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.

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

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

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

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

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

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

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

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

Rl ( Considering 4 sq.mm Cu conductor Cable, 200M length)

=

2 x 5.54 x 0.2

R relay

0.1

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

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 )

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

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

=
=

CT secondary current
CT winding resistance

=
=

Cable resistance between CT & Relay

1
5

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 (From site plan)

2 x
2.77

5.54
Ohms

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

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

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

Cable resistance between CT & Relay

48200
5 ohm

A
(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

200

2 x
2.22

5.54
Ohms

0.2

Considering 1000/1 ratio,

Vk required =
=

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

372.104 Volts

Selection of Class of CT
By choosing 5P20 ,

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

=
=
=
=

1000 /1 Amp
5P20
20 VA
4 sq.mm copper cable

6) Calculation of CT parameters for ( 51 & 51N UAT) on HV side

CT Ratio Selection :
UAT Rating

Full Load current on 13.8kV side

=
=

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

1.0713 kA

0.48 /

13.8

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

Formula for calculating Knee point voltage requirement :

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

If
Rct
Rl

=
=

CT Primary fault current in Amps

CT winding resistance

Cable resistance between CT & Relay

1071.3 A
1 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

200

2 x
2.22

5.54
Ohms

2
x

0.2

Considering 100/1 ratio,

Vk required =
=

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

39.85 Volts

Selection of Class of CT
By choosing 5P20 ,

10 VA CT,

Vk available = 20 x 1.0 (1+10) V

=
220
Volts
Hence, the following CT Parameters are selected
CT RATIO
CT Class
VA burden required
CT Cable to be used

=
=
=
=

100 /1 Amp
5P20
10 VA
4 sq.mm copper cable

7) Calculation of CT parameters for ( 51& 51N) on LV side of UAT

CT Ratio Selection :
UAT Rating

Full Load current on 0.48kV side

=
=

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

=
=

CT Primary fault current in Amps

CT winding resistance

Cable resistance between CT & Relay

50000
6 ohm

A
(Assumed)

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

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

200

2 x
2.22

5.54
Ohms

0.2

Considering 2000/1 ratio,

Vk required =
=

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

218
Volts

Selection of Class of CT
By choosing 5P20 ,

15 VA CT,

Vk available = 20 x 1.0 (5+15) V

=
400
Volts
Hence, the following CT Parameters are selected
CT RATIO
CT Class
VA burden required
CT Cable to be used

=
=
=
=

2000 /1 Amp
5P20
15 VA
4 sq.mm copper cable

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

Rl ( Considering 4 sq.mm Cu conductor Cable, 200M length)

=

2 x 5.54 x 0.2

R relay

0.1

Winding res. of CT sec Rct for 2000/1A

Therefore

=
=

Vr

VA (Assumed))
Ohms

25 *(6+2.216+ 0.1)
207.9
Volts

Selection of Class of CT
By choosing 5P20 ,

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

TCE CONSULTING ENGINEERS LIMITED

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

07/7/2004

Sh.

Of 10

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

=
=

CT secondary current
CT winding resistance

=
=

Cable resistance between CT &

1
6

Amp
ohm

(Assumed)

Relay

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

2 x
2.77

5.54
Ohms

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

Voltage Transducer

###

25

###

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

TCE CONSULTING ENGINEERS LIMITED

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

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.