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
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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
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
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
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
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
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
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
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):
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
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
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
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
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 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
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
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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
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
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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
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
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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
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
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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
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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
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 :
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
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
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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
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
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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
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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
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
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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
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
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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
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
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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
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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
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
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
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
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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
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
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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
(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
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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
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
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
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
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:
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)
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
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
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
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
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
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
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
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
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
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
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
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
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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 = 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):
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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
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
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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
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
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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
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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 )
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
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
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
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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
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
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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 )
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)
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
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
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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)
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
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