INTRODUCTION TO CURRENT TRANSFORMER PERFORMANCE ANALYSIS Hands on workshop developed for field relay techs practical approach.

INTRODUCTION

TO

CURRENT TRANSFORMER

PERFORMANCE ANALYSIS

Hands on workshop developed for field relay techs practical approach

Yellow Brick Road

• INTRODUCTION

• DEFINITIONS

• PERFORMANCE CALCULATIONS

• RATIO SELECTION CONSIDERATIONS

• VARIOUS TOPICS

• TEST

Z = V/I --- accurate value of I

DISTANCE ~ Z

INTRODUCTION• IEEE Standard Requirements for Instrument

Transformers C57.13

• IEEE Guide for the Application of Current Transformers Used for Protective Relaying Purposes C37.110

INTRODUCTION• Bushing, internal to Breakers and

Transformers

• Free standing, used with live tank breakers.

• Slipover, mounted externally on breaker/transformers bushings.

• Window or Bar - single primary turn

• Wound Primary

• Optic

MAGNETO-OPTIC CT

• Light polarization passing through an optically active material in the presence of a magnetic field .

• Passive sensor at line voltage is connected to substation equipment by fiber cable.

• Low energy output used for microprocessor relays

• Eliminates heavy support for iron.

DEFINITIONS

• EXCITATION CURVE

• EXCITATION VOLTAGE

• EXCITATION CURRENT

• EXCITATION IMPEDANCE

DEFINITIONS

• EQUIVALENT CIRCUIT/DIAGRAM

• POLARITY

• BURDEN

• TERMINAL VOLTAGE

• CLASSIFICATIONS T AND C

DEFINITIONS

• KNEE POINT

• RELAY ACCURACY CLASS

• MULTI-TAPS ACCURACY

• SATURATION ERROR - RATIO/ANGLE

EXCITATION CURVE

f

Ip

Ie Ze

XpRp e Rs

Sec

g

h

c

d

Pri

Is

EQUIVALENT DIAGRAM

Ve = EXCITATION VOLTAGE Vef Ie = CURRENT (read a few values)Ze = IMPEDANCEVt = TERMINAL VOLTAGE VghPOLARITY - next

TYPICAL EXCITATION BBC CURRENT vs VOLTAGE

V (volts) Ie(amps) Ze(ohms)3.0 0.004 7507.5 0.007 107115 0.011 136442 ------ -----85 ------ -----180 ------ ------310 ------ 3100400 0.25 1600425 ------ ------450 ------ ------500 5.0 100.0520 10.0 52.0

CURRENT vs VOLTAGE

V (volts) Ie(amps) Ze(ohms)3.0 0.004 7507.5 0.007 107115 0.011 136442 0.02 210085 0.03 2833180 0.05 3600310 0.1 3100400 0.25 1600425 0.5 850450 1.00 450

500 5.0 100.0520 10.0 52.0

N1N2

I1 Ze

Ie

I2Rsec

RB

LB

EXTERNALBURDEN {

Ie+I2

Zint

POLARITY

I1

DEFINITIONS

• EXCITATION CURVE

• EXCITATION VOLTAGE

• EXCITATION CURRENT

• EXCITATION IMPEDANCE

• EQUIVALENT CIRCUIT/DIAGRAM

• BURDEN - NEXT

BURDEN

• The impedances of loads are called BURDEN

• Individual devices or total connected load, including sec impedance of instrument transformer.

• For devices burden expressed in VA at specified current or voltage, the burden impedance Zb is:

• Zb = VA/IxI or VxV/VA

RB

LB

BURDEN=

VA / I² {

EXTERNAL BURDENBurden: 0.27 VA @ 5A = …….. Ohms

2.51 VA @ 15A = …….. Ohms

I2

RB

CT winding resistance = 0.3 ohmsLead length = 750 ft # 10 wireRelay burden = 0.05 ohms

QUIZ

DEFINITIONS

• CLASSIFICATIONS T AND C

ANSI/IEEE STANDARD FOR CLASSIFICATION T & C

• CLASS T: CTs that have significant leakage flux within the transformer core - class T; wound CTs, with one or more primary-winding turns mechanically encircling the core. Performance determined by test.

