69445325 CT Sizing Calculation of 11kV System Rev0 Ver3

March 20, 2018 | Author: rajinipre-1 | Category: Relay, Electrical Resistance And Conductance, Transformer, Quantity, Electrical Equipment
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THIS SUBMISSION IS DECLARED TO BE STRICTLY INACCORDANCE 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 GERMANY SIEMENS LIMITED 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 Super1,Ain Hamad, Ain Khalid south. PROJECT DRAWING NUMBER PH7-3B-10-15-C001 SUBCONTRACTOR / SUPPLIER 0 REV 29-11-2007 DATE DRAWING/DOCUMENT DESCRIPTION FIRST ISSUE MODIFICATION R.K. DRAWN SCALE DESIGN REPORT FOR CURRENT TRANSFORMER (CT) & VOLTAGE N.A. TRANSFORMER (VT) SIZING FOR 11kV SYSTEM Document No.:- TOTAL NO. OF PAGES VER SIEMENS V.H CHECK GTC123-BN00-AQA-10001 91 PAGES 3 V.A. APPRD SIZE A4 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 1.0 PROJECT:GTC/123/2006 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 %age impedance at principal tapping is assumed as 9.5kA.58%. %age impedance at principal tapping is assumed as 12. PH7-3B-10-15-C001..61) / 1000 = 0. 3.61 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. %age impedance at principal tapping is assumed as 6. %age impedance at principal tapping is assumed as 12. C: Resistance at 20 deg. 4. C = 4.2 = 0. Power Transformer 25/30MVA.415kV Earthing Transformer.5/10MVA.785 * 1.60% 6. 11/0. C for copper = 0. %age impedance at principal tapping is assumed as 16. C = 4. Rev 0 Page 4 of 91 . 66/11kV. (clause 12).00 % 7.58%. %age impedance at principal tapping is assumed as 12.61 (1 + 0. C Resistance at 75 deg.785 ohms Considering 20% safety margin as per contract requirement. 33/11kV.SIEMENS 3. 2000kVA. the fault level considered is 31. Calculation for resistance at 75 deg. have been done for a cable length of 50mts.0 DESIGN CRITERIA: KNEE POINT VOLTAGE Apart from rated short time rating of the system.415kV Earthing Transformer. to arrive at minimum knee point voltage. 66/11kV. 11/0. 5.942 ohms All CT sizing calculations. C (in ohms/kM) = (2 x 70 x 5. = 0. 33/11kV. coefficient) at 20 deg. 1000kVA. This is the maximum length possible.0% 4. value for steady state through fault current values needs to be determined.415kV Earthing Transformer. Considering the rated capacity of the bus bars of 11kV. %age impedance at principal tapping is assumed as 12%.00393 / deg. 2. Power Transformer 32/40MVA.61 ohms / Km Value of Alpha (Temp. 500kVA. 11/0.00393 (75-20)) = 5.33%. Power Transformer 20/25MVA.0 PROJECT:GTC/123/2006 ASSUMPTIONS : 1. PARAMETERS FOR CABLE BETWEEN CT & RELAY PANEL Cross section taken is 4 mm2. Power Transformer 7. Annexure # 1: CT Knee point voltage calculations for 11kV feeders 2.0 RESULT OF STUDY: Calculation results show that selected parameters for CT/VT are adequate to meet the minimum requirements. Annexure #5: LSOH Power Cable Data Sheet PH7-3B-10-15-C001. Annexure # 3: VT burden (in VA) calculations for 11kV feeders 4. Rev 0 Page 5 of 91 . Annexure # 4: Relay back-up sheets 5. Annexure # 2: CT burden (in VA) calculations for 11kV feeders 3. 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.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. 7.0 ATTACHMENTS: 1. 942 Ohms (I)² x 0.MIC-2. =A28 Note: .=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.40 MVA LV side Core-4 is not applicable for NDQ.732 Ohms/Km 0. Muraikh North.0 Rct≤8 10VA To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) Core-1 750/1 Class-PX Vk≥250 Rct≤6 Io<25ma 15VA Core-1 750/1 Class-5P20 15VA To Back-up earth fault (7SJ61) Core-4 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma 11kV Bus Core-2 50/1 Class-5P20 15VA Core-2 50/1 Class-PX Vk≥100 Rct≤1. Ain Khalid south 66kV side Core-1 50/1 Class-PX Vk≥100 Rct≤1.MIC-3.QRE.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18.A28) Type:1 Configuration 40MVA. South West Wakrah-1. Rev 0 2500 /1 1 amp.00 Page 6 of 91 .Al Wadi.942 = 9.2 Ohms/Km 6.Abu Hamour central.Al Dhahiya West.942 9.2 Io<30ma To REF protection for Earthing transformer (7SJ61) To REF protection for 11kV side (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Feeder =A18. NBK-2 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.610 Ohms/Km 5. 66/11kV transformer.61x1.2 Io<30ma at Vk 66/11KV 32/40MVA ONAN/ONAF 11/0.Earthing/auxiliary transformer rating of 500kVA Applicable substations:Mosemeer. Al Soudan.Khore Community. NBK-2. 70 meter 5.415KV 500KVA Cable and trafo differential protection (7SD52) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To partial busbar differential protection (7SJ61) Core-2 2500/1 Class-5P20/1. 5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.1388 Vk If = If = 40 1.042 VA (0. Rev 0 Page 7 of 91 .1388 15.100 VA 1.3 = 154.1262 kA 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0.942 )x2 > 2500 > 180.1633 Taking a negative tolerance of 15% = 0.732 x 11 x 0. a current of 1.46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942+0.2x(9+0.042) x 1 1.49 volts = 185.5x15126.39 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.1) Ihigh set point x I2N x (Ri+RBC) 1.3 x Ipn Knee Point voltage required 20% margin on Vk value = 20 x (9+1. the formula used is : (For determining the Voltage developed across CT.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1. 15 1.732 x 11 x 0.05 = RBC = (Loop Burden+ Relay Burden) = 0.20 168.042 + 9) x 1 x 18.1388 ISSC = ISSC = 40 1.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .2 x 3 K'SSC to be considered for calculations = 2500 18.1388 15.1262 kA K'SSC = 15126.24 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.1633 Taking negative tolerance of 15% = 0.3 Knee Point voltage required = (1.042 VA (0.3 Knee Point voltage required 20% margin on Vk value = = 140.15 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer.942+0. Rev 0 Page 8 of 91 .100 VA 1.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0. 100 VA 1. Hence selected CT is OK PH7-3B-10-15-C001.942+0.81 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).1) = K'SSC = (10+8) x20 (1.05 RBC = (Loop Burden+ Relay Burden) = 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.042 VA = 8.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA Relay Burden = 2*0. Rev 0 Page 9 of 91 .042+8) Calculated Value for K' SSC = 39.0 VA (0. 042 VA (0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18.732 0.00 Ohms CT Internal Burden in VA = (I)² x 9 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC = (Relay will be mounted in Relay panel) 0.05 = = RBC = (Loop Burden+ Relay Burden) b) Knee point voltage : 0. Rev 0 Page 10 of 91 .=A28 (connected across core-3 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 2500 1 amp.1) Ihigh set point x I2N x (Ri+RBC) The calculated value of knee point voltage is = Knee Point voltage required = 20% margin on Vk value = = 1.610 6.3 x Ipn 20 x (1.942 = 9. Length of cable between CT and Relay Cable Resistance for mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = 70 5.942+0.000 atleast IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942 Loop Burden in VA Current Transformer resistance (R ct) = = meter Ohms/Km Ohms/Km Ohms (I)² x 0.100 VA 1.942 9.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.3 154.042+9) x 1 1.49 185. 1388 ISSC = K'SSC 15.042+9) x1x7.1633 Taking negative tolerance of 15% = 0.732 x 11 x 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SD5 relay for 11kV incoming feeders =A18. =A28(connected across Core-4) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.2 2500 K'SSC to be considered for calculations = Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.05 = = RBC = (Loop Burden+ Relay Burden) 7.30 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.942+0.000 1.610 Ohms/Km 6.2x1. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = 70 meter 5.942 9.1262 kA = 15126.042 VA (0.26 1.100 VA 1.942 Ohms = Formula Used a) Transformer dimensioning factor (k td): For line differential protection ktd (I)² x 0.1388 = 40 ISSC 1.3 Knee Point voltage required 20% margin on Vk value = = 56.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.732 Ohms/Km 0.00 Ohms (I)² x 9 = = (Relay will be mounted in Relay panel) 0.08 67.942 = 9.20 b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I SCC x ktd I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11kV. the maximum through fault current on 40 MVA trafo would be : %age impedance at 40 MVA = 0.3 Knee Point voltage required = (1. Rev 0 Page 11 of 91 .26 0. 1) Ihigh set point x I2N x (Ri+RBC) 1.39 volts 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.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.942 )x2 > 2500 > 180.=A28 (connected across 11kV Neutral side CTs of 500KVA) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.610 Ohms/Km 5.042 VA (0.61x1.3 154.1262 kA 1.942 = 9.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.732 Ohms/Km (As per contract document) 0.2 6.00 Ohms (I)² x 9 = = = 0.1388 Vk If = If = 40 1.00 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942 9.732 x 11 x 0.942+0. a current of 1.042) x 1 1.942 Ohms = (I)² x 0.100 VA 1.1388 15.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.3 x Ipn 20 x (9+1. the formula used is : (For determining the Voltage developed across CT.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.2x(9+0. Relay Normal Current (IN) Length of cable between CT and Relay = Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = Loop Resistance ( 2R L ) = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA (Relay will be mounted in Relay panel) 70 meter 5.5x15126.49 185.1633 Taking a negative tolerance of 15% = 0. Rev 0 Page 12 of 91 .46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18. 732 0.the maximum through fault current on 500 KVA transformer would be: = = %age impedance at 500 KVA Overall Impedance ( taking 15% negative tolerance) If = 500 1.042 VA (0.0816 As the through fault current could be ground fault current which can be max 750A.942 Ohms = (I)² x 0.49 41.3 x Ipn 1.942+0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = (Relay will be mounted in Relay panel) 70 meter 5. Vk > 1.942 ) x 2 50 > 96. Rev 0 Page 13 of 91 .2 x (1.0816 If = 321.610 Ohms/Km 6.942 = 1.942 = 1.3 34.5 x 750 x (1. the stablility is being checked for this max value.5(I f x (Rct + 2RL ) x 2) Where I f is magnitude of through fault current Considering infinite source.6172 A 0.2 + 0.2 0.2 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.20 Ohms (I)² x 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs) Current Transformer Ration (CTR) = 50 /1 = 1 amp.732 x 11 x 0.1) Ihigh set point x I2N x (Ri+RBC) 1.100 VA 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on vk value = = 0.39 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042) x 1 1.0960 0.39 volts The minimum CT knee point Voltage shall be selected more than above Stability check for REF protection for through fault condition Vk > 1.2+1. 992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 500 KVA HV Neutral side CTs) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp.06 6. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+6) x20 6+0. 1 amp.992 Calculated Value for K' SSC = 60.050 VA 0. Hence selected CT is OK PH7-3B-10-15-C001. Rev 0 Page 14 of 91 .0 VA 15 VA 0. 992 Calculated Value for K' SSC = 147.2 VA 15 VA 0.2+0. 