

I3 = = 25.90 Amp

.072

0.00187 x 1000

I2 = = 14.96 Amp

.124

0.00187 x 1000

I1 = = 8.13 Amp

.230

Practical Values

Procedure as take before.

The measured current So for 23 x 10 = 230 V

values are at 23V :-

I3

I

2

I

1

= 2.52 A

= 1.53 A

= 0.91 A

I3

I2

I1

= 2.52 x 10 = 25.2 A

= 1.53 x 10 = 15.3 A

= 0.91 x 10 = 9.1 A

Theoretical Value :-

I3

I2

I1

= 25.90 A

= 14.96 A

= 8.13 A

Hence the practical value & theoretical values are nearly same and verified.

PRECAUTIONS

1.

time.

2.

time.

Result:

The voltage at 230 volts side of transformer (T) should not exceed 250 volts at any

The voltage at 125 volts side of transformer (t) should not exceed 140 volts at any

EXPERIMENT NO. 10

THREE PHASE FAULT

Aim : To Study Three Phase Fault as practical application in transmission lines.

INSTRUMENTS REQUIRED

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

Three Phase Earth Fault Relay, Static Type

Digital AC Voltmeter 0-600 V,

Push Button

Rotary Switch

Indicating Light

TP Isolator

DP Switch

Insulating Terminals

Transformers

Line Impedances

Digital Clamp on Meter

-

-

1 No.

2 No.

- 1 No

-

-

2 No.

1 No.

WHY THREE PHASE OCCUR ON TRANSMISSION LINES?

Three Phase occurs in 11 KV and 33 KV due to :-

(a)

(b)

(c)

Tree falling

By Birds

Oftenly occurred due to hotline wire broken.

NEED OF FAULT STUDY CALCULATIONS

1. The point of state electricity supply advice the fault occur (either three phase or single line

to ground fault) must be isolated from the transmission line supply so that the rest of the

transmission supply line should be normal without any hindrance. For this purpose 11 KV fault

point should be isolated immediately and 33 KV system at 11 KV fault should be tripped after 0.5

second.

2. First the chance is to be given at 11 KV protection and if that fails 33 KV protection will

operate and isolate the fault zone. Similarly 132 KV, 220 KV and 400 KV protection will operate

after giving the chance at lower voltage side.

3. A definite time relay is used to give the proper time of lower voltage side protection. For

operation of protections the fault current at various points should be known, hence after knowing

the minimum & maximum fault current various protections are used hence this study is required.

THEORY

The three phase fault comes under Shunt type of fault (Short Circuit) this fault occur due

to one broken conductor on transmission line. The shunt faults are characterize by increase in

voltage & frequency and fall in current in the faulted phase.

The need for line fault study is because these fault mostly occur in transmission lines.

These are broadly classified as 3 Phase fault & 1 Ph to ground fault. Fault point which may occur

in transmission line should be isolated from rest supply system after the fault occurs. The rest

transmission line should remain normal as usual without hindrance.

Here to study the above fault in the lab we will be creating mock faults at various points

and first calculating the theoretical values of current at the point where the fault occurs and then

practically verifying the calculated values.

For this purpose we have considered a three-phase (four wire) system. Operating voltage

for MAIN TRANSFORMER referred to as (T) is 400/220 V and the transmission line impedance

to be 2+2 Ohms each phase and the secondary transformer referred to as (t) as 220/133 V. Also

we have provided an Instantaneous Earth Fault Relay (Static Type) showing the tripping of line at

the time when fault occurs.

BASIC BLOCK DIAGRAM

WORKING OF RELAY

1.Whenever breakdown or fault occurs on a transmission line some kind of protection is

essential in order to isolate the faulty point from the rest of the transmission line. The

Instantaneous Earth Fault Relay is one of the protective relays.

2.

3.

This relay is generally used in the distributing lines such as 33 KV, 11 KV and 0.4 KV etc.

As soon as a fault occurs on any line tripping command issues from this relay to breakers,

24 V DC is extended through this Relay to tripping coils of the 33 KV or 11 KV Breakers. In the

Breakers of 0.4 KV, AC/DC 220V/110V is given as per design and the line is again charged. In

our lab experiment we are using a Contactor as a Breaker. This tripping Relay shorts the coils of

an other Auxiliary relay thereby coil denergises and the supply to contactor coil is disconnected.

