CT Sizing Calculation of 11kV System Rev0 Ver3
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Transcript of CT Sizing Calculation of 11kV System Rev0 Ver3
THIS SUBMISSION IS DECLARED TO BE STRICTLY IN
ACCORDANCE WITH THE REQUIREMENTS OF THE CONTRACT
SIGNATURE
QATAR POWER TRANSMISSION SYSTEM EXPANSION
PHASE VII (Substations)
CONTRACT NO. GTC/123/2006
SIEMENS CONSORTIUM-GTC/123/2006
SIEMENS AG SIEMENS LIMITED GERMANY INDIA
SUBSTATION NAME / CIRCUIT NAME
Mosemeer, Abu Hamour central, Al Soudan, Ain Khalid South, Al Wadi, MIC-2, MIC-3,QRE,EDS,Khore Community, Al Dhahiya West, Al Jumailyah, Khore Junction, NDQ, Muraikh North, South West Wakrah-1, NBK-2, Abu Thaila Modification, RLF-3, Al Dhahiya, Al Waab Super, MIC Super, Wakrah-2, Lusail Development Super-1,Ain Hamad, Ain Khalid south.
PROJECT DRAWING NUMBER
PH7-3B-10-15-C001
SUBCONTRACTOR / SUPPLIER
0 29-11-2007 FIRST ISSUE R.K. V.H V.A. REV DATE MODIFICATION DRAWN CHECK APPRD
SCALE SIZE DRAWING/DOCUMENT DESCRIPTION DESIGN REPORT FOR CURRENT TRANSFORMER (CT) & VOLTAGE TRANSFORMER (VT) SIZING FOR 11kV SYSTEM
N.A.
A4
TOTAL NO. OF PAGES VER
SIEMENS
Document No.:- GTC123-BN00-AQA-10001
3
91 PAGES
SIEMENS PROJECT:GTC/123/2006 Index
S.No. DESCRIPTION PAGE NUMBER
1 PURPOSE 1
2 DESIGN INPUT 1
3 ASSUMPTIONS 1
4 DESIGN CRITERIA 2
5 CALCULATIONS 2
6 RESULT OF STUDY 3
7 ATTACHMENTS 3
ANNEXURE 1 6-80
ANNEXURE 2 81-82
ANNEXURE 3 83
ANNEXURE 4 84-90
ANNEXURE 5 91
PH7-3B-10-15-C001, Rev 0 Page 2 of 91
SIEMENS PROJECT:GTC/123/2006 1.0 PURPOSE:
This document is intended to establish the minimum sizes of
- Current transformer in terms of Knee point voltage & Rated burden
- Voltage transformer in terms of Rated burden for various feeders
of 11KV for the following mentioned substations:
Mosemeer
Abu Hamour Central
Al Soudan
Al Wadi
Al Jumailyah
Khore Junction
MIC -2
MIC -3
QRE
EDS
Khore Community
Al Dhahiya West
Muraikah North
South West Wakrah 1
NDQ
NBK -2
RLF-3
MIC Super
Al Dhahiya
Al Waab Super
Lusail Development Super 1
Wakrah 2
Abu Thaila substation modification
Ain Hamad
Ain Khalid south
2.0 DESIGN INPUT : 1. Project contract document
2. Relay catalogue for relay burden
PH7-3B-10-15-C001, Rev 0 Page 3 of 91
SIEMENS PROJECT:GTC/123/2006
3.0 ASSUMPTIONS : 1. Power Transformer 32/40MVA, 66/11kV, %age impedance at principal
tapping is assumed as 16.33%.
2. Power Transformer 20/25MVA, 66/11kV, %age impedance at principal tapping is assumed as 12.58%..
3. Power Transformer 7.5/10MVA, 33/11kV, %age impedance at principal tapping is assumed as 12%.
4. Power Transformer 25/30MVA, 33/11kV, %age impedance at principal tapping is assumed as 12.58%.
5. 500kVA, 11/0.415kV Earthing Transformer, %age impedance at principal tapping is assumed as 9.60%
6. 1000kVA, 11/0.415kV Earthing Transformer, %age impedance at principal tapping is assumed as 6.00 %
7. 2000kVA, 11/0.415kV Earthing Transformer, %age impedance at principal tapping is assumed as 12.0%
4.0 DESIGN CRITERIA: KNEE POINT VOLTAGE
Apart from rated short time rating of the system, to arrive at minimum knee point voltage, value for steady state through fault current values needs to be determined.
Considering the rated capacity of the bus bars of 11kV, the fault level considered is 31.5kA.
PARAMETERS FOR CABLE BETWEEN CT & RELAY PANEL
Cross section taken is 4 mm2.
Calculation for resistance at 75 deg. C:
Resistance at 20 deg. C = 4.61 ohms / Km Value of Alpha (Temp. coefficient) at 20 deg. C for copper = 0,00393 / deg. C Resistance at 75 deg. C = 4.61 (1 + 0.00393 (75-20)) = 5.61 ohms / kM
Calculation of Loop resistance of the cable between CT & Relay panel:
Taking length of cable between CT & Relay panel as 50 meter Loop resistance = 2 x Length of cable (in kM) X resistance at 75 deg. C (in ohms/kM)
= (2 x 70 x 5.61) / 1000 = 0.785 ohms
Considering 20% safety margin as per contract requirement, (clause 12), = 0.785 * 1.2 = 0.942 ohms
All CT sizing calculations, have been done for a cable length of 50mts. This is the maximum length possible.
PH7-3B-10-15-C001, Rev 0 Page 4 of 91
SIEMENS PROJECT:GTC/123/2006
RATED BURDEN Rated burden selected for a CT/VT shall be more than the sum of relay/metering burden connected across CT/VT.
5.0 CALCULATIONS: Calculations performed for the CT/VT parameters are enclosed in following annexure:
Annexure # 1: CT Knee point voltage calculations for the 11kV feeders
Annexure # 2: CT burden (in VA) calculations for the 11kV feeders
Annexure # 3: VT burden (in VA) calculations for 11kV the feeders
6.0 RESULT OF STUDY:
Calculation results show that selected parameters for CT/VT are adequate to meet the minimum requirements.
7.0 ATTACHMENTS:
1. Annexure # 1: CT Knee point voltage calculations for 11kV feeders
2. Annexure # 2: CT burden (in VA) calculations for 11kV feeders
3. Annexure # 3: VT burden (in VA) calculations for 11kV feeders
4. Annexure # 4: Relay back-up sheets
5. Annexure #5: LSOH Power Cable Data Sheet
PH7-3B-10-15-C001, Rev 0 Page 5 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
Core-1Core-1 Core-2 750/1
66/11KV 50/1 50/1 Class-PX32/40MVA Class-PX Class-5P20 Vk≥250ONAN/ONAF Vk≥100 15VA 11/0.415KV Rct≤6
Rct≤1.2 500KVA Io<25maIo<30ma at Vk 15VA
Core-1750/1
Core-4 Class-5P202500/1 15VAClass-PXVk≥250 Core-2Rct≤9 50/1Io<30ma Class-PX
Vk≥100Rct≤1.2
Core-3 Io<30ma 2500/1Class-PX Core-3Vk≥250 2500/1Rct≤9 Class-PXIo<30ma
Rct≤9Core-2 Io<30ma 2500/1 Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥250Rct≤9Io<30ma
Note: - 40 MVA LV side Core-4 is not applicable for NDQ, Muraikh North, South West Wakrah-1, NBK-2
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
(Relay will be mounted in Relay panel)
Applicable substations:Mosemeer,Abu Hamour central,Al Wadi,MIC-2,MIC-3,QRE,Khore Community,Al Dhahiya West, NBK-2, Al Soudan, Ain Khalid south
40MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 500kVA
Vk≥250
Incomer Feeder (Typical bay no: A18,A28) Type:1
66kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
To Back-up earth fault(7SJ61)
To REF protection for Earthing transformer(7SJ61)
To REF protection for 11kV side(7SJ61)
Cable and trafo differential protection(7SD52)
PH7-3B-10-15-C001, Rev 0 Page 6 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
401.732 x 11 x 0.1388
Ihigh set point x I2N x (Ri+RBC)
The minimum CT knee point Voltage shall be selected more than above
1.320 x (9+1.042) x 1
Where If is magnitude of through fault current
Stability check for REF protection for through fault condition
The minimum CT knee point Voltage shall be selected more than above
1.5x15126.2x(9+0.942 )x22500
a) Effective symmetrical short-circuit current factor (K' SSC):
I2N = Relay Normal Current
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
PH7-3B-10-15-C001, Rev 0 Page 7 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 40 MVA = 0.1633Taking negative tolerance of 15% = 0.1388
ISSC =
ISSC = 15.1262 kA
K'SSC = 15126.2 x 32500
K'SSC to be considered for calculations = 18.15
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 140.20 volts20% margin on Vk value = 168.24 volts
ISSC = symmetrical short-circuit current
a) Transient dimensioning factor (ktd):
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
1.732 x 11 x 0.1388
IPN = CT rated primary current
The minimum CT knee point Voltage shall be selected more than above
(1.042 + 9) x 1 x 18.15
40
b) Effective symmetrical short-circuit current factor (K' SSC):
ISN = CT rated secondary current
(RBC+ Ri ) X ISN X K'SSC
1.3
Considering infinite source at 11 kV ,the maximum through fault current on 40 MVA transformer would be:
PH7-3B-10-15-C001, Rev 0 Page 8 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
(10+8) x20
RBC = Connected Burden across CT in VA
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
PH7-3B-10-15-C001, Rev 0 Page 9 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :
The calculated value of knee point voltage is
=1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
Ri = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
The minimum CT knee point Voltage shall be selected more than above
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
Feeders =A18,=A28 (connected across core-3 )CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection
PH7-3B-10-15-C001, Rev 0 Page 10 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000
For line differential protection ktd = 1.20
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 40 MVA = 0.1633Taking negative tolerance of 15% = 0.1388
ISSC = 40
ISSC = 15.1262 kA
K'SSC = 15126.2x1.22500
K'SSC to be considered for calculations = 7.26Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VAc) Knee point voltage :
The calculated value of knee point voltage is
=
Knee Point voltage required = (1.042+9) x1x7.261.3
Knee Point voltage required = 56.08 volts20% margin on Vk value = 67.30
RBC = Connected Burden across CT in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
The minimum CT knee point Voltage shall be selected more than above
Ri = Internal CT burden in VA
Considering infinite source at 11kV, the maximum through fault current on 40 MVA trafo would be :
IPN = CT rated primary currentISSC = symmetrical short-circuit current
Formula Used
b) Effective symmetrical short-circuit current factor (K' SSC):
1.732 x 11 x 0.1388
ISN = CT rated secondary current
a) Transformer dimensioning factor (k td):
(Relay will be mounted in Relay panel)
CT Knee point voltage calculation for 7SD5 relay for 11kV incoming feeders =A18, =A28(connected across Core-4)
PH7-3B-10-15-C001, Rev 0 Page 11 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61x1.2
= 6.732 Ohms/Km
Loop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point =
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
The minimum CT knee point Voltage shall be selected more than above
I2N = Relay Normal Current
The minimum CT knee point Voltage shall be selected more than above
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
1.732 x 11 x 0.138840
Stability check for REF protection for through fault condition
20 x (9+1.042) x 1
Ihigh set point x I2N x (Ri+RBC)
25001.5x15126.2x(9+0.942 )x2
Where If is magnitude of through fault current
(As per contract document)
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 500KVA)
PH7-3B-10-15-C001, Rev 0 Page 12 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 50 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732Loop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 1.20 OhmsCT Internal Burden in VA = (I)² x 1.2 = 1.2
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 34.49 volts20% margin on vk value = 41.39
Vk > 1.5(If x (Rct + 2RL ) x 2)
= 0.0960 = 0.0816
If =
If = 321.6172 A
Vk >
> 96.39 volts
50
As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value.
