CT4_4

INTRODUCTIONTOCURRENT TRANSFORMERPERFORMANCE ANALYSISHands on workshop developed for field relay techs practical approach

Yellow Brick RoadINTRODUCTIONDEFINITIONSPERFORMANCE CALCULATIONSRATIO SELECTION CONSIDERATIONSVARIOUS TOPICSTEST

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

DISTANCE ~ Z

INTRODUCTIONIEEE Standard Requirements for Instrument Transformers C57.13

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

INTRODUCTIONBushing, internal to Breakers and TransformersFree standing, used with live tank breakers.Slipover, mounted externally on breaker/transformers bushings.Window or Bar - single primary turnWound PrimaryOptic

MAGNETO-OPTIC CTLight 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 relaysEliminates heavy support for iron.

DEFINITIONSEXCITATION CURVE

EXCITATION VOLTAGE

EXCITATION CURRENT

EXCITATION IMPEDANCE

DEFINITIONSEQUIVALENT CIRCUIT/DIAGRAM

POLARITY

BURDEN

TERMINAL VOLTAGE

CLASSIFICATIONS T AND C

DEFINITIONSKNEE POINT

RELAY ACCURACY CLASS

MULTI-TAPS ACCURACY

SATURATION ERROR - RATIO/ANGLE

EXCITATION CURVE

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

TYPICAL EXCITATION BBC CURRENT vs VOLTAGEV (volts)Ie(amps)Ze(ohms)3.00.0047507.50.0071071150.011136442-----------85-----------180------------310------31004000.251600425------------450------------5005.0100.052010.052.0

CURRENT vs VOLTAGEV (volts)Ie(amps)Ze(ohms)3.00.0047507.50.0071071150.0111364420.022100850.0328331800.0536003100.131004000.2516004250.58504501.00450500 5.0100.052010.052.0

N1N2I1ZeIeI2RsecRBLBEXTERNALBURDEN{Ie+I2ZintPOLARITYI1

DEFINITIONSEXCITATION CURVEEXCITATION VOLTAGE EXCITATION CURRENT EXCITATION IMPEDANCEEQUIVALENT CIRCUIT/DIAGRAMBURDEN - NEXT

BURDENThe 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

RBLBBURDEN=VA / I{EXTERNAL BURDENBurden:0.27 VA @ 5A = .. Ohms2.51 VA @ 15A = .. Ohms

I2RBCT winding resistance = 0.3 ohmsLead length = 750 ft # 10 wireRelay burden = 0.05 ohmsQUIZ

DEFINITIONS

CLASSIFICATIONS T AND C

ANSI/IEEE STANDARD FOR CLASSIFICATION T & CCLASS 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 CCTs with very minimal leakage flux in the core, such as the through, bar, and bushing types. Performance can be calculated.

KNEE POINT

DEFINITIONSKNEE POINT IEEE IEC - effective saturation point

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

ANSI/IEEE KNEE POINTExcitation VoltsKnee Point Volts45 LINE QUIZ: READ THE KNEE POINT VOLTAGE

KNEE POINT OR EFFECTIVE POINT OF SATURATIONANSI/IEEE: as the intersection of the curve with a 45 tangent lineIEC 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 POINTANSI/IEE KNEE POINTEX: READ THE KNEE POINT VOLTAGE

DEFINITIONSEQUIVALENT CIRCUIT/DIAGRAMEXCITATION VOLTAGE, CURRENT, IMPEDANCETERMINAL VOLTAGEBURDENCLASSIFICATIONS T AND CEXCITATION CURVEKNEE POINT IEEE IECACCURACY 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 BURDENACCURACY 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

TAPSKNEE 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

TAPSKNEE POINT EFFECTIVE ACCURACY

2000/5590800

1500/5390600

1100/5 120440 500/5 132200

300/5 78120

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

TAPSKNEE 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

TAPSKNEE POINT EFFECTIVE ACCURACY

2000/5220400

1500/5 170300

1100/5 125220

500/5 55100

300/5 3260

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 C400STANDARD BURDEN FOR C400: (4.0 )SECONDARY CURRENT RATING 5 A20 TIMES SEC CURRENT: 100 AMPSSEC. VOLTAGE DEVELOPED: 400VMAXIMUM RATIO ERROR: 10%IF BURDEN 2 , FOR 400V, IT CAN SUPPLY MORE THAN 100 AMPS SAY 200 AMPS WITHOUT EXCEECING 10% ERROR.

