CHAPTER FOUR

CHAPTER FOUR

RELAY COODINATION

1.0INTRODUCTION

Co-ordination of relays is an integral part of the overall system protection and is absolutely necessary to:

(a)Isolate only the faulty circuit or apparatus from the system.

(b)Prevent tripping of healthy circuits or apparatus adjoining the faulted circuit or apparatus.

(c)Prevent undesirable tripping of other healthy circuits or apparatus elsewhere in the system when a fault occurs somewhere else in the system.

(d)Protect other healthy circuits and apparatus in the adjoining system when a faulted circuit or apparatus is not cleared by its own protection system.

2.0Methods of Relay Co-ordination

A correct relay co-ordination can be achieved by one or other or all of the following methods:

Current graded systems

Time graded systems or Discriminative fault protection

Operate in a time relation in some degree to the thermal capability of the equipment to be protected.

A combination of time and current grading.

A common aim of all these methods is to give correct discrimination or selectivity of operation. That is to say that each protective system must select and isolate only the faulty section of the power system network, leaving the rest of the healthy system undisturbed. This selectivity and co-ordination aims at choosing the correct current and time settings or time delay settings of each of the relays in the system network.

3.0Co-ordination Procedure

3.1The correct application and setting of a relay requires knowledge of the fault current at each part of the power system network. The following is the basic data required for finding out the settings of a relay.

(a) A single line diagram of the power system.

(b) The impedance of transformers, feeders, motors etc. in ohms, or in p.u. or % ohms.

(c) The maximum peak load current in feeders and full load current of transformers etc, with permissible overloads.

(d) The maximum and minimum values of short circuit currents that are expected to flow.

(e) The type and rating of the protective devices and their associated protective transformers.

(f) Performance curves or characteristic curves of relays and associated protective transformers.

3.2The following are the guidelines for correct relay co-ordination:

(a) Whenever and wherever possible, use relays with the same characteristics in series with each other.

(b) Set the relay farthest from the source at the minimum current settings.

(c) For succeeding relays approaching the source, increase the current setting or retain the same current setting. That is the primary current required to operate the relay in front is always equal to or less than the primary current required to operate the relay behind it.

3.3Time Graded Systems3.3.1In this method, selectivity is achieved by introducing time intervals for the relays. The operating time of the relay is increased from the farthest side to the source towards the generating source. This is achieved with the help of definite time delay over current relays. When the number of relays in series increases, the operating time increases towards the source. Thus the heavier faults near the generating source are cleared after a long interval of time, which is definitely a draw back of this system of co-ordination. However, its main application is in systems where the fault levels at successive locations do not vary greatly.

3.3.2The diagram below represents the principle of a time graded over current system of protection for a radial feeder.

Protection is provided at sections A, B, and C. The relay at C is set at the shortest time delay in order to allow the fuse to blow out for a fault in the secondary of the distribution Transformer D. If 0.3 secs is the time delay for relay at C, then for a fault at F1, the relay will operate in 0.3 secs.

Relays at A, B and S do not operate, but these relays only act as back up Protection relays. For a fault at F2, the fuses blow out in say 0.1 secs and if they fail to blow out then the relay at C operates to clear the fault in 0.3 secs. It may be noted that between successive relays at C, B, and A etc there is an interval of time difference. This is known as Time Delay Step, which varies from 0.3 to 0.8 secs.

3.4Current Graded Systems.3.4.1This principle is based on the fact that the fault current varies with the position of the fault because of the difference in impedance values between the source and the fault. The relays are set to pick up at progressively higher currents towards the source. This current grading is achieved by high set over current relays and with different current tap positions in the over current relays. Since their selectivity is based solely on the magnitude of the current, there must be a substantial difference (preferably a ratio of 3:1) in the short circuit currents between two relay points to make them selective.

3.4.2A simple current graded scheme applied to the system as shown in fig 1 above will consist of high set over current relays at S, A, B and C such that the relay at S would operate for faults between S and A; the relay at A would operate for faults between A and B and so on.

3.4.3In practice the following difficulties are experienced with the application of purely current graded systems:

(a) The relay cannot differentiate between faults that are very close to, but are on each side of B, since the difference in current would be very small.

(b) The magnitude of the fault current cannot be accurately determined since all the circuit parameters may not be known exactly and accurately.

(c) There may be variations in the fault level depending upon the source generation, thereby necessitating the frequent change in the settings of the relay.

3.4.4 Thus discriminating by current grading alone is not a practical proposition for exact grading. As such current grading alone is not used, but may be used to advantage along with a Time Graded System.

3.5 Time and Current Graded System

3.5.1 The limitations imposed by the independent use of either time or current graded systems are avoided by using a combination of time and current graded systems.

