CT & PT Calculations

By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004

TCE CONSULTING ENGINEERS LIMITED Client : AL TOUKHI COMPANY

Project : Generation expansion at ARAR P/P Doc. No. Subject : CT & PT Calculation for Gen & UAT bay

Contract No.- 30421007-3CL-00003

Sh.

1

CALCULATIONS FOR CT & PT PARAMETERS IN ARAR CPS EXTENSION PROJECT REFERENCE DATA 1) ABB CATALOGUE FOR RET 316, (Transformer differential protection). 2) CATALOGUES FOR VARIOUS RELAYS & METERS 3) SHORT CIRCUIT STUDY 4) GENERAL SITE LAYOUT - Dwg No KA-667182 5) SEC -EOA's SPECIFICATION : PTS-1022 ASSUMPTIONS : 1)All Relay and meter VA burdens are assumed 2)Presuming Numerical relays for differential protection, Aux CTs are not used.

Of 10

Rev. No.

P0A

TCE CONSULTING ENGINEERS LIMITED

By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004

Project : Generation expansion at ARAR P/P Doc. No.

Client : AL TOUKHI COMPANY

Subject : CT & PT Calculation for Gen & UAT bay Contract No.- 30421007-3CL-00003

Sh.

Of 10

2

Rev. No.

P0A

1) Calculation of CT parameters for Generator Transformer Feeder Differential Protection (87GT FDR) CT Ratio Selection : Generator Transformer Rating

=

60/80 MVA @ 50deg.C, ONAN/ONAF 132/13.8kV, Ynd1, Z=12.5% at ONAN rating

Therefore full load currrent on 132kV Side

= = =

(MVAx 1000)/(1.732 x kV) 80 x 1000 / 1.732 x 349.92 Amps

132

Multi ratio CT is selected to meet the requirement of Bus bar CT 2000-400/1 Amp CT ratio selected on 132kV side is = (Normally CT has a overload capacity of 20%. Therefore at lower temperature the maximum output will be taken care by this 20%. However this would be confirmed with CT manufacturer) Knee Point Voltage Calculation (For 87GT FDR) : Formula for calculating Knee point voltage requirement for differential protection Vk = 40 x I (Rct + 2 Rl+Rrelay) Where

I Rct Rl

= =

CT secondary current CT winding resistance

= =

=

Cable resistance between CT & Relay

1 3

Amp ohm

(Assumed)

Considering 4 sq.mm copper conductor cable between CT and relay, resistance of 4 sq.mm copper cable = 5.54 Ohms / km Maximum cable length Cable Lead Resistance for the loop Rl

=

250

M

=

2 x 2.77

5.54 Ohms

=

(from site plan) x

0.25

Therefore CT knee point requirement = 40 x1x(3 + 2 x 5.54 x 0.25 + 0.1) (Relay Burden is assumed as 0.1 VA, same as transformer differential protection as per ABB catalogue RET 316) Therefore, Since the secondary is 1A, RRelay = =

0.1

ohms

234.8

Volts

We choose Vk >= 300V to account for variation in cable length, variation in CT, relay types etc. Hence,

Vk >=300V

Magnetising Current (Im) Calculation (For 87GT FDR): Formula for calculating CT magnetising current Im. Im = Where

n PFSC Ir Im

= = = =

Im = = =

1/3[ 70 x ( 1/400) - 0.1] 0.025 Amp 25 mA

Hence Im at Vk/2 = 25/2 =

1/n [PFSC x (1/CT ratio) - Ir]

Nos. of CTs in parallel Primary Fault Setting Current Relay current at set point in Amp Magnetising Current in Amps

12.5 mA.

Im at Vk/2 <= 15 mA

is selected

= = =

3 70 0.1

Amp (Assuming 20% of Ifull load) Amp


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004




Sh.

3

Of 10

Rev. No.

P0A

Selection of Class of CT As the CT is meant for Transformer differential protection, Class X CT is selected. CT Class = X Therefore, CT parameters for 87GT FDR are CT RATIO = Knee Point Voltage Vk = CT Wdg. Res. Rct = Mag. Current Im at Vk/2 CT Class = CT Cable to be used =

2000-400/1 Amp Vk >=300V <= 3 <= 15 mA X 4 sq.mm copper cable

2) Calculation of parameters for 132kV GT Transformer bushing CT & Neutral side CT used for Restricted earth fault protection 87GTG.