1.5 x 750 x (1.2 + 0.942 ) x 2
The minimum CT knee point Voltage shall be selected more than above
%age impedance at 500 KVAOverall Impedance ( taking 15% negative tolerance)
5001.732 x 11 x 0.0816
Where If is magnitude of through fault current
1.2 x (1.2+1.042) x 1
Considering infinite source,the maximum through fault current on 500 KVA transformer would be:
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
I2N = Relay Normal Current
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
Ihigh set point x I2N x (Ri+RBC)
PH7-3B-10-15-C001, Rev 0 Page 13 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 6.0 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =6+0.992
Calculated Value for K'SSC = 60.06 > 20
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 500 KVA HV Neutral side CTs)
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+6) x20
PH7-3B-10-15-C001, Rev 0 Page 14 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 50 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 1.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =1.2+0.992
Calculated Value for K'SSC = 147.80 > 20
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+1.2) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 15 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 6.00 OhmsCT Internal Burden in VA = (I)² x 6 = 6.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 108.34 volts20% margin on Vk value = 130.01
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.0960 = 0.0816
If =
If = 8.5248 kA
Vk >
> 236.72 volts
CT Knee point voltage calculation for 7SJ61 type used for REF protection of 433 V side of Earthing transformer (Applicable for Neutral side CTs)
5001.732 x 0.415 x 0.0816
1.5x8524.8x(6+0.942 )x2
Overall Impedance ( taking 15% negative tolerance)%age impedance at 500 KVA
Stability check for REF protection for through fault condition
Where If is magnitude of through fault current
The minimum CT knee point Voltage shall be selected more than above
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
20 x (6+1.042) x 1
Considering infinite source, the maximum through fault current on 500 KVA transformer would be:
Ri = Internal CT burden in VAI2N = Relay Normal Current
The minimum CT knee point Voltage shall be selected more than above
Ihigh set point x I2N x (Ri+RBC)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VA
750
PH7-3B-10-15-C001, Rev 0 Page 16 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
Core-1Core-1 Core-2 750/1
66/11KV 50/1 50/1 Class-PX20/25MVA Class-PX Class-5P20 Vk≥250ONAN/ONAF Vk≥100 15VA 11/0.415KV Rct≤6
Rct≤1.2 500KVA Io<25maIo<30ma at Vk 15VA
Core-1750/1
Core-4 Class-5P202500/1 15VAClass-PXVk≥250 Core-2Rct≤9 50/1Io<30ma Class-PX
Vk≥100Rct≤1.2
Core-3 Io<30ma 2500/1Class-PX Core-3Vk≥250 2500/1Rct≤9 Class-PXIo<30ma
Rct≤9Core-2 Io<30ma 2500/1 Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥250Rct≤9Io<30ma
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Vk≥250
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
(Relay will be mounted in Relay panel)
25MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 500kVA
Applicable substations:Al Jumailyah, Khore Junction
Incomer Feeder (Typical bay no: A18,A28) Type:2
66kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
To Back-up earth fault(7SJ61)
To REF protection for Earthing transformer(7SJ61)
To REF protection for 11kV side(7SJ61)
Cable and trafo differential protection(7SD52)
PH7-3B-10-15-C001, Rev 0 Page 17 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 25 MVA = 0.1258Taking a negative tolerance of 15% = 0.1069
If =
If = 12.2750 kA
Vk >
> 146.45 volts
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 11.3
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 25 MVA transformer would be:
251.732 x 11 x 0.1069
1.5x12275x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 18 of 91
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Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 25 MVA = 0.1258Taking negative tolerance of 15% = 0.1069
ISSC =
ISSC = 12.2750 kA
K'SSC = 12275 x 32500
K'SSC to be considered for calculations = 14.73
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 113.78 volts20% margin on Vk value = 136.54 volts
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 25 MVA transformer would be:
251.732 x 11 x 0.1069
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
(1.042 + 9) x 1 x 14.73
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 19 of 91
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The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(10+8) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 20 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
The minimum CT knee point Voltage shall be selected more than above
Ri = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18,=A28 (connected across core-3 )
PH7-3B-10-15-C001, Rev 0 Page 21 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000
For line differential protection ktd = 1.20
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 25 MVA = 0.1258Taking negative tolerance of 15% = 0.1069
ISSC = 25
ISSC = 12.2750 kA
K'SSC = 12275x1.22500
K'SSC to be considered for calculations = 5.89Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VAc) Knee point voltage :
The calculated value of knee point voltage is
=
Knee Point voltage required = (1.042+9) x1x5.891.3
Knee Point voltage required = 45.50 volts20% margin on Vk value = 54.60
(Relay will be mounted in Relay panel)
Formula Useda) Transformer dimensioning factor (k td):
b) Effective symmetrical short-circuit current factor (K' SSC):
CT Knee point voltage calculation for 7SD5 relay for 11kV incoming feeders =A18, =A28(connected across Core-4)
(RBC+ Ri ) X ISN X K'SSC
1.3
The minimum CT knee point Voltage shall be selected more than above
1.732 x 11 x 0.1069
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11kV, the maximum through fault current on 25 MVA trafo would be :
PH7-3B-10-15-C001, Rev 0 Page 22 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61x1.2
= 6.732 Ohms/Km
Loop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point =
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 25 MVA = 0.1258Taking a negative tolerance of 15% = 0.1069
If =
If = 12.2750 kA
Vk >
> 146.45 volts
CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 500KVA)
(Relay will be mounted in Relay panel)
(As per contract document)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault conditionTo establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 25 MVA transformer would be:
251.732 x 11 x 0.1069
1.5x12275x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 23 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 50 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732Loop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 1.20 OhmsCT Internal Burden in VA = (I)² x 1.2 = 1.2
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 34.49 volts20% margin on vk value = 41.39
Vk > If x (Rct + 2RL ) x 2
= 0.0960 = 0.0816
If =
If = 321.6172 A
Vk >
> 64.26 volts
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
1.2 x (1.2+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 500 KVA transformer would be:
%age impedance at 500 KVAOverall Impedance ( taking 15% negative tolerance)
500
50
The minimum CT knee point Voltage shall be selected more than above
1.732 x 11 x 0.0816
As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value.
750 x (1.2 + 0.942 ) x 2
PH7-3B-10-15-C001, Rev 0 Page 24 of 91
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Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 6.0 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =6+0.992
Calculated Value for K'SSC = 60.06 > 20
Adequecy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 500 KVA HV Neutral side CTs)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+6) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 25 of 91
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Current Transformer Ration (CTR) = 50 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 1.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =1.2+0.992
Calculated Value for K'SSC = 147.80 > 20
(15+1.2) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
PH7-3B-10-15-C001, Rev 0 Page 26 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 6.00 OhmsCT Internal Burden in VA = (I)² x 6 = 6.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 108.34 volts20% margin on Vk value = 130.01
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.0960 = 0.0816
If =
If = 8.5248 kA
Vk >
> 236.72 volts
CT Knee point voltage calculation for 7SJ61 type used for REF protection of 433 V side of Earthing transformer (Applicable for Neutral side CTs)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (6+1.042) x 1
Considering infinite source, the maximum through fault current on 500 KVA transformer would be:
%age impedance at 500 KVA
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
750
Overall Impedance ( taking 15% negative tolerance)
5001.732 x 0.415 x 0.0816
1.5x8524.8x(6+0.942 )x2
Where If is magnitude of through fault current
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 27 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
Core-1Core-2 Core-1 1500/1
66/11KV 100/1 100/1 Class-PX32/40MVA Class-5P20 Class-PX Vk≥500ONAN/ONAF 15VA Vk≥100 11/0.415KV Rct≤7.5
Rct≤0.4 1000KVA Io<25maIo<30ma at Vk/2
Core-3400/1 Core-1
Class-PX 750/1Vk≥250 Class-5P20
Rct≤9 15VAIo<30ma at Vk
Core-2100/1Class-PXVk≥100Rct≤0.4
Core-3 Io<30ma at Vk/2
2500/1Class-PX Core-3Vk≥250 2500/1Rct≤9 Class-PXIo<30ma
Rct≤9Core-2 Io<30ma 2500/1 Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥250Rct≤9Io<30ma
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Incomer Feeder (Typical bay no: A18,A28) Type:3
Applicable substations : NDQ, Murraikh North, South west wakrah,Lusail Development Super 1
40MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 1000kVA
Vk≥250
(Relay will be mounted in Relay panel)
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
66kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
To Back-up earth fault(7SJ61)
To REF protection for Earthing transformer(7SJ61)
To REF protection for 11kV side(7SJ61)
PH7-3B-10-15-C001, Rev 0 Page 28 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
a) Effective symmetrical short-circuit current factor (K' SSC):
I2N = Relay Normal Current
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
1.5x15126.2x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
Ihigh set point x I2N x (Ri+RBC)
The minimum CT knee point Voltage shall be selected more than above
1.320 x (9+1.042) x 1
Where If is magnitude of through fault current
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
401.732 x 11 x 0.1388
PH7-3B-10-15-C001, Rev 0 Page 29 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 40 MVA = 0.1633Taking negative tolerance of 15% = 0.1388
ISSC =
ISSC = 15.1262 kA
K'SSC = 15126.2 x 32500
K'SSC to be considered for calculations = 18.15
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 140.20 volts20% margin on Vk value = 168.24 volts
ISN = CT rated secondary current
(RBC+ Ri ) X ISN X K'SSC
1.3
Considering infinite source at 11 kV ,the maximum through fault current on 40 MVA transformer would be:
b) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
The minimum CT knee point Voltage shall be selected more than above
(1.042 + 9) x 1 x 18.15
1.732 x 11 x 0.1388
IPN = CT rated primary currentISSC = symmetrical short-circuit current
a) Transient dimensioning factor (ktd):
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
40
Ri = Internal CT burden in VA
PH7-3B-10-15-C001, Rev 0 Page 30 of 91
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The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
(10+8) x20
RBC = Connected Burden across CT in VA
PH7-3B-10-15-C001, Rev 0 Page 31 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meter#REF! = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :
The calculated value of knee point voltage is
=1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection
Ri = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
Feeders =A18,=A28 (connected across core-3 )
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 32 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61x1.2
= 6.732 Ohms/Km
Loop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 Ohms
CT Internal Burden in VA = (I)² x 9 = 9.00
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point =
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 1000KVA)
The minimum CT knee point Voltage shall be selected more than above
I2N = Relay Normal Current
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
Where If is magnitude of through fault current
(As per contract document)
Ihigh set point x I2N x (Ri+RBC)
25001.5x15126.2x(9+0.942 )x2
The minimum CT knee point Voltage shall be selected more than above
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
1.732 x 11 x 0.138840
Stability check for REF protection for through fault condition
20 x (9+1.042) x 1
PH7-3B-10-15-C001, Rev 0 Page 33 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 100 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732Loop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 0.40 OhmsCT Internal Burden in VA = (I)² x 0.4 = 0.4
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 22.18 volts20% margin on vk value = 26.62
Vk > 1.5(If x (Rct + 2RL ) x 2)
= 0.0600 = 0.0510
If =
If = 1029.1751 A
Vk >
> 30.20 volts
Ihigh set point x I2N x (Ri+RBC)
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
I2N = Relay Normal Current
20 x (0.4+1.042) x 1
Considering infinite source,the maximum through fault current on 1000 KVA transformer would be:
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value.