3.5.2 It is for this purpose that over current relays with inverse time characteristics are used. In such relays the time of operation is inversely proportional to the fault current level and the actual characteristics is a function of both time and current settings. The most widely used is the IDMT characteristic where grading is possible over a wide range of currents and the relay can be set to any value of definite minimum time required. There are other inverse relay characteristics such as very inverse and extremely inverse, which are also sometimes employed. If the fault current reduces substantially as the fault position moves away from the source, very inverse or extremely inverse type relays are used instead of IDMT relays.

3.5.3 There are two basic adjustable settings on all inverse time (IDMT) relays. One is the TMS (Time Multiplier Setting) and the other is the current setting, which is usually called the PSM (Current Plug Setting Multiplier)

Time Multiplier Setting (TMS) = TTM

Where T=required time of operation

TM=time obtained from the standard IDMT curve at TMS = 1.

Plug Setting Multiplier (PSM)=Primary Current________________Relay operating current x C.T. ratio

3.5.4 As per B.S., there are two types of IDMT relays, namely 3.0 secs and 1.3 secs relays. This only means that with TMS = 1.0 and PSM = 10, the relay operates at the time of 3.0 secs or 1.3 secs as the case may be.

3.5.5 The time interval of operation between two adjacent relays depends upon a number of factors. These are:

(a) The fault current interrupting time of the circuit breaker.

(b) The overshoot time of the relay.

(c) Variation in measuring devices - Errors.

(d) Factor of Safety.

3.5.6Circuit breaker interruption time

It is the total time taken by the circuit breaker from the opening of the contacts to the final extinction of the arc and energization of the relay. Modern circuit breakers have an operating time or tripping time of 3 to 5 cycles in the EHV ranges and up to 8 cycles in the H.V and M.V ranges.

3.5.7Overshoot

When the relay is de-energised, operation may continue for a little longer until any stored energy has been dissipated. This is predominant only in electromagnetic relays but not in static relays.

3.5.8Errors

All devices such as relays, CTs etc are subject to some degree of error. Relay grading is carried out by assuming the accuracy of the measuring device or by allowing a margin for errors.

3.5.9Factor of SafetySome safety margin is intentionally introduced to account for errors and delays in breaker operating time.

The Phase-to-Phase fault current should be considered for phase fault relays and the phase to earth fault current for earth fault relays.

The setting for phase fault element (OCR) may be kept as high as 150 to 200% of full load current. Normally the minimum operating current is set not to exceed 130% of the setting i.e.

I setting=Minimum short circuit current1.3

The setting also depends upon the practices followed by a Power Authority and may be limited to 100% as in NEPA. In the examples that follow, we shall limit ourselves to 100% setting and it is advisable that we dont exceed this value most especially for transformer protection.

4.0 Examples on relay Co-ordination

4.1Data: Required to calculate relay settings of an IDMT 3 secs relay to operate in 2 secs on a short circuit current of 8000A. Connected C.T. ratio is 400/5A.

Normal full load current is 400A.

Relay Plug settings available 2.5, 3.75, 5, 6.25, 7.5, 8.75, 10

TMS: 0.1 to 1.0 in multiples of 0.1.

SOLUTION

Secondary value of short circuit current=8000 x 5

400

=100 A

Full load current = 400A

Secondary value of full load current=400 x 5

400

=5A

With 100% current setting IR =5A

Therefore Plug setting =5.0

Fault current of 100 A corresponds to 20 times IR i.e.

MPS=100 = 20

5

Looking into the relay characteristic curve, the time of operation for this value is 2.2 seconds at Unity TMS. If the relay is to operate in 2.0 sec., then

TMS= 2.0 = 0.9

2.2

i.e. from formulae Tu = To TMS

Or TO= TU x TMS

Alternatively:

TO = 0.14

TMS = 1 for 3 secs relays

MPS0.02 - 1

4.2 Data: Given a radial feeder with fault current and C.T. ratios at substations A, B, and C as indicated. Full load current at C = 100A.

Available relay is 1DMT 3 secs. Relay.

Find out the current setting P.S and TMS at each substation.

SOLUTION

We proceed from the farthest station towards the source.

Substation C

Secondary value of fault current =2000 x 5 _=50A

200

Full load current =100A

Secondary value of full load current =100 x 5 _=2.5A

200

For 100% setting our Plug set =2.5A =IRFault current of 50 A corresponds to: 50 =20 times IR2.5

Time of operation of the relay at 20 times IR with TMS = 1 is 2.2 secs (from relay characteristic curve)

Now the time of operation of relay at C has to be the lowest.

