Testing Low Impedance

8/3/2019 Testing Low Impedance

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Testing Low ImpedanceBus Differential Relays

International ProTesT User Group Meeting

Vancouver, BC



What You’ll Learn

• Basics of bus differential protection• Differential Protection Methods

• Guidelines for testing low impedance

bus differential relays



Basics of Differential Protection

• Based on Kirchoff’s Current Law (KCL) – The sum of currents entering and exiting a

node must equal 0

– Think of a bus as a node



Simple Bus – Normal Flow

52 52

I1 I2

i1 i2

I1 = 1∠0°I2 = 1∠180°

I1 + I2 = 0, per KCL



Simple Bus – External Fault

52 52

I1 I2

i1 i2

I1 = 3∠0°I2 = 3∠180°

I1 + I2 = 0, per KCL



Simple Bus – Internal Fault

I1 = 3∠0°I2 = 3∠0°

I1 + I2 = 6, >0!

• I1 + I2 = ID, the

differential current

52 52

I1 I2

i1 i2



Differential Protection

• Looks for the presence of differentialcurrent

• Reliable protection concept

• Several different techniques



Bus Fault Protection Requirements

• High speed – Bus faults are typically high-magnitude,

damaging events

• Secure – Incorrect tripping a bus can drop a significant

part of the system



Bus Protection Techniques

• Overcurrent• High Impedance Differential

• Low Impedance Differential



Overcurrent Bus Protection

• Uses an overcurrentelement to detect ID

• ID = i1 + i2 = 0, or

does it? – CT replication error

– CT Saturation

52 52

I1 I2

i1 i2

50

ID



CT Replication Error

• CT performancerated ±10% (per ANSI)

• I1 + I2 = 0• i1 + i2 ≠ 0

– As much as 20%error

• +10% on i1• -10% on i2

• 50 element must beset less sensitive!

52 52

I1 I2

i1 i2

50

ID



CT Saturation

• Saturated CTproduces no currentoutput

• I1 + I2 = 0• i1 + i2 ≠ 0

– i2 = 0 due tosaturated CT

• ID = i1• 50 element must be

extremely un-sensitive

52 52

I1 I2

i1 i2

50

ID



High Impedance DifferentialI1

I2

ID

52

52

87

+ V -

R

• Actually anovervoltage relay

– Relay operates on

voltage across

internal resistancefrom ID



High Impedance Differential

• Pluses – Clever solution to

CT saturation

• High impedance

forces differential

current through

other CTs

• Voltage

developed is less

than that of internal fault

– Reliable

• Minuses – Dedicated CTs

– Matched

performance class

CTs – Identical CT ratios,

tapped at full ratio



Low Impedance Differential

• Mathematically sumscurrents

• Uses restraint to

maintain security

• No special CTrequirements

– Different ratios,

performance class

• Can provide

waveform capture,

and communications

52 52

I1 I2

i1

87 ID= i1 + i2

i2



Low Impedance Characteristic

S2

S1

ID

IR

Operate

w/ Restraint

IRs

IOmin

High

Current

Setting

Restrain

Operate

w/o Restraint

21 ii ID +=

21 ii IR +=



Low Impedance – Load Flow

I1 = 1∠0°I2 = 1∠180°

ID = |i1 + i2| = 0

IR = |i1| + |i2| = 2

S2

S1

ID

IR

Operate

w/ Restraint

IRs

IOmin

HighCurrent

Setting

Restrain

Operate

w/o Restraint

52 52

I1 I2

i1

87ID= |i1 + i2|

IR

= |i1| + |i2|

i2



Low Impedance – External Fault

I1 = 3∠0°I2 = 3∠180°

ID = |i1 + i2| = 0

IR = |i1| + |i2| = 6

S2

S1

ID

IR

Operate

w/ Restraint

IRs

IOmin

High

Current

Setting

Restrain

Operate

w/o Restraint

52 52

I1 I2

i1

87ID= |i1 + i2|

IR

= |i1| + |i2|

i2



Low Impedance – Internal Fault

I1 = 3∠0°I2 = 3∠0°

ID = |i1 + i2| = 6

IR = |i1| + |i2| = 6

S2

S1

ID

IR

Operate

w/ Restraint

IRs

IOmin

HighCurrent

Setting

Restrain

52 52

I1 I2

i1

87ID= |i1 + i2|

IR

= |i1| + |i2|

i2



CT Error: Low Impedance

S2

S1

ID

Operatew/ Restraint

IOmin

High

Current

Setting

Restrain

Load flow w/

CT Error External fault

w/ CT Error

External fault w/

CT Saturations



Low Impedance Relays

• NxtPhase B-PRO• GE B-30

• SEL 487B

• All use similar operating characteristic

• All use 6 inputs



Low Impedance Applications

52 52 52

52 52 52

I1 I2 I3

I4 I5 I6

B-PRO

NxtPhase

• (6) 3-phase inputs• 87B function

• Possible 27, 59, 81

• 50/51, possible 67

for each input

• 50BF for each input

• Possible multiple

protection zones





Testing a Low Impedance Bus

Differential Relay



4 Pieces of Knowledge

• How is the operating characteristicdefined?

