1 BASIC FOR CALCULATION 1 In the calculation only three phase symmetrical fault has been considered. 2 3 The PU system is adopted for fault calculations as it simplifies the calculations 4 5 Motor Contribution has been considered as 2 times the rated current of HT motors and negligible in case of LT motors. 6 For claculation,the impedances of cables has been neglected. 7 The distribution transformers in different load centres of the plant shall operate in isolation and not in parallel. 8 The utility fault level has been considered as 600 MVA. 2 DESIGN DATA : a Utility fault level at plant grid = 600 MVA b Base MVA chosen = 25 MVA c) Distribution Transformer 1 Rated capacity = 3.15 MVA 2 Voltage at HT side = 11 KV 3 Voltage at LT side(full load) = 0.380 KV 4 Percentage impedance = 8 % d) Rating of HT motors‐ 1 HT motor rating = 4150 Kw 2 Motor Power factor = 0.85 red 3 Motor efficiency = 0.95 % 4 Maximum starting torque considered for = 200 % FLT the motor at 11kv 5 Maximum Running torque considered for = 100 % FLT the motor at 11kv Cable Size XLPE,Armoured 240 Sqmm Cable no of run 5 No. 5 HT motor fed from cable = 0.23 ohm/km 5Rx3x240Sqmm XLPE CU (Z) 6 HT motor fed from cable length = 150 ohm/km e) Rating of LT motors‐ 1 LT motor rating = 350 Kw 2 Motor Power factor = 0.85 red 3 Motor efficiency = 0.90 % 4 Maximum starting torque considered for = 200 % FLT the motor at 11kv 4 Maximum Running torque considered for = 100 % FLT the motor at 11kv Cable Size XLPE,Armoured 1.5 Sqmm Cable no of run 1 No. 5 LT motor fed from cable = 0.020 ohm/km 3Rx4x300Sqmm XLPE CU (Z) 6 LT motor fed from cable Length = 70 mtr A simplified procedure for calculating the symmetrical short circuit current called the E/X method is used, which disregards all resistance,all static load and all prefault current. Since X/R ratio of the effective impedance at the bus is unknown,It si considered that the calculated current should not more than 80% of the allowed valve for the breaker at the existing bus voltage.
A FAULT CURRENT CALCULATION FOR HT (UP TO NEAREST MOTOR TERMINAL FROM SOURCE) 1 FAULT CURRENT AT 33KV SWITCHGEAR AT APTRANSCO END 1.1 Calculation for source impedance (Zo) System Voltage (KV) = 33 KV Sorce Fault MVA = 500 MVA Source Impedance (Z0) in Ohm System KV2 Short Circuit MVA Square (33) 500 ZO = 2.178 ohm 1.2 Fault Current at 33KV Switchgear at APTRANSCO End (IF0) 1.1 X KV √ 3 X ZO 1.1 x 33 √ 3 X 2.178 IFO = 9.62 kA 2 FAULT CURRENT AT 33KV SWITCHGEAR AT PROPOSED SUBSTATION 2.1 Resistance of cable ( R ) (Resistance / Km) = 0.130 ohm/Km Length of Cable (L) = 1500 Meter No.of Run (N) = 1 No. Actual Resistance of Cable (Ra) R x L (N x 1000) Ra = 0.195 ohm 2.2 Reactance of cable (X) (Reactance/Km) = 0.100 ohm/Km Actual Reactance of Cable (Xa) X x L (N x 1000) Xa = 0.150 ohm 2.3 Impedance of Cable (Za) = √(Ra2 + Xa2) ohm Za = 0.2460 ohm 2.4 Total Impedance at 33KV SWGR at Substation‐1(Z1) = Za + Zo Z1 = 2.425 ohm 2.5 Fault Current at 33KV SWGR at Proposed Substation (IF1) 1.1 X KV √ 3 X Z1 ( 1.1 X 33) (√ 3 X 2.425) Selected Fault KA at 33KV Bus IF1 = 8.643 kA 25 kA 3 FAULT CURRENT AT PRIMARY OF 8MVA, 33/6.9KV TRANSFORMER AT PROPOSED SUBSTATION 33KV, 1R X 3C X 300Sqmm (E) Aluminium armoured Cable is connected between 33KV HT Panel & 8 MVA Transformer 3.1 Resistance of cable (Resistance / Km) = 0.130 ohm/km Length of Cable (L) = 25 Meter No.of Run (N) = 1 No. Actual Resistance of Cable (Rb) R x L (N x 1000) Rb = 0.00325 ohm 3.2 Reactance of cable (X) (Reactance/Km) = 0.100 ohm/Km Actual Reactance of Cable (Xb) X x L (N x 1000) Xb = 0.0025 ohm 3.3 Impedance of Cable (Zb) = √(Rb2 + Xb2) ohm kA = ohm ohm = ohm ohm kA = = = = Main Incoming power supply through 33kV, 1R X 3C X 300Sq.mm. Aluminium armoured Cable from APTRANSCO Substation to Plant Substation kA kA = = ohm = ohm
