Co-Ordination of Over Current Relays in Distribution System

7/27/2019 Co-Ordination of Over Current Relays in Distribution System

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Co-ordination of over current

relays in distribution system

(Prepared by: N.Shahul HameedB.E, EE/P&C,TNEB)


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110KV

2600MVA

D

8

7

110/33KV

10MVAZ% = 10.8

5 6

C

B

A

4

3

2 1

2.5MVA 2.5MVA

33/11KV

5MVA

Z% = 6.5Dy1

15.7KM

J0.625 /km

Yy0

33KV

33KV

11KV


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The fault MVA of each component For the system (1) MVA = 2600 = 2600 MVA

1 For the 10MVA transformer (2), MVA = 10 = 92.57 MVA 0.108 For theline (3) MVA = 332 = 111 MVA 15.7 x 0.625 For the 5MVA transformer (4), MVA = 5 = 77 MVA

0 .065 Fault level of Bus D = 2600 MVA Fault level of Bus C = 1 = 89.28 MVA 1 + 1 . 2600 92.59 Fault level of Bus B = 1 = 49.48 MVA

1 + 1 + 1 . 2600 92.59 111 Fault level of Bus A = 1 = 49.48 MVA 1 + 1 + 1 + 1 . 2600 92.59 111 77


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1

2

3

4

2600MVA (Source)

92.59MVA

111MVA

77MVA

30.12 MVA

49.48 MVA

89.28 MVA

2600 MVAD

C

B

A


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ISCA = 30.12 = 1.581 KA

3 x 11 ISCB = 49.48 = 0.8657 KA 3 x 33 ISCC = 89.28 = 1.564 KA 3 x 33

ISCD = 2600 = 13.647 KA 3 x 110


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SELECTION FOR CT RATIO :- The CT ratio is selected using the higher of the

following two currents.

the nominal current the maximum short circuit current for which no

saturation is occurred (0.05 x Isc)

Relay 1 &2 : Inom 1 = Inom 2 = Pnom 1 3 x V1 = 2.5 x106 . 3 x 11 x 103

= 131.22 A (referred to 11KV)


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Relay No Pnom Isc 0.05 x Isc Inom CT Ratio MVA A A A

1 & 2 2.5 1581.21 79.06 131.22 150/5

3 5 1581.21 79.06 262.44 300/5

4 5 866.37 43.32 87.48 100/5

5 5 1564.42 78.22 87.48 100/5

6 5 1564.42 78.22 87.48 100/5

7 10 1564.42 78.22 174.96 200/5

8 10 13646.41 682.32 52.49 700/5


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SETTING OF INSTANTANEOUS UNITS OF OVERCURRENT RELAYS :-.

Lines and sub-stations :- The setting of instantaneous units is carried out by

taking at least 125% of the maximum fault level at thenext sub-station. The procedure must be started from the

farthest sub-station, and continued by moving backtowards the source. When the characteristic of tworelays cross at a particular system fault level, thusmaking it difficult to obtain correct co-ordination, it isnecessary to set the instantaneous unit at the sub-

station that is farthest away from the source to such avalue that the relay operates for slightly lower level ofcurrent. 25% margin avoids overlapping the downstreaminstantaneous unit.


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Distribution lines:-

Since the distribution lines are the end of the system anyone of the following criteria may be adopted.

i) 50% of the maximum fault current at the point of

connection of the CT supplying relay.

ii) Between six and ten times of the maximum loadcurrent.


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Transformer units:- The instantaneous units of over current relays on the

primary side of the transformer should be set between125 and 150% of the short circuit current at the bus bar

on the secondary side, referred to the primary side. If the

instantaneous units of the transformer secondary and

feeder relays are subjected to same short circuit levelthen the transformer secondary instantaneous over

current unit needs to be overridden to avoid loss of co-

ordination.


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CALCULATION OF INSTANTANEOUS SETTING:

Relays 1 & 2

Iinst = 0.5 x Isc = 0.5 x1581.21 = 790.6 A (Primary)

ie = 790.6 x 5/150 = 26.35A (Secondary)

Set the relay at 27A secondary amps equivalent to 810Aprimary amps.

Relay 3 The instantaneous unit is overridden to avoid thepossibility of loss of co-ordination.


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Relay 4

The setting is based on 125% of the short circuitcurrent that exists at the busbar on the low voltage sideof the transformer, referred to the high voltage side.

Iinst = 1.25 x1581.21 x (11/33) = 658.84 A

primary ampsie =658.84 x(5/100) = 32.94A

secondary amps

Set 33 amps equivalent to 660 primary amps


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Relay 5 The setting is calculated on the basis of 125% of the

current for the maximum fault level that exists at the nextdownstream substation.