1.5 x 750 x (0.4 + 0.942 ) x 2
The minimum CT knee point Voltage shall be selected more than above
100
%age impedance at 1000 KVAOverall Impedance ( taking 15% negative tolerance)
10001.732 x 11 x 0.051
Where If is magnitude of through fault current
PH7-3B-10-15-C001, Rev 0 Page 34 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 6.0 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =6+0.992
Calculated Value for K'SSC = 60.06 > 20
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
(15+6) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBN+ Ri x KSSC
RBC + Ri
(Applicable for 1000 KVA HV Neutral side CTs)Adequacy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer
PH7-3B-10-15-C001, Rev 0 Page 35 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 100 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 1.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =1.2+0.992
Calculated Value for K'SSC = 147.80 > 20
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+1.2) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer
PH7-3B-10-15-C001, Rev 0 Page 36 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPNWhere:
%age impedance at 1 MVA = 0.0600Taking negative tolerance of 15% = 0.0510
ISSC =
ISSC = 1.0292 kA
K'SSC =400
= 7.72K'SSC to be considered for calculations = 7.72
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 59.62 volts20% margin on Vk value = 71.55 volts
The minimum CT knee point Voltage shall be selected more than above
1
(1.042 + 9) x 1 x 7.72
1.732 x 11 x 0.051
1029.17505444336 x 3
ISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 1 MVA transformer would be:
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit current
RBC = Connected Burden across CT in VA
Ri = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
IPN = CT rated primary current
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 1000 KVA EAT-(connected across core-3 )
a) Transient dimensioning factor (ktd):
PH7-3B-10-15-C001, Rev 0 Page 37 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 1500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 7.50 OhmsCT Internal Burden in VA = (I)² x 7.5 = 7.500Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 131.42 volts20% margin on Vk value = 157.70
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.0600 = 0.0510
If =
If = 27.2793 kA
Vk >
> 460.58 volts
1500
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VA
Considering infinite source, the maximum through fault current on 1000 KVA transformer would be:
Ri = Internal CT burden in VAI2N = Relay Normal Current
Stability check for REF protection for through fault condition
Where If is magnitude of through fault current
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
20 x (7.5+1.042) x 1
Ihigh set point x I2N x (Ri+RBC)
(Relay will be mounted in Relay panel)
Overall Impedance ( taking 15% negative tolerance)%age impedance at 1000 KVA
CT Knee point voltage calculation for 7SJ61 type used for REF protection of 433 V side of Earthing transformer (Applicable for Neutral side CTs)
10001.732 x 0.415 x 0.051
1.5x27279.3x(7.5+0.942 )x2
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 38 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
Core-1Core-2 Core-3 1500/1
66/11KV 100/1 400/1 Class-PX20/25MVA Class-5P20 Class-PX Vk≥500ONAN/ONAF 15VA Vk≥250 11/0.415KV Rct≤7.5
Rct≤9 1000KVA Io<25maIo<30ma
Core-1 Core-1100/1 750/1
Class-PX Class-5P20Vk≥100 15VARct≤0.4
Io<30ma Core-2100/1Class-PXVk≥100Rct≤0.4
Core-3 Io<30ma 2500/1Class-PX Core-3Vk≥250 2500/1Rct≤9 Class-PXIo<30ma
Rct≤9Core-2 Io<30ma 2500/1 Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥250Rct≤9Io<30ma
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
25MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 1000kVA
Applicable substations:Abu Thaila Substation Modification
Incomer Feeder (Typical bay no: A18,A28) Type:4
Vk≥250
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
(Relay will be mounted in Relay panel)
66kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
To Back-up earth fault(7SJ61)
To REF protection for Earthing transformer(7SJ61)
To REF protection for 11kV side(7SJ61)
PH7-3B-10-15-C001, Rev 0 Page 39 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 25 MVA = 0.1258Taking a negative tolerance of 15% = 0.1069
If =
If = 12.2750 kA
Vk >
> 146.45 volts
2500
The minimum CT knee point Voltage shall be selected more than above
251.732 x 11 x 0.1069
1.5x12275x(9+0.942 )x2
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 25 MVA transformer would be:
1.3
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
PH7-3B-10-15-C001, Rev 0 Page 40 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 25 MVA = 0.1258Taking negative tolerance of 15% = 0.1069
ISSC =
ISSC = 12.2750 kA
K'SSC = 12275 x 32500
K'SSC to be considered for calculations = 14.73
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 113.78 volts20% margin on Vk value = 136.54 volts
(1.042 + 9) x 1 x 14.73
The minimum CT knee point Voltage shall be selected more than above
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
ISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 25 MVA transformer would be:
251.732 x 11 x 0.1069
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit currentIPN = CT rated primary current
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
PH7-3B-10-15-C001, Rev 0 Page 41 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
(10+8) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
PH7-3B-10-15-C001, Rev 0 Page 42 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meter#REF! = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
Feeders =A18,=A28 (connected across core-3 )
(Relay will be mounted in Relay panel)
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 43 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61x1.2
= 6.732 Ohms/Km
Loop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point =
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 25 MVA = 0.1258Taking a negative tolerance of 15% = 0.1069
If =
If = 12.2750 kA
Vk >
> 146.45 volts
The minimum CT knee point Voltage shall be selected more than above
1.732 x 11 x 0.1069
1.5x12275x(9+0.942 )x22500
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 25 MVA transformer would be:
25
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault conditionTo establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 1000KVA)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(Relay will be mounted in Relay panel)
(As per contract document)
PH7-3B-10-15-C001, Rev 0 Page 44 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 100 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732Loop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 0.40 OhmsCT Internal Burden in VA = (I)² x 0.4 = 0.4
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 22.18 volts20% margin on vk value = 26.62
Vk > If x (Rct + 2RL ) x 2
= 0.0600 = 0.0510
If =
If = 1029.1751 A
Vk >
> 20.13 volts
The minimum CT knee point Voltage shall be selected more than above
1.732 x 11 x 0.051
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 1000 KVA transformer would be:
%age impedance at 1000 KVA
As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value.