CLASS C

• CTs with very minimal leakage flux in the core, such as the through, bar, and bushing types. Performance can be calculated.

KNEE POINT

DEFINITIONS

• KNEE POINT IEEE IEC - effective saturation point

• Quiz- read a few knee point voltages and also at 10 amps Ie.

ANSI/IEEE KNEE POINT

Exc

itat

ion

Vol

ts

Kne

e P

oint

Vol

ts

45° LINE

QUIZ: READ THE KNEE POINT VOLTAGE

KNEE POINT OR EFFECTIVE POINT OF SATURATION

• ANSI/IEEE: as the intersection of the curve with a 45 tangent line

• IEC defines the knee point as the intersection of straight lines extended from non saturated and saturated parts of the excitation curve.

• IEC knee is higher than ANSI - ANSI more conservative.

IEC KNEE POINT

ANSI/IEE KNEE POINT

EX: READ THE KNEE POINT VOLTAGE

DEFINITIONS

• EQUIVALENT CIRCUIT/DIAGRAM

• EXCITATION VOLTAGE, CURRENT, IMPEDANCE

• TERMINAL VOLTAGE

• BURDEN

• CLASSIFICATIONS T AND C

• EXCITATION CURVE

• KNEE POINT IEEE IEC

• ACCURACY CLASS

CT ACCURACY CLASSIFICATION

The measure of a CT performance is its ability to reproduce accurately the primary current in secondary amperes both is wave shape and in magnitude. There are two parts:

• Performance on symmetrical ac component.• Performance on offset dc component. Go over the

paper

ANSI/IEEE ACCURACY CLASS • ANSI/IEEE CLASS DESIGNATION C200:

INDICATES THE CT WILL DELIVER A SECONDARY TERMINAL VOLTAGE OF 200V

• TO A STANDARD BURDEN B - 2 (2.0 ) AT 20 TIMES THE RATED SECONDARY CURRENT

• WITHOUT EXCEEDING 10% RATIO CORRECTION ERROR. Pure sine wave

Standard defines max error, it does not specify the actual error.

ACCURACY CLASS CSTANDARD BURDEN

• ACCURACY CLASS: C100, C200, C400, & C800 AT POWER FACTOR OF 0.5.

• STANDARD BURDEN B-1, B-2, B-4 AND B-8 THESE CORRESPOND TO 1, 2, 4 AND 8.

• EXAMPLE STANDARD BURDEN FOR C100 IS 1 , FOR C200 IS 2 , FOR C400 IS 4 AND FOR C800 IS 8 .

• ACCURACY CLASS APPLIES TO FULL WINDING, AND ARE REDUCED PROPORTIONALLY WITH LOWER TAPS.

• EFFECTIVE ACCURACY =

TAP USED*C-CLASS/MAX RATIO

AN EXERCISE• 2000/5 MR C800 tap used*c-class/max

ratio

TAPS KNEE POINT EFFECTIVE ACCURACY

2000/5 ……………….. ……………...

1500/5 ……………….. ……………...

1100/5 ……………….. ……………...

500/5 ……………….. ……………...

300/5 ……………….. ……………...

AN EXERCISE• 2000/5 MR C800 tap used*c-class/max ratio

TAPS KNEE POINT EFFECTIVE ACCURACY

2000/5 590 800

1500/5 390 600

1100/5 120 440 500/5 132 200

300/5 78 120

AN EXERCISE• 2000/5 MR C400 tap used*c-class/max ratio

TAPS KNEE POINT EFFECTIVE ACCURACY

2000/5 ……………….. ……………...

1500/5 ……………….. ……………...

