c.t dimensioning

PART I - Current Transformers Dimensioning

[IEC 60044 – 1]

Prepared by : Mohammad Subhi Abdel -–HalimSenior Substation Design EngineerPBME – AMMAN Office / JORDAN

1

Definitions :

Current transformerAn instrument transformer in which the secondary current, in normal conditions of use, is substantiallyproportional to the primary current and differs in phase from it by an angle which is approximately zerofor an appropriate direction of the connections

Current error (ratio error)The error which a transformer introduces into the measurement of a current and whicharises from the fact that the actual transformation ratio is not equal to the rated transformation ratio.Current Error : is the ratio of the exciting current, which supplies the eddy current and hysteresis lossesand magnetizes the core to the Primary Current. This current flows in the primary winding only andtherefore, is the cause of the transformer errors.Generally expressed as a percentage of the r.m.s value of the primary current and is given by theformula:

p

psn

I100IIK

or%CurrentErr

whereKn is the rated transformation ratio;Ip is the actual primary current;Is is the actual secondary current when Ip is flowing, under the conditions of measurement.

Phase displacementThe difference in phase between the primary and secondary current vectors, the direction of the vectorsbeing so chosen that the angle is zero for a perfect transformer.The phase displacement is said to be positive when the secondary current vector leads the primarycurrent vector. It is usually expressed in minutes or centiradians.NOTE This definition is strictly correct for sinusoidal currents only.

Accuracy Limit Factor (ALF ) Kssc :The Ratio of the rated Accuracy Limit Primary Current ( Max. Fault Current ) to the rated PrimaryCurrent .A current transformer is designed to maintain its ratio within specified limits up to a certainvalue of primary current, expressed as a multiple of its rated primary current. This multiple is known asthe current transformer’s rated accuracy limit factor (ALF).

Dimensioning factor (Kx) :A factor assigned by the purchaser to indicate the multiple of rated secondary current (Isn) occurringunder power system fault conditions, inclusive of safety factors, up to which the transformer is requiredto meet performance requirements"

Composite errorUnder steady-state conditions, the r.m.s. value of the difference between:a) the instantaneous values of the primary current; andb) the instantaneous values of the actual secondary current multiplied by the rated transformation ratio,the positive signs of the primary and secondary currents corresponding to the convention for terminalmarkingsThe composite error Ec is generally expressed as a percentage of the r.m.s. values of the primarycurrent according to the formula:


[IEC 60044 – 1]


2

T

0

2psn

pc dtiiK

T1

I100

whereKn is the rated transformation ratio;Ip is the r.m.s. value of the primary current;ip is the instantaneous value of the primary current;is is the instantaneous value of the secondary current;T is the duration of one cycle.

For more details and the use of the composite error , please refer to Annex – A .

Rated knee point e.m.f. (Ek) :That minimum sinusoidal e.m.f. (r.m.s.) at rated power frequency when applied to thesecondary terminals of the transformer, all other terminals being open-circuited, which when increasedby 10 % causes the r.m.s. exciting current to increase by no more than 50 %.Some clients require the calculations to be based on the corrected VA and not the rated VA of the C.Twhere a margin of 25 % shall be taken into account for rated VA calculations for 5P class cores for theRated Knee Point Voltage and Operating ALF conversions as per the following formulas which will giveaccurate dimensioning :

1.3

RatedALFIResWdg.VACorrectedVoltagePointKneeRated sn

Since “The rated equivalent limiting secondary e.m.f.” Eal or The Rated Knee Point Voltage Vk inaccording with the IEC 60044-6 standard to specify the CT requirements for different protectionequipment which shall be higher than the required limiting secondary e.m.f EalReq. or the required kneepoint voltage Vkreq. Which is to be calculated as per all the Protection Relay Manufacturersrecommendations and compliance with IEC 60044-6 standard .NOTE The Rated knee point e.m.f. will be the Required knee point e.m.f. which will be in accordingwith the protection relay manufacturer recommendation .

Rated equivalent limiting secondary e.m.f. (Eal)That r.m.s. value of the equivalent secondary circuit e.m.f. of rated frequency necessary to satisfy thespecified duty cycle and derived from the following: Eal = Kssc. Ktd (Rct + Rb) Isn (V, r.m.s.)

Rated symmetrical short-circuit current factor (Kssc)Kssc = which is equal to the Required Accuracy Limit Factor (ALF req. ) . Kssc = Ipsc/IpnIpsc r.m.s. value of primary symmetrical short-circuit currentIpn Rated primary currentKtd Rated transient dimensioning factor

Current transformers according to IEC 60044-1, class P, PR :A CT according to IEC 60044-1 is specified by the secondary limiting e.m.f. E2max. The value of theE2max is approximately equal to the corresponding Eal according to IEC 60044-6. Therefore, the CTsaccording to class P and PR must have a secondary limiting e.m.f. E2max that fulfills the following:


[IEC 60044 – 1]


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Current transformers according to IEC 60044-1 class PX,IEC 60044-6 class TPS (and old British Standard, class X) :CTs according to these classes are specified approximately in the same way by a rated knee-pointe.m.f. Eknee (Ek or Vk) for class PX, EkneeBS for class X and the limiting secondary voltage Ual for TPS).The value of the Eknee is lower than the corresponding Eal according to IEC 60044-6. It is not possible togive a general relation between the Eknee and the Eal but normally the Eknee is approximately 80% of theEal. Therefore, the CTs according to class PX, X and TPS must have a rated knee-point e.m.f. Eknee thatfulfills the following:

Eknee Ek EkneeBS U al > 0.8 . ( maximum of E alreq) .

Class X or PX CTs are mostly used for high impedance circulating current protection which requireshigh Equivalent Secondary e.m.f (E) .This can be economically achieved with Class X or PX which is ofLow Leakage Reactance and can have a High magnetizing Current at Vk . Three factors will influencethe emf “E”. It’s the number of secondary turns “N”, the core area “A” and the induction in Wb/m2 “B”.

2j

Si

NAf2

EB

where:A = core area in m2B = ux density in Tesla (T)f = frequencyN2 = number of secondary turns

The induction is dependent of the core material, which influences the size of the magnetizing current.For a certain application the secondary turns and the core area are thus selected to give the requiredemf output.

C.T – AS 60044 – 1 Class PXThe rated knee point e.m.f. is generally determined as follows:

EK ( VK ) = KX . ( Rct + Rb ) . I snWhere ,

Ek = rated knee point emf .Kx = multiple of Isn at which C.T must perform satisfactorilyIsn = rated secondary current .Rb = rated resistive burden .Rct = winding dc resistance at 75º C .

CT types:Generally, there are three different types of CTs from the Construction Point of view :

• High remanence type CT• Low remanence type CT• Non remanence type CT

1) High remanence CTs – This is mostly commonly used .The high remanence type has no given limit for the remanent flux. The CT has a magnetic core withoutany air gaps and the remanent flux might remain for almost infinite time. The remanent flux can be up to70-80% of the saturation flux. Typical examples of high remanent type CTs are class P, PX, TPS,TPX according to IEC 60044 and non-gapped class C according to ANSI/IEEE.

2) Low remanence CTsThe low remanence type has a specified limit for the remanent flux. The magnetic core is provided withsmall air gaps to reduce the remanent flux to a level that does not exceed 10% of the saturation flux.Examples are class TPY according to IEC 60044-6 and class PR according to IEC 60044-1.


[IEC 60044 – 1]


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3) Non remanence CTsThe non remanence type CT has practically negligible level of remanent flux. The magnetic core hasrelatively large air gaps in order to reduce the secondary time constant of the CT (to lower the neededtransient factor) which also reduces the remanent flux topractically zero level. An example is class TPZ according to IEC 60044-6.Types of C.T’s from the Function point of view are : The output required from a current transformerdepends on the application and the type of load connected to it :

1. Measuring Current Transformers :Metering equipment or instruments, like KW, KVar, A instruments or KWh or KVArh meters, aremeasuring under normal load conditions. These metering cores require high accuracy, a low burden(output) and a low saturation voltage. They operate in the range of 5-120% of rated current according toaccuracy classes 0.2 or 0.5 (IEC) or 0.3 or 0.6 (IEEE).In each current transformer a number of different cores can be combined. Normally one or two coresare specified for metering purposes and two to four cores for protection purposes.Metering cores :To protect the instruments and meters from being damaged by high currents during faultconditions, a metering core must be saturated typically between 5 and 20 times the rated current.Normally energy meters have the lowest withstand capability, typically 5 to 20 times rated current. Therated Instrument Security Factor (FS) indicates the overcurrent as a multipleof the rated current at which the metering core will saturate. It is thus limiting the secondary current toFS times the rated current. The safety of the metering equipment is greatest when the value of FS issmall. Typical FS factors are 5 or 10. It is a maximum value and only valid at rated burden. A higheroutput from a core will also result in a bigger and more expensive core, especially for cores with highaccuracy (class 0.2).Class 0.2 for Cores with billing values metering or Class 0.5 or 1 for measuring/instrumentation without billing or Class 3 or 5 for no accurate measuring .Example : in 0.2 SFS10 CT,0.2 : Accuracy Class % current ratio error at % of rated currentS : stands for sec. It means that the C.T shall withstand for 0.2 sec a Current equal to 20 times I max.with a relative tolerance of 0% - 10 % .10 : Rated Burden .Accuracy Class 0.2 : has 0.2 % current error at 100 % - 120 % of the Rated Current .Accuracy Class 0.2S : means 0.2 % current error at 20 % - 120 % of the Rated Current .Same for Class 0.5 and 0.5S . This means the Class with S is more accurate .

2. Protection Current Transformers :For protection relays and disturbance recorders information about a primary disturbance must betransferred to the secondary side. Measurement at fault conditions in the overcurrent range requireslower accuracy, but a high capability to transform high fault currents to allow protection relays tomeasure and disconnect the fault. Typical relay classes are 5P, 10P or TP (IEC) or C 100-800 (IEEE).Protection Cores :5P/10P are commonly used for Over Current/ Earth Fault, and Class PX CT Class PX is the definitionin IEC 60044-1 formerly covered by class X of BS 3938 type CTs are used for high impedancecirculating current protection and are also suitable for most other protection schemes.We will elaborate only on the most commonly used Classes P & PX . For further details for the choiceof C.T class for transient performance TPS , TPX , TPY & TPZ , please refer to the IEC 60044 – 6 .Class PX protective current transformer :A transformer of low leakage reactance for which knowledge of the transformer secondary excitationcharacteristic, secondary winding resistance, secondary burden resistance and turns ratio is sufficientto assess its performance in relation to the protective relay system with which it is to be used.