750 x (0.4 + 0.942 ) x 2
Overall Impedance ( taking 15% negative tolerance)
1000
100
Stability check for REF protection for through fault condition
Ri = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
0.4 x (0.4+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VA
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs)
(Relay will be mounted in Relay panel)
PH7-3B-10-15-C001, Rev 0 Page 45 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 6.0 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =6+0.992
Calculated Value for K'SSC = 60.06 > 20
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+6) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequecy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 1000 KVA HV Neutral side CTs)
PH7-3B-10-15-C001, Rev 0 Page 46 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 100 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 1.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =1.2+0.992
Calculated Value for K'SSC = 147.80 > 20
RBC = Connected Burden across CT in VA
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
(15+1.2) x20
PH7-3B-10-15-C001, Rev 0 Page 47 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPNWhere:
%age impedance at 1 MVA = 0.0600Taking negative tolerance of 15% = 0.0510
ISSC =
ISSC = 1.0292 kA
K'SSC =400
K'SSC to be considered for calculations = 7.72
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point =
Knee Point voltage required =1.3
Knee Point voltage required = 59.62 volts20% margin on Vk value = 71.55 volts
1029.17505444336 x 3
RBC = Connected Burden across CT in VA
The minimum CT knee point Voltage shall be selected more than above
Ri = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
(1.042 + 9) x 1 x 7.72
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit current
11.732 x 11 x 0.051
IPN = CT rated primary current
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 1000 KVA EAT-(connected across core-3 )
a) Transient dimensioning factor (ktd):
ISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 1 MVA transformer would be:
PH7-3B-10-15-C001, Rev 0 Page 48 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 1500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 7.50 OhmsCT Internal Burden in VA = (I)² x 7.5 = 7.500Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 131.42 volts20% margin on Vk value = 157.70
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.0600 = 0.0510
If =
If = 27.2793 kA
Vk >
> 460.58 volts
The minimum CT knee point Voltage shall be selected more than above
1.732 x 0.415 x 0.051
1.5x27279.3x(7.5+0.942 )x21500
Considering infinite source, the maximum through fault current on 1000 KVA transformer would be:
%age impedance at 1000 KVAOverall Impedance ( taking 15% negative tolerance)
1000
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Ihigh set point x I2N x (Ri+RBC)
20 x (7.5+1.042) x 1
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
CT Knee point voltage calculation for 7SJ61 type used for REF protection of 433 V side of Earthing transformer (Applicable for Neutral side CTs)
(Relay will be mounted in Relay panel)
PH7-3B-10-15-C001, Rev 0 Page 49 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
Core-1Core-2 Core-3 3000/1
66/11KV 200/1 400/1 Class-PX32/40MVA Class-5P20 Class-PX Vk≥450ONAN/ONAF 15VA Vk≥100 11/0.415KV Rct≤15
Rct≤9 2000KVA Io<25maIo<30ma at Vk/2
Core-1200/1 Core-1
Class-PX 750/1Vk≥100 Class-5P20Rct≤0.6 15VA
Io<30ma at Vk/2Core-2200/1Class-PXVk≥100Rct≤0.6
Core-3 Io<30ma 2500/1Class-PX Core-3Vk≥250 2500/1Rct≤9 Class-PXIo<30ma
Rct≤9Core-2 Io<30ma 2500/1 Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥250Rct≤9Io<30ma
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Incomer Feeder (Typical bay no: A18,A28) Type:5
40MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 2000kVA
Applicable substations:MIC Super , Al Dhahiya,Al Waab Super, Wakrah 2, EDS
(Relay will be mounted in Relay panel)
Vk≥250
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
66kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
To Back-up earth fault(7SJ61)
To REF protection for Earthing transformer(7SJ61)
To REF protection for 11kV side(7SJ61)
PH7-3B-10-15-C001, Rev 0 Page 50 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
a) Effective symmetrical short-circuit current factor (K' SSC):
1.3
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
401.732 x 11 x 0.1388
1.5x15126.2x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 51 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 40 MVA = 0.1633Taking negative tolerance of 15% = 0.1388
ISSC =
ISSC = 15.1262 kA
K'SSC = 15126.2 x 32500
K'SSC to be considered for calculations = 18.15
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 140.20 volts20% margin on Vk value = 168.24 volts
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 40 MVA transformer would be:
401.732 x 11 x 0.1388
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
(1.042 + 9) x 1 x 18.15
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 52 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(10+8) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 53 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
The minimum CT knee point Voltage shall be selected more than above
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18,=A28 (connected across core-3 )
PH7-3B-10-15-C001, Rev 0 Page 54 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61x1.2
= 6.732 Ohms/Km
Loop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point =
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 40 MVA = 0.1633Taking a negative tolerance of 15% = 0.1388
If =
If = 15.1262 kA
Vk >
> 180.46 volts
CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 2000KVA)
(Relay will be mounted in Relay panel)
(As per contract document)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault conditionTo establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 40 MVA transformer would be:
401.732 x 11 x 0.1388
1.5x15126.2x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 55 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 200 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732Loop Resistance ( 2RL ) = 0.942 Ohms
Loop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 0.60 OhmsCT Internal Burden in VA = (I)² x 0.6 = 0.6
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 25.26 volts20% margin on vk value = 30.31
Vk >1.5( If x (Rct + 2RL ) x 2)
= 0.1200 = 0.1020
If =
If = 1029.1751 A
Vk >
> 17.34 volts
CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs)
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (0.6+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 2000 KVA transformer would be:
%age impedance at 2000 KVAOverall Impedance ( taking 15% negative tolerance)
20001.732 x 11 x 0.102
As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value.
750 x (0.6 +, 0.942 ) x 2x1.5200
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 56 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =Where
Ri = Internal CT burden in VA = 6.0 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =6+0.992
Calculated Value for K'SSC = 60.06 > 20
(15+6) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBC = Connected Burden across CT in VA
Adequacy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 2000 KVA HV Neutral side CTs)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
PH7-3B-10-15-C001, Rev 0 Page 57 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 200 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =Where
Ri = Internal CT burden in VA = 1.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =1.2+0.992
Calculated Value for K'SSC = 147.80 > 20
RBC = Connected Burden across CT in VA
a) Effective symmetrical short-circuit current factor (K' SSC):
(15+1.2) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer
RBN+ Ri x KSSC
PH7-3B-10-15-C001, Rev 0 Page 58 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 2 MVA = 0.1200Taking negative tolerance of 15% = 0.1020
ISSC =
ISSC = 1.0292 kA
K'SSC =400
K'SSC to be considered for calculations = 7.72
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 59.62 volts20% margin on Vk value = 71.55 volts
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 2000 KVA EAT-(connected across core-3 )
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K' SSC):
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault 40 current on 2 MVA transformer would be:
21.732 x 11 x 0.102
RBC = Connected Burden across CT in VA
1029.17505444336 x 3
Ri = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
(1.042 + 9) x 1 x 7.72
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 59 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 3000 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 15.00 OhmsCT Internal Burden in VA = (I)² x 15 = 15.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 246.80 volts20% margin on Vk value = 296.16
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.1200 = 0.1020
If =
If = 27.2793 kA
Vk >
> 434.89 volts
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
CT Knee point voltage calculation for 7SJ61 type used for REF protection of 433 V side of Earthing transformer (Applicable for Neutral side CTs)
(Relay will be mounted in Relay panel)
Ihigh set point x I2N x (Ri+RBC)
20 x (15+1.042) x 1
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source, the maximum through fault current on 2000 KVA transformer would be:
%age impedance at 2000 KVA
3000
The minimum CT knee point Voltage shall be selected more than above
1.5x27279.3x(15+0.942 )x2
Overall Impedance ( taking 15% negative tolerance)
20001.732 x 0.415 x 0.102
PH7-3B-10-15-C001, Rev 0 Page 60 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration:
33 KV SIDE Core-2 Core-1750/1 600-300/1Class - 5P20 Class - PX15 VA Vk≥ -
33/11 KV Rct≤9 - 4.57.5/10MVA Io=30mA at Vk/2ONAN/ONAF
Core-3800/1Class - PXVk≥250Rct≤9Io=30mA at Vk/2
Core-2600-300/1Class - 5P20/1.020 - 10 VA
Core-1600-300/1Class - PXVk≥600 - 300 Rct≤9 - 4.5
Io=30mA at Vk/2
Current Transformer Ration (CTR) = 600 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meter#REF! = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.730 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
(Relay will be mounted in Relay panel)
10MVA, 33/11kV transformer
Applicable substations:Ain Hamad
Incomer Feeder (Typical bay no: A18,A28) Type:6
Partial Bus Bar protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer REF (11kV side) & Main differential protection(7SJ61 & 7UT613)
11kV Bus Feeder =A18, =A28
11kV side REF Protection(7SJ61)
Backup earth fault(7SJ61)
8.47ohm750A, 30sNER
PH7-3B-10-15-C001, Rev 0 Page 61 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942 + 0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts 85.26 volts
20% margin on Vk value = 185.39 102.31
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 10 MVA = 0.1200Taking a negative tolerance of 15% = 0.102
If =
If = 5.146 kA
Ratio 600/1 Ratio 300/1
Vk > 1.5 x 5146 x (9 + 0.942 ) x 2 1.5 x 5146 x (4.5 + 0.942 ) x 2600 300
> 255.81 volts 280.05 volts
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VA
20 x (9 + 1.042) x 11.3
The minimum CT knee point Voltage shall be selected more than above
Ri = Internal CT burden in VA
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
Ratio 600/1 Ratio 300/120 x (4.5 + 1.042) x 1
Considering infinite source,the maximum through fault current on 10 MVA transformer would be:
101.732 x 11 x 0.102
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 62 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3.00
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 10 MVA = 0.1200Taking negative tolerance of 15% = 0.1020
ISSC =
ISSC = 5.146 kA
Ratio 600/1 Ratio 300/1
K'SSC = 5146 x 3 5146 x 3600 300
K'SSC to be considered for calculations = 25.73 51.46
Where
Relay Burden = 2*0.05 = 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942 + 0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Ratio 600/1
Knee Point voltage required = (1.042 + 9) x 1 x 25.73 (1.042 + 4.5) x 1 x 51.461.3 1.3
Knee Point voltage required = 198.75 volts 219.38 volts20% margin on Vk value = 238.51 volts 263.25 volts
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 10 MVA transformer would be:
CT knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K' SSC):
10.001.732 x 11 x 0.102
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
Ratio 300/1
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 63 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Relay Burden = 2*0.05 = 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942 + 0.1) 1.042 VA
Ratio 600/1 Ratio 300/1K'SSC = (20 + 8) x 20 (10 + 4) x 20
(1.042 + 8) (1.042 + 4)
Calculated Value for K'SSC = 61.93 55.53
> 20 20
RBN = Nominal Burden of CT in VA
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
PH7-3B-10-15-C001, Rev 0 Page 64 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 800Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18,=A28 (connected across core-3 )
RBC = Connected Burden across CT in VARi = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 65 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
The required K'SSC = I High set point = 20 IN atleast
Relay Burden =0.05VA 0.05 VARBC = (Loop Burden+ Relay Burden) = (0.942 + 0.05) 0.992 VA
Ratio 600/1 Ratio 300/1
Knee Point voltage required = (0.992 + 9) x 1 x 20 (0.992 + 4.5) x 1 x 201.3 1.3
Knee Point voltage required = 153.72 84.4920% margin on Vk value = 184.47 101.39(As per contract document)
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 10 MVA = 0.1200Taking a negative tolerance of 15% = 0.1020
If =
If = 5.146 kA
Ratio 600/1 Ratio 300/1
Vk >600 300
> 255.81 280.05
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
CT Knee point voltage calculation for REF protection with 7SJ61 relay (connected across LV neutral core-1)
a) Effective symmetrical short-circuit current factor (K' SSC):
I2N = Relay Normal Current
Considering infinite source,the maximum through fault current on 10 MVA transformer would be:
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
The minimum CT knee point Voltage shall be selected more than above
101.732 x 11 x 0.102
1.5 x 5146 x (9 + 0.942 ) x 2 1.5 x 5146 x (4.5 + 0.942 ) x 2
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
PH7-3B-10-15-C001, Rev 0 Page 66 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 amp.Relay Normal Current (IN) = 1 amp.