1100/5 ……………….. ……………...

500/5 ……………….. ……………...

300/5 ……………….. ……………...

AN EXERCISE• 2000/5 MR C400 tap used*c-class/max ratio

TAPS KNEE POINT EFFECTIVE ACCURACY

2000/5 220 400

1500/5 170 300

1100/5 125 220

500/5 55 100

300/5 32 60

CT SELECTIONACCURACY CLASS

POINT OF SATURATION : KNEE POINT IT IS DESIRABLE TO STAY BELOW OR VERY CLOSE TO KNEE POINT FOR THE AVAILABLE CURRENT.

Recap

ANSI/IEEE ACCURACY CLASS C400

• STANDARD BURDEN FOR C400: (4.0 )• SECONDARY CURRENT RATING 5 A

• 20 TIMES SEC CURRENT: 100 AMPS

• SEC. VOLTAGE DEVELOPED: 400V

• MAXIMUM RATIO ERROR: 10%

• IF BURDEN 2 , FOR 400V, IT CAN SUPPLY MORE THAN 100 AMPS SAY 200 AMPS WITHOUT EXCEECING 10% ERROR.

N1N2

I1 Ze

Ie <10

Isec = 100 Rsec

RB

LB

EXTERNALBURDEN

Ie+Isec

Zint

ACCURACY ACLASS: C200 RATED SEC CURRENT = 5 AEXTERNALBURDEN = STANDARD BURDEN = 2 .0 OHMSVe=200 V Isec = 100 A Ie <10 Amps.

I1

PERFORMANCE

CALCULATIONS

BUTTHE REST OF US

“SHOW US THE DATA”

PERFORMANCE CRITERIA

• THE MEASURE OF A CT PERFORMANCE IS ITS ABILITY TO REPRODUCE ACCURATELY THE PRIMARY CURRRENT IN SECONDARY AMPERES - BOTH IN WAVE SHAPE AND MAGNITUDE …. CORRECT RATIO AND ANGLE.

CT SELECTION AND PERFORMANCE EVALUATION FOR PHASE FAULTS

600/5 MR Accuracy class C100 is selected Load Current= 90 AMax 3 phase Fault Current= 2500 AMin. Fault Current=350 A

STEPS: CT Ratio selectionRelay Tap SelectionDetermine Total Burden (Load)CT Performance using ANSI/IEEE StandardCT Performance using Excitation Curve

PERFORMANCE CALCULATION

STEPS: CT Ratio selectionRelay Tap SelectionDetermine Total Burden (Load)CT Performance using ANSI/IEEE StandardCT Performance using Excitation Curve

STEPS: CT Ratio selection- within short time and continuous current – thermal limits- max load just under 5ALoad Current= 90 ACT ratio selection : 100/5

PERFORMANCE CALCULATION

STEP: Relay Tap SelectionO/C taps – min pickup , higher than the max. load167%, 150% of specified thermal loading.Load Current= 90 A for 100/5 CT ratio = 4.5 A sec.Select tap higher than max load say = 5.0

How much higher – relay characteristics, experience and judgment.

Fault current: min: 350/20 = 17.5Multiple of PU = 17.5/5 = 3.5Multiple of PU = 17.5/6 = 2.9

PERFORMANCE CALCULATION

STEP: Determine Total Burden (Load)

Relay: 2.64 VA @ 5 A and 580 VA @ 100 ALead: 0.4 Ohms

Total to CT terminals:

(2.64/5*5 = 0.106) + 0.4 = 0.506 ohms @ 5A

(580/100*100 = 0.058) + 0.4 = 0.458 ohms @ 100 A

PERFORMANCE CALCULATION

STEPS: CT Ratio selectionRelay Tap SelectionDetermine Total Burden (Load)CT Performance using ANSI/IEEE StandardCT Performance using Excitation Curve

PERFORMANCE CALCULATION

STEP: CT Performance using ANSI/IEEE Standard

Ip

Ie Ze

XpRp e Rs

Sec

g

h

c

d

Pri

Is

Determine voltage @ max fault current CT must develop across its terminals gh

PERFORMANCE CALCULATION

STEP: Performance – ANSI/IEEE StandardVgh = 2500/20 * 0.458 = 57.25

600/5 MR C100 CT used at tap 100/5 -- effective accuracy class

(100/600) x 100 = ?