1 amp. Rev 0 Page 15 of 91 .05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+1. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0. Hence selected CT is OK PH7-3B-10-15-C001.2) x20 1.992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).80 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer Current Transformer Ration (CTR) Relay Normal Current (IN) = = 50 amp.050 VA 0. 1) Ihigh set point x I2N x (Ri+RBC) 1.8x(6+0.3 x Ipn 20 x (6+1.000 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.5248 kA > 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.732 Ohms/Km 0.942 Ohms = (I)² x 0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = (Relay will be mounted in Relay panel) 70 meter 5.3 108. the maximum through fault current on 500 KVA transformer would be: = = %age impedance at 500 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk 0.0816 500 1.942 = 6.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 750 /1 = 1 amp. a current of 1.34 130.042) x 1 1.042 VA (0.00 Ohms CT Internal Burden in VA = (I)² x 6 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC 0.942 )x2 750 > 236.100 VA 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source. Rev 0 Page 16 of 91 .942+0.732 x 0. the formula used is : (For determining the Voltage developed across CT.01 volts 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.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.942 = 6.72 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.5x8524.415 x 0.610 Ohms/Km 6.0816 8.0960 0. 942 = 9.942 Ohms (I)² x 0. =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.00 Page 17 of 91 . Rev 0 2500 /1 1 amp.A28) Type:2 Configuration 25MVA.2 Ohms/Km 6.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.2 Io<30ma To REF protection for Earthing transformer (7SJ61) To REF protection for 11kV side (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Feeder =A18.2 Io<30ma at Vk 66/11KV 20/25MVA ONAN/ONAF 11/0.415KV 500KVA Cable and trafo differential protection (7SD52) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To partial busbar differential protection (7SJ61) Core-2 2500/1 Class-5P20/1.61x1.0 Rct≤8 10VA To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) Core-1 750/1 Class-PX Vk≥250 Rct≤6 Io<25ma 15VA Core-1 750/1 Class-5P20 15VA To Back-up earth fault (7SJ61) Core-4 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma 11kV Bus Core-2 50/1 Class-5P20 15VA Core-2 50/1 Class-PX Vk≥100 Rct≤1. Khore Junction 66kV side Core-1 50/1 Class-PX Vk≥100 Rct≤1. 66/11kV transformer.942 9.610 Ohms/Km 5.732 Ohms/Km 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18.=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.Earthing/auxiliary transformer rating of 500kVA Applicable substations:Al Jumailyah. 70 meter 5. 5x12275x(9+0.3 = 154.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0. the formula used is : (For determining the Voltage developed across CT.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1. Rev 0 Page 18 of 91 .2750 kA > 1.942 )x2 2500 > 146.100 VA 1.1069 Vk If = If = 25 1.1) Ihigh set point x I2N x (Ri+RBC) 1.1258 Taking a negative tolerance of 15% = 0.042 VA (0.042) x 1 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942+0.3 x Ipn Knee Point voltage required 20% margin on Vk value = 20 x (9+1.45 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.49 volts = 185. a current of 1.732 x 11 x 0.39 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.1069 12.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0. 1258 Taking negative tolerance of 15% = 0.73 1. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .042 + 9) x 1 x 14. Rev 0 Page 19 of 91 .78 136.54 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.1069 12.732 x 11 x 0.1069 ISSC = ISSC = 25 1.05 = RBC = (Loop Burden+ Relay Burden) = 0.942+0.=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer.3 Knee Point voltage required = (1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18.3 Knee Point voltage required 20% margin on Vk value = = 113.2750 kA K'SSC = 12275 x 3 K'SSC to be considered for calculations = 2500 14.100 VA 1.042 VA (0.73 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0. PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.100 VA 1.042+8) Calculated Value for K' SSC = 39.05 RBC = (Loop Burden+ Relay Burden) = 0.942+0.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA Relay Burden = 2*0.0 VA (0. Rev 0 Page 20 of 91 .042 VA = 8.81 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20). Hence selected CT is OK PH7-3B-10-15-C001.1) = K'SSC = (10+8) x20 (1. 00 Ohms Current Transformer resistance (R ct) CT Internal Burden in VA = (I)² x 9 = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.1) Ihigh set point x I2N x (Ri+RBC) 1.732 Ohms/Km = 0.49 185.=A28 (connected across core-3 ) Current Transformer Ration (CTR) = 2500 Relay Normal Current (IN) = 1 amp.610 Ohms/Km 20% margin on cable resistance = 6.942 9.3 x Ipn 20 x (1. Length of cable between CT and Relay = 70 meter Cable Resistance for mm²at 75°C = 5.042 VA (0.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.000 atleast 0.100 VA 1.042+9) x 1 1.942 Ohms Loop Resistance ( 2R L ) Loop Burden in VA = (I)² x 0.942+0.942 = = 9.05 = = RBC = (Loop Burden+ Relay Burden) b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = (Relay will be mounted in Relay panel) 0. Rev 0 Page 21 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18.3 154. 1069 ISSC = K'SSC 12. =A28(connected across Core-4) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.732 x 11 x 0. the maximum through fault current on 25 MVA trafo would be : %age impedance at 25 MVA = 0.942 = 9.00 Ohms (I)² x 9 = = (Relay will be mounted in Relay panel) 0.89 1.042+9) x1x5.2750 kA = 12275x1.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.1258 Taking negative tolerance of 15% = 0.05 = = RBC = (Loop Burden+ Relay Burden) 5.610 Ohms/Km 6.20 b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I SCC x ktd I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11kV.042 VA (0.100 VA 1.1069 = 25 ISSC 1.000 1.942 9.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SD5 relay for 11kV incoming feeders =A18.2 2500 K'SSC to be considered for calculations = Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = 70 meter 5.942+0.732 Ohms/Km 0. Rev 0 Page 22 of 91 .50 54.3 Knee Point voltage required = (1.942 Ohms = Formula Used a) Transformer dimensioning factor (k td): For line differential protection ktd (I)² x 0.60 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.3 Knee Point voltage required 20% margin on Vk value = = 45.89 0. 61x1.39 volts 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.732 x 11 x 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5x12275x(9+0.942+0. a current of 1.1) Ihigh set point x I2N x (Ri+RBC) 1.610 Ohms/Km 5.45 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.100 VA 1.=A28 (connected across 11kV Neutral side CTs of 500KVA) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.732 Ohms/Km (As per contract document) 0.2750 kA > 1.1258 Taking a negative tolerance of 15% = 0.3 154.042) x 1 1.1069 12. Rev 0 Page 23 of 91 . Relay Normal Current (IN) Length of cable between CT and Relay = Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = Loop Resistance ( 2R L ) = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA 70 meter (Relay will be mounted in Relay panel) 5.1069 Vk If = If = 25 1.942 = 9.00 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0. the formula used is : (For determining the Voltage developed across CT.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.942 )x2 2500 > 146.00 Ohms (I)² x 9 = = = 0.942 9.942 Ohms = (I)² x 0.2 6.49 185.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.042 VA (0.3 x Ipn 20 x (9+1. 942 = 1.0816 As the through fault current could be ground fault current which can be max 750A.0960 0.732 x 11 x 0.732 0.39 volts The minimum CT knee point Voltage shall be selected more than above Stability check for REF protection for through fault condition Vk > If x (Rct + 2RL ) x 2 Where I f is magnitude of through fault current Considering infinite source.942+0.2 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.610 Ohms/Km 6.3 34.3 x Ipn 1. Vk 750 x (1.942 ) x 2 > 50 > 64.2+1.042) x 1 1.the maximum through fault current on 500 KVA transformer would be: %age impedance at 500 KVA Overall Impedance ( taking 15% negative tolerance) = = If = 500 1.2 + 0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on vk value = = 0. the stablility is being checked for this max value.6172 A 0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = 70 meter 5.1) Ihigh set point x I2N x (Ri+RBC) 1.26 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001. Rev 0 Page 24 of 91 .0816 If = 321.042 VA (0.100 VA 1.49 41.2 x (1.20 Ohms (I)² x 1.942 Ohms = (Relay will be mounted in Relay panel) (I)² x 0.2 0.942 = 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs) Current Transformer Ration (CTR) = 50 /1 = 1 amp. 050 VA 0.0 VA 15 VA 0.992 Calculated Value for K' SSC = 60. Hence selected CT is OK PH7-3B-10-15-C001.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+6) x20 6+0. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0. 1 amp.06 6. Rev 0 Page 25 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 500 KVA HV Neutral side CTs) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp.992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20). 992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).992 Calculated Value for K' SSC = 147. Hence selected CT is OK PH7-3B-10-15-C001.2) x20 1.2+0. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0.2 VA 15 VA 0.050 VA 0.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+1. 1 amp.80 1. Rev 0 Page 26 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer Current Transformer Ration (CTR) Relay Normal Current (IN) = = 50 amp. 01 volts 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.3 108.3 x Ipn 20 x (6+1.5248 kA > 1.042) x 1 1.5x8524.942 = 6. the maximum through fault current on 500 KVA transformer would be: %age impedance at 500 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk = = 0.0816 8.000 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942 Ohms = (Relay will be mounted in Relay panel) (I)² x 0. Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = 70 meter 5.1) Ihigh set point x I2N x (Ri+RBC) 1.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0. Rev 0 Page 27 of 91 .942 = 6.732 x 0.0960 0.34 130.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.8x(6+0. the formula used is : (For determining the Voltage developed across CT.942 )x2 750 > 236. a current of 1.72 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.415 x 0.042 VA (0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 750 /1 Relay Normal Current (IN) = 1 amp.0816 500 1.732 Ohms/Km 0.00 Ohms CT Internal Burden in VA = (I)² x 6 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 0.610 Ohms/Km 6.100 VA 1.942+0. 4 Io<30ma at Vk/2 11/0.Lusail Development Super 1 66kV side Core-2 100/1 Class-5P20 15VA 66/11KV 32/40MVA ONAN/ONAF Core-3 400/1 Class-PX Vk≥250 Rct≤9 Io<30ma at Vk Core-1 100/1 Class-PX Vk≥100 Rct≤0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18. =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18. 