ABB

Application Note

Rel356 AN-92L-02

Substation Automation and Protection Division

Current Differential Relay REL356 Current Pickup Calculation Introduction This note describes how to calculate the current pick-up level for different types of faults.

IT current REL356 uses sequence filters to obtain positive, negative and zero sequence currents. These currents are then combined into one quantity:

I T = −C1 ⋅ I 1 + C 2 ⋅ I 2 + C 0 ⋅ I 0 The positive, negative and zero sequence current is computed from the phase currents in conventional manner:

I A + aI B + a 2 I C I A + I B ∠120° + I C ∠ − 120° = 3 3 2 I + a I B + aI C I A + I B ∠ − 120° + I C ∠120° I2 = A = 3 3 I + I B + IC I0 = A 3

I1 =

Note that all sequence component currents are referenced to phase A current.

General calculation of IT Based on input currents and C-settings, the IT current is calculated as follows:

Settings C1 C2 C0 Input currents IA IB IC IT

Current Differential Relay REL356 Current Pickup Calculation. AN-92L-02

I T = −C1 ⋅ I 1 + C 2 ⋅ I 2 + C 0 ⋅ I 0 = IT =

− C1 ( I A + I B ∠120° + I C ∠ − 120°) + C 2 ( I A + I B ∠ − 120° + I C ∠120°) + C 0 ( I A + I B + I C ) 3

Example 1A, phase A to ground fault With settings

C1 = 0.1

C 2 = 0.7 C 0 = 1.0 and input currents

I A = 5.0∠0° A IB = 0 IC = 0

IT becomes:

− C1 ( I A + I B ∠120° + I C ∠ − 120°) + C 2 ( I A + I B ∠ − 120° + I C ∠120°) + C 0 ( I A + I B + I C ) 3 − 0 .1 ⋅ 5 .0 + 0 .7 ⋅ 5 .0 + 1 .0 ⋅ 5 .0 IT = = 2.67 A 3 IT =

Example 1B, phase B to ground fault With settings

C1 = 0.1

C 2 = 0.7 C 0 = 1.0 and input currents

IA = 0

I B = 5.0∠ − 120° IC = 0

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Current Differential Relay REL356 Current Pickup Calculation AN-92L-02

IT becomes:

− C1 ( I A + I B ∠120° + I C ∠ − 120°) + C 2 ( I A + I B ∠ − 120° + I C ∠120°) + C 0 ( I A + I B + I C ) 3 − 0.1(5.0∠ − 120° + 120°) + 0.7(5.0∠ − 120° − 120°) + 1.0(5.0∠ − 120°) IT = = 1.64 A 3 IT =

That the IT current is different for a phase B to ground fault compared to phase A to ground is due to the fact that the symmetrical component computations are made referenced to phase A.

REL356 Trip Criterion REL356 operation equation is:

OP − 0.7 RES > OTH where OP =| I L + I R | RES =| I L | + | I R | IL = local IT current, IR = remote IT current and OTH is set operating threshold.

Pickup calculation To determine the theoretical pickup current for different types of fault, we need to determine that the output from the trip criterion exceeds the set operating threshold. Set-up in loop-back or back-to-back is assumed so that IR = IL = IT, i.e. the infeed current from both ends are equal in magnitude and in phase. This represents an internal fault. Then

OP =| I L + I R |= 2 ⋅ I T RES =| I L | + | I R |= 2 ⋅ I T OP − 0.7 RES = 0.6 ⋅ I T > OTH IT >

OTH 0 .6

In order to determine the required current threshold for operation for different types of faults the expressions above for IT and sequence currents need to be entered into the formula, solving the phase current(s).

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Current Differential Relay REL356 Current Pickup Calculation. AN-92L-02

Phase A to ground fault Input currents I A = I a ∠0° A IB = 0 IC = 0 Pickup current phase A IT >

OTH 0.6

OTH 0.6 − C1 I a + C 2 I a + C 0 I a OTH > 3 0.6 3 I a > OTH ⋅ 0.6(−C1 + C 2 + C 0 )

− C1 I 1 + C 2 I 2 + C 0 I 0 >

Phase B to ground fault Input currents IA = 0 I B = I b ∠ − 120° IC = 0 Pickup current phase B

IT >

OTH 0.6

OTH 0.6 − C1 I B ∠120° + C 2 I B ∠ − 120° + C 0 I B OTH > 3 0.6 − C1 I b + C 2 I b ∠ − 240° + C 0 I b ∠ − 120° OTH > 3 0.6 3 I b > OTH ⋅ 0.6(−C1 + C 2 ∠ − 240° + C 0 ∠ − 120°)

− C1 I 1 + C 2 I 2 + C 0 I 0 >

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Current Differential Relay REL356 Current Pickup Calculation AN-92L-02

Phase C to ground fault Input currents IA = 0 IB = 0 I C = I c ∠120° Pickup current phase C

