Power System Protection Manual

Note:

This manual is in the formative stage. Not all the experiments have been covered here though they are operational in the laboratory. When the full manual is ready, we will make it available here. Electrical Engineering Department, Birla Vishvakarma Mahavidyalaya (BVM) Engineering College, Vallabh Vidyanagar, Gujarat, India Pin:388120.

Experiment No-1

Radial Feeder Protection Theoretical Background: Whole of the power system can be subdivided in to number of radial feeders fed from one end. Generally such radial feeders are protected by over current and earth fault relays used as primary relays for 11 kV and 66 kV lines. For lines of voltage rating beyond 66 kV, distance protection is applied as a primary protection whereas over current and earth fault relays are used as back up relays. A simplified radial feeder network without transformers (in actual practice transformers do exist at substations) is shown in single line diagram of fig. 1.1 below. SoSo

R1

urce

Fuse

ToLoad

R2 R3

STransmissionLine

Section I Section II Section III

A B C D

R1

urce

Fuse

ToLoad

R2 R3

STransmissionLine

Section I Section II Section III

A B C D

Fig. 1.1. A Typical Radial Transmission Line If the fault occurs in distribution network, fuse should isolate the faulty section. Should the fuse fail, relay R3 shall give back-up protection. Relays R1, R2, and R3 act as primary relays for faults in section I, section I, and section III respectively. If fault in section III is not cleared by relaying scheme at relaying point R3, relay R2 will act as a back-up. Similarly back-up protection is provided by relay R1 for faults in section II. A,B, C and D are substations in fig. 1.1. Generally Inverse time overcurrent relays with Definite Minimum Time feature (IDMT relays) are used in practice. There are many types of such relays available in relay-market, viz. normal inverse relays, very inverse relays and extremely inverse relays. The characteristics of these relays are shown in fig. 1.2. The other types of o/c relays are 3 second relay and 1.3 second relay. This means the time of operation of the relay is either 3 or 1.3 second at Plug Setting Multiplier (PSM) equal to 10. Long time inverse relays are used for o/c cum overload application. Voltage restrains o/c relays have their own application. Very inverse relays are less prone to the ratio ZS/ZL. Extremely inverse relays are yet better. Very inverse relays are faster in operation for close-in faults yet maintaining the discrimination with fuse and other relays. Extremely inverse relays are more meritorious

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in this aspect too. Instantaneous o/c relays are not immune to ZS/ZL ratio. Definite time o/c relays are 100 % immune to this ratio. Very inverse relays can be used with an additional advantage while protecting a machine or a transformer as they match with the heating characteristic of equipment better than their normal inverse equivalent. Extremely inverse relays can best co-ordinate with the fuse characteristic. The aim of this

experiment is to reveal these facts experimentally.

Fig. 1.2 Normal, Very and Extremely Inverse Characteristics

aboratory Simulations:L 1.3. a live model of a radial feeder fed from one end can be Referring to a.c. circuit of fig.

self-understood.

Fig. 1.3 Main AC Circuit in the Experiment.

A

V

R1 R2 R3230 V, 50 Hz1-phase, ac supply

C1-1 C2-1 C3-110/5 10/5 10/59Ω 9Ω

MCB

550Ω(Load)

18Ω(Fault resistance)

S1 S2 S3

A

V

R1 R2 R3230 V, 50 Hz1-phase, ac supply

C1-1 C2-1 C3-110/5 10/5 10/59Ω 9Ω

MCB

550Ω(Load)

18Ω(Fault resistance)

S1 S2 S3

Section - I Section - II Section - IIISection - I Section - II Section - III

2

This is only a single phase version of a radial feeder. Transmission lines are simulated by

Fig. 1.4 Control Circuit in the Experiment.