We assume this time equal to the sum of operating time of the fuse say 0.1 sec. and a time delay (of 0.16sec.) to allow the fuse to blow.

Actual time of operation of the relay at C is

=0.1 + 0.16=0.26 secs

TMS=0.26=0.12

2.2

At C:

P.S=2.5

TMS=0.12

Substation BThe relays at B must act at a time grading higher than that of relays at C.

Therefore we assume a time grading of 0.35 secs. (in our own case)

Relay operating time at B for a fault at C (i.e. a fault current of 2000A) is

=0.26 + 0.35=0.61 secs

The current setting at B must be increased when compared to that at C. We shall set this at 130% of that at C. This is in order to allow for load increases.

Current setting of the relay at B = 1.3 times current setting at C

=1.3 x 2.5=3.25

We choose a plug setting of 3.75

Secondary value of short circuit current at B is

=2000 x 5 =33.33A

300

Multiples of plug setting=33.33=8.88

3.75

The time of operation of the relay at MPS = 8.88 with TMS = 1 is 3.2 secs (from the relay characteristic curve)

TMS at 0.61 secs.=0.61=0.19

3.2

Secondary value of fault current at B

=3000 x 5__

300

=50A

But our Plug Setting PS =3.75

MPS =50 =13.33

3.75

The time of operation of the relay at MPS = 13.33 with TMS = 1 is 2.6 secs (from the relay characteristic curve)

But TMS chosen for the relay at B is 0.19

Actual operating time of the relay at B for a fault current of 3000A (a fault very close to B) is equal to:

To=Tu x TMS

=0.19 x 2.6

=0.49 secs.

Substation at A

Required operating time for relay at A for a fault current at B is:

=0.49 + 0.35 =0.84sec

Assume that PS at A = PS at B i.e. 3.75

Secondary value of fault current at B for relay at A:

=3000 x 5__ 300

=50A

Multiples of Plug Setting=50 _3.75

=13.33

With TMS = 1, operating time for this value of MPS = 13.33 is given as 2.6 sec.

TMS for the operating time of 0.84 secs

=0.84 =0.32

2.6

TMS at A =0.32

For a fault close to A, secondary value of fault current

=5000 x 5_ =83.33A

300

MPS

=83.33

3.75

=22.22

Time of operation of relay at 22.22 times IR at TMS = 1.0 is 2.2 secs (using 20 MPS available on the graph)

Actual time of operation of the relay at A is

=0.32 x 2.2 secs

=0.7 secs

SUBSTATIONCTRP.SActual Operating time of relays

A300/53.750.7 secs

B300/33.750.49 secs

C200/52.500.26 secs

4.3Given data on a 33 KV transmission line and substation as shown below. Determine the relay settings at the substations.

Fault level at station A =37.17MVA

Transmission Line constants for 29Kms:

Z1 =19.58 + j12.86 ohms

ZO =23.89 + j38.37 ohms

SOLUTIONAssume base MVA = 100

Source impedance at station A =Base MVA

Fault MVA

Zs =100__37.17

=2.69 p.u

Transmission line constants on base MVA in p.u

Z1 =([(19.58) 2 + (12.86) 2 ]

=23.43 ohms

Zp.u =Z1 x MVA

(KV) 2=23.43 x 100 (33) 2=2.15 p.u

Z0 =([(23.89) 2 + (38.37) 2 ]

=45.19 ohms

Zp.u =45.19 x 100

332

=4.15 p.u

Impedance of transformer at station B on 100 MVA base

Zp.u =%Z x base MVA_______Transformer MVA

=6.5 x 1001005

Zt =1.3 p.u

Total fault impedance at station B in p.u is:

Zf =Zs + Z1 + Zt

=2.69 + 2.15 + 1.3

=6.14 p.u.

Assuming a 3-phase fault on 11KV at station B

Fault MVA=Base MVA Zf=1006.14

=16.29MVA

Fault current =16.29 x 106(3 x 11 x 103=855A

RELAY CO-ORDINATION FOR 11KV FEEDER BREAKER OVER CURRENT RELAYFeeder CT ratio =100/5

Secondary value of fault current

=855 x 5__ 100

=42.75A

Assuming a full load current of 100A on the feeder

We have secondary value of full load current

=100 x 5__100

IR =5A

Hence we choose a P.S of 5.0

Fault current of 42.75A corresponds to 42.75 = 8.55 MPS

5

Time of operation for 8.55 times IR with TMS = 1 is given as 3.25 secs.

Now the time of operation of the feeder has to be the lowest.

Time of operation of relay =0.1 + 0.16 =0.26 secs.

