TO CURRENT TRANSFORMER PERFORMANCE ANALYSIS 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

Ve = EXCITATION VOLTAGE Vef Ie = CURRENT (read a few values) EQUIVALENT DIAGRAM Ip Ie Ze Xp Rp e Rs Sec g h c d Pri Is f Ve = EXCITATION VOLTAGE Vef Ie = CURRENT (read a few values) Ze = IMPEDANCE Vt = TERMINAL VOLTAGE Vgh POLARITY - next

TYPICAL EXCITATION BBC CURRENT vs VOLTAGE V (volts) Ie(amps) Ze(ohms) 3.0 0.004 750 7.5 0.007 1071 15 0.011 1364 42 ------ ----- 85 ------ ----- 180 ------ ------ 310 ------ 3100 400 0.25 1600 425 ------ ------ 450 ------ ------ 500 5.0 100.0 520 10.0 52.0

V (volts) Ie(amps) Ze(ohms) CURRENT vs VOLTAGE V (volts) Ie(amps) Ze(ohms) 3.0 0.004 750 7.5 0.007 1071 15 0.011 1364 42 0.02 2100 85 0.03 2833 180 0.05 3600 310 0.1 3100 400 0.25 1600 425 0.5 850 450 1.00 450 500 5.0 100.0 520 10.0 52.0

Rsec Zint I2 I1 Ie+I2 { Ie RB EXTERNAL BURDEN N1 N2 I1 Ze LB POLARITY

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

{ EXTERNAL BURDEN RB BURDEN = VA / I² LB Burden: 0.27 VA @ 5A = …….. Ohms 2.51 VA @ 15A = …….. Ohms { RB BURDEN = VA / I² LB

RB QUIZ I2 CT winding resistance = 0.3 ohms Lead length = 750 ft # 10 wire Relay burden = 0.05 ohms

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.

QUIZ: READ THE KNEE POINT VOLTAGE 45° LINE ANSI/IEEE KNEE POINT Excitation Volts Knee Point Volts 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.

EX: READ THE KNEE POINT VOLTAGE 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 C STANDARD 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 SELECTION ACCURACY 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.

Rsec RB EXTERNAL BURDEN LB Zint Isec = 100 I1 Ie <10 N1 N2 I1 Ze Ie+Isec Ie <10 RB EXTERNAL BURDEN N1 N2 I1 Ze LB ACCURACY ACLASS: C200 RATED SEC CURRENT = 5 A EXTERNALBURDEN = STANDARD BURDEN = 2 .0 OHMS Ve=200 V Isec = 100 A Ie <10 Amps.

PERFORMANCE CALCULATIONS

THE REST OF US “SHOW US THE DATA” BUT THE 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 A Max 3 phase Fault Current= 2500 A Min. Fault Current=350 A STEPS: CT Ratio selection Relay Tap Selection Determine Total Burden (Load) CT Performance using ANSI/IEEE Standard CT Performance using Excitation Curve

PERFORMANCE CALCULATION STEPS: CT Ratio selection Relay Tap Selection Determine Total Burden (Load) CT Performance using ANSI/IEEE Standard CT Performance using Excitation Curve STEPS: CT Ratio selection - within short time and continuous current – thermal limits - max load just under 5A Load Current= 90 A CT ratio selection : 100/5

PERFORMANCE CALCULATION STEP: Relay Tap Selection O/C taps – min pickup , higher than the max. load 167%, 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 = 17.5 Multiple of PU = 17.5/5 = 3.5 Multiple of PU = 17.5/6 = 2.9

PERFORMANCE CALCULATION STEP: Determine Total Burden (Load) Relay: 2.64 VA @ 5 A and 580 VA @ 100 A Lead: 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

Determine Total Burden (Load) CT Performance using ANSI/IEEE Standard PERFORMANCE CALCULATION STEPS: CT Ratio selection Relay Tap Selection Determine Total Burden (Load) CT Performance using ANSI/IEEE Standard CT Performance using Excitation Curve

PERFORMANCE CALCULATION STEP: CT Performance using ANSI/IEEE Standard Ip Ie Ze Xp Rp 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 Standard Vgh = 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 A Relay 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 ohms Total burden at CT terminals = 1.56 + 0.4 = 1.96 Vgh = 2500/80 * 1.96 = 61.25 600/5 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 CALCULATION STEP: 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 = 0.024 Ip = (1.5+0.024) * 80 = 123 A well below the min If = 350 A (350/123=2.84 multiple of pick up)

PERFORMANCE CALCULATION STEP: CT Performance using Excitation Curve For max fault current Burden = CT secondary resistance + lead resistance + relay burden Burden = 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.

TEST RB Determine CT performance using Excitation Curve method: I2 CT winding resistance = 0.3 ohms Lead length = 750 ft # 10 wire Relay burden = 0.05 ohms as constant Fault current = 12500A/18000A CT CLASS = C400/C800 2000/5 MR current transformer CT RATIO = 800/5

AN EXAMPLE – C400 CT RESISTANCE 0.3 OHMS LEAD RESISTANCE 1.5 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 RESISTANCE 1.5 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 RATINGS Transmission 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 Ratio Insulation test This test checks the polarity of the CT Ratio Insulation test