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- Words: 3,239
- Pages: 10
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 S 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 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.
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
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
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
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
=
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 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
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
= =
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
=
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 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
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
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
x
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.
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
= =
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
M
=
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.
8
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
M
=
2 x 2.22
5.54 Ohms
=
x
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
=
6
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 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
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
= =
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
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 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
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.
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 S 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 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.
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
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
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
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
=
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 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
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
= =
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
=
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 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
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
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
x
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.
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
= =
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
M
=
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.
8
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
M
=
2 x 2.22
5.54 Ohms
=
x
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
=
6
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 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
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
= =
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
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 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
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.
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