Testing Low Impedance

Testing Low Impedance Bus Differential Relays International ProTesT User Group Meeting Vancouver, BC

What You’ll Learn • Ba Basi sics cs of bu bus s di diff ffer eren enti tial al pr prot otec ecti tion on • Di Diff ffer ere ent ntia iall Pro Prote tect ctio ion n Met Meth hod ods s • Guid Guidel elin ines es fo forr te test stin ing g lo low w im impe peda danc nce e bus differential relays

Basics of Differential Protection • Based on Kirchoff’s Current Law (KCL)  – The sum of cur curren rents ts ente enterin ring g and and exi exitin ting ga node must equal 0  – Th Thin ink k of of a bu bus s as as a no nod de

Sim impl ple e Bus Bus – Nor orm mal Flo Flow w I1 = 1∠0° I2 = 1∠180°

I1

I2 I1 + I2 = 0, per KCL

52

i1

52

i2

Simp Si mple le Bu Bus s – Ex Exte tern rnal al Fa Faul ultt I1 = 3∠0° I2 = 3∠180°

I1

I2 I1 + I2 = 0, per KCL

52

i1

52

i2

Simp Si mple le Bu Bus s – In Inte tern rnal al Fa Faul ultt I1 = 3∠0° I2 = 3∠0°

I1

I2 I1 + I2 = 6, >0!

52

52

•

i1

i2

I1 + I2 = ID, the differential current

Differential Protection • Look Looks s fo forr th the e pr pres esen ence ce of di diff ffer eren enti tial al current • Re Relliable pr protection co concept • Several di different tec techniques

Bus Fault Protection Requirements • High speed  – Bus fau faults lts are are typi typical cally ly high high-m -mag agni nitu tude de,, damaging events

• Secure  – Incorre Incorrect ct trip trippin ping g a bus can dro drop p a sig signif nifican icantt part of the system

Bus Protection Techniques • Overcurrent • High Im Impedance Di Differential • Low Impedance Differential

Overcurrent Bus Protection

I1

•

Uses an overcurrent element to detect ID

•

ID = i1 + i2 = 0, or  does it?

I2

52

52

 – CT re rep pli lica cati tion on err rror  or  i1

i2 ID 50

– CT Sa Saturation

CT Replication Error 

I1

•

CT performance rated ±10% (per  ANSI)

• •

I1 + I2 = 0 i1 + i2 ≠ 0

I2

52

52

i1

– As much as 20% error 

i2 ID 50

• +10% on i1 • -10% on i2

•

50 element must be set less sensitive!

CT Saturation

I1

•

Saturated CT produces no current output

• •

I1 + I2 = 0 i1 + i2 ≠ 0

I2

52

52

i1

– i2 = 0 due to saturated CT

i2 ID 50

• •

I D = i1 50 element must be extremely unsensitive

High Impedance Differential I1 52

I2 52

ID + V R 87

•

Actually an overvoltage relay – Relay operates on voltage across internal resistance from ID

High Impedance Differential •

Pluses – Clever so solution to to CT saturation • High impedance forces differential current through other CTs • Voltage developed is less than that of  internal fault

– Reliable

•

Minuses – Dedicated CT CTs – Matched performance class CTs  – Iden Identi tica call CT ra rattio ios, s, tapped at full ratio