Locking and also over current is recovered the Instantaneous Earth Fault Relay becomes normal

and Auxiliary Relay whose coil was short circuited by the Instantaneous Earth fault Relay is again

operated, therefore our system again comes in the normal manually start position. If it happens

practically on site, the patrolling is making in Fault area, fault locates and removes.

4. It is a static definite time relay whose definite time is almost 1.5 to 2.0 seconds as soon as

out of two lines, any line current will increase the preset amp values at relay and remains for

almost 2 seconds, the relay will issue the trip command and contactor will open in case the failure

of auxiliary supply or low voltage less then 300 Volt the protection will fail so the relay will also

issue the trip command to open the contact point. The relay is manual reset type and tripping

indication will appear at the relay. After resetting the relay can be reused.

5. First we will create the fault which is having high fault impedance so that our system can

tolerate in our lab i.e. the fault current should be in permissible limit so that our equipments and

supply should able to bear the fault current values.

6. During fault condition our source should be able to supply fault current so that fault

condition occurs by short circuiting 133 V side of transformer (t). Both the line resistance should

be in fault. In this condition the current at 400 side is approx 12 Amp and our relay can tolerate

upto 5 amp and relay will issue a trip command thus the relay is operated.

WORKING OF PANEL

At first we have to see that any fault has not been framed on the panel. If it is so then we

have to make it free from any such fault by disconnecting all the leads.

1. VOLTAGE RATIO TEST

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Turn the knob of the three phase variac to bring it to zero indication.

Variac O/P is connected to the Transformer (T) I/P.

Connect the transformer (T) Input & Output to the Voltmeter.

Press the 'ON' push button.

Increase the Variac voltage upto 400 volts & the secondary voltage should be 220

volts.

Feed the Output from Transformer (T) to the Input of Transformer (t).

The Output of Transformer (t) shall be 133 volts.

Voltage Ratio Test 1 st Transformer (T) Star/Star

Input - 440 Volts

Output - 220 Volts

nd

Voltage Ratio Test 2 Transformer (t) Delta/Star

Input - 220 Volts

Output - 132 Volts

2. PERCENTAGE IMPEDANCE OF TRANSFORMER (%Z)

Full load short ckt voltage

Formula =

Applied Voltage

for Transformer (T)

MVA - 0.003

x 100%

(a)

Voltage Ratio - 400/226 V

Phase to Neutral voltage secondary side is - 127 V

Full load secondary current = 7.87 A (T)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)



Connect the Voltmeter to Variac output.



Increase the voltage by Variac gradually so that the current in tong tester reaches

7.87 A.

(viii) Read the voltage at output of Variac it should be approx 13 V.

Primary voltage at secondary full load short ckt current - 13 V

13

% Z T

= x 100 = 3.25

400

(b) for Transformer (t)

MVA - 0.0015

Voltage Ratio - 220/132 V

Phase to Neutral voltage secondary side is - 76.21 V

Full load secondary current - 6.56 A (T)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)



Connect the transformer (T) 220 Volts side of transformer (t).

Connect the voltmeter to 220 volts of transformer (T).



(viii) Increase the voltage by Variac gradually so that the current in tong tester reaches

6.56 A.

(ix) Read the voltage at output of Transformer (T) it should be 9 Volts.

Primary voltage at secondary full load short ckt current - 9 V

9

% Z t = x 100 = 4.09

220

% IMPEDANCE OF TRANSFORMER 'Z' ON 100 MVA BASE

% Z x 100

Formula : %Z on base MVA =

MVA (Transformer Capacity)

For Transformer (T) %Z on base MVA

3.25 x 100

= = 108333.33 %

0.003

For Transformer (t) %Z on base MVA

4.09 x 100

= = 272726.66 %

0.0015

LINE IMPEDANCE ON 100 MVA BASE

Line Impedance x 100 x 100

Line Impedance On 100 MVA Base = 2

(KV)

Line Impedance = 2 Ohms

KV = 0.220

2 x 100 x 100

= = 413223.14 2

(.220)

Framing of Network for faults

Jumpers are to be connected as network design or samples as given below.

