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    TENDER NO. G1408 FIVE (5) NEW 33/11KV PRIMARY SUBSTATIONS WITH

    ASSOCIATED 33KV OVERHEAD LINES & 33KV CABLE LINES

    Doc-No:G1408-AA-2030-J-0-101 Page 2 of 39Rev - A

    Document Title : CT / VT calculations for 11kV switchgear forADWEA contract no G1408

    TABLE OF CONTENTS

    SECTION TITLE PAGE NO

    1.0 OBJECTIVE......................................................................................................3

    2.0 REFERENCES..................................................................................................3

    3.0 SCOPE..............................................................................................................3

    4.0 CT SIZING CALCULATIONS............................................................................3

    4.1 33/11.55kV, 15MVA Transformer Incomer Feeder........................................................ 34.1.1 CT's for Metering and Transducers.................................................................................. 34.1.2 CT's for Differential Protection & REF protection.............................................................. 44.1.3 CT for Standby Earth Fault(SBEF) protection ................................................................... 74.1.4 CT's for OverCurrent and Earth Fault protection ............................................................ 104.2 33/11.55kV, 20MVA Transformer Incomer Feeder...................................................... 134.2.1 CT's for metering and Transducers ................................................................................. 134.2.2 CT's for Differential Protection & REF protection............................................................ 144.2.3 CT for Standby Earth Fault(SBEF) protection ................................................................. 174.2.4 CT's for Overcurrent and Earth fault protection .............................................................. 194.3 33/11.55kV, 15MVA Transformer Bus sect ion Feeder ............................................... 22

    4.3.1 CT's for Metering and Protection..................................................................................... 224.3.2 CT's for Transducers ....................................................................................................... 244.4 33/11.55kV, 20MVA Transformer Bus sect ion Feeder ............................................... 254.4.1 CT's for Metering and Protection..................................................................................... 254.4.2 CT's for Transducers ....................................................................................................... 264.5 Outgoing 11/0.433kV, 500kVA Transformer Feeder................................................... 274.5.1 CT's for Metering and Protection..................................................................................... 274.5.2 CT's for Transducers ....................................................................................................... 294.6 Outgoing 11kV, 5MVAr Capacitor Feeder ................................................................... 304.6.1 CT's for Metering and Protection..................................................................................... 304.6.2 CT's for Transducers ....................................................................................................... 334.7 11kV Outgoing Feeder .................................................................................................. 344.7.1 CT's for Metering and Protection..................................................................................... 34

    4.7.2 CT's for Transducers ....................................................................................................... 37

    5.0 VT SIZING CALCULATIONS.......................................................................... 38

    6.0 CONCLUSION ................................................................................................ 39

    7.0 SUMMARY OF CURRENT TRANSFORMERS(Annexure-1)..........................39

    8.0 SUMMARY OF VOLTAGE TRANSFORMERS(Annexure-2)..........................39

    9.0 Relavent pages of Catalogues for Burden datas(Annexure-3) ................... 39

    NOTE

    Refer DTS no 0012 dated 15.11.2003 for Relay Catalogues.

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    TENDER NO. G1408 FIVE (5) NEW 33/11KV PRIMARY SUBSTATIONS WITH

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    Document Title : CT / VT calculations for 11kV switchgear forADWEA contract no G1408

    1.0 OBJECTIVE

    To establish Current transformer(CT) and Voltage transformer parameters at 11 kVlevel for all the 5 new substations.

    2.0 REFERENCES

    a) Bay control unit REF542 Plus catalogue

    b) 11kV switchgear vendor (ABB FJH) Single line diagram no. N611844/1 toN611844/8

    c) Ducab cable catalogue

    d) ESI standard 48-3 and Clients/Consultant Recommendations

    e) Relay catalogues

    3.0 SCOPE

    To establish the requirements of CT/VT parameters such as CT/VT ratio, VA burden,knee point voltage, accuracy class and magnetising current for CTs on the 11 kVswitchgear at all the sub-stations.

    4.0 CT SIZING CALCULATIONS

    4.1 33/11.55kV, 15MVA Transformer Incomer Feeder

    4.1.1 CTs for Metering and Transducers

    CT Ratio:

    800-900 / 1A

    Type of Metering:

    It is proposed to use Bay Control Unit(BCU) type REF542 Plus.

    Transformer full load current on secondary side ( considering tap setting at 15% )

    = 15000/(3*11.55*0.85)

    = 882.12 Amps

    Hence it is proposed to provide 800-900 / 1A CT for metering.

    Total burden of bay control unit = 0.1 VA (Refer REF542 Plus catalogue)

    Maximum burden of Current Transducer = 2 VA (Actual is 0.2VA)Maximum burden of Power Transducer = 2 VAMaximum burden of Power factor Transducer = 2 VAThe transducers are located in LDC cubicle.

    Minimum length of cable = 40 meters(As per equipment layout)between CT and transducers

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    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.42

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.42 x 12

    = 0.42 VA

    Total burden = Burden of BCU+Burden of transducers+Burden due to lead resistance= 0.1 + 6 + 0.42 = 6.52 VA

    Considering 25% future margin

    Total burden required =1.25 x 6.52 = 8.15 VA

    Hence a standard burden rating of 15 VA is chosen for a Tap of 800A and 30VAchosen for a Tap of 900A.

