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Basic Information

O/C E/F Relay & Time Coordination 1

O/C E/F Relay &Time Coordination

Basic

Information

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General Circuit Diagram200/1 Amp

R Ph O/C (51R)

E/F (51N)

B Ph O/C (51B)

150 Amp

150 Amp

150 Amp

0.75 Amp

0.75 Amp

0.75 Amp

0.0 Amp

C11

C31

C51

C71

S1

S1

S1

S2

P1 P2

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O/C E/F Relay & Time Coordination 31S1R

1S2R

1S3R

1S1Y

1S2Y

1S3Y1S1B

1S2B

1S3B

2S1R

2S2R

2S3R

2S1Y2S2Y

2S3Y

2S1B

2S2B

2S3B

3S1R

3S2R3S3R

3S1Y

3S2Y

3S3Y

3S1B

3S2B

3S3B

R Ph CT

Y Ph CT

B Ph CT

Core-1Core-2Core-3

Core-1Core-2Core-3

Core-1Core-2Core-3

A11

A31

A51

A71

C11

C31

C51

C71

D71

D11

D31

D51

Yard MB Wiring

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O/C E/F Relay & Time Coordination 41S1R

1S2R

1S3R

1S1Y

1S2Y

1S3Y1S1B

1S2B

1S3B

2S1R

2S2R

2S3R

2S1Y2S2Y

2S3Y

2S1B

2S2B

2S3B

3S1R

3S2R3S3R

3S1Y

3S2Y

3S3Y

3S1B

3S2B

3S3B

R Ph CT

Y Ph CT

B Ph CT

Core-1Core-2Core-3

Core-1Core-2Core-3

Core-1Core-2Core-3

A11

A31

A51

A71

C11

C31

C51

C71

D71

D11

D31

D51

Yard MB Wiring

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Terminal Diagram of MiComP141


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Single Line to Ground Fault200/1 Amp

R Ph O/C (51R)

E/F (51N)

B Ph O/C (51B)

1500 Amp

7.5 Amp

7.5 Amp

C11

C31

C51

C71

S1

S1

S1

S2

P1 P2

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Electromagnetic Induction relays

50%75%100%125%150%

200%

1 2

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Relay Operation Time - 1


E/F PSM 30% i.e. 0.3 Amp

E/F Relay Current 7.5 AmpE/F Relay Current is 7.5/0.3 = 25 Times

its operating current

From Graph for 25 Times relay operating

current for TMS = 0.15 relay time of

operation would be @ 0.35 Sec

O/C PSM 100%

O/C Relay Current 7.5 Amp

It is 7.5 times relay operating current

From graph for 7.5 Times relay operating

current and for TMS = 0.1 time of

operation for the relay would be 0.35 Sec

( Zoom out Graph)
http://iec-a%20normal%20inverse%20jpeg.jpg/http://iec-a%20normal%20inverse%20jpeg.jpg/

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Actually our problem is to decide relay settings and not relay time of

operations as shown previously

Hence Unknowns are

Relay PSM

Relay TMS

Whereas known facts areRelay placement and purpose of use

Relay current during fault ( i.e. CT secondary current during fault. )

Relay desired time of operation.

General Steps

1) Decide PSM2) Find out fault current

3) Find out multiple of relay set current as per decided PSM in step-1

4) Find out time of operation for above multiple of current and TMS=1 using

relay characteristic curve

5) Decide relay time of operation as per protection needs

6) Find out TMS = Required Time of operation /Time of operation with TMS =1

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Basic Information Selection of PSM


E/F PSM generally selected as 30% ( Other than 30% settings may also be selected but about thisdiscussed somewhere else in the presentation)

For O/C PSM is selection depends upon place and purpose of use for example

1.Transformer O/C protection

a) Transformer HV or LV side O/C relay PSM settings should be in commensuration with transformer

full load current and respective CT ratio such that PSM = T/F Full load current / CT ratio

( Generally expressed in %)

b) For example for a 25 MVA transformer HV side full load current is 109 A if HV CT ratio is 200/1

