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Current Transformer Sizing (August 19 th , 2010) By: Siva Singupuram James Trinh

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Page 1: CT Sizing

Current Transformer Sizing

(August 19th, 2010)

By: Siva SingupuramJames Trinh

Page 2: CT Sizing

Agenda

• Current Transformer (CT) introduction• Mathematical Modeling• Saturated Waveforms• CT saturation (Hysteresis Curve)• Remnant Flux, High Fault Currents, DC offset• Saturation Voltage, • CT ratings, Selection and Applications• ElectroMagnetic Transients Program (EMTP)/

PSCAD

Page 3: CT Sizing

Introduction CT types

Donut

Bus Bar

Hall Effect CT Magneto-Optical

Page 4: CT Sizing

Typical Current Transformers

Page 5: CT Sizing

Introduction Application

• Current Transformers (CT’s) are instrument transformers that are used to supply a reduced value of current to metering, protective relays, and other instruments. CT’s provide galvanic isolation from the high voltage primary, permit grounding of the secondary for safety, and step-down the magnitude of the measured current to a value that can be safely handled by the instruments (high fault currents).

• To achieve the above goals CT contains1) Iron Core2) Secondary windings3) Primary conductor4) External insulation

Page 6: CT Sizing

CT Symbols• When current flows in the CT from the H1 lead polarity (±) lead through,

the burden (load), and return to the secondary X2 non-polarity lead. The next half cycle the current will reverse.

ANSI IEC

Page 7: CT Sizing

Current Transformer (CT) Principle

Electrical Schematic 3D-Representation

Page 8: CT Sizing

CT Connections

Page 9: CT Sizing

ANSI Standard Letter Designation• B-The CT is a metering type• C-The CT has low leakage flux and the excitation characteristic can be

used directly to determine performance. (Accuracy can be calculated before manufacturing)

• K-Same as the C rating, but the knee point voltage must be at least 70% of the secondary terminal voltage rating.

• T-The CT can have significant leakage flux (Accuracy must be determined by testing at the factory)

• H-The CT accuracy is applicable within the entire range of secondary currents from 5 to 20 time the nominal CT rating. (Typically wound CTs)

• L-Same as C class except there were two accuracy classes 2.5% and 10%. The ratio accuracy can be up to four times greater than the listed value, depending on connected burden and fault current. (Typically window, busing, or bar-type CTs)

• L and H ANSI rating are applicable to old CT that where manufactured before 1954

Page 10: CT Sizing

ANSI Standard Terminal Voltage Class C CTs

C Class ZSTD (Ω) VSTD (V)

C100 1.0 100

C200 2.0 200

C400 4.0 400

C800 8.0 800

STDNSTD ZIV 20

For IN = 5A secondary

Page 11: CT Sizing

Metering Accuracy Classifications• Metering Accuracy CT’s are used where a high degree of accuracy is

required from low-load values up to full-load of a system. These are utilized by utility companies for revenue metering.

Available in Maximum Ratio Error Classes of: ±0.3%, ±0.6%, ±1.2%, ±2.4%

For Burdens (Loads) of: 0.1Ω, 0.2Ω, 0.5Ω, 0.9Ω, 1.8Ω

Volt-amperes (va) equivalent: 2.5va, 5.0va, 12.5va, 22.5va, 45va

Typical Number

0.3 B 0.2

Burden Ohms (Burden)Max Ratio Error ±%

Page 12: CT Sizing

Relaying Accuracy Classifications• Relaying Accuracy CT’s are used for supplying current to protective relays.

In this application, the relays do not normally operate in the normal load range, but they must perform with a reasonable degree of accuracy at very high overload and fault-current levels which may reach 20x the full-load amplitude.

Class C (C for calculated) is low leakage reactance type – typical of donut units – Formerly Class L (L for Low Leakage)

Typical Number

10 C 800

Low Leakage UnitMax secondary voltage developed at 20x rated current without exceeding the +10% ratio error

10% Max Ratio Error at 20x Rated Current

Class T (T for tested) is high leakage reactance type – typical of bar-type units – Formerly Class H (H for High Leakage)

Available secondary voltages: 100V, 200V, 400V, 800V

Burden support (ZSTD): 1.0Ω, 2.0Ω, 4.0Ω, 8.0Ω

Page 13: CT Sizing

CT Knee point

ANSI Method

Page 14: CT Sizing

CT Knee point

ANSI MethodIEC Method

Page 15: CT Sizing

Shorting of CT when not used

• CT’s can be very dangerous devices! Stepping down the current means that the voltage is automatically stepped up the same ratio. The secondary's of the CT’s are, therefore; always kept shorted when not in use and/or when no low impedance ammeter is connected! On the open-circuited secondary of a CT, just a few volts on the primary voltage would become a lethal thousands of volts on the secondary.