    The CT Accuracy Class selected is CL. 0.5S as per the specification requirements.

    The instrument security factor (ISF) selected is less than or equal to Five(5) as perthe specification requirements.

    Abstract

    Therefore a CT of CL. 0.5,15VA is proposed for 800-900/1A and CL. 0.5,30VA isproposed for 800-900/1A.The CTs are with a factor of safety less than or equal to 5.

    4.1.2 CTs for Differential Protection and REF Protection:

    CT Ratio:

    900/1

    Type of relay: RET 316*4 (ABB make)

    Knee Point Voltage for Calculation for Differential Protection:

    Fault current rating of 11 kV switchgear = 31.5 kA

    As per ESI Standard 48-3 enclosed the formulae for calculating Knee Point Voltage is

    VK = If * N * (Rct+Rl)

    Where If= fault current

    N = Turns Ratio

    Rct= CT secondary resistance

    Rl= Lead resistance

    Maximum fault current carrying capacity of 11 kV switchgear is 31.5 kA ,which can be

    considered for knee point calculation.

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    If= 31500 Amps

    N = 1/900

    Rct = 7(As per Manufacturers Recommendations)

    The differential relay is mounted on 33kV relay panel which is located in control room.

    Minimum length of cable = 40 meters(As per equipment layout)between CT and relay

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/kmTotal resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.42

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.42 x 12

    = 0.42 VA

    Therefore, substituting the above values in the formula for Vk,

    VK= 31500* (1/900) * (7 + 0.42 )

    VK= 259.7V

    VK> 400V

    Therefore a CT of 900/1A, CL. X, VK> 400V is proposed. Imag shall be 30mA @Vk/2

    Recommendations from Consultant/Client for Differential Core requirement :

    Further, to avoid mal-operation on energization of power transformer and inconnection with fault current that passes through power transformer, the ratedsecondary voltage has to satisfy the following conditions:

    Condition-1

    The core may not saturate for current lower than 30 times the PowerTransformer rated current at connected burden. This ensures stability alsowith heavy DC saturation.

    Vk 30 * Int* (Rct+ Rl+ Rr/ Ir2)

    Where,

    Int = Main CT secondary current corresponding to rated primary

    current of power transformer

    Rct = CT secondary resistance

    Rl = Lead resistanceRr = Relay burden

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    Ir = Nominal relay current

    Full load current of the transformer = (15000 / 1.732 * 11) = 787 Amps.

    Vk 30 * (787/900) * (7+0.42 + 0.1 / 12)

    Vk 187V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

    Condition-2

    The core may not saturate for current lower than 4 times the maximumthrough fault current at connected burden.

    Vk 4 * Ift* (Rct+ Rl+ Rr/ Ir2)

    Where,

    Ift = Maximum secondary side through fault current

    Rct = CT secondary resistance

    Rl = Lead resistance

    Rr = Relay burden

    Ir = Nominal relay current

    Full load current of the transformer = (15000 / 1.732 * 11) = 787 Amps.

    Impedance of transformer = 10%

    Through fault current = 787 / 0.1 = 7870 Amps

    Secondary side through fault current = 7870 / 900 = 8.75 Amps.

    Hence Vk 4 * (7870/900) * (7+ 0.42 + 0.1 / 12)

    Vk 263 V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

    Abstract

    Therefore a CT of Vk > 400V is proposed for 900/1A.Imag shall be 30mA @ Vk/2

    REF Protection Core Knee Point Calculations

    CT Ratio:

    900/1

    Type of relay: SPAJ 115C (ABB make)

    Fault current rating of 11 kV switchgear = 31.5 kA

    The REF relay is mounted on 33kV relay panel, which is located in control room.

    The knee-point voltage Vk should be 2 times higher than the stabilising voltage Vs

    required in through fault conditions:Vk = 2 x Vs = 2 x Ikmaxx (Rct+ Rl) / n

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    Where,

    Vs = stabilising voltage

    Ikmax = the maximum through fault current

    According to ESI Standard 48, the voltage set on the relay is chosen such that thestability of the protection during a through fault being taken as 16 times of the ratedtransformer current.

    Ikmax = 16 x Transformer rated secondary current = 16 x 882.12 = 14114 amp

    Rct = Internal resistance of the current transformer = 7 ohms

    Rl = total lead resistance = 0.42 ohms

    n = turns ratio of the current transformer = 900

    Substituting the above values in the equation the stabilising voltage,

    Vs = 16 x 882.12 x (7 + 0.42) / 900= 116.36 Volts

    Hence, Vk = 2 x Vs

    Vk = 2 x 116.36 V = 232.72 V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

    Abstract

    Since the CT Requirement of the REF and Differential Core is the same, we proposeto use a common core for both. Therefore a CT of Vk > 400V is proposed for 900/1A.

    Imag shall be 30mA @ Vk/2

    4.1.3 CT for Standby Earth Fault (SBEF) Protect ion:

    CT Ratio:

    800-900/1(As calculated in 4.1.1 above)

    This CT is located on the transformer neutral.

    Type of relay: REJ 521 (ABB make)

    Fault current rating of 11kV switchgear = 31.5kA

    The SBEF relay is mounted on 33kV relay panel which is located in control room.

