55523949 Current Transformer Slide

8/3/2019 55523949 Current Transformer Slide

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CURRENTTRANSFORMER

B Y

ENG .Yehia Tag Eldin


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The main tasks of instrumenttransformer are:

To transform current, or voltages, from a high value to a valueeasy to handle for relays and instruments.

21 NI=

Insulate secondary circuits from the primary.

permit the use of standard current ratings forsecondary equipment.

12


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APPLICATION

Current transformers (CT,s) are instrumenttransformers that are used to supply a reducedvalue of current to protective relays , meters and

o er ns rumen s. CT,s provide isolation from the high voltage

primary , permit grounding of the secondary

windings for safety , and step down the magnitudeof the measured current to a value that can besafely handled by the instruments


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RATIO

The most common CT secondary full load currentis 1A or 5A.

CT ratio are expressed as a ratio of rated primary

current to t e rate secon ary current . Example

a 1000/1 A CT will produce 1A of secondary

current when 1000 A flows through the primary .As the primary current changes the secondarycurrent will vary accordingly.


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POLARITY

All CT,s are subtractive polarity . Polarity refers to the instantaneous direction of the

primary current with respect to the secondary

current an is etermine y t e way t etransformer leads are brought out of the case .

On subtractive polarity transformers the H1

primary lead and the X1 secondary lead will be onthe same side of the transformer.


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CT POLARITY

P1

S1


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Terminal markings

GENERAL RULESThe terminal markings shall identify: the primary andsecondary windings; the winding sections, if any; the

the intermediate tapings, if any


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GRAPHIC SYMBOLS OF CURRENTTRANSFORMERS


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CT RING TYPE


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CT


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SUMMATION CT


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CT EQUIVALENTCIRCUIT


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EXCITATIONCURVE


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1 Non saturated - zone

2 intermediate - zone

3 saturated -zone


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TEST RESULT

current(I) voltage(V)

0.01 9

0.04 9

0.10 9

0.12 90.14 9

0.20 9

0.30 9

0.40 940.0 9


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SECONDARY EXCITINGCURRENT


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CT 1200 / 5 ARs 0.0024 ohm/turn

V = 1200*785.1

= 1.785 x 240

= 428.4 v


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A C.T consists essentially of an iron core with two windings. Onewinding is connected in the circuit whose current is to be measured.

The flow of current in the primary winding produces an alternatingflux in the core and this flux induces an e.m.f in the secondarywinding which results in the flow of secondary current when thisw n ng s connecte to an externa c ose c rcu t .

The magnetic effect of the secondary current , in accordance withfundamental principles , is in opposition to that of the primary and thevalue of the secondary current automatically adjust itself to such a

value , that the resultant magnetic effect of the primary and secondarycurrents , produce a flux required to induce the e.m.f. necessary to drivethe secondary current through the impedance of the secondary.


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TERMS & SPECIFICATIONS

Thermal continuous current ratingThe thermal continuous current rating

(r.m.s.value in operates) 1.2 times , or in extended-range

current trans ormers 1.2 or 2.0 t mes , t e rate current.


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Thermal short time current

Ith

Ith is the value of current quoted on the rating platewith a duration of 1 sec. whose heating effect thecurrent transformer can withstand without damage

. .

KA)Ith =

Ik =

(50/f)0.05(tIk +

Un*3Ssc


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Example

MVA SC = 5000 MVA V = 380 KV

I k =5000

I k = 7.597 KA

Ith =

I dyn = 2.5 Ith

)0.05(50/(1Ik +


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Dynamic current rating I dyn

I dyn is the h ighest amplitude of current whosemechanical effects the CT can withstand , with the

,(peak value in KA)

I dyn = 2.5 3 I th


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Burden

Burden = The impedance of the secondary circuit inohms and power factor. The burden is usually

-absorbed at a specified power-factor at the ratedsecondary current.


