C.T Dimensioning

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Definitions : Current transformer An instrument transformer in which the secondary current, in normal conditions of use, is i s substantially proportional to the primary current and differs in phase from it by an angle which is approximately zero for an appropriate direction of the connections Current error (ratio error) The error which a transformer introduces into the measurement of a current and which arises 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 losses and magnetizes the core to the Primary Current. This current flows in the primary winding only and therefore, 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 the formula:

CurrentError% 

K n I s  I p  100 Ip

where Kn Ip Is

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

Phase displacement The difference in phase between the primary and secondary current vectors, the direction of the vectors being so chosen that the angle is zero for a perfect transformer. The phase displacement is said to be positive when the secondary current v ector leads the primary current 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 Primary Current .A current transformer is designed to m aintain its ratio within specified limits up to a certain value of primary current, expressed as a multiple of its rated primary current. This multiple is known as the current transformer’s rated accuracy accu racy limit factor (ALF). Dimensioning factor (Kx) : A factor assigned by the purchaser to indicate the multiple of rated secondary current (Isn) occurring under power system fault conditions, conditions, inclusive of safety factors, up to which the transformer is required to meet performance requirements" Composite error Under steady-state conditions, the r.m.s. value of the difference between: a) the instantaneous values of the primary current; and b) 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 terminal markings The composite error Ec is generally expressed as a percentage of the r.m.s. values of the primary current according to the formula: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

1

PART I - Current Transformers Dimensioning [IEC 60044 – 1]



c



100 Ip

1 T

T

  K 0

i  i p  dt 2

n s

where Kn Ip ip is T

is the rated transformation ratio; is the r.m.s. value of the prim ary current; is the instantaneous value of the primary current; is the instantaneous value of the secondary current; 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 the secondary terminals of the transformer, t ransformer, 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 %. Some clients require the calculations to be based on the corrected VA and not the rated VA of the C.T where a margin of 25 % shall be taken into account for rated VA calculations for 5P class cores for the Rated Knee Point Voltage and Operating ALF conversions as per the f ollowing formulas which will give accurate dimensioning :











Since “The rated equivalent limiting secondary e.m.f.” Eal or The Rated Knee Point Voltage Vk in according with the IEC 60044-6 standard to specify the CT requirements for different protection equipment which shall be higher than the required limiting secondary e.m.f EalReq. or the required knee point voltage Vkreq. Which is to be calculated as per all the Protection Relay Manufacturers recommendations and compliance with IEC 60044-6 standard . NOTE The Rated Rated knee point e.m.f. e.m.f. will will be  the Required knee point e.m.f. which will be in according with the protection relay manufacturer recommendation . Rated equivalent limiting secondary e.m.f. (E al) That r.m.s. value of the equivalent secondary circuit e.m.f. of rated frequency necessary to satisfy the specified duty cycle and derived from the following: E al (V, r.m.s.) al = K ssc ssc . K td td (R ct ct + R b b ) I sn sn Rated symmetrical short-circuit current factor (Kssc) Kssc = which is equal to the Required Accuracy Limit Factor (ALF req. ) . K ssc ssc = Ipsc/Ipn Ipsc r.m.s. value of primary symmetrical symmetrical short-circuit current Ipn Rated primary current Ktd 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 the E2max is approximately equal to the corresponding Eal according to IEC 60044-6. Therefore, the CTs according to class P and PR must have a secondary limiting e.m.f. E2max that fulfills the following:

Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

2




c



100 Ip

1 T

T

  K 0

i  i p  dt 2

n s

where Kn Ip ip is T

is the rated transformation ratio; is the r.m.s. value of the prim ary current; is the instantaneous value of the primary current; is the instantaneous value of the secondary current; 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 the secondary terminals of the transformer, t ransformer, 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 %. Some clients require the calculations to be based on the corrected VA and not the rated VA of the C.T where a margin of 25 % shall be taken into account for rated VA calculations for 5P class cores for the Rated Knee Point Voltage and Operating ALF conversions as per the f ollowing formulas which will give accurate dimensioning :











Since “The rated equivalent limiting secondary e.m.f.” Eal or The Rated Knee Point Voltage Vk in according with the IEC 60044-6 standard to specify the CT requirements for different protection equipment which shall be higher than the required limiting secondary e.m.f EalReq. or the required knee point voltage Vkreq. Which is to be calculated as per all the Protection Relay Manufacturers recommendations and compliance with IEC 60044-6 standard . NOTE The Rated Rated knee point e.m.f. e.m.f. will will be  the Required knee point e.m.f. which will be in according with the protection relay manufacturer recommendation . Rated equivalent limiting secondary e.m.f. (E al) That r.m.s. value of the equivalent secondary circuit e.m.f. of rated frequency necessary to satisfy the specified duty cycle and derived from the following: E al (V, r.m.s.) al = K ssc ssc . K td td (R ct ct + R b b ) I sn sn Rated symmetrical short-circuit current factor (Kssc) Kssc = which is equal to the Required Accuracy Limit Factor (ALF req. ) . K ssc ssc = Ipsc/Ipn Ipsc r.m.s. value of primary symmetrical symmetrical short-circuit current Ipn Rated primary current Ktd 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 the E2max is approximately equal to the corresponding Eal according to IEC 60044-6. Therefore, the CTs according to class P and PR must have a secondary limiting e.m.f. E2max that fulfills the following:


2

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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-point e.m.f. Eknee (Ek or Vk) for class PX, E kneeBS 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 to give a general relation between the Eknee and the Eal but normally the Eknee is approximately approximately 80% of of the Eal. Therefore, the CTs according to class PX, X and TPS must have a rated knee-point e.m.f. Eknee that fulfills the follow f ollowing: ing: E knee knee  E k k  E kneeBS kneeBS  U al > 0.8 . ( maximum of E alreq) . Class X or PX CTs are mostly used for high impedance circulating current protection which requires high Equivalent Secondary e.m.f (E) .This can be economically achieved with Class X or PX which is of Low Leakage Reactance and can have a High magnetizing Current at Vk . Three factors will influence the emf “E”. It’s the number of secondary turns “N”, the core area “A” and the induction in Wb/m2 “B”.





 



where: A B f N2

= = = =

core area in m2  ux density in Tesla (T) frequency number of secondar y 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 t hus selected to give the required emf output. C.T – AS 60044 – 1 Class PX The rated knee point e.m.f. is generally determined as follows: E K K ( V K K ) = K X X . ( R ct ct + R b b ) . I sn Where , Ek = rated knee point emf . Kx = multiple of Isn at which C.T must perform satisfactorily Isn = 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 without any air gaps and the remanent r emanent flux might remain for almost infinite time. time. The remanent flux can be up to 70-80% of the saturation 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. ANSI/IEEE. 2) Low remanence CTs The low remanence type has a specified limit for the remanent flux. The magnetic core is provided with small air gaps to reduce the remanent flux flux to a level level that does not exceed 10% 10% of the saturation flux. Examples are class TPY according to IEC 60044-6 and class PR according to IEC 60044-1. Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

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PART I - Current Transformers Dimensioning [IEC 60044 – 1] 3) Non remanence CTs The non remanence type CT has practically negligible level of remanent flux. The magnetic core has relatively large air gaps in order to reduce the secondary time constant of the CT (to lower the needed transient factor) which also reduces the remanent flux to practically 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 transformer depends 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, are measuring 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 to accuracy 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 cores are 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 fault conditions, 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. The rated Instrument Security Factor (FS) indicates the overcurrent as a m ultiple of the rated current at which the metering core will saturate. It is thus limiting the secondary current to FS times the rated current. The safety of the metering equipment is greatest when the value of FS is small. Typical FS factors are 5 or 10. It is a maximum value and only valid at rated burden. A higher output from a core will also result in a bigger and more expensive core, especially for cores with high accuracy (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 current S : 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 be transferred to the secondary side. Measurement at fault conditions in the overcurrent range requires lower accuracy, but a high capability to transform high fault currents to allow protection relays to measure and disconnect the f ault. 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 definition in IEC 60044-1 formerly covered by class X of BS 3938 type CTs are used for high impedance circulating current protection and are also suitable f or most other protection schemes. We will elaborate only on the most commonly used Classes P & PX . For further details for the choice of 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 f or which knowledge of the transformer secondary excitation characteristic, secondary winding resistance, secondary burden resistance and turns ratio is sufficient to assess its performance in relation to the protective relay system with which it is to be used.