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 2.2 VARBN = Nominal Burden of CT in VA = 15 VARelay Burden = 0.05 = 0.050 VARBC = (Loop Burden+ Relay Burden) = 0.992 VA
K'SSC =
2.2+0.992
Calculated Value for K'SSC = 107.76 > 20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(15+2.2) x20
Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer -Core-2
a) Effective symmetrical short-circuit current factor (K' SSC):
PH7-3B-10-15-C001, Rev 0 Page 67 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Configuration
33/11KV25/30MVAONAN/ONAF
Core-175/1
Class-PXVk≥250Rct≤0.4
Io<30ma at Vk
Core-32500/1Class-PXVk≥250Rct≤9Io<30ma
Core-22500/1Class-5P20/1.0Rct≤810VA
Core-12500/1Class-PXVk≥ 250Rct≤9Io<30ma
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on Cable resistance = 5.61x1.2 Ohms/Km
= 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.00
30MVA, 33/11kV transformer,Earthing/auxiliary transformer rating of 1000kVA
Applicable substations:RLF-3
CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
(Relay will be mounted in Relay panel)
Incomer Feeder (Typical bay no: A18,A28) Type:7
33kV side
To partial busbar differential protection(7SJ61)
To Directional Overcurrent & earth fault / Metering(6MD6 & 7SJ62)
To Transformer Main differential protection( 7UT613)
11kV Bus Feeder =A18, =A28
NER
PH7-3B-10-15-C001, Rev 0 Page 68 of 91
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Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point voltage is
=
1.3 x Ipn
Knee Point voltage required =
= 154.49 volts
20% margin on Vk value = 185.39
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
%age impedance at 30 MVA = 0.1258Taking a negative tolerance of 15% = 0.1069
If =
If = 14.7300 kA
Vk >
> 175.73 volts
a) Effective symmetrical short-circuit current factor (K'SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 11.3
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
Where If is magnitude of through fault current
Considering infinite source,the maximum through fault current on 30 MVA transformer would be:
301.732 x 11 x 0.1069
1.5x14730x(9+0.942 )x22500
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 69 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Formula Used
For transformer, ktd = 3
The required K'SSC = I SCC x ktd
IPN
Where:
%age impedance at 30 MVA = 0.1258Taking negative tolerance of 15% = 0.1069
ISSC =
ISSC = 14.7300 kA
K'SSC = 14730 x 32500
K'SSC to be considered for calculations = 17.68
Where
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
c) Knee point voltage :The calculated value of knee point voltage is
=
Knee Point voltage required =1.3
Knee Point voltage required = 136.57 volts20% margin on Vk value = 163.89 volts
CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 )
a) Transient dimensioning factor (ktd):
b) Effective symmetrical short-circuit current factor (K'SSC):
ISSC = symmetrical short-circuit currentIPN = CT rated primary currentISN = CT rated secondary current
Considering infinite source at 11 kV ,the maximum through fault current on 30 MVA transformer would be:
301.732 x 11 x 0.1069
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
(RBC+ Ri ) X ISN X K'SSC
1.3
(1.042 + 9) x 1 x 17.68
The minimum CT knee point Voltage shall be selected more than above
PH7-3B-10-15-C001, Rev 0 Page 70 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA = 8.0 VARBN = Nominal Burden of CT in VA = 10 VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) = 1.042 VA
K'SSC =(1.042+8)
Calculated Value for K'SSC = 39.81 > 20
Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-2 )
a) Effective symmetrical short-circuit current factor (K'SSC):
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(10+8) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 71 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 2500Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meter (Relay will be Cable Resistance for mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Useda) Effective symmetrical short-circuit The required K'SSC = I High set point = 20 atleast
IN
RBC = Connected Burden across CT in Ri = Internal CT burden in VAI2N = Relay Normal CurrentRelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage : Ihigh set point x I2N x The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (1.042+9) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
The minimum CT knee point Voltage
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18,=A28 (connected across core-3 )
PH7-3B-10-15-C001, Rev 0 Page 72 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Core-1400-300/1Class-PXVk≥250-300Rct≤1.75-1.25Io<30ma
Core-2400-300/1Class-5P20/1.0Rct≤3-2VA Burden -10 VA
Current Transformer Ration (CTR) = 300 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 5.61 x 1.2
= 6.732 Ohms/KmLoop Resistance (2RL ) = 0.942 Ohms
Loop burden in VA; (I)2 x 2RL = 0.942 VACurrent Transformer resistance (Rct) = 1.25 Ohms
Short time rating of 11kV system (I sc) = 31.50 kA
Formula UsedKnee point voltage requirement Vk >
If = Primary current under maximum steady state through fault cnditions.
RL = Lead resistance of single lead from relay to current transformerRct = Secondary resistance Lead resistance of single lead from relay to current transformer If = Maximum through fault current = 50 x In 50Guard relay burden per element (RG) = 0.05 ohm
Knee point voltage requirement Vk >
Required parameters for CTs Ratio 300/1 Ratio 400/1
> 50/1+50/1 x (1.25+0.942+2x0.05) 50/1+50/1 x (1.75+0.942+2x0.05)
> 164.60 189.60
20% margin on Vk value > 197.52 227.52
Outgoing Feeder (Typical bay no: A10,A20) Type:1
Applicable substations:All substations with 11kV switchgear except Ain Hamad
CT Knee point voltage calculation for SOLKOR type relay used for Pilot wire protection Applicable for 11kV Outgoing Feeders
(Relay will be mounted in Relay panel)
50 / In + If / In * (Rct + 2Rl)Where:In = Rated current, amps. = 1A
N = CT ratio = 400-300/1A
50 / In + If / In * (Rct + 2R L + 2RG)
The minimum CT knee point Voltage shall be selected more than above
11kV Bus
To Feeder Overcurrent & earth fault protection / metering(7SJ61 + 6MD6)
To Pilot wire differential, Cable overload protection(SOLKOR)
PH7-3B-10-15-C001, Rev 0 Page 73 of 91
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The required K'SSC = 20 atleast
Also K'SSC =
Where
RBN = Nominal Burden of CT in VA Ri = Internal CT burden in VARelay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VARequired parameters for CTs Ratio 300/1 Ratio 400/1
K'SSC = (10+2) x20 (+3) x20(1.042+2) (1.042+3)
Calculated Value for K'SSC = 78.90 14.84
> 20 20
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
Adequacy checking calculation for 7SJ61 & 6MD6 relay for 11kV Cable feeders (connected across core-2)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBN+ Ri x KSSC
RBC + Ri
Since the calculated K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
PH7-3B-10-15-C001, Rev 0 Page 74 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Outgoing Feeder (A30,A40) Type:2
Core-1100/1Class - 5P1010VA
Core-2100/1Class - 1.0 Core-110VA 750/1
Class - PXVk≥Rct≤6
11/0.415 kV Io<25mA500 KVAONAN/ONAF
Core-1750/1Class - PXVk≥Rct≤6Io<25mA
Current Transformer Ration (CTR) = 100 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meter
Cable Resistance for 4.0mm2 at 750C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 1.500 OhmsCT Internal Burden in VA = (I)² x 1.5 = 1.50
The required K'SSC = 20 atleast
Also K'SSC =
Where
Relay Burden = 2*0.05 = 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
K'SSC =(1.042 + 1.5)
Calculated Value for K'SSC = 90.48 > 20.00
Applicable substations:Ain Hamad
Adequacy checking calculations for 7SJ61 relay for 11kV Incomer Feeders (connected across core-1)
RBN+ Ri x KSSC
RBC + Ri
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VARi = Internal CT burden in VA RBN = Nominal Burden of CT in VA
(10 + 1.5) x 20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
11kV Bus
To Over current / Earth fault (7SJ61)
415V side REF Protection(7SJ61)
415V side REF Protection(7SJ61)
Metering
PH7-3B-10-15-C001, Rev 0 Page 75 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Current Transformer Ration (CTR) = 750 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4.0mm2 at 750C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 6.000 OhmsCT Internal Burden in VA = (I)² x 6 = 6.000Formula Used
The required K'SSC = IHigh set point = 20 atleast IN
Where
Relay Burden = .1 VA 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =
= 108.34 volts20% margin on Vk value = 130.01
(For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible throughfault current has been considered)Vk > 2 x 1.5 x If x (Rct + 2RL )
= 0.0960 = 0.082
If =
If = 8.525 kA
Vk >
> 236.72 volts
Overall Impedance ( taking 15% negative tolerance)
500
The minimum CT knee point Voltage shall be selected more than above
(1.732 x415 x 0.0816)
1.5 x 8525 x (6 +0.942) x 2750
Where If is magnitude of through fault current
Considering infinite source, the maximum through fault current on 500 KVA transformer would be:
%age impedance at 500 KVA
The minimum CT knee point Voltage shall be selected more than above
Stability check for REF protection for through fault condition
To establish that CT rated knee point volatge is atleast 2 times that voltage developed across the CT in case of through fault condition with one of the CT saturated, the formula used is :
1.3
Ri = Internal CT burden in VAI2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (6 + 1.042) x 1
a) Effective symmetrical short-circuit current factor (K' SSC):
Value 20 will be selected for calculations
RBC = Connected Burden across CT in VA
CT knee point voltage calculations for 7SJ61 type used for REF protection of 415 V side of 315kVA transformer (Applicable for Phase and Neutral side CTs)
PH7-3B-10-15-C001, Rev 0 Page 76 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Core-1 Core-22500/1 2500/1Class-PX Class-PXVk≥250 Vk≥250Rct≤9 Rct≤9Io<30ma Io<30ma
Core-12500/1Class-5P20/1.0Rct≤8.510 VA
Current Transformer Ration (CTR) = 2500 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4 mm²at 75°C = 5.610 Ohms/Km20% margin on cable resistance = 6.732 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required = 20 x (9+1.042) x 11.3
= 154.49 volts20% margin on Vk value = 185.39 volts
Applicable substations: All substations with 11kV switchgear except Ain Hamad
Bus Coupler (Typical bay no.A12) Type:1
(Relay will be mounted in Relay panel)
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VA
CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar differential protection
The minimum CT knee point Voltage shall be selected more than above
Ri = Internal CT burden in VA
Applicable for I/C and B/C
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
11kV Bus-1 11kV Bus-2
To partial busbar differential protection(7SJ61)
To partial busbar differential protection(7SJ61)
To Overcurrent/Earth fault and Metering(6MD6 + 7SJ61)
PH7-3B-10-15-C001, Rev 0 Page 77 of 91
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The required K'SSC = 20 atleast
Also K'SSC =
Where
Ri = Internal CT burden in VA
RBN = Nominal Burden of CT in VA Relay Burden = 2*0.05 = 0.100 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
K'SSC =(1.042+8.5)
Calculated Value for K'SSC = 38.78 > 20.00
a) Effective symmetrical short-circuit current factor (K' SSC):
Adequacy checking calculation for 7SJ61 & 6MD6 relay (connected across core-2 )
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VA
(10+8.5) x20
PH7-3B-10-15-C001, Rev 0 Page 78 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
Core-2Core-1 800/1800/1 Class-PXClass-PX Vk≥250Vk≥250 Rct≤9Rct≤9 Io<30mA at Vk/2Io<30mA at Vk/2
Core-1800/1Class-5P20/1.0Rct≤8.515 VA
Current Transformer Ration (CTR) = 800 /1Relay Normal Current (IN) = 1 amp.Length of cable between CT and Relay = 70 meterCable Resistance for 4.0mm2 at 750C = 5.610 Ohms/Km20% margin on cable resistance = 6.730 Ohms/KmLoop Resistance ( 2RL ) = 0.942 OhmsLoop Burden in VA = (I)² x 0.942 = 0.942Current Transformer resistance (Rct) = 9.00 OhmsCT Internal Burden in VA = (I)² x 9 = 9.000Formula Used
The required K'SSC = I High set point = 20 atleast IN
Relay Burden = 0.1 VA 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
b) Knee point voltage :The calculated value of knee point = 1.3 x Ipn
Knee Point voltage required =1.3
= 154.49 volts20% margin on Vk value = 185.39 volts
The minimum CT knee point Voltage shall be selected more than above
I2N = Relay Normal Current
Ihigh set point x I2N x (Ri+RBC)
20 x (9+1.042) x 1
a) Effective symmetrical short-circuit current factor (K' SSC):
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
CT knee point voltage calculations for 7SJ61 type used for Partial Bus bar differential protectionApplicable for I/C and B/C
Bus Coupler (A12) Type:2
Applicable substations:Ain Hamad
11kV Bus-1 11kV Bus-2
To partial busbar differential protection(7SJ61)
To partial busbar differential protection(7SJ61)
To Overcurrent/Earth fault and Metering(6MD6 + 7SJ61)
PH7-3B-10-15-C001, Rev 0 Page 79 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 1
The required K'SSC = 20 atleast
Also K'SSC =
Where
Relay Burden = 2*0.05 = 0.10VA 0.1 VARBC = (Loop Burden+ Relay Burden) = (0.942+0.1) 1.042 VA
K'SSC =(1.042+8.5)
Calculated Value for K'SSC = 49.26 > 20.00
RBN = Nominal Burden of CT in VA
(15+8.5) x20
Since the selected K'SSC is more than Minimum required K'SSC (20), Hence selected CT is OK
RBN+ Ri x KSSC
RBC + Ri
RBC = Connected Burden across CT in VARi = Internal CT burden in VA
Adequacy checking calculation for 7SJ61 & 6MD6 relay (connected across core-1 )
a) Effective symmetrical short-circuit current factor (K' SSC):
PH7-3B-10-15-C001, Rev 0 Page 80 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 2
Sr. No.