Page 16: CT Sizing

CT Equivalent Circuit

LOADsssss ZIjXRIV )(

es

ps I

N

II

Page 17: CT Sizing

Hysteresis/Residual FluxReferring to figure on the left, if H is gradually increased from zero to H1, the magnetization follows path oa. Decreasing H back to zero will leave the material magnetized at a B-value corresponding to point b. Distance ob on the B axis is called the “Residual Flux Density”. To bring B back to zero requires a negative value of H corresponding to point c on the cure. Distance oc on the H-axis is called the Coercive Force. Decreasing H to H2=-H1 results in the B-value of point d. Finally increasing H back to the original H1 now causes the magnetization to traverse path defa. The latter path is nearly equivalent to path abcd, but flipped about the B and H axes. Path abcdefa is called a hysteresis loop which occurs when an AC voltage is applied to a coil on a ferromagnetic core. As the voltage and current cycle between their max and min values, the core magnetization repeatedly traverses a hysteresis loop.

Page 18: CT Sizing

Remanent Flux (Information Only)

Remanent Flux % Percentage of cts

0-20 39

21-40 18

41-60 16

61-80 27

Based on IEEE survey of 141 cts on a 230kV system

Page 19: CT Sizing

DC offset

Saturation Occurs

Page 20: CT Sizing

CT Errors impact

Errors to in currents magnitude and angle will have a significant effect on protection relays particularly (directional (67), distance (21), and differentials (87))

Page 21: CT Sizing

t tL

R

SBF

tdtetRZIvdtBANN

00))(cos(2

R

Le

R

LtRZI

tL

R

SBF

sin1

2

R

Xe

R

Xt

RZI tX

RSBF

sin

)(2

R

Xe

R

XtRZIBAN

tX

R

SBF

sin)(2

R

Le

R

Lt

RZI tL

RSBF

sin)(2

CT terminal Voltage (Vgh)

Page 22: CT Sizing

maxmax SVBAN

R

XRZIBAN SBF 1)(2max

:Recall

R

XRZIV SBFS 1)(2max

R

XZIV BFgh 12max_

R

XZI

VBF

gh 12max_

CT terminal Voltage (Vgh)

Page 23: CT Sizing

R

XZI

VBF

gh 12max_

STDNSTDgh ZIV

V20

2max_

STDBF VR

XZI

1

STDNSTD ZIV 20:Recall

STDNBF ZIR

XZI 201

201

R

X

Z

Z

I

I

STD

B

N

F

CT terminal Voltage (Vgh)

Page 24: CT Sizing

201

R

X

Z

Z

I

I

STD

B

N

F 201

bf ZIR

X

CT primary rating if IF is given in primary Amps orNI

Standard Burden = 1, 2, 4, 8 Ω (at 60°)STDZ

Max Fault in per unit of CT secondary nominal CurrentfI

Burden in per unit of standard burdenbZ

CT secondary rating if IF is given in secondary Amps

Max Fault CurrentFI Burden from the CT’s point of view (CTwdg + Cable Leads + Relays)BZ

To Avoid CT saturation

Page 25: CT Sizing

Limitation of CT Selection Criterion

• We face high X/R ratio and high fault currents near generation sites. This becomes impractical to size the CT to avoid saturation during a asymmetrical fault. The formula derived earlier should not be used for this study and we must size the CT with reasonable sensitivity for line end faults

• For this situation we should analyze in further detail via EMTP simulation.

Page 26: CT Sizing

Example #1

C8002000/5 (full ratio)If = 3.07 puZb = 0.5 puX/R = 12

19.955 ≤ 20

201

bf ZIR

X

(1+12)(3.07)(0.5) ≤ 20 ?

CT OK!

Page 27: CT Sizing

Example #1

C8002000/5 (full ratio)If = 3.07 puZb = 0.5 puX/R = 12

201

bf ZIR

X

19.955 ≤ 20

Example #2

C8002000/5 (full ratio)If = 7.69 puZb = 0.5 puX/R = 12

49.985 > 20

201

bf ZIR

X

(1+12)(3.07)(0.5) ≤ 20 ? (1+12)(7.69)(0.5) ≤ 20 ?

