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Co-ordination of over current relays in distribution system(Prepared by: N.Shahul Hameed B.E, SE/P&C(Retired),TNEB)

The single line diagram of the network considered for relay calculation shown infig 1

Fig 1

The fault MVA of each component

For the system (1) MVA = 2600 = 2600 MVA1

For the 10MVA transformer (2), MVA =

10

= 92.57 MVA0.108

For the line (3) MVA = 332 = 111 MVA15.75 x 0.625

For the 5MVA transformer (4), MVA = 5 = 77 MVA0 .065

110KVVVVVVKKKV

2600MVADE

8

7

110/33KV10MVAZ% = 10.8

5 6C

B

A

4

3

2 1

2.5MVA 2.5MVA

33/11KV5MVAZ% = 6.5

Dy1

15.7KMJ0.625 /km

Yy0

33KV

33KV

11KV


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Fault level of Bus D = 2600 MVA

Fault level of Bus C = 1 = 89.28 MVA1

+

1 .

2600 92.59

Fault level of Bus B = 1 = 49.48 MVA1

+1

+1 .

2600 92.59 111

Fault level of Bus A = 1 = 49.48 MVA1

+1

+1

+1 .

2600 92.59 111 77

Fig 2

1

2

3

4

2600MVA (Source)

92.59MVA

AA

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

SELECTION FOR CT RATIO :-

The CT ratio is selected using the higher of the following two currents.

i) the nominal currentii) the maximum short circuit current for which no saturation is

occurred (0.05 x Isc)

Relay 1 &2 : Inom 1 = Inom 2 =P

nom 1

3 x V1

= 2.5 x106 .

3 x 11 x 103

= 131.22 A (referred to 11KV)

Relay 3 : Inom 3 =P

nom 3

3 x V3

=5 x106 .

3 x 11 x 103



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Relay 4 : Inom 4 =P

nom 4

3 x V4

= 5 x106 .

3 x 33 x 103


Relay 5 &6 : Inom 5 = Inom 6 = 87.48 A (referred to 33KV)

Relay 7 : Inom 7 =P

nom 7

3 x V7

= 10 x106 .

3 x 33 x 103


Relay 8 : Inom 8 =P

nom 8

3 x V8

= 10 x106 .

3 x 110 x 103


The short circuit current, nominal load current and CT ratio inline with thecriteria for selection of CTs are tabulated in the table below :

Relay No Pnom

Isc 0.05 x Isc Inom CT Ratio

MVA A A A1 & 2 2.5 1581.21 79.06 131.22 150/53 5 1581.21 79.06 262.44 300/54 5 866.37 43.32 87.48 100/55 5 1564.42 78.22 87.48 100/5

6 5 1564.42 78.22 87.48 100/57 10 1564.42 78.22 174.96 200/58 10 13646.41 682.32 52.49 700/5


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SETTING OF INSTANTANEOUS UNITS OF OVER CURRENT RELAYS :-

The setting of instantaneous units for various system elements such as linebetween sub- stations, distribution lines and transformers are adopted as per followingcriteria.

Lines between sub-stations :-The setting of instantaneous units is carried out by taking at least 125% of the maximumfault level at the next sub-station. The procedure must be started from the farthest sub-station, and continued by moving back towards the source. When the characteristic oftwo relays cross at a particular system fault level, thus making it difficult to obtain correctco-ordination, it is necessary to set the instantaneous unit at the sub-station that is farthestaway from the source to such a value that the relay operates for slightly lower level ofcurrent. 25% margin avoids overlapping the downstream instantaneous unit.

Distribution lines:-

Since the distribution lines are the end of the system any one of the following criteriamay 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 load current.

Transformer units:-

The instantaneous units of over current relays on the primary side of the transformershould be set between 125 and 150% of the short circuit current at the bus bar on thesecondary side, referred to the primary side. If the instantaneous units of the transformersecondary and feeder relays are subjected to same short circuit level then the transformersecondary instantaneous over current unit needs to be overridden to avoid loss of co-ordination.

CALCULATION OF INSTANTANEOUS SETTING:

Relays 1 & 2Iinst = 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 810A primary amps.

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

Relay 4The setting is based on 125% of the short circuit current that exists at the busbaron the low voltage side of the transformer, referred to the high voltage side.


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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

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

fault level that exists at the next downstream substation.

Iinst = 1.25 x 866.37 = 1082.96 A primary ampsie = 1082.96 x (5/100) = 54.15A secondary amps

Set 55 amps equivalent to 1100 primary amps.

Relay 6

Iinst = 1000 A primary ampsie = 50A secondary amps

Relay7The instantaneous unit is overridden to avoid the possibility of loss of co-

ordination.

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

maximum fault level that exists at the next downstream substation referred to the highvoltage side

Iinst = 1.25 x 1564.42 x (33/110) = 586.66 A primary ampsie = 586.66 x (5/700) = 4.19A secondary amps

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

PICK-UP SETTING:The pick-up setting or plug setting is determined by the following expression.

Pickup setting = OLF x Inom

CTR

OLF = over load factor, CTR = CT ratio

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

For phase fault relays:

For motor 1.05For lines, generator, and transformers 1.25 to 1.5


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For distribution feeders under emergency condition 2

For earth fault relays:For lines, generator, and transformers 0.2For transmission lines 0.1

For distribution feeders 0.3Calculation 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 7Relay 3 PU 3 =1.5 x 262.44 x (5/300) =6.56; set at 7Relay 4 PU 4 =1.5 x 87.48 x (5/100) =6.56; set at 7Relay 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 7Relay 7 PU 7 =1.5 x 174.96 x (5/200) =6.56; set at 7Relay 8 PU 8 =1.5 x152.49 x (5/700) =0.56; set at 1

TIME MULTIPLIER SETTING :

The criteria procedure for calculating the time multiplier setting (TMS) to obtainappropriate protection and co-ordination for the system are given below.