CT ratio chosen

=

400/1 A

Generator Transformer terminal fault current on 132kV side Zgt on 60 MVA base = 12.5 Zgt on 80 MVA base = 16.6667 Vr = Relay operating Voltage Ifs = fault level on CT secondary

Vr = Ifs =

= =

If / (% Zgt ) 349.92 / 0.167 2.1

Ifs (Rct + Rlead + Rrelay) (2100/400/1) =

5.25

A

2.77

Ohms

Rl ( Considering 4 sq.mm Cu conductor Cable, 250M length) =

2 x 5.54 x 0.25

=

R relay

=

1 VA

(Assumed)

Winding res. of CT sec Rct for 400/1A

=

3 Ohms

(Assumed)

Therefore

=

5.25 *(3+2.77+ 1)

=

35.5425

Volts

Vk = 2 Vr

=

71.085

Volts

Taking margin, keep Vk

>=

120

Volts

Vr

Therefore CT chosen

=

400/1A Vk>=120V Rct<=3 Ohms

Magnetising Current (Im) Calculation Formula for calculating CT magnetising current Im. Im = Where

n PFSC Ir Im

= = = =

Im = = =

1/4[ 140 x ( 1/400) - 0.1] 0.063 Amp 63 mA

Hence Im at Vk/2 = 63/2 =

1/n [PFSC x (1/CT ratio) - Ir]

Nos. of CTs in parallel Primary Fault Setting Current Relay current at set point in Amp Magnetising Current in Amps

32 mA.

Im at Vk/2 <=

30

is selected

= = =

4 140 Amp (Assuming 40% of Ifull load) 0.1 Amp


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004




Sh.

Of 10

4

Rev. No.

P0A

Selection of Class of CT As the CT is meant for Restricted E/F protection, Class X CT is selected. CT Class = X Therefore, parameters for 87GTG protection CTs are CT RATIO = Knee Point Voltage Vk = CT Wdg. Res. Rct = Mag. Current Im at Vk/2 CT Class = CT Cable to be used =

2000-400 /1 Amp >= 120 Volts 3 Ohms <= 30 X 4 sq.mm copper cable

3) Calculation of parameters for 132kV Generator Transformer Neutral side CT used for 51GTT protection. CT ratio chosen

=

400/1 A

132kV Bus Design Fault Level Vr If = fault level on CT secondary

=

40 kA

=

If(Rct + Rlead + Rrelay)

=

40000/400/1

=

100

=

2.22


2 x 5.54 x 0.2

R relay

=

0.1


=

3

Therefore

=

Vr

VA (Assumed) Ohms

100 *(3+2.216+ 0.1)

=

531.6

Volts

Selection of Class of CT By choosing 5P20 ,

30 VA CT,

Vk available = 20 x 1.0 (3+30) V = 660 Volts

To take into account variations in CT parameters etc.,

Hence, the following CT Parameters are selected Therefore CT chosen for 51GTT protection CT RATIO CT Class Rated VA of CT CT Cable to be used

= = = =

400 /1 Amp 5P20 30 VA 4 sq.mm copper cable

UAT CT Sizing 4) Calculation of parameters for 13.8kV side UAT bushing ( For Overall Differential Protection 87OA). UAT Rating

=

1.5 MVA

Therefore full load currrent on 132kV Side

= = =

(MVAx 1000)/(1.732 x kV) x 1000 / 1.732 x 1.5 62.76 Amps

However CT ratio selected on 13.8kV side is

=

1000 /1 Amp

13.8

( To ensure non-saturation of CT for faults )


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004




Sh.

Of 10

5

Rev. No.