CT OK! CT Saturates

Page 28: CT Sizing

RX

K

TT

s

s

1

1ln1 R

XT

1

primary system time constant1T the saturation factor Vx/Vs, where Vx is the excitation voltage at 10A and, Vs is the saturation voltage IF x ZB

sK

Resistance of the primary system to the point of the faultR

Natural log functionln time to saturatesT

Time to Saturate

Reactance of the primary system to the point of the faultX2πf, where f is the system frequency

Page 29: CT Sizing

Waveforms

Page 30: CT Sizing

Waveforms

Page 31: CT Sizing

Multi-Ratio CTs

1200/5, C800 tapped at 600/5C400 effective rating4.0 Ω effective standard burden

4008001200

600atingEffectiveR

481200

600__ BurdenStdEffective

Page 32: CT Sizing

CT winding Resistance

Typical turn ratio resistance = 2.5mΩ/turn

0.15.25

20005/2000 mFor example a 2000:5 ratio CT;

Page 33: CT Sizing

Lead Resistance CalculationFormula to determine Copper lead resistanceLead resistance, Ω/1000’ = e0.232G-2.32

where, G = AWG number

For a 10AWG cable at (305m = 1000’)R = e0.232(10) -2.32 = e0 = 1Ω/1000’

R = 1Ω/1000’x500’ = 0.5Ω10AWG cable at (152m = 500’) will have a resistance of

Similarly for 8AWG cable at (305m = 1000’)R = e0.232(8) -2.32 = 0.629Ω/1000’

Page 34: CT Sizing

CT Application

ConnectionType of Fault

3 ph or ph-to-ph Ph-to-ground

Wye (connect at CT) Z = RS + RL + ZR Z = RS + 2RL + ZR

Wye (connected at switchhouse) Z = RS + 2RL + ZR Z = RS + 2RL + ZR

Delta (connected at switchhouse) Z = RS + 2RL +3 ZR Z = RS + 2RL +3 ZR

Delta (connected at CT) Z = RS + 3RL +3 ZR Z = RS + 2RL +2 ZR

Z is the effective impedance seen by the CTRS is the CT secondary winding resistance and CT lead resistance; also

includes any relay impedance that is inside the delta connection (Ω)RL is the circuit one-way lead resistance (Ω)ZR is the relay impedance in the CT secondary current path (Ω)

Page 35: CT Sizing

IEEE typical X/R Ratios

Page 36: CT Sizing

IEEE typical X/R Ratios

Page 37: CT Sizing

IEEE typical X/R Ratios

Page 38: CT Sizing

IEEE typical X/R Ratios

Page 39: CT Sizing

IEEE typical X/R Ratios

Page 40: CT Sizing

Parameters for X/R & SCC Example Given ParametersSCC = 7316A @ V=240kVX/R = 12Transformer size = 50MVA; 240/138kVZxmfr = 8.0%

Solve for X/R at 138kV.Solve for SCC at 138kV.

Page 41: CT Sizing

Example for X/R

%644.110009.204,041,3/000,50% xkVAkVAZ source

Step #1: Calculate the equivalent source impedance

SCCkVKVA LLitshortcircu 3

100/% xKVAKVAZ itshortcircurtransformesource

kVAAkVKVA itshortcircu 09.204,041,373162403

Page 42: CT Sizing

Example for X/RStep #2: Source R and X

644.1% sourceZ

12/ RX

24.8512tan 1

%638.1%)644.1(*)24.85( SinX

%138.0%)644.1(*)18.85( CosR

R

X1tan ZSinX *ZCosR *

Page 43: CT Sizing

Example for X/RStep #3: Transformer R and X

0.8% xmfrZ

15/ RX

19.8615tan 1

%982.7%)0.8(*)19.86( SinX

%531.0%)0.8(*)19.86( CosR

(From IEEE typical value)

Page 44: CT Sizing

Example for X/RStep #4: Total Resistance and Reactance at 138kV

%669.0%531.0%138.0 xmfrsourcetotal RRR

%62.9%982.7%638.1 xmfrsourcetotal XXX

Step #5: X/R at138kV

4.14%669.0

%622.9

total

total

R

X

R

X

Page 45: CT Sizing

Example for SCC

%644.110009.204,041,3/000,50% xkVAkVAZ source

Step #1: Calculate the equivalent source impedance

SCCkVKVA LLitshortcircu 3

100/% xKVAKVAZ itshortcircurtransformesource

kVAAkVKVA itshortcircu 09.204,041,373162403

Page 46: CT Sizing

Example for SCCStep #2: Calculate the 138kV side full load current

LL

xmfrLV

kV

kVAFLA

3

AkV

kVAFLALV 18.209

)138(3

000,50

Page 47: CT Sizing

Example for SCCStep #3: Calculate the short circuit current on the 138kV side

)%(%

100

sourcexmfr

LVLV ZZ

FLASCC

AA

SCCLV 06.169,2%)644.1%0.8(

10018.209

Page 48: CT Sizing

Global Example

C200400/5 (full ratio)IF = 7,316A (HV side)X/R = 15 (HV side faults)One-way RL length = 152m (500’)Determine what CT Class and Ratio is validon the 138kV

240kV 138kV

Page 49: CT Sizing

Global Example

4.14%669.0

%622.9

total

total

R

X

R

X

Case #1) From previous slide for X/R and SCC on the 138kV side.