4

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Class P current transformers Are defined so that, at rated frequency and with rated burden connected, the current error, phase displacement and composite error shall not exceed the values given in the table below : Accuracy Class

Current Error at Rated Primary Current

5P 10P

±1% ±3%

Composite Error at Rated Accuracy Limit Primary Current 5% 10%

The primary difference is that the measuring current transformer(FS) is required to retain a specified accuracy 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 fraction of full load to many times full load. CT – Measuring vs Protection :





Measuring CT Under over current, the CT should limit the secondary current to avoid thermal overload of the connected equipment At rated burden (cos= 0.8) and rated security factor, the composite error should be>10%

Protection CT In case of over current, the CT should maintain the secondary current within the rated limit to enable proper protection operation At rated burden (cos= 0.8) and rated accuracy limit factor 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 secondary terminal voltage UANSI is specified for a CT of class C. UANSI is the secondary terminal voltage the CT will deliver to a standard burden at 20 times rated secondary current without exceeding 10 % ratio correction. There are a number of standardized UANSI values e.g. UANSI is 400 V for a C400 CT. A corresponding rated equivalent limiting secondary e.m.f. EalANSI can be estimated as follows:

E alANSI  20  I sn  RCT  U ANSI  20  I sn  RCT  20  I sn  Z bANSI where 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. EalANSI that fulfills the following:

E alANSI  Maximum of E alreq A C.T according to ANSI/IEEE is also specified by the knee-point voltage UkneeANSI that is graphically defined from an excitation curve. The knee-point voltage UkneeANSI normally has a lower value than the knee-point e.m.f. according to IEC and BS. UkneeANSI can approximately be estimated to 75 % of the corresponding Eal according to IEC 60044-6. Therefore, the CTs according to ANSI/IEEE must have a knee-point voltage UkneeANSI that fulfills the following: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

5

PART I - Current Transformers Dimensioning [IEC 60044 – 1] E alANSI  0.75  ( Maximum of E alreq ) 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 the rating is specified differently. The IEEE C class standard voltage rating required will be lower than an IEC knee-point voltage. This is because the IEEE voltage rating is defined in terms of useful output voltage at the terminals of the CT, whereas the IEC knee-point voltage includes the voltage 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 determine the equivalent knee-point voltage (Vk) according to IEC. The equivalence formula is:

V k   C 1.05   K SSC  I n  RCT 

 C 1.05  100  RCT  Note: IEEE C.Ts are always 5A secondary rated, i.e. In =5A, and are defined with an accuracy limit factor 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 for the calculation of the Knee Point Voltage ( Vk ) f or Feeder Protection , Transformer differential & Restricted E/F protection .Since every protection relays manufacturer has his own recommendation and C.T requirements which shall be met and must be submitted by the contractors and must be followed .



Some clients require the calculations to be based on the corrected VA and not the rated VA of the C.T where a margin of 25 % shall be taken into account for rated VA calculations for 5P class cores for the Rated Knee Point Voltage and Operating ALF conversions as per the following formulas which will give accurate dimensioning which we will follow in our examples :

Rated Knee Po int Voltage 

     Corrected VA Wdg s I ALF    . Re sec .       1.3

The Operating Accuracy Limit Factor K OALF :

(Corrected VA  Wdg . Re s ) K OALF  RATED ALF 

(Connected VA  Wdg . Re s ) 

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



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


6








burden have to be considered. For earth faults in solidly earthed systems it is important to consider 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 in different positions. Maximum fault current will occur for three-phase faults or in some cases for single-phase-to-earth faults. The current for a single phase-to-earth fault will exceed the current for a three-phase fault when the zero sequence impedance in the total f ault loop is less than the positive sequence impedance. This can be the case in solidly earthed systems 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 will saturate. In most cases the CTs will have some remanence, which can increase the saturation rate of the CTs. If a CT has been saturated the secondary current will not recover until the DC-component in the primary fault current has subsided. This can cause a delay in the relay operation. Depending on the design of the relay and to consider a certain amount of DC-component, a factor has been used in the Rated Knee Point Voltage calculation Equation . This safety factor gives a satisfactory operation. The following sample calculations will assist you in quick reviewing and approximate verification of the different contractors submittals as a minimum requirement to be met for the C.T sizing calculations , the calculations will be based on the most commonly used 132/11 KV substations with 2000 A bus bar , 40 KA which is applicable for all Voltage levels and both the Outdoor AIS Switchgear or Indoor GIS where the different data and C.T ratio 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.T requirements as per the different Manufacturer recommendations .

A. 132 KV Cable Feeder :

Core Design T1L

Core No. 1

Ratio

Accuracy Class PX

Vkn (V)

RCT ()

Io (mA)







1500-1000/1A

Burden VA -

1950/1300

5.5/3.67

1500-1000/1A

-

PX

1950/1300

5.5/3.67

650 650

2/1.33 8 8.5 8.5

50/100 at Vkn 50/100 at Vkn 15 at Vkn 15 at Vkn

T1L

2

T1L 3 1500-1000/1A 22.5/15 5P20/0.5 T2L 1 3000/1A 30 5P20 T2L 2 3000/1A PX T2L 3 3000/1A PX Notes : 1. Rated Continuous Thermal Current is 150% In

2. 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 .


7

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 2. Over Current & Earth Fault Protection : Mostly for O/C & E/F protection , class 5P C.T is widely used and the v erification of the Operating Accuracy Limiting Factor (KOALF ) is to be equal or more than the required ALF is enough to v erify if the proposed C.T is adequate or not. ALF op.



ALF req.

2

For cables of 4 mm 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 relevant manufacturer’s catalogues

Burden (VA) Bay Control Unit , REF545-ABB

0.1

(Synchro Check Function) 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

Re quired ALF 

40000 1500

Re quired ALF 

 26.6

40000 1000

For 1500/1 A CT ratio

 40

For 1000/1 A CT ratio

The Operating Accuracy Limit Factor (KOALF ) (Corrected VA  Wdg. Re s) K OALF  RATED ALF 

(Connected VA  Wdg. Re s )

Rated VA Corrected VA 

1.25




22.5 1.25

 18VA , and

8

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 15 1.25

 20 

18  2 1.322  2

 20 

 12VA  120.4  26.6

12  1.33 1.322  1.33

 100.52  40

For CT ratio 1500/1 A

For CT ratio 1000/1 A

Since , the operating accuracy limiting factor KOALF=120.4/100.52 being higher than the required accuracy limiting factor 26.6/40 for CT ratios 1500/1000/1 A , the proposed 5P20, 22.5/15VA current transformer is adequate The O/C & E/F protection functions are mostly incorporated in the Bay Control Unit ( BCU ) or the Pilot Wire Differential Protection along with the Breaker Failure function . However , the same procedure to be followed for the loose relays if used. If the proposed BCU incorporate other protection functions such as Breaker Failure where as per the manufacturer recommendation that the CTs must have 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 low impedance relays. Differential protection uses the principal of Kirchhoff’s first law to sum the current from the CTs covering the protected zone. The relay detects the summation current and will trip the associated 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 on common secondary cores of the current transformers. It has to allow the use of different class and type of 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 44-6 Class TPX, TPY or TPZ).The protection has to be stable for all types of external faults and in particular under CT saturation conditions. This saturation has to be able to be detected in less than 2 ms. The differential bus bar protection has to implement also an effective protection against circuit breaker failures. Example for Low Impedance with SIEMENS Relay Type 7SS52 C.T Data : CT ratio: Class:

