55523949 current transformer slide
TRANSCRIPT
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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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RATIO CONSIDERATIONS
RATIOS SHOULD NOT BE SO HIGH AS TO,
ACCOUNT AVAILABLE RANGES.
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RATIO CONSIDERATIONS
THE MAXIMUM SECONDARY CURRENT SHOULDNOT EXCEED 20 TIMES RATED CURRENT. (100 A
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RATIO CONSIDERATIONS
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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RATIO CONSIDERATIONS
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