IT >

OTH 0.6

OTH 0.6 − C1 I C ∠ − 120° + C 2 I C ∠120° + C 0 I C OTH > 3 0.6 − C1 I c ∠0° + C 2 I c ∠240° + C 0 I c ∠120° OTH > 3 0.6 3 I c > OTH ⋅ 0.6(−C1 + C 2 ∠240° + C 0 ∠120°)

− C1 I 1 + C 2 I 2 + C 0 I 0 >

Phase A to B fault Input currents I A = I ab ∠0° I B = I ab ∠180° IC = 0

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Current Differential Relay REL356 Current Pickup Calculation. AN-92L-02

Pickup current phases A and B

IT >

OTH 0.6

OTH 0.6 − C1 ( I A + I B ∠120°) + C 2 ( I A + I B ∠ − 120°) + C 0 ( I A + I B ) OTH > 3 0.6 − C1 ( I ab + I ab ∠180 + 120°) + C 2 ( I ab + I ab ∠180 − 120°) + C 0 ( I ab + I ab ∠180°) OTH > 3 0.6 3 I ab > OTH ⋅ 0.6[− C1 (1 + 1∠300) + C 2 (1 + 1∠60°)]

− C1 I 1 + C 2 I 2 + C 0 I 0 >

Phase B to C fault Input currents IA = 0 I B = I bc ∠ − 120° I C = I bc ∠60° Pickup current phases B and C

IT >

OTH 0.6

OTH 0.6 − C1 ( I B ∠120° + I C ∠ − 120°) + C 2 ( I B ∠ − 120° + I C ∠120°) + C 0 ( I B + I C ) OTH > 3 0.6 − C1 ( I bc + I bc ∠60 − 120°) + C 2 ( I bc ∠ − 240° + I bc ∠60 + 120°) + C 0 ( I bc ∠ − 120 + I bc ∠60°) OTH > 3 0.6 3 I ab > OTH ⋅ 0.6[− C1 (1 + 1∠ − 60°) + C 2 (1∠ − 240° + 1∠180°)]

− C1 I 1 + C 2 I 2 + C 0 I 0 >

Phase C to A fault Input currents

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Current Differential Relay REL356 Current Pickup Calculation AN-92L-02

I A = I ca ∠ − 60 IB = 0 I C = I ca ∠120° Pickup current phases C and A

IT >

OTH 0.6

OTH 0.6 − C1 ( I A + I C ∠ − 120°) + C 2 ( I A + I C ∠120°) + C 0 ( I A + I C ) OTH > 3 0.6 − C1 ( I ca ∠ − 60° + I ca ) + C 2 ( I ca ∠ − 60° + I ca ∠240°) + C 0 ( I ca ∠ − 60 + I ca ∠120°) OTH > 3 0.6 3 I ca > OTH ⋅ 0.6[− C1 (1∠ − 60° + 1) + C 2 (1∠ − 60° + 1∠240°)]

− C1 I 1 + C 2 I 2 + C 0 I 0 >

Three phase ABC fault Input currents I A = I abc ∠0° I B = I abc ∠ − 120° I C = I abc ∠120° Pickup current phases A, B and C

IT >

OTH 0.6

OTH 0.6 − C1 ( I A + I B ∠120° + I C ∠ − 120°) + C 2 ( I A + I B ∠ − 120° + I C ∠120°) + C 0 ( I A + I B + I C ) OTH > 3 0.6 − C1 ( I abc + I abc + I abc ) OTH > 3 0.6 1 I ab > OTH ⋅ − C1 ⋅ 0.6

− C1 I 1 + C 2 I 2 + C 0 I 0 >

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Current Differential Relay REL356 Current Pickup Calculation. AN-92L-02

Example 2A, Phase A to ground fault With input currents:

I A = I a ∠0° A IB = 0 IC = 0

and settings:

OTH = 0.5 C1 = 0.1 C 2 = 0 .7 C 0 = 1 .0

the required phase A current becomes:

I a = OTH ⋅

3 3 = 0 .5 ⋅ = 1.56 A 0.6(−C1 + C 2 + C 0 ) 0.6(−0.1 + 0.7 + 1.0)

Example 2B, Phase B to ground fault With input currents:

IA = 0

I B = I b ∠ − 120° IC = 0 and settings:

OTH = 0.5 C1 = 0.1 C 2 = 0 .7 C 0 = 1 .0

the required phase B current becomes:

I b = OTH ⋅

3 3 = 0 .5 ⋅ = 2.54 A 0.6(−C1 + C 2 ∠ − 240° + C 0 ∠ − 120°) 0.6(− 0.1 + 0.7∠ − 240° + 1.0∠ − 120°)

That the pickup current is higher for a phase B to ground fault compared to phase A to ground is due to the fact that the symmetrical component computations of IT are made referenced to phase A.

Contributed by: Solveig Ward Revision 0, 03/15/02

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