3000/1A

VA:

30

RCT 8

5P20

Burden ( VA ) BBP relay Type 7SS52 , SIEMENS

0.1

2×100m cable of 4 mm²

1.122

Total

1.222 < 30 VA


9

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Verification of the Accuracy Limit Factor : As per Manufacturer C.T requirement : CT requirement for Low Impedance Busbar Protection Type 7SS52-SIEMENS(M2BZP)

  K td  K SSC

I SSC max( ext . fault ) I pn

  K SSC  K SSC

  100(measuring range) and K SSC

RCT  Rb RCT  Rb

Where

K SSC

Rated Symmetrical Short Circuit Current Factor =20

 K SSC

Effective Symmetrical Short Circuit Current Factor

Rb

Rated Resistive Burden (30 )

R CT

Secondary Winding Resistance (8 )

R L

CT Secondary Loop Resistance =1.8  For Loop Resistance ,2×100 m cable of 2.5 sqmm is c onsidered

R R

Relay Burden (0.1 )

Rb

Connected Resistive Burden

I pn

CT Rated Primary Current

RLead + RRelay =1.8+0.1=1.9 

Transient Dimensioning Factor =0.5

K td I SSC max( ext . f

Max. short circuit current(40kA)

V k

Knee Point Voltage

  0. 5  K SSC

40000 3000

 6.667(required )

And

  K SSC  K SSC   20  K SSC

RCT  Rb RCT  Rb

8  30 8  1.222

(effective)

 82.4  6.667

As the effective symmetrical short circuit current factor of proposed CT is above the required effective symmetrical short circuit current factor . Hence the proposed CT is adequate Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

10

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Further , as per manufacture catalogue Vk for relay type 7SS (measuring ) is calculated as follows :

V k 

100

V k 

100

1.3

1.3

  RCT  R   I sn  8  1.222   1  709.4V

    Corrected VA  Wdg. Re s   I sn  ALF     Rated Knee Po int Voltage  1.3

Corrected VA = 30/1.25= 24VA



 24  8  1  20 1.3

 492.3V

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 Class of Accuracy CT Resistance Burden Corrected Burden

3000/1A 5P20 RCT > 15  50 VA 40 VA

Burden of REB670 SR Burden of CT Leads *

= 0.02 VA = 1.12 VA

Total Burden on the Core (PL) =1.14 2 * CT Lead Resistance for 100 m with a 4 mm cable As per the relay manual , the CT requirement are as follows :

 V

V k available

V k cal

k cal

       0.5  I t max   I sn   RCT  R L   S R 2     I pn    I R  


11

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Calculation for Ealreq Max primary fundamental fault Current CT primary Current CT secondary Current Relay Current CT Resistance CT Leads Resistance Relay Burden

V k cal

= Itmax

=40000A

=Ipn =Isn =IR =RCT =RL =SR

=3000A =1A =1A =15.00  =1.12  =0.02VA

       0.5  I t max   I sn   RCT  R L   S R 2     I pn    I R  



 0.5  40000  1

     15  1.1212903   0.02    3000 1 



2

 

=108 v As per the relay manual , since the CT requirement are mentioned for BS Class PX class . Hence final class 5P CT parameters are CV The design values according to BS can be approximately transferred in to IEC-60044 std using the following formula:

Rated Knee Po int Voltage 

    Corrected VA  Wdg . Re s   I sn .ratedALF     1.3

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 is because the other CTs will provide a lower-impedance path as compared with the high relay input impedance. Traditionally high impedance schemes have provided greater stability during through faults. This type of protection requires all CTs must have the same transformation ratio, are class x and dedicated to the protection scheme. The current transformers used in high impedance circulating current differential protection systems must be equal turns ratio and have reasonably low secondary winding resistance. Current transformers


12

PART I - Current Transformers Dimensioning [IEC 60044 – 1] of similar magnetizing characteristic with low reactance construction such as IEC60044 Class PX, or similar, are preferred. The relay requirements are based upon a calculation of the required knee-point voltage with the IEC definition 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 scheme depends upon the operate voltage setting being greater than the maximum voltage which can appear across the relay under a given through fault condition. An external series stabilizing resistor and shunt non-linear resistor per phase complete the scheme. The series resistor value is determined by the voltage level required for stability and the value of relay current calculated to provide the required primary fault setting. Non-linear resistors protect the relay circuit from high over-voltages. C.T Data : CT Ratio Class of Accuracy CT Resistance Knee Point Voltage Magnetizing Current Burden of DAD-N Burden of CT Leads

3000/1A PX RCT< 15  Vk > 1500 V Io< 40 mA at Vk /2 = 0.10 VA =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 during saturation of the CTs on an external fault, the actual stability voltage Us must be higher than the voltage (Ustab ) produced by the maximum secondary through fault current , flowing

U s  U stab U s  I k . max .thr  ( I sn / I pn )  ( RCT  Rwire ) In addition to this, the knee point voltage must be higher than twice the actual stability voltage:

U Knee  2  U s Where: Us Ustab Uknee Ik,max,thr Iscc,max,thr Rwire RCT Ipn Isn

Relay Setting Voltage Minimum Stability Voltage Knee Point Voltage of CT Max. symmetrical short-circuit current for external f C Max. symmetrical short-circuit current for internal f a’ Cable Burden Max. internal burden of CT at 75  CT primary nominal current CT secondary nominal current


13

=40000A 40000A =1.12  =15  =3000A =1A

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Rb If

Relay Burden Secondary Fault Current

 40000  (1 / 3000 )  (15  1.1212903)  215V

U stab

=0.10  =13.33A

Since

U s  U stab is considered as Us= 225 V Therefore

U Knee  2  U s

 2  225  450V Criteria is Proposed Vk Required Vk 1500  450 Hence CT is : Suitably dimensioned Calculation of maximum sensitivity

 I I P   pn 

  ( I s. min  n  I e ) I sn 

Where

I s. min

Minimum relay setting

=0.005A

n

Number of CTs in parallel with relay including future bays Mag. Current at relay setting voltage (225.00)V

=30

I e

I P

 I    pn I   ( I s. min  n  I e ) sn    3000 1  (0.005  30  0.012)





=0.0120A (Assumed)

=36.5 % of CT rated current

= 1095

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


14

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Power Transformer Rating Power Transformer Percentage Impedance (%Z) Rated Voltage Trafo. Full Load Current

= 50 =30.5 = 132

 

MVA % kV

MVA

3  Voltage 50  1000 1.732  132

=218.69 A Effective Sensitivity Calculation : As per ESI 48-3 , the fault setting shall be between 10% and 30% of minimum fault current available . DEWA ( Dubai Electricity & Water Authority ) specify the minimum and m aximum primary operating current as follows which will be f ollowed in the calculations : IP min= 1100 A IP max= 1400 A For a desired increased sensitivity of 2000 A primary setting . The relay setting shall be

I s set (min)

I s set (min)

   I  p 

 I sn    U S     I    n  I e    U knee   des pn      





 



 1100  1  30  0.012  225 3000 1500



= 0.310 A

I s set (max)

I s set (max)

   I  p 

des



 I sn    U S      n  I e       U knee    I pn   



 



 1400  1  30  0.012  225 3000 1500



= 0.410 A Calculation of Stabilizing Resistor The proper value of stabilizing resistor Rstab is required to ensure stability and is calculated by using the formula

R stab (max)