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EXTERNAL BURDEN

RB

LB

BURDEN=

VA / I{


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To protect instrument and meters from high fault currents the meteringcores must be saturated 10-40 times the rated current depending of the typeof burden.

The instrument security factor Fs

PP +

sib PP

n+

=


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The main characteristics of protection CT cores are:

Lower accuracy than for measuring transformer .

High saturation voltage.

Little , or no turn correction at all.

5P and 10P The error is then 5 and 10 at the specified ALF and at

rated burden.


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The Accuracy Limit Factor indicates the over current as a multiple times the

rated current , up to which the rated accuracy (5P or 10P) is fulfilled (withthe rated burden connected).

ALFPpPPnib

in *+

+=


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No. of primary turns = 1 turn

No. of secondary turns = N turn

Ip = N * Is

or s o ow roug ere mus e someea rans ormer

potential Es = The E.M.F

Es = Is * R

Es is produced by an alternating flux in the core.

dt

dE s


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Flux required to produce Es

BsCTss zIRIE ** +=

AB*=

Where= ux ens ty n t e core

A = cross-sectional area of core

NAfBEk ****44.4=

( )LCTBss zzzIE ++= Required sk EE f

Equ. 1

Equ. 2


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CT 2000/5 , Rs =0.31, Imax =40 KA , MaX Flux density =1.6 Tesla

Find maximum secondary burden permissible if no saturation is to occur.

Solution

N=2000/5 = 400Turns

Is max = 40000/400 = 100Amps

From Equ.1Vk = 4.44*1.6*20*60*(400/10000) = 340 Volt

Max burden = 340/100 = 3.4 ohms

Max connected burden = 3.4 - 0.31 = 3.09


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CT ratio are selected to match the maximum load current requirements.

i.e. the maximum design load current should not exceed the CT ratedcurrent.

The CT ratio should be large enough so that the CT secondary current doesnot exceed 20 times rated current under the maximum symmetrical primaryfault current.


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It is customary to place CT,s on both sides of the breaker. So that theprotection zones will overlap.

The protection Engineer can determine which side of the breaker is best forCT location .

All possibilities of fault position should be considered .


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The overlap should occur across a C.B, so the C.B lies in both zones for thisarrangement it is necessary to install C.Ts on both sides of the C.B.

C.T,s mounted on both sides of breaker nounprotected region

No region un protected


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Current transformers mounted on C.B sideonly of breaker fault shown not cleared by

bus bar protection.


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Current transformers mounted on bus barside only of breaker fault shown not

cleared circuit protection.

C,B will open by line protection but faultwill last.


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IEC standardPROTECTION RATIO 2000/5 A

POWER 20 VA

CLASS 5P20


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IEC standardMEASURING RATIO 2000/5 A

POWER 20 VA

CLASS 0.5SF5


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CTclass X THE FOLLOWING INFRMATION IS REQUIRED

Turns Ratio

Knee Point Voltage

Maximum Excitation Current Secondary Circuit Resistance


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TPX, TPY AND TPZ current transformers

CTs of class P, models were developed for CTs of classTPX (closed-core), TPYand TPZ (nonclosed-core). Allmodels are based on known rated values of the CTs.

,

no additional measurements of the parameters of theCTs are needed.


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TPX High remanence type CT

The high remanence type has no limit for theremanence flux. This CT has a magnetic core without

almost infinite time. In this type of transformers theremanence flux can be up to 70-80% of the saturationflux.

Typical examples of high remanence type CT are classP, TPS, TPX according to IEC,class P, X according to BS(British Standard) and non gapped class C, K

according to


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TPY Low remanence type CT

The low remanence type has a specified limit for theremanence flux. This CT is made with a small air gap

exceed 10% of the saturation flux. The small air gap has only very limited influence on

the other properties of the CT. Class TPY according to

IEC is a low remanence type CT.