    Burden of relay = 0.1VA (As per relay catalogue)

    Minimum length of cable = 50 meters(As per equipment layout)between CT and relay

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

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    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 50 / 1000

    (Lead & Return conductors) = 0.5244

    Burden due to lead resistance = Total resistance x (CT secondary current)

    2

    = 0.5244 x 12

    = 0.5244 VA

    Total burden = Burden of the relay + Burden due to leadresistance

    = 0.1 + 0.5244

    = 0.6244 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.6244 = 0.78 VA

    Hence a standard burden rating of 10 VA is chosen for a Tap of 800A and a burdenof 15 VA is chosen for a Tap of 900A.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 800A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 800A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 10 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 3Ohm (Assumed)

    = 12x 3

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    = 3 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.5244 = 0.6244 VA

    10 + 3

    Koalf = 20 x0.6244 + 3

    = 71

    Iscc/ Ipn = 31500 / 800

    = 40

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 800 900 / 1, 5P20, 10VA is adequate.For a Tap of 900A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 900A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 4Ohm (As per Manufacturers recommendations)

    = 12x 4

    = 4 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.5244 = 0.6244 VA

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    15 + 4

    Koalf = 20 x0.6244 + 4

    = 82.17Iscc/ Ipn = 31500 / 900

    = 35

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 800 900 / 1, 5P20, 15VA is adequate

    Abstract

    Therefore a CT of CL. 5P20,10VA is proposed for 800-900/1A and CL. 5P20,15VA isproposed for 800-900/1A.

    4.1.4 CTs for Overcurrent and Earth Fault Protection:

    CT Ratio:

    800-900/1

    Type of relay: SPAJ 140C (ABB make)

    Fault current rating of 11kV switchgear = 31.5kA

    The relay is mounted on 11kV switchgear.

    Burden of relay = 0.1VA(As per relay catalogue)

    Minimum length of cable = 5 meters

    between CT and relay

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) ) = 8.43 ohm/km55 deg C, consideringtemperature correction factorTotal resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.0843

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12

    = 0.0843 VA

    Total burden = Burden of the relay + Burden due to leadresistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.1843 = 0.23 VA

    Hence a standard burden rating of 10 VA is chosen for a Tap of 800A and a burdenof 15 VA is chosen for a Tap of 900A .

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    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 800A

    To ensure correct operation of the connected relay in case of faults, the CT must be

    able to transform the maximum symmetrical short circuit current without saturation.To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 800A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalf x

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 10 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 3Ohm (Assumed)

    = 12x 3

    = 3VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.0843 = 0.1843 VA

    15 + 3

    Koalf = 20 x0.1843 + 3

    = 113

    Iscc/ Ipn = 31500 / 800

    = 40

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 800 900 / 1, 5P20, 10VA is adequate.

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    For a Tap of 900A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 900A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 4

    = 12x 4

    = 4 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.0843 = 0.1843 VA

    15 + 4

    Koalf = 20 x

    0.1843 + 4= 90.81

    Iscc/ Ipn = 31500 / 900

    = 35

    Koalf> Iscc/ Ipn

    Hence the selected CT wi th parameters 800 900 / 1, 5P20, 15VA is adequate.

    Abst ract

    Therefore a CT of CL. 5P20,10VA is proposed for 800-900/1A and CL. 5P20,15VA isproposed for 800-900/1A.

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    4.2 33/11.55kV, 20MVA Transformer Incomer Feeder

    4.2.1 CTs for Metering and Transducers

    CT Ratio:900-1200/1A

    Type of Metering:

    It is proposed to use Bay Control Unit(BCU) type REF542 Plus for the purpose ofmetering.

    Transformer full load current on secondary side ( considering tap setting at 15% )

    = 20000/(3*11.55*0.85)

    = 1176.16 Amps

    Hence it is proposed to provide 900-1200/1A CT for metering.

    Total burden of bay control unit = 0.1 VA (Refer REF542 Plus catalogue)

    Maximum burden of Current Transducer = 2 VA (Actual is 0.2VA)Maximum burden of Power Transducer = 2 VAMaximum burden of Power factor Transducer = 2 VAThe transducers are located in LDC cubicle.

    Minimum length of cable = 60 meters(As per equipment layout)between CT and Transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.422 ohm/km

    Total resistance = 2 x 5.422 x 60 / 1000

    (Lead & Return conductors) = 0.63

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.63 x 12

    = 0.63 VA

    Total burden = Burden of BCU+Burden of transducers+Burden due to lead resistance= 0.1 + 6 + 0.63 = 6.73 VA

    Considering 25% future margin

    Total burden required =1.25 x 6.73 = 8.41 VA

    Hence burden of 10VA is chosen for 900A Tap and 15VA is chosen for 1200A Tap.

    The instrument security factor (ISF) selected is less than or equal to 5, as per the

    specification requirements.

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    Abstract

    Therefore a CT of CL. 0.5,10VA is proposed for 900-1200/1A and CL. 0.5,15VA isproposed for 900-1200/1A.The CTs are with a factor of safety less than or equal to 5.