Iinst = 1.25 x 866.37 =1082.96 A primary amps

ie = 1082.96 x (5/100) = 54.15A secondary amps Set 55 amps equivalent to 1100 primary

amps.

Relay 6 Iinst = 1000 A primary amps ie = 50A secondary amps


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Relay7 The instantaneous unit is overridden to avoid

the possibility of loss of co-ordination.

Relay 8 The setting is calculated on the basis of 125% of the

current for the maximum fault level that exists at the nextdownstream substation referred to the high voltage side

Iinst = 1.25 x 1564.42 x (33/110)

= 586.66 A primary amps

ie = 586.66 x (5/700)

= 4.19A secondary amps

Set at 6 secondary amps (the minimum setting)equivalent to 840 primary amps.


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PICK-UP SETTING:

The pick-up setting or plug setting is determined by thefollowing expression.

Pickup setting = OLF x Inom

CTR

OLF = over load factor, CTR = CT ratio


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The overload factor(OLF) recommended for variouscircuits are given below.

For phase fault relays: For motor 1.05 For lines, generator, and transformers 1.25 to 1.5

For distribution feeders under emergency condition 2

For earth fault relays: For lines, generator, and transformers 0.2

For transmission lines 0.1 For distribution feeders 0.3


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Calculation of pick-up settings: PU = OLF x Inom

CTR Relay 1 &2 PU 1,2 =1.5 x 131.22 x (5/150) =6.56;

set at 7

Relay 3 PU 3 =1.5 x 262.44 x (5/300) =6.56;set at 7

Relay 4 PU 4 =1.5 x 87.48 x (5/100) =6.56;set at 7

Relay 5 PU 5 =1.5 x 87.48 x (5/100) =6.56;set at 7

Relay 6 PU 6 =1.5 x 87.48 x (5/100) =6.56;set at 7

Relay 7 PU 7 =1.5 x 174.96 x (5/200) =6.56;set at 7

Relay 8 PU 8 =1.5 x152.49 x (5/700) =0.56;set at 1


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TIME MULTIPLIER SETTING :

The criteria procedure for calculating the timemultiplier setting (TMS) to obtain appropriate protectionand co-ordination for the system are given below.

determine the required operating time t1 of the relay

farthest away from the source by choosing the lowertime multiplier setting and considering the fault level forwhich the instantaneous unit of this relay picks up. Thismay be higher if it is necessary to co-ordinate withdevices installed downstream. eg. Fuses or reclosersand condition like cold load pick-up.


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determine the required operating time of the relayassociated with the next upstream breaker t2a = t1 +tmargin. Use the same fault level that used to

determine t1 of the relay associated with the previousbreaker for this calculation also.

For the same fault current as in 1 above and knowingt2a and pick-up value for relay 2, calculate time

multiplier setting of the relay 2.

Determine the operating time t2b of relay 2, but nowusing the fault level just before the operation of itsinstantaneous unit.

Continue with the sequence, starting from the secondstage.


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Time discrimination margin :

A discrimination margin between twosuccessive time characteristic of the order of 0.2to 0.4s should be typically used.

The operating time of the relay can beobtained from the operating characteristic onlog-log paper or from the mathematical formulagiven below.

t = k + L (I/Is) - 1

Where I = fault current, Is = pick-up setting. K = time multiplier setting, L a constant t = relay operating time in sec


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By knowing the PSM and operating time theequation can be solved for TMS

Curve description standard LModerately inverse IEEE 0.02 0.0515 0.114Very inverse IEEE 2 19.16 0.491Extremely inverse IEEE 2 28.2 0.1217inverse CO8 2 5.95 0.18

Short time inverse CO2 0.02 0.0239 0.0169Standard inverse IEC 0.02 0.14 0Very inverse IEC 1 13.5 0Extremely inverse IEC 2 80 0Long time inverse IEC 1 120 0

By knowing the PSM and operating time theequation can be solved for TMS


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CALCULATION OF TIME MULTIPLIER SETTIG

Characteristic used - IEC very inverse time Constants used for operating time of inversecharacteristic is

= 13.5, = 1, and L = 0

Substituting the constants in the characteristic reducesto

t = k x 13.5 + 0

(I/Is)1 - 1

k x 13.5

PSM - 1


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Relay 3 The relay backs up relay 1 and 2

Use a grading margin of 0.4sec There fore the required operating time is t3a =0.24+0.4 = 0.64