CT is capable of developing 16.6 volts. Severe Saturation. Cannot be used.

PERFORMANCE CALCULATION

STEP: Performance – ANSI/IEEE Standard

For microprocessor based relay:Burden will change from 0.458 to o.4

Vgh = 2500/20 * 0.4 = 50.0

600/5 MR C100 CT used at tap 100/5 -- effective accuracy class(100/600) x 100 = ?

CT is capable of developing 16.6 volts. Severe Saturation. Cannot be used.

PERFORMANCE CALCULATION

STEP: Performance – ANSI/IEEE Standard

Alternative: use 400/5 CT tap:Max Load = 90 ARelay Tap = 90/80 = 1.125 Use: 1.5 relay tap.Min Fault Multiples of PU=(350/80=4.38, 4.38/1.5= 2.9)Relay burden at this tap = 1.56 ohmsTotal burden at CT terminals = 1.56 + 0.4 = 1.96Vgh = 2500/80 * 1.96 = 61.25600/5 MR C100 CT used at tap 400/5-- effective accuracy class is = (400/600) x 100 = ?CT is capable of developing 66.6 volts. Within CT capability

PERFORMANCE CALCULATIONSTEP: CT Performance using Excitation Curve

ANSI/IEEE ratings “ballpark”. Excitation curve method provides relatively exact method. Examine the curve

Burden = CT secondary resistance + lead resistance + relay burden

Burden = 0.211 + 0.4 + 1.56 = 2.171

For load current 1.5 A:Vgh = 1.5 * 2.171 = 3.26 V Ie = 0.024Ip = (1.5+0.024) * 80 = 123 A well below the min If = 350 A (350/123=2.84 multiple of pick up)

PERFORMANCE CALCULATIONSTEP: CT Performance using Excitation Curve

For max fault currentBurden = CT secondary resistance + lead resistance + relay burdenBurden = 0.211 + 0.4 + 1.56 = 2.171

Fault current 2500/80 = 31.25 A:

Vgh = 31.25 * 2.171 = 67.84 V Ie = 0.16

Beyond the knee of curve, small amount 0.5% does not significantly decreases the fault current to the relay.

I2

RB

CT winding resistance = 0.3 ohmsLead length = 750 ft # 10 wireRelay burden = 0.05 ohms as constantFault current = 12500A/18000ACT CLASS = C400/C8002000/5 MR current transformerCT RATIO = 800/5

TEST

Determine CT performance using Excitation Curve method:

AN EXAMPLE – C400

• CT RESISTANCE 0.3 OHMS

• LEAD RESISTANCE1.5 OHMS

• IMPEDANCE OF VARIOUS DEVICES 0.05 OHMS

• FAULT CURRENT 12500 AMPS

• CT RATIO 800/5

• ACCURACY CLASS C400

• supply curves C400/800

CALCULATIONS for 12500 A – C400

• BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)

• Ve = (1.5 + 0.3 + 0.05 ) 12500/160

• Ve = 144.5 VOLTS Plot on curve

• Plot on C400

CALCULATIONS for 18000 –C400

• BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)

• Ve = (1.5 + 0.3 + 0.05 ) 18000/160

• Ve = 209 VOLTS Plot on curve

• Plot on C400

ANOTHER EXAMPLE C800

• CT RESISTANCE 0.3 OHMS

• LEAD RESISTANCE1.5 OHMS

• IMPEDANCE OF VARIOUS DEVICES 0.05 OHMS

• FAULT CURRENT 12500 AMPS

• CT RATIO 800/5

• ACCURACY CLASS C800

• supply curves C400/800

CALCULATIONS for 12500 A – C800

• BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)