   U    s I   Rb  s set   


15






 225 0.41  0.1 = 726  Minimum Power Rating Pstab

Pstab  I 2 s

set

 Rstab

 0.412  726 =122 W Voltage developed across resistor during internal fault

  1500



V f  V k  Rstab  I f 3

3

1

4

 1.3

 726  13.34

1

 1.3

4

 3108

V Half second rating of stabilizing resistor P half =



V 1

2

Rstab

  3108  / 726  13306W  13kW 2

Hence stabilizing resistor of 0-5000  variable , having a power rating of 350 W and short time rating of 15kW will be selected Calculation for max. voltage at relay terminal The theoretical Voltage which may occur at the relay terminals is

 I

U SCC max int

SCC max

    I sn    RCT  R wire  Rstab  int  I pn 





 40000  13000  15  1.1212903  726 =9895 V

U max relay

U max relay

       2  2  U knee  U  U knee     SCC max int       2   2  1500   9895  1500  = 10037 V


16

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Metrosil is required if

 1500V

U max relay

10037  1500V Hence Metrosil is required The type of metrosil required is chosen by its thermal rating as defined by the formula

P   4 / 3.14  I f  V k

  4 / 3.14  13.34  1500  25477.71 J / S  25.48 kJ / S Select a metrosil with C=900 , = 0.2 Another example with another Type which is also widely used AREVA - MCAG34 High Impedance Protection Relay for 11 KV Busbar Protection . 11 kV Busbar Protection Discriminating and check Zone The relay is MCAG34,AREVA CT Ratio : 3200/1A Class 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 a current transformer , then VS shall be not less than :

V s  I F ( A  D  2G )T Where : IF

= Fault Current Corresponding to the rated stability limit (100 percentage of switchgear short circuit rating as per clause 5.6 of ESI 4 8-3) =25kA =Resistance of Wiring plus current transformer winding = RCT+RL=10+0.9=10.9 2 RL=0.9 , L=100 m, A=2.5 mm , r= 9  /km

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

V s 

2500 3200

 10.9  0.9  92.19V

V s  92.19V a.2) Also Consider a phase –to –phase or 3 phase external short circuit and assume complete saturation of a current transformer , then VS shall not be less than : Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

17

PART I - Current Transformers Dimensioning [IEC 60044 – 1] V s  I F ( A  G )T (V ) V s 

2500 3200

 10.9  85.16V

V s  85.16V b)

Knee Point Voltage (Vk)

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

V kn min

 2  V s  2  95  200V  200V

V kn min

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

I fs   I s  nI t  I sr  I m  

1 T

Is

= Selected Relay Setting Current

=0.05A

n

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

=66

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

=24

Magnetizing Current of CT

= 1.8mA at Vs (Average)

Maximum Magnetizing Current

=66 x 1.8=118.8 mA

I1

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

= Impedance of busbar supervision relay 2

=

U s B



600 2 4

=90000 

 90000 = 90000

Considering the worst case as R s= 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


18

PART I - Current Transformers Dimensioning [IEC 60044 – 1] For check zone

100    0.003  3200  553.3 A I fs  0.05  66  0.0018  90000





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 ca assumed that average current is approximately 1mA, reconsidering the above for

100    0.003  3200  384.4 A I fs  0.05  66  0.001  90000





Which is approximately 19.2% , of the min , fault current at the 11kV Bus For Disc. zone

100    0.003  3200  311.4 A I fs  0.05  24  0.0018  90000





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 ca assumed that average current is approximately 1mA, reconsidering the above for

100    0.003  3200  249.95 A I fs  0.05  24  0.001  90000   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 protecti Therefore, the relay current setting will be provided with range (5%-20%) in s equal 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 series resistance is calculated as follows :

RS 

V s I r



VA I r 2

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


19

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Assumed current setting is for check zone and disc. Zone

I r  0.05 A RS 

100 0.05

 2000

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 be less than 2000. Hence selected is 0-2700  (variable).

The resistors incorporate in the scheme must be capable of withstanding the associated thermal conditions The continuous Power Rating of a resistor is defined as Pcon

  I con   R 2

Where Pcon

= Resistor Continuous Power Rating

Icon

=Continuous Resistor Current Power Rating

R

=Resistance

For Check 2

Pcon

 553.3    2000  3200 

= 59.8 Watts For Disc 2

Pcon

 311.4     2000  3200  =18.9 Watts

The voltage developed across a resistor for a maximum internal fault condit is defined as Vf

 2 V k 3  R  I fs  1.3 

830.32

2000  344.7Watts


20

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Where Vf Ifs

= rms voltage across resistor = max. fault current 25kA Secondary current is

2

Vf

Phalf

=

 220

3



25000 3200

 7.813

 2000  7.813  1.3

 830.3V

Thus the half second power rating is given by :

Phalf 

 

V f 2 R V f 2 R

830.3 2

2000  344.7Watts As per the manufacturer catalogue, short time rating of stabilizing resistor used in MCAG34 AREVA relay is : ZB9016783 e) Requirement of Metrosil The maximum voltage in the absence of CT saturation is :

V f  I F T RCT  2 R L  RS  RRe lay 



25000 3200

10  2  0.9  2000  1  15725

Peak to peak voltage developed across the relay is:

V P  2  2  V k  V f  V k 

Where Vk=200 V Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

21

V

PART I - Current Transformers Dimensioning [IEC 60044 – 1] V P  2  2  220  15725  220 

= 5223.8 V Since the peak voltage developed across the relay is more than 3k V , metrosils are required . Selected metrosil 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 di sadvantages of a 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



3.

Advantages of numerical protection technology(e.g. fault recording, communication, etc.) not available



Check zone needs separate CT cores



Isolator replica requires switching of CT secondaries , additional check zone obligatory. 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 the currents entering the protected zone is zero. In the case of a power system transmission line, that may not be entirely true because of the capacitive charging current of the line. For short overhead transmission lines, the charging current can be treated as a small unknown error. In that case, the error due to the line charging current is covered by the percentage restraint characteristic of the current differential scheme. For long transmission lines and cables, the charging current may be too large to treat as an unknown error. In that case, it is often necessary to desensitize the current differential protection to prevent mis-operations due to the line charging current which shall be considered in the Relay setting :

The sensitive differential set point I-DIFF> is calculated according the charge current of the cable or set to a minimum value, which results from the CT´s transient behavior. The charge current caused Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

22

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 manufacturer recommendation . The charge current is calculated as follows:

I C  3.63  106  U N  f N  C OP  1 IC

Primary charge current in A

UN

Nominal voltage of the line/cable in kV

f

Rated frequency in Hz

N

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

Class :

RCT 5.5/3.67 

PX

Ie 50/100mA at Vk

To ensure correct operation of the connected relay , CT to remain stable under all through fault conditions and the rated knee point voltage should be greater than the calculated knee point voltage Since, effective CT accuracy limiting factor,

K OALF 

K OALF 

I SSC , max

and K OALF  30

I N

40000 A 1500 A

 26.67 ,

40000 A 1000 A

 40

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

I V k  K td 

SSC max( Ext . fault )

I PN

    RCT  Rb   I SN  

Where Ktd

= 3.75 Transient Dimensioning F actor (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 (R B+2RL)

RL

One way lead resistance from CT to relay =0.9  (for Lead resistance ,100 m cable of 2.5 sqmm is c onsidered )

UKN

Rated Knee Point Voltage


23

PART I - Current Transformers Dimensioning [IEC 60044 – 1] ISN

CT secondary current 1A

IPN

CT primary current 1500/1000A

Using the above data VK can be calculated :

U KN  3.75 

U KN  3.75 

40000 5.5  0.05  1.8  1 1500

 735V

40000 3.67  0.05  1.8  1 1000

 828V

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

I V k  K td ( a ) 

SSC max( close.in. fault )

1.3  I PN

 RCT  R L  R R 

I V k  K td ( b) 

SSC zone1.end . fault )

1.3  I PN

 RCT  R L  R R 

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


24

PART I - Current Transformers Dimensioning [IEC 60044 – 1] RL

CT Secondary Loop Resistance =1.8  (For loop resistance ,2X100 m cable of 2.5 sqmm is considered )

