dist-cord(r)

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 in fig 1 110KV VVVV V KKKV

2600MVA 8 Yy0 7

D E

110/33KV 10MVA Z% = 10.8

33KV

C 5 15.7KM J0.625 Ώ/km

33KV

6 B

4 33/11KV 5MVA Z% = 6.5

Dy1 3

11KV 2 2.5MVA

A

1 2.5MVA Fig 1

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

MVA

For the 10MVA transformer (2), MVA

= 2600 1 10

=

= 2600 MVA

= 92.57 MVA

0.108 For the line (3)

MVA

=

332

= 111 MVA

15.75 x 0.625 For the 5MVA transformer (4), MVA

=

5 0 .065

= 77 MVA

Fault level of Bus D

= 2600 MVA

Fault level of Bus C

1

= 1

= 89.28 MVA

2600 Fault level of Bus B

92.59

1

= 1

= 49.48 MVA 1

+

2600 Fault level of Bus A

1 .

+

+

1 .

92.59

111

1

1

1

= 1

= 49.48 MVA +

2600

+

92.59

+

111

1 . 77

2600MVA (Source)

1 D

2600 MVA 92.59MVA AA

2 C

89.28 MVA 3

111MVA

B

49.48 MVA 4

77MVA 30.12 MVA

A Fig 2

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) ii)

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

P

=

nom 1

√3 x V1 =

2.5 x106

. 3

√3 x 11 x 10

= 131.22 A (referred to 11KV) Relay 3

: Inom 3

=

P

nom 3

√3 x V3 =

5 x106

.

√3 x 11 x 103 = 262.44 A (referred to 11KV)

Relay 4

: Inom 4

=

P

nom 4

√3 x V4 =

5 x106

.

√3 x 33 x 103 = 87.48 A (referred to 33KV) Relay 5 &6 : Inom 5 = Inom 6 Relay 7

= 87.48 A (referred to 33KV)

: Inom 7

=

P

nom 7

√3 x V7 =

10 x106

.

√3 x 33 x 103 = 174.96 A (referred to 33KV) Relay 8

: Inom 8

P

=

nom 8

√3 x V8 =

10 x106

.

√3 x 110 x 103 = 52.49 A (referred to 110KV) The short circuit current, nominal load current and CT ratio inline with the criteria for selection of CTs are tabulated in the table below : Relay No

P

1 & 2 3 4 5 6 7 8

MVA 2.5 5 5 5 5 10 10

nom

Isc

0.05 x Isc

Inom

CT Ratio

A 1581.21 1581.21 866.37 1564.42 1564.42 1564.42 13646.41

A 79.06 79.06 43.32 78.22 78.22 78.22 682.32

A 131.22 262.44 87.48 87.48 87.48 174.96 52.49

150/5 300/5 100/5 100/5 100/5 200/5 700/5

SETTING OF INSTANTANEOUS UNITS OF OVER CURRENT RELAYS :The setting of instantaneous units for various system elements such as line between sub- stations, distribution lines and transformers are adopted as per following criteria. Lines between sub-stations :The setting of instantaneous units is carried out by taking at least 125% of the maximum fault level at the next sub-station. The procedure must be started from the farthest substation, and continued by moving back towards the source. When the characteristic of two relays cross at a particular system fault level, thus making it difficult to obtain correct co-ordination, it is necessary to set the instantaneous unit at the sub-station that is farthest away from the source to such a value that the relay operates for slightly lower level of current. 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 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 load current. Transformer units:The instantaneous units of over current relays on the primary side of the transformer should be set between 125 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 level then the transformer secondary instantaneous over current unit needs to be overridden to avoid loss of coordination. 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 810A primary amps. Relay 3 The instantaneous unit is overridden to avoid the possibility of loss of coordination. Relay 4 The setting is based on 125% of the short circuit current that exists at the busbar on the low voltage side of the transformer, referred to the high voltage side.

Iinst ie

= 1.25 x1581.21 x (11/33) = 658.84 A primary amps =658.84 x(5/100) = 32.94A secondary amps

Set 33 amps equivalent to 660 primary amps Relay 5 The setting is calculated on the basis of 125% of the current for the maximum fault level that exists at the next downstream substation. Iinst ie

= 1.25 x 866.37 = 1082.96 x (5/100)

= 1082.96 A primary amps = 54.15A secondary amps

Set 55 amps equivalent to 1100 primary amps. Relay 6 Iinst ie

= 1000 A primary amps = 50A secondary amps

Relay7 The instantaneous unit is overridden to avoid the possibility of loss of coordination. Relay 8 The 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 high voltage side Iinst ie

= 1.25 x 1564.42 x (33/110) = 586.66 x (5/700)

= 586.66 A primary amps = 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. = OLF x Inom CTR OLF = over load factor, CTR = CT ratio Pickup setting

The overload factor(OLF) recommended for various circuits are given below. For phase fault relays: For motor For lines, generator, and transformers

1.05 1.25 to 1.5

For distribution feeders under emergency condition 2 For earth fault relays: For lines, generator, and transformers For transmission lines For distribution feeders

