58889354-Oc-Relay-ion

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O/C E/F Relay &

Time Coordination

Basic

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200/1 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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1S3R 1S1Y 1S2Y 1S3Y 1S1B 1S2B 1S3B 2S1R 2S2R 2S3R 2S1Y 2S2Y 2S3Y 2S1B 2S2B 2S3B 3S1R 3S2R 3S3R 3S1Y 3S2Y 3S3Y 3S1B 3S2B 3S3B Y Ph CT B Ph CT Core-1 Core-2 Core-3 Core-1 Core-2 Core-3 Core-1 Core-2 Core-3 A31 A51 A71 C11 C31 C51 C71 D71 D11 D31 D51

Yard MB Wiring

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1S3R 1S1Y 1S2Y 1S3Y 1S1B 1S2B 1S3B 2S1R 2S2R 2S3R 2S1Y 2S2Y 2S3Y 2S1B 2S2B 2S3B 3S1R 3S2R 3S3R 3S1Y 3S2Y 3S3Y 3S1B 3S2B 3S3B Y Ph CT B Ph CT Core-1 Core-2 Core-3 Core-1 Core-2 Core-3 Core-1 Core-2 Core-3 A31 A51 A71 C11 C31 C51 C71 D71 D11 D31 D51

Yard MB Wiring

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200/1 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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50% 75% 100% 125% 150% 200% Φ 1 Φ 2

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E/F PSM 30% i.e. 0.3 Amp E/F Relay Current 7.5 Amp

E/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)

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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 are

Relay 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 PSM

2) 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

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E/F PSM generally selected as 30% ( Other than 30% settings may also be selected but about this discussed 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 from above 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 feeder

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

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400/1 A

132 kV 33 kV

400/1 A

25 MVA

33kV Bus fault level

1Ph 170 MVA , 3Ph 210 MVA

Relay 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

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O/C E/F Relay &

Time Coordination

More

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• Fuse wire is simplest protection

• Fusing ampere of copper wire of diameter ‘d’

expressed in ‘Cm’ is given by the formula A =

2530*d

3/2

• Time taken by fuse to blow off depends up on

fusing amperes

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• For a wire of length L carrying current I and diameter d heat produced is

• H = I2_R

• H = I2_{σ (L/A)}

• H = I2_{σ ( L/(πd}2_/4))

• Heat dissipated = K’ (πd)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_{σ ( L/(πd}2_{/4)) = K’ (πd)L}

• I2_{σ ( 1/(d}2_{/4)) = K’ d}

• I2 ₌_{K’’ d}3

• 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 Fusing Amp Fusing Amp by Formula 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

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For T/F with normal load of 100 Amp Fuse Transformer Current Fusing Time Current Safe Operatio n Time as per IEEE Safe Operation Time With FOS 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

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• These

characteristic

graphs are

generally double

log graph

• This is due to

including from very

small to very large

values on both axis

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• Log scale graph are

use full tool where

range of values varies

very widely

• This variation in range

is generally 10,000

times

• It does not affect

overall accuracy of

selecting proper value

manually

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• General mathematical formula for time

characteristic of the relay as per IEC

Standards

K

Time Of Operation =

(

( I

_s

/I

_b

)

α

- 1

)

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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

α

K

Normal Inverse

0.02

0.14 Very Inverse

1

13.5 Extremely Inverse

2

80

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• Damages to the equipment due to fault current flowing through it are mainly due to heating effect of the current ( I2_Rt)

• Hence fuse time characteristic initially suited very well to the equipments in the power system

• This figure shows protection of transformer with the help of relay and breaker

• This also indicates how inverse characteristic of O/C Relay is suitable to protection of power system equipments • ( More about Transformer Damage Curves)

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• Transformer damage curve as per IEEE 57.109 for class – III transformers ( 5 MVA to 30 MVA )

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Trafo Damage Curve Long Time Inverse Extremely Inverse Normal Inverse

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After understanding basics of relay

characteristic curves and its selection

according to protection needs we will

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

relaying

This allied information will prove helpful

in overall understanding about

development of protective relays and its

use in power system

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O/C E/F Relay &

Time Coordination

Allied

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• 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

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• This simple device (Fuse) played a very

vital role during early development of

power systems

• As the complexity of power system

increased other technique get introduced

like breaker, relay DC battery etc. (

How?

)

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• History of power system protection dates back nearly to the start of development of power system it self

• In real sense power system started growing due to invention of incandescent lamp by Edison during 1880

• Edison was promoter of DC power system ( Why ? )

• General Electric founded by him was main supplier of electricity in Newyork.

• Washington first introduced AC system with the advancement in transformer during 1887

• During 1890 charls introduced symmetrical component analysis which helped in analyzing 3 ph. Power system and there by possible to design larger machines and power systems.

• Modern day power system came into existence from 1890 • One of the patent of fuse is in the name of Edison

• Development of relays breakers and instrument transformers took place during 1890 to 1920 and modern day protection system came into existence.

• And during last century development of power system continuous to be there however main principles of power system protection are 3S and 1R remained same.

• Development of relays breakers and instrument transformers took place during 1890 to 1920 and modern day protection system came into existence.

• And during last century development of power system

continuous to be there however main principles of power system protection are 3S and 1R remained same.

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• 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 above

areas) other protective devices get developed

with the advancement of power system

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• Sensitivity is that property of protection system which enables it to distinguish between fault and no fault condition very correctly.

• As if we say that some animals are more sensitive than humans to natural disasters like earthquake.

• Where as selectivity is that property of the power system which enables it to isolate only the faulty part from healthy part.

• In this sense differential protection is most selective protection • Once the fault detected by SENSITIVE system and area to be

disconnected detected by SELECTIVE system then there comes the SPEED.

• This faulty section should be get cleared as early as possible.

• For EHV system Faults are once in blue moon. Hence this all above said things should happen RELIABELY even after 5-10 years from design and commissioning of the protection system.

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• 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 have some working knowledge of these relays for better

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F u n cti o n s A va ila b le in N u m e ric a l O /C R e la y

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R3 R2 R1

A B C

110 ms 350 ms

500 ms

1) Consider a representative part of a power system as shown above. 2) It is being protected by over current relay

3) Typical expected time of operation for over current relays are as shown 4) In next couple of hour we will see

a) What is mean by relay characteristics curve

b) How relay characteristic curve suites our protection needs c) How it helps us in deciding relay time of operation

d) Workout relay settings so that they shall operate at expected time

e) Methodology being adopted for selective tripping by over current relay including directional relay

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R3 R2 R1 A B C 10 sec. 25 sec. 40 sec. R3 R2 R1 A B C 200 ms 220 ms 180 ms R3 R2 R1 110 ms 350 ms 500 ms

S

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• Over Current and Earth Fault Protection is

used for

– Protecting a equipment

– Selective tripping of faulty section of the

power system

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• It is obvious that over current protective system should act and interrupt the fault current before to damage of equipment due to fault current through it.

• Power system equipments include Line, Isolator, CT, Breaker, Transformer

• Obviously Transformer is most costliest and delicate (for fault 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

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• 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