# Current Transformer f2

September 18, 2017 | Author: MohammedSaadaniHassani | Category: Transformer, Electrical Equipment, Electronic Engineering, Electronics, Power Engineering

#### Description

Current Transformer :By

‫ جججج ججج‬/ ‫ججججج‬ ‫ججججج‬

APPLICATION Current

transformers (CT,s) are instrument transformers that are used to supply a reduced value of current to protective relays , meters and other instruments.

CT,s provide isolation from the high voltage

primary , permit grounding of the secondary windings for safety , and step down the magnitude of the measured current to a value that can be safely handled by the instruments

Ratio The most common CT secondary full load current is

1A or 5A. CT ratio are expressed as a ratio of rated primary

current to the rated secondary 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 secondary .current will vary accordingly

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

the primary current with respect to the secondary current and is determined by the way the transformer leads are brought out of the case On subtractive polarity transformers the H1

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

CT POLARITY

P 1 S 1

Terminals Marking general rules

The terminal markings shall identify: the primary and secondary windings; the winding sections, if any; the relative polarities of windings and winding sections; the intermediate tapings, if any

Graphic symbols of current transformers

The main tasks of instrument transformer are: To transform current, or voltages, from a high value to a value easy to handle for relays and instruments. •

Insulate secondary circuits from the primary. •

• permit the use of standard current ratings for secondary equipment.

I1 N2 = I2 N1

CT Ring Type

CT

Summation CT

CT equivalent circuit

Excitation Curve

intermediate zone Non saturated zone

TEST RESULT Volt (V)

Current (mA)

Volt (V)

Current (mA)

8.7

0.9

595

27.7

11.3

1.2

603

28.8

15.1

1.4

625

32.3

21.5

1.9

645

35.8

30.2

2.4

650

37.6

51.7

3.4

680

43.7

103.36

5.6

690

45.9

215.34

9.2

710

53.3

267

10.7

730

60.6

301.48

11.8

750

96.7

387.6

14.5

770

150.5

430.68

16

785

204.3

473.75

18.3

800

287.0

516.8

20.6

805

379.8

560

23.8

820

573.4

595

27.7

840

1111.0

603

28.8

860

2150

SECONDARY EXCITING CURRENT M AG CURVE 1000

M AG CURVE 100

10

1 0 .1

1 .0

1 0 .0

1 0 0 .0

1 0 0 0 .0

1 0 0 0 0 .0

A C.T consists essentially of an iron core with two windings. One winding is connected in the circuit whose current is to be measured. The flow of current in the primary winding produces an alternating flux in the core and this flux induces an e.m.f in the secondary winding which results in the flow of secondary current when this winding is connected to an external closed circuit . The magnetic effect of the secondary current , in accordance with fundamental principles , is in opposition to that of the primary and the value of the secondary current automatically adjust itself to such a value , that the resultant magnetic effect of the primary and secondary currents , produce a flux required to induce the e.m.f. necessary to drive the secondary current through the impedance of the secondary.

TERMS & SPECIFICATIONS

Thermal continuous current rating

The thermal continuous current rating (r.m.s.value in operates) 1.2 times , or in extended-range current transformers 1.2 or 2.0 times , the rated current.

Thermal short –time current Ith Ith is the value of current quoted on the rating

plate with a duration of 1 sec. whose heating effect the current transformer can withstand without damage with the secondary winding short circuited (r.m.s in KA) Ith = Ik (t + 0.05 (50/f) Ik

=

Ssc 3 * Un

Example MVA SC = 5000 MVA V = 380 KV 5000 Ik =

3 * 380

Ik

= 7.597 KA

Ith

=

I dyn

= 2.5 Ith

Ik (1+ 0 .0 5 (5 0 /6 ) 0

Dynamic current rating

I dyn

I dyn is the h ighest amplitude of current whose mechanical effects the CT can withstand , with the secondary winding short circuited , without damage (peak value in KA) I dyn =

2.5 … 3 I th

Burden Burden = The impedance of the secondary

circuit in ohms and power factor. The burden is usually expressed as the apparent power (S) in volt-amperes absorbed at a specified power-factor at the rated secondary current.

EXTERNAL BURDEN

BURDEN = VA / I²

RB LB

To protect instrument and meters from high fault currents the metering cores must be saturated 10-40 times the rated current depending of the type of burden. ”The instrument security factor “Fs

Pn + Pi n= * Fs Pb + Pi

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.

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 (with the rated burden connected).