0.2 0.1 0.3

Calculation of pick-up settings: PU Relay 1 &2 Relay 3 Relay 4 Relay 5 Relay 6 Relay 7 Relay 8

= OLF x Inom CTR PU 1,2 PU 3 PU 4 PU 5 PU 6 PU 7 PU 8

=1.5 x 131.22 x (5/150) =1.5 x 262.44 x (5/300) =1.5 x 87.48 x (5/100) =1.5 x 87.48 x (5/100) =1.5 x 87.48 x (5/100) =1.5 x 174.96 x (5/200) =1.5 x152.49 x (5/700)

=6.56; set at =6.56; set at =6.56; set at =6.56; set at =6.56; set at =6.56; set at =0.56; set at

7 7 7 7 7 7 1

TIME MULTIPLIER SETTING : The criteria procedure for calculating the time multiplier setting (TMS) to obtain appropriate protection and co-ordination for the system are given below. 1) determine the required operating time t1 of the relay farthest away from the source by choosing the lower time multiplier setting and considering the fault level for which the instantaneous unit of this relay picks up. This may be higher 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 next upstream breaker t2a = t1 + tmargin. Use the same fault level that used to determine t1 of the relay associated with the previous breaker for this calculation also. 3) For the same fault current as in 1 above and knowing t2a and pick-up value for relay 2, calculate time multiplier setting of the relay 2. 4) Determine the operating time t2b of relay 2, but now using the fault level just before 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 order of 0.2 to 0.4s should be typically used. The operating time of the relay can be obtained from the operating characteristic on log-log paper or from the mathematical formula given below.

t = kβ +L (I/Is)α - 1 Where I = fault current, Is = pick-up setting. K = time multiplier setting, L a constant t = relay operating time in sec The consants α, β and L of relays of various standard charactricstic are given below. Curve description

standard

α

β

L

Moderately inverse Very inverse Extremely inverse inverse Short time inverse Standard inverse Very inverse Extremely inverse Long time inverse

IEEE IEEE IEEE CO8 CO2 IEC IEC IEC IEC

0.02 2 2 2 0.02 0.02 1 2 1

0.0515 19.16 28.2 5.95 0.0239 0.14 13.5 80 120

0.114 0.491 0.1217 0.18 0.0169 0 0 0 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 time Constants used for operating time of inverse characteristic is β = 13.5, α = 1, and L = 0 Substituting the constants in the characteristic reduces to t = k x 13.5 (I/Is)1 - 1

+0

k x 13.5 PSM - 1 For relays associated with breakers 1 & 2 choose the smaller multiplier setting and calculate the operating time. Relays 1 & 2 Choose time multiplier setting k = 0.05 PSM = Iinst.sec x (1/PU1,2)

= 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 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 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 is used and multiplied by the factor 0.86 in order to cover the delta-star transformer arrangement. 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 3.238 - 1

= 0.3sec

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 0.7 k

= (3.764 – 1) x 0.7 13.5

= 0.14

It is now necessary to calculate the operating time of relay 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 = 0.14 x 13.5 4.714 - 1

t4b

= 0.51sec

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 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 13.5

= 0.25

It is now necessary to calculate the operating time of relay 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

With time multiplier setting 0.25 and PSM5b 7.857 t5b Relay 6

= 0.25 x 13.5 7.857 - 1

= 0.492sec

Time multiplier setting = 0.3 (assume slightly higher setting than relay 5 for clarity of problem) Relay 7 Relay 7 backs up relay 5 & 6 and should be coordinated with the slower of these two 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 relays should be calculated for 1000 primary amps current. Relay 5; PSM5 = 1000 x (5/100) x 1/7) = 7.14 times With PSM = 7.14 and time dial setting 0.25 = 0.25 x 13.5 7.14 - 1

t5

= 0.55sec

Relay 6; PSM6 = 1000 x (5/100) x 1/7) = 7.14 times With PSM = 7.14 and time dial setting 0.3 = 0.3 x 13.5 7.14 - 1

t6

= 0.66sec

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 k

=3.57 and backup time of 1.06 sec

= (3.57 – 1) x 10.6 13.5

= 0.2

As the instantaneous unit is overridden, the multiplier 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 With time multiplier setting 0.2 and PSM7b 5.5 T7b

= 0.2 x 13.5 5.5 - 1

= 0.6sec

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 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 times With PSM8a = 3.35 and backup time t8a = 1 k

= (3.35 – 1) x 1 13.5

= 0.17

Table 2 Summery of setting Relay no

CT Ratio

pickup

Time dial

1&2 3 4 5 6 7 8

130/5 300/5 100/5 100/5 100/5 200/5 700/5

7 7 7 7 7 7 1

0.05 0.05 0.14 0.25 0.3 0.2 0.17

Ref:- Protection of Electricity Distribution Networks By Juan Gers and Edward Holmes IEEE Power and Energy Series

Instantaneous setting primary secondary 810 27 overridden 600 33 1100 55 1000 50 overridden 840 6