    4.2.2 CTs for Differential Protection and REF Protection:

    CT Ratio:

    Ratio is considered as 1200/1(As calculated in 4.2.1 above)

    Type of relay: RET 316*4 (ABB make)

    Knee Point Voltage for Calculation for Differential Protection:

    Fault current rating of 11 kV switchgear = 31.5 kA

    As per ESI Standard 48-3 enclosed the formulae for calculating Knee Point Voltage is

    VK = If * N * (Rct+Rl)

    Where If= fault current

    N = Turns Ratio

    Rct= CT secondary resistance

    Rl= Lead resistance

    Maximum fault current carrying capacity of 11 kV switchgear is 31.5 kA which can beconsidered for knee point calculation.

    If= 31500 AmpsN = 1/1200

    Rct = 12(As per manufacturers recommendations)

    The differential relay is mounted on 33kV relay panel which is located in control room.

    Minimum length of cable = 50 meters(As per equipment layout)between CT and relay

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 50 / 1000

    (Lead & Return conductors) = 0.5244

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.5244 x 12

    = 0.5244 VA

    Therefore, substituting the above values in the formula for Vk,

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    VK= 31500* (1/1200) * (12 + 0.5244 )

    VK= 329V

    VK> 400V

    Therefore a CT of 1200/1A, CL. X, VK> 400V is proposed. Imag shall be 30mA @Vk/2

    Recommendations from Consultant/Client for Differential Core requirement :

    Further, to avoid mal-operation on energization of power transformer and inconnection with fault current that passes through power transformer, the ratedsecondary voltage has to satisfy the following conditions:

    Condition-1

    The core may not saturate for current lower than 30 times the Power

    Transformer rated current at connected burden. This ensures stability alsowith heavy DC saturation.

    Vk 30 * Int* (Rct+ Rl+ Rr/ Ir2)

    Where,

    Int = Main CT secondary current corresponding to rated primary

    current of power transformer

    Rct = CT secondary resistance

    Rl = Lead resistance

    Rr = Relay burdenIr = Nominal relay current

    Full load current of the transformer = (20000 / 1.732 * 11) = 1050Amps.

    Vk 30 * (1050/1200) * (10.5+0.5244 + 0.1 / 12)

    Vk 292V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

    Condition-2

    The core may not saturate for current lower than 4 times the maximumthrough fault current at connected burden.

    Vk 4 * Ift* (Rct+ Rl+ Rr/ Ir2)

    Where,

    Ift = Maximum secondary side through fault current

    Rct = CT secondary resistance

    Rl = Lead resistance

    Rr = Relay burden

    Ir = Nominal relay currentFull load current of the transformer = (20000 / 1.732 * 11) = 1050 Amps.

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    Impedance of transformer = 10%

    Through fault current = 1050 / 0.1 = 10500 Amps

    Secondary side through fault current = 10500 / 1200 = 8.75 Amps.

    Hence Vk 4 * (10500/1200) * (10.5+ 0.5244 + 0.1 / 12)

    Vk 389 V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

    Abstract

    Therefore a CT of Vk > 400V is proposed for 1200/1A.Imag shall be 30mA @ Vk/2

    REF Protection Core Knee Point Calculations

    CT Ratio:

    1200/1

    Type of relay: SPAJ 115C (ABB make)

    Fault current rating of 11 kV switchgear = 31.5 kA

    The REF relay is mounted on 33kV relay panel, which is located in control room.

    The knee-point voltage Vk should be 2 times higher than the stabilising voltage Vsrequired in through fault conditions:

    Vk = 2 x Vs = 2 x Ikmaxx (Rct+ Rl) / nWhere,

    Vs = stabilising voltage

    Ikmax = the maximum through fault current

    According to ESI Standard 48, the voltage set on the relay is chosen such that thestability of the protection during a through fault being taken as 16 times of the ratedtransformer current.

    Ikmax = 16 x Transformer rated secondary current = 16 x 1050 = 16796 amp

    Rct = Internal resistance of the current transformer = 10.5 ohms

    Rl = total lead resistance = 0.524 ohms

    n = turns ratio of the current transformer = 1200

    Substituting the above values in the equation the stabilising voltage,

    Vs = 16 x 1050 x (10.5 + 0.5244) / 1200

    = 154 Volts

    Hence, Vk = 2 x Vs

    Vk = 2 x 154 V = 308V

    The knee point voltage proposed by us is 400V.

    Hence this condition is Verified.

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    Abstract

    Since the CT Requirement of the REF and Differential Core is the same, we proposeto use a common core for both. Therefore a CT of Vk > 400V is proposed for1200/1A.Imag shall be 30mA @ Vk/2 .

    4.2.3 CT for Standby Earth Fault(SBEF) Protect ion:

    CT Ratio:

    Ratio is considered as 900-1200/1(As calculated in 4.2.1 above)

    Type of relay: REJ 521 (ABB make)

    The CT is mounted on the transformer neutral.

    Fault current rating of 11kV switchgear = 31.5kA

    The SBEF relay is mounted on 33kV relay panel which is located in control room.