[IEC 60044 – 1]


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Class P current transformersAre defined so that, at rated frequency and with rated burden connected, the current error, phasedisplacement and composite error shall not exceed the values given in the table below :

Accuracy Class Current Error at Rated Primary Current Composite Error at Rated Accuracy Limit Primary Current

5P ±1% 5%10P ±3% 10%

The primary difference is that the measuring current transformer(FS) is required to retain a specifiedaccuracy over the normal range of load currents, whereas the protective current transformer (5P , PX )must be capable of providing an adequate output over a wide range of fault conditions, from a fractionof full load to many times full load.CT – Measuring vs Protection :

Measuring CT Protection CT Under over current, the CT In case of over current, the

should limit the secondary CT should maintain the current to avoid thermal secondary current within overload of the connected the rated limit to enable equipment proper protection operation

At rated burden (cos = 0.8) At rated burden (cos = 0.8) and rated security factor, the and rated accuracy limit factor composite error should be>10% the composite error should be < 5% or 10%

Current transformers according to ANSI/IEEE

Current transformers according to ANSI/IEEE are partly specified in different ways. A rated secondaryterminal voltage UANSI is specified for a CT of class C. UANSI is the secondary terminal voltage the CTwill deliver to a standard burden at 20 times rated secondary current without exceeding 10 % ratiocorrection. There are a number of standardized UANSI values e.g. UANSI is 400 V for a C400 CT. Acorresponding rated equivalent limiting secondary e.m.f. EalANSI can be estimated as follows:

bANSIsnCTsnANSICTsnalANSI ZIRIURIE 202020where

ZbANSI : The impedance (i.e. complex quantity) of the standard ANSI burden for the specific C class ( ).

UANSI : The rated secondary terminal voltage for the specific C class (V).

The C.Ts according to class C must have a calculated rated equivalent limiting secondary e.m.f. EalANSIthat fulfills the following:

alreqalANSI EofMaximumE

A C.T according to ANSI/IEEE is also specified by the knee-point voltage UkneeANSIthat is graphically defined from an excitation curve. The knee-point voltage UkneeANSI normally has alower value than the knee-point e.m.f. according to IEC and BS. UkneeANSI can approximately beestimated to 75 % of the corresponding Eal according to IEC 60044-6. Therefore, the CTs according toANSI/IEEE must have a knee-point voltage UkneeANSI that fulfills the following:


[IEC 60044 – 1]


6

)(75.0 alreqalANSI EofMaximumE

ANSI/IEEE CTs as specified in the IEEE C57.13 standard. The applicable class for protection is class"C", which specifies a non air-gapped core. The CT design is identical to IEC class P but therating is specified differently. The IEEE C class standard voltage rating required will be lowerthan an IEC knee-point voltage. This is because the IEEE voltage rating is defined in terms of usefuloutput voltage at the terminals of the CT, whereas the IEC knee-point voltage includes thevoltage drop across the internal resistance of the CT secondary winding added to the useful output..Where IEEE standards are used to specify CTs, the C class voltage rating can be checked to determinethe equivalent knee-point voltage (Vk) according to IEC. The equivalence formula is:

CTnSSCk RIKCV 05.1

CTRC 10005.1Note: IEEE C.Ts are always 5A secondary rated, i.e. In =5A, and are defined with an accuracy limitfactor of 20, i.e. Kssc =20.

C.T Dimensioning (Sizing Calculations) :

Introduction :

In general , there is no single rule/equation to apply for the C.T sizing calculations specially forthe calculation of the Knee Point Voltage ( Vk ) for Feeder Protection , Transformer differential& Restricted E/F protection .Since every protection relays manufacturer has his ownrecommendation and C.T requirements which shall be met and must be submitted by thecontractors and must be followed .

Some clients require the calculations to be based on the corrected VA and not the rated VAof the C.T where a margin of 25 % shall be taken into account for rated VA calculations for5P class cores for the Rated Knee Point Voltage and Operating ALF conversions as per thefollowing formulas which will give accurate dimensioning which we will follow in ourexamples :

3.1

.secRe.int

ALFIsWdgVACorrectedVoltagePoKneeRated

The Operating Accuracy Limit Factor KOALF :

)Re.(

)Re.(

sWdgVAConnected

sWdgVACorrectedALFRATEDKOALF

Also , some clients specify two Main Protections from different manufacturers for each typeof protection on the 400 KV and 132 KV Substations . In our calculated examples , we willelaborate on the different protection relays which are most commonly used for each circuiton the 132 KV system (cable feeder , OHL , Transformer & Bus coupler / Section ) whichwill cover both ( One Protection or Two Protection schemes ).

The CT saturation is directly affected by the voltage at the CT secondary terminals. Thisvoltage is developed in a loop containing the conductors and the relay burden. For three-phase faults, the neutral current is zero, and only the phase conductor and relay phase


[IEC 60044 – 1]


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burden have to be considered. For earth faults in solidly earthed systems it is important toconsider the loop containing both the phase and the neutral conductors.

In most cases the CT requirements are based on the maximum fault current for faults indifferent positions. Maximum fault current will occur for three-phase faults or in some casesfor single-phase-to-earth faults. The current for a single phase-to-earth fault will exceed thecurrent for a three-phase fault when the zero sequence impedance in the total fault loop isless than the positive sequence impedance. This can be the case in solidly earthedsystems and therefore both fault types have to be considered.

If the fault current contains a DC-component there is a considerable risk that the CTs willsaturate. In most cases the CTs will have some remanence, which can increase thesaturation rate of the CTs. If a CT has been saturated the secondary current will notrecover until the DC-component in the primary fault current has subsided. This can cause adelay in the relay operation. Depending on the design of the relay and to consider a certainamount of DC-component, a factor has been used in the Rated Knee Point Voltagecalculation Equation . This safety factor gives a satisfactory operation.

The following sample calculations will assist you in quick reviewing and approximateverification of the different contractors submittals as a minimum requirement to be met forthe C.T sizing calculations , the calculations will be based on the most commonly used132/11 KV substations with 2000 A bus bar , 40 KA which is applicable for all Voltagelevels and both the Outdoor AIS Switchgear or Indoor GIS where the different data and C.Tratio if any to be considered , 4 mm² as a lead conductors will be used in the calculations ,types of the protection relays used in the calculations are only examples for different C.Trequirements as per the different Manufacturer recommendations .

A. 132 KV Cable Feeder :

CoreDesign

CoreNo.

Ratio BurdenVA

AccuracyClass

Vkn (V) RCT ( ) Io (mA)

T1L 1 1500-1000/1A - PX 1950/1300 5.5/3.67 50/100 atVkn

T1L 2 1500-1000/1A - PX 1950/1300 5.5/3.67 50/100 atVkn

T1L 3 1500-1000/1A 22.5/15 5P20/0.5 - 2/1.33 -T2L 1 3000/1A 30 5P20 - 8 -T2L 2 3000/1A - PX 650 8.5 15 at VknT2L 3 3000/1A - PX 650 8.5 15 at Vkn

Notes : 1. Rated Continuous Thermal Current is 150% In2. Rated Short Thermal Current (1 sec ) is 40kA for all cores

1. Measuring ( Metering ) Cores :

The following should be verified : In general and quick verification :

Rated Burden (VA) > Total Connected Burden.> 25 % of the Rated Burden .

Total Connected Burden (VA ) = P ammeters + P wires .. etc.

Mostly included in the BCU ( Bay Control Unit ) and do not have separate core .


[IEC 60044 – 1]


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2. Over Current & Earth Fault Protection : Mostly for O/C & E/F protection , class 5P C.T iswidely used and the verification of the Operating Accuracy Limiting Factor (KOALF ) is to beequal or more than the required ALF is enough to verify if the proposed C.T is adequate ornot.

ALF op. ALF req.

For cables of 4 mm2 used in secondary circuits:

Lead resistance 4sqmm at 20 C: RL20=0.00461 /m

Temperature Coefficient: =0.00393 1/K

Lead Resistance 4sqmm at 75 C:

RL75= RL20 * (1+ *(75-20))= 0.00461*(1+0.00393*55)

= 0.00561 /m

The burdens of each protection relay and meter are considered as per relevantmanufacturer’s catalogues

Burden (VA)

Bay Control Unit , REF545-ABB

(Synchro Check Function)

0.1

2 × 100m Cable of 4 mm² 1.122

Digital Fault Recorder 0.1

Total 1.322

Verification of the Accuracy Limit Factor :

Required ALF ( Kssc) = Ipsc/Ipn

6.26150040000Re ALFquired For 1500/1 A CT ratio

40100040000Re ALFquired For 1000/1 A CT ratio

The Operating Accuracy Limit Factor (KOALF )

)Re.(

)Re.(

sWdgVAConnected

sWdgVACorrectedALFRATEDKOALF

VAVARated

VACorrected 1825.1

5.2225.1

, and


[IEC 60044 – 1]


9

VA1225.1

15

6.264.1202322.1

21820 For CT ratio 1500/1 A

4052.10033.1322.1

33.11220 For CT ratio 1000/1 A

Since , the operating accuracy limiting factor KOALF=120.4/100.52 being higher than the requiredaccuracy limiting factor 26.6/40 for CT ratios 1500/1000/1 A , the proposed 5P20, 22.5/15VA currenttransformer is adequate

The O/C & E/F protection functions are mostly incorporated in the Bay Control Unit ( BCU ) or the PilotWire Differential Protection along with the Breaker Failure function . However , the sameprocedure to be followed for the loose relays if used. If the proposed BCU incorporate other protectionfunctions such as Breaker Failure where as per the manufacturer recommendation that the CTs musthave a rated equivalent secondary e.m.f. Eal that is larger than or equal to the required secondary e.m.f.Ealreq which must be verified also .

3. Bus Bar Protection :

This is a form of bus bar protection using either unbiased high input impedance relays or biased lowimpedance relays. Differential protection uses the principal of Kirchhoff’s first law to sum the currentfrom the CTs covering the protected zone. The relay detects the summation current and will trip theassociated breakers if it indicates a fault within the zone.It shall be verified with all circuits connected to the Bus Bar Protection .

1. Bus Bar Protection - Low Impedance :

Protection must have a low burden to enable it to be installed in series with other equipment oncommon secondary cores of the current transformers. It has to allow the use of different class and typeof CTs made by different manufacturers. In particular, it will have to be able to accept mixes of plant,satisfying different standards (e.g. British Standard 3938: Class X, IEC 185: Class 5P20, IEC 44-6Class TPX, TPY or TPZ).The protection has to be stable for all types of external faults and in particularunder CT saturation conditions. This saturation has to be able to be detected in less than 2 ms. Thedifferential bus bar protection has to implement also an effective protection against circuit breakerfailures.