Feeder Description CT Ratio used CT Knee Point Voltage (V)
CT Resistance (ohms)
VA burden due to Secondary
leads
Total VA burden
Remarks Applicable substations
1 Bay =A18,=A28 (Incomer feeder)
Core-1, Cl. PX 2500/1 250 9 7SJ61 7UT613 0.050 0.050 0.942 1.042Core-2, 5P20/1.0 2500/1 6MD6 7SJ62 0.050 0.050 0.942 1.292 Rated Burden selected= 10 VA
Ammeter 0.250
Core-3, Cl. PX 2500/1 250 9 7SJ61 0.050 0.942 0.992
Core-4, Cl. PX 2500/1 250 9 7SD52 0.050 0.942 0.992 Mosemeer,Abu Hamour central,Al Wadi,MIC-2,MIC-3,QRE,Khore Community,Al Dhahiya West, NBK-2, Al Soudan, Ain Khalid south,Al Jumailyah, Khore Junction
2 Earthing transformer (500KVA)
Core-1, Cl. PX 50/1 100 1.20 7SJ61 0.050 0.942 0.992Core-2, 5P20/1.0 50/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VA
Trafo Neutral, Cl. PX (11kV side) 2500/1 250 9.0 7SJ61 0.050 0.942 0.992Trafo Neutral, Cl. PX (11kV side) 50/1 100 1.20 7SJ61 0.050 0.942 0.992
Trafo Neutral, 5P20/1.0 (11kV side) 750/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VATrafo Neutral, Cl. PX (415 V side) 750/1 250 6 7SJ61 0.050 0.942 0.992
3 Earthing transformer (1000KVA)
Core-1, Cl. PX 100/1 100 0.40 7SJ61 0.050 0.942 0.992Core-2, 5P20/1.0 100/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VA
Trafo Neutral, Cl. PX (11kV side) 2500/1 250 9.0 7SJ61 0.050 0.942 0.992Trafo Neutral, Cl. PX (11kV side) 100/1 100 0.40 7SJ61 0.050 0.942 0.992
Trafo Neutral, 5P20/1.0 (11kV side) 750/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VATrafo Neutral, Cl. PX (415 V side) 1500/1 500 7.50 7SJ61 0.050 0.942 0.992
4 Earthing transformer (2000KVA)Core-1, Cl. PX 200/1 100 0.60 7SJ61 0.050 0.942 0.992
Core-2, 5P20/1.0 200/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VACore-3, Cl. PX 400/1 100 9.00 7UT613 0.050 0.942 0.992
Trafo Neutral, Cl. PX (11kV side) 2500/1 250 9.0 7SJ61 0.050 0.942 0.992Trafo Neutral, Cl. PX (11kV side) 200/1 100 0.40 7SJ61 0.050 0.942 0.992
Trafo Neutral, 5P20/1.0 (11kV side) 750/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 15 VATrafo Neutral, Cl. PX (415 V side) 3000/1 450 15.00 7SJ61 0.050 0.942 0.992
6 Bay =A12 (Bus sectionalizer)T1, Core-1, 5P20/1.0 2500/1 7SJ61 6MD6 0.050 0.050 0.942 1.292 Rated Burden selected= 10 VA
Ammeter 0.250
T1, Core-2, Cl. PX 2500/1 9 7SJ61 0.050 0.942 0.992
T2, Core-1, Cl. PX 2500/1 9 7SJ61 0.050 0.942 0.992
Mosemeer,Abu Hamour central,Al Wadi,MIC-2,MIC-3,QRE,Khore Community,Al Dhahiya West, NBK-2, Al Soudan, Ain Khalid south,Al Jumailyah, Khore Junction
Common for all substations with 11kV switchgear except Ain Hamad
MIC Super , Al Dhahiya,Al Waab Super, Wakrah 2, EDS
Relays connected Relay Burden (in VA)
Al Jumailyah, Khore Junction,Mosemeer,Abu Hamour central,Al Wadi,MIC-2,MIC-3,QRE,Khore Community,Al Dhahiya West, NBK-2, Al Soudan, Ain Khalid south,Abu
Thaila Substation Modification,NDQ, Murraikh North, South west wakrah,Lusail Development Super 1,MIC
Super , Al Dhahiya,Al Waab Super, Wakrah 2, EDS,RLF-3
NDQ, Murraikh North, South west wakrah,Lusail Development Super 1,Abu Thaila Substation
Modification.
PH7-3B-10-15-C001, Rev 0 Page 81 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 2
Sr. No.
Feeder Description CT Ratio used CT Knee Point Voltage (V)
CT Resistance (ohms)
VA burden due to Secondary
leads
Total VA burden
Remarks Applicable substationsRelays connected Relay Burden (in VA)
7 Outgoing Feeders =Typ. A10,A20T1, Core-1, Cl. PX 400-300/1 1.75-1.25 SOLKOR 3.500 0.942 4.442
T2, Core-1, 5P20/1.0 400-300/1 7SJ61 6MD6 0.050 0.050 0.942 1.292 Rated Burden selected= 15 VAAmmeter 0.250
8 Bay =A18,=A28 (Incomer feeder)Core-1, Cl. PX 600-300/1 600-300 9-4.5 7SJ61 7UT613 0.050 0.050 0.942 1.042
Core-2, 5P20/1.0 600-300/1 6MD6 7SJ62 0.050 0.050 0.942 1.292 Rated Burden selected= 10 VAAmmeter 0.250
Core-3, Cl. PX 800/1 250 9 7SJ61 0.050 0.942 0.992
9 Bay =A18 & =A28 (LV neutral)Core-1, Cl. PX 600-300/1 9-4.5 7SJ61 0.050 0.942 0.992Core-2, 5P20 750/1 Rated Burden selected= 15 VA
10 Bay =A30 & =A40 (315kVA transformer feeders)
Core-1, 5P10/1.0 100/1 7SJ61 0.050 0.942 0.992 Rated Burden selected= 10 VACore-2, Cl. 1.0 100/1 Rated Burden selected= 10 VA
Cl. PX (415V side) 750/1 6.00 7SJ61 0.050 0.942 0.992
11 Bay =A12,=A22 (Bus sectionalizer)T1, Core-1, CL5P20/1.0 800/1 8.5 7SJ61 6MD6 0.050 0.050 0.942 1.092 Rated Burden selected= 15 VA
T1, Core-2, Cl. PX 800/1 250 9 7SJ61 0.050 0.942 0.992T2, Core-1, Cl. PX 800/1 250 9 7SJ61 0.050 0.942 0.992
12 Outgoing Feeders =A90, =A70, =A50, =A10,= A20, =A60, =A80, =A100,
T1, Core-1, Cl. PX 400-300/1 300-250 1.25-1.75 SOLKOR 3.500 0.942 4.442T2, Core-2, 5P20/1.0 400-300/1 7SJ61 6MD6 0.050 0.050 0.942 1.292 Rated Burden selected= 15 VA
Ammeter 0.250
Ain Hamad
Ain Hamad
Ain Hamad
Common for all substations with 11kV switchgear except Ain Hamad
Ain Hamad
Ain Hamad
PH7-3B-10-15-C001, Rev 0 Page 82 of 91
SIEMENS PROJECT:GTC/123/2006ANNEXURE 3
Sr. No.
Feeder Description VT Ratio used Total VA burden
Remarks
1 Bay =A14 -T15Winding-1, 1.0/3P 11/ √3 : 0.11/√3 6MD6 0.050 1.450 Rated Burden selected= 30 VA
Voltmeter 1.400
2 Bay =A24 -T16Winding-1, 1.0/3P 11/ √3 : 0.11/√3 6MD6 0.050 1.450 Rated Burden selected= 30 VA
Voltmeter 1.400.