9 ohms resistors as we are studying only the steady state behavior of the relays and the network. Circuit breakers are simulated by contactors. Distributor is protected by a 0.5 Amp M.C.B. Semaphore indicators on the panel show the status of the contactor(whether ON or OFF). Visual neon lamp indictors are also used. Faults in different sections can be created by switches S1, S2 and S3. Fault limiting resistance of 18 ohms is used for practical purposes only, as otherwise the source would get shorted for a fault at start of the first section. For fault in distributor, Ohmic value of load rheostats can be decreased. MCB simulates fuses or MCCB. In actual practice C.T. secondary rated current (1 Amp or 5 Amp) and relay rating should be same. Here C.T. secondary rating is 5 Amp and relay rating is 1 Amp. This is contradicting the practice for for practical purpose. 1111 DCDC SupplySupply

(a)

C1

PB-1

PB-2

C1-2

A1-1

C2

PB-1

PB-2

C2-2

A2-1

C3

PB-1

PB-2

C3-2

A3-1

A1

R1-1A1-2

PB-3

A2

R2-1A2-2

PB-3T2

A3

R3-1A3-2

PB-3T3

0 V

L1 L2 L3C1

PB-1

PB-2

C1-2

A1-1

C2

PB-1

PB-2

C2-2

A2-1

C3

PB-1

PB-2

C3-2

A3-1

A1

R1-1A1-2

PB-3

A2

R2-1A2-2

PB-3T2

A3

R3-1A3-2

PB-3T3

0 V

L1 L2 L3

(b)

A1-3

A2-3

A3-3

Buzzer

bulb110 VDC Supply

C1-3 C2-3 C3-3

S-1 S-2 S-3

S-1, S-2, S-3 are semaphore indicators

A1-3

A2-3

A3-3

Buzzer

bulb110 VDC Supply

C1-3 C2-3 C3-3

S-1 S-2 S-3

S-1, S-2, S-3 are semaphore indicators

3

Referring to control circuit of figure 1.4(a) a d (b), any section can be manually charged

Fig. 1.4 Control Circuit in the Experiment.

bservations and calculations:

nor made off using “start” (PB1) and “stop” (PB2) push buttons, which are spring loaded. On occurrence of fault, the corresponding section relay will operate and the concerned auxiliary relay A1, A2 or A3 will energize giving signal to the concerned contactor and making it off. This will also activate the buzzer and bulb which can be reset using “Accept” pushbutton PB3. Back up can be shown by using switches T1 and T2 on the panel. Time of operation of relays can be measured by a time –interval counter connected as shown in fig. 1.4 (c). S1S1S1 A1-4A1-4A1-4

(c)

S2

S3

To timer “Start”

terminals

A2-4

A3-4

To timer “Stop”

terminals

S2

S3

To timer “Start”

terminals

S2

S3

To timer “Start”

terminals

A2-4

A3-4

To timer “Stop”

terminals

A2-4

A3-4

To timer “Stop”

terminals

O extreme faults in sections I, II and III by adjusting the

Fault Location Fault Current (A)

1. Measure the fault currents forcorresponding rheostat in minimum (zero resistance) and maximum (full resistance) positions and using the corresponding fault-switch S1, S2, or S3 (refer fig. 1.3). Maximum fault currents in sections I, II and III are denoted by F1, F2, F3 respectively and the minimum fault currents by F1’, F2’, F3’ respectively. Deactivate the relays for this purpose. Record the readings in table 1:

F1 F1’ F2 F2’ F3 F3’

Table:1.1 Measured fault currents for extreme faults in each section

. Calculate the plug –settings (or Tap Value) of relays R1, R2 and R3. (Plug settings will

ent

2be same irrespective of type of relays ). For this purpose, assume the number of distributors each of 0.5 Amp rating from the following possibilities (The experim

4

is a simulation and hence the distributor current is 0.5 Amp. In actual practice it may be 500 Amp or more or less): Number of distributors: 1, 2, 3 or 4.

l be done w.r.t. following considerations.) t

varies from 1.05 to 1.3 times the plug setting of the relay.