Zb = 0.0041 ohm 3.4 Total Impedance at 33KV SWGR at Substation‐1(Z1) = Zb + Z1 ohm Z2 = 2.430 ohm 3.5 Fault Current at 33KV SWGR at Proposed Substation (IF1) 1.1 X KV √ 3 X Z2 ( 1.1 X 0.15) (√ 3 X 2.43) IF2 = 8.625 kA 4 FAULT CURRENT AT SECONDARY OF 8MVA TRANSFORMER AT SUBSTATION Power Transformer 8MVA, 33/6.6 KV , 8.35% Impedance (as per IS‐2026) Primary Voltage of Transformer Vp = 33 KV Secondary Voltage of Transformer Vs = 6.6 KV Transformer Capacity = 8000 KVA % Impedance = 8.35 % 4.1 Transformer Impedance (Zc) (% Imp/100) x KVs2 (KVA / 1000) Zc (8.35/100) x 6.6) 2 ohm (8000/1000) Zc = 0.455 ohm 4.2 Total Impedance on secondary side (Z3 ) = {((Vs)²/ (Vp)²)x Z2 }+ Zc ohm = ((6.6) / (33) X 2.43) + 0.455 ohm Z3 = 0.553 ohm 4.3 Fault Current at Secondary side of 8MVA Transformer(IF3) 1.1 X KVs √ 3 X Z3 ( 1.1 X X x L) (√ 3 X 0.553) IF3 = 7.580 KA 5 FAULT CURRENT AT 6.6 KV HT PANEL AT PROPOSED SUBSTATION Incoming power supply through 6.6kV, 3R X 3C X 300Sq.mm. Aluminium armoured Cable from 6.6KV Terminal of Power Transformer 5.1 Resistance of cable (Resistance / Km) = 0.129 ohm/km Length of Cable (L) = 25 Meter No.of Run (N) = 3 No. Actual Resistance of Cable (Rd) R x L (N x 1000) Rd = 0.001075 ohm 5.2 Reactance of cable (X) (Reactance/Km) = 0.0830 ohm/Km Actual Reactance of Cable (Xd) X x L (N x 1000) Xd = 0.000691667 ohm 5.3 Impedance of Cable (Zd) = √(Rd2 + Xd2) ohm Zd = 0.0013 ohm 5.4 Total Impedance at 6.6KV HT Panel (Z4) = Zd + Z3 Z4 = 0.555 ohm 5.5 Fault Current at Proposed Substation 6.6KV HT Panel (IF4) 1.1 X KV √ 3 X Z4 ( 1.1 X ) (√ 3 X 0.555) IF4 = 7.553 kA Selected Fault KA at 6.6KV Bus 26.3 kA kA kA = ohm kA kA KA = = = = = = = = ohm ohm kA =
6 FAULT CURRENT AT MOTOR TERMINAL (IBC‐3 ‐ nearest) 6.6KV, 1R X 3C X 150Sqmm Aluminium armoured Cable is connected between 6.6kV HT Panel & HT Motor terminal 6.1 Resistance of cable (Resistance / Km) = 0.264 ohm/km Length of Cable (L) = 180 Meter No.of Run (N) = 1 No. Actual Resistance of Cable (Re) R x L (N x 1000) Re = 0.04752 ohm 6.2 Reactance of cable (X) (Reactance/Km) = 0.0890 ohm/Km Actual Reactance of Cable (Xe) X x L (N x 1000) Xe = 0.01602 ohm 6.3 Impedance of Cable (Ze) = √(Re2 + Xe2) ohm Ze = 0.0501 ohm 6.4 Total Impedance at HT Motor terminal (Z5) = Ze + Z4 Z5 = 0.606 ohm 6.5 Fault Current at IBC‐3 Motor Terminal (IF5) 1.1 X KV √ 3 X Z5 ( 1.1 X X x L) (√ 3 X 0.606) IF5 = 6.917 kA = kA ohm = ohm = kA =
3 PU IMPEDANCES OF VARIOUS EQUIPMENT/SYSTEMS: 1 Base MVA chosen = 25 MVA 2 PU impedance of Utility 25 600 = 0.041667 Pu (Base MVA) x (Percentage Impedance) 3 PU Impedance of equipment 25 x ( 8/ 100 ) i) Distribution transformer(3.15 MVA) = 0.634921 Pu 4 FAULT LEVEL AT 11 KV BUS Effective impedance of utility = 0.041667 Pu Total impedance at 11 kV bus = 0.041667 Pu Fault level corresponding to above value 25 0.042 = 600 MVA Contribution of HT motors = Contribution of HT motors assuming a p.f. 0.9 = and efficiency 95% = 