Example for Low Impedance with SIEMENS Relay Type 7SS52 C.T Data :

CT ratio: 3000/1A VA: 30

Class: 5P20 RCT 8

Burden ( VA )

BBP relay Type 7SS52 , SIEMENS 0.1

2×100m cable of 4 mm² 1.122

Total 1.222 < 30 VA


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10

Verification of the Accuracy Limit Factor :

As per Manufacturer C.T requirement :

CT requirement for Low Impedance Busbar Protection Type 7SS52-SIEMENS(M2BZP)

pn

faultextSSCtdSSC I

IKK ).max( and SSCK 100(measuring range)

bCT

bCTSSCSSC RR

RRKK

Where

SSCK Rated Symmetrical Short Circuit Current Factor =20

SSCK Effective Symmetrical Short Circuit Current Factor

Rb Rated Resistive Burden (30 )

RCT Secondary Winding Resistance (8 )

RL CT Secondary Loop Resistance =1.8

For Loop Resistance ,2×100 m cable of 2.5 sqmm is considered

RR Relay Burden (0.1 )

bR Connected Resistive Burden

RLead + RRelay =1.8+0.1=1.9

pnI CT Rated Primary Current

tdK Transient Dimensioning Factor =0.5

.max( faultextSSCI Max. short circuit current(40kA)

Vk Knee Point Voltage

)(667.63000400005.0 requiredKSSC

And

)(effectiveRRRRKK

bCT

bCTSSCSSC

667.64.82222.1830820SSCK

As the effective symmetrical short circuit current factor of proposed CT is above the requiredeffective symmetrical short circuit current factor .

Hence the proposed CT is adequate


[IEC 60044 – 1]


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Further , as per manufacture catalogue Vk for relay type 7SS (measuring ) is calculated as follows:

snCTk IRRV3.1

100

VVk 4.7091222.183.1

100

3.1

Re.int

ALFIsWdgVACorrectedVoltagePoKneeRated

sn

Corrected VA = 30/1.25= 24VA

V3.4923.1

201824

Therefore , Rated knee point voltage < Required Vk

Therefore , C.T is adequate and suitably dimensioned

Other Example for Low Impedance with ABB Relay Type REB 670

CT Ratio 3000/1AClass of Accuracy 5P20CT Resistance RCT > 15Burden 50 VACorrected Burden 40 VA

Burden of REB670 SR = 0.02 VABurden of CT Leads * = 1.12 VA

Total Burden on the Core (PL) =1.14* CT Lead Resistance for 100 m with a 4 mm2 cable

As per the relay manual , the CT requirement are as follows :

calkavailablekVV

2max5.0R

RLCT

pn

snt

calk ISRRI

IIV


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Calculation for EalreqMax primary fundamentalfault Current

= Itmax =40000A

CT primary Current =Ipn =3000ACT secondary Current =Isn =1ARelay Current =IR =1ACT Resistance =RCT =15.00 CT Leads Resistance =RL =1.12 Relay Burden =SR =0.02VA

2max5.0R

RLCT

pn

snt

calk ISRRI

IIV

2102.01212903.1153000

1400005.0

=108 v

As per the relay manual , since the CT requirement are mentioned for BS Class PX class . Hencefinal class 5P CT parameters are CV

The design values according to BS can be approximately transferred in to IEC-60044 std usingthe following formula:

3.1

.Re.int

ratedALFIsWdgVACorrectedVoltagePoKneeRated

sn

Corrected VA = 50 /1.25 = 40 VA .

Rated Knee Point Voltage = ( 40 +15 ) × 1× 20 /1.3 = 846 V .

846V > 108 V.

Rated Knee Point Voltage ( available Vk) > Vkcal

Therefore , C.T is suitably dimensioned .

2. Bus Bar Protection - High Impedance :

High-impedance protection responds to an equivalent voltage across the relay. During external faults(with severe saturation of some of the CTs) the voltage does not rise above certain level. This isbecause the other CTs will provide a lower-impedance path as compared with the high relay inputimpedance.Traditionally high impedance schemes have provided greater stability during through faults. This type ofprotection requires all CTs must have the same transformation ratio, are class x and dedicated to theprotection scheme.

The current transformers used in high impedance circulating current differential protection systemsmust be equal turns ratio and have reasonably low secondary winding resistance. Current transformers


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of similar magnetizing characteristic with low reactance construction such as IEC60044 Class PX, orsimilar, are preferred.

The relay requirements are based upon a calculation of the required knee-point voltage with the IECdefinition of the knee-point voltage .

Example with (VA TECH – Reyrolle Siemens )DAD – N , 7SG12 High Impedance Relay :

Typically applied to provide 3 – phase high impedance differential protection of busbar,connections, auto-transformers, reactors and motors. The stability of the high impedance schemedepends upon the operate voltage setting being greater than the maximum voltage whichcan appear across the relay under a given through fault condition. An external series stabilizingresistor and shunt non-linear resistor per phase complete the scheme. The series resistor valueis determined by the voltage level required for stability and the value of relay current calculatedto provide the required primary fault setting. Non-linear resistors protect the relay circuit from highover-voltages.

C.T Data :

CT Ratio 3000/1AClass of Accuracy PXCT Resistance RCT< 15 Knee Point Voltage Vk > 1500 VMagnetizing Current Io< 40 mA at Vk /2Burden of DAD-N = 0.10 VABurden of CT Leads =1.12 VA

Total Burden of the Core (PL) =1.22 VA

CT requirement for 7SG12 relay (high impedance application)

All CTs must have the same transformation ratio . To prevent maloperation of the relay duringsaturation of the CTs on an external fault, the actual stability voltage Us must be higher than thevoltage (Ustab ) produced by the maximum secondary through fault current , flowing

stabs UU

)()/(.max. wireCTpnsnthrks RRIIIU

In addition to this, the knee point voltage must be higher than twice the actual stability voltage:

sKnee UU 2Where:

Us Relay Setting VoltageUstab Minimum Stability VoltageUknee Knee Point Voltage of CTIk,max,thr Max. symmetrical short-circuit current for external fC =40000AIscc,max,thr Max. symmetrical short-circuit current for internal fa’ 40000ARwire Cable Burden =1.12 RCT Max. internal burden of CT at 75 =15 Ipn CT primary nominal current =3000AIsn CT secondary nominal current =1A


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Rb Relay Burden =0.10 If Secondary Fault Current =13.33A

stabU V215)1212903.115()3000/1(40000

Since

stabs UU is considered as

Us= 225 V

Therefore

sKnee UU 22252V450

Criteria is

Proposed Vk Required Vk

1500 450

Hence

CT is : Suitably dimensioned

Calculation of maximum sensitivity

)( min. essn

pnP InII

II

Where

min.sI Minimum relay setting =0.005A

n Number of CTs in parallel with relayincluding future bays

=30

eI Mag. Current at relay setting voltage(225.00)V

=0.0120A (Assumed)

PI)( min. es

sn

pn InIII

)012.030005.0(13000

= 1095

=36.5 % of CT rated current

Fault Setting

The primary operating current of busbar protection is normally set to less than 30% of theminimum expected fault current. Unless otherwise specified. Further as per specificationrequirement, the setting shall be above 125% of nominal full load current of transformer.


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Power Transformer Rating = 50 MVAPower Transformer Percentage Impedance (%Z) =30.5 %Rated Voltage = 132 kVTrafo. Full Load Current

VoltageMVA

3

132732.1100050

=218.69 A

Effective Sensitivity Calculation :As per ESI 48-3 , the fault setting shall be between 10% and 30% of minimum fault currentavailable .

DEWA ( Dubai Electricity & Water Authority ) specify the minimum and maximum primary operatingcurrent as follows which will be followed in the calculations :

IP min= 1100 AIP max= 1400 A

For a desired increased sensitivity of 2000 A primary setting . The relay setting shall be

knee

Se

pn

sndespsets U

UInIIII

(min)

1500225012.0303000

11100(min)sets

I

= 0.310 A

knee

Se

pn

sndespsets U

UInIIII

(max)

1500225012.0303000

11400(max)sets

I

= 0.410 A

Calculation of Stabilizing Resistor

The proper value of stabilizing resistor Rstab is required to ensure stability and is calculated byusing the formula

bsets

sstab

RIUR

(max)


[IEC 60044 – 1]


16

1.041.0225

= 726

Minimum Power Rating Pstab

stabsetsstab RIP 2

72641.0 2

=122 WVoltage developed across resistor during internal fault

3.1413

fstabkf IRVV

3.134.137261500 413

3108 VHalf second rating of stabilizing resistor Phalf =

kWW

RV

stab

1313306

726/3108 2

21

Hence stabilizing resistor of 0-5000 variable , having a power rating of 350 W and shorttime rating of 15kW will be selected

Calculation for max. voltage at relay terminal

The theoretical Voltage which may occur at the relay terminals is

stabwireCTpn

snSCCSCC

RRRIIIU

intmaxintmax

7261212903.1153000140000

=9895 V

kneeSCC

kneerelay

UUUUintmaxmax

22

15009895150022max relay

U

= 10037 V


[IEC 60044 – 1]


17

Metrosil is required if

V

VUrelay

150010037

1500max

Hence Metrosil is required

The type of metrosil required is chosen by its thermal rating as defined by the formula

kf VIP 14.3/4150034.1314.3/4

SJ /71.25477

SkJ /48.25

Select a metrosil with C=900 , = 0.2Another example with another Type which is also widely used AREVA - MCAG34 High ImpedanceProtection Relay for 11 KV Busbar Protection .

11 kV Busbar Protection Discriminating and check Zone

The relay is MCAG34,AREVACT Ratio : 3200/1AClass of Accuracy :CL.X

a) Stability Voltage (VS)As per clause 7.3.1 and fig 18 indicated in ESI :48:3 the following formulas could be applied :a.1) Consider and external phase –to – earth short circuit and assume complete saturation of acurrent transformer , then VS shall be not less than :

TGDAIV Fs )2(

Where :

IF = Fault Current Corresponding to the rated stability limit (100 percentage of switchgear short circuitrating as per clause 5.6 of ESI 48-3)=25kA=Resistance of Wiring plus current transformer winding= RCT+RL=10+0.9=10.9RL=0.9 , L=100 m, A=2.5 mm2 , r= 9 /km

Taking into account the above data , VS is calculated as follows

VVs 19.929.09.1032002500

VVs 19.92

a.2) Also Consider a phase –to –phase or 3 phase external short circuit and assume completesaturation of a current transformer , then VS shall not be less than :


[IEC 60044 – 1]


18

)()( VTGAIV Fs

VVs 16.859.1032002500

VVs 16.85

b) Knee Point Voltage (Vk)

As per ESI 48-3 , the minimum current transformer knee point voltage is :

VVV skn

2009522min

VVkn

200min

c) Magnetizing Current (Ie)

As per ESI 48-3 , the fault setting shall be between 10% and 30% of minimum fault current available

As per DEWA requirement 2kA shall be considered as minimum fault current at 11kV bus

The fault setting shall be between 200A and 600 A

Effective primary fault settings

TIInIII msrtsfs

1

Is = Selected Relay Setting Current =0.05A

n = Max. number of CTs in parallel with relay for check Zone

= Max. number of CTs in parallel with relay for Disc. Zone

=66

=24

I1 Magnetizing Current of CT

Maximum Magnetizing Current

(Ien & Iet at Vs =100 V is considered for 3200/1A from the vendor documents)

= 1.8mA at Vs (Average)

=66 x 1.8=118.8 mA

Rs = Impedance of busbar supervision relay

= 900004

60022

BU s

= 90000

Considering the worst case as Rs= 90000

=90000

U = Rated voltage of relay = 600 V (AREVA Catalogue) = 600 V (AREVA Catalogue)

B = Relay Burden =4VA (AREVA catalogue)

Taking into account the above data , Ifs is calculated as follows


[IEC 60044 – 1]


19

For check zone

AI fs 3.5533200003.0900001000018.06605.0

Which is approximately 27.6% , of the min , fault current at the 11kV Bus .