3. e characteristic of relay R3 with that of an

1.0 second between two characteristics. This will

ating R2 with R3, use the worst possible current to decide TMS of R2 (F3

ormal inverse Relays.

P.S. TMS

(Calculations of the plug setting shal(I) The plug setting shall be more than or equal to the maximum full-load currenpassing through the relay. (II) The pick up of the relay(III) For back up, relay R1 shall reach for the fault F’2 and R2 for the fault F’3. Time Settings. (Normal Inverse Relays) For deciding time-settings, co-ordinate thMCB.( Refer figures 1.5 and 1.6.) Use discriminating time-interval ofdecide TMS of R3. Why discrimination time of 1.0 second? (Try answer to this question.) For coordinin table:1.1). Similarly decide TMS of R1 by using F2 and setting of R2. Use discriminating time interval between two successive relays as 0.4 seconds (why 0.4 seconds?) for these calculations. Tabulate the results as follows:

NRelay

R1 R2 R3

Table:1.2 Calculated relay settings.

. Time-settings (Very Inverse Relays)

can be carried out for very inverse relays also.

e 1.5 and 1.7 for deciding TMS of very inverse relays. 5.

ays for extreme faults in

7. ries

8. III. For

9.

4Similar exercise as at Sr. No. 3 aboveThe results are to be tabulated as in the case of normal inverse relays as per calculated settings. Use figurSet the normal inverse relays as per the settings in table 1.2.

6. Calculate the time of operation of the main and back-up releach section using the relay settings in table:1.2 and the fault current readings recorded in table:1.1. Enter these in table:1.3 as “Calculated Time of Operation”. Vary the 550 ohms load resistance such that the current in the radial feeder vafrom 0.5 Amp to about 4 Amp. See that MCB trips and relay R3 does not trip. Now create extreme faults (one by one) in each section starting from section each fault, measure the time of operation of the main and the back-up relay (to measure time of operation of back-up relay, the main relay has to be deactivated using switches T2 or T3 (as the case may be) in fig. 1.4). Record these in table 1.3 as “Measured Time of Operation”. Calculate the error between the calculated and the measured time of operation for each fault and record it in table: 1.3. Derive the table 1.4 from table 1.3 as follows:

5

Table 1.4 Difference in time of operation of relays for extreme

faults.

1R2 II R3 III

Relays

Section Observed Calculated

R I

10. Replace the normal inverse relays by very inverse relays.

uestions: e circuit of the experiment.

cator? wer than the settings of the overcurrent

4. a.c. circuit and d.c. control circuit for two overcurrent and one earth-fault

5. by time discrimination?

s? in remote back-up protection.

r

10. Why are the IDMT relays popular in practice? ng settings of phase relays and ground

11. Repeat steps at Sr. No. 5 to 9 for very inverse relays. 12. Draw your own conclusion. Q1. Explain th2. What is the function of semaphore indi3. Why are the settings of the earth-fault relays lo

relays? Draw anscheme of protection of a feeder used in practice. How does our experimental scheme differ from that? Why? What do you understand

6. What do you understand by overshoot of a relay? 7. What is the significance of resetting time of a relay8. What do you understand by back-up protection? Expla9. How does the source impedance affect the choice of relay to be used in radial feede

protection?

11. What are the factors to be considered for decidirelays?

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7

8 8

EXPERIMENT NO. 2

PARELLEL FEEDER PROTECTION

Theoretical Background: In power systems, power is also fed by parallel feeders. A simplified single-line diagram of parallel feeder network (without transformers) is shown in fig. 1. Referring to the fig. 1 let the fault occur at F1. There are two paths for the fault current. (I) AF1 and (II) DCBF1

Fig. 1 Parallel feeder network t is required that for fault at F1, CB 1 and 2 should trip. Similarly CB 3, 4 should trip for

f fault occurs at F1, R3 should operate and R4 should give back-up to relay R3. This

o solve this problem we have to introduce an additional directional feature in the relays

ow, for the fault at F2 since the current through R3 is in the non-operative direction, the