• Ve = (1.5 + 0.3 + 0.05 ) 12500/160

• Ve = 144.5 VOLTS Plot on curve

• Plot on C800

CALCULATIONS for 18000 A –C800

• BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)

• Ve = (1.5 + 0.3 + 0.05 ) 18000/160

• For 18,000 A (Ve =209 V) Plot on curve

• Plot on C800

FAULT CURRENT MAGNITUDES

• 25 -33 KA 8

• 20 - 25 KA 10

• 12.5 -20 KA 46

• 20 - 25 KA 35

• 10 -12.5 KA 35

• <10 KA +150

REFER TO PAGE 6 OF PAPER

RED DELICIOUS

C400

ZONE1

Z = V/A

DISTANCE ~ Z

STANDARD DATA FROM MANUFACTURER

• ACCURACY:– RELAY CLASS C200– METERING CLASS, USE 0.15%– 0.3%, 0.6% & 1.2% AVAIALABLE BUT NOT

RECOMMENDED– 0.15% MEANS +/- 0.15% error at 100%

rated current and 0.30% error at 10% of rated current ( double the error)

STANDARD DATA FROM MANUFACTURER

• CONTINUOUS (Long Term) rating– Primary

– Secondary, 5 Amp ( 1Amp)

– Rating factor (RF) of 2.0 provides Twice Primary and Secondary rating continuous at 30degrees

STANDARD DATA FROM MANUFACTURER

• SHORT TIME TERMINAL RATINGSTransmission Voltage Applications– One Second Rating = 80% Imax Fault, based

on IxIxT=K where T=36 cycles & I=Max fault current

Distribution Voltage Applications

One Second Rating = Maximum Fault Current level

RATIO CONSIDERATIONS

• CURRENT SHOULD NOT EXCEED CONNECTED WIRING AND RELAY RATINGS AT MAXIMUM LOAD. NOTE DELTA CONNECTD CT’s PRODUCE CURRENTS IN CABLES AND RELAYS THAT ARE 1.732 TIMES THE SECONDARY CURRENTS

RATIO CONSIDERATIONS

• SELECT RATIO TO BE GREATER THAN THE MAXIMUM DESIGN CURRENT RATINGS OF THE ASSOCIATED BREAKERS AND TRANSFORMERS.

RATIO CONSIDERATIONS

• RATIOS SHOULD NOT BE SO HIGH AS TO REDUCE RELAY SENSITIVITY, TAKING INTO ACCOUNT AVAILABLE RANGES.

RATIO CONSIDERATIONS

• THE MAXIMUM SECONDARY CURRENT SHOULD NOT EXCEED 20 TIMES RATED CURRENT. (100 A FOR 5A RATED SECONDARY)

RATIO CONSIDERATIONS

• HIGHEST CT RATIO PERMISSIBLE SHOULD BE USED TO MINIMIZE WIRING BURDEN AND TO OBTAIN THE HIGHEST CT CAPABILITY AND PERFORMANCE.

RATIO CONSIDERATIONS

• FULL WINGING OF MULTI-RATIO CT’s SHOULD BE SELECTED WHENEVER POSSIBLE TO AVOID LOWERING OF THE EFFECTIVE ACCURACY CLASS.

TESTING• Core Demagnetizing

– The core should be demagnetized as the final test before the equipment is put in service. Using the Saturation test circuit, apply enough voltage to the secondary of the CT to saturate the core and produce a cecondary currrent of 3-5 amps. Slowly reduce the voltage to zero before turning off the variac.

TESTING• Saturation

– The saturation point is reached when there is a rise

in the test current but not the voltage.

TESTING• Flashing• This test checks the polarity of the CT

• Ratio

• Insulation test