However based on the data provided by manufacturer for similar type of CTs it can beassumed that average current is approximately 1mA, reconsidering the above formula

AI fs 4.3843200003.090000100001.06605.0

Which is approximately 19.2% , of the min , fault current at the 11kV Bus

For Disc. zone

AI fs 4.3113200003.0900001000018.02405.0

Which is approximately 15.56 % , of the min , fault current at the 11kV Bus .

However based on the data provided by manufacturer for similar type of CTs it can beassumed that average current is approximately 1mA, reconsidering the above formula

AI fs 95.2493200003.090000100001.02405.0

Which is approximately 12.5 % , of the min , fault current at the 11kV

Since the fault setting is between 10% and 30% of the minimum fault current

(as per ESI 48-3 and tender documents) and DEWA’s requirement is fulfilled,

the selected magnetizing current 1.5 mA at Vk/2 is suitable for the system protection.

Therefore, the relay current setting will be provided with range (5%-20%) in sevenequal steps, which is confirmed by above calculation as adequate.

d) Stabilizing Resistor

To assure stability for through faults a stabilizing resistor will be required . The value of seriesresistance is calculated as follows :

2rr

sS I

VAIVR

Where : RS Value of the stabilizing resistor

VS Minimum required stability voltage i.e. setting voltage

Ir Relay setting current is negligible

VA 1VA AREVA Catalogue for MCAG 34


[IEC 60044 – 1]


20

Assumed current setting is for check zone and disc. Zone

AIr 05.0

200005.0

100SR

And it is 74.1% of 2000 , Hence selected is 0-2700 (variable)

However the actual VS based on the site tests will be less than 100V. Accordingly the set value will beless than 2000 . Hence selected is 0-2700 (variable).

The resistors incorporate in the scheme must be capable of withstanding the associated thermalconditions

The continuous Power Rating of a resistor is defined asPcon RIcon

2

Where

Pcon = Resistor Continuous Power Rating

Icon =Continuous Resistor Current Power Rating

R =Resistance

For Check

Pcon 20003200

3.553 2

= 59.8 Watts

For Disc

Pcon 20003200

4.311 2

=18.9 Watts

The voltage developed across a resistor for a maximum internal fault conditionis defined as

Vf 3.12 3fsk IRV

Watts7.3442000

3.830 2


[IEC 60044 – 1]


21

Where

Vf = rms voltage across resistor

Ifs = max. fault current 25kA

Secondary current is 813.7320025000

Vf =

V3.8303.1813.720002202 3

Phalf Thus the half second power rating is given by :

RV

P fhalf

2

Watts

RV f

7.3442000

3.830 2

2

As per the manufacturer catalogue, short time rating of stabilizing resistor used in MCAG34 AREVArelay is : ZB9016783

e) Requirement of Metrosil

The maximum voltage in the absence of CT saturation is :

laySLCTFf RRRRTIV Re2

V15725120009.0210320025000

Peak to peak voltage developed across the relay is:

kfkP VVVV 22

Where

Vk=200 V


[IEC 60044 – 1]


22

2201572522022PV

= 5223.8 V

Since the peak voltage developed across the relay is more than 3kV , metrosils are required . Selectedmetrosil is three phase 6”,600A/S3/S802,C=450, =0.25

Comparison of High and Low-Impedance Bus bar Protection :

Nowadays high impedance protection is still widely used, because it is considered “cheap and

easy”. But most users only look at the relay price itself, without considering the other disadvantages ofa high impedance schemes:

All CTs must have the same ratio

Class X for all CT cores

Bus sectionalizers with circuit-breaker must be equipped with two CTs

Separate CT cores for busbar protection

Advantages of numerical protection technology(e.g. fault recording, communication, etc.) notavailable

Check zone needs separate CT cores

Isolator replica requires switching of CT secondaries , additional check zone obligatory.

3. Pilot Wire Current Differential Protection :

CHARGING CURRENT COMPENSATION

The basic premise for the operation of differential protection schemes in general is that the sum of thecurrents entering the protected zone is zero. In the case of a power system transmission line, that maynot be entirely true because of the capacitive charging current of the line. For short overheadtransmission lines, the charging current can be treated as a small unknown error. In that case, the errordue to the line charging current is covered by the percentage restraint characteristic of the currentdifferential scheme. For long transmission lines and cables, the charging current may be too large totreat as an unknown error. In that case, it is often necessary to desensitize the current differentialprotection to prevent mis-operations due to the line charging current which shall be considered in theRelay setting :

The sensitive differential set point I-DIFF> is calculated according the charge current of the cable orset to a minimum value, which results from the CT´s transient behavior. The charge current caused


[IEC 60044 – 1]


23

by the capacitance of the line/cable is a permanent differential current during normal operation. I-DIFF>should be set 2,5-3 times of this steady state charge current or as per the relay manufacturerrecommendation . The charge current is calculated as follows:

11063.3 6OPNNC CfUI

IC Primary charge current in A

U N Nominal voltage of the line/cable in kV

f N Rated frequency in Hz

COp’ Capacitance of the line or cable in nF/km (typ. line 8 nF/km, cable 250 nF/km)

l Length of the line/cable in km

Example : Pilot Wire Current Differential Protection with SIEMENS Relay Type 7SD522 :

CT Ratio: 1500-1000/1A Vk 1950/1300V Ie 50/100mA at Vk

Class : PX RCT 5.5/3.67

To ensure correct operation of the connected relay , CT to remain stable under all through faultconditions and the rated knee point voltage should be greater than the calculated knee point voltage

Since, effective CT accuracy limiting factor,

30max,OALF

N

SSCOALF Kand

II

K

40100040000,67.26

150040000

AA

AAKOALF

The knee point voltage as per the revised formula given by DEWA :

SNbCTPN

faultExtSSC

tdk IRRI

I

KV).max(

Where Ktd = 3.75 Transient Dimensioning Factor

(Considering 75% remanance in the CT core as suggested by DEWA)

ISSC,max(ext.fault current) =40000A

RCT CT internal Resistance

R’b Connected Resistance (RB+2RL)

RL One way lead resistance from CT to relay =0.9

(for Lead resistance ,100 m cable of 2.5 sqmm is considered )

UKN Rated Knee Point Voltage


[IEC 60044 – 1]


24

ISN CT secondary current 1A

IPN CT primary current 1500/1000A

Using the above data VK can be calculated :

VU KN 7351500

18.105.05.54000075.3

VU KN 8281000

18.105.067.34000075.3

Hence the calculated knee point voltages for ratios 1500/1000/1 A are satisfied as their values are lessthan the rated CT knee point voltages , Vk=1950/1300V > UKN=735/828 V respectively

Hence the proposed CT cores for ratios 1500/1000/1A are acceptable

There are no specific requirements on magnetizing current

CT Requirements for Main-2 Backup distance protection:

The CT requirement to ensure correct operation of the distance protection relay

RLCTPN

faultincloseSSC

atdk RRRI

IKV

3.1

)..max(

)(

RLCTPN

faultendzoneSSC

btdk RRRI

IKV

3.1

)..1

)(

Where

Ktd(a) = 4 (for TS< 200ms)- Transient Dimensioning Factor

For close – in faults

Ktd(b) = 5 (for TS< 200ms)- Transient Dimensioning Factor

For zone-1 end fault

RCT Secondary winding Resistance for each star connected CT


[IEC 60044 – 1]


25


(For loop resistance ,2X100 m cable of 2.5 sqmm is considered )



ISSC,max(close in fault ) Max short circuit current for faults close to the relay

ISSC,max(zone-1 end fault ) Max short circuit current for faults at zone-1 reach

For Close – in faults :

RLCTPN

faultincloseSSC

atdk RRRI

IKV

3.1

..

)(

For CT ratio 1500/1A

05.08.15.515003.1

400004kV

= 603.07 V< 1950 V


05.08.167.310003.1

400004kV

= 679.38V< 1300 V

For Zone 1 end fault :

RLCTPN

faultendzoneSSC

btdk RRRI

IKV

3.1

)..1

)(


[IEC 60044 – 1]


26


05.08.15.515003.1

400005kV

= 753.85 V< 1950 V


05.08.167.310003.1

400005kV

= 849.23V< 1300 V

Hence the calculated knee point voltages for ratios 1500/1000/1A are satisfied as their valuesare less than the rated CT knee point voltages


There are no specific requirements on magnetizing currents

Remote End C.T Dimensioning :

For proper protection scheme operation , the Protection Relay at the remote end shall be the sameSIEMENS Type 7SD522 , same all above C.T requirement equations and procedure shall be followed :

C.T Data :

C.T Ratio Vk Class Rct at 75º C RL at 75º C Rb

1500-1000-500/1 1950 – 1300 – 650 V X 4.284 – 2.783 – 1.365 0.75 0.05

RL is calculated as 0.00375 ohm/meter for (A = 6 mm²X100 meter length = 0.00375x100x2 = 0.75

To ensure correct operation of the connected relay , CT to remain stable under all through faultconditions and the rated knee point voltage should be greater than the calculated knee point voltage

Since, effective CT accuracy limiting factor,

30max,OALF

N

SSCOALF Kand

II

K

40100040000,67.26

150040000

AA

AAKOALF

500 /1 ratio result can be ignored . Since , no 500/1 tap at the remote end and will not be used .