3 Bay =A18, =A28 (Incomer feeder)Winding-1, 1.0/3P 11/ √3 : 0.11/√3 6MD6 7SJ62 0.050 0.050 2.500 Rated Burden selected= 30 VA
Voltmeter 1.400AVR 1.000
4 Outgoing Feeders (Typ.) =A10,A20Winding-1, 1.0/3P 11/ √3 : 0.11/√3 6MD6 0.050 0.050 1.500 Rated Burden selected= 30 VA
Voltmeter 1.400
Relays connected Relay Burden (in VA)
PH7-3B-10-15-C001, Rev 0 Page 83 of 91
Siemens SIP · 2006
2 Overview
2
2/53
CT design according to BS 3938/IEC 60044-1 (2000)
The design values according to IEC 60044 can beapproximately transfered into the BS standarddefinition by following formula:
CT dimensioning formulae
K'ssc = Kssc · R R
R Rct b
ct b
++ '
(effective)
with K'ssc W Ktd ·I
ISCC max
pn
(required)
The effective symmetrical short-circuit currentfactor K'ssc can be calculated as shown in the tableabove.
The rated transient dimensioning factor Ktd de-pends on the type of relay and the primary DCtime constant. For relays with a required satura-tion free time from w 0.4 cycle, the primary (DC)time constant TP has only little influence.
Table 2/1 CT requirements
Relay type Transient dimensioning factor Ktd Min. required sym. short-circuit current factor K’ssc
Min. required kneepoint voltage VK
Overcurrent-time protection7SJ511, 512, 5317SJ45, 46, 607SJ61, 62, 63, 64
– K'ssc =I
I
High set point
pn
at least: 20
VK =I
I
High set point
pn13. •· (Rct + R'b) · Isn
at least:20
13.· (Rct + R'b) · Isn
Line differential protection(pilot wire)7SD600
– K'ssc = I
Iscc max (ext. fault)
pn
and:
3
4
4
3=
••
=( )
( )
K'
K'
ssc pn end1
ssc pn end2
I
I
VK = I
Iscc max (ext. fault)
pn13. •· (Rct + R'b) · Isn
and:
3
4
4
3=
••
=( / )
( / )
V
V
K pn sn end1
K pn sn end2
I I
I I
Line differential protection(without distance function)7SD52x, 53x, 610 (50 Hz)7SD52x, 53x, 610 (60 Hz)
Transformer1.21.6
Busbar /Line1.21.6
Gen. /Motor1.21.6
K'ssc =
Ktd ·I
Iscc max (ext. fault)
pn
and (only for 7SS):
K'ssc = 100 (measuring range)
VK =
Ktd ·I
Iscc max (ext. fault)
pn13. •· (Rct + R'b) · Isn
and (only for 7SS):
VK = 100
13.· (Rct + R'b) · Isn (measuring range)
Transformer / Generatordifferential protection7UT6127UT613, 633, 6357UM62
Transformer434
Busbar /Line43–
Gen. /Motor555
Busbar protection7SS5, 7SS600
for stabilizing factors k = 0.50.5
Distance protection7SA522, 7SA6, 7SD5xx*)*) with distance function
primary DC time constant TP [ms] K'ssc =Ktd (a) ·
I
Iscc max (close -in fault)
pn
and:
Ktd (b) ·I
Iscc max (zone 1- end fault)
pn
VK =Ktd (a) ·
I
Iscc max (close -in fault)
pn13. •· (Rct + R'b) · Isn
and:
Ktd (b) ·I
Iscc max (zone 1-end fault)
pn13. •· (Rct + R'b) · Isn
Ktd (a)Ktd (b)
= 30 = 50 = 100 = 200
14
25
45
45
VK =( )R R Ib ct sn sscK+ • •
13.
Example:IEC60044:
600/1, 5P10, 15 VA, Rct = 4 .
IEC PX or BS:( )
VK1.3
V V=+ • •
=15 4 1 10
146
Rct = 4 .
For CT design according to ANSI/IEEE C 57.13please refer to page 2/56
A
PH7-3B-10-15-C001, Rev 0 Page 84 of 91
2 Overview
Protection Coordination
Siemens SIP · 2006
2
2/56
Example 2:Stability-verification of the numerical busbarprotection relay 7SS52
I
ISCC.max.
pn
=30 000
60050
, A
A=
According to table 2/1 on page 2/53 Ktd = ½)
K'ssc = 1
250 25• =
Rb =15
15VA
1 A 2= .
Rrelay = 0.1 .
Rlead =2 0 0175 50
60 3
• • =.. .
R'b = Rl + Rrelay = 0.3 . + 0.1 . = 0.4 .
K'ssc =R R
R Rct b
ct bsscK
++
• = ++
• =' .
.4 15
4 0 410 43 2
. .. .
Result:
The effective K'ssc is 43.2, the required K'ssc is25. Therefore the stability criterion is fulfilled.
Relay burden
The CT burdens of the numerical relays ofSiemens are below 0.1 VA and can therefore beneglected for a practical estimation. Exceptionsare the busbar protection 7SS60 and the pilot-wirerelays 7SD600.
Intermediate CTs are normally no longer neces-sary as the ratio adaptation for busbar and trans-former protection is numerically performed in therelay.
Analog static relays in general have burdens belowabout 1 VA.
Mechanical relays, however, have a much higherburden, up to the order of 10 VA.
This has to be considered when older relays areconnected to the same CT circuit.
In any case, the relevant relay manuals should al-ways be consulted for the actual burden values.
Burden of the connection leads
The resistance of the current loop from the CT tothe relay has to be considered:
Rlead =2 • •. l
A
l = single conductor length from the CTto the relay in m.
Specific resistance:
. = 0.0175. • mm
m
2(copper wires) at 20 °C
A = conductor cross-section in mm2
Fig. 2/93
CT design according to ANSI/IEEE C 57.13
Class C of this standard defines the CT by ist sec-ondary terminal voltage at 20 times rated current,for which the ratio error shall not exceed10 %. Standard classes are C100, C200, C400 andC800 for 5 A rated secondary current.
This terminal voltage can be approximately calcu-lated from the IEC data as follows:
Vs.t.max = 20 •5 A •Rb •Kssc
20
withRb =
P
II Ab
sn
Nsnand we get2
5= ,
Vs.t.max =Pb sscK
A
•5
Example:IEC60044:
600/5, 5P20, 25 VA
ANSIC57.13: Vs.t.max =
( )25 20
5
VA
A
•= 100 V, acc. to class C100
ANSI CT definition
A
PH7-3B-10-15-C001, Rev 0 Page 85 of 91
4 Technical Data
268 7SJ61 Manual
C53000-G1140-C118-7
4.1 General Device Data
4.1.1 Analog Inputs
Current Inputs
1) only in models with input for sensitive ground fault detection (see ordering data in Appendix
A.1)
4.1.2 Auxiliary Voltage
DC Voltage
Nominal Frequency fNom 50 Hz or 60 Hz (adjustable)
Nominal Current INom 1 A or 5 A
Ground Current, Sensitive INs = linear range 1.6 A 1)
Burden per Phase and Ground Path
- at INom = 1 A
- at INom = 5 A
- for sensitive ground fault detection at 1 A
Approx. 0.05 VA
Approx. 0.3 VA
Approx. 0.05 VA
Current overload capability
- Thermal (rms)
- Dynamic (peak value)
100· INom for 1 s
30· INom for 10 s
4· INom continuous
250· INom (half-cycle)
Current overload capability for high-sensitivity input INs 1)
- Thermal (rms)
- Dynamic (peak value)
300 A for 1 s
100 A for 10 s
15 A continuous
750 A (half-cycle)
Voltage supply using integrated converter
Rated auxiliary DC VAux 24/48 VDC 60/110/125 VDC
Permissible Voltage Ranges 19 to 58 VDC 48 to 150 VDC
Rated auxiliary DC VAux 110/125/220/250 VDC
Permissible Voltage Ranges 88 to 300 VDC
Permissible AC ripple voltage,
Peak to Peak, IEC 60 255-1115 % of the auxiliary voltage
Power Input Quiescent Approx. 3 W
Energized Approx. 7 W
Bridging Time for Failure/Short Circuit,
IEC 60255–11
(in not energized operation)
= 50 ms at V = 110 VDC
= 20 ms at V = 24 VDC
A
- at INom = 1 A Approx. 0.05 VA
Current Inputs
PH7-3B-10-15-C001, Rev 0 Page 86 of 91
4.1 General
5197SD5 Manual
C53000-G1176-C169-1
4.1 General
4.1.1 Analog Inputs
Current Inputs
Requirements for current transformers
Voltage inputs
Nominal frequency fN 50 Hz or 60 Hz (adjustable)
Nominal current IN 1 A or 5 A
Power Consumption per Phase and Earth Path
- at IN = 1 A Approx. 0.05 VA
- at IN = 5 A Approx. 0.3 VA
- for sensitive earth fault detection at 1A Approx. 0.05 VA
Current Overload Capability per Current Input
- thermal (rms) 100 · IN for 1 s
30 · IN for 10 s
4 · IN continuous
- dynamic (pulse current) 250 · IN (half-cycle)
Current Overload Capability for Sensitive Earth Current Input
- thermal (rms) 300 A for 1 s
100 A for 10 s
15 A continuous
- dynamic (pulse current) 750 A (half-cycle)
1st Condition:
For a maximum fault current the current transformers must not be
saturated under steady-stateconditions
2nd Condition:
The operational accuracy limit factor n' must be at least 30 or a non-
saturated period of t'AL of at least 1/4 AC cycle after fault inception
must be ensured
n' = 30
or
t'AL = 1/4cycle
3 rd Condition:
Maximum ratio between primary currents of current transformers at
the ends of the protected object
Nominal voltage UN 80 V to 125 V (adjustable)
Measuring range 0 V to 218.5 V (rms)
Power consumption At 100 V = 0.1 VA
Voltage overload capability per phase
- thermal (rms) 230 V continuous
A
- at IN = 1 A Approx. 0.05 VA
Current Inputs
Voltage inputs
Power consumption = 0.1 VA
PH7-3B-10-15-C001, Rev 0 Page 87 of 91
4
Page 10 of 10
A
PH7-3B-10-15-C001, Rev 0 Page 88 of 91
A
PH7-3B-10-15-C001, Rev 0 Page 89 of 91
AVR TECHNICAL DATA SHEETAVR TECHNICAL DATA SHEETA
PH7-3B-10-15-C001, Rev 0 Page 90 of 91
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PA
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ICU
LA
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AB
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Sr
LV
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BL
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UN
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AD
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DO
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OF
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4C
X4
4C
X6
4C
X10
4C
X16
4C
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4C
X50
4C
X120
4C
X150
1C
X240
1C
X120
1C
X95
2C
X4
2C
X10
2C
X6
2C
X16
2C
X25
2C
X35
3C
X4
3C
X6
3C
X10
3C
X16
3C
X35
5C
X10
5C
X6
1L
V P
OW
ER
CA
BL
ES
1.1
Man
ufactu
rerO
MA
N C
AB
LE
SO
CI
1.2
Cab
le type
Cu/X
LP
/SW
A/L
SO
HC
u/X
LP
/SW
A/L
SO
HC
u/X
LP
/SW
A/L
SO
H
1.3
Stan
dard
sB
S 6
724
BS
6724
1.4
Rated
Voltag
eV
600/1
000
600/1
000
600/1
000
1.5
Test V
oltag
e (1 m
in)
VA
C/m
in4000
4000/m
in3500 (5
min
)
1.6
Conducto
r
1) C
ross sectio
nal area (N
om
inal)
mm
²4
610
16
25
50
120
150
240
120
95
410
616
25
35
46
10
16
35
10
6
2) M
aterial Copper
Cu
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
Copper
3) D
esign (S
tranded
, sectoral ect)
Stran
ded
Stan
ded
/Circu
larnded
/Circ
nded
/Circ
randed
/Sec
randed
/Sec
randed
/Sec
randed
/Sec
randed
/Sec
nded
/Circ
nded
/Circ
nded
/Circ
Stan
ded
/Circu
larnded
/Circ
anded
/Circ
anded
/Circ
randed
/Sec
randed
/Sec
anded
/Circ
anded
/Circ
anded
/Circ
anded
/Circ
tranded
/Sectan
ded
/Circ
anded
/Circu
lar
4) O
verall d
iameter (A
ppro
x)
mm
2.5
23.1
24.0
2N
.A.