Source

ToLoadS

B

CD

R1

R2

R3

R4

A F1

F2

1

3

2

4Source

ToLoadS

B

CD

R1R1R1

R2R2R2

R3R3

R4R4

A F1

F2

1

3

2

4

2

4

Ifault at F2. In this scheme if we use all overcurrent relays, the following problem arises: Imeans, time of operation of R4 should be more than that of R3, current being same. Let the settings be done according to this criterion. Now, for the fault at F2, it is required that relay R4 operate and relay R3 operates to provide backup. This fact requires that relay R3 should operate later than R4. But the relay settings are not satisfying this condition. Hence by using simple IDMT o/c relays it is impossible to fulfill both the criteria. TR3 and R4. These relays will operate only if the fault current is flowing away from the bus. Relays R1 and R2 are kept as non-directional IDMT o/c relays. Nrelay doest not operate. Relay R4 operates as the directional criterion is satisfied. Hence for fault at F2, R2 and R4 operate and if R4 fails to operate R1 provides the backup. Relays operate on similar lines for fault at F1.

1

2

aboratory Simulations:L arallel feeders fed from one end is self explanatory. This is

ance of 9Ω, 12 A is

bservations and Calculations:

Referring fig. 2, a model for pa single-phase version of the parallel feeders. One feeder is simulated by two 9 Ω, 12 A rheostats in series. Load is simulated using a 185 Ω, 1.1 A rheostat. Faults are simulated using switches S1 and S2 and a limiting resistused to limit the fault current. The control circuit is similar to that in the radial feeder simulation (Refer to fig. 3, 4 and 5) except for the additional PT to feed directional overcurrent relay. Remote backup can be seen by closing a CT secondary shorting switch at the rear of the panel (thus, making the corresponding relay non-operational). O

ap Value T.V.) and T.M.S. of relays theoretically

R NO Relay P.S. T.M.S.

Calculate the Plug Setting P.S (or Tsuch that if one line trips the other healthy line should be able to feed the full power till the load is shed. Remember relay R1 should provide backup to R4 and R2 to R3. Select some arbitrary T.M.S. for R3 and R4 and find the T.M.S. of R1 and R2 by keeping proper discrimination time arrival. Tabulate the settings as follows: S1 R1 2 R2 3 R3 4 R4 Procedure:

nections as per the circuit diagram.

are operating in desired manner.

uestions:

1. Do the con2. Set the relays according to your calculations. 1. 3 Create the faults and observe whether relays3. Observe the backup protection using C.T. shorting switch at rear of the panel. Q

ast three single line diagrams of power system network where Directional

2. ms of Directional relays. Line, Maximum Torque Line, MTA

4. you prefer? und directional relays.

1. Draw at lefeature is required. What are the proble

3. What do you understand by Zero Torquecharacteristic, Dead Zone, Characteristic Angle? Out of 300 , 600 and 900connection, which would

5. Draw figures for phase advancing techniques in phase and gro

EXPERIMENT NO.3 Principle of Reverse Power Protection:

Motivation: Reversal of power takes place when turbine does not get steam or water (in thermal/hydro plants). When alternator is connected to infinite bus and its turbine fails, machine works as synchronous motor driving the turbine as a pump; thus direction of current reverses. This experiment is designed to understand and demonstrate all possible conditions of prime-mover failure. Theoretical Background: When prime mover fails (steam input to turbine gets lost), generator works as a synchronous motor driving the turbine as the pump but obviously drawing a very small current w.r.t. its rated value. This condition is not dangerous for the alternator. The damage occurs to the prime mover. In modern power plants there are low forward power relays rather than reverse power relays. This protection is very important as not only it protects the prime mover, but also all class-B protections are routed through this protection (low forward power protection). Experimental Model: The main circuit is shown in fig. 1. The normal operating condition is achieved by keeping the switches S1 and S2 open. In this case, there are two parallel paths, each with 9 Ω resistance feeding a load simulated by a 185 Ω rheostat. The two paths model two 240V generators in parallel feeding a load of 185 Ω. The generator currents are measured through CT1 and CT2 and fed to relays D1 and D2 respectively. D1 and D2 are reverse power relays (in fact, low forward power relays to be precise), and hence are also provided with a voltage input through a Voltage Transformer (VT) shown in fig. 1. It is important to note here that D1 and D2 remain energized during normal condition. Closing of switch S1 will make the current through CT1 equal zero. Closing of S2 will make the current through CT2 equal zero. Either of these conditions simulates prime-mover failure and relay will de-energize in response. S1 and S2 are Double Pole Double Throw (DPDT) switches used simultaneously in the main (fig. 1) and control (fig. 2) circuits. The control circuit is shown in fig. 2. It shows that the contactors can be energized by push buttons PB1 and de-energized by PB2, same as for other experiments. The switch S3 simulates the steam valve interlock. This interlock does not allow the generator to start if the steam valve feeding steam to the turbine is closed. S3 should be kept open to bypass the lock, i.e., to let contactors close. Procedure: 1. Connect the required circuit. Keep S3 open. This means A1 and A2 can never energize.

So, A1-1 and A2-1 are always closed. The steam valve interlock is bypassed. 2. Make the contactors ON with pushbuttons. Circuit will energize and relays D1 and D2

will operate. Now close S3. This means that in the control circuit of fig. 2, D1-1 and D2-1 will close energizing D1X/D2X. Hence contacts D1X-1 or D2X-1 will open out, not allowing auxiliary relays A1 and A2 to operate. Thus, contactors remain ON.

1

3. Now to simulate prime-mover failure, make switch S1 on. Relay D1 will de-energize, hence D1X, deenrgizes. D1x-1 will close and the auxiliary relay A1 will operate. A1-1 will isolate the faulty turbo–alternator from bus. Buzzer, bulb and semaphore indicators operate the same way as for other experiments.

4. Open S1. Then open S3. Now make contactor C1 On with push button. 5. Close S2 to simulate zero power in alternator-2. On similar lines to what is explained

in steps 1-3, contactor C2 will open. 6. Draw your own conclusion.

2

3 3

EXPERIMENT NO: 4

DIFFERENTIAL PROTECTION OF TRANSFORMER Theoretical Background: The transformer is one of the most important devices and a very expensive equipment of a power system. The internal fault current due to short-circuit of winding within a transformer results in enormous damage to the winding and the core. Hence we use the most reliable, fast and the selective protective scheme, the differential protection scheme, which operates only for the internal faults. Some problems envisaged with differential protection scheme of transformer are listed below with their remedies: 1. The voltage ratings of both sides of transformer are different; hence the CTs used are

not identical. So there may be some “spill” current flowing through the relay that may cause mal-operation. Remedy: Use biased differential relay.

2. The full load current of transformer primary and secondary are different. CT should be selected such that the pilot wire current is the same on both sides otherwise mal-operation of relay would occur. To achieve this, the CT ratios required on primary and secondary side are different from the standard ratio. Remedy: CTs of standard ratios are employed in conjunction with interposing CTs.

3. At no load, the transformer secondary current is zero but the magnetizing current flows in the primary. Hence mal-operation of relay may occur due to the presence of spill current. Remedy: As the magnetizing current is only 1-2% of the rated current the basic setting of relay is done such that it does not operate at no-load.

4. During heavy external faults, CT ratio error will result in significant spill current flowing through the operating coil of the relay. Hence the relay can mal-operate. Remedy: Use biased differential relay.

5. 5.The power transformers are always provided with a tap-changing arrangement to regulate output voltage as demanded by the load. But CTs are chosen with respect to the normal tap. Hence, tap changing results in a spill current through the relay causing mal-operation. Remedy: Use biased differential relay.