PERFORMANCE

CALCULATIONS

BUT

THE REST OF US

SHOW US THE DATA

PERFORMANCE CRITERIATHE 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 FAULTS600/5 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 CALCULATIONSTEPS: CT Ratio selectionRelay Tap SelectionDetermine Total Burden (Load)CT Performance using ANSI/IEEE StandardCT Performance using Excitation CurveSTEPS: CT Ratio selection- within short time and continuous current thermal limits- max load just under 5ALoad Current= 90 ACT ratio selection : 100/5

PERFORMANCE CALCULATIONSTEP: 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 CALCULATIONSTEP: 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 CALCULATIONSTEP: CT Performance using ANSI/IEEE StandardDetermine voltage @ max fault current CT must develop across its terminals gh

PERFORMANCE CALCULATIONSTEP: 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 CALCULATIONSTEP: Performance ANSI/IEEE StandardFor 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 CALCULATIONSTEP: Performance ANSI/IEEE StandardAlternative: 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 CurveANSI/IEEE ratings ballpark. Excitation curve method provides relatively exact method. Examine the curveBurden = 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.

I2RBCT 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 = 800/5TESTDetermine CT performance using Excitation Curve method:

AN EXAMPLE C400CT RESISTANCE0.3 OHMSLEAD RESISTANCE1.5 OHMSIMPEDANCE OF VARIOUS DEVICES 0.05 OHMSFAULT CURRENT 12500 AMPSCT RATIO 800/5 ACCURACY CLASSC400

supply curves C400/800

CALCULATIONS for 12500 A C400BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)Ve = (1.5 + 0.3 + 0.05 ) 12500/160Ve = 144.5 VOLTS Plot on curvePlot on C400

CALCULATIONS for 18000 C400 BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)Ve = (1.5 + 0.3 + 0.05 ) 18000/160Ve = 209 VOLTS Plot on curvePlot on C400

ANOTHER EXAMPLE C800CT RESISTANCE0.3 OHMSLEAD RESISTANCE1.5 OHMSIMPEDANCE OF VARIOUS DEVICES 0.05 OHMSFAULT CURRENT 12500 AMPSCT RATIO 800/5 ACCURACY CLASSC800

supply curves C400/800

CALCULATIONS for 12500 A C800BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)Ve = (1.5 + 0.3 + 0.05 ) 12500/160Ve = 144.5 VOLTS Plot on curvePlot on C800

CALCULATIONS for 18000 A C800BURDEN = ( Z-LEAD + Z - CT SEC + D - DEVICES)Ve = (1.5 + 0.3 + 0.05 ) 18000/160For 18,000 A (Ve =209 V) Plot on curvePlot on C800

RED DELICIOUSC400ZONE1

Z = V/A

DISTANCE ~ Z

STANDARD DATA FROM MANUFACTURERACCURACY:RELAY CLASS C200METERING CLASS, USE 0.15%0.3%, 0.6% & 1.2% AVAIALABLE BUT NOT RECOMMENDED0.15% MEANS +/- 0.15% error at 100% rated current and 0.30% error at 10% of rated current ( double the error)

STANDARD DATA FROM MANUFACTURERCONTINUOUS (Long Term) ratingPrimary

Secondary, 5 Amp ( 1Amp)

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

STANDARD DATA FROM MANUFACTURERSHORT TIME TERMINAL RATINGS

Transmission Voltage ApplicationsOne Second Rating = 80% Imax Fault, based on IxIxT=K where T=36 cycles & I=Max fault current

Distribution Voltage ApplicationsOne Second Rating = Maximum Fault Current level

RATIO CONSIDERATIONSCURRENT SHOULD NOT EXCEED CONNECTED WIRING AND RELAY RATINGS AT MAXIMUM LOAD. NOTE DELTA CONNECTD CTs PRODUCE CURRENTS IN CABLES AND RELAYS THAT ARE 1.732 TIMES THE SECONDARY CURRENTS

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

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

RATIO CONSIDERATIONSFULL WINGING OF MULTI-RATIO CTs SHOULD BE SELECTED WHENEVER POSSIBLE TO AVOID LOWERING OF THE EFFECTIVE ACCURACY CLASS.

TESTINGCore DemagnetizingThe 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.

TESTINGSaturationThe saturation point is reached when there is a rise in the test current but not the voltage.

TESTINGFlashingThis test checks the polarity of the CTRatioInsulation test