N.A
.N
.A.
N.A
.N
.A.
18.4
13
11.4
2.5
24.0
23.1
24.8
N.A
.N
.A.
2.5
23.1
24.0
24.8
N.A
.4.0
23.1
2
5) N
o. o
f cores
44
44
44
44
11
12
22
22
23
33
33
55
1.7
Insu
lation
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
XL
PE
1.8
Fillers
Material - P
oly
pro
pylen
eP
oly
pro
pylen
e fillers - Wherev
er necessary
Poly
pro
pylen
e fillers - Wherev
er necessary
1.9
Arm
our b
eddin
g
Type
Poly
meric L
SO
H P
oly
meric L
SO
H
Nom
inal T
hick
ness
mm
0.8
0.8
0.8
0.8
11
1.4
1.4
11
0.8
0.8
0.8
0.8
0.8
0.8
10.8
0.8
0.8
0.8
10.8
0.8
1.1
0A
rmour
Type o
f wire
Galv
anized
Steel W
ire Arm
our
AW
A G
alvan
ised S
teel Wire
Num
ber o
f wires
No.
27
30
35
37
36
44
44
48
44
31
28
23
29
25
33
32
30
22
24
32
36
34
39
33
Diam
eter of w
ires - Nom
inal
mm
1.2
5*
1.2
51.2
51.2
51.6
1.6
2.5
2.5
1.6
1.6
*1.6
*1.2
5*
1.2
5*
1.2
5*
1.2
51.2
51.6
1.2
5*
1.2
5*
1.2
51.2
51.6
1.2
51.2
5
* W
ire size hig
her th
an sp
ecified in
BS
6724
1.1
1O
uter co
verin
gL
SO
HL
SO
H T
ype L
TS
-1 L
SO
H T
ype L
TS
1
Material
LS
OH
Type L
TS
-1 L
SO
H T
ype L
TS
1
Nom
inal T
hick
ness
mm
1.4
1.5
1.5
1.6
1.7
1.9
2.3
2.4
1.8
1.6
1.6
1.4
1.5
1.4
1.5
1.6
1.7
1.4
1.4
1.5
1.6
1.8
1.6
1.5
Min
. thick
ness
mm
0.9
21.0
01.0
01.0
81.1
61.3
21.6
41.7
21.2
41.0
81.0
80.9
21.0
00.9
21.0
01.0
81.1
60.9
20.9
21.0
01.0
81.2
41.0
81.0
0
1.1
2C
om
pleted
cable
Overall d
iameter - A
ppro
xm
m16.4
18.7
21.1
21.3
26.1
32.0
47.1
51.4
32.8
32.8
22.3
14.7
18
15.9
20.4
20.4
23.3
15.3
16.6
19.5
21.6
25.7
22.9
20
Weig
ht p
er meter - A
ppro
xkg
600
750
960
1195
1850
2860
6900
8190
2740
1460
1210
460
715
550
905
985
1395
515
625
840
1025
1830
1130
860
Max
imum
dru
m len
gth
m1000
1000
1000
1000
1000
1000
500
500
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1000
1.1
3C
able d
rum
s
Overall d
iameter - A
ppro
xm
1000
1200
1200
1200
1400
1600
1800
2000
1600
1800
1200
1000
1200
1000
1200
1200
1200
1000
1000
1200
1200
1400
1200
1200
Wid
th - A
ppro
xm
710
740
1040
1040
1040
1070
1130
1135
1070
1130
1040
700
740
700
1040
1040
1040
710
710
740
1040
1040
1040
1040
Weig
ht lo
aded
- Appro
xkg
690
880
1110
1345
2030
3150
3710
4395
3030
1720
1360
560
845
650
1055
1135
1545
615
725
970
1175
2010
1280
1010
1.1
4C
ontin
uous cu
rrent carry
ing cap
acity
based
on th
e conditio
ns
Laid
in th
e gro
und (T
ouch
ing each
oth
er)
a) One circu
itA
47
59
79
102
131
187
312
349
480
332
292
56
94
70
121
157
188
47
59
79
102
157
79
59
b) T
wo circu
itsA
38.0
747.7
963.9
982.6
2106.1
1151.4
7252.7
2282.6
9388.8
0268.9
2236.5
245.3
676.1
456.7
098.0
1127.1
7152.2
838.0
747.7
963.9
982
.62
127.1
763.9
947.7
9
c) Three circu
itsA
32.9
041.3
055.3
071.4
091.7
0130.9
0218.4
0244.3
0336.0
0232.4
0204.4
039.2
065.8
049.0
084.7
0109.9
0131.6
032.9
041.3
055.3
07
1.4
0109.9
055.3
041.3
0
Draw
n in
to d
ucts
a) One circu
itA
39
48
65
83
107
152
258
291
421
315
281
46
77
58
99
127
153
39
48
65
83
128
65
48
b) T
wo circu
itsA
31.2
038.4
052.0
066.4
085.6
0121.6
0206.4
0232.8
0336.8
0252.0
0224.8
036.8
061.6
046.4
079.2
0101.6
0122.4
031.2
038.4
052.0
066.4
0102.4
052.0
038.4
0
c) Three circu
itsA
27.3
033.6
045.5
058.1
074.9
0106.4
0180.6
0203.7
0294.7
0220.5
0196.7
032.2
053.9
040.6
069.3
088.9
0107.1
027.3
033.6
045.5
058
.10
89.6
045.5
033.6
0
In air
One circu
it at 50 °C
A37
46
64
83
109
163
293
335
510
328
282
43
74
55
98
128
158
37
46
64
83
134
64
46
1.1
5M
ax co
nducto
r temp
°C90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
90
1.1
6C
onducto
r short circu
it curren
t
carryin
g cap
acity fo
r 1 sec.
Cab
le loded
as above p
rior to
kA
0.5
70.8
61.4
32.2
93.5
87.1
517.1
621.4
534.3
217.1
613.5
90.5
71.4
30.8
62.2
93.5
85.0
10.5
70.8
61.4
32.2
95.0
11.4
30.8
6
short ck
t & fin
al conducto
r temp o
f°C
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
1.1
7M
in rad
ius o
f ben
d aro
und w
hich
cable can
be laid
a) Laid
direct
m6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
6 x
OD
6 x
OD
b) In
ducts
m6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
6 x
OD
6 x
OD
c) In air
m6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
8 x
OD
6 x
OD
6 x
OD
6 x
OD
6 x
OD
8 x
OD
6 x
OD
6 x
OD
1.1
8D
ucts
Nom
inal in
ternal d
iameter o
f pip
es or
mm
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
100
ducts th
rough w
hich
cable can
be p
ulled
1.1
9M
axim
um
dc resistan
ce
per k
m o
f cable at 2
0 °C
of co
nducto
rohm
/Km
4.6
13.0
81.8
31.1
50.7
27
0.3
87
0.1
53
0.1
24
0.0
754
0.1
53
0.1
93
4.6
11.8
33.0
81.1
50.7
27
0.5
24
4.6
13.0
81.8
31.1
50.5
24
1.8
33.0
8
1.2
0M
axim
um
ac resistance
of co
nducto
r per k
m o
f cable at m
ax.
ohm
/Km
5.8
83.9
32.3
31.4
70.9
27
0.4
94
0.1
97
0.1
60.0
968
0.1
97
0.2
47
5.8
82.3
33.9
31.4
70.9
27
0.6
68
5.8
83.9
32.3
31.4
70.6
68
2.3
33.9
3
conducto
r temp - 9
0 °C
1.2
1in
sulatio
n resistan
ce
per k
m o
f cable p
er core
Mohm
Will b
e calculated
form
the in
sulatio
n resistan
ce constan
t at 90 °C
I.e. 3.6
7 m
egao
hm
.km
Will b
e calculated
form
the in
sulatio
n resistan
ce constan
t at 90 °C
I.e. 3.6
7 m
egao
hm
.km
1.2
2M
anufactu
rer quality
assuran
ce accord
ing
CO
MP
LY
CO
MP
LY
to IS
O 9
000, 9
001, 9
003 &
9004
1.2
3T
ype test certificate to
be issu
ed b
y
CO
MP
LY
CO
MP
LY
indep
enden
t laborato
ry
A
PH7-3B-10-15-C001, Rev 0 Page 91 of 91