6. When the transformer is switched on there is a large magnetizing inrush which depends on the instant of switching. This current is 8-10 times the normal rated current. Hence relay will mal-operate and will not allow the switching of transformer. Remedy: Use of DTIM-11 relay, or any comparable relay that provides second harmonic restraint.

7. The modern transformers are designed to operate near the knee point voltage of the B-H curve of their core. Hence 20% voltage rise would cause the rise in magnetizing current by 400%. This may cause mal-operation of relay. As the wave shape of the magnetizing current under this condition is rich of 5th harmonic, the 5th harmonic bypass circuit of the relay avoids the mal-operation.

1

In early days, a time delay was used to deal with the magnetizing inrush. This compromised relay performance when the transformer energizes with a fault already present in the system. The present remedies are: 1. Even harmonic cancellation. 2. Harmonic restraint. 3. Harmonic blocking. 4. De-biasing. Laboratory Simulation Fig. 1 shows the main circuit for demonstrating the biased differential protection. The function of PB1, PB2 and PB3 is to start, stop and to reset the circuit respectively. The switches S1, S2, and S3 are provided to make the fault at different locations. S1 is for an external fault, S2 is for an internal fault, and S3 is for a severe internal fault. The transformer has primary taps for 230V, 199.8 V, 115 V, 53.3 V and secondary taps for 110 V, 95.48 V, 55 V and 25.5 V. Procedure: PART-A Application of DTTM-11 for protection of single phase transformer 1. From the rating of the transformer (230/115 V, 1 kVA), the CT ratings (10/5 A on

primary, 20/5A on secondary), calculate the basic and bias settings of the relay, as well as the setting of the high-set unit.

2. Set the DTTM-11 relay according to calculations. Connect the circuit as shown. 3. Keep the secondary circuit open-circuited. Push “PB1” button to energize the

transformer with no load. Observe that the transformer can be energized without the relay operating. Repeat this process several times and observe the relay never operates while energizing the transformer. This is because DTTM-11 has the second harmonic restraint feature. This obviates the necessity of time delay during starting.

4. After switching the circuit off, complete the secondary circuit and make the circuit ON. Observe that relay does not operate for normal condition.

5. Make S1 ON, and observe that the relay does not operate for external faults as proved in the calculations. Switch S1 OFF.

6. Energize the circuit. Make S2 ON (internal fault on secondary side), and observe that relay 87 operates as proved in the calculations. Switch S2 OFF.

7. Energize the circuit. Make S3 ON (internal severe fault) and observe that unit 87 and the high set unit both operate as proved in the calculations. Switch S3 OFF.

8. Introduce a variable resistance across the secondary of CT on the secondary as shown in fig. 1. This simulates CT saturation or CT mismatch. Energize the circuit. Create an external fault using S1. Observe if the relay mal-operates. Decrease the CT secondary resistance till the relay mal-operates. Now switch off the supply, increase the bias setting and switch on the circuit again. Observe that for the same external fault, the relay does not operate.

PART-B Effect of tap-changing and effect of inrush current: 1. Shift the transformer to another table that has similar set-up but has ABB-make type

T relay. This relay does not have second harmonic restraint capability. Set this relay

2

with the same settings as for DTTM-11 relay. However, simulate the load resistance with an 18 Ω rheostat.

2. Repeat step 3 of part-A above. Observe that the relay mal-operates several times. The intensity of inrush depends on the instantaneous value of the input voltage when the transformer is energized, hence the relay does not mal-operate all the times. Now repeat the same process with increased time delay setting (this setting is provided with this relay). Observe the mal-operation ceases. However, this remedy is dangerous if the transformer is energized with an existing fault, especially in a part of system that has high fault levels. Keep the time delay in effect for the rest of the experiment.

3. Switch off the circuit and connect the secondary circuit. Switch on the transformer and observe the relay not operating for normal operation.

4. Change the primary tap to 199.8 V and repeat step-3. Observe the relay mal-operates. 5. Increase the bias setting and switch on the circuit. Observe that the relay does not

operate.

3

4

5 5

smbrahma

Fig. 1