    Burden of relay = 0.1VA(As per relay catalogue)

    Minimum length of cable = 50 meters(As per equipment layout)between CT and relay

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, considering

    temperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 50 / 1000

    (Lead & Return conductors) = 0.5244

    Total burden = Burden of the relay + Burden due to leadresistance

    = 0.1 + 0.5244

    = 0.6244 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.6244 = 0.78 VA

    Hence burden rating of 10 VA is chosen for 900A Tap and 15VA is chosen for 1200A.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 900A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 900A

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    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limiting

    factor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 10VA

    Pi = Internal CT burden= V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 3Ohm(Assumed)

    = 12x 3

    = 3 VA

    Pb = Total connected burden

    = Pr+ PL= 0.1 + 0.5244 = 0.6244 VA

    10 + 3

    Koalf = 20 x0.6244 + 3

    = 71

    Iscc/ Ipn = 31500 / 900

    = 35

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 900 1200 / 1, 5P20, 10VA is adequate.

    For a Tap of 1200A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 1200A

    Iscc

    = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

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    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ Pi

    Koalf = KnalfxPb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I

    2

    * RctRct = 5(As confirmed by manufacturer)

    = 12x 5

    = 5 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.5244 = 0.6244 VA

    15 + 5

    Koalf = 20 x0.6244 + 5

    = 71.11

    Iscc/ Ipn = 31500 / 1200

    = 26.25

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 900 1200 / 1, 5P20, 15VA is adequate.

    Abstract

    Therefore a CT of CL. 5P20,10VA is proposed for 900-1200/1A and CL. 5P20,15VA

    is proposed for 900-1200/1A.The same is as per the specification.

    4.2.4 CTs for Overcurrent and Earth Fault Protection:

    CT Ratio:

    900-1200/1(As calculated in 4.2.1 above)

    Type of relay: SPAJ 140C (ABB make)

    Fault current rating of 11kV switchgear = 31.5kA

    The relay is mounted on 11kV switchgear.

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    Burden of relay = 0.1VA(As per relay catalogue)

    Minimum length of cable = 5 metersbetween CT and relay

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) ) = 8.43 ohm/km55 deg C, consideringtemperature correction factor

    Total resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.0843

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12

    = 0.0843 VA

    Total burden = Burden of the relay + Burden due to leadresistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.1843 = 0.23 VA

    Hence burden rating of 10 VA is chosen for 900A Tap and 15VA is chosen for 1200A.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 900ATo ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 900A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 10VA

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    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 3Ohm(Assumed)

    = 12x 3

    = 3 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.0843

    = 0.1843 VA

    10 + 3

    Koalf = 20 x0.1843+ 3

    = 82

    Iscc/ Ipn = 31500 / 900

    = 35

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 900 1200 / 1, 5P20, 10VA is adequate.

    For a Tap of 1200ATo ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 1200A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limiting

    factor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

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    = I2* Rct

    Rct = 5(As confirmed by manufacturer)

    = 12x 5

    = 5 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.0843

    = 0.1843 VA

    15 + 5

    Koalf = 20 x0.1843 + 5

    = 77.15Iscc/ Ipn = 31500 / 1200

    = 26.25

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 900 1200 / 1, 5P20, 15VA is adequate.

    Abstract :

    Therefore a CT of CL. 5P20, 10VA is proposed for 900-1200/1A and CL. 5P20,15VAis proposed for 900-1200/1A.

    4.3 33/11.55kV, 15MVA Transformer Bus-section Feeder

    4.3.1 CTs for Metering and Protect ion

    CT Ratio:

    1600 /1A

    Type of Metering and Protection:

    It is proposed to use Bay Control Unit(BCU) type REF542 Plus for the same

    Total burden of bay control unit = 0.1 VA(Refer REF542 Plus catalogue)

    Minimum length of cable = 5 metersbetween CT and BCU

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 8.43 ohm/km

    Total resistance = 2 x 8.43 x 5 / 1000(Lead & Return conductors) = 0.0843

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    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12

    = 0.0843 VA

    Total burden = Total burden of the BCU + Burden due to leadresistance

    = 0.1 + 0.0843 = 0.1843 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.1843 = 0.23 VA

    Hence burden rating of 15 VA is chosen.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    Therefore a CT of 1600/1A, CL. 5P20, 15VA is proposed.

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 1600A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 8Ohm(As confirmed by the manufacturer)

    = 12x 8

    = 8 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 +0.0843

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    = 0.1843 VA

    15 + 8

    Koalf = 20 x

    0.1843 + 8= 56.2

    Iscc/ Ipn = 31500 / 1600

    = 19.68

    Koalf> Iscc/ Ipn

    Abstract

    Therefore a CT of CL. 5P20,15VA is proposed with for 1600/1A

    4.3.2 CTs for Transducers

    CT Ratio:

    1600 /1A CT

    Type of Metering:

    We propose to use Transducers for the same

    Maximum burden of current transducer = 2 VA (Actual Burden is 0.2VA only)

    The transducers are located in LDC cubicle.

    Minimum length of cable = 40 metersbetween CT and transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg CResistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.419 Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.419 x 12

    = 0.419 VA

    Total burden = Total burden of the transducer + Burden due to leadresistance

    = 2 + 0.419 = 2.419 VA

    Considering 25% future margin

    Total burden required =1.25 x 2.419 = 3 VAHence burden rating of 15 VA is chosen.

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    The CT Accuracy Class selected is CL. 0.5 as per the specification requirements.