P0A

Knee Point Voltage Calculation (For 87OA) Formula for calculating Knee point voltage requirement for differential protection Vk = 40 x I (Rct x 2 Rl+Rrelay) Where

I Rct Rl

= =


= =

=


1 5

Amp ohm

(Assumed)


=

250

M (From site plan)

=

2 x 2.77

5.54 Ohms

=

x

0.25

Therefore CT knee point requirement = 40 x1(5 + 2 x 5.54 x 0.25 + 0.1) (Relay burden is considered as 0.1 VA as per catalogue for RET 316, of ABB make) Therefore, Since the secondary is 1A, RRelay = =

0.1

ohms

314.8

Volts

We choose Vk >= 400V to account for variations in cable length, variation in CT, relay types etc. Hence,

Vk >=400V

Selection of Class of CT As the CT is meant for Overall differential protection, Class X CT is selected. CT Class = X Therefore, parameters for 87OA are CT RATIO Knee Point Voltage Vk CT Wdg. Res. Rct Mag. Current Im at Vk/2 CT Class CT Cable to be used

= = = = =

1000 /1 Amp 400 Volts 5 ohm <= 15 mA X 4 sq.mm copper cable >=

5) Calculation of CT parameters for ( 50UAT) on HV side For the operation of Instantaneous protection, the CT will see the higher value of subtransient current.Due to this high fault current the CT may get saturated if lower ratio is used. To avoid CT saturation and to keep knee point voltage low, a separate CT of higher ratio is selected for Instantaneous(50) protection. CT Ratio Selection : UAT Rating

=

Full Load current on 13.8kV side

= =

1.5

MVA

(MVAx 1000)/(1.732 x kV) 62.76 A 1000 /1 Amp

However CT ratio selected on 13.8kV side is

=

Relay Burden :

( Assumed)

0.5

( To ensure non-saturation of CT for faults )

Since the secondary Current is 1A, the above burden can be taken as the total burden in ohms. i.e., Rrelaytotal in ohms = Rrelaytotal in VA/(Is2) , where Is = Secondary Current of CT. Therefore Rrelaytotal in ohms = =

0.5 0.5

ohms

Formula for calculating Knee point voltage requirement :

If x 1/CTratio x (Rct + 2 Rl+Rrelay)

Vk = For a fault at 13.8kV side of the UAT


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004



Subject : CT & PT Calculation for Gen & UAT bay Sh. 6 Of 10 (From ARAR Short circuit study

Contract No.- 30421007-3CL-00003

Contribution from the Generator =

22.1

Contribution from system through GT Total fault current

=

Where

If Rct Rl

=

kA

26.1 48.2

kA

(From ARAR Short circuit study

kA

= =

CT Primary fault current in Amps CT winding resistance

=


=

48200 5 ohm

A (Assumed)


=

200

M

=

2 x 2.22

5.54 Ohms

=

x

0.2

Considering 1000/1 ratio, Vk required = =

48200 x 1/1000(5 + 2.22 + 0.5) 372.104 Volts


20 VA CT,

Vk available = 20 x 1.0 (5+20) V = 500 Volts Hence, the following CT Parameters are selected CT RATIO CT Class VA burden required CT Cable to be used

= = = =


6) Calculation of CT parameters for ( 51 & 51N UAT) on HV side CT Ratio Selection : UAT Rating

=


= =

Therefore CT ratio selected on 13.8kV side is

1.5

MVA

(MVAx 1000)/(1.732 x kV) 62.76 A =

100 /1 Amp

Relay Burden : 0.5 ( Assumed) Since the secondary Current is 1A, the above burden can be taken as the total burden in ohms. i.e., Rrelaytotal in ohms = Rrelaytotal in VA/(Is2) , where Is = Secondary Current of CT. Therefore Rrelaytotal in ohms = =

0.5 0.5

ohms

For a fault on HV side of UAT, the fault is cleared by 50UAT, 87OA. Therefore 51UAT acts as a back up for UAT LV side faults and as such CT for 51UAT will be designed for the LV fault. Fault on LV side of the UAT =

30.8

kA

(From short circuit study)

Fault on LV side of UAT as seen from HV side of UAT = =

30.8

x

1.0713 kA

0.48 /

13.8

Rev. No.

P0A


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004




Sh.

