By Momo Malis May 01, 2009

By Momo Malis May 01, 2009

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Overview Current Transformer (CT) Basics Terms in Which a CT Rating Shall be Expressed Accuracy, Accuracy Class, Burden Differences between Metering & Protection CT Classes CT Tests Saturation, Knee-Point Voltage Effects of Remanent Flux Altalink CT selection requirements Relay Manuf. Recommendations Against CT Saturation

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Purpose of CT Electrical Isolation (Between primary

Voltage System and Metering & Relaying) Reduction in Magnitude of the Primary Current Metering CT - Accuracy required for 5 A or

less Relaying CT - Correct operation required from low to very high fault currents

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Construction • • • • • • •

Bushing (BCT) Window Bar Wound Free-Standing Auxiliary CT etc

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CT - Principles of Operation

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CT Equivalent Circuit

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Terms in Which a CT Rating Shall Be Expressed • Basic impulse insulation level (BIL) voltage • Nominal system voltage or maximum system voltage • Frequency (in Hertz) • Rated primary and secondary currents • Accuracy classes at standard burden • Continuous thermal current rating factor based on 30 °C average ambient air temperature • Short-time mechanical current rating and short-time thermal current rating

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BIL Requirements

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Standard CT Ratio Ratings

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Accuracy Classes and Burden of a CT

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Metering Metering Accuracy Accuracy xx-B-xx Accuracy Accuracy ••0.3 0.3 0.3% 0.3%@ @100%In 100%In 0.6% 0.6%@ @10% 10%In In ••0.6 0.6 0.6% 0.6%@ @100%In 100%In 1.2% 1.2%@ @10% 10%In In ••1.2 1.2 1.2% 1.2%@ @100%In 100%In 2.4% 2.4%@ @10% 10%In In

1.012 1.009 1.006 1.003

Ratio 1.000 Correction Factor 0.997 0.994 0.991 0.988 -60

-45

-30

-15

lagging

0

+15

+30

+45

+60

leading

Phase Angle (minutes) Ref. ANSI/IEEE 57.13 11

Metering Metering Accuracy Accuracy (cont.) (cont.)  A CT accuracy of 0.3 means that the CT is certified by the manufacturer to

be accurate to within 0.3% of its rated ratio value for a primary current of 100% of rated ratio (100%In). For primary currents of 10% of rated ratio the specified accuracy for a given classification is double that of the 100% value or 0.6%.  The accuracy is relatively linear between these two points.  For example a CT with a rated ratio of 200/5 A and with accuracy class of 0.3 would operate within 0.45% of its rated ratio value for a primary current of 100 A. Or to be more explicit, for a primary current of 100 A, the CT is certified to produce a secondary current between 2.489 A and 2.511 A.

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Metering CT - ANSI/IEEE Standard Burden

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Protection CT - ANSI/IEEE Standard Accuracy

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Typical Accuracy Class of Bushing CTs

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Protection CT - ANSI/IEEE Standard Burden

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CT Excitation Curves

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Determining “C” Class from Excitation Curves Quick Arithmetic Method (assumes worst case) Ie / Is = o.1 (10% error i.e. Is = 100 A (20*5 A) ) => Ie = o.1*100 = 10 A => Vs ≈ 470 V (from curve, for 1200/5, Ie= 10 A) => Vb = Vs – Is*Rs = 470-61 V = 409 V

Therefore Class C400

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CT Testing Routine Tests at the Factory Applied Potential Tests between Windings and Ground Induced Potential Tests Partial Discharge Tests Accuracy Tests Polarity Tests

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CT Testing (cont.) Type (Design) Tests (IEEE Std) - for at least one CT in each

design group Measurement and Calculation of Ratio and Phase Angle Demagnetization Impedance and Excitation Measurements Polarity Resistance Measurements Short-Time Characteristics Temperature Rise Characteristics Dielectric Tests Measurement of Open-Circuit Voltage of CTs

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CT Testing (cont.) In-Service Testing

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CT Testing (cont.) In Service Testing (cont.)