    The instrument security factor (ISF) selected is less than or equal to Five(5) as perthe specification requirements.

    Abstract

    Therefore a CT of 1600/1A, CL. 0.5FS5, 15VA is proposed with factor of safety lessthan or equal to 5.

    4.4 33/11.55kV, 20MVA Transformer Bus-section Feeder

    4.4.1 CTs for Metering and Protect ion

    CT Ratio:

    2500 /1A

    Type of Metering and Protection:

    It is proposed to use Bay Control Unit(BCU) type REF542

    It is proposed to provide 2500/1A CT for metering and protection.

    Total burden of bay control unit = 0.1 VA(Refer REF542 Plus catalogue)

    Minimum length of cable = 5 metersbetween CT and BCU

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)

    km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 8.43 ohm/km

    Total resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.08429

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12

    = 0.0843 VATotal burden = Total burden of the BCU + Burden due to lead

    resistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.1843 = 0.23 VA

    Hence burden rating of 15 VA is chosen.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    Therefore a CT of 2500/1A, CL. 5P20, 15VA is proposed.

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    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ IpnWhere, Ipn = CT primary nominal current = 2500A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalf x

    Pb+ Pi

    Where,

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    = 12x 10

    = 10 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 +0.0843 = 0.1843 VA

    15 + 10

    Koalf = 20 x0.1843 + 10

    = 49

    Iscc/ Ipn = 31500 / 2500

    = 12.6Koalf> Iscc/ Ipn

    Abstract

    Therefore a CT of CL. 5P20,15VA is proposed with for 2500/1A

    4.4.2 CTs for Transducers

    CT Ratio:

    2500 /1A CT

    Type of Metering:

    We propose to use Transducers for the same

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    Maximum burden of current transducer = 2 VA (Actual is 0.2VA )

    The transducers are located in LDC cubicle.Minimum length of cable = 40 metersbetween CT and transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.419

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.419 x 12

    = 0.419 VA

    Total burden = Total burden of the transducer + Burden due to leadresistance

    = 2 + 0.419 = 2.419 VA

    Considering 25% future margin

    Total burden required =1.25 x 2.419 = 3 VA

    Hence burden rating of 15 VA is chosen.

    The CT Accuracy Class selected is CL. 0.5 as per the specification requirements.

    The instrument security factor (ISF) selected is less than Five(5) as per thespecification requirements.

    Abstract

    Therefore a CT of 2500/1A, CL. 0.5FS5, 15VA is proposed with factor of safety lessthan or equal to 5.

    4.5 Outgoing 11/0.433kV, 500kVA Transformer Feeder

    4.5.1 CTs for Metering and Protect ion

    CT Ratio:

    30 /1A

    Type of Metering and Protection:

    It is proposed to use Bay Control Unit(BCU) type REF542

    Transformer full load current on primary side = 500/(3*11)

    = 26.24 Amps

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    Hence 30/1A CT is selected for metering and protection as per the specificationrequirements.

    Total burden of bay control unit = 0.1 VA(Refer REF542 Plus catalogue)

    Minimum length of cable = 5 metersbetween CT and BCU

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 8.43 ohm/kmTotal resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.0843

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12 = 0.0843 VA

    Total burden = Total burden of the BCU + Burden due to leadresistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required =1.25 x 0.1843 = 0.23 VA

    Hence burden rating of 6 VA is chosen.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    Therefore a CT of 30/1A, CL. 5P20, 6VA is proposed.

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 30A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

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    Knalf = 20

    Pn = Nominal CT burden = 6 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 0.1(As confirmed by manufacturer)

    = 12x 0.1

    = 0.1 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 +0.0843 = 0.1843 VA

    6+ 0.1

    Koalf = 20 x0.1843 + 0.1

    = 430

    Transformer Short Circuit current is limited by the 5% impedence of theTransformer,

    Hence,

    Iscc = 500/(3*110.05)

    = 530 Amps

    Iscc/ Ipn = 530 / 30X 0.05

    = 17

    Koalf> Iscc/ Ipn

    Thus from the above calculations its also evident that,the actual short Circuitwithstand of the CT is 430 x 30 =12.9kA

    Note: As the ratio of CT for auxiliary transformer is very low i.e 30 / 1A as per

    the specification requirement, the best CT size realized to be accommodatedwill be 30/1A, 5P20, 6VA with 31.5kA for 1sec short circuit level. Due tolimitation on CT dimension imposed by low CT ratio and CT short circuitwithstand capacity restr icted to 31.5kA for 1 sec., We request you kind approvalfor the same.

    Abstract

    Therefore a CT of CL. 5P20,6VA is proposed with for 30/1A

    4.5.2 CTs for Transducers

    CT Ratio:

    30 /1A CT

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    Type of Metering:

    We propose to use Transducers for the same

    Maximum burden of current transducer = 2 VA (Actual is 0.2VA )

    The transducers are located in LDC cubicle.Minimum length of cable = 40 metersbetween CT and transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg CResistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.419

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.419 x 12

    = 0.419 VA

    Total burden = Total burden of the transducer + Burden due to leadresistance

    = 2 + 0.419 = 2.419 VA

    Considering 25% future margin

    Total burden required =1.25 x 2.419 = 3 VA

    Hence burden rating of 7.5 VA is chosen.