Rone-way = 1Ω/1000’x500’ = 0.5Ω

For a C200: ZSTD = 2.0Ω

ZB = RS + 2RL + ZR

ZB = 0.2Ω + 2(0.5Ω) + 0.1Ω=1.3

Note: 0.1Ω = digital & 1.0 Ω for electromechanical

puZ

ZZ

STD

Bb 7.0

0.2

3.1

Page 50: CT Sizing

Global Example

AA

SCCLV 06.169,2%)644.1%0.8(

10018.209

?201

bf ZIR

X

From previous slide for X/R and SCC on the 138kV side.

puA

A

I

II

N

Ff 5.5

400

169,2

607.05.54.141 CT Saturates

Page 51: CT Sizing

Global Example

RX

K

TT

s

s

1

1ln1

sR

XT 038.0

377

4.141

33.63.35

230

V

V

V

VK

s

Xs

VVx 230

VA

ZIV BFS 3.35)3.1(5/400

2169

Taken from excitation Curve at IE = 10A (full ratio)

)1(19

15.8

1ln038.0 cyclemsRXTs

Page 52: CT Sizing

Global ExampleCase #2 Choosing C400 (2000/5 tapped at 600/5).

puZ

ZZ

STD

Bb 18.1

2.1

41.1

2.142000

600__ BurdenStdEffective

puA

A

I

II

N

Ff 62.3

600

169,2

?201

bf ZIR

X

6618.162.34.141 CT Saturates

ZB = 1.4Ω due to 600/5 ratio CT winding resistance.

Page 53: CT Sizing

Global Example

RX

K

TT

s

s

1

1ln1

sR

XT 038.0

377

4.141

8.55.25

150

V

V

V

VK

s

Xs

VVx 150

VA

ZIV BFS 5.25)41.1(5/600

2169

Taken from exication Curves and at IE = 10A (600/5)

cyclemsRXTs 111

18.5

1ln038.0

Page 54: CT Sizing

Global ExampleCase #3 Choosing C400 (2000/5 full ratio).

puZ

ZZ

STD

Bb 525.0

0.4

1.2

puA

A

I

II

N

Ff 08.1

2000

169,2

?201

bf ZIR

X

73.8525.008.14.141 CT OK

For a C400: ZSTD = 4.0Ω

ZB = 2.1Ω due to 2000/5 ratio CT winding resistance.

Page 55: CT Sizing

Global ExampleCase #4 Choosing C400 (2000/5 full ratio with 80% remenant flux).

?4%)801(*201

bf ZIR

X

73.8525.008.14.141 CT Saturates

Case #4 Choosing C400 (2000/5 full ratio with 50% remenant flux).

?10%)501(*201

bf ZIR

X

73.8525.008.14.141 CT OK

Page 56: CT Sizing

IEEE PSRC CALCULATOR

Vs

a m p s rm s

slope= 1/S

lo g - lo g p lo t ,e q u a l

d e c a d es p a c in g

vo l tsrm s

I e

mf gr 'sd a taV e

1 0

Saturation Curve

iee VIS

V loglog1

log

The Saturation Curve above is represented by the following equation

Where Vi is the value of Ve for Ie=1, that is for (log Ie=0)

Page 57: CT Sizing

Measuring the ‘S’

0

1

2

3

4

5

6

7

8

9

1

0

slope = 0.7/9.3 = 1/S

S = 9.3/0.7 = 13

Ruler, inches or centimeters

Use a ruler to determine the slope.Do NOT read currents or voltages for this purpose.(We assume equal decade spacing of the axes)

HOW TO MEASURE THE SLOPE

Page 58: CT Sizing

IEEE PSRC CALCULATOR EXAMPLECase #1) From previous slide for X/R and SCC on the 138kV side.

Page 59: CT Sizing

EMTP/PSCAD SIMULATION

• The CTs are subject to saturation during DC transient of fault current , hence there is growing interest in simulations like EMTP/PSCAD.

• The IEEE-PSRC CT calculator is only accurate at high fault currents and burden , hence to get better idea of CT performance at low fault currents/ burden we need to model in PSCAD/EMTP.

• The primary aim of PSCAD simulation is to obtain digitized records, which can be recognized as secondary analog signal using D/A conversion and amplification, for the purpose of relay testing.

Page 60: CT Sizing

EMTP/PSCAD SIMULATION

Page 61: CT Sizing

PSCAD -CT Modeling

Page 62: CT Sizing

QUESTIONS?