Amp then PSM =109/200 55% ( exact value 54.5%)

c) For old type numerical relay it was not possible to go as near as possible to value calculated fromabove formula due to large steps available

d) Under such condition it is decision as per local condition to select higher or lower nearest PSM

e) In above example it is customary to select 50%, however due to this selection there is apparent

loss of about 10% capacity of the T/F

f) It is also possible to select 75% but load on transformer is to be monitored carefully ( and manually

)2.For 220-132 kV feeder

Here generally it is customary to select relay PSM as per-

a) Line conductor allowable loading limit

b) CT primary normal current

c) Substations capacity/normal load feed by the line

d) Considering above facts it is very common to select 100% PSM for 132kV lines with CT ratio 400/1

Amp

e) For 220kV lines with CT ratio 800/1 amp and conductor 0.4 ACSR or 0.525 AAAC it is 100%

a)For 33-11kV feedera) As per local feeder condition, load pattern and needs ranging between 50% to 100%

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Desired time of operation will depend upon

a) Equipment being protected

b) Time discrimination from down stream protection (150 ms 250 ms)

c) Time of operation of main protection etc.

For transformer LV side protection it is common to adopt 250 ms as operating

time. This is so as to have 150 ms time discrimination from 100 ms relay time of

operation for lower (feeder) protection.

When relays are used as backup protection of 132kV lines its time of

operation shall be equal to Z-2 time of operation (300 350 ms).

Once these two things decided there remains only mathematical part


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Worked out Example


400/1 A

132 kV 33 kV400/1 A

25 MVA

33kV Bus fault level

1Ph 170 MVA , 3Ph 210 MVARelay current during fault

1Ph 7.43 Amp, 3 Ph 9.18 Amp

Relay PSM

E/F 30%,O/C 100 %

Multiple of relay current

E/F 25, O/C 9.

Time of operation with TMS = 1

E/F 2.2 s, O/C 3.0 Sec

Desired time of operation

E/F 250 ms, O/C 250 ms

TMS

E/F 0.114, O/C 0.083

Roundup to

E/F 0.125, O/C 0.1

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More Information



More

Information

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Introduction

Fuse wire is simplest protection

Fusing ampere of copper wire of diameter d

expressed in Cm is given by the formula A =

2530*d3/2

Time taken by fuse to blow off depends up onfusing amperes


15

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Introduction

For a wire of length L carrying current I

and diameter d heat produced is H = I2R

H = I2 (L/A)

H = I2 ( L/(d2/4))

Heat dissipated = K (d)L ( i.e.

proportional to surface area where K

is constant of proportionality) Temperature will be steady state if heat

generated is equal heat dissipated or

I2 ( L/(d2/4)) = K (d)L

I2 ( 1/(d2/4)) = K d

I2 =K d3

I = K d 3/2

And by experiments for normal ambient

temperature value of K for copper is

determined as 2530 for d expressed in

Cm.


SWG D in mm D in Inch Amp FusingAmp

Fusing Amp byFormula

40 0.122 0.0048 1.5 3 3.41

39 0.132 0.0052 2.5 4 3.84

38 0.152 0.006 3 5 4.74

37 0.173 0.0681 3.5 6 5.76

36 0.193 0.0076 4.5 7 6.78

35 0.213 0.0084 5 8 7.86

34 0.234 0.00921 5.5 9 9.06

33 0.254 0.01 6 10 10.24

32 0.274 0.0108 7 11 11.47

31 0.29464 0.0116 8 12 12.80

30 0.315 0.0124 8.5 13 14.14

29 0.345 0.0136 10 16 16.21

28 0.376 0.0148 12 18 18.45

27 0.416 0.0164 13 23 21.47

26 0.457 0.018 14 27 24.72

25 0.508 0.02 15 30 28.97

24 0.559 0.022 17 33 33.44

23 0.61 0.024 20 38 38.12 More

16
http://9%20annexure.doc/http://9%20annexure.doc/http://9%20annexure.doc/

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Protection of transformer by a fuse


For T/F with normal load of 100 AmpFuse Transformer

Current FusingTime

Current SafeOperation Timeas perIEEE

SafeOperationTime WithFOS 2.5

200 10000 200 1800 720

430 5 300 300 120

1200 0.4 475 60 24

1800 0.2 630 30 12

2800 0.1 1130 10 4

2500 2 0.8

17

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Simplest Protection Fuse

Thesecharacteristicgraphs aregenerally double

log graph This is due to

including from verysmall to very largevalues on both axis


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19

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General mathematical formula for timecharacteristic of the relay as per IEC