    The CT Accuracy Class selected is CL. 0.5 as per the specification requirements.

    The instrument security factor (ISF) selected is less than or equal to Five(5) as perthe specification requirements.

    Abstract

    Therefore a CT of 30/1A, CL. 0.5FS5, 7.5VA is proposed with factor of safety lessthan or equal to 5.

    4.6 Outgoing 11kV, 5MVAr Capacitor Feeder

    4.6.1 CTs for Metering and Protect ion

    CT Ratio:

    300-600 / 1A

    Type of Metering and Protection:

    It is proposed to use Bay Control Unit(BCU) type REF542

    Capacitor full load current = MVAr / ( 3 * kV * sin)

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    = 5000 / (3 * 11 * 0.6) = 437.38 Amps

    Hence it is proposed to provide 300-600 / 1A CT for metering and protection.

    Total burden of bay control unit = 0.1 VA (Refer REF542 Plus catalogue)

    Minimum length of cable = 5 metersbetween CT and BCU

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 8.43 ohm/km

    Total resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.0843

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.0843 x 12

    = 0.0843 VA

    Total burden = Total burden of the BCU + Burden due to leadresistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required = 1.25 x 0.1843 = 0.23 VA

    Hence burden range of 7.5VA is proposed for 300A Tap and 15VA is proposed for600A Tap.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 300A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 300A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

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    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ Pi

    Koalf = KnalfxPb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 7.5 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I

    2

    * RctRct = 1.5Ohm (Assumed)

    = 12x 1.5

    = 1.5 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.08429 = 0.1843 VA

    7.5 + 1.5

    Koalf = 20 x0.1843 + 1.5

    = 109

    Iscc/ Ipn = 31500 / 300

    = 10.5

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 300-600 / 1, 5P20, 7.5VA is adequate.

    For a Tap of 600A

    To ensure correct operation of the connected relay in case of faults, the CT must be

    able to transform the maximum symmetrical short circuit current without saturation.To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 600A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

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    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 3 Ohm

    = 12x 3

    = 3VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.08429 = 0.1843 VA

    15 + 3

    Koalf = 20 x0.1843 + 3

    = 113

    Iscc/ Ipn = 31500 / 600

    = 52.5

    Koalf> Iscc/ Ipn

    Hence the selected CT wi th parameters 300-600 / 1, 5P20, 15VA is adequate

    Abstract

    Therefore a CT of CL. 5P20,7.5VA is proposed for 300-600/1A and CL. 5P20,15VA isproposed for 300-600/1A.

    The same is as per the specification.4.6.2 CTs for Transducers

    CT Ratio:

    300 600 /1A

    Type of Measurement:

    We Propose to use Transducers form the same

    Maximum burden of Current transducer = 2 VA (Actual is 0.2VA)

    Maximum burden of Power factor transducer = 2 VA

    Maximum burden of MVAr transducer = 2 VA

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    Total burden of transducers = 6VA

    The transducers are located in LDC cubicle.

    Minimum length of cable = 40 metersbetween CT and transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000(Lead & Return conductors) = 0.419

    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.419 x 12

    = 0.419 VA

    Total burden = Total burden of transducers + Burden due to leadresistance

    = 6 + 0.419 = 6.419 VAConsidering 25% future margin

    Total burden required =1.25 x 6.419 = 8 VA

    Hence burden range of 7.5VA is proposed for 300A Tap and 15VA is proposed for600A Tap.

    The CT Accuracy Class selected is CL. 0.5 as per the specification requirements.

    The instrument security factor (ISF) selected is equal to or less than Five(5) as perthe specification requirements.

    Abstract

    Therefore a CT of CL. 0.5,7.5VA is proposed for 300-600/1A and CL. 0.5,15VA isproposed for 300-600/1A.The CTs are with a factor of safety less than or equal to 5.

    4.7 11kV Outgoing Feeder

    4.7.1 CTs for Metering and Protect ion

    CT Ratio:

    300-600 / 1A

    Type of Metering and Protection:

    It is proposed to use Bay Control Unit(BCU) type REF542Total burden of bay control unit = 0.1 VA (Refer REF542 Plus catalogue)

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    Minimum length of cable = 5 metersbetween CT and BCU

    Size of cable = 2.5 sqmm

    Resistance of cable per = 7.41 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 7.41(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 8.43 ohm/km

    Total resistance = 2 x 8.43 x 5 / 1000

    (Lead & Return conductors) = 0.0843

    Burden due to lead resistance = Total resistance x (CT secondary current)

    2

    = 0.0843 x 12

    = 0.0843 VA

    Total burden = Total burden of the BCU + Burden due to leadresistance

    = 0.1 + 0.0843

    = 0.1843 VA

    Considering 25% future margin

    Total burden required = 1.25 x 0.1843 = 0.23 VAHence burden range of 7.5VA is proposed for 300A Tap and 15VA is proposedfor 600A Tap.

    The CT Accuracy Class selected is CL. 5P20 as per the specification requirements.

