beyond the nameplate – selecting transformer...

Copyright © PowerServices, Inc., FirstEnergy Corporation, and SEL 2017

Beyond the Nameplate –Selecting Transformer Compensation

Settings for Secure Differential ProtectionBarker Edwards

PowerServices, Inc.

David G. WilliamsFirstEnergy Corporation

Ariana Hargrave, Matthew Watkins, and Vinod K. Yedidi Schweitzer Engineering Laboratories, Inc.

What Do You Need to Set a Transformer Relay?

Percentage-Restrained Differential

+= 1 2I I

IRTk

= + 1 2IOP I I

Compensating for Transformer

1TAP1

Compensation Matrices[ ]

=

compensated

compensated

compensated

Ia IaIb Compensation Matrix • Ib

IcIc

1 0 00 1 00 0 1

0

− − −

− − −

− − −

1 1 01 • 0 1 13 1 0 1

0 1 11 • 1 0 13 1 1 0

1 0 11 • 1 1 03 0 1 1

1

3

5

− − −

− − − − − −

− − −

1 2 11 • 1 1 23

2 1 1

1 1 21 • 2 1 13

1 2 1

2 1 11 • 1 2 13

1 1 2

2

4

6

− − −

− − −

− − −

1 1 01 • 0 1 13 1 0 1

0 1 11 • 1 0 13 1 1 0

1 0 11 • 1 1 03 0 1 1

7

9

11

− − − − − −

− − −

− − − − − −

1 2 11 • 1 1 23

2 1 1

1 1 21 • 2 1 13

1 2 1

2 1 11 • 1 2 13

1 1 2

8

10

12

• Nameplate phase shift is only valid when System has ABC

phase sequence

Phase-to-bushing connections are ABC

• Phase shift that relay sees is influenced by external factors CTs in delta or wye

CT polarity

CT-to-relay connections

“I Thought the Nameplate Was All I Needed”

Entire installation must be taken into account when setting relay!

1. Derive phase shift seen by relay2. If delta winding exists, choose it as reference

and select Matrix 03. If delta winding does not exist, select Matrix 11 for

one of wye windings and appropriate matrix for other 4. Avoid even-numbered matrices

Rules for Selecting Correct CompensationGoal: W1 180 Degrees Out of Phase With W2

Applying the Rules

On System At Relay

Compensation?(0, 11)

• Nameplate used to set compensation to (0, 1)

• Relay tripped twice

Field Case: Incorrect Compensation Settings30 MVA DABY Transformer, 69 kV / 12.47 kV150

–150–50

50

Cur

rent

Cur

rent

500

0

–500

–15,000–5,000

5,00015,000

Vol

tage

1:87R1:TRIP

0 5 10 15 20

1:IAW1_A1:IBW1_A1:ICW1_A


Time (cycles)

1:VA_V1:VB_V1:VC_V

0.00.20.40.6

Cur

rent

0.00.20.40.6

Cur

rent

0.00.20.40.6

Cur

rent

1:87R1:TRIP

8 10 12 14 16 18

1:IRT1=0.4611:IOP1=0.246

1:IRT2=0.5021:IOP2=0.210

1:IRT3=0.4581:IOP3=0.250

Time (cycles)

1:IOP1_pu1:IRT1_pu

1:IOP2_pu1:IRT2_pu

1:IOP3_pu1:IRT3_pu

• Correct compensation per rules is (0, 1)

• Compensation set to (11,12) –correct pair for phase angle compensation, but arbitrary selection of reference winding

• Relay tripped on external A-B fault (wye side)

• 7% CT error from saturation

Field Case: New Rules Add Security22 MVA DABY Transformer, 72 kV / 13 kV,

Standard Connections

1,500

500

–500

–1,500

Cur

rent

8,000

0

–8,000

Cur

rent

1:87R11:87R21:87R3

1:87R1:TRIP

4,000

–4,000

0 2 4 6 8 10 12 14Time (cycles)



Compensated Currents

With (11, 12)

Input Currents Measured by Relay (A primary)

Compensated Currents (pu of tap)

IAW1 912 ∠0˚ 4.21 ∠–1.4˚IBW1 414 ∠–175˚ 4.27 ∠–178˚ICW1 395 ∠175˚ 0.23 ∠74.8˚IAW2 4,620 ∠178˚ 4.66 ∠179˚IBW2 4,320 ∠2˚ 4.38 ∠1.43˚ICW2 323 ∠–35.9˚ 0.36 ∠–39.2˚

c

b

a

C

B

A

IF

IF

2IF

0

IF

0

IF

IF

H3

H2

H1

X3

X2

X1

22 MVA73 kV / 13 kV

IF X0

Original Compensation (11, 12)