70 meter 5.2 Ohms/Km 6.732 Ohms/Km 0.0 Rct≤8 10VA To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) 11kV Bus Core-1 1500/1 Class-PX Vk≥500 Rct≤7.942 Ohms (I)² x 0.=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.00 Page 28 of 91 .415KV 1000KVA Core-1 750/1 Class-5P20 15VA To Back-up earth fault (7SJ61) Core-2 100/1 Class-PX Vk≥100 Rct≤0. South west wakrah. 66/11kV transformer.A28) Type:3 Configuration 40MVA. Murraikh North.61x1.942 9.942 = 9.Earthing/auxiliary transformer rating of 1000kVA Applicable substations : NDQ. Rev 0 2500 /1 1 amp.610 Ohms/Km 5.4 Io<30ma at Vk/2 To REF protection for Earthing transformer (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To partial busbar differential protection (7SJ61) Core-2 2500/1 Class-5P20/1.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.5 Io<25ma To REF protection for 11kV side (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Feeder =A18. 1262 kA 1. Rev 0 Page 29 of 91 .942+0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5x15126.2x(9+0.732 x 11 x 0.1388 15.46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.49 volts = 185.042) x 1 1. the formula used is : (For determining the Voltage developed across CT.942 )x2 > 2500 > 180.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.3 x Ipn Knee Point voltage required 20% margin on Vk value = 20 x (9+1.3 = 154.1388 Vk If = If = 40 1.39 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.1) Ihigh set point x I2N x (Ri+RBC) 1.042 VA (0. a current of 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0.1633 Taking a negative tolerance of 15% = 0.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.100 VA 1. 942+0.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.042 VA (0.2 x 3 K'SSC to be considered for calculations = 2500 18.1388 15.15 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.1388 ISSC = ISSC = 40 1.100 VA 1. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18.15 1.05 = RBC = (Loop Burden+ Relay Burden) = 0.1262 kA K'SSC = 15126. Rev 0 Page 30 of 91 .3 Knee Point voltage required = (1.24 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.1633 Taking negative tolerance of 15% = 0.732 x 11 x 0.20 168.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer.3 Knee Point voltage required 20% margin on Vk value = = 140.042 + 9) x 1 x 18. 05 RBC = (Loop Burden+ Relay Burden) = 0.100 VA 1.942+0.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA Relay Burden = 2*0. Rev 0 Page 31 of 91 .042+8) Calculated Value for K' SSC = 39.042 VA = 8.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.1) = K'SSC = (10+8) x20 (1.0 VA (0. Hence selected CT is OK PH7-3B-10-15-C001.81 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20). 942 9. Rev 0 Page 32 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18.000 atleast IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.732 0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : 0.3 154.942 Loop Burden in VA Current Transformer resistance (R ct) = = meter Ohms/Km Ohms/Km Ohms (I)² x 0.100 VA 1.1) Ihigh set point x I2N x (Ri+RBC) The calculated value of knee point voltage is = Knee Point voltage required = 20% margin on Vk value = = 1.49 185.00 Ohms CT Internal Burden in VA = (I)² x 9 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC = (Relay will be mounted in Relay panel) 0.=A28 (connected across core-3 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 2500 1 amp.610 6.942+0.042 VA (0.942 = 9. Length of cable between CT and Relay #REF! 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = 70 5.3 x Ipn 20 x (1.042+9) x 1 1.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001. 05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.=A28 (connected across 11kV Neutral side CTs of 1000KVA) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.042 VA (0.46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.00 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942 = 9.1) Ihigh set point x I2N x (Ri+RBC) 1.61x1.042) x 1 1.1388 15.732 x 11 x 0.942 = 9.942 )x2 > 2500 > 180.942 Ohms = (I)² x 0.2x(9+0.1388 Vk If = If = 40 1.732 Ohms/Km (As per contract document) 0.1633 Taking a negative tolerance of 15% = 0.3 154.00 Ohms (I)² x 9 0.610 Ohms/Km 5.49 185. the formula used is : (For determining the Voltage developed across CT.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.2 6. a current of 1.39 volts 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.1262 kA 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source. Rev 0 Page 33 of 91 . Relay Normal Current (IN) Length of cable between CT and Relay = Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = Loop Resistance ( 2R L ) = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = (Relay will be mounted in Relay panel) 70 meter 5.5x15126.the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.3 x Ipn 20 x (9+1.100 VA 1.942+0. 942+0.0600 0.4 0.20 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042 VA (0.1) Ihigh set point x I2N x (Ri+RBC) 1.942 ) x 2 > 100 > 30.942 = 0.042) x 1 1.0510 1000 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on vk value = = 0.18 26. Rev 0 Page 34 of 91 . Vk 1.610 Ohms/Km 6.40 Ohms (I)² x 0.3 22.62 volts The minimum CT knee point Voltage shall be selected more than above Stability check for REF protection for through fault condition Vk > 1.5 x 750 x (0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = (Relay will be mounted in Relay panel) 70 meter 5.732 x 11 x 0.051 1029.3 x Ipn 20 x (0.732 0.942 Ohms = (I)² x 0.5(I f x (Rct + 2RL ) x 2) Where I f is magnitude of through fault current Considering infinite source.942 = 0. the stablility is being checked for this max value.4+1.1751 A As the through fault current could be ground fault current which can be max 750A.4 + 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs) Current Transformer Ration (CTR) = 100 /1 = 1 amp.4 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.100 VA 1.the maximum through fault current on 1000 KVA transformer would be: %age impedance at 1000 KVA Overall Impedance ( taking 15% negative tolerance) If = If = = = 0. 1 amp.050 VA 0. Hence selected CT is OK PH7-3B-10-15-C001.06 6. Rev 0 Page 35 of 91 .992 Calculated Value for K' SSC = 60.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+6) x20 6+0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 1000 KVA HV Neutral side CTs) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0.0 VA 15 VA 0.992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20). 80 1.992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).2) x20 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer Current Transformer Ration (CTR) Relay Normal Current (IN) = = 100 amp.2 VA 15 VA 0. Rev 0 Page 36 of 91 .992 Calculated Value for K' SSC = 147.050 VA 0. Hence selected CT is OK PH7-3B-10-15-C001. 1 amp.2+0.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+1. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0. 72 = 7.3 Knee Point voltage required 20% margin on Vk value = = 59.051 = ISSC = K'SSC = K'SSC to be considered for calculations 1.0292 kA 1029. k td b) Effective symmetrical short-circuit current factor (K' SSC): I SCC x ktd The required K'SSC = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .1) Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.042 VA volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.55 0.942+0.62 71.17505444336 x 3 400 7.72 = (0.100 VA 1.72 1.0510 ISSC = 1 1. Rev 0 Page 37 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 1000 KVA EAT-(connected across core-3 ) Formula Used a) Transient dimensioning factor (k td): = 3 For transformer.the maximum through fault current on 1 MVA transformer would be: %age impedance at 1 MVA = 0.05 RBC = (Loop Burden+ Relay Burden) = c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.0600 Taking negative tolerance of 15% = 0.732 x 11 x 0.3 Knee Point voltage required = (1.042 + 9) x 1 x 7. 942+0.051 27.5+1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.5+0.3x(7.5x27279.042) x 1 1.1) Ihigh set point x I2N x (Ri+RBC) 1.732 x 0.100 VA 1.58 volts Page 38 of 91 .3 131.042 VA (0.942 = 7.415 x 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.3 x Ipn 20 x (7.0600 0.50 Ohms CT Internal Burden in VA = (I)² x 7. a current of 1.942 = 7. the formula used is : (For determining the Voltage developed across CT.942 Ohms = (I)² x 0.610 Ohms/Km 6.0510 1000 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 1500 /1 Relay Normal Current (IN) = 1 amp.5 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 0.942 )x2 > 1500 > PH7-3B-10-15-C001.42 157. the maximum through fault current on 1000 KVA transformer would be: %age impedance at 1000 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk = = 0.70 volts 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. Rev 0 460.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.732 Ohms/Km 0. Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = (Relay will be mounted in Relay panel) 70 meter 5.2793 kA 1.500 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0. 0 Rct≤8 10VA 11kV Bus Core-3 400/1 Class-PX Vk≥250 Rct≤9 Io<30ma To REF protection for 11kV side (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Feeder =A18.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18.610 Ohms/Km 5.5 Io<25ma Core-1 750/1 Class-5P20 15VA To Back-up earth fault (7SJ61) Core-2 2500/1 Class-5P20/1.4 Io<30ma Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma 11/0. =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.4 Io<30ma To REF protection for Earthing transformer (7SJ61) To partial busbar differential protection (7SJ61) To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) Core-1 1500/1 Class-PX Vk≥500 Rct≤7.942 9.732 Ohms/Km 0.415KV 1000KVA Core-2 100/1 Class-PX Vk≥100 Rct≤0.942 Ohms (I)² x 0. 70 meter 5.942 = 9.A28) Type:4 Configuration 25MVA. 66/11kV transformer.00 Page 39 of 91 .2 Ohms/Km 6.Earthing/auxiliary transformer rating of 1000kVA Applicable substations:Abu Thaila Substation Modification 66kV side Core-2 100/1 Class-5P20 15VA 66/11KV 20/25MVA ONAN/ONAF Core-1 100/1 Class-PX Vk≥100 Rct≤0. Rev 0 2500 /1 1 amp.61x1. 1069 Vk If = If = 25 1.1258 Taking a negative tolerance of 15% = 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.39 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.100 VA 1. a current of 1. the formula used is : (For determining the Voltage developed across CT.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0.5x12275x(9+0.1) Ihigh set point x I2N x (Ri+RBC) 1.942+0.732 x 11 x 0.042) x 1 1.45 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.3 = 154.49 volts = 185.1069 12.942 )x2 2500 > 146.3 x Ipn Knee Point voltage required 20% margin on Vk value = 20 x (9+1.042 VA (0.2750 kA > 1. Rev 0 Page 40 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0. 1258 Taking negative tolerance of 15% = 0. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .732 x 11 x 0.73 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.1069 12.05 = RBC = (Loop Burden+ Relay Burden) = 0.1069 ISSC = ISSC = 25 1.3 Knee Point voltage required = (1.942+0.100 VA 1.2750 kA K'SSC = 12275 x 3 K'SSC to be considered for calculations = 2500 14.73 1.042 + 9) x 1 x 14.78 136.54 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042 VA (0.=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18.3 Knee Point voltage required 20% margin on Vk value = = 113.