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TPZ Non remanence type CT

The non remanence type CT has practically negligiblelevel of remanence flux. This type of CT has relatively

practically zero level. At the same time, these air gapsminimize the influence of the DC-component fromthe primary fault current.

The air gaps will also reduce the measuring accuracy inthe non-saturated region of operation. Class TPZaccording to IEC is a non remanence type CT.


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As a matter of safety, the secondary circuits of acurrent transformer should never be opened underload, because these would then be no secondary mmf

,

current would become exciting current and thus mightinduce a very high voltage in the secondary.


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General

As a matter of safety, the secondary circuits of acurrent transformer should never be opened under

,

to oppose the primary mmf, and all the primary current would become exciting current and thus mightinduce a very high voltage in the secondary.


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Ip

Ie Ze

XpRp e Rs

Sec

g

h

c

Pri

Is

EQUIVALENT DIAGRAM

fVe = EXCITATION VOLTAGE VefIe = CURRENT

Ze = IMPEDANCEVt = TERMINAL VOLTAGE Vgh


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KNEE POINT OR EFFECTIVE POINT OFSATURATION

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

tangent line IEC defines the knee point as the intersection of

strai ht lines extended from non saturated and

saturated parts of the excitation curve. IEC knee is higher than ANSI - ANSI more

conservative.


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45 LINE

ANSI/IEEEKNEE POINT

Ex

citation

Volts

Knee

PointVolts


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IEC KNEE POINT

ANSI/IEEKNEE POINT

EX: READ THE KNEE POINT VOLTAGE


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RATIO CONSIDERATIONS

CURRENT SHOULD NOT EXCEED CONNECTEDWIRING AND RELAY RATINGS AT MAXIMUMLOAD. NOTE DELTA CONNECTD CTs PRODUCE

1.732 TIMES THE SECONDARY CURRENTS


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RATIO CONSIDERATIONS SELECT RATIO TO BE GREATER THAN THE

MAXIMUM DESIGN CURRENT RATINGS OF THEASSOCIATED BREAKERS AND TRANSFORMERS.


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RATIOS SHOULD NOT BE SO HIGH AS TO,

ACCOUNT AVAILABLE RANGES.


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THE MAXIMUM SECONDARY CURRENT SHOULDNOT EXCEED 20 TIMES RATED CURRENT. (100 A


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HIGHEST CT RATIO PERMISSIBLE SHOULD BEUSED TO MINIMIZE WIRING BURDEN AND TO

BTAIN THE HI HE T T APABILITY AND

PERFORMANCE.


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FULL WINDING OF MULTI-RATIO CTsSHOULD BE SELECTED WHENEVER POSSIBLETO AVOID LOWERING OF THE EFFECTIVEACCURACY CLASS.


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Core Demagnetizing The core should be demagnetized as the final test

before the equipment is put in service. Using theSaturation test circuit a l enou h volta e to the

secondary of the CT to saturate the core and producea cecondary current of 3-5 amps. Slowly reduce thevoltage to zero before turning off the variac.

TESTING


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Saturation The saturation point is reached when there is a rise in the

test current but not the voltage.

Burden

TESTING


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Flashing This test checks the polarity of the CT

Insulation test


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SATURATIONAbnormal high primary current

High secondary burden

Combination of the above two factors will result in thecreat on o g ux ens ty n t e current

transformer iron core.

When this density reaches or exceeds the design limitof the core , saturation results.


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SATURATION The accuracy of the CT becomes very poor.

The output wave form distorted.

The result secondary current lower in magnitude.

The greatest dangerous is loss of protective devicecoordination


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SATURATION


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list of CT problems usually found at site:

Shorted CT secondaries

Open-circuited CT secondaries Miswired CTs

CTs that had not been wired

CTs installed backwards

Incorrect CTs

Defective CTs

CTs with incorrect ratios or on the wrong taps

Mind you, this was just at one site and had been


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THANKSYehia Tag ELdin

Documents

55523949 Current Transformer Slide