RR


IPN


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 :

I V k  K td ( a ) 

SSC close.in . fault

1.3  I PN

 RCT  R L  R R 

For CT ratio 1500/1A

V k  4 

40000 1.3  1500

5.5  1.8  0.05

= 603.07 V< 1950 V


V k  4 

40000 1.3  1000

3.67  1.8  0.05

= 679.38V< 1300 V For Zone 1 end fault :

I V k  K td ( b ) 



1.3  I PN

25

 RCT  R L  R R 

PART I - Current Transformers Dimensioning [IEC 60044 – 1] For CT ratio 1500/1A

V k  5 

40000 1.3  1500

5.5  1.8  0.05

= 753.85 V< 1950 V For CT ratio 1000/1A

V k  5 

40000 1.3  1000

3.67  1.8  0.05

= 849.23V< 1300 V Hence the calculated knee point voltages for ratios 1500/1000/1A are satisfied as their values are less than the rated CT knee point voltages Hence the proposed CT cores for ratios 1500/1000/1A are acceptable 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 same SIEMENS Type 7SD522 , same all above C.T requirement equations and procedure shall be f ollowed : C.T Data : C.T Ratio

Vk

1500-1000-500/1

Class

1950 – 1300 – 650 V

Rct at 75º C 4.284 – 2.783 – 1.365 

X

RL at 75º C 0.75 

Rb 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 fault conditions and the rated knee point voltage should be greater than the calculated knee point voltage Since, effective CT accuracy limiting factor,

K OALF 

K OALF 

I SSC , max I N

40000 A 1500 A

and K OALF  30

 26.67 ,

40000 A 1000 A

 40

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


26

PART I - Current Transformers Dimensioning [IEC 60044 – 1] I V k  K td 

SSC max( Ext . fault )

I PN

    RCT  Rb   I SN  

Where Ktd

= 3.75 Transient Dimensioning F actor (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 (R B+2RL)

RL

One way lead resistance from CT to relay =0.75  (for Lead resistance ,100 m cable of 2.5 sqmm is c onsidered )

UKN

Rated Knee Point Voltage

ISN

CT secondary current 1A

IPN


U KN  3.75 

U KN  3.75 

40000 4.284  0.05  0.75  1 1500 40000 2.783  0.05  0.75  1 1000

 508.4V

 537.45V

CT Requirements for Main-2 Backup distance protection: The CT requirement to ensure correct operation of the distance protection relay

I V k  K td ( a ) 

SSC max( close.in. fault )

1.3  I PN

 RCT  R L  R R 

I V k  K td ( b) 


1.3  I PN

 RCT  R L  R R 

Where Ktd(a)

= 4 (for TS< 200ms)- Transient Dimensioning Factor For close – in faults


27

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Ktd(b)

= 5 (for TS< 200ms)- Transient Dimensioning Factor For zone-1 end fault

RCT

Secondary winding Resistance for each star connected CT

RL

CT Secondary Loop Resistance =0.75  (For loop resistance ,2X100 m cable of 2.5 sqmm is considered )

RR


IPN


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 :

I V k  K td ( a ) 

SSC close.in . fault

1.3  I PN

 RCT  R L  R R 


V k  4 

40000 1.3  1500

 4.284  0.75  0.05

= 417.15 V< 1950 V Acceptable . For CT ratio 1000/1A

V k  4 

40000 1.3  1000

 2.783  0.75  0.05

= 383.3V< 1300 V Acceptable . For Zone 1 end fault :

I V k  K td ( b ) 


1.3  I PN



28

 RCT  R L  R R 

PART I - Current Transformers Dimensioning [IEC 60044 – 1] V k  5 

40000 1.3  1500

 4.284  0.75  0.05

= 521.4 V< 1950 V Acceptable . For CT ratio 1000/1A

V k  5 

40000 1.3  1000

 2.783  0.75  0.05

= 551.2 V< 1300 V Acceptable . Hence the proposed CT cores for ratios 1500/1000/1A are 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 and better 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 precise calculation of Issc.max ( zone1 - end fault ) and and Ts (tp) , an example with ABB relay type RED 670 is detailed below for information :

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


29



30



31



32


Same must be repeated for the 1000 / 1 A ratio considering the Rct for the 1000/1 A ratio : Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

33




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


34

PART I - Current Transformers Dimensioning [IEC 60044 – 1] a) Relay Type SEL 311 L :

Same as above if I F = 40000 A is used, it will give better safety . Other wise , the actual Zs and Z L shall be calculated .


35

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 :


36


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 , the C.T must fulfill the Manufacturer recommendation , the following example with ABB REC 561 control and protection unit As per ABB REC 561 manual with Circuit breaker failure protection in REC 561 , CT core must meet the following requirement:

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


37


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


38

REL 511




Line distance protection Function in REL 511

According to ABB application manual for REL 511 the CT core must fulfill two requirements. The current transformer must have a rated equivalent secondary e.m.f. E al that is larger than or equal to the maximum of the required secondary e.m.f. E alreq (formulas 1 and 2.):

Ealreq calculated values with formula 1:


39


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 Earth Fault (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 for the Feeder circuit is applicable : 1. Differential Protection : A differential relay, of high impedance type should be used as main protection. CT’s should be specified at both the phase and the neutral side of each phase and a three phase protection should be used as a three phase protection gives a higher sensitivity for internal faults. The general requirement on the function values of the high impedance differential protection is that at maximum through f ault current for an 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 a through fault current equal to rated current, at an external earth-fault. The maximum inrush current for a reactor is approximately two times the rated current. If no specific requirement concerning at what current the relay should be stable exists, five times the rated current is used when operating voltage is selected. The function value of the relay should be chosen: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

40

PART I - Current Transformers Dimensioning [IEC 60044 – 1] U function  5I n  R ct  2Rl  where “In” is the reactors rated current at th e secondary side of the CT, “Rct” is CT secondary resistance at 75 °C and “RL” is the lead resistance, from the CT t o 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.

10  10  103

=

3  132

A  437.39 A

Corresponding secondary current is:



437.4 150

 2.916

The operating accuracy limit factor (K OALF):

 20.

 

 

KOALF=

15  0.75 2.02  0.75

 

 113.5

Hence, the operating accuracy limit factor > required accuracy limit factor Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

41

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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. V kn

  R ct  2Rl  R b  .I F

Where: Rct

= Resistance of the CT secondary circuit 0.75

Rl

= is the resistance of the secondary winding (longest lead l=120m, A=2.5mm , 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

2

48-3

10.S r

I F 



10  10  103

3 .u k .U n

3  132

A  437.38 A

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



437.38 150

 2.92

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

V kn  5  0.75  2.16  0.1  15.1 V kn  15.1V 2. V kn

 0.25 . I f  R ct  2Rl  R b 

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. Corresponding secondary current is:



40000 150

 266.67 A

V kn  0.25  266.67 0.75  2.16  0.1  200.67V V kn  200.67 V for 150A tap From the above calculation 1 and 2, the required knee point voltage shall be considered as follows:

V kn  220V Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

42

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 Class of Accuracy

150/1 A CLX

Introduction : The MICOM P122 restricted earth fault relay is a high impedance differential scheme which balances zero sequence current flowing in the transformer neutral against zero sequence current flowing in the transformer phase windings. Any unbalance for in-zone fault will result in an increasing voltage on the CT secondary and thus will activate the REF protection. This scheme is very sensitive and can then protect against low levels of fault current in resistance grounded systems where the earthing impedance and the fault voltage limit the fault current. In addition, this scheme can be used in a solidly grounded system. It provides a more sensitive protection, even though the overall differential scheme provides a protection for faults over most of the windings. The high impedance differential technique ensures that the impedance of the circuit is of sufficiently high impedance such that the differential voltage that may occur under external fault conditions is lower than the voltage required to drive setting current through the relay. This ensures stability against external f ault conditions and then the relay will operate only for faults occurring inside the protected zone. High impedance schemes are used in a differential configuration where one current transformer is completely saturated and the other CTs are healthy.