Standards

K

Time Of Operation = ---------------------

( ( Is/I

b) - 1 )


O/C / & C 20

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General mathematical formula for time characteristic

of the relay shown on previous slide, with parameter

values for different curves are shown here


Characteristic K

Normal Inverse 0.02 0.14

Very Inverse 1 13.5

Extremely Inverse 2 80

Long Time Inverse 1 120

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O/C E/F R l & Ti C di ti 22

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Use of Log Scale-2



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Use of Log Scale-3



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Use of Log Scale-4


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Transformer Protection Damage Curve

Transformer damage curve as per IEEE

57.109 for class III transformers ( 5 MVA to

30 MVA )



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Protection of Transformer by O/C Relay


Trafo

Damage

Curve

Long Time

Inverse

Extremely Inverse

Normal Inverse


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End of More Information


After understanding basics of relaycharacteristic curves and its selection

according to protection needs we will

turn to allied information about O/C E//Frelaying

This allied information will prove helpful

in overall understanding aboutdevelopment of protective relays and its

use in power system


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Basic Information



Allied

Information


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Disadvantages of fuses

Though simple less accurate ( If Rewirable)

Because of previous heating effect

Ambient Temperature

In consistencies in material

Limitations for breaking capacities hence suitable for LV and to

some extent MV HRC Fuses

More accurate

Higher rupturing capacities

Requires time for replacement

Suitable for LV and to some extent MV

O/C E/F Relay & Time Coordination

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General Requirements of Protective Scheme

For any protective device following Functional

Characteristic are important.

Sensitive

Selectivity

Speed

Reliability

( Note:- 3 S & 1 R)

As a improvement over simple fuses (in aboveareas) other protective devices get developed

with the advancement of power system

http://link-functional%20characteristics.jpg/http://link-functional%20characteristics.jpg/

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Changing Trend In Protective Relaying

Protection relay is a tool for

protection engineer

During last 30 years relay

operating principles changed

very drastically

Electromagnetic Relays

Static Relays

Digital Relays

Numerical Relays

Though it is not required to

design a relay or repair a relay

at site it is customary to havesome working knowledge of

these relays for better

understanding and use of it


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Electromagnetic Induction relays


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Static Relays

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Numerical Relay

Functions

Available

inNum

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Introduction


R3 R2 R1

A B C

10 sec.25 sec.40 sec.

R3 R2 R1

A B C

200 ms220 ms180 ms

R3 R2 R1

110 ms350 ms500 ms

S

S

S


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Study of Time Co-ordination and its role in design of protection scheme.

Over Current and Earth Fault Protection is

used for

Protecting a equipment

Selective tripping of faulty section of the

power system

Backing up the main protection



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Role of Over Current Relay in Protecting the Equipment

It is obvious that over current protective system should

act and interrupt the fault current before to damage ofequipment due to fault current through it.

Power system equipments include Line, Isolator, CT,

Breaker, Transformer

Obviously Transformer is most costliest and delicate (forfault currents) equipment first we will consider its

damage curve and decide parameters of protection

system so that it should act fast enough to protect the

transformer This can be ascertained with the help of Damage Curve

of the transformer and time-current curve of the

protective system



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Role of Over Current Protection in Selective Tripping

It is obvious that only that part of the power system

should get disconnected where the fault exists Hence proper time co-ordination should be there so as to

let the down stream protection should act fast enough

and up-stream protection should give sufficient time for

down stream protection to act

Otherwise un-necessary larger area get affected



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Backup Protection

When ever main protection fails to separate the

faulty section backup protection take up this role As such there is inherent time delay in operation

of backup protection This backup protection can be employed in main

protection itself as additional function, butinvariably it is employed as a separate relay toensure its operation even if failure ofquantities/links which are common to bothfunctions such as- DC Source PT supply Relay hardware Main CTs

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y

Backup Relay Time Coordination

A C

E

F

X Y

Z

M

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