    For a Tap of 300A

    To ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 300AIscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limitingfactor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

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    Pn = Nominal CT burden = 7.5 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

    = I2* Rct

    Rct = 1.5Ohm (Assumed)

    = 12x 1.5

    = 1.5 VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.08429 = 0.1843 VA

    7.5 + 1.5

    Koalf = 20 x0.1843 + 1.5

    = 109

    Iscc/ Ipn = 31500 / 300

    = 10.5

    Koalf> Iscc/ Ipn

    Hence the selected CT with parameters 300-600 / 1, 5P20, 7.5VA is adequate.

    For a Tap of 600ATo ensure correct operation of the connected relay in case of faults, the CT must beable to transform the maximum symmetrical short circuit current without saturation.

    To satisfy the above, following condition to be checked:

    Koalf Iscc/ Ipn

    Where, Ipn = CT primary nominal current = 600A

    Iscc = max. symmetrical short circuit current = 31.5kA

    Koalf = Operating accuracy limiting factor

    The operating accuracy limiting factor (Koalf) depends on the nominal accuracy limiting

    factor (Knalf), the nominal CT burden (Pn), the internal CT burden (P i) and the totalconnected burden (Pb).

    Pn+ PiKoalf = Knalfx

    Pb+ Pi

    Where,

    Knalf = 20

    Pn = Nominal CT burden = 15 VA

    Pi = Internal CT burden

    = V * I

    = (I * Rct) * I

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    = I2* Rct

    Rct = 3 Ohm

    = 12x 3

    = 3VA

    Pb = Total connected burden

    = Pr+ PL

    = 0.1 + 0.08429 = 0.1843 VA

    15 + 3

    Koalf = 20 x0.1843 + 3

    = 113

    Iscc/ Ipn = 31500 / 600= 52.5

    Koalf> Iscc/ Ipn

    Hence the selected CT wi th parameters 300-600 / 1, 5P20, 15VA is adequate

    Abstract

    Therefore a CT of CL. 5P20,7.5VA is proposed for 300-600/1A and CL. 5P20,15VA isproposed for 300-600/1A.

    4.7.2 CTs for Transducers

    CT Ratio:

    300 600 /1A

    Type of Metering:

    It is proposed to use Transducers.

    Maximum burden of current transducer = 2 VA (Actual Burden is 0.2 only)

    The transducers are located in LDC cubicle.

    Minimum length of cable = 40 metersbetween CT and transducer

    Size of cable = 4 sqmm

    Resistance of cable per = 4.61 Ohms (as per Ducab cable catalogue)km at 20 deg C

    Resistance of cable at = 4.61(1+0.00393 (55-20) )55 deg C, consideringtemperature correction factor

    = 5.244 ohm/km

    Total resistance = 2 x 5.244 x 40 / 1000

    (Lead & Return conductors) = 0.419

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    Burden due to lead resistance = Total resistance x (CT secondary current)2

    = 0.419 x 12

    = 0.419 VA

    Total burden = Total burden of the transducer + Burden due to leadresistance

    = 2 + 0.419 = 2.419 VA

    Considering 25% future margin

    Total burden required =1.25 x 2.419 = 3 VA

    Hence burden range of 7.5VA is proposed for 300A Tap and 15VA is proposed for600A Tap.

    Abstract

    Therefore a CT of CL. 0.5,7.5VA is proposed for 300-600/1A and CL. 0.5,15VA isproposed for 300-600/1A.The CTs are with a factor of safety less than or equal to 5.

    5.0 VT SIZING CALCULATIONS

    The VTs considered are as follows:

    Line VT

    The devices connected on winding 1 of line VTs are BCU and Voltage Transducer

    for line feeder.

    The burden of the same are as given below:

    BCU: 0.25VA 2VA Maximum ConsideredVoltage Transducers: 5VA max. 5VA Maximum Considered

    Total 7VA max.

    The devices connected on winding 2 of line VTs Synchronising devices for line

    feeder. As per the manufacturer the maximum burden for the same is 15VA.

    Hence the VT considered for line is of following parameters.

    11kV/

    3 / 110V/

    3 / 110V/

    3, Winding 1 - 30VA , 0.5SWinding 2 - 30VA , 3P

    Bus VT

    The devices connected on winding 1of Bus VTs are BCU, Voltage Transducer andLoad shedding relay.

    The burden of the same are as given below,

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    BCU: 0.25VA 2VA Maximum ConsideredVoltage Transducers: 5VA max. 5VA Maximum ConsideredLoad Shedding relay: 0.1VA max. 2VA Maximum Considered

    Total 9VA max.

    The devices connected on winding 2 of line VTs Synchronising devices for Busfeeder. As per the manufacturer the maximum burden for the same is 15VA.

    Hence the VT considered for BUS VT is of following parameters .

    11kV/

    3 / 110V/

    3 / 110V/

    3, Winding 1 - 30VA , 0.5SWinding 2 - 30VA , 3P

    6.0 CONCLUSION

    The CT/VT parameters recommended are as calculated above. All the values at theUsage Tap are Confirmed by the Manufacturer..

    7.0 SUMMARY OF CURRENT TRANSFORMERS

    The summary of the Current Transformers for all the Sub-stations is given inAnnexure 1.

    8.0 SUMMARY OF VOLTAGE TRANSFORMERS

    The summary of the Voltage Transformers for all the Sub-stations is given inAnnexure 2.

    9.0 Relevant pages of Relay Catalogues

    The same is made available in Annexure 3.