Voltage across relay circuit Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

43

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Vs  I R  2R L f CT Stabilizing resistor relay circuit

V

R If Where RR

 s - R

ST I s

R

= Maximum secondary through fault current = 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:

V  I R  2R s f CT L 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 of the relay under external fault conditions. This resistor will limit the spill current under Is. Vs = Is × ( RS T ) 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 :

Vs  I  A  C  T f

V

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 =

10 10 103 3 132

A  437.39 A


44

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 =

1 150

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

Vs  437.39   0.75  1.08  1.08 

1 150

 8.49V

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



  

 

Vs  437.39   0.75  1.08  1.08 

1 150

 8.49V

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



 

 

Vs  437.39   0.75  1.08 

1 150

 5.34V

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



 

Vs  437.39   0.75  1.08 

  1 150

 5.34V

From above calculations, Itemsa.1, a.2, a.3 and a.4 refer, the stability voltage setting should be considered as follows: Hence, Vs  10V For can be considered respectively f)

Knee Point Voltage (Vk)

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


45

PART I - Current Transformers Dimensioning [IEC 60044 – 1] V min  2 Vs  2 10V , V min  20V kn kn 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% and 20% of the minimum current available for an earth fault at the reactor terminals. Rated current at 132kV side 

Sn

3 132

10000



3  132

 43.74 A

The fault setting shall be between 4.374A and 8.748A The primary operating current (Ipo) for relay setting:

 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=

1 150

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 kneepoint voltage is : The value of series resistance is calculated as follows:

 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 Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

46

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 thermal conditions. The continuous power rating of the resistor is defined as







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







Where: Vf = rms voltage across resistor Ifs = Maximum secondary fault current which can be calculated from the circuit breaker rating, I ch, if the maximum internal fault current is not given. The maximum internal f ault current is usually the same as the maximum through fault current. Ifs = The maximum three phase through f ault current = 10 times rated current of t he protected winding

Hence, Ifs = The max. three phase through fault current.


47

PART I - Current Transformers Dimensioning [IEC 60044 – 1] =

10  10  103 3  132

A  437.39 A

Corresponding secondary current is:

437.4 = 150

 2.916

V f  4  450  333.3  2.92 1.3  709.43 3

OR

V f  4  450  1000  2.92  1.3  933.7 3

Thus the half second power rating is given by:

 = 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 :

V  I T  R  2R L  RS  R relay   CT f F   = 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 :











Where Vk ( Actual Vk of CT ) =2x

2  450 980.4  450 = 1381.8 V

C. Transformer Feeder : 1.

Differential Protection : Example using AREVA MICOM P633:

CT Data for T1L CORE-1:


48

PART I - Current Transformers Dimensioning [IEC 60044 – 1] CT ratio: 500-300/1A Class:

VA: 25/15

5P20

Rct:

 0.8/0.5

T2LN CORE-2 (TRF HV NCT): CT ratio:

300/1A

Class:

5P20

VA: 20 Rct:

 1.0

Verification of the accuracy limit factor: Required limiting factor

=

40000 300

 133.33

=

40000 500

 80

for CT ratio 500/1 A


For CT cables size 4sqmm used in secondary circuits: CT

Description

Burden (VA)

M1IDTP relay type MiCOM P633, AREVA

0.1

(87T1, HV 64REF1) T1L CORE-1

2×100m cable of 4sqmm

1.122

TOTAL

1.222<25/15 VA


 

 

KOALF=

 

Corrected VA = rated VA/1.25 = 25/1.25 = 20VA = 15/1.25 = 12VA

 20.  20.

20  0.8 1.222  0.8 12  0.5 1.222  0.5

 205.7  80


 145.18  133.33


Based on the AREVA recommendation for the relay type P633, the following formulae for the knee point voltage are applied: 1. For through faults:

V kn  I thr  R ct  2Rl  R b  Where: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

49

PART I - Current Transformers Dimensioning [IEC 60044 – 1] RCT

= CT secondary resistance

Rl

= CT secondary lead resistance 2

(Lead resistance from manufacturer catalogue: 5.62 ohm/km for A=4mm ) = 5.62×0.06=0.3372  (for lead length 60m) Rb

= Relay burden

= Ithr

VA



2 nom

I

0.1 1 2

 0.1 

= Max. three phase fault current

For 500/1A:

I thr 

S r

.

1

3 .u k .U n CT ratio



50.103 3  0.315  132



1 500

 1.39 A  1.5 A

For 300/1A:

I thr 

S r

.

1




50.103 3  0.315  132



1 300

 2.31 A  2.5 A

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

V kn  1.5 0.8  0.6744  0.1  2.362V

(for 500/1 A)

V kn  2.5 0.5  0.6744  0.1  3.186V

(for 300/1 A)

2. For internal faults:

V kn  0.25 . I f  R ct  2Rl  R b  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

I f  16.

S r

.

1

3 .U n CT ratio

 16.

 16.

50.103 3  132 50.10

3

3  132





1 500 1

300





3499 500

3499 300

 7 A

 11.66 A


V kn  0.25  7   0.8  0.6744  0.1  2.76V

(for 500/1 A)

V kn  0.25  12   0.5  0.6744  0.1  3.82V

(for 300/1 A)


50

(for 500/1 A)

(for 300/1 A)

PART I - Current Transformers Dimensioning [IEC 60044 – 1] ii) Max. fault current

I f 



I 3 FMax CT ratio

40000 300



40000 500

 80 A

 133.33 A

(for 500/1 A)

(for 300/1 A)

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

V kn  0.25  800.8  0.6744  0.1  31.49V

(for 500/1 A)

V kn  0.25  133.33 0.5  0.6744  0.1  42.48V

(for 300/1 A)

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

 70V


 100V


The rated knee point voltage =

 =

=





 20  0.8 

1  20

1.3

12  0.5 

1  20

1.3

=320 V


=192.31 V


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 for differential protection. As the line CTs for low impedance REF protection are the same for those used for differential protection, the differential CT requirements cover both differential and low impedance REF applications. CT Adequacy check for TRF LV side CTs T3L (TRF LV side CT) and T3LN CORE-2(11kV NCT) CT ratio: 3200/1A Class:

5P20

VA: 20 Rct:

 15.5

CT requirements: Based on the AREVA recommendations for the relay type P63X, the following formulae for the knee point voltage are applied:

1.

For through faults:

V kn  I thr  R ct  2Rl  R b  Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

51

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Where: RCT

= CT secondary resistance

Rl

= CT secondary lead resistance 2

(Lead resistance from manufacturer catalogue: 5.62 ohm/km for A=4mm ) = 5.62×0.1=0.562 (for lead length 100m) Rb

= Relay burden

= Ithr

VA 2 nom

I



0.1 1 2

 0.1 

= Max. three phase through fault current

S r

I thr 

1

.




50.103 3  0.315  12



1 3200



7637.13 3200

 2.387 A

To be on safety side, 2.5 A is considered further in calculation: Taking into account the above data, required Vkn is calculated as f ollows:

V kn  2.515.5  1.124  0.1  41.81V 2. For internal faults:

V kn  0.25 . I f  R ct  2Rl  R b  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

I f  16.

S r

.

1

3 .U n CT ratio

 16.

50.103 3  12



1 3200



3849.15 3200

 12.03 A


V kn  0.25  12.03  15.5  1.124  0.1  50.3V


52


For impedance calculations the following equations are applied: Source impedance calculation:

I 3 FMax 

1.1 . U n

1.1 . U n

(1)  Z HV 

(1) 3 . Z HV



3 . I 3 FMax

1.1  132 3  21.6

 3.88 

Power transformer impedances:









 

34.0 132 2   118.48 . 100 50

Total impedance at LV side:

   

  



2

118.48  12       3.88     0.358 3 132    

Max. fault current at 11kV bus:

I 3 FMax 

1.1 . U n (1) 3 . Z LV



1.1  11 3  0.358

 19.57kA

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

I f 

2

.