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CT Polarity

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SATURATION

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CT Saturation

CT Saturation

CT Saturation Calculator

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Remanent Flux When the primary current is interrupted on a

saturated CT, a substantial portion of the flux present in the core just prior to interruption remains in the core. This flux does not dissipate (or decay) under normal operating conditions. CT can be de-rated for remanent flux CTs are available that have low remanent flux (lower grades of steel, gapped cores) 29

Criterion to Avoid Saturation (Protection CT) Rated CT Terminal Voltage > Calculated Secondary Terminal Voltage To avoid saturation with a DC component in the primary wave and with a pure resistive burden, the required saturation voltage shall be:

Rated CT Terminal Voltage > Isc*(1+X/R)*(RCT +RR+2*RL)/(CTratio) Where: RL - Cable Lead Resistance RR - Relay Resistance RCT - CT resistance Isc – Short Circuit current fault X and R are the primary system reactance and resistance up to the point of fault 30

Criterion to Avoid Saturation -Example Application : Protection CTs Length (L) of cable lead = 70m. Cross Section of Cable (A) = 8.37 mm2 (8-AWG size) Cable Resistance (RL) = ρL/A = 0.02128 * 70/8.37 = 0.1779 Ω at 75 °C

Criteria: Rated CT Terminal Voltage > Calculated Secondary Terminal Voltage To avoid saturation with a DC component in the primary wave and with a pure resistive burden, the required saturation voltage is (IEEE C37.110-1996): Rated CT Terminal Voltage > Isc* (1+X/R) * (RCT + RR + 2*RL)/(CT ratio) Where: RL = Cable Lead Resistance (=0.1779 Ω at 75 °C), RR = Relay Resistance (= 0.008), RCT = CT resistance (= 0.00169 Ω /turn*40 turn = 0.0676 Ω) Isc – Short Circuit current fault (= 2900 A), X and R are the primary system reactance and resistance up to the point of fault (X/R=3). Calculated Secondary Terminal Voltage = 2900 *4* (0.0676 + 0.008 + 2*(0.1779)/40 = 125 V for 200/5 tap. Therefore the CT 200/5MR ( C200 ) is adequately sized for tap 200/5. If the tap is 150/5 the rated CT Terminal Voltage is 150/200*200=150V and Calculated Secondary Terminal Voltage = 2900 *4* (0.0507 + 0.008 + 2*(0.1779)/150/5=160 V Therefore the CT 200/5MR ( C200 ) is not adequately sized for 150/5 tap position but a solution could be to increase the cable lead size: If Cross Section of Cable (A) = 13.3 mm2 (6-AWG) then the cable resistance (RL) = ρL/A = 0.02128 * 70/13.3 = 0.1147 Ω at 75 °C Calculated Secondary Terminal Voltage = = 2900 *4* (0.0507 + 0.008 + 2*(0.1147)/150/5=111 V < 150 V (rated CT Terminal Voltage at tap 150/5)

Burden Verification (Metering CT) CT Rated Burden > Secondary Connected Burden Application : Metering Equipment PML Meter Type: 7650 , Burden (PM) : 0.05 VA CT Specification Manufacturer : Instrument Transformer Inc. (ITI) Model : 780-801, Ratio : 800/5A, Class : 0.3 B 1.8, Burden (Pn) : 45 VA Cable Lead Resistance (RL) : 0.5 Ω CT SIZING CALCULATION Burden Verification Criteria: CT Rated burden > Secondary Connected Burden (Pb) Pb = PM + 2 *(In) 2 RL Pb = 0.0625 + 2*(5)2 *0.5 = 25.0625 VA Hence 45 VA > 25.0625 VA Therefore the CT is adequately sized.

Application Guidelines Specifying CTs that will experience no saturation

for fully offset fault currents can result in unreasonably large and expensive CT. As a rule of thumb, CT performance will be satisfactory if the CT secondary maximum symmetrical external fault current multiplied by the total secondary burden in ohms is less than half the “C" voltage rating of the CT. (i.e. 2Vs ≤ Vx) Verify relay manufacture requirements

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Relay Manufacture CT saturation requirements –ABB Relays

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Relay Manufacture CT saturation requirements –SEL-351 & SEL-551 Family Relays

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Relay Manufacture CT saturation requirements –GE Relays

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Optical Current Transducers OCTs do not experience saturation problems Measure magnetic field associated with

current flow by measuring phase shift in polarized light

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References 1. CSA C13 - Instrument Transformers 2. ANSI C57.13 - Instrument Transformers 3. IEC 60044-1 Instrument transformers - Part

1: Current transformers

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Questions?