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0. Rev 0 Page 41 of 91 . 942+0.05 RBC = (Loop Burden+ Relay Burden) = 0.042 VA = 8.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA Relay Burden = 2*0. Rev 0 Page 42 of 91 .100 VA 1.0 VA (0. Hence selected CT is OK PH7-3B-10-15-C001.81 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).042+8) Calculated Value for K' SSC = 39.1) = K'SSC = (10+8) x20 (1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18. 610 Ohms/Km 20% margin on cable resistance = 6.942+0. Length of cable between CT and Relay = 70 meter #REF! = 5.=A28 (connected across core-3 ) Current Transformer Ration (CTR) = 2500 Relay Normal Current (IN) = 1 amp.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042 VA (0.1) Ihigh set point x I2N x (Ri+RBC) 1.00 Ohms Current Transformer resistance (R ct) CT Internal Burden in VA = (I)² x 9 = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.49 185.100 VA 1.3 154.05 = = RBC = (Loop Burden+ Relay Burden) b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = (Relay will be mounted in Relay panel) 0.942 Ohms Loop Resistance ( 2R L ) Loop Burden in VA = (I)² x 0.000 atleast 0.042+9) x 1 1. Rev 0 Page 43 of 91 .942 = = 9.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18.3 x Ipn 20 x (1.732 Ohms/Km = 0.942 9. 1069 Vk If = If = 25 1.the maximum through fault current on 25 MVA transformer would be: %age impedance at 25 MVA = 0.1) Ihigh set point x I2N x (Ri+RBC) 1.2750 kA > 1. Rev 0 Page 44 of 91 .3 154.49 185.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.610 Ohms/Km 5.732 Ohms/Km (As per contract document) 0.2 6.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.39 volts 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.942 = 9.942 Ohms = (I)² x 0.042) x 1 1.1069 12.00 Ohms (I)² x 9 = = = 0.942 )x2 2500 > 146.00 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.942 9.042 VA (0.61x1. Relay Normal Current (IN) Length of cable between CT and Relay = Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = Loop Resistance ( 2R L ) = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA (Relay will be mounted in Relay panel) 70 meter 5.1258 Taking a negative tolerance of 15% = 0. a current of 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.=A28 (connected across 11kV Neutral side CTs of 1000KVA) Current Transformer Ration (CTR) = 2500 /1 = 1 amp.732 x 11 x 0.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source. the formula used is : (For determining the Voltage developed across CT.942+0.3 x Ipn 20 x (9+1.5x12275x(9+0.45 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.100 VA 1. Vk 750 x (0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = (Relay will be mounted in Relay panel) 70 meter 5.942 Ohms = (I)² x 0.4 x (0.942 = 0.942 = 0.1751 A As the through fault current could be ground fault current which can be max 750A. the stablility is being checked for this max value.18 26.1) Ihigh set point x I2N x (Ri+RBC) 1.942 ) x 2 > 100 > 20.732 x 11 x 0.3 22.4+1.042) x 1 1.610 Ohms/Km 6.0600 0.3 x Ipn 0.13 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.0510 1000 1.100 VA 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs) Current Transformer Ration (CTR) = 100 /1 = 1 amp.4 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.40 Ohms (I)² x 0.051 1029.732 0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on vk value = = 0. Rev 0 Page 45 of 91 .942+0.4 + 0.62 volts The minimum CT knee point Voltage shall be selected more than above Stability check for REF protection for through fault condition Vk > If x (Rct + 2RL ) x 2 Where I f is magnitude of through fault current Considering infinite source.4 0.042 VA (0.the maximum through fault current on 1000 KVA transformer would be: %age impedance at 1000 KVA Overall Impedance ( taking 15% negative tolerance) If = If = = = 0. 992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).992 Calculated Value for K' SSC = 60. 1 amp.06 6.050 VA 0.0 VA 15 VA 0.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+6) x20 6+0. Rev 0 Page 46 of 91 . a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0. Hence selected CT is OK PH7-3B-10-15-C001.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 1000 KVA HV Neutral side CTs) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp. 2) x20 1.80 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer Current Transformer Ration (CTR) Relay Normal Current (IN) = = 100 amp. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0.05 RBC = (Loop Burden+ Relay Burden) = = = = K'SSC = (15+1. Rev 0 Page 47 of 91 . Hence selected CT is OK PH7-3B-10-15-C001. 1 amp.2+0.992 Calculated Value for K' SSC = 147.992 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).2 VA 15 VA 0.050 VA 0. PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 1000 KVA EAT-(connected across core-3 ) Formula Used a) Transient dimensioning factor (k td): = 3 For transformer.100 VA 1.05 = RBC = (Loop Burden+ Relay Burden) = 1 1. k td b) Effective symmetrical short-circuit current factor (K' SSC): I SCC x ktd The required K'SSC = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .55 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.942+0.042 + 9) x 1 x 7.62 71.0292 kA 1029.3 Knee Point voltage required 20% margin on Vk value = = 59.1) c) Knee point voltage : The calculated value of knee point = (RBC+ Ri ) X ISN X K'SSC Knee Point voltage required = (1.0600 Taking negative tolerance of 15% = 0.the maximum through fault current on 1 MVA transformer would be: %age impedance at 1 MVA = 0.0510 ISSC = ISSC = K'SSC = K'SSC to be considered for calculations = Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.042 VA (0.051 1.732 x 11 x 0.17505444336 x 3 400 7.72 1. Rev 0 Page 48 of 91 .72 0. 50 Ohms CT Internal Burden in VA = (I)² x 7.942 Ohms = (Relay will be mounted in Relay panel) (I)² x 0.58 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.415 x 0.5x27279.042 VA (0.3x(7.732 x 0.042) x 1 1.70 volts 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 : (For determining the Voltage developed across CT.100 VA 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.5 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC 0.942+0.1) Ihigh set point x I2N x (Ri+RBC) 1. a current of 1.500 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 1500 /1 = 1 amp.942 )x2 1500 > 460.0600 0. the maximum through fault current on 1000 KVA transformer would be: = = %age impedance at 1000 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk 0.942 = 7.5+1.2793 kA > 1.3 x Ipn 20 x (7.3 131.5+0.0510 1000 1. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = 70 meter 5.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.42 157.942 = 7.610 Ohms/Km 6.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0. Rev 0 Page 49 of 91 .051 27.732 Ohms/Km 0. PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18,A28) Type:5 Configuration 40MVA, 66/11kV transformer,Earthing/auxiliary transformer rating of 2000kVA Applicable substations:MIC Super , Al Dhahiya,Al Waab Super, Wakrah 2, EDS 66kV side Core-2 200/1 Class-5P20 15VA 66/11KV 32/40MVA ONAN/ONAF Core-1 200/1 Class-PX Vk≥100 Rct≤0.6 Io<30ma at Vk/2 Core-3 400/1 Class-PX Vk≥100 Rct≤9 Io<30ma at Vk/2 11/0.415KV 2000KVA Core-1 750/1 Class-5P20 15VA To Back-up earth fault (7SJ61) Core-2 200/1 Class-PX Vk≥100 Rct≤0.6 Io<30ma To REF protection for Earthing transformer (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To partial busbar differential protection (7SJ61) Core-2 2500/1 Class-5P20/1.0 Rct≤8 10VA To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) 11kV Bus Core-1 3000/1 Class-PX Vk≥450 Rct≤15 Io<25ma To REF protection for 11kV side (7SJ61) Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Feeder =A18, =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001, Rev 0 2500 /1 1 amp. 70 meter 5.610 Ohms/Km 5.61x1.2 Ohms/Km 6.732 Ohms/Km 0.942 Ohms (I)² x 0.942 = 9.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.942 9.00 Page 50 of 91 PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0.100 VA 1.042 VA (0.942+0.1) Ihigh set point x I2N x (Ri+RBC) 1.3 x Ipn Knee Point voltage required 20% margin on Vk value = 20 x (9+1.042) x 1 1.3 = 154.49 volts = 185.39 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 : (For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source,the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.1633 Taking a negative tolerance of 15% = 0.1388 Vk If = If = 40 1.732 x 11 x 0.1388 15.1262 kA > 1.5x15126.2x(9+0.942 )x2 2500 > 180.46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001, Rev 0 Page 51 of 91 PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18,=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer, k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV ,the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.1633 Taking negative tolerance of 15% = 0.1388 ISSC = ISSC = 40 1.732 x 11 x 0.1388 15.1262 kA K'SSC = 15126.2 x 3 K'SSC to be considered for calculations = 2500 18.15 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.05 = RBC = (Loop Burden+ Relay Burden) = 0.100 VA 1.042 VA (0.942+0.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.3 Knee Point voltage required = (1.042 + 9) x 1 x 18.15 1.3 Knee Point voltage required 20% margin on Vk value = = 140.20 168.24 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001, Rev 0 Page 52 of 91 81 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA Relay Burden = 2*0.1) = K'SSC = (10+8) x20 (1.05 RBC = (Loop Burden+ Relay Burden) = 0. Hence selected CT is OK PH7-3B-10-15-C001.042+8) Calculated Value for K' SSC = 39.942+0. Rev 0 Page 53 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.042 VA = 8.100 VA 1.0 VA (0. PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18,=A28 (connected across core-3 ) Current Transformer Ration (CTR) = 2500 Relay Normal Current (IN) = 1 amp. Length of cable between CT and Relay = 70 meter Cable Resistance for mm²at 75°C = 5.610 Ohms/Km 20% margin on cable resistance = 6.732 Ohms/Km = 0.942 Ohms Loop Resistance ( 2R L ) Loop Burden in VA = (I)² x 0.942 = = 9.00 Ohms Current Transformer resistance (R ct) CT Internal Burden in VA = (I)² x 9 = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.05 = = RBC = (Loop Burden+ Relay Burden) b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = (Relay will be mounted in Relay panel) 0.942 9.000 atleast 0.100 VA 1.042 VA (0.942+0.1) Ihigh set point x I2N x (Ri+RBC) 1.3 x Ipn 20 x (1.042+9) x 1 1.3 154.49 185.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001, Rev 0 Page 54 of 91 PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 relay for 11kV Incomer Feeders =A18,=A28 (connected across 11kV Neutral side CTs of 2000KVA) Current Transformer Ration (CTR) = 2500 /1 = 1 amp. Relay Normal Current (IN) Length of cable between CT and Relay = Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = Loop Resistance ( 2R L ) = Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA 70 meter (Relay will be mounted in Relay panel) 5.610 Ohms/Km 5.61x1.2 6.732 Ohms/Km (As per contract document) 0.942 Ohms = (I)² x 0.942 = 9.00 Ohms (I)² x 9 = = = 0.942 9.00 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.100 VA 1.042 VA (0.942+0.1) Ihigh set point x I2N x (Ri+RBC) 1.3 x Ipn 20 x (9+1.042) x 1 1.3 154.49 185.39 volts 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 : (For determining the Voltage developed across CT, a current of 1.