1.

2.

3.

Only one transformer & Line.

One transformer line & other transformer.

Line & other transformer etc.

SAMPLE - I

As a sample-I we are considering the given design & fault at point F. Considering zero line

impedance.


I1 I2 I3

108333.33 272726.66

Fault Impedance = 108333.33 + 272726.66 = 381059.99

100 x 100

Formula : Fault MVA =

Fault MVA =

Fault Impedance at 100 MVA Base

100 x 100

381059.99

= 0.02624

Fault current at transformer (t) output

Fault MVA x 1000

•

Formula I =

1.732 x KV

0.02624 x 1000

1.732 x 0.132

I3 = = 114.00 Amp

• Fault current at transformer (T) output

I2 =

0.02624 x 1000

1.732 x 0.220

= 68.8 Amp

• Fault current at transformer (T) input

0.02624 x 1000

I1 =

1.732 x 0.4

= 37.8 Amp

Practical Value

(i)

(ii)

(iii)


Design the N/W as per design on panel.

Apply the voltage 20 Volts at port 400 Volts side of Transformer (T).

The measured current values These current values at 400 V Theoretical Value :-

are at 20V :-

I3

I

2

I

1

= 5.44 A

= 3.58 A

= 1.91 A

shall be :-

I3

I2

I1

= 5.44 x 20 = 108.8 A

= 3.58 x 20 = 71.6 A

= 1.91 x 20 = 38.2 A

I3

I2

I1

= 114.0 A

= 68.8 A

= 37.8 A

Hence the practical value & theoretical values are nearly same and little difference due to lead

impedance only.

SAMPLE - II

As a sample II we are considering the given design fault at point F & line impedance 2 Ohms.

Line Impedance = 2

KV = 0.220

2 x 100 x 100

= 2

(.220)

= 413223


I1 I2 I2 I3

108333.33 413223.00+413223.00 272726

Fault Impedance = 1207505

•

100 x 100

Fault MVA =

1207505

Fault current at transformer (t) output

= 0.0082815

Formula I =

•

Fault MVA x 1000

1.732 x KV

0.0082815 x 1000

I3 =

1.732 x 0.132

Fault current at transformer (T) output

0.0082815 x 1000

I2 =

1.732 x 0.220

Fault current at transformer (T) input

= 36.22 Amp

= 21.7 Amp

•

0.0082815 x 1000

I1 = = 11.95 Amp

1.732 x 0.4

Practical Values

Procedure as take before.

The measured current values are These current values at 400 V Theoretical Value :-

at 400 Volt Applying 40 V :- shall be :-

I3

I

2

I

1

= 1.31 A

= 2.45 A

= 3.64 A

I3

I2

I1

= 1.31 x 10 = 13.1 A

= 2.45 x 10 = 24.5 A

= 3.64 x 10 = 36.4 A

I3

I2

I1

= 11.95 A

= 21.7 A

= 36.22 A

Hence the practical value & theoretical values are nearly same and verified.

PRECAUTIONS

1.

2.

3.

The voltage at 400 volts side of transformer (T) should not exceed 440 volts at any time.

The voltage at 220 volts side of transformer (t) should not exceed 230 volts at any time.

Neutral is to be connected on input side.

MPORTANT FORMULAS

Transformer

%Z x 100

%Z on 100 MVA Base =

MVA

%Z =

Full Load Short Ckt x 100

Applied Voltage

Line Impedance

Line Impedance on 100 MVA Base =

Z x 100 x 100

2 (KV)

Fault MVA =

Fault Current I f =

100 x 100

Fault Impedance

Fault MVA x 1000

1.732 x KV

220 KV line Imp/KM = 0.418 Ohms 79.1O

132 Line Imp/KM = 0.16 + 0.408

or Magnitude Value = 0.44 Ohms

33 KV Line Imp/KM = 0.484 Ohms

100 x 100

Fault MVA =

Zf

Fault MVA x 1000

Fault MVA =

KV (Ph to N)

After T/F Delta/Star Ph to Ph fault feet then fault current

=

Fault MVA x 100

KV (Ph to Ph)

`

Documents

Power System Lab Manual Original and Modified