[IEC 60044 – 1]


27

SNbCTPN

faultExtSSC

tdk IRRI

I

KV).max(

Where Ktd = 3.75 Transient Dimensioning Factor

(Considering 75% remanance in the CT core as suggested by DEWA)

ISSC,max(ext.fault current) =40000A

RCT CT internal Resistance

R’b Connected Resistance (RB+2RL)

RL One way lead resistance from CT to relay =0.75

(for Lead resistance ,100 m cable of 2.5 sqmm is considered )

UKN Rated Knee Point Voltage

ISN CT secondary current 1A


VU KN 4.5081500

175.005.0284.44000075.3

VU KN 45.5371000

175.005.0783.24000075.3

CT Requirements for Main-2 Backup distance protection:

The CT requirement to ensure correct operation of the distance protection relay

RLCTPN

faultincloseSSC

atdk RRRI

IKV

3.1

)..max(

)(

RLCTPN

faultendzoneSSC

btdk RRRI

IKV

3.1

)..1

)(

Where

Ktd(a) = 4 (for TS< 200ms)- Transient Dimensioning Factor

For close – in faults


[IEC 60044 – 1]


28

Ktd(b) = 5 (for TS< 200ms)- Transient Dimensioning Factor

For zone-1 end fault

RCT Secondary winding Resistance for each star connected CT


(For loop resistance ,2X100 m cable of 2.5 sqmm is considered )



ISSC,max(close in fault ) Max short circuit current for faults close to the relay

ISSC,max(zone-1 end fault ) Max short circuit current for faults at zone-1 reach

For Close – in faults :

RLCTPN

faultincloseSSC

atdk RRRI

IKV

3.1

..

)(


05.075.0284.415003.1

400004kV

= 417.15 V< 1950 V Acceptable .


05.075.0783.210003.1

400004kV

= 383.3V< 1300 V Acceptable .

For Zone 1 end fault :

RLCTPN

faultendzoneSSC

btdk RRRI

IKV

3.1

)..1

)(



[IEC 60044 – 1]


29

05.075.0284.415003.1

400005kV

= 521.4 V< 1950 V Acceptable .


05.075.0783.210003.1

400005kV

= 551.2 V< 1300 V Acceptable .


The 500 /1 ratio can be ignored . Since , no 500/1 tap at the remote end and will not be used .

Notes :

40000 A is used for both Issc.max ( close - in fault ) and Issc.max ( zone1 - end fault ) which will give higher Vk andbetter safety margin . Since , mostly 40000 A (Issc.max = 40 KA – Switchgear S/C level ) is >Issc.max ( zone1 - end fault ) and ( Earth Fault Current for Zone – 1 end fault ) . For precisecalculation of Issc.max ( zone1 - end fault ) and and Ts (tp) , an example with ABB relay type RED670 is detailed below for information :

Other Example for the Pilot Wire Differential Protection with ABB relay Type RED 670 :


[IEC 60044 – 1]


30


[IEC 60044 – 1]


31


[IEC 60044 – 1]


32


[IEC 60044 – 1]


33

Same must be repeated for the 1000 / 1 A ratio considering the Rct for the 1000/1 A ratio :


[IEC 60044 – 1]


34

Vk requirement equations for other types of relays which are also widely used are asfollows :


[IEC 60044 – 1]


35

a) Relay Type SEL 311 L :

Same as above if IF = 40000 A is used, it will give better safety . Other wise , the actual Zs and ZL shallbe calculated .


[IEC 60044 – 1]


36

b) AREVA – MICOM Relay Type P543 :

B. 132 KV OHL Feeder:

Example : Length : 30 km , 132 KV , 40 KA OHL with ABB relays :

CT data :


[IEC 60044 – 1]


37

1. O/C & E/F Protection :

Same as the Cable feeder . the CT core must fulfill the following requirements:

ALF op. ALF req

ALF req. ( Kssc) = Ipsc/Ipn

If a separate BCU ( Bay Control & Protection Unit ) is used with Circuit breaker failure protection , theC.T must fulfill the Manufacturer recommendation , the following example with ABB REC 561 controland protection unit

As per ABB REC 561 manual with Circuit breaker failure protection in REC 561 , CT core must meetthe following requirement:

The CT core is connected 1600/1A and has Eknee = 300V


[IEC 60044 – 1]


38

With maximum value for Ereq = 159,3V and Eknee = 300V Ereq < Eknee

Thus , CT core is adequate and suitably dimensioned.

2. Bus Bar Protection : same as the cable feeder .

3. Distance Protection : Example with ABB distance Protection Relay Type REL 511protection

Short Circuit calculation for Zone – 1 three phase & phase to earth fault currents :


[IEC 60044 – 1]


39

Line distance protection Function in REL 511According to ABB application manual for REL 511 the CT core must fulfill tworequirements. The current transformer must have a rated equivalent secondary e.m.f. Ealthat is larger than or equal to the maximum of the required secondary e.m.f. Ealreq(formulas 1 and 2.):

Ealreq calculated values with formula 1:


[IEC 60044 – 1]


40

Ealreq calculated values with formula 2:

With maximum value for Ealreq = 476V and Eknee = 1000V Ealreq < Eknee

Core 2 data for REL 511 are thus satisfactory.

Circuit breaker failure protection Function in REL 511 :

(same requirement as REC 561)

The CT core is connected 1600/1A and has Eknee = 1000V

C. 132 KV Feeder with Reactor : For reactor detail , please refer to Annex – C :

A. Line Protection : The same as the above Feeder Protection (OHL or Cable Feeder )

B. Reactor Protection : We will elaborate on the Reactor Differential Protection , Restricted EarthFault (REF) and the Stand – by E/F protection which mostly will cover the Transformer Protection .Since , if separate back up Over – Current and E/F are used , the same previous dimensioning forthe Feeder circuit is applicable :

1. Differential Protection :

A differential relay, of high impedance type should be used as main protection. CT’s should be specifiedat both the phase and the neutral side of each phase and a three phase protection should be used as athree phase protection gives a higher sensitivity for internal faults. The general requirement on thefunction values of the high impedance differential protection is that at maximum through fault current foran external faults the relay wont mal-operate even with one CT fully saturated.

For a reactor the dimensioning criteria will be for the inrush current, since a reactor only will give athrough fault current equal to rated current, at an external earth-fault. The maximum inrush current for areactor is approximately two times the rated current. If no specific requirement concerning at whatcurrent the relay should be stable exists, five times the rated current is used when operating voltage isselected. The function value of the relay should be chosen:


[IEC 60044 – 1]


41

lctnfunction 2RR5IU

where“In” is the reactors rated current at the secondary side of the CT,

“Rct” is CT secondary resistance at 75 °C and

“RL” is the lead resistance, from the CT to the summation point.

In order to protect the protection relay from over voltages at internal faults, non-linear

resistors are connected, in parallel with the relay, at each phase .

REACTOR DATA:

Rated voltages of windings 132/145kV

Rated Current 43.7/48A

Connection symbol YN

Rating 10/12.1MVAr

Cooling ONAN

Example : The Relay proposed is AREVA Type MICOM P632

Current Transformer Data

Line CT ratio: 150/1A

Neutral CT ratio: 150/1A

The C.T must fulfill the following requirement

ALF op. ALF req

ALF req. ( Kssc) = Ipsc/Ipn

Ipsc = 10 times the rated current of the protected winding.

= AA 39.4371323

3101010

Corresponding secondary current is:

916.2150

4.437

The operating accuracy limit factor (KOALF):

KOALF=Wdg.ResAConnectedV

Wdg.ResACorrectedVRatedALF.

5.11375.002.2

75.015.20

Hence, the operating accuracy limit factor > required accuracy limit factor


[IEC 60044 – 1]


42

113.5 > 2.916

The C.T is suitably dimensioned.

Based on AREVA recommendation , the following formulas for the knee – point voltage are applied :

1. Fblctkn .IR2RRV

Where:

Rct = Resistance of the CT secondary circuit 0.75

Rl = is the resistance of the secondary winding (longest lead l=120m, A=2.5mm2, Rl=1.08 )

Rb = Relay burden = 0.1 VA

IF = 10 times the rated current of the protected winding as per clause 5.5.2(ii) of ESI

48-3

AA 38.4371323

3101010

nk

rF .U.u3

10.SI

Corresponding secondary current is (CT Ratio: 150/1A as per Tender documents):

92.2150

38.437

To be on safety side 5A is considered further in calculation.

1.155 0.12.160.75Vkn

V1.15knV

2. blctfkn R2RRI.0.25V

If = 40000A

Max. non-offset fault current for an internal fault is taken 40kA (the same as 132kV switchgear rating)which is on safety side, considering the level of max. fault current on 132kV given by DEWA.


A67.266150

40000

VVkn 67.2001.016.275.067.26625.0

67.200knV V for 150A tap

From the above calculation 1 and 2, the required knee point voltage shall be considered as follows:

VVkn 220


[IEC 60044 – 1]


43

There are no specific requirements on the magnetizing current.

2. Restricted E/F Protection of 132 KV Reactor – HV Side and Neutral CT Cores :

Example :

The relay type MICOM P122 , AREVA

CT Ratio 150/1 A

Class of Accuracy CLX

Introduction :

The MICOM P122 restricted earth fault relay is a high impedance differential scheme which balanceszero sequence current flowing in the transformer neutral against zero sequence current flowing in thetransformer phase windings. Any unbalance for in-zone fault will result in an increasing voltage on theCT secondary and thus will activate the REF protection. This scheme is very sensitive and canthen protect against low levels of fault current in resistance grounded systems where the earthingimpedance and the fault voltage limit the fault current. In addition, this scheme can be used in asolidly grounded system. It provides a more sensitive protection, even though the overalldifferential scheme provides a protection for faults over most of the windings. The high impedancedifferential technique ensures that the impedance of the circuit is of sufficiently high impedancesuch that the differential voltage that may occur under external fault conditions is lower than the voltagerequired to drive setting current through the relay. This ensures stability against external fault conditionsand then the relay will operate only for faults occurring inside the protected zone. High impedanceschemes are used in a differential configuration where one current transformer is completelysaturated and the other CTs are healthy.

Voltage across relay circuit


[IEC 60044 – 1]


44

L2RCTRfIVs

Stabilizing resistor relay circuit

RR -sIsV

STR

If = Maximum secondary through fault current

Where RR = Relay Burden

RCT = Current transformer secondary winding resistance

RL = Resistance of a single lead from the relay to the current transformer

The voltage applied across the relay is:

L 2RCTRfIsV

If : Maximum secondary external fault current

RCT : Resistance of the current transformer secondary winding

RL : Resistance of a sigle wire from the rely to the CT

A stabilizing resistor RST can be used in series with the relay circuit in order to improve the stability ofthe relay under external fault conditions. This resistor will limit the spill current under Is.