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source,the maximum through fault current on 40 MVA transformer would be: %age impedance at 40 MVA = 0.1633 Taking a negative tolerance of 15% = 0.1388 Vk If = If = 40 1.732 x 11 x 0.1388 15.1262 kA > 1.5x15126.2x(9+0.942 )x2 2500 > 180.46 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001, Rev 0 Page 55 of 91 PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT calculation for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for Phase and Neutral side CTs) Current Transformer Ration (CTR) = 200 /1 = 1 amp. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = CT Internal Burden in VA = 70 meter 5.610 Ohms/Km 6.732 0.942 Ohms = (Relay will be mounted in Relay panel) (I)² x 0.942 = 0.60 Ohms (I)² x 0.6 0.942 = 0.6 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on vk value = = 0.100 VA 1.042 VA (0.942+0.1) Ihigh set point x I2N x (Ri+RBC) 1.3 x Ipn 20 x (0.6+1.042) x 1 1.3 25.26 30.31 volts The minimum CT knee point Voltage shall be selected more than above Stability check for REF protection for through fault condition Vk >1.5( I f x (Rct + 2RL ) x 2) Where I f 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 Overall Impedance ( taking 15% negative tolerance) If = If = = = 0.1200 0.1020 2000 1.732 x 11 x 0.102 1029.1751 A As the through fault current could be ground fault current which can be max 750A, the stablility is being checked for this max value. Vk 750 x (0.6 +, 0.942 ) x 2x1.5 > 200 > 17.34 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001, Rev 0 Page 56 of 91 PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy check for 7SJ61 type used for REF protection of 11kV side of Earthing transformer (Applicable for 2000 KVA HV Neutral side CTs) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp.992 Calculated Value for K' SSC = 60. Hence selected CT is OK PH7-3B-10-15-C001.050 VA 0.992 VA K'SSC = 6. Rev 0 Page 57 of 91 . a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = = Also K'SSC 20 atleast RBN+ Ri x KSSC Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 15 VA Relay Burden = 0.0 VA (15+6) x20 6+0.05 RBC = (Loop Burden+ Relay Burden) = = 0. 1 amp.06 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20). 1 amp.05 RBC = (Loop Burden+ Relay Burden) = = 0.992 Calculated Value for K' SSC = 147. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = = Also K'SSC 20 atleast RBN+ Ri x KSSC Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 15 VA Relay Burden = 0. Hence selected CT is OK PH7-3B-10-15-C001.992 VA K'SSC = 1.050 VA 0.2+0. Rev 0 Page 58 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer Current Transformer Ration (CTR) Relay Normal Current (IN) = = 200 amp.80 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).2 VA (15+1.2) x20 1. 3 Knee Point voltage required 20% margin on Vk value = = 59.3 Knee Point voltage required = (1.72 1.1200 Taking negative tolerance of 15% = 0. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I SCC x ktd I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .0292 kA 1029.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1. Rev 0 Page 59 of 91 .042 VA (0.62 71.72 0.the maximum through fault 40 current on 2 MVA transformer would be: %age impedance at 2 MVA = 0.102 ISSC = K'SSC = K'SSC to be considered for calculations = Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.05 = RBC = (Loop Burden+ Relay Burden) = 1.1020 ISSC 2 = 1.042 + 9) x 1 x 7.100 VA 1.55 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.732 x 11 x 0.17505444336 x 3 400 7.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders to 2000 KVA EAT-(connected across core-3 ) Formula Used a) Transient dimensioning factor (k td): 3 For transformer.942+0. 1020 2000 1.2793 kA 1.89 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.3 246.942 = 15.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.942+0.942 = 15.3 x Ipn 20 x (15+1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 3000 /1 = 1 amp.00 Ohms CT Internal Burden in VA = (I)² x 15 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 0.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.000 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0. a current of 1.942 Ohms = (Relay will be mounted in Relay panel) (I)² x 0.942 )x2 > 3000 > 434.042) x 1 1.3x(15+0.100 VA 1.5x27279. the maximum through fault current on 2000 KVA transformer would be: = = %age impedance at 2000 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk 0.732 x 0. the formula used is : (For determining the Voltage developed across CT.1) Ihigh set point x I2N x (Ri+RBC) 1. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = 70 meter 5.415 x 0.1200 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.102 27.80 296.732 Ohms/Km 0.16 volts 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. Rev 0 Page 60 of 91 .610 Ohms/Km 6.042 VA (0. 610 Ohms/Km 5.PX Vk≥250 Rct≤9 Io=30mA at Vk/2 Core-2 600-300/1 Class .0 20 .730 Ohms/Km 0.300 Rct≤9 .5 Io=30mA at Vk/2 11kV side REF Protection (7SJ61) Partial Bus Bar protection (7SJ61) To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) To Transformer REF (11kV side) & Main differential protection (7SJ61 & 7UT613) Io=30mA at Vk/2 11kV Bus Feeder =A18.PX Vk≥ Rct≤9 .5P20/1.5 Core-1 600-300/1 Class . 30s NER 33/11 KV 7.A28) Type:6 Configuration: 10MVA.=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 600 /1 1 amp.942 Ohms Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.61x1.5/10MVA ONAN/ONAF Backup earth fault (7SJ61) Core-3 800/1 Class .942 9.2 Ohms/Km 6. =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.4.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18.47ohm 750A.4. Length of cable between CT and Relay #REF! 20% margin on Cable resistance Loop Resistance ( 2R L ) = = = = = 70 meter 5.10 VA Core-1 600-300/1 Class .00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.5P20 15 VA 8.00 Page 61 of 91 . Rev 0 (I)² x 0. 33/11kV transformer Applicable substations:Ain Hamad 33 KV SIDE Core-2 750/1 Class .942 = 9.PX Vk≥600 . 3 = 154.042 VA (0.1200 Taking a negative tolerance of 15% = 0. Rev 0 Page 62 of 91 .05 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042) x 1 1.26 volts = 185.102 5.3 Ratio 300/1 20 x (4.1 VA 1.942 ) x 2 1.39 102.5 + 0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5 x 5146 x (9 + 0.5 + 1.942 + 0.05 = RBC = (Loop Burden+ Relay Burden) = b) Knee point voltage : The calculated value of knee point voltage is = 0.31 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.102 If = If = 10 1.042) x 1 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source. a current of 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = The required K'SSC 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.81 300 volts 280. the formula used is : (For determining the Voltage developed across CT.732 x 11 x 0.1) Ihigh set point x I2N x (Ri+RBC) 1.49 volts 85.3 x Ipn Knee Point voltage required 20% margin on Vk value = Ratio 600/1 20 x (9 + 1.the maximum through fault current on 10 MVA transformer would be: %age impedance at 10 MVA = 0.146 kA Ratio 600/1 Vk > Ratio 300/1 1.942 ) x 2 600 > 255.5 x 5146 x (4. Rev 0 Page 63 of 91 .46 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18.00 For transformer.46 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.73 300 51.38 volts 263.5) x 1 x 51.1020 ISSC = ISSC = 10.=A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (k td): 3.1200 Taking negative tolerance of 15% = 0.942 + 0.1 VA 1.25 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.3 219.042 + 4.3 Ratio 600/1 Knee Point voltage required = (1. k td = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .00 1.102 5.75 volts 238.146 kA Ratio 600/1 K'SSC K'SSC to be considered for calculations = 5146 x 3 5146 x 3 = 600 25.1) 0.732 x 11 x 0.3 Knee Point voltage required 20% margin on Vk value = = 198.the maximum through fault current on 10 MVA transformer would be: %age impedance at 10 MVA = 0.042 + 9) x 1 x 25.05 = = RBC = (Loop Burden+ Relay Burden) c) Knee point voltage : The calculated value of knee point voltage is = Ratio 300/1 (0.51 volts Ratio 300/1 (1.73 1.042 VA (RBC+ Ri ) X ISN X K'SSC 1. Hence selected CT is OK PH7-3B-10-15-C001.942 + 0.1 VA 1.1) 20 atleast Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 2*0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.042 + 8) Calculated Value for K' SSC = 61.53 > 20 20 Since the selected K' SSC is more than Minimum required K' SSC (20).042 VA Ratio 300/1 (10 + 4) x 20 (1.042 + 4) 55. Rev 0 Page 64 of 91 .05 = RBC = (Loop Burden+ Relay Burden) K'SSC = Ratio 600/1 (20 + 8) x 20 (1.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = RBN+ Ri x KSSC RBC + Ri = (0.93 0. 942 9.000 atleast 0.1) Ihigh set point x I2N x (Ri+RBC) 1. Rev 0 Page 65 of 91 .610 Ohms/Km 20% margin on cable resistance = 6.49 185.042+9) x 1 1.100 VA 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18.942 Ohms Loop Resistance ( 2R L ) Loop Burden in VA = (I)² x 0.05 = = RBC = (Loop Burden+ Relay Burden) b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = (Relay will be mounted in Relay panel) 0. Length of cable between CT and Relay = 70 meter Cable Resistance for mm²at 75°C = 5.00 Ohms Current Transformer resistance (R ct) CT Internal Burden in VA = (I)² x 9 = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): = I High set point = 20 The required K'SSC IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.042 VA (0.3 154.3 x Ipn 20 x (1.732 Ohms/Km = 0.942 = = 9.=A28 (connected across core-3 ) Current Transformer Ration (CTR) = 800 Relay Normal Current (IN) = 1 amp.942+0.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001. 992 + 9) x 1 x 20 = = 0.102 5.49 101. a current of 1. Rev 0 Page 66 of 91 .5) x 1 x 20 1.942 ) x 2 Ratio 300/1 1.5 x 5146 x (4.3 153.05 The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.05VA RBC = (Loop Burden+ Relay Burden) 0.942 + 0.39 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.1200 Taking a negative tolerance of 15% = 0.1020 If = If = 10 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for REF protection with 7SJ61 relay (connected across LV neutral core-1) Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = IN 20 atleast RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden =0.the maximum through fault current on 10 MVA transformer would be: %age impedance at 10 MVA = 0.72 184. the formula used is : (For determining the Voltage developed across CT.732 x 11 x 0.5 + 0.05) Ratio 600/1 Knee Point voltage required = Knee Point voltage required 20% margin on Vk value (As per contract document) (0.05 VA = (0.5 x 5146 x (9 + 0.3 1.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.47 84.81 280.992 VA Ratio 300/1 (0.146 kA Ratio 600/1 Vk > > 1.942 ) x 2 600 300 255.992 + 4. 2+0.76 > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).050 VA 0.992 VA (15+2. a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequecy check for 7SJ61 type used for O/C protection of 11kV side of Earthing transformer -Core-2 Current Transformer Ration (CTR) Relay Normal Current (IN) = = 750 amp.992 Calculated Value for K' SSC = 107. 1 amp. Hence selected CT is OK PH7-3B-10-15-C001.2) x20 2. Rev 0 Page 67 of 91 .05 RBC = (Loop Burden+ Relay Burden) K'SSC 20 atleast = = = = = 2.2 VA 15 VA 0. 