( )1 1ICW1C • • –IBW1 ICW1CTR1• TAP1 3

= +

≅Simplifies to ICW1C 0

( )1 1ICW2C • • –IAW2 –IBW2 2• ICW2CTR2• TAP2 3

= + +

( )= ≅1 1Simplifies to ICW2C • • 2 • ICW2 0

CTR2• TAP2 3

Differential Characteristic Current Differential Element 1 Differential Element 2 Differential Element 3

IOP (pu) 0.452 0.115 0.337IRT (pu) 8.868 8.650 0.584

IOP / IRT (%) 5.1 1.3 57.7

=

1 0 –11Matrix 11 –1 1 03 0 –1 1

− = − − −

2 1 –11Matrix 12 –1 2 13

1 –1 2

–2

26

10

Cur

rent

–2

2

6

10C

urre

nt

–2

2

6

10

Cur

rent

1:87R30 2 4 6 8 10 12 14

1:IOP11:IRT1

1:IOP21:IRT2

1:IOP31:IRT3

Time (cycles)

1:IOP1=0.4521:IRT1=8.868

1:IOP2=0.1151:IRT2=8.650

1:IOP3=0.3371:IRT3=0.584

Correct Compensation (0, 1)

Both W1 and W2 compensated C-phase currents now contain fault current contribution, increasing restraint current

( )1ICW1C • ICW1CTR1• TAP1

=

( )1 1ICW2C • • –IAW2 ICW2CTR2• TAP2 3

= +

Differential Characteristic Current Differential Element 1 Differential Element 2 Differential Element 3

IOP (pu) 0.131 0.068 0.652IRT (pu) 10.307 4.687 5.062

IOP / IRT (%) 1.3 1.5 12.9

=

1 0 0Matrix 0 0 1 0

0 0 1

=

1 –1 01Matrix 1 0 1 –13 –1 0 1

–2

2

6

10

Cur

rent

–2

2

6

10C

urre

nt

–2

2

6

10

Cur

rent

0 2 4 6 8 10 12 14

1:IRT1_P1:IOP1_P

1:IOP2_P1:IRT2_P

1:IRT3_P1:IOP3_P

Time (cycles)

1:IRT1_P=10.3071:IOP1_P=0.131

1:IOP2_P=0.0681:IRT2_P=4.687

1:IRT3_P=5.0621:IOP3_P=0.652

Original vs. Correct Compensation

• Correct compensation per rules is (11, 11)

• Compensation set to (12, 12) –correct pair for phase angle compensation (common in autotransformers)

• Relay tripped on external A-B fault (115 kV side)

• 12% CT error from saturation

New Rules Add Security for Autotransformers100 MVA Autotransformer, 230 kV / 115 kV,

Standard Connections

3,0002,0001,000

0–1,000

–3,000–2,000

Cur

rent

6,0004,0002,000

0–2,000

–6,000–4,000

Cur

rent

1:87R11:87R21:87R3

40.954000 41.004000 41.054000 41.104000 41.154000



Compensated Currents for (12, 12) and (11, 11)

InputCurrents

Measured by Relay (A primary)

Compensated Currents for Pair (12, 12)

(A secondary pu of tap)

Compensated Currents for Pair (11, 11)

(pu of tap)

IAW1 1,910 ∠0˚ 7.05 ∠9.16˚ 4.46 ∠–0.5˚

IBW1 1,640 ∠–148˚ 6.62 ∠–160˚ 7.85 ∠–165˚

ICW1 33.8 ∠151˚ 1.41 ∠123˚ 3.73 ∠32.8˚

Operate and

Restraint Currents

Differential Characteristic

Current

Differential Element 1



IOP (pu) 1.32 2.64 1.31

IRT (pu) 14.71 14.19 1.51

IOP / IRT (%) 9 18 87

Differential Characteristic

Current




IOP (pu) 0.00 2.29 2.28

IRT (pu) 8.93 16.65 8.06

IOP / IRT (%) 0 14 28

Compensation (12,12)

Compensation (11,11)

Original vs. Correct Compensation

00

0.81.62.43.2

4

4 8 12 16 20Restraint Current (IRT)

Trip Characteristic

Ope

rate

Cur

rent

(IO

P)

87R1 Correct

87R1 Original87R2 Original87R3 Original

87R2 Correct87R3 Correct

• Follow 4 simple rules to select compensation matrices

• Consider CT saturation in transformer applications• Recognize likelihood of operations

Conclusions

1 Derive phase shift seen by relay

2 If delta winding exists, choose it as reference and select Matrix 0

3 If delta winding does not exist, select Matrix 11 for one of wye windings and appropriate matrix for other

4 Avoid even-numbered matrices

Questions?

beyond the nameplate – selecting transformer...

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