Vs = Is × (RST)

Is : Current relay setting

Vs : Stability Voltage setting

e) Stability Voltage (Vs)

As per clause clause 7.9.1 in ESI /48-3 , the following formulas can be applied :

a.1) Consider and external phase to earth short circuit and assume complete saturation of a line CT ,then Vs shall not be less than :

VTCAfIVs

Where is:

IF = Max. primary current for which stability is required

= 10 times rated current of protected winding as per clause 5.5.2 (ii) of ESI 48-3

= AA 39.4371323

3101010


[IEC 60044 – 1]


45

A, B = Resistance of winding plus CT secondary winding for line neutral CTs respectively

= RCT + RL = 0.75 + 1.08 = 1.83

RL = 1.08 , l = 120 m, A = 2.5 mm2, r = 9.0 /km

(For resistance of wiring refer also to Item 1. of this document)

C, D = Resistance of wiring

= 1.08

T = Turns ratio

=150

1

Taking into account the above data, Vs is calculated as follows:

VVs 49.8150

108.108.175.039.437

a.2) Consider an external phase to earth short circuit and assume complete saturation of the neutralcurrent transformer, then Vs shall be not less than

VTDBf

IVs

VVs 49.8150

108.108.175.039.437

a.3)Considering an external phase to phase short circuit and assume complete saturation of a line CT,then Vs shall be not less than

VTAf

IVs

VVs 34.5150

108.175.039.437

a.4)Considering an external 3 phase short circuit and assume complete saturation of a line CT, then Vsshall be not less than

VTBf

IVs

VVs 34.5150

108.175.039.437

From above calculations, Itemsa.1, a.2, a.3 and a.4 refer, the stability voltage setting should beconsidered as follows:

Hence, VVs 10 For can be considered respectively

f) Knee Point Voltage (Vk)

As per ESI 48-3, the minimum current transformer knee point voltage is:


[IEC 60044 – 1]


46

VknVVVsknV 20min,1022min

g) Calculation of the Magnetizing Current

As per DEWA’s spec. 1.5.1.5.9.01 Rev-11 clause 10.2, the fault setting shall be between 10% and20% of the minimum current available for an earth fault at the reactor terminals.

Rated current at 132kV side ASn 74.431323

100001323

The fault setting shall be between 4.374A and 8.748A

The primary operating current (Ipo) for relay setting:

T

1m

Il

3In

Is

Ipo

I

Is = Selected relay setting current as Vs=10V, = 0.02A

In = Neutral CT magnetizing current at Vs=10V, <50mA at Vk/2

3Il = Sum of line CTs magnetizing currents at Vs=10V

Im = Metrosil Current at Vs=10V

T = Turns ratio=150

1

Hence Vk > 450 V and Vs = 10 V

Im = Metrosil current = 1mA

Il & In = Line & Neutral CT Magnetizing current at Vs = < 2.5mA

IOP = [0.03 + 0.0025 + 3*0.005 + 0.001] x 150 = 6.15A

Approximately 14.1 % of the rated current of the protected winding.

Considering Relay setting current = 0.015 A

IOP = [0.01 + 0.01 + 3*0.01 + 0.001] x 150 = 7.65A.

Approximately 17.5 % of the rated current of the protected winding.

Hence, relay current setting will be from 0.002 A to 1.0 A (Adjustable range)

h) Stabilizing Resistor :

As per ESI Standard 48-3 , the required minimum knee-point voltage for CT ratio 150/1 A is VkMin. 2 Vs = 2 x 10 = 20 V. Hence the required stabilizing resistor to assure the minimum required knee-point voltage is :

The value of series resistance is calculated as follows:

Is

VsRs

The burden of the relay is a small value and it is negligible. Therefore

Rs =Relay Circuit impedance at setting

Is =Selected relay setting current at Vs=10V


[IEC 60044 – 1]


47

Assumed current setting is

Is = 0.03 A

Rs = 10/0.03 = 333.3

0r

= 10/0.01 = 1000

Selected, Rsr = 0 – 1000 (Variable)

Selected Stabilizing resistor 0 – 1000 is adequate.

The resistors incorporate in the scheme must be capable of withstanding the associated thermalconditions.

The continuous power rating of the resistor is defined as

.R2con

Icon

P

Where:

Pcon = Resistor continuous power rating

Icon = Continuous resistor current i.e operating current of the relay

R = Resistance

Pcon = (6.15/150)² x 333.3 = 0.56 Watts

Or

= ( 7.65/150)² x 1000 = 2.6 Watts.

The rms voltage developed across a resistor for a maximum internal fault condition is defined

1.34fs

.R.I3k

Vf

V

Where:Vf = rms voltage across resistor

Ifs = Maximum secondary fault current which can be calculated from the circuit breaker rating, Ich, if the maximuminternal fault current is not given. The maximum internal fault current is usually the same as the maximumthrough fault current.

Ifs = The maximum three phase through fault current

= 10 times rated current of the protected winding

Hence,

Ifs = The max. three phase through fault current.


[IEC 60044 – 1]


48

= AA 39.4371323

3101010


=916.2

1504.437

43.7093.192.23.3334504 3fV

OR

7.9333.192.210004504 3fV

Thus the half second power rating is given by:

/R2fVhalfP

= 709² / 333.3 = 1510.02 Watts

Or = 933² / 1000 = 871.8 Watts

Hence selected stabilizing resistor is adequate .

i) Requirement of Metrosil :

The maximum voltage in the absence of CT saturation :

relayRSRL2RCTRTFIfV

= 437.4/150 ( 0.75 + 2*1.08 + 333.3+0 ) = 980.4 V

OR = 437.4/150 ( 0.75 + 2*1.08 + 1000 +0 ) = 2924.5 V

The peak to peak voltage developed across relay :

kV

fV

kV22Vp

Where Vk ( Actual Vk of CT )

= 2 x 4504.9804502 = 1381.8 V

C. Transformer Feeder :

1. Differential Protection : Example using AREVA MICOM P633:

CT Data for T1L CORE-1:


[IEC 60044 – 1]


49

CT ratio: 500-300/1A VA: 25/15

Class: 5P20 Rct: 0.8/0.5

T2LN CORE-2 (TRF HV NCT):


Class: 5P20 Rct: 1.0

Verification of the accuracy limit factor:

Required limiting factor = 80500

40000 for CT ratio 500/1 A

= 33.133300


For CT cables size 4sqmm used in secondary circuits:

CT Description Burden (VA)

M1IDTP relay type MiCOM P633, AREVA

(87T1, HV 64REF1)0.1

2×100m cable of 4sqmm 1.122T1L CORE-1

TOTAL 1.222<25/15 VA




Corrected VA = rated VA/1.25 = 25/1.25 = 20VA

= 15/1.25 = 12VA

807.2058.0222.1

8.020.20 for CT ratio 500/1 A

33.13318.1455.0222.1

5.012.20 for CT ratio 300/1 A

Based on the AREVA recommendation for the relay type P633, the following formulae for the kneepoint voltage are applied:

1. For through faults:

blctthrkn R2RRIV

Where:


[IEC 60044 – 1]


50

RCT = CT secondary resistance

Rl = CT secondary lead resistance

(Lead resistance from manufacturer catalogue: 5.62 ohm/km for A=4mm2)

= 5.62×0.06=0.3372 (for lead length 60m)

Rb = Relay burden

= 0.110.1

IVA

22nom

Ithr = Max. three phase fault current

For 500/1A:

AA 5.139.1500

1132315.03

10.50 3

ratioCT1.

.U.u3SI

nk

rthr

For 300/1A:

AA 5.231.2300

1132315.03

10.50 3

ratioCT1.

.U.u3SI

nk

rthr

Taking into account the above data, required Vkn is calculated as follows:

VVkn 362.21.06744.08.05.1 (for 500/1 A)

VVkn 186.31.06744.05.05.2 (for 300/1 A)

2. For internal faults:

blctfkn R2RRI.0.25V

Where: If = Max. non-offset fault current for an internal fault

Two currents will be considered:

i) Inrush current

If = 16 times rated current of protected winding

A7500

3499500

11323

10.50.16.163

ratioCT1.

.U3SI

n

rf (for 500/1 A)

A66.113003499

3001

132310.50.16

3

(for 300/1 A)


VVkn 76.21.06744.08.0725.0 (for 500/1 A)

VVkn 82.31.06744.05.01225.0 (for 300/1 A)


[IEC 60044 – 1]


51

ii) Max. fault current

AratioCT

II FMaxf 80

500400003 (for 500/1 A)

A33.133300

40000 (for 300/1 A)


VVkn 49.311.06744.08.08025.0 (for 500/1 A)

VVkn 48.421.06744.05.033.13325.0 (for 300/1 A)


Required VVkn 70 for CT ratio 500/1 A

V100 for CT ratio 300/1 A

The rated knee point voltage =1.3

ALFRated..IsnResWdgVACorrected

=31

2018020.

.=320 V for CT ratio 500/1 A

=31

2015012.

.=192.31 V for CT ratio 300/1 A

Therefore, Rated knee point voltage > Required Vk.

Hence the proposed CT cores for ratios 500/300/1 A are adequate.

The CT requirements for low impedance REF protection are generally lower than those fordifferential protection. As the line CTs for low impedance REF protection are the same for thoseused for differential protection, the differential CT requirements cover both differential and lowimpedance REF applications.

CT Adequacy check for TRF LV side CTs

T3L (TRF LV side CT) and T3LN CORE-2(11kV NCT)



CT requirements:

Based on the AREVA recommendations for the relay type P63X, the following formulae for the kneepoint voltage are applied:

1. For through faults:

blctthrkn R2RRIV


[IEC 60044 – 1]


52

Where:RCT = CT secondary resistance

Rl = CT secondary lead resistance

(Lead resistance from manufacturer catalogue: 5.62 ohm/km for A=4mm2)

= 5.62×0.1=0.562 (for lead length 100m)

Rb = Relay burden

= 0.110.1

IVA

22nom

Ithr = Max. three phase through fault current

A387.23200

13.76373200

112315.03

10.50 3

ratioCT1.

.U.u3SI

nk

rthr

To be on safety side, 2.5 A is considered further in calculation:


VVkn 81.411.0124.15.155.2

2. For internal faults:

blctfkn R2RRI.0.25V

Where: If = Max. non-offset fault current for an internal fault

Two currents will be considered:

i) Inrush current

If = 16 times rated current of protected winding

A03.123200

15.38493200

1123

10.50.16.163

ratioCT1.

.U3SI

n

rf


VVkn 3.501.0124.15.1503.1225.0


[IEC 60044 – 1]


53

For impedance calculations the following equations are applied:

Source impedance calculation:

88.36.213

1321.1.3

.1.1.3.1.1

3

)1()1(3

FMax

nHV

HV

nFMax I

UZZUI

Power transformer impedances:

... 4811850

132100

034 2

r

2nk(1)

IDT3(1)IDT2

(1)IDT1

(1)IDT

S

U.

100

%uZZZZ

Total impedance at LV side:

358.013212

348.11888.3

22(1)IDT(1)

HV(1)LV 132

12

3

ZZZ

Max. fault current at 11kV bus:

kAZUI

LV

nFMax 57.19

358.03111.1

.3.1.1

)1(3

Maximum fault current at LV side for the internal fault at 11 kV terminals:

AratioCT

II FMaxf 5.4

320021600.

32.

32 3


VVkn 8.181.0124.15.155.425.0



[IEC 60044 – 1]


54

VVkn 70



=31

20151516.