2 Ohms/Km 6.00 Ohms (I)² x 9 = (Relay will be mounted in Relay panel) 0.4 Io<30ma at Vk Core-3 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma To partial busbar differential protection (7SJ61) Core-2 2500/1 Class-5P20/1. 70 meter 5.942 Ohms (I)² x 0. 33/11kV transformer.942 = 9.=A28 (connected across core-1 ) Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on Cable resistance Loop Resistance ( 2RL ) = = = = = Loop Burden in VA Current Transformer resistance (Rct) CT Internal Burden in VA = = = PH7-3B-10-15-C001.0 Rct≤8 10VA To Directional Overcurrent & earth fault / Metering (6MD6 & 7SJ62) Core-1 2500/1 Class-PX Vk≥ 250 Rct≤9 Io<30ma To Transformer Main differential protection ( 7UT613) 11kV Bus Feeder =A18.00 Page 68 of 91 . Rev 0 2500 /1 1 amp.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Incomer Feeder (Typical bay no: A18.A28) Type:7 Configuration 30MVA. =A28 CT knee point voltage calculation for 7SJ61 relay for 11kV Incomer Feeders =A18.732 Ohms/Km 0.942 9.610 Ohms/Km 5.Earthing/auxiliary transformer rating of 1000kVA Applicable substations:RLF-3 33kV side 33/11KV 25/30MVA ONAN/ONAF NER Core-1 75/1 Class-PX Vk≥250 Rct≤0.61x1. 1258 Taking a negative tolerance of 15% = 0.1) Ihigh set point x I2N x (Ri+RBC) 1.39 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.042 VA (0.3 = 154.942 )x2 > 2500 > 175.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = I High set point = 20 atleast IN Value 20 will be selected for calculations RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.the maximum through fault current on 30 MVA transformer would be: %age impedance at 30 MVA = 0.49 volts = 185. Rev 0 Page 69 of 91 .1069 If = If = Vk 30 1. a current of 1.732 x 11 x 0.5 x If x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.1069 14.100 VA 1.05 RBC = (Loop Burden+ Relay Burden) = = b) Knee point voltage : The calculated value of knee point voltage is = 0.3 x Ipn Knee Point voltage required = 20% margin on Vk value 20 x (9+1.5x14730x(9+0.042) x 1 1.7300 kA 1.73 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001. the formula used is : (For determining the Voltage developed across CT.942+0. =A28 (connected across core-1 ) Formula Used a) Transient dimensioning factor (ktd): For transformer.7300 kA K'SSC = 14730 x 3 K'SSC to be considered for calculations = 2500 17.1069 ISSC = ISSC = 30 1.89 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.042 VA (0.042 + 9) x 1 x 17.05 = = RBC = (Loop Burden+ Relay Burden) 0.1) c) Knee point voltage : The calculated value of knee point voltage is = (RBC+ Ri ) X ISN X K'SSC 1.1258 Taking negative tolerance of 15% = 0.3 Knee Point voltage required = (1.57 163.732 x 11 x 0.3 Knee Point voltage required 20% margin on Vk value = = 136.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7UT613 relay for 11kV Incomer Feeders =A18.1069 14.68 Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0. Rev 0 Page 70 of 91 .100 VA 1.the maximum through fault current on 30 MVA transformer would be: %age impedance at 30 MVA = 0.68 1. ktd 3 = b) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC I SCC x ktd = I PN Where: ISSC = symmetrical short-circuit current IPN = CT rated primary current ISN = CT rated secondary current Considering infinite source at 11 kV .942+0. 100 VA 1. Rev 0 Page 71 of 91 .PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculations for 7SJ62 and 6MD6 relay for 11kV Incomer Feeders =A18.1) = 0.042 VA Relay Burden = 2*0. Hence selected CT is OK PH7-3B-10-15-C001.042+8) Calculated Value for K'SSC = 39.0 VA > 20 Since the selected K' SSC is more than Minimum required K' SSC (20).05 RBC = (Loop Burden+ Relay Burden) = K'SSC = (10+8) x20 (1.942+0.=A28 (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = 20 atleast RBN+ Ri x KSSC RBC + Ri Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA = RBN = Nominal Burden of CT in VA = 10 VA = (0.81 8. 3 x Ipn 20 x (1.49 185.05 RBC = (Loop Burden+ Relay Burden) = = = = = = = = = = 2500 1 amp.942 Ohms (I)² x 0.100 VA 1.942+0.000 atleast 0. Rev 0 Page 72 of 91 .39 volts volts The minimum CT knee point Voltage PH7-3B-10-15-C001.610 Ohms/Km 6.042+9) x 1 1.00 Ohms (I)² x 9 = I High set point = IN 20 = = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = (Relay will be 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar protection Feeders =A18. 70 meter 5.942 = 9.=A28 (connected across core-3 ) Current Transformer Ration (CTR) Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2RL ) Loop Burden in VA Current Transformer resistance (Rct) CT Internal Burden in VA Formula Used a) Effective symmetrical short-circuit The required K'SSC RBC = Connected Burden across CT in Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.732 Ohms/Km 0.1) Ihigh set point x I2N x 1.042 VA (0.942 9.3 154. 05 ohm Guard relay burden per element (R G) = Knee point voltage requirement V k > Ratio 300/1 Required parameters for CTs 20% margin on Vk value 50 / In + If / In * (Rct + 2R L + 2RG) Ratio 400/1 > 50/1+50/1 x (1.75+0.05) 50/1+50/1 x (1.610 5.25+0.60 > 197.25 Ohms 31.61 x 1.732 0.50 kA 50 / In + If / In * (Rct + 2Rl) Where: In = Rated current.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Outgoing Feeder (Typical bay no: A10.A20) Type:1 Applicable substations:All substations with 11kV switchgear except Ain Hamad 11kV Bus Core-1 400-300/1 Class-PX Vk≥250-300 To Pilot wire differential. = 1A If = Primary current under maximum steady state through fault cnditions.05) > 164. Rev 0 Page 73 of 91 . (I) 2 x 2RL Current Transformer resistance (R ct) = Short time rating of 11kV system (I sc) = Formula Used Knee point voltage requirement V k > 300 /1 1 amp.75-1.52 The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.942+2x0.2 6. N = CT ratio = 400-300/1A RL = Lead resistance of single lead from relay to current transformer Rct = Secondary resistance Lead resistance of single lead from relay to current transformer If = Maximum through fault current = 50 x In 50 0. 70 5.60 189.0 Rct≤3-2 VA Burden -10 VA To Feeder Overcurrent & earth fault protection / metering (7SJ61 + 6MD6) CT Knee point voltage calculation for SOLKOR type relay used for Pilot wire protection Applicable for 11kV Outgoing Feeders Current Transformer Ration (CTR) Relay Normal Current (IN) = = Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance = = = = = Loop Resistance (2RL ) Loop burden in VA.942+2x0. Cable overload protection (SOLKOR) Rct≤1.25 Io<30ma Core-2 400-300/1 Class-5P20/1.942 VA = 1.52 227. amps.942 meter Ohms/Km (Relay will be mounted in Relay panel) Ohms/Km Ohms 0. 100 VA 1. Rev 0 Page 74 of 91 . Hence selected CT is OK PH7-3B-10-15-C001.042+3) 14.90 > 20 Ratio 400/1 (+3) x20 (1.042+2) Calculated Value for K' SSC = 78.1) 20 atleast Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Ri = Internal CT burden in VA Relay Burden = 2*0.942+0.042 VA Ratio 300/1 Required parameters for CTs K'SSC = (10+2) x20 (1.84 20 Since the calculated K' SSC is more than Minimum required K' SSC (20).PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculation for 7SJ61 & 6MD6 relay for 11kV Cable feeders (connected across core-2) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = RBN+ Ri x KSSC RBC + Ri = (0.05 RBC = (Loop Burden+ Relay Burden) = 0. A40) Type:2 Applicable substations:Ain Hamad 11kV Bus Core-1 100/1 Class .5) x 20 Also K'SSC 0.732 Ohms/Km 0.415 kV 500 KVA ONAN/ONAF 415V side REF Protection (7SJ61) Core-1 750/1 Class .05 = RBC = (Loop Burden+ Relay Burden) = (0.0mm 2 at 750C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = 5.942 = 1.942+0. Rev 0 Page 75 of 91 .942 Ohms Loop Burden in VA Current Transformer resistance (R ct) CT Internal Burden in VA = = = (I)² x 0.042 + 1.042 VA (1.PX Vk≥ Rct≤6 Io<25mA 11/0.5 = a) Effective symmetrical short-circuit current factor (K' SSC): = The required K'SSC = RBN+ Ri x KSSC RBC + Ri Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 2*0. Length of cable between CT and Relay = 70 meter Cable Resistance for 4.5) Calculated Value for K' SSC = 90.500 Ohms (I)² x 1.00 Since the selected K' SSC is more than Minimum required K' SSC (20).PX Vk≥ Rct≤6 Io<25mA 415V side REF Protection (7SJ61) Adequacy checking calculations for 7SJ61 relay for 11kV Incomer Feeders (connected across core-1) Current Transformer Ration (CTR) Relay Normal Current (IN) = = 100 /1 1 amp.610 Ohms/Km 6.1) K'SSC = (10 + 1.5P10 10VA To Over current / Earth fault (7SJ61) Core-2 100/1 Class . Hence selected CT is OK PH7-3B-10-15-C001.50 20 atleast Where RBC = Connected Burden across CT in VA 0.1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Outgoing Feeder (A30.0 10VA Metering Core-1 750/1 Class .942 1.1 VA 1.48 > 20. the maximum through fault current on 500 KVA transformer would be: = = %age impedance at 500 KVA Overall Impedance ( taking 15% negative tolerance) If = If = Vk 0.000 20 atleast IN Value 20 will be selected for calculations Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = .942+0.72 volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.525 kA > 1.1) Ihigh set point x I2N x (Ri+RBC) 1. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4.732 x415 x 0.1 VA 1.5 times of the maximum possible through fault current has been considered) Vk > 2 x 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 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) = 750 /1 = 1 amp.042) x 1 1.942 Ohms (I)² x 0.732 Ohms/Km 0. the formula used is : (For determining the Voltage developed across CT.0960 0.3 108. Rev 0 Page 76 of 91 .5 x I f x (Rct + 2RL ) Where If is magnitude of through fault current Considering infinite source.01 volts 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.610 Ohms/Km 6.34 130.5 x 8525 x (6 +0.082 500 (1.042 VA (0.942) x 2 750 > 236.942 = 6.1 VA RBC = (Loop Burden+ Relay Burden) = b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20% margin on Vk value = = 0.0mm 2 at 750C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = = 70 meter 5.3 x Ipn 20 x (6 + 1.942 = 6.000 Ohms CT Internal Burden in VA = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = (I)² x 6 IHigh set point = 0. a current of 1.0816) 8. PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Bus Coupler (Typical bay no.732 0.49 185. Rev 0 Page 77 of 91 .3 20% margin on Vk value = = 154.1) 1.942 = 9.942 = (I)² x 0.00 Ohms CT Internal Burden in VA = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = meter Ohms/Km Ohms/Km Ohms (I)² x 9 I High set point = = 20 (Relay will be mounted in Relay panel) 0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4 mm²at 75°C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = 70 5.042) x 1 1.042 VA Ihigh set point x I2N x (Ri+RBC) 1.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.3 x Ipn b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20 x (9+1.5 10 VA To Overcurrent/Earth fault and Metering (6MD6 + 7SJ61) CT Knee point voltage calculation for 7SJ61 type used for Partial Bus bar differential protection Applicable for I/C and B/C Current Transformer Ration (CTR) = 2500 /1 = 1 amp.000 atleast IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 2*0.610 6.942 9.100 VA = (0.05 = RBC = (Loop Burden+ Relay Burden) 0.A12) Type:1 Applicable substations: All substations with 11kV switchgear except Ain Hamad 11kV Bus-2 11kV Bus-1 Core-1 2500/1 Class-PX Vk≥250 Rct≤9 Io<30ma Core-2 2500/1 To partial busbar differential protection (7SJ61) Class-PX To partial busbar differential protection (7SJ61) Vk≥250 Rct≤9 Io<30ma Core-1 2500/1 Class-5P20/1.0 Rct≤8.942+0. 