Where 0.1sec =Fuse operation time on 11KV side

0.16sec =Time delay to allow fuse to blow

TMS =0.263.25

=0.08

For 11 KV feeder:P.S=5.0

TMS=0.08

RELAY CO-ORDINATION FOR 11 KV TRANSFORMER BREAKER OVER CURRENT RELAY (OCR)

Transformer bank C.T. ratio =300/5

Secondary value of fault current

=855 x 5__300=14.25A

Transformer secondary full load current

=5 x 106 ____

(3 x 11 x 103=262.5A

Secondary value of full load current

=262.5 x 5__ 300

=4.375A

Choose a P.S = 5.0

Fault current of 14.25A corresponds to 14.25 = 2.85MPS and with TMS = 1, the

5

time of operation = 6.29 secs.

Operating time required for the transformer breaker

=Relay operating time of feeder + time step delay

=0.26 + 0.35 =0.61 secs

TMS=0.61

6.29

=0.096=0.10

For 11 KV Transformer breaker:

P.S=5.0

TMS=0.1

33KV Line breaker relay co-ordination at station A (OCR)

Fault current on 33KV=855 A (by transformer ratio)

3

=285 A

CTR=100/5

Secondary value of fault current

=285 x 5_ 100

=14.25 A

33KV Transformer full load current

=5 x 106 _____

(3 x 33 x 103=87.5A

Secondary value of full load current

=87.5 x 5__100

=4.375A

We choose a P.S = 5A

MPS=14.25

5

=2.85

With Unity TMS, operation time = 6.29 secs

The operating time required is:

= Relay operating time of 11KV transformer breaker + step delay

=0.61 + 0.3

=0.91 secs.

TMS=0.916.29

=0.1446=0.15

For 33KV breaker at station A:

P.S=5.0

TMS=0.15

STATION B

11KV main breaker

11KV feeder breaker

STATION A

33KV line breakerP.STMSCTRRELAY

5.00.10300/5OCR

5.00.08100/5OCR

5.00.15100/5OCR

Earth Fault Relay Co-ordination

For transmission line and transformer Z1 = Z2Z0 of transmission line= 4.15 p.u

Z1 = 2.15 + 1.3 = 3.45 = Z2Z0 of transformer = 80% of Zt

= 0.8 x 1.3=1.04 p.u

Z0 =4.15 + 1.04 =5.19 p.u

Assume a single line to ground fault then:

Earth fault impedance=Zs + Z1 + Z2 + Z0 3

=2.69 + 3.45 + 3.45 + 5.19 3

=2.69 + 12.09

3

=6.72 p.u

Earth fault MVA on 11KV at station B

=Base MVA Zf

=100

6.72

=14.88 MVA

Earth fault current=14.88 x 106

(3 x 11 x 103

=781 A

Feeder CTR=100/5

Secondary value of Earth fault current

=781 x 5

100

=39.05A

For earth fault the P.S is kept at the lowest setting for the feeder and so also the operating time at the minimum say, 0.1 sec.

Therefore, P.S = 1.0

A fault current of 39.05 A corresponds to an MPS of 39.05 = 39.05 which

1.0

operating time at Unity TMS is given as 1.84 secs.

TMS=0.1

1.84

=0.05

Earth Fault Relay setting for the 11KV feeder is given as:

P.S=1.0

TMS=0.05

Transformer breaker CTR=300/5

Secondary value of fault current is:

=781 x 5__300

=13.02 A

P.S is again kept at the lowest value of 1.0 (IR)

The relay operating time will be

=EFR operating time of feeder + Time step delay

=0.1 + 0.3 = 0.4 secs

Fault current of 13.02 A will give an MPS of 13.02 = 13.02

1.0

With Unity TMS, Operating time = 2.66 secs.

TMS =0.4

2.66

=0.15

Earth Fault Relay setting for 11KV Transformer breaker is:

P.S=1.0

TMS=0.15

On 33KV bus at station A, 33KV line breaker CTR = 100/5

Secondary value of earth fault current

=781 x 5__

100

=39.05 A

Fault current on 33KV =39.05 3

=13.02 A (by transformation ratio)

With P.S = 1.0, MPS = 13.02 = 13.02 and at Unity TMS,

1

Operating time = 2.66secs

Relay operation time will be:

= EFR operating time + time step delay for transformer breaker

= 0.4 + 0.3 = 0.7 sec.

TMS=0.7

2.66

=0.26

Therefore Earth Fault Relay setting of 33KV line panel at station A is:

P.S =1.0

TMS=0.26

STATION BP.STMSCTRRELAY

11KV Feeder1.00.05100/5EFR

11KV Transformer breaker1.00.15300/5EFR

STATION A

33KV Line breaker1.00.26100/5EFR

PAGE 58