1) determine the required operating time t1 of the relay farthest away from thesource by choosing the lower time multiplier setting and considering the fault

level for which the instantaneous unit of this relay picks up. This may behigher if it is necessary to co-ordinate with devices installed downstream. eg.Fuses or reclosers and condition like cold load pick-up.

2) determine the required operating time of the relay associated with the nextupstream breaker t2a = t1 + tmargin. Use the same fault level that used todetermine t1 of the relay associated with the previous breaker for thiscalculation also.

3) For the same fault current as in 1 above and knowing t2a and pick-up value forrelay 2, calculate time multiplier setting of the relay 2.

4) Determine the operating time t2b of relay 2, but now using the fault level justbefore the operation of its instantaneous unit.

5) Continue with the sequence, starting from the second stage.

Time discrimination margin :

A discrimination margin between two successive time characteristic of the orderof 0.2 to 0.4s should be typically used.

The operating time of the relay can be obtained from the operating characteristicon log-log paper or from the mathematical formula given below.


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

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

The consants , and L of relays of various standard charactricstic are givenbelow.

Curve description standard L

Moderately inverse IEEE 0.02 0.0515 0.114Very inverse IEEE 2 19.16 0.491

Extremely 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 0

Very inverse IEC 1 13.5 0Extremely inverse IEC 2 80 0Long time inverse IEC 1 120 0

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

CALCULATION OF TIME MULTIPLIER SETTIG :

Characteristic used - IEC very inverse timeConstants used for operating time of inverse characteristic is

= 13.5, = 1, and L = 0

Substituting the constants in the characteristic reduces to

t = k x 13.5 + 0(I/Is)1 - 1

k x 13.5

PSM - 1

For relays associated with breakers 1 & 2 choose the smaller multiplier settingand calculate the operating time.

Relays 1 & 2

Choose time multiplier setting k = 0.05PSM = Iinst.sec x (1/PU1,2)


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= 27 x (1/7)= 3.857 times

t1 =0.05 x 13.5

3.857 - 1= 0.236 sec or say 0.24 sec

Relay 3

The relay backs up relay 1 and 2Use a grading margin of 0.4secThere fore the required operating time is t3a = 0.24+0.4 = 0.64

PSM is based on 810 primary amps associated the instantaneous 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 Or say0.05 (minimum setting)13.5

Since in this case the instantaneous unit is overridden the short circuit current isused and multiplied by the factor 0.86 in order to cover the delta-star transformerarrangement.

So PSM3b = 0.86 x IscA x (1/CTR) x (1/PU3)= 0.86 x 1581.21 x (5/300) x (1/7) = 3.238 times

With time multiplier setting 0.05 and PSM3b of 3.238 the operating time of the relay

t3b =0.05 x 13.5 = 0.3sec3.238 - 1

Relay 4

The relay 4 backs up relay 3Use a grading margin of 0.4secHence 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


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With PSM4a = 3.764 and backup time of 0.7

k = (3.764 1) x 0.7 = 0.1413.5

It is now necessary to calculate the operating time of relay 4 just before theoperation 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

Relay 5

The relay 5 backs up relay 4Use a grading margin of 0.4secHence the required operating time is t5a = 0.51+0.4 = 0.91

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

So PSM5a = 660 x (5/100) x (1/7)= 4.714 times

With PSM5a = 4.714 and backup time of 0.91

k = (4.714 1) x 0.91 = 0.2513.5

It is now necessary to calculate the operating time of relay 5 just before theoperation 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

Relay 6


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Time multiplier setting = 0.3 (assume slightly higher setting than relay 5 forclarity of problem)

Relay 7

Relay 7 backs up relay 5 & 6 and should be coordinated with the slower of thesetwo relays.Relay 5 has an instantaneous setting of 1100primary amps and relay 6 has an

instantaneous setting of 1000 primary amps. Therefore the operating of both the relaysshould be calculated for 1000 primary amps current.

Relay 5; PSM5 = 1000 x (5/100) x 1/7) = 7.14 timesWith 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 timesWith 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

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

t7a = 0.66+0.4 = 1.06 sec

PSM7a is based on the 1000 primary amps of the CT associated 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 sec

k = (3.57 1) x 10.6 = 0.213.5

As the instantaneous unit is overridden, the multiplier PSM7b is calculated usingthe current for a short circuit on busbar C

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


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= 1564.42 x (5/200) x (1/7)= 5.5 times


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

Relay 8The relay 8 backs up relay 7Use a grading margin of 0.4secHence the required operating time is t8a = 0.6+0.4 = 1.0 sec

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

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

= 3.35 timesWith PSM8a = 3.35 and backup time t8a = 1

k = (3.35 1) x 1 = 0.1713.5

Table 2

Summery of setting

Relay no CT Ratio pickup Time dial Instantaneous settingprimary secondary

1 & 2 130/5 7 0.05 810 273 300/5 7 0.05 overridden4 100/5 7 0.14 600 335 100/5 7 0.25 1100 556 100/5 7 0.3 1000 507 200/5 7 0.2 overridden8 700/5 1 0.17 840 6

Ref:- Protection of Electricity Distribution NetworksBy Juan Gers and Edward HolmesIEEE Power and Energy Series

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