7

Of 10

Formula for calculating Knee point voltage requirement : Vk = If x 1/CTratio x (Rct x 2 Rl+Rrelay) Where

If Rct Rl

= =


=


1071.3 A 1 ohm (Assumed)

=


=

200

M

=

2 x 2.22

5.54 Ohms

=

2 x

0.2


1071.3 x 1/100(1 + 2.22 + 0.5) 39.85 Volts


10 VA CT,


= = = =


7) Calculation of CT parameters for ( 51& 51N) on LV side of UAT CT Ratio Selection : UAT Rating

=


= =

CT ratio selected on 13.8kV side is

1.5

(MVAx 1000)/(1.732 x kV) 1804.22 A =

Relay Burden :

0.5

Total Burden

0.5

MVA

2000 /1 Amp

VA

Since the secondary Current is 1A, the above burden can be taken as the total burden in ohms. i.e., Rrelaytotal in ohms = Rrelaytotal in VA/(Is2) , where Is = Secondary Current of CT. Therefore Rrelaytotal in ohms = =

(0.5 /1*1) 0.5 ohms

Formula for calculating Knee point voltage requirement : Vk = If x 1/CTratio x (Rct x 2 Rl+Rrelay) Fault on 0.480 kV side = Where

50 If Rct Rl

kA

= =


=


=

50000 6 ohm

A (Assumed)

Rev. No.

P0A


By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004




Sh.

8

Of 10


=

200

M

=

2 x 2.22

5.54 Ohms

=

x

0.2


50000 x 1/2000(6 + 2.22 + 0.5) 218 Volts


15 VA CT,


= = = =


8) Calculation of parameters for UAT Neutral side CT used for 51G protection.

CT ratio chosen

=

2000/1 A

480V Bus Design Fault Level Vr If = fault level on CT secondary

=

50 kA

=

If(Rct + Rlead + Rrelay)

=

50000/2000/1

=

25

=

2.22


2 x 5.54 x 0.2

R relay

=

0.1


=

6

Therefore

= =

Vr

VA (Assumed)) Ohms

25 *(6+2.216+ 0.1) 207.9 Volts


15 VA CT,

Vk available = 20 x 1.0 (6+15) V = 420 Volts

To take into account variations in CT parameters etc.,

Hence, the following CT Parameters are selected Therefore CT chosen for 51GTT protection CT RATIO CT Class Rated VA of CT CT Cable to be used

= = = =

2000 /1 Amp 5P20 15 VA 4 sq mm

Rev. No.

P0A

By

DG

Date

07/7/2004

Chd.

SJH

Date





07/7/2004

Sh.

Of 10

9

Rev. No.

9) Calculation of parameters for UAT LV side 87GT CT protection CT ratio selected on is

=

2000 /1 Amp

Knee Point Voltage Calculation (For 87GT) Formula for calculating Knee point voltage requirement for differential protection Vk = 40 x I (Rct x 2 Rl+Rrelay) Where

I Rct Rl

= =


= =

=

Cable resistance between CT &

1 6

Amp ohm

(Assumed)

Relay


=

250

M

=

2 x 2.77

5.54 Ohms

=

x

0.25

Therefore CT knee point requirement = 40 x1(6 + 2 x 5.54 x 0.25 + 0.1) (RRelay is considered as 0.1 VA as per catalogue for RET 316, of ABB make) Therefore, Since the secondary is 1A, RRelay = =

0.1

ohms

354.8

Volts

We choose Vk >= 400V to account for variations in cable length, variation in CT, relay types etc. Hence,

Vk >=400V

Selection of Class of CT As the CT is meant for Overall differential protection, Class X CT is selected. CT Class = X Therefore, parameters for 87GT are CT RATIO Knee Point Voltage Vk CT Wdg. Res. Rct Mag. Current Im at Vk/2 CT Class CT Cable to be used

= = =

2000 /1 Amp 400 Volts 6 ohm <= 15 mA X 4 sq.mm copper cable >=

= =

10) Calculation of parameters for 480V VT on LV side of UAT VT ratio chosen

=

480/ 3 /115/ 3

VT will be with dual accuracy

=

0.5 / 3P class

Equipment

VA Burdens

Frequency Transducer

8

Voltage Transducer

8

###

25

###

3

Total

44

P0A

Therefore PT chosen =

By

DG

Date

07/7/2004

Chd.

SJH

Date

07/7/2004

480/ 3 /115/ 3 0.5/3P 75 VA

V





Sh.###

Of 10

Rev. No.

P0A

11) UAT LV side metering CT Presuming Numerical Meters to be provided. The VA burden of the multipurpose meter is assumed to be less than 10VA Hence CT VA burden selected is Therefore CT chosen

10VA =

2000 /1 A. 10 VA CL.0.5 ISF <5.

CT & PT Calculations

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