042+8.05 RBC = (Loop Burden+ Relay Burden) = K'SSC = (10+8.942+0.00 Since the selected K' SSC is more than Minimum required K' SSC (20).5) Calculated Value for K' SSC = 38. Rev 0 Page 78 of 91 .78 0.042 VA > 20. Hence selected CT is OK PH7-3B-10-15-C001.100 VA 1.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculation for 7SJ61 & 6MD6 relay (connected across core-2 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = RBN+ Ri x KSSC RBC + Ri = (0.1) 20 atleast Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 2*0.5) x20 (1. 3 x Ipn b) Knee point voltage : The calculated value of knee point = Knee Point voltage required = 20 x (9+1.0mm 2 at 750C 20% margin on cable resistance Loop Resistance ( 2R L ) = = = = Loop Burden in VA Current Transformer resistance (R ct) = = 70 meter 5.942 = 9. Rev 0 Page 79 of 91 .730 Ohms/Km 0.042 VA Ihigh set point x I2N x (Ri+RBC) 1.942 9.042) x 1 1.1 VA 0.39 volts volts The minimum CT knee point Voltage shall be selected more than above PH7-3B-10-15-C001.000 atleast IN RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA I2N = Relay Normal Current Relay Burden = 0.1) 1.610 Ohms/Km 6.942+0.5 15 VA To Overcurrent/Earth fault and Metering (6MD6 + 7SJ61) CT knee point voltage calculations for 7SJ61 type used for Partial Bus bar differential protection Applicable for I/C and B/C Current Transformer Ration (CTR) = 800 /1 = 1 amp.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Bus Coupler (A12) Type:2 Applicable substations:Ain Hamad 11kV Bus-2 11kV Bus-1 Core-2 800/1 Core-1 800/1 Class-PX Vk≥250 Rct≤9 Io<30mA at Vk/2 To partial busbar differential protection (7SJ61) Class-PX To partial busbar differential protection (7SJ61) Vk≥250 Rct≤9 Io<30mA at Vk/2 Core-1 800/1 Class-5P20/1.942 Ohms (I)² x 0.3 20% margin on Vk value = = 154.00 Ohms CT Internal Burden in VA = Formula Used a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = (I)² x 9 I High set point = = 20 0.0 Rct≤8.49 185.1 VA RBC = (Loop Burden+ Relay Burden) = (0. Relay Normal Current (IN) Length of cable between CT and Relay Cable Resistance for 4. 05 = 0.PROJECT:GTC/123/2006 SIEMENS ANNEXURE 1 Adequacy checking calculation for 7SJ61 & 6MD6 relay (connected across core-1 ) a) Effective symmetrical short-circuit current factor (K' SSC): The required K'SSC = Also K'SSC = RBN+ Ri x KSSC RBC + Ri = (0.042+8.10VA RBC = (Loop Burden+ Relay Burden) K'SSC = (15+8.1) 20 atleast Where RBC = Connected Burden across CT in VA Ri = Internal CT burden in VA RBN = Nominal Burden of CT in VA Relay Burden = 2*0.1 VA 1. Hence selected CT is OK PH7-3B-10-15-C001.042 VA > 20.942+0. Rev 0 Page 80 of 91 .5) x20 (1.5) Calculated Value for K' SSC = 49.26 0.00 Since the selected K' SSC is more than Minimum required K' SSC (20). 40 450 15. PX (11kV side) Trafo Neutral.942 0.050 0.Mosemeer. Cl.942 0.050 0.292 Ammeter 0.992 0. 5P20/1. Cl.050 0.Al Dhahiya West.Abu Hamour central. PX (11kV side) Trafo Neutral. 5P20/1. Cl.942 0.Al Jumailyah.40 250 100 9.MIC-2.050 0.992 0.PROJECT:GTC/123/2006 SIEMENS Sr.050 0.0 0. Rev 0 Mosemeer.992 PH7-3B-10-15-C001.QRE.0 1.Lusail Development Super 1. 5P20/1.050 0.050 0.992 0.992 0. Wakrah 2. PX 2500/1 9 7SJ61 0.942 0.40 500 7.50 100 0. Cl.050 0. PX (11kV side) Trafo Neutral.Khore Community. 1 2 Feeder Description 4 6 CT Ratio used CT Knee Point CT Resistance Voltage (V) (ohms) Relays connected Relay Burden (in VA) VA burden due to Secondary leads Total VA burden Remarks Applicable substations Rated Burden selected= 10 VA Al Jumailyah.050 0.250 T1.0 0.942 0.942 0. Khore Junction Page 81 of 91 . Cl.050 0. PX (415 V side) 100/1 100/1 2500/1 100/1 750/1 1500/1 Earthing transformer (2000KVA) Core-1. Al Soudan. NBK-2.0 (11kV side) Trafo Neutral. Ain Khalid south. PX Core-2.050 0. PX (11kV side) Trafo Neutral.942 0.0 100 0.050 0. PX Core-2.942 0. Ain Khalid south.992 T2.NDQ.050 0.992 0.992 50/1 50/1 2500/1 50/1 750/1 750/1 100 1. 5P20/1.050 0. PX (415 V side) 3 ANNEXURE 2 Rated Burden selected= 15 VA Rated Burden selected= 15 VA Earthing transformer (1000KVA) Core-1.0 Core-3. No.Al Waab Super. PX 2500/1 9 7SJ61 0.992 0.050 0. Cl.Lusail Development Super 1.942 1. Cl.60 100 250 100 9. Al Soudan.0 (11kV side) Trafo Neutral.050 0.0 2500/1 250 9 7SJ61 7UT613 0. Cl.MIC-3.942 0. Core-2.Abu Hamour central.Abu Thaila Substation Modification. PX Trafo Neutral.Al Wadi.050 0.050 Rated Burden selected= 15 VA NDQ.=A28 (Incomer feeder) Core-1.992 0.992 Core-4.292 7SJ61 6MD6 Ammeter 0. Cl.20 250 100 9. 5P20/1.Khore Community. 5P20/1. Cl.942 0.942 0.050 0. PX Core-2.050 0.942 0.992 Mosemeer. NBK-2.Al Dhahiya West.992 0. Al Dhahiya.942 0.250 Core-3. 5P20/1.050 0.992 0.Al Dhahiya West.20 250 6 7SJ61 7SJ61 7SJ61 7SJ61 7SJ61 7SJ61 0.942 1. NBK-2.MIC-2. EDS Rated Burden selected= 15 VA Rated Burden selected= 10 VA Common for all substations with 11kV switchgear except Ain Hamad 0. South west wakrah.QRE.942 0.050 0.RLF-3 Bay =A18. PX (11kV side) Trafo Neutral.MIC Super . Cl.0 Trafo Neutral. Core-1.050 0.050 0.050 0.MIC3.942 0. Cl.00 9.00 2500/1 7SJ61 7SJ61 7SJ61 7SJ61 7SJ61 7SJ61 0.992 7SJ61 7SJ61 7UT613 7SJ61 7SJ61 7SJ61 7SJ61 0. 5P20/1. Murraikh North. Cl. Cl.050 0. Cl. PX 2500/1 Core-2. Cl.942 0.050 0. Core-1. Wakrah 2.Al Jumailyah. PX 2500/1 250 9 7SJ61 0.942 0. Ain Khalid south. Khore Junction.050 0.942 0.Al Wadi.942 0.942 0. Al Soudan.942 1. PX (415 V side) 200/1 200/1 400/1 2500/1 200/1 750/1 3000/1 Bay =A12 (Bus sectionalizer) T1.MIC3. Cl.Khore Community.Abu Thaila Substation Modification. Cl.992 0. PX (11kV side) Trafo Neutral.942 0.992 0.QRE.992 0. PX 2500/1 250 9 7SD52 0.992 0.942 0. South west wakrah.942 0. Al Dhahiya.992 0.Abu Hamour central.992 0.Al Waab Super. EDS.050 0.992 0.992 0.Al Wadi.0 (11kV side) Trafo Neutral. Murraikh North.0 Trafo Neutral. Rated Burden selected= 15 VA Rated Burden selected= 15 VA MIC Super .042 6MD6 7SJ62 0. Khore Junction Earthing transformer (500KVA) Core-1.MIC-2. 942 0. 5P20/1.942 0.0 400-300/1 400-300/1 Bay =A18.092 0. PX Core-2.75-1.992 300-250 1. Core-1.A20 T1. =A10.0 T1.942 0. =A60.942 1.5 250 9 7SJ61 6MD6 Ammeter 7SJ61 9-4.25-1. PX 9 ANNEXURE 2 Bay =A18 & =A28 (LV neutral) Core-1.050 Ain Hamad Ain Hamad Rated Burden selected= 15 VA Rated Burden selected= 10 VA Rated Burden selected= 10 VA Ain Hamad Rated Burden selected= 15 VA Ain Hamad 12 Outgoing Feeders =A90. Cl. =A100. Core-1. 5P10/1.942 0.992 0. Core-1. Core-2.942 0.992 SOLKOR 7SJ61 600-300 9-4. PX 800/1 800/1 800/1 VA burden due to Secondary leads Total VA burden 0. Core-1.5 9 9 7SJ61 7SJ61 7SJ61 0. 5P20 CT Knee Point CT Resistance Voltage (V) (ohms) 100/1 100/1 750/1 11 Bay =A12. PX T2. 5P20/1.050 0.0 Cl.292 Rated Burden selected= 15 VA Common for all substations with 11kV switchgear except Ain Hamad 0. Cl.442 1. Cl.75 SOLKOR 7SJ61 0. Cl.050 0. 7 8 Feeder Description CT Ratio used Outgoing Feeders =Typ. T1. Cl. 1.25 600-300/1 750/1 10 Bay =A30 & =A40 (315kVA transformer feeders) Core-1.942 0.5 7SJ61 0.050 0. PX T2. Core-1.992 7SJ61 0. CL5P20/1.050 0. PX Core-2. No.0 Core-2.050 Relay Burden (in VA) Applicable substations 4.250 0.= A20.050 0.0 400-300/1 400-300/1 PH7-3B-10-15-C001.042 1.442 1.050 0. Cl. PX (415V side) Relays connected 0.992 6. =A70.PROJECT:GTC/123/2006 SIEMENS Sr.=A28 (Incomer feeder) Core-1.942 0.942 0. Cl.050 0.942 0. Core-2.050 0. =A50.992 250 250 8.=A22 (Bus sectionalizer) T1.050 Rated Burden selected= 15 VA Ain Hamad Page 82 of 91 .0 600-300/1 600-300/1 800/1 Core-3.942 1.050 0.050 0.050 0. A10. Rev 0 6MD6 Ammeter 3. =A80.00 7SJ61 0.250 Remarks 1.050 0.292 6MD6 Ammeter 7UT613 7SJ62 6MD6 3.942 4.250 0. 5P20/1. PX T2. Cl.292 Rated Burden selected= 10 VA 0.500 0.500 0.050 0.942 0. 000 0.050 1.500 Rated Burden selected= 30 VA 0.400 0.050 1.11/√3 6MD6 Voltmeter AVR 4 Outgoing Feeders (Typ.050 2.) =A10. 1.11/√3 6MD6 Voltmeter 0.0/3P 3 Bay =A18.050 1.PROJECT:GTC/123/2006 SIEMENS Sr. 0.500 Rated Burden selected= 30 VA Page 83 of 91 .A20 Winding-1.400 1.400 1. Feeder Description No.450 Rated Burden selected= 30 VA 11/ √3 : 0.11/√3 6MD6 Voltmeter PH7-3B-10-15-C001. Rev 0 Relay Burden (in VA) 7SJ62 . 1. 1.0/3P 11/ √3 : 0.0/3P 11/ √3 : 0.400 1.11/√3 6MD6 Voltmeter 0.0/3P ANNEXURE 3 VT Ratio used Relays connected Total VA Remarks burden 11/ √3 : 0.450 Rated Burden selected= 30 VA 2 Bay =A24 -T16 Winding-1. 1.050 1. =A28 (Incomer feeder) Winding-1.050 1. 1 Bay =A14 -T15 Winding-1. 6 Busbar / Line 1. 5P10. For relays with a required saturation free time from w 0. VK = (Rb + Rct ) •I sn •K ssc 2 13 . / Motor 1. required sym.5 0.13 please refer to page 2/56 Table 2/1 CT requirements Relay type Transient dimensioning factor Ktd Overcurrent-time protection 7SJ511. •I pn Page 84 of 91 2/53 . 635 7UM62 Transformer 4 3 4 Busbar / Line 4 3 – Gen.6 Transformer / Generator differential protection 7UT612 7UT613. 53x. shortcircuit current factor K’ssc Min. 53x. the primary (DC) time constant TP has only little influence. IEC PX or BS: VK = (15 + 4) •1 •10 V = 146 V 1. fault) · (Rct + R'b) · Isn 13 . fault) I pn VK = and (only for 7SS): Ktd · I scc max (ext. •I pn and: and: 3 ( K' ssc •I pn )end1 4 = = 4 ( K' ssc •I pn )end2 3 3 (VK •I pn / I sn )end1 4 = = 4 (VK •I pn / I sn )end2 3 K'ssc = I Ktd · scc max (ext. 15 VA. fault) · (Rct + R'b) · Isn 13 .2 1. 64 – Line differential protection (pilot wire) 7SD600 – K'ssc = K'ssc = Transformer 1.4 cycle. 46.5 primary DC time constant TP [ms] = 30 Ktd (a) 1 Ktd (b) 4 PH7-3B-10-15-C001. Rev 0 Siemens SIP · 2006 K'ssc = = 50 = 100 = 200 Ktd (a) · 2 5 4 5 4 5 I scc max (close -in fault) I pn and: and: Ktd (b) · VK = Ktd (a) · I scc max (zone 1. 610 (50 Hz) 7SD52x. •I pn at least: I scc max (ext. 60 7SJ61. 512. Rct = 4 . 62. Example: IEC 600/1. •I pn and (only for 7SS): 100 · (Rct + R'b) · Isn (measuring range) 13 . 633.2 1. 63.3 Rct = 4 . •I pn I scc max (zone 1-end fault) · (Rct + R'b) · Isn 13 . 7SS600 Min.A 2 Overview CT dimensioning formulae CT design according to BS 3938/IEC 60044-1 (2000) K'ssc = Kssc · Rct + Rb (effective) Rct + R' b with K'ssc W Ktd · The design values according to IEC 60044 can be approximately transfered into the BS standard definition by following formula: I SCC max (required) I pn The effective symmetrical short-circuit current factor K'ssc can be calculated as shown in the table above. 60044: The rated transient dimensioning factor Ktd depends on the type of relay and the primary DC time constant. 610 (60 Hz) Busbar protection 7SS5. fault) I pn VK = · (Rct + R'b) · Isn 20 · (Rct + R'b) · Isn 13 . K'ssc = 100 (measuring range) VK = for stabilizing factors k = 0. required kneepoint voltage VK VK = I High set point 13 .end fault) I pn Ktd (b) · I scc max (close -in fault) · (Rct + R'b) · Isn 13 . 7SA6. 531 7SJ45.2 1. 7SD5xx*) *) with distance function I High set point I pn at least: 20 Line differential protection (without distance function) 7SD52x. I scc max (ext. / Motor 5 5 5 Distance protection 7SA522. For CT design according to ANSI/IEEE C 57.6 Gen. = 0.2.max = (25 VA •20) = 100 V. to class C100 5A PageSiemens 85 of 91 SIP · 2006 . K'ssc = = l = single conductor length from the CT to the relay in m. Rct + Rb •K ssc = 4 . CT design according to ANSI/IEEE C 57.4 .A 2 Overview Protection Coordination Relay burden Example 2: Stability-verification of the numerical busbar protection relay 7SS52 The CT burdens of the numerical relays of Siemens are below 0.t. • mm 2 (copper wires) at 20 °C m . 