I 3 FMax

3 CT ratio

2 21600

 .

3 3200

 4.5 A

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

V kn  0.25  4.5  15.5  1.124  0.1  18.8V From the above calculation 1 and 2, the required knee point voltage shall be considered as follows: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

53

PART I - Current Transformers Dimensioning [IEC 60044 – 1] V kn  70V The rated knee point voltage =

=





16  15.5  1  20 1.3



=484.62 V

Therefore, Rated knee point voltage > Required Vk. 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 Data for T1L CORE-1: CT ratio: 500-300/1A Class:

VA: 25/15

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 for transformer differential protection must have a rated equivalent secondary e.m.f Eal that is larger the maximum of the required secondary e.m.f Ealreq below: Equation 1:

I . nt . sn E al  E alreq  30 I I pn

 S   RCT  R L  R2  I r  

Equation 2:

I . tf . sn E al  E alreq  2 I I pn

 S   RCT  R L  R2  I r  

Equation 3:

I  S  E al  E alreq  I f . sn  RCT  R L  R2  I pn  I r  Where: Int

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

=

50.103 3  132

= 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)


54

PART I - Current Transformers Dimensioning [IEC 60044 – 1] = 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 s econdary loop r esistance ( ) =0.75 (For loop resistance, 2×100m cable of 6sqmm is considered)

Equation 1:


 S   RCT  R L  R2  I r  

E al  E alreq  30  218.69 

0.02 1   0.8  0.75  2 500  1

   


0.02 1   0.5  0.75  2 300  1

   


0.02 1   0.8  0.75  2 500  1

   


0.02 1   0.5  0.75  2 300  1

   


= 20.6 V

E al  E alreq  30  218.69  =27.7 V Equation 2:


 S   RCT  R L  R2  I r  

E al  E alreq  2  3499.09  = 21.9 V

E al  E alreq  2  3499.09  =29.6 V Equation 3:

I  S  E al  E alreq  I f . sn  RCT  R L  R2  I pn  I r 

E al  E alreq  40000 

0.02 1   0.8  0.75  2 500  1

= 125.6 V Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

55

   


PART I - Current Transformers Dimensioning [IEC 60044 – 1] E al  E alreq  40000 

0.02 1   0.5  0.75  2 300  1

   


=169.3 V Required knee point voltage, V kn

=


 169.3V







=

 125.6V

 20  0.8 

1  20

1.3

12  0.5 

1  20

1.3

 =320 V


=192.31 V


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 for differential protection. As the line CTs for low impedance REF protection are the same for those used for differential protection, the differential CT requirements cover both differential and low impedance REF applications. Verification of the accuracy limit factor: Required limiting factor =

40000 300

 133.33



Description

Burden (VA)

M2IDTP relay type RET670, ABB

0.02

(HV 64REF2 protection) T2LN CORE-1


1.122

TOTAL

1.142<30 VA


 

 

KOALF=

 

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

 20.

16  1 1.142  1


 158.7  133.33

56


PART I - Current Transformers Dimensioning [IEC 60044 – 1] Since, the operating accuracy limiting factor is 158.7 being higher than the required accuracy limiting factor 133.3 for ratio of 300/1A. The proposed 5P20, 20VA current transformer is adequate.

11 KV Side : Proposed current transformer data: CT ratio: 3200/1A Class:

VA: 20

5P20

Rct:

 15.5

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

Equation 1:


 S   RCT  R L  R2  I r  

Equation 2:


 S   RCT  R L  R2  I r  

Where: Int

= the rated primary current of the power transformer (A) 3

50.10

=

3  12

= 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 s econdary loop r esistance ( ) =1.8 (For loop resistance, 2×100m cable of 2.5sqmm is considered)

SR

= the rated burden=0.02VA

Equation 1: E al

 E alreq

E al  E alreq  30  2406 

 0.02  15.5  1.8  2  3200  1  1

= 390.6 V


57

for CT ration 3200/1 A

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Equation 2: E al

 E alreq E al  E alreq  2  2406 

 0.02  15.5  1.8  2   3200  1  1

for CT ration 3200/1 A

= 26.04 V Required knee point voltage, Vk  390.6V There 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: Impedanc e: 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 Line Feeder. 3. Breaker Failure Protection: Protection: It was not shown as separate protection in the Line Feeder Protection; it is included as a Function in the Pilot Wire Differential Protection Protection and the Distance Protection. Example 1 : Using AREVA MICOM P142 – Breaker Failure Relay: Relay: CT Data for T1L CORE-3: CT ratio: 3000/1A Class:

VA: 30

5P20

Rct:

 

Verification of the accuracy limit factor: Required limiting factor =

40000 3000

 13.33

For CT cables size 4sqmm used in secondary circuits: CT T2LN CORE-3

Description M1BCP relay type P142, AREVA (50BF1 protection)


58

Burden (VA) 0.15

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 2×100m cable of 4sqmm

1.122

TOTAL

1.272<30 VA


 

 

KOALF=

 


 20.

24  8 1.272  8

 69  13.33


Since, the operating accuracy limiting factor is 69 being higher than the required accuracy limiting factor 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 MiCOM P142, AREVA. Non-directional/directional Non-directional/directional DT/IDMT overcurrent and earth fault protection.

   

Time-delayed phases overcurrent

  

Time-delayed earth fault overcurrent

       

Where Rl= Connected lead resistance=1.8 Ifp=Max. secondary fault current=

40000 3000

 13.33 A

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

 13.33    8  1.8  0.04  65.58V  2 

V k  

Earth fault overcurrent


59





 





PART I - Current Transformers Dimensioning [IEC 60044 – 1]  13.33    8  1.8  0.04   65.58V  2 

V k  








Corrected VA = 30/1.25 = 24 VA =

 24  8 

1  20

1.3

=492.3 V

Therefore, Rated knee point voltage > Required Vk. Hence , C.T is adequate and suitably sized. Example 2: using SIEMENS Type 7SJ64: with the same above C.T Data:

Verification of the accuracy limit factor: Required limiting factor =

40000 3000

 13.33


Description

Burden (VA)

M2IDTP relay type 7SJ64, SIEMENS (50BF2, 50/51 protection) T2L CORE-3


0.05 1.122

TOTAL

1.172<30 VA


 

 

KOALF=

 


 20.

24  8 1.172  8

 69.7  13.33


Since, the operating accuracy limiting factor is 69.7 being higher than the required accuracy limiting factor 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 = Vk 

I High set po int

1.3. I pn

Where: Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

60

 Rct  R' b . I sn

PART I - Current Transformers Dimensioning [IEC 60044 – 1] 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 k 

40000 1.3  3000

 8  1.85  1  101.02V








Corrected VA = 30/1.25 = 24 VA =

 24  8 

1  20

1.3

=492.3 V

Therefore, Rated knee point voltage > Required Vk. 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 the rated burden • For classes 0.1, 0.2 and 0.2S with rated burden < 15VA, an extended range of burden can be specified (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) occurring under power system fault conditions, inclusive of safety factors, up to which the transformer is required to meet performance requirements". CT – Magnetizing Curve Rated knee point e.m.f. (Ek) : That minimum sinusoidal e.m.f. (r.m.s.) at rated power frequency when applied to the secondary terminals 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) × Isn Where (Kx) is the dimensioning factor . Secondary voltage Esi: The secondary induced voltage Esi can be calculated from