.=484.62 V


Hence the proposed CT core for ratios 3200/1 A is adequate.

The same above note regarding the Low Impedance REF C.T is applicable.

Example 2: using ABB Differential Protection Relay Type RET 670:


CT ratio: 500-300/1A VA: 25/15

Class: 5P20 Rct: 0.8/0.5

1) CT Requirements for Main-2 Transformer differential protection (87T2):

As per manufacturer’s recommendation for relay type RET670-ABB, the current transformers fortransformer differential protection must have a rated equivalent secondary e.m.f Eal that is larger themaximum of the required secondary e.m.f Ealreq below:

Equation 1:

2..30r

RLCT

pn

snntalreqal I

SRRIIIEE

Equation 2:

2..2r

RLCT

pn

sntfalreqal I

SRRIIIEE

Equation 3:

2.r

RLCT

pn

snfalreqal I

SRRIIIEE

Where:

Int = the rated primary current of the power transformer (A)

=1323

10.50 3

= 218.69 A

Itf = max. primary fundamental frequency current that passes two main CTs and the power transformer (A)

= 16×218.69=3499.09A

If = max. primary fundamental frequency current that passes two main CTs without passing the power transformer (A)


[IEC 60044 – 1]


55

= 40000A

Ipn = the rated primary CT current (A)

Isn = the rated secondary CT current (A)

RCT = the secondary resistance of the CT ( )

SR = the rated burden=0.02VA

RL = the CT secondary loop resistance ( ) =0.75

(For loop resistance, 2×100m cable of 6sqmm is considered)

Equation 1:

2..30r

RLCT

pn

snntalreqal I

SRRIIIEE

2102.0

75.08.0500

169.21830alreqal EE for CT ratio 500/1 A

= 20.6 V

2102.0

75.05.0300


=27.7 V

Equation 2:

2..2r

RLCT

pn

sntfalreqal I

SRRIIIEE

2102.0

75.08.0500


= 21.9 V

2102.0

75.05.0300


=29.6 V

Equation 3:

2.r

RLCT

pn

snfalreqal I

SRRIIIEE

2102.0

75.08.0500

140000alreqal EE for CT ratio 500/1 A

= 125.6 V


[IEC 60044 – 1]


56

2102.0

75.05.0300

140000alreqal EE for CT ratio 300/1 A

=169.3 V

Required knee point voltage, VVkn 6.125 for CT ratio 500/1 A

V3.169 for CT ratio 300/1 A



=31

2018020.

.=320 V for CT ratio 500/1 A

=31

2015012.

.=192.31 V for CT ratio 300/1 A


Hence the proposed CT cores for ratios 500/300/1 A are adequate.

The CT requirements for low impedance REF protection are generally lower than those for differentialprotection. As the line CTs for low impedance REF protection are the same for those used fordifferential protection, the differential CT requirements cover both differential and low impedance REFapplications.


Required limiting factor = 33.133300




M2IDTP relay type RET670, ABB

(HV 64REF2 protection)0.02

2×100m cable of 4sqmm 1.122T2LN CORE-1

TOTAL 1.142<30 VA




Corrected VA = rated VA/1.25 = 20/1.25 = 16 VA

33.1337.1581142.1

116.20 for CT ratio 300/1 A


[IEC 60044 – 1]


57

Since, the operating accuracy limiting factor is 158.7 being higher than the required accuracy limitingfactor 133.3 for ratio of 300/1A. The proposed 5P20, 20VA current transformer is adequate.

11 KV Side :

Proposed current transformer data:



As per manufacturer’s recommendation for relay type RET670-ABB, the current transformers fortransformer differential protection must have a rated equivalent secondary e.m.f Eal that is larger themaximum of the required secondary e.m.f Ealreq below:

Equation 1:

2..30r

RLCT

pn

snntalreqal I

SRRIIIEE

Equation 2:

2..2r

RLCT

pn

sntfalreqal I

SRRIIIEE

Where:

Int = the rated primary current of the power transformer (A)

=123

10.50 3

= 2406 A

Itf = max. primary fundamental frequency current that passes two main CTs and the power transformer (A)

Ipn = the rated primary CT current (A)

Isn = the rated secondary CT current (A)

RCT = the secondary resistance of the CT ( )

RL = the CT secondary loop resistance ( ) =1.8

(For loop resistance, 2×100m cable of 2.5sqmm is considered)

SR = the rated burden=0.02VA

Equation 1: alreqal EE

2102.0

8.15.153200

1240630alreqal EE for CT ration 3200/1 A

= 390.6 V


[IEC 60044 – 1]


58

Equation 2: alreqal EE

2102.0

8.15.153200

124062alreqal EE for CT ration 3200/1 A

= 26.04 V

Required knee point voltage, VVk 6.390There are no specific requirements on the magnetizing current.

The proposed 5P20, 20VA current transformer is adequate.

2. Bus Bar Protection :

A. Low Impedance: Same as the above Line Feeder.

B. High Impedance: Same as the above Line Feeder.

D. Bus Coupler:1. Bus Bar Protection :

A. Low Impedance: Same as the above Line Feeder.

B. High Impedance: Same as the above Line Feeder.

2. O/C & E/F Protection: included in the Bay Control Unit – same as the above LineFeeder.

3. Breaker Failure Protection: It was not shown as separate protection in the LineFeeder Protection; it is included as a Function in the Pilot Wire Differential Protectionand the Distance Protection.

Example 1 : Using AREVA MICOM P142 – Breaker Failure Relay:



Class: 5P20 Rct:





T2LN CORE-3M1BCP relay type P142, AREVA

(50BF1 protection)0.15


[IEC 60044 – 1]


59

2×100m cable of 4sqmm 1.122

TOTAL 1.272<30 VA





33.13698272.1

824.20 for CT ratio 3000/1 A

Since, the operating accuracy limiting factor is 69 being higher than the required accuracylimiting factor 13.3 for ratio of 3000/1A. The proposed 5P20, 30VA current transformer isadequate.

As per manufacturer’s recommendation the knee point voltage requirement for relay typeMiCOM P142, AREVA.

Non-directional/directional DT/IDMT overcurrent and earth fault protection.

Time-delayed phases overcurrent rplctfp

k RRR2

IV

Time-delayed earth fault overcurrent rnlctfn

k RRR2

IV

Where

Rl= Connected lead resistance=1.8

Ifp=Max. secondary fault current= A33.13300040000

Ifn=Max. secondary fault current=13.33A

Rct=Internal resistance=8

Rrp=Relay burden=0.04 VA

Rrn= Relay burden=0.04 VA

Hence,

Phase overcurrent

V....Vk 586504081823313

Earth fault overcurrent


[IEC 60044 – 1]


60

V....Vk 586504081823313



Corrected VA = 30/1.25 = 24 VA

=31

201824.

=492.3 V


Hence , C.T is adequate and suitably sized.

Example 2: using SIEMENS Type 7SJ64: with the same above C.T Data:





M2IDTP relay type 7SJ64, SIEMENS

(50BF2, 50/51 protection)0.05

2×100m cable of 4sqmm 1.122T2L CORE-3

TOTAL 1.172<30 VA





33.137.698172.1

824.20 for CT ratio 3000/1 A

Since, the operating accuracy limiting factor is 69.7 being higher than the required accuracy limitingfactor 13.3 for ratio of 3000/1A. The proposed 5P20, 30VA current transformer is adequate.

As per manufacturer’s recommendation the knee point voltage requirement for relay type 7SJ64,SIEMENS:

Required knee point voltage = snctpn

set poHigh IbRRI

IVk .'

.3.1int

Where:


[IEC 60044 – 1]


61

Vk = Knee point voltage

Isn = CT secondary current=1A

Ipn = CT primary current=3000A

Rct = Internal resistance=8

Rlead = Connected lead resistance=1.8

Rrelay = Relay burden = 0.05VA

IHigh set point = 40kA (As the maximum fault current is 40kA, value of IHigh set point will be lower than the 40kA)

R’b = Connected resistive burden (R lead+Rrelay)

=1.8+0.05=1.85

Hence,

V.Vk 02.1011851830003.1

40000



Corrected VA = 30/1.25 = 24 VA

=31

201824.

=492.3 V


Hence, C.T is adequate and suitably sized.Choice of Measuring CT

• For classes 0.1 – 1, the accuracy is based on a total connected burden between 25% to 100% of therated burden

• For classes 0.1, 0.2 and 0.2S with rated burden < 15VA, an extended range of burden can bespecified (from 1VA to 100% of the rated burden)

Dimensioning factor (Kx) :a factor assigned by the purchaser to indicate the multiple of rated secondary current (Isn) occurringunder power system fault conditions, inclusive of safety factors, up to which the transformer is requiredto meet performance requirements".CT – Magnetizing CurveRated knee point e.m.f. (Ek) :That minimum sinusoidal e.m.f. (r.m.s.) at rated power frequency when applied to the secondaryterminals of the transformer, all other terminals being open-circuited, which when increased by 10 %causes the r.m.s. exciting current to increase by no more than 50 %NOTE The actual knee point e.m.f. will be the rated knee point e.m.f.NOTE The rated knee point e.m.f. is generally determined as follows:Ek = Kx × (Rct + Rb) × IsnWhere (Kx) is the dimensioning factor .

Secondary voltage Esi:The secondary induced voltage Esi can be calculated from

Z2IsiE


[IEC 60044 – 1]


62

2bX2

bRiRZ

The inductive flux density necessary for inducing the voltage Esi can be calculated from

2NiAf2siE

B

where:A = core area in m2B = ux density in Tesla (T)f = frequencyN2 = number of secondary turns

Transient Dimensioning Factor Ktd :Since the total permissible error limit is 10 %, the transient dimensioning factor shall be consideredconjunctively with the secondary circuit time constant:

%10sTf2

tdK100

Where , (Ts) is the Rated secondary loop time constant value of the time constant of the secondaryloop of the current transformer obtained from the sum of the magnetizing and the leakage inductance(Ls) and the secondary loop resistance (Rs) Ts = Ls/RsFlux overcurrent factor (nf)


[IEC 60044 – 1]


63

tdKsscKnfThe secondary core will be designed on the basis of the Kssc (AC- ux) and Ktd (DC ux).Typical value ofKtd is 10 - 25.For special measurement of fault current (transient current,) including both a.c. andd.c. components, IEC 60044-6 defines the accuracy classes TPX, TPY and TPZ.The cores must be designed according to the transient current:• TPX cores have no requirements for remanence flux and have no air gaps.• TPY cores have requirements for remanence flux and are provided with small air gaps.• TPZ cores have specific requirements for phase displacement and the air gaps will be large.Typical secondary time constants are for

• TPX core 5 - 20 seconds (no air gaps)• TPY core 0.5 - 2 seconds (small air gaps)• TPZ core ~ 60 msec. (phase displacement 180 min.+/- 10%) (large air gaps)Air gaps in thecore give a shorter Ts.