10.28 MVA Fault Level corresponding to impedance at = 11kV bus + Contribution of HT motor = 610 MVA Hence Fault current at 11kv Bus 610 (√ 3 x 11) = 32.03233 kA The 11kv Switchgears with fault with stand capacity of up to 40kA is feasible a) Conclusion: Fault Level at 11 kv Bus is 32.0KA 5 FAULT LEVEL AT 0.380 KV BUS 1 Base MVA chosen = 25 MVA 2 The pu impedance up to 11kV bus to 11kV bus Hight fault level at 11kV bus 25 610 = 0.041 pu Effiective impedance of distribution = 0.635 pu transformer Fault level at 0.380 Kv bus = 37 MVA Hence fault current at 0.380Kv bus = 37 (√ 3 x 0.38) = 56.19 MVA b) Conclusion: Hence , the fault level at 0.380kV bus = 37 MVA rounded of to the nearest integer Value 6 CALCULATION FOR VOLTAGE DROP = ( Equipment MVA) = = 3.15 (Base MVA) (Utility MVA) = (Base MVA) = Total Pu Impedance = (1000 x pf x efficiency) (Total kW of HT motors x 2) ( 4150 x 2 ) (1000 x 0.85 x 0.95) 600 + 10.28 = Base MVA = = = (The pu imedance up to 11kV bus + inpedance of distribution Transformer) Base MVA 25 (0.041 + 0.635) =
The per unit impedance of the system components are calculated as follows : i) Per Unit impedance of the system Components: Sr.No. i) pu impedance of utility = 25 600 ii) HT main motor fed from cable Impedance = 5Rx3x240Sqmm XLPE CU for 150mm length iii) Starting of HT motor impedance 4150kW = (4150 / 1000 ) x (200/ 100 ) iii) Running of HT motor impedance 4150kW = (350 / 1000 ) x (100/ 100 ) iv) Distribution transformer Impedance 3.15MVA =
v) LT main motor fed from cable Impedance = 3Rx4x300Sqmm XLPE CU for 70mm length vi) Starting of LT motor impedance 350kW = vi) Running of LT motor impedance 350kW = 100 Total Pu Impedance = (0.0417 + 0.0024 + 2.4323) = 2.4764 Pu 7 11kV ,4150kW System Component % Voltage Drop i) Utility System 0.0417 2.476 ii) 5‐3x240mm2 XLPE Cable 0.0024 2.4764
iii) Voltage Drop upto 11KV Bus = 1.6839 % = 1.68 %
iv) Voltage drop upto 4150kW motor terminals= (1.6839 + 0.097) = 1.78 %
v) Voltage avaliable to the motor = 100 ‐ 1.7809 = 98.22 %
vi) Hence maximum starting torque normally = avaliable with 10.81 Kv
Conclusion
Higher starting torque,however,can be achived by adjustment of electrolyte strength of LRS.
For starting of 4150 kw motor,the starting surrent have to be resticted to less than 200 % Even with above assumption the voltage drop is 1.78 % 8 150kV ,0.9kW System Component % Voltage Drop i) Utility System 0.0417 2.476 ii) 5‐3x240mm2 XLPE Cable 0.0024 2.4764
iii) Voltage Drop upto 11KV Bus = 1.6839 % = 1.68 %
iv) Voltage drop upto 4150kW motor terminals= (1.6839 + 0.097) = 1.78 %
v) Voltage avaliable to the motor = 100 ‐ 1.7809 = 98.22 % % = x 100% = 0.097 % (25 x 0.85 x 0.95) = 57.679 Voltage requirement for stable Running of HT motor = x 100% = 1.68 (25 x 0.85 x 0.9) = 27.32 Pu P.U.Impedance 0.02 x (70/ 1000 ) x 25 = 0.121 Pu 2 x (0.38^ 2 ) (25 x 0.85 x 0.95) = 2.4323 Pu Pu pu (25 x ( 8/ 100) 3.15 = 0.635 pu pu = 0.0417 0.23 x (150/ 1000 ) x 25 = 0.0024 System Component 3 x (11^ 2 ) (25 x 0.85 x 0.9) = Voltage requirement for stable starting of HT motor (200 x 10.81 ^ 2 ) (11^2 ) = x 100% 54.64 Pu (350 / 1000 ) x (200/ 100 ) (350 / 1000 ) x (100/ 100 ) 193.15 % = 0.097 % = x 100% = 1.68 % =
vi) Hence maximum starting torque normally = avaliable with 10.81 Kv
Conclusion
Higher starting torque,however,can be achived by adjustment of electrolyte strength of LRS.