E  I  Z si 2 Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

61




Z  R  R i b

 2  X b 2

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

E si B    2  f  A  N i 2 where: A B f N2

= = = =

core area in m2 ux density in Tesla (T) frequency number of secondary turns

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

100 K td  2  f T s



 10 %

Where , (Ts) is the Rated secondary loop time constant value of the time constant of the secondary loop of the current transformer obtained from the sum of the magnetizing and the leakage inductance (Ls) and the secondary loop resistance (Rs) Ts = Ls/Rs Flux overcurrent factor (nf) Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

62

PART I - Current Transformers Dimensioning [IEC 60044 – 1] nf  K ssc  K td The secondary core will be designed on the basis of the Kssc (AC-ux) and Ktd (DC ux).Typical value of Ktd is 10 - 25. For special measurement of fault current (transient current,) including both a.c. and d.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 the core 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 Saturation value increases



B-H characteristic of core



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



Fault current & DC offset : If a CT has been saturated the secondary current will not recover until 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 error Under steady-state conditions, the r.m.s. value of the difference between: a) the instantaneous values of the primary current; and b) 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 terminal markings The composite error Ec is generally expressed as a percentage of the r.m.s. values of the primary current according to the formula:

 



100 I p

1

T

 K i T 

n s

 i p  2 .dt

0

Use of composite error The numeric value of the composite error will never be less than the vector sum of the current error and the phase displacement (the latter being expressed in centiradians). Consequently, the composite error always indicates the highest possible value of current error or phase displacement. Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

63

PART I - Current Transformers Dimensioning [IEC 60044 – 1] The current error is of particular interest in the operation of overcurrent relays, and the phase displacement 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 transformers involved which must be considered. An additional advantage of a limitation of composite error is the resulting limitation of the harmonic content 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:

S n  R ct  I sn 2

n  F s 

S  R ct  I sn 2

where Sn = S Isn Rct

rated burden in VA = actual burden in VA = rated secondary current in A = internal resistance at 75ºC in ohm

To fulll high accuracy classes (e.g. class 0.2, IEC) the magnetizing current in the core must be kept at a low value. The consequence is a low ux density in the core. High accuracy and a low number of ampere-turns result in a high saturation factor (FS). To fulll high 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 the rated burden (IEC). With modern meters and instruments with low consumption the total burden can be lower than 25% of the rated burden (see Figure 2.1). Due to turns correction and core material the error may increase at lower burdens. To fulll accuracy requirements the rated burden of the metering core shall thus be relatively well matched to the actual burden connected. The minimum error is typically at 75% of the rated burden. The best way to optimize the core regarding accuracy is consequently 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 the connected 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 must be maintained within a small range, typically, from 0.95 to 1.05 pu of rated value. Due to the load variations, shunt reactors and capacitors have been applied in power systems to compensate excess Prepared by : Mohammad Subhi Abdel -–Halim Senior Substation Design Engineer PBME – AMMAN Office / JORDAN

64

PART I - Current Transformers Dimensioning [IEC 60044 – 1] reactive power (inductive for heavy load conditions, and capacitive for light load conditions). Shunt reactors are commonly used to compensate the capacitive reactive power of transmission and distribution systems and thereby to keep the operating voltages within admissible levels. Shunt reactors are used in high voltage systems to compensate for the capacitive generation of long overhead lines or extended 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 the networks, a higher level of active power can be transferred in the network. Reactors to limit overvoltages are most needed in weak power systems, i.e. when network short-circuit power is relatively low. Voltage increase in a system due to the capacitive generation is: U(%) 

Q c  100 S shc

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 of compensation to limit over-voltages will be less accentuated. Reactors to achieve reactive power balance in the different part of the network are most needed in heavy loaded networks where new lines cannot be built because of environmental reasons. Reactors for this purpose mostly are Thyristor controlled in order to adapt fast to the reactive power required. Especially in industrial areas with arc furnaces the reactive power demand is fluctuating between each half cycle. In such applications there are 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 the momentary demand. Four leg reactors also can be used for extinction of the secondary arc at singlephase 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 single phase reactor in the neutral the secondary arc can be extinguished and the single-phase auto-reclosing successful. The calculation of optimum ratings and points of connection of shunt reactors is generally done by means 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, with or without a circuit-breaker (see gure 1).


65


The shunt reactor is normally connected to the transmission line to avoid excess over voltage appearance when the load decreases at night and off days. Since the substations in the urban areas mostly have the transmission lines of small short-circuit capacities, the line voltages fluctuate largely when the shunt reactor of a large capacity is connected or di sconnected. Shunt reactors carry out different types of tasks: 





They compensate the capacitive reactive power of the transmission cables, in particular in networks with only light loads or no load. They reduce system-frequency over-voltages when a sudden load drop occurs or there is no load. 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-circuit power is relatively low. Mostly problems occur on gas-insulated circuit breaker for shunt reactor switching. The frequent switching 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 of inductive current switching caused re-striking phenomenon and incomplete tripping. These two mai n characteristics make the SF6 CB used for shunt reactor un-expectantly damaged . Therefore “ Separate Type Test Report from Accredited Testing Laboratory shall be submitted for 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 due to inductive switching can be depressed to reasonable and safely level by equipping suitable arrestor. The maintenance policy for the breakers should be planned based on a fixed period maintenance schedule or conditional basic maintenance schedule. Finally, the circuit breaker should be re-flashed after every 500 switching operations to maintain the power system normally Typically, the voltage variation at the high voltage bus bar after switching of a shunt reactor shall not be higher than 2 to 3% of rated voltage. MAIN CALCULATION OF SHUNT REACTORS: All the shunt reactor data and rating will be given in the contract specifications, this summary is just for information. For the calculation of the positive sequence reactance and the current requirements of a shunt reactor, it is necessary to know only the rated three-phase reactive power and the rated system voltage and frequency, as summarized in the table below.


66

PART I - Current Transformers Dimensioning [IEC 60044 – 1] Rating Reactance

WYE Connection

X R 

Delta Connection

2 U N

S R3



2 U N

X R  3 

3  S R1

2 U N

S R3



2 U N

3  S R1

Rayed Current

I N 

S R3 3  U N

Maximum Continuous Curre (Design Current Parameters



I MAX 

S R1 U N



3 U MAX U N

U N 3  X R

 I N

I N 

S R3 3  U N

I MAX 



S R1 U N

U MAX U N



U N X R

 I N

XR=Rated reactance per phase (positive sequence. SR3=Rated three phase reactive power SR1=Rated reactive power per phase UN=Rated system voltage UMAX=Maximum system operating voltage IN=Rated current IMAX=Maximum continuous current

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

  ZY L  B   .V 2  V 1   1   .V 2 V 1  A'.V 2  V 1   A   X 2 L  

R





R



Application Example Consider a lossless radial transmission line, frequency 60 Hz, length  = 350 km, and parameters z = j 0,32886  /km and bC = j 5,097 µS/km. To estimate the reactive power of shunt reactors to be installed in the transmission line to provide a maximum operating voltage of 1.05 pu at the open-circuited terminal (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

Z  j(z. )  j115.1 Y  j(b c . )  j1783.95 S Parameter A:

A  1 

ZY 2

 0.8973

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

V 2 

1 0.8973

.V 1  1.1144.V 1

Calculation of the shunt reactor reactance:


67

PART I - Current Transformers Dimensioning [IEC 60044 – 1]  ZY L   .V 2 V 1   1   2 L  R  

1.0   0.8973 



XR 

115.1  .1.05 X R 

115.1 1.0 1.05

 2090

 0.8973

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

S R3 

2 U N

X R



5252 2090

 132 MVAr

Calculation of the three-phase reactive power of the shunt reactor: 2 QC3  U N .bC .  5252  5.079  350  490 MVAr

So, compensation degree is:

K SH  2.

S R3 QC3

 54%

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

The shunt reactor rating is given by:

V 2 

V 1 S R3

1

 V  V   S R3  S CC  1 2   V 2  

S CC

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


68

C.T Dimensioning

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