– Curve equations

• How does the relay calculate ID and IR?• Does the characteristic work in amps or

per unit

• Relay settings



GE B-30

• ID = |i1+i2+i3+i4+i5+i6|• IR = max (I1, I2, I3, I4, I5, I6)

• Per unit. Base is maximum primary

current on an input



SEL 487B

• ID = |i1+i2+i3+i4+i5+i6|• IR = |i1|+…|i6|

• Per unit. Base is max CT ratio



NxtPhase B-PRO

• ID = |i1+i2+i3+i4+i5+i6|• IR = (|i1|+…+|i6|)/2

• Per unit base on Bus MVA / Bus Voltage



Testing Issues

• Do I have to test 3-phase? – No! Differential protection is single-phase

element

• Do I have to test all 6 inputs at the same

time?

– No! No current into an input is 0 current.

Differential characteristic still performs

correctly.



B-PRO Characteristic

S2

S1

ID

IR

Operate

w/ Restraint

IRs

IOmin

High

Current

Setting

Restrain

Test Zone 1 Test Zone 3

(0, IOmin)

min,

1

100min* IO

S

IO ( )( ) IRsS IRs *1001,

( )

( )

−−

I HighS

IRsS S

I High

,

1002

100

21

( ) IRS IO *1001=

( ) b IRS IO += *100

2

( ) IRs

S S b

100

21−=

2

654321 iiiiii IR

+++++=

654321 iiiiii ID +++++=



Test Plan

• Test Obvious External Fault – Verifies that test setup is correct

• Test Obvious Internal Fault

– Verify relay operation, test setup• Test characteristic performance



Possible Test Setup

Differential Relay

Input 2

A Phase

Input 3

A Phase

Input 1

A Phase

Input 5

A Phase

Input 6

A Phase

Input 4

A Phase

TestSource 1

0o

TestSource 2

180o



Possible Test Setup

Divide all calculated test currents by 3!

Differential Relay

Input 2

A Phase

Input 3

A Phase

Input 1

A Phase

Input 5

A Phase

Input 6

A Phase

Input 4

A Phase

TestSource 1

0o

TestSource 2

180o





Base Current

• Base Current is defined by Bus MVA,Bus Voltage

pri Base

Base

A I

kV

MVA Base I

kV Voltage

MVA Base

20002303

10007963

1000

230

796

=×

×=

×

×=

=

=



Test Point – Load Flow

unit per ii IR

unit per iiii ID

unit per i

unit per i

o

o

12

11

2

21

02121

18012

011

=+=+=

=−=+=∠=

∠=



Test Point – Load Flow

( )

( )

o pri

o pri

pri Base

pri Base

A A

Current SourceTest

A ACurrent SourceTest

CTR Input

CTR Input

A I unit per

A I

[email protected]

53000

20002

0@55

200020001

5:30002

5:20001

2000200011

2000

sec

sec

==

==

=

=

=×=×

=



Operating Quantity Display

Bus Differential (87B) A Phase B Phase C Phase

------------------------------ ---------- ---------- ----------

Operating Current, IO (PU) 0.0 0.0 0.0

Restraint Current, IR (PU) 1.0 0.0 0.0

Note: 1 PU = 796.0 MVA for 87B



Test Point – Internal Fault

unit per ii IR

unit per iiii ID

unit per i

unit per i

o

o

1211

221

22121

012

011

=+=+=

=+=+=∠=

∠=



Test Point – Internal Fault

( )

( )

o pri

o pri

pri Base

pri Base

A A

Current SourceTest

A A

Current SourceTest

CTR Input

CTR Input

A I unit per

A I

[email protected]

53000

20002

0@55

2000

20001

5:30002

5:20001

2000200011

2000

sec

sec

==

==

=

=

=×=×

=





------------------------------ ---------- ---------- ----------






Testing the characteristic

• Why can’t you start with an externalfault, and vary 1 current until the relay

operates?

• Answer…



You can…

• ID and IR vary withchanging current

• You must calculate

to determine that i1

and i2 from testsource match

characteristic

• Must verify this is on

characteristic!IRs

IOmin

Initial test point

Possible TripPoints



A better way

• Determine ID and IRfor a specific test

point

• Calculate i1 and i2

• Test, varying slightlyaround this region

IRs

IOmin

Initial test point

(IR, IO)



Calculations( )

unit per IO IB IB

unit per IB

IO IB IB

IR IO IB

IB IR IO

IB IB IR

IB IB IO

IB IB IR

IB IB IO

unit per ID IRat Test

exampleContinuing

inout

in

inout

in

in

out in

out in

out in

out in

875.025.0125.1

125.12

00.1225.0

2

2

22

2

2

)25.0,00.1(,

...

=−=−=

=×+

=

−=

×+=

×=×+

+=×

−=

+=

−=

=



Calculations

( )

( )o

pri pri

o

pri pri

out

in

ACTR

Aprii

A Aunit per

ACTR

Aprii

A Aunit per

iunit per IB

iunit per IB

[email protected]

53000

17501

17502000875.0

[email protected]

52000

22501

22502000125.1

2875.0

1125.1

sec

sec

==

=×

==

=×

==

==





------------------------------ ---------- ---------- ----------






Summary

• Low impedance bus differential is easyto apply

• Testing

– 1 phase okay – 2 inputs only okay

– Must understand operating characteristic

– Have to remember per unit calculations!

Documents

Testing Low Impedance