For Running of 0.9 kw motor,the starting surrent have to be resticted to less than 0 % Even with above assumption the voltage drop is 1.78 % 8 0.38kV ,350kW Total Pu Impedance = (0.0417 + 0.635 + 0.1212+27.3215) = 28.119 Pu System Component % Voltage Drop i) Utility System 0.0417 28.119 ii) Distribution transformer 3.15MVA 0.6350 28.119 iii) 3Rx4x300Sqmm XLPE CU Cable 0.1212 28.119
iv) Voltage Drop upto 0.38 KV Bus = 0.15 % = 0.15 %
v) Voltage drop upto 350kW motor terminals = (0.15 + 0.44 +2.26) = 2.85 % vi) Voltage avaliable to the motor = 100 ‐ 2.85 = 97.15 % 368.6 vii) Hence maximum starting torque normally = avaliable with 369 V viii) Higher staring torque, however can be achieved by VVVF drive 380 8 kV ,3.15kW Total Pu Impedance = (0.0417 + 0.635 + 0.1212+54.6429) = 54.81 Pu System Component % Voltage Drop i) Utility System 0.0417 54.806 ii) Distribution transformer MVA 0.6350 54.806 iii) 3Rx4x300Sqmm XLPE CU Cable 0.1212 54.806
iv) Voltage Drop upto KV Bus = 0.08 % = 0.08 %
v) Voltage drop upto 3.15kW motor terminals = (0.08 + 0.23 +1.16) = 1.47 % vi) Voltage avaliable to the motor = 100 ‐ 1.47 = 98.53 % vii) Hence maximum starting torque normally = avaliable with 374.41 V viii) Higher staring torque, however can be achieved by VVVF drive CALCULATION FOR SELECTION OF CABLE SIZES FOR VARIOUS RATING OF LT MOTORS PU impedance of Utility = 0.0417 = 0.0417 pu Distribution transformer Impedance 3.15MVA = 0.6350 = 0.6350 pu Therefor ,pu impedance upto 0.38 kV Bus = 0.0417+0.635 = 0.6767 pu (100 x 10.81 ^ 2 ) = 96.58 % (11^2 ) (200 x 369.17 ^ 2 ) = 189 % ((0.38 x 100 )^ 2) = 1.16 % = 0.230 % 94 % 0.15 % = x 100% = 0.440 % = x 100% = 2.26 % x 100% = x 100% = (100 x 374.414 ^ 2 ) = ((0.38 x 100 )^ 2) Voltage requirement for stable starting of LT motor squirrel cage induction motor with DOL starter. = Voltage requirement for stable Running of LT motor squirrel cage induction motor with DOL starter. = x 100% = 0.08 % x 100% =
Base MVA = 25 MVA = 25 MVA Motor Rating = 350 kw Motor Voltage rating = 380 v Motor full load current = Motor full load current 695.13 A Motor Efficiency 0.90
Power Factor Cos θ = 0.85 rad
Sin θ = 0.53 rad Cable From MCC to Motor Cable Size Al = 1.5 Sqmm Cable Impedance z% Z % = 0.020 ohm/mtr Cable Length (L) = 70 mtr No.of Run of the Cable (N) = 1 No. AC resistance of the cable at 90°C = 0.02317 Ω /mtr Reactance of the cable at 50Hz = 0.000107 Ω /mtr The pu impedance of 150m (typical) cable of different sizes is calculated as follows : = %Z of Cable x Base MVA x Length of cable in kms = ( Z% Ohm/km x 12.12) Pu Impedance of the Cable = 242.40 pu 695.13 Voltage drop of the selected cable = 1 = 16.43 V Voltage drop of the selected cable = 16.43 (380 / √3 ) = 2.50 % The starting KVA of motors is calculated as: For squirrel cage induction motors = Starting KVA = (k W x 6) (0.9 x 0.85) = kw X 7.84 = 2745.098 KVA For Slip ring induction motors = Starting KVA = (k W x 2.5) (0.9 x 0.85) = kw X 3.27 = 1143.791 KVA For squirrel cage induction motors = Running KVA = (k W x 1) (0.9 x 0.85) = kw X 1.31 = 457.5163 KVA (350 x 1000) (√ 3 x (380 x 1000) x 0.85 x 0.95) (kV)2 (380x 1000^2) ( x 0.02 x 70) = ( % Z Ohm/km x 25 x 70/ 1000 ) = 12.12 x 100 Kw x 6 Efficiency x Power factor Kw x 2.5 Efficiency x Power factor Kw x 1 Efficiency x Power factor