PSM is based on 810 primary amps associated theinstantaneous setting of relays 1 & 2

So PSM3a = 810 x(1/CTR3) x (1/PU3)= 810 x (5/300) x(1/7) = 1.928 times

k = (PSM 1) x t3a

13.5= (1.928 1) x 0.64 = 0.04 say0.0513.5 (minimum setting)


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Relay 4

The relay 4 backs up relay 3 Use a grading margin of 0.4sec Hence the required operating time is t4a = 0.3+0.4 =0.7 PSM4a is based on 1581.21 primary amps of the CT

associated with relay 3

So PSM4a = 1581.21 x (11/33) x (5/100) x (1/7)= 3.764 times

With PSM4a = 3.764 and backup time of 0.7k = (3.764 1) x 0.7 = 0.1413.5


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It is now necessary to calculate the operating time ofrelay 4 just before the operation of the instantaneous

unit.

PSM4b = Iinst.prim4 x (1/CTR4) x (1/PU4)

= 660 x (5/100) x (1/7) = 4.714 times

With time multiplier setting 0.14 and PSM4b 4.714

t4b = 0.14 x 13.5 = 0.51sec

4.714 - 1


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Relay 5

The relay 5 backs up relay 4 Use a grading margin of 0.4sec

Hence the required operating time is t5a=0.51+0.4=0.91

PSM5a is based on 660 primary amps of the CTassociated with relay 4

So PSM5a = 660 x (5/100) x (1/7)

= 4.714 times

With PSM5a = 4.714 and backup time of 0.91k = (4.714 1) x 0.91 = 0.25

13.5


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It is now necessary to calculate the operating time ofrelay 5 just before the operation of the instantaneous

unit.

PSM5b = Iinst.prim5 x (1/CTR5) x (1/PU5)

= 1100 x (5/100) x (1/7) = 7.857 times


t5b = 0.25 x 13.5 = 0.492sec

7.857 - 1


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Relay 6

Time multiplier setting = 0.3 (assume slightlyhigher setting than relay 5 for clarity of problem


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Relay 7

Relay 7 backs up relay 5 & 6 and should becoordinated with the slower of these two relays.

Relay 5 has an instantaneous setting of 1100primaryamps and relay 6 has an instantaneous setting of 1000primary amps. Therefore the operating of both the relays

should be calculated for 1000 primary amps current.


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Relay 5; PSM5 = 1000 x (5/100) x 1/7) = 7.14 times With PSM = 7.14 and time dial setting 0.25

t5 = 0.25 x 13.5 = 0.55sec

7.14 - 1

Relay 6; PSM6 = 1000 x (5/100) x 1/7) = 7.14 times

With PSM = 7.14 and time dial setting 0.3

t6 = 0.3 x 13.5 = 0.66sec

7.14 - 1

The slower relay is relay 6


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Therefore, in order to be slower than relay 6, the back uptime of relay 7 should be

t7a = 0.66+0.4 = 1.06 sec

PSM7a is based on the 1000 primary amps of the CTassociated with relay 6

So, PSM7a = Iinst.prim7 x (1/CTR7) x (1/PU7)= 1000 x (5/200) x (1/7)

= 3.57 times

With PSM7a = 3.57 and backup time of 1.06 seck = (3.57 1) x 10.6 = 0.2

13.5


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As the instantaneous unit is overridden, themultiplier PSM7b is calculated using the current for a

short circuit on busbar C

So, PSM7b = Isc x (1/CTR7) x (1/PU7)

= 1564.42 x (5/200) x (1/7)

= 5.5 times


T7b = 0.2 x 13.5 = 0.6sec5.5 - 1


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Relay 8

The relay 8 backs up relay 7 Use a grading margin of 0.4sec Hence the required operating time is t8a = 0.6+0.4

= 1.0 sec

PSM8a is based on 1564.42, the short circuit current ofbusbar C referred to 110 KV bus, since theinstantaneous unit of relay 7 is overridden.

So PSM8a = 1564.42 x (33/110) x (5/700) x (1/1)= 3.35 times

With PSM8a = 3.35 and backup time t8a = 1k = (3.35 1) x 1 = 0.17

13.5


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Table 2 Summery of setting

Rno CT R pickup Tms Instpri sec

1&2 130/5 7 0.05 810 27

3 300/5 7 0.05 overridden

4 100/5 7 0.14 600 335 100/5 7 0.25 1100 55

6 100/5 7 0.3 1000 50

7 200/5 7 0.2 overridden

8 700/5 1 0.17 840 6

Documents

Co-Ordination of Over Current Relays in Distribution System