Low Impedance Differential I1

I2

52

•

Mathematically sums currents

•

Uses restraint to maintain security

•

No special CT requirements

52

i1

i2

– Different ra ratios, performance class

• 87

ID= i1 + i2

Can provide waveform capture, and communications

Low Impedance Characteristic ID

Operate w/o Restraint

High Current Setting

Operate w/ Restraint

 ID = i1 + i 2

Restrain

S2

IOmin

 IR = i1 + i 2

S1

IRs

IR

Low Lo w Im Impe peda danc nce e – Lo Load ad Fl Flow ow I1

I2

52

52

i1

I1 = 1∠0°

i2

87

I2 = 1∠180°

ID= |i1 + i2| IR = |i1| + |i2|

ID

Operate w/o Restraint

High Current Setting

ID = |i1 + i2| = 0 IR = |i1| + |i2| = 2

Operate w/ Restraint

Restrain

S2

IOmin

S1

IRs

IR

Low Lo w Im Impe peda danc nce e – Ex Exte tern rnal al Fa Faul ultt I1

I2

52

52

i1

I1 = 3∠0°

i2

87

I2 = 3∠180°

ID= |i1 + i2| IR = |i1| + |i2|

ID

Operate w/o Restraint

High Current Setting

ID = |i1 + i2| = 0 IR = |i1| + |i2| = 6

Operate w/ Restraint

Restrain

S2

IOmin

S1

IRs

IR

Low Lo w Impe Impeda danc nce e – In Inte tern rnal al Fau Fault lt I1

I2

52

52

i1

I1 = 3∠0°

i2

87

I2 = 3∠0°

ID= |i1 + i2| IR = |i1| + |i2|

ID = |i1 + i2| = 6

ID High Current Setting

IR = |i1| + |i2| = 6 Operate w/ Restraint

Restrain

S2

IOmin

S1

IRs

IR

CT Error: Low Impedance ID High Current Setting External fault w/ CT Saturations

Load flow w/ CT Error 

IOmin

Operate w/ Restraint

S2

S1

Restrain

External fault w/ CT Error 

Low Impedance Relays • NxtPhase B-PRO • GE B-30 • SEL 487B • Al Alll use use si simi mila larr ope opera rati ting ng ch char arac acte teri rist stic ic • All use 6 inputs

Low Impedance Applications I1

52

I2

52

I3

52

•

(6) 3-phase inputs

•

87B function

•

Possible 27, 59, 81

•

50/51, possible 67 for each input

•

50BF for each input

•

Possible mu multiple protection zones

B-PRO  NxtPhase

52

52

I4

52

I5

I6

Testing a Low Impedance Bus Differential Relay

4 Pieces of Knowledge • How How is is th the op oper era ati ting ng ch cha ara ract cte eri rist stiic defined? – Curve equations

• Ho How w doe does s th the rel relay ay ca callcu cullat ate e ID ID an and IR? IR? • Does Does th the e ch char arac acte teri rist stic ic wo work rk in am amps ps or  per unit • Relay settings

GE B-30 • ID = |i1+i2+i3+i4+i5+i6| • IR = max (I (I1, I2 I2, I3 I3, I4 I4, I5 I5, I6 I6) • Per Per uni nit. t. Ba Base se is max maxiimum pri rim mar ary y current on an input

SEL 487B • ID = |i1+i2+i3+i4+i5+i6| • IR = |i1|+…|i6| • Per unit. Base is ma max CT CT ratio

NxtPhase B-PRO • ID = |i1+i2+i3+i4+i5+i6| • IR = (|i1|+…+|i6|)/2 • Pe Perr uni unitt bas base e on on Bus Bus MV MVA A / Bus Vol olta tag ge

Testing Issues • Do I have to test 3-phase?  – No! Dif Differ ferent ential ial pro protect tection ion is sing singlele-pha phase se element

• Do I hav have e to to tes testt all all 6 inp input uts s at at the the sa same me time?  – No! No cur curren rentt into into an inp input ut is 0 curr current ent.. Differential characteristic still performs correctly.

B-PRO Characteristic ID High Current Setting

Operate w/ Restraint Test Zone 1

Test Zone 3

(

)

 IO = S 1 * IR 100

S2

(

 IO = S 2

IOmin

(0, IOmin)

Restrain

S1

  IO min* 100   , IO min   S 1    

)

* IR + b 100 (S 1 − S 2) b=  IRs 100

IRs

IR

( IRs, (S 1100)* IRs )

     (S 1 − S 2 )    IRs    High  I  −     100   , High  I      S 2 100      

 ID = i1 + i 2 + i3 + i 4 + i5 + i 6  IR =

i1 + i 2 + i3 + i 4 + i5 + i 6 2

(

)

Test Plan • Test Ob Obvious Ex External Fault  – Ve Veri rifi fies es tha thatt test test set setup up is cor corre rect ct

• Test Ob Obvious In Internal Fault  – Ve Veri rify fy rel relay ay ope opera ratio tion, n, tes testt setu setup p

• Te Test st ch cha ara ract cter eris isti tic c pe perf rfo orm rman ance ce

Possible Test Setup Test Source 1 0o

Input 1 A Phase

Test Source 2 180o

Input 2 A Phase

Input 3 A Phase

Input 4 A Phase

Input 5 A Phase

Input 6 A Phase

Differential Relay

Possible Test Setup Test Source 1 0o

Input 1 A Phase

Input 2 A Phase

Test Source 2 180o

Input 3 A Phase

Input 4 A Phase

Input 5 A Phase

Input 6 A Phase

Differential Relay

Divide all calculated test currents by 3!