Factors affecting CT saturation

CT ratio : Over Dimensioning of the C.T ratio

Core cross-sectional area

Core design

Connected burden : If the Load Burden is lower than the Rated Burden (<25% ) , the Saturationvalue increases

B-H characteristic of core

Amount of remanent flux : In most cases the CTs will have some remanence, which canincrease the saturation rate of the CTs.

Fault current & DC offset : If a CT has been saturated the secondary current will not recoveruntil the DC-component in the primary fault current has subsided.

System time constant

CT primary may be in earthed position. CT internal winding may be short circuit and Winding resistance is less than designed value.

Annex – A : Composite errorUnder steady-state conditions, the r.m.s. value of the difference between:a) the instantaneous values of the primary current; andb) the instantaneous values of the actual secondary current multiplied by the rated transformation ratio,the positive signs of the primary and secondary currents corresponding to the convention for terminalmarkingsThe composite error Ec is generally expressed as a percentage of the r.m.s. values of the primarycurrent according to the formula:

T

0

2psn

p

.dtiiKT1

I100

Use of composite errorThe numeric value of the composite error will never be less than the vector sum of the current error andthe phase displacement (the latter being expressed in centiradians).Consequently, the composite error always indicates the highest possible value of current error or phasedisplacement.


[IEC 60044 – 1]


64

The current error is of particular interest in the operation of overcurrent relays, and the phasedisplacement in the operation of phase sensitive relays (e.g. directional relays).In the case of differential relays, it is the combination of the composite errors of the current transformersinvolved which must be considered.An additional advantage of a limitation of composite error is the resulting limitation of the harmoniccontent of the secondary current which is necessary for the correct operation of certain types of relays.

Annex – B : Metering Cores :At lower burdens than the rated burden, the saturation value increases approximately to n:

2snct

2snctn

s IRSIRSFn

whereSn = rated burden in VAS = actual burden in VAIsn = rated secondary current in ARct = internal resistance at 75ºC in ohm

To ful ll high accuracy classes (e.g. class 0.2, IEC) the magnetizing current in thecore must be kept at a low value. The consequence is a low ux density in thecore. High accuracy and a low number of ampere-turns result in a high saturation factor (FS). To ful llhigh accuracy with low saturation factor the core is usually made of nickel alloyed steel.NOTE! The accuracy class will not be guaranteed for burdens above rated burden or below 25% of therated burden (IEC).With modern meters and instruments with low consumption the total burden canbe lower than 25% of the rated burden (see Figure 2.1). Due to turns correction and core material theerror may increase at lower burdens. To ful ll accuracy requirements the rated burden of the meteringcore shall thus be relatively well matched to the actual burden connected. The minimum error istypically at 75% of the rated burden. The best way to optimize the core regarding accuracy isconsequently to specify a rated burden of 1.5 times the actual burden.It is also possible to connect an additional burden, a “dummy burden”, and in this way adapt theconnected burden to the rated burden. However, this method is rather inconvenient.

Annex – C : Shunt Reactors :Introduction :

During normal operation of an electrical power system, the transmission and distribution voltages mustbe maintained within a small range, typically, from 0.95 to 1.05 pu of rated value. Due to the loadvariations, shunt reactors and capacitors have been applied in power systems to compensate excess


[IEC 60044 – 1]


65

reactive power (inductive for heavy load conditions, and capacitive for light load conditions). Shuntreactors are commonly used to compensate the capacitive reactive power of transmission anddistribution systems and thereby to keep the operating voltages within admissible levels. Shunt reactorsare used in high voltage systems to compensate for the capacitive generation of long overhead lines orextended cable networks. The reasons for using shunt reactors are mainly two:

1. to limit the over-voltages

2. to limit the transfer of reactive power in the network.

If the reactive power transfer is minimized i. e. the reactive power is balanced in the different part of thenetworks, a higher level of active power can be transferred in the network. Reactors to limit over-voltages are most needed in weak power systems, i.e. when network short-circuit power is relativelylow.

Voltage increase in a system due to the capacitive generation is:

shc

c

S100QU(%)

where “Qc” is the capacitive input of reactive power to the network

and “Sshc” is the short circuit power of the network.

With increasing short circuit power of the network the voltage increase will be lower and the need ofcompensation to limit over-voltages will be less accentuated. Reactors to achieve reactive powerbalance in the different part of the network are most needed in heavy loaded networks where new linescannot be built because of environmental reasons. Reactors for this purpose mostly are Thyristorcontrolled in order to adapt fast to the reactive power required. Especially in industrial areas with arcfurnaces the reactive power demand is fluctuating between each half cycle. In such applications thereare usually combinations of Thyristor controlled reactors (TCR) and Thyristor switched capacitor banks(TSC).These together makes it possible to both absorb, and generate reactive power according to themomentary demand. Four leg reactors also can be used for extinction of the secondary arc at single-phase reclosing in long transmission lines. Since there always is a capacitive coupling between phases,this capacitance will give a current keeping the arc burning, a secondary arc. By adding one singlephase reactor in the neutral the secondary arc can be extinguished and the single-phase auto-re-closing successful.

The calculation of optimum ratings and points of connection of shunt reactors is generally done bymeans of extensive load-flow studies, taking into account all possible system configurations.

CONNECTIONS IN THE SUBSTATION

The reactors can be connected to the bus bar, a transformer tertiary winding or directly to the line, withor without a circuit-breaker (see gure 1).


[IEC 60044 – 1]


66

The shunt reactor is normally connected to the transmission line to avoid excess over voltageappearance when the load decreases at night and off days. Since the substations in the urbanareas mostly have the transmission lines of small short-circuit capacities, the line voltages fluctuatelargely when the shunt reactor of a large capacity is connected or disconnected.

Shunt reactors carry out different types of tasks:

They compensate the capacitive reactive power of the transmission cables, in particular innetworks with only light loads or no load.

They reduce system-frequency over-voltages when a sudden load drop occurs or there is noload.

They improve the stability and efficiency of the energy transmission.

Reactors to limit over-voltages are most needed in weak power systems, i.e. when network short-circuitpower is relatively low.

Mostly problems occur on gas-insulated circuit breaker for shunt reactor switching. The frequentswitching of the SF6 CB for shunt reactors degraded gas insulation level. The melted contacts, in turn,could not clear the current prospectively. Meanwhile, the high rise rate of transient recovery voltage ofinductive current switching caused re-striking phenomenon and incomplete tripping. These two maincharacteristics make the SF6 CB used for shunt reactor un-expectantly damaged .

Therefore “ Separate Type Test Report from Accredited Testing Laboratory shall be submittedfor the proposed C.B Type for Shunt Reactor or Shunt Capacitor banks Switching .”

The unavoidable high frequency transient recovery voltage (TRV) existed in circuit breakers dueto inductive switching can be depressed to reasonable and safely level by equipping suitablearrestor. The maintenance policy for the breakers should be planned based on a fixed periodmaintenance schedule or conditional basic maintenance schedule. Finally, the circuit breakershould be re-flashed after every 500 switching operations to maintain the power system normallyTypically, the voltage variation at the high voltage bus bar after switching of a shunt reactor shall not behigher than 2 to 3% of rated voltage.

MAIN CALCULATION OF SHUNT REACTORS: All the shunt reactor data and rating will begiven in the contract specifications, this summary is just for information.For the calculation of the positive sequence reactance and the current requirements of a shuntreactor, it is necessary to know only the rated three-phase reactive power and the rated systemvoltage and frequency, as summarized in the table below.


[IEC 60044 – 1]


67

Rating WYE Connection Delta ConnectionReactance

R1

2N

R3

2N

R S3U

SUX

R1

2N

R3

2N

R S3U

SUX 3

Rayed Current

R

N

N

R1

N

R3N X3

U

3U

SU3

SI

R

N

N

R1

N

R3N X

UUS

U3S

I

Maximum Continuous Current(Design Current N

N

MAXMAX I

UUI N

N

MAXMAX I

UUI

Parameters XR=Rated reactance per phase (positive sequence.SR3 =Rated three phase reactive powerSR1 =Rated reactive power per phaseUN=Rated system voltageUMAX=Maximum system operating voltageIN=Rated currentIMAX=Maximum continuous current

So, the relation between the ending voltages of the transmission line is given by:

2R

12R

121 .VLL

2ZY1V.V

XBAVA'.VV

Application Example

Consider a lossless radial transmission line, frequency 60 Hz, length = 350 km, and parameters z = j0,32886 /km and bC = j 5,097 µS/km. To estimate the reactive power of shunt reactors to be installedin the transmission line to provide a maximum operating voltage of 1.05 pu at the open-circuitedterminal (receiving ending), when the line is energized with 1.0 pu in the sending ending.

Solution:

Total impedance and admittance of the non-compensated transmission line

1.115j)j(z.Z

Sj 95.1783).j(bY c

Parameter A:

8973.02

ZY1A

Operating voltage at the receiving ending of the non- compensated transmission line

112 .1144.1.8973.01 VVV

Calculation of the shunt reactor reactance:


[IEC 60044 – 1]


68

2R

1 .VLL

2ZY1V

.1.05X

115.10.89731.0R

20908973.0

1.051.0

1.115XR

Calculation of the three-phase reactive power of the shunt reactor:

MVAr13220905252

R

2N

R3 XUS

Calculation of the three-phase reactive power of the shunt reactor:

MVAr490350079.55252..bUQ C2NC3

So, compensation degree is:

%54C3

R3SH Q

S2.K

A practical circuit is used to simplify the analysis of voltage control (see picture below). Thedetermination of the shunt reactor to provide a required voltage variation in the bus bar can calculatedthrough the short-circuit power of system at the bus bar where the reactor will be connected.

The shunt reactor rating is given by:

2

21CCR3

CC

R3

12 V

VVSS

SS

1

VV

Application Example:

To estimate the reactive power of shunt reactors to be installed in the 34.5kV busbar in order toreduce the voltage level from 1.02 to 0.99pu, considering a fault current of 25kA (or short circuitpower of 1495MVA).


[IEC 60044 – 1]


69

Solution

Calculation of the three phase reactive power of the shunt reactor:

45MVAr0.99

0.991.021485R3S

Reference Documents :IEC 60044 – 1 & IEC 60044 – 6 .

ESI Standard 48-3 : Instantaneous High Impedance Differential Protection , December1977 .

ABB : Instrument Transformers Application Guide .

Siemens : C.T Dimensioning .

AREVA : Tech. News : Shunt Reactors in Power Systems .

Different Contractors Submittals for Different Projects in different countries which werereviewed by us , commented , revised & re-submitted till approved .

c.t dimensioning

Documents

actual secondary current

fault current

primary current andb

eddy current

current flows

secondary current vectors

current factor kssckssc

rated primarycurrent