Pn + Pi n= * ALF pb + Pi

No. of primary turns = 1 turn No. of secondary turns = N turn

Ip = N * Is Ideal transformer for (Is) to flow through R there must be some potential Es = The E.M.F Es = Is * R

Es

dφ ∝ dt

.Es is produced by an alternating flux in the core

Es = I s * RCT + I s * z B Flux required to produce Es

φ = B* A Where

B = Flux density in the core A = cross-sectional area of core

Ek = 4.44* B * f * A * N

E s = I s ( z B + zC T + z L )

Equ. 1

Equ. 2

Required

Ek  E s

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.1 Vk = 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

CT ratio are selected to match the maximum load current requirements. i.e. the maximum design load current should not exceed the CT rated current. The CT ratio should be large enough so that the CT secondary current does not exceed 20 times rated current under the maximum symmetrical primary fault current.

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

The protection Engineer can determine which side of the breaker is best for CT location . All possibilities of fault position should be considered .

The overlap should occur across a C.B, so the C.B lies in both zones for this arrangement it is necessary to install C.Ts on both sides of the C.B. C.T,s mounted on both sides of breaker no unprotected region  No region un protected

Current transformers mounted on C.B side only of breaker fault shown not cleared by bus bar protection.

Current transformers mounted on bus bar side only of breaker fault shown not cleared circuit protection.

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

PROTECTION RATIO

IEC standard

2000/5 A POWER 20 VA CLASS 5P20

MEASURING RATIO

2000/5 A POWER 20 VA CLASS 0.5SF5

IEC standard

CT

Class X

THE FOLLOWING INFRMATION IS REQUIRED Turns Ratio Knee Point Voltage Maximum Excitation Current  Secondary Circuit Resistance

TPX, TPY AND TPZ Current Transformers CTs of class P, models were developed for CTs

of class TPX (closed-core), TPY and TPZ (nonclosed-core). All models are based on known rated values of the CTs. This is an advantage of the presented method, because no additional measurements of the parameters of the CTs are needed.

TPX High remanence type CT The high remanence type has no limit for the remanence

flux. This CT has a magnetic core without any air gap and a remanence flux might remain for almost infinite time. In this type of transformers the remanence flux can be up to 70-80% of the saturation flux. Typical examples of high remanence type CT are class P, TPS, TPX according to IEC,class P, X according to BS (British Standard) and non gapped class C, K according to ANSI/IEEE.

TPY Low remanence type CT The low remanence type has a specified limit for the

remanence flux. This CT is made with a small air gap to reduce the remanence flux to a level that does not 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.

TPZ Non remanence type CT The non remanence type CT has practically

negligible level of remanence flux. This type of CT has relatively big air gaps in order to reduce the remanence flux to practically zero level. At the same time, these air gaps minimize the influence of the DC-component from the primary fault current. The air gaps will also reduce the measuring accuracy in the non-saturated region of operation. Class TPZ according to IEC is a non remanence type CT.

As a matter of safety, the secondary circuits

of a current transformer should never be opened under load, because these would then be no secondary mmf to oppose the primary mmf, and all the primary current would become exciting current and thus might induce a very high voltage in the secondary.

General As a matter of safety, the secondary circuits

of a current transformer should never be opened under load, because these would then be no secondary mmf to oppose the primary mmf, and all the primary current would become exciting current and thus might induce a very high voltage in the secondary.

EQUIVALENT DIAGRAM Ip

Rp

Xp

Rs

e

Is g

c Pri

Ie d

Ze

Sec h

f

Ve = EXCITATION VOLTAGE Vef Ie = CURRENT Ze = IMPEDANCE Vt = TERMINAL VOLTAGE Vgh

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.

Knee Point Volts

Excitation Volts

LINE 45°

ANSI/IEEE KNEE POINT

IEC KNEE POINT ANSI/IEE KNEE POINT

EX: READ THE KNEE POINT VOLTAGE

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 WINDING OF MULTI-RATIO

CT’s SHOULD BE SELECTED WHENEVER POSSIBLE TO AVOID LOWERING OF THE EFFECTIVE ACCURACY CLASS.

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 secondary current 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.

Burden

TESTING Polarity This test checks the polarity of the CT

Ratio Insulation test

SATURATION Abnormal high primary current High secondary burden Combination of the above two

factors will result in the creation of high flux density in the current transformer iron core. When this density reaches or exceeds the design limit of the core , saturation results.

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 device coordination

SATURATION

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 •

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