Test Procedure • Collect B-PRO settings • Ca Callculate re relay ba base cu currrent • Dete Determ rmin ine e tes testt poi point nts s fro from m di diff ffer eren enti tial al characteristic • Ca Callculate se seco con ndary cu currents • Test

Base Current • Base Base Cu Curr rre ent is de defi fine ned d by by Bus Bus MV MVA, A, Bus Voltage

 Base MVA = 796

Voltage = 230 kV   I  Base =  I  Base =

 Base MVA × 1000 3 × kV  796 × 1000 3 × 230

= 2000  A pri

Tes estt Poi oin nt – Lo Load ad Flo low w i1 = 1∠0  per  unit  o

i 2 = 1∠180 o  per  unit   ID = i1 + i 2 = i1 − i 2 = 0  per  unit   IR =

i1 + i 2 2

=

1+1 2

= 1  per  unit 

Tes estt Poi oin nt – Lo Load ad Flo low w  I  Base = 2000  A pri 1 per  unit × I  Base = 1 × 2000 = 2000  A pri  Input 1 CTR = 2000 : 5  Input  2 CTR = 3000 : 5 Test  Source 1 Current  =

2000  A pri

Test  Source 2 Current  =

2000  A pri

(2000 5 )

= 5  Asec @ 0o

(3000 5 )

= 3.33  Asec @ 180o

Operating Quantity Display

Bus Differential Differentia l (87B) -----------------------------Operating Current, IO (PU) Restraint Current, IR (PU) Note: 1 PU =

796.0 MVA for 87B

A Phase ---------0.0 1.0

B Phase ---------0.0 0.0

C Phase ---------0.0 0.0

Test Te st Po Poin intt – In Inte tern rnal al Fa Faul ultt i1 = 1∠0o  per  unit  i 2 = 1∠0  per  unit  o

 ID = i1 + i 2 = i1 + i 2 = 2  per  unit   IR =

i1 + i 2 2

=

1+1 2

= 1  per  unit 

Test Te st Po Poin intt – In Inte tern rnal al Fa Faul ultt  I  Base = 2000  A pri 1 per  unit × I  Base = 1× 2000 = 2000  A pri  Input 1 CTR = 2000 : 5  Input  2 CTR = 3000 : 5 Test  Source 1 Current  =

2000  A pri

Test  Source 2 Current  =

2000  A pri

(2000 5 )

= 5  Asec @ 0o

(3000 5 )

= 3.33  Asec @ 0o

Operating Quantity Display

Bus Differential (87B) -----------------------------Operating Current, IO (PU) Restraint Current, IR (PU) Note: 1 PU =

796.0 MVA for 87B

A Phase ---------2.0 1.0

B Phase ---------0.0 0.0

C Phase ---------0.0 0.0

Testing the characteristic • Why Why can can’’t yo you sta start rt wi with th an ext exte ern rna al fault, and vary 1 current until the relay operates? • Answer…

You can… Possible Trip Points

•

ID and IR vary with changing current

•

You must calculate to determine that i1 and i2 from test source match characteristic

•

Must verify this is on characteristic!

IOmin

IRs Initial test point

A better way (IR, IO)

IOmin

IRs Initial test point

•

Determine ID and IR for a specific test point

•

Calculate i1 and i2

•

Test, varying slightly around this region

Calculations Continuing  example... Test  at  ( IR,  ID ) = (1.00, 0.25)  per  unit   IO =  IBin −  IBout   IR =

 IBin +  IBout  2

 IO =  IBin −  IBout  2 × IR =  IBin +  IBout   IO + 2 × IR = 2 × IBin  IBin =

 IO + 2 × IR 2

 IBout  =  IBin −  IO

 IBin =

0.25 + 2 × 1.00

= 1.125  per  unit  2  IBout  =  IBin −  IO = 1.125 − 0.25 = 0.875  per  unit 

Calculations  IBin = 1.125  per  unit  = i1  IBout  = 0.875  per  unit  = i 2 1.125  per  unit × 2000  A pri = 2250  A pri

i1 =

2250  Apri

(2000 5 ) CTR

= 5.625  Asec @ 0o

0.875  per  unit × 2000  A pri = 1750  A pri

i1 =

1750  Apri

(3000 5 ) CTR

= 2.92  Asec @ 180 o

Operating Quantity Display

Bus Differential (87B) -----------------------------Operating Current, IO (PU) Restraint Current, IR (PU) Note: 1 PU =

796.0 MVA for 87B

A Phase ---------0.3 1.0

B Phase ---------0.0 0.0

C Phase ---------0.0 0.0

Summary • Low Low im impe peda danc nce e bus bus di diff ffer eren enti tial al is ea easy sy to apply • Testing – 1 phase okay – 2 in inputs only ok okay  – Mus Mustt und unders erstan tand d ope operat rating ing cha charac racter teristi istic c  – Ha Have ve to rem remem embe berr per per unit unit calc calcul ulat atio ions ns!!