5P20.1 . •l A Rlead .3 . 25 VA 60044: ANSI C57. = 30 .0175 •50 = 0.3 .max.0175 A = conductor cross-section in mm2 •10 = 43. Fig. = 0. have a much higher burden. 6 4 .t. acc.max = 2 K ssc 20 and I Nsn = 5 A . Specific resistance: 2 •0. the relevant relay manuals should always be consulted for the actual burden values. Standard classes are C100. Exceptions are the busbar protection 7SS60 and the pilot-wire relays 7SD600.2 Result: The effective K'ssc is 43. up to the order of 10 VA. we get Pb •K ssc 5A Example: IEC 600/5. 1 A2 The resistance of the current loop from the CT to the relay has to be considered: = Rrelay = 0. the required K'ssc is 25. + 0. Mechanical relays.13 Class C of this standard defines the CT by ist secondary terminal voltage at 20 times rated current.4.000 A = 50 600 A Burden of the connection leads According to table 2/1 on page 2/53 Ktd = ½) K'ssc Rb 1 •50 = 25 2 15 VA = = 15 . + 0.t. Rct + R' b 2 •.13: 2/56 PH7-3B-10-15-C001.1 VA and can therefore be neglected for a practical estimation. however. 2/93 I SCC. Rlead = R'b = Rl + Rrelay = 0. 2 Intermediate CTs are normally no longer necessary as the ratio adaptation for busbar and transformer protection is numerically performed in the relay. This terminal voltage can be approximately calculated from the IEC data as follows: ANSI CT definition Vs. I pn In any case. This has to be considered when older relays are connected to the same CT circuit. C200. Therefore the stability criterion is fulfilled. Analog static relays in general have burdens below about 1 VA. for which the ratio error shall not exceed 10 %.max = 20 •5 A •Rb • with Rb = Pb I sn Vs. C400 and C800 for 5 A rated secondary current.1 . + 15. Rev 0 Vs. at INom = 1 A .Thermal (rms) 100· INom for 1 s 30· INom for 10 s 4· INom continuous 250· INom (half-cycle) .2 only in models with input for sensitive ground fault detection (see ordering data in Appendix A.at INom = 5 A .Thermal (rms) 300 A for 1 s 100 A for 10 s 15 A continuous 750 A (half-cycle) .6 A 1) Burden per Phase and Ground Path .1. 3 W Energized Approx. IEC 60 255-11 15 % of the auxiliary voltage Power Input Quiescent Approx.05 VA Approx.1) Auxiliary Voltage DC Voltage 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.Dynamic (peak value) 1) 4.1 Analog Inputs Current Inputs Nominal Frequency fNom 50 Hz or 60 Hz Nominal Current INom 1 A or 5 A Ground Current. IEC 60255–11 (in not energized operation) 268 PH7-3B-10-15-C001. 0.1 General Device Data 4. 0. Sensitive INs = linear range 1.Dynamic (peak value) Current overload capability for high-sensitivity input INs 1) .05 VA Current overload capability . 7 W Bridging Time for Failure/Short Circuit. Rev 0 = 50 ms at V = 110 VDC = 20 ms at V = 24 VDC 7SJ61 Manual Page 86 of 91 C53000-G1140-C118-7 . 0.3 VA Approx. Peak to Peak.for sensitive ground fault detection at 1 A (adjustable) Approx.1.A 4 Technical Data 4. 0.A 4.thermal (rms) 7SD5 Manual PH7-3B-10-15-C001.1.3 VA .1 VA Voltage overload capability per phase . 0.for sensitive earth fault detection at 1A Approx.thermal (rms) 300 A for 1 s 100 A for 10 s 15 A continuous .at IN = 1 A Approx.thermal (rms) 100 · IN for 1 s 30 · IN for 10 s 4 · IN continuous .1 General 4.dynamic (pulse current) 250 · IN (half-cycle) Current Overload Capability for Sensitive Earth Current Input .dynamic (pulse current) 750 A (half-cycle) Requirements for current transformers 1st Condition: For a maximum fault current the current transformers must not be saturated under steady-stateconditions n' = 30 2nd Condition: The operational accuracy limit factor n' must be at least 30 or a non.1 Analog Inputs Nominal frequency fN 50 Hz or 60 Hz (adjustable) IN 1 A or 5 A Current Inputs Nominal current Power Consumption per Phase and Earth Path .at IN = 5 A Approx. C53000-G1176-C169-1 230 V continuous Rev 0 Page 87 of 91 519 .5 V (rms) Power consumption At 100 V = 0.05 VA Current Overload Capability per Current Input .05 VA . 0.1 General 4.or saturated period of t'AL of at least 1/4 AC cycle after fault inception t'AL = 1/4cycle must be ensured 3 rd Condition: Maximum ratio between primary currents of current transformers at the ends of the protected object Voltage inputs Nominal voltage UN 80 V to 125 V (adjustable) Measuring range 0 V to 218. Rev 0 Page 88 of 91 .A 4 Page 10 of 10 PH7-3B-10-15-C001. Rev 0 Page 89 of 91 .A PH7-3B-10-15-C001. AVR TECHNICAL DATA SHEET PH7-3B-10-15-C001. Rev 0 A Page 90 of 91 . 193 5.93 3.86 250 55 90 58 46.3 1395 1000 0.A.17 Min radius of bend around which cable can be laid a) Laid direct m b) In ducts m c) In air m 1.88 Will be calculated form the insulation resistance constant at 90 °C I.Wherever necessary 16.93 44 1.40 180.08 2.32 250 510 90 421 336.5 1.61 COMPLY Will be calculated form the insulation resistance constant at 90 °C I.30 121 98.8 N.8 4 Copper Standed/Circular 2.61 2.7 750 1000 1.57 250 37 90 39 31.60 1200 1040 1545 23.6* 0.29 250 83 90 83 66.02 N.668 0.7 1.30 2000 1135 4395 51.9 550 1000 1.20 69.3 515 1000 LSOH Type LTS1 LSOH Type LTS 1 1. ohm/Km conductor temp .61 3. N.99 55. Diameter of wires .8 1000 710 690 59 47.km COMPLY 5.40 1200 1040 1175 21.4 905 1000 1.8 4 6 10 16 25 50 120 150 240 120 95 Copper Copper Copper Copper Copper Copper Copper Copper Copper Copper Copper Standed/Circular anded/Circuanded/Circutranded/Sectranded/Sectranded/Sectranded/Sectranded/Sectanded/Circuanded/Circuanded/Circu 2.52 2 XLPE OCI Cu/XLP/SWA/LSOH BS 6724 600/1000 3500 (5 min) 2CX4 TECHNICAL PARTICULARS .60 70 56.25 0.153 0.67 megaohm.01 250 158 90 153 122.12 Completed cable Overall diameter .16 0.6* 1 100 6 x OD 6 x OD 6 x OD 13.20 27.40 58.70 480 388.6 Conductor 1) Cross sectional area (Nominal) mm² 2) Material Copper 3) Design (Stranded.LSOH POWER CABLES OFFERED 4CX25 4CX50 DATA 4CX16 Polypropylene fillers .52 3.08 AWA 31 1.20 27.70 49.90 1600 1070 3150 32.16 250 293 90 258 206.93 3.16 0.8 47 38.70 292 236.90 131 106.Wherever necessary 3CX10 3CX16 3CX35 5CX10 5CX6 10 6 16 25 35 4 6 10 16 35 10 6 Copper Copper Copper Copper Copper Copper Copper Copper Copper Copper Copper Copper anded/Circuanded/Circuanded/Circutranded/Sectranded/Secanded/Circ t uanded/Circuanded/Circuanded/Circutranded/Sectanded/Circuanded/Circular 4.927 0.8 100 6 x OD 6 x OD 6 x OD 1.40 1800 1130 3710 47.07 32.7 1830 1000 1. 18.4 Rated Voltage V 1.60 74.23 Type test certificate to be issued by independent laboratory REQUIRED 4CX4 4CX6 4CX10 4CX120 4CX150 1CX240 1CX120 1CX95 2CX10 2CX6 2CX16 2CX25 DATA OFFERED 2CX35 3CX4 3CX6 Polypropylene fillers . 2.25* 0.4 4 4 4 4 4 4 4 4 1 1 1 XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE OMAN CABLES Cu/XLP/SWA/LSOH Cu/XLP/SWA/LSOH BS 6724 600/1000 4000/min 600/1000 4000 Cu Stranded XLPE LSOH 90 100 6 x OD 6 x OD 6 x OD 4.47 130.00 220.60 59 47.0754 0.11 Outer covering Material Nominal Thickness mm Min.47 1.A.6 1025 1000 1.6 0.33 1.8 1200 1040 1110 21.40 89.15 0.30 1000 710 725 16.90 48 38.72 218.2 Cable type 1.6 625 1000 1.25* 0.40 1800 1130 1720 32.00 1000 700 650 15.08 107 85.00 102 82.59 250 282 90 281 224. 3.25* 1.80 32.40 33.524 2.727 0.30 1200 1040 1280 22.8 100 8 x OD 8 x OD 8 x OD 3.25 0.4 985 1000 1.40 58.6 1.4 100 6 x OD 6 x OD 6 x OD 34.16 250 328 90 315 252.12 4.A Sr LV CABLES UNIT 1 LV POWER CABLES 1.52 3.18 Ducts Nominal internal diameter of pipes or mm ducts through which cable can be pulled 1.40 40.3 Standards 1. sectoral ect) 4) Overall diameter (Approx) mm 5) No.50 332 268. 9001.90 1400 1040 2010 25.Approx kg 1.86 250 100 6 x OD 6 x OD 6 x OD 3. Rev 0 .4 1.12 2 2 2 2 2 3 3 3 3 3 5 5 XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE XLPE 1.07 32.08 32 1.21 insulation resistance per km of cable per core Mohm 1.33 1.A.33 1.58 250 100 8 x OD 8 x OD 8 x OD Polymeric LSOH 0.8 100 6 x OD 6 x OD 6 x OD 0.25 0.5 1.50 83 90 3.7 1.Nominal mm * Wire size higher than specified in BS 6724 1.25 0.0968 0.25* 0.15 0.50 79 63.14 Continuous current carrying capacity based on the conditions Laid in the ground (Touching each other) a) One circuit A b) Two circuits A c) Three circuits A Drawn in to ducts a) One circuit A b) Two circuits A c) Three circuits A In air One circuit at 50 °C A 1.9 Armour bedding Type Nominal Thickness mm 1.40 107.124 0.4 600 1000 1200 740 880 79 63.5 1.668 0.25 1.36 39.67 megaohm.14 65.93 3.8 100 8 x OD 8 x OD 8 x OD 5.32 Galvanized Steel Wire Armour 35 37 36 1.45 250 335 90 291 232.494 0.90 1200 1040 1135 20.11 91.5 1.08 28 1.30 65 52.8 Fillers Material .9 1.92 Galvanised Steel Wire 30 22 24 1.25* 1 100 6 x OD 6 x OD 6 x OD 100 6 x OD 6 x OD 6 x OD 1. N.6 1 100 6 x OD 6 x OD 6 x OD 1.524 5.80 203.40 1200 1040 1360 22.86 250 46 90 48 38.64 44 2.A.6 1.4 13 11.70 1200 1040 1055 20.6 163 90 17.90 °C 1.92 23 1.00 45.83 1.15 0.13 Cable drums Overall diameter .17 109.33 1.1 1850 1000 152 121.12 4.Polypropylene 1.02 3.30 47 38.29 250 98 90 99 79.28 131.16 1.70 349 282.01 250 134 90 128 102.A.1 6900 500 2.01 84.8 2740 1000 1.e.10 109 90 7.40 33.60 88. N.43 250 74 90 77 61.00 45.727 0.99 55.4 1.8 N.83 3.57 250 43 90 46 36.86 250 46 90 48 38.43 250 100 8 x OD 8 x OD 8 x OD 0.5 840 1000 1.4 0.5 1.30 46 90 1. thickness mm 1.60 64 90 2.7 460 1000 1.Approx kg Maximum drum length m 1.8 100 6 x OD 6 x OD 6 x OD 2.22 Manufacturer quality assurance according to ISO 9000.80 294.247 0.62 71. N.00 30 1.8 100 6 x OD 6 x OD 6 x OD 0.15 Max conductor temp °C 1.387 0.79 41. N.80 1200 740 845 18 715 1000 1.24 34 1. Cable loded as above prior to kA short ckt & final conductor temp of °C 1.10 Armour Type of wire Number of wires No.60 53.60 106.4 100 8 x OD 8 x OD 8 x OD 21.8 18.4 8190 500 2.83 1.Approx m Weight loaded .08 39 1.153 0.8 100 6 x OD 6 x OD 6 x OD 0.00 29 1.20 56 45.6 1 100 6 x OD 6 x OD 6 x OD 17.Approx mm Weight per meter .10 188 152.90 1000 710 615 15.79 41.25* 0.99 55.6 1.15 250 100 8 x OD 8 x OD 8 x OD 3.25 0.5 1.A.08 0.6 1.30 1200 740 970 19.52 204.08 2.e.4 0.9 1130 1000 1.20 Maximum ac resistance of conductor per km of cable at max.00 33 1.47 1.197 0.30 1200 1040 1010 20 860 1000 1.57 250 100 6 x OD 6 x OD 6 x OD 0.92 232.1 Manufacturer 1.24 44 1.58 250 128 90 127 101.83 3.40 33.8 1.8 1460 1000 1.00 33 1.29 250 100 8 x OD 8 x OD 8 x OD Polymeric LSOH 1 1 39 31.40 187 151.90 157 127.Approx m Width .92 100 6 x OD 6 x OD 6 x OD 0.A.6 1.8 1.47 1.30 83 66.60 312 252.92 25 1.4 0.80 196.00 1600 1070 3030 32.72 48 2.8 100 6 x OD 6 x OD 6 x OD 2.3 1.7 Insulation 1.4 0.02 3.88 4.92 27 1.70 1400 1040 2030 26.43 250 64 90 65 52.3 1195 1000 1.00 45.5 1.69 244.62 71.25 0.43 250 64 90 65 52.40 1200 1040 1345 21.00 32 1.10 102 82.A.8 100 8 x OD 8 x OD 8 x OD 5.08 36 1.16 Conductor short circuit current carrying capacity for 1 sec.5 Test Voltage (1 min) VAC/min 1. 3.927 0.90 94 76.5 1.60 157 127.50 79 63. 4.17 109.3 1210 1000 1.6 1.79 41.197 0.08 1.12 4. of cores 1.25 0.6 1.88 4.80 336.60 59 47.0 2860 1000 LSOH Type LTS-1 LSOH Type LTS-1 1.1 960 1000 1.20 1000 700 560 14.19 Maximum dc resistance per km of cable at 20 °C of conductor ohm/Km 1.02 4.8 37 90 0. 9003 & 9004 1.25 1.km COMPLY COMPLY Page 91 of 91 PH7-3B-10-15-C001.
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