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Fault level calculations
- 1. Today’s Session • FaultLevel Calculations – Power-system Overview – PU Concept – 3 Ph. Fault Level Calculations – 1 Ph Fault Level Calculations • O/C E/F Relay Time Coordination – General information about relay characteristic – Worked out example • Directional Relay – What is MTA – How to confirm direction of relay reach while in service What is ahead?
- 2.
- 3. Fault Level Calculations3 What is ahead? • Overview of transmission network. • What is mean by fault level • PU Concept • Network Modeling • Fault Level Calculation & Worked out example – Three Phase Fault – SLG Fault
- 4. Fault Level Calculations4 Step-1 Overview of power system grid and transmission network
- 5. Fault Level Calculations5 Typical State Level Power System & our area of interest
- 6. Fault Level Calculations6 Typical State Level Power System & our area of interest
- 7. Fault Level Calculations7 400kV 220kV 220kV 132kV 33kV 33kV220kV 132kV 11kV33kV S/S - A S/S - F S/S - E S/S - B S/S - C S/S - D Local Network – or – Sub-Transmission System
- 8. Fault Level Calculations8 Step-2 What is mean by fault level
- 9. Fault Level Calculations9 What is fault level ABC Transmission Co. Ltd. 3 Ph Fault Levels For 2010-2011 14530 MVA 18290 MVA 9700 MVA 8400 MVA 11300 MVA 4630 MVA 5800 MVA 6400 MVA 7200 MVA 2700 MVA 1185 MVA 3670 MVA 9730 MVA S/S - B S/S - A S/S - C 14530 MVA 8400 MVA 9730 MVA S/S - B
- 10. Fault Level Calculations10 What is fault level-Continued…. 14530 MVA 9730 MVA S/S - B 2200 MVA 2200 MVA 2300 MVA 2450 MVA 2450 MVA 1465 MVA1465 MVA 8700 MVA 1030 MVA
- 11. Fault Level Calculations11 Step-3 Understanding Per-Unit Measurements
- 12. Fault Level Calculations12 PU Concept • PU impedance values are calculated by – For Transformers – For Lines %ImpedanceNew Base MVA = % ImpedanceName Plate Base MVA Capacity in MVA %Impedance = Ohmic Value Base MVA kV 2
- 13. Fault Level Calculations13 PU Concept • PU impedance values are calculated by – For Transformers • %Impedance = Ohmic Value * ( Base MVA / Name Plate MVA ) – For Lines • %Impedance = Ohmic Value * ( Base MVA / BasekV2 )
- 14. Fault Level Calculations14 PU Concept 400/220 kV 15% 315 MVA 12 Ohm Sec DSec CSec B 0.4 ohm/Km 200 Km 0.4 ohm/Km 50 Km220/132 kV 14% 200 MVA 132/33 kV 9.33% 25 MVA 0.4 ohm/Km 30 Km Sec A Base MVA = 200 MVA 400kV 33kV132kV220kV T1 L1 T2 L2 T3 L3
- 15. Fault Level Calculations15 Step-4 Network Modeling
- 16. Fault Level Calculations16 PU Concept 400/220 kV 15% 315 MVA L3T3L2T2L1T1 0.4 ohm/Km 200 Km 0.4 ohm/Km 50 Km220/132 kV 14% 200 MVA 132/33 kV 9.33% 25 MVA 0.4 ohm/Km 30 Km
- 17. Fault Level Calculations17 Short Circuit Current • 3 Phase Short Circuit MVA at fault point Base MVA x 100 = ---------------------------------------------------- Percentage Impedance (from source to fault point)
- 18. Fault Level Calculations18 PU Concept-Continued 9.52% 33.05% 14% 22.96% 74.64% 220% XT1 XL1 XL2 XL3XT2 XT3
- 19. Fault Level Calculations19 PU Concept 400/220 kV 15% 315 MVA 12 Ohm 65 Ohm20 Ohm34 Ohm80 Ohm75 Ohm 0.4 ohm/Km 200 Km 0.4 ohm/Km 50 Km220/132 kV 14% 200 MVA 132/33 kV 9.33% 25 MVA 0.4 ohm/Km 30 Km
- 20. Fault Level Calculations20 Step-5 Finding Source Impedance, Worked out Example
- 21. Fault Level Calculations21 Typical State Level Power System & our area of interest 4630 Base MVA= Xs(pu) MVA 100 Xs(pu) 4630 Xs (pu) = 0.0216 Or Xs% = 2.16%
- 22. Fault Level Calculations22 Solved Example 2850 MVA 8 Km 0.4 Ohms / Km 5 Km 0.4 Ohms / Km 50 MVA 10.21% 50 MVA 10.88 % 25 MVA 9.73 % 25 MVA 10.04 % 50 MVA 10.21% 50 MVA 10.88 % 25 MVA 9.73 % 25 MVA 10.04 % 50 MVA 10.03% 25 MVA 10.52 % 25 MVA 10.21 % 25 MVA 9.84 % 220/132kV Padegaon 132/33-11 kV Harsool 132/33-11 kV Chikalthana 132kV 33kV 11kV 132kV 33kV 11kV
- 23. Fault Level Calculations23 Solved Example 2850 MVA 668 MVA 2257 MVA 220/132kV Padegaon 132/33-11 kV Harsool 132/33-11 kV Chikalthana 132kV 33kV 11kV 132kV 33kV 11kV 1998 MVA 538 MVA
- 24. Fault Level Calculations24 Unsymmetrical Faults-Sequence Network • Up till now we have discussed 3 Phase faults which are rear. Most of the fault in power system are single line to ground faults. • To find out the short circuit MVA due to single phase to ground fault it is necessary to use Sequence Network • Let us have a closer look of sequence network
- 25. Fault Level Calculations25 Unsymmetrical Faults-Sequence Network M1 M2 T2 L GT1 Xg1XT11XL1XT21 XM21XM11 Xg2XT12XL2 XT22 XM22XM12 Xg0XT10XL0XT20 XM20XM10 F FF SLD +Ve Sequence Network -Ve Sequence NetworkZero Sequence Network
- 26. Fault Level Calculations26 Single Line to Ground Fault-Sequence Network Connections Xg1XT11XL1XT21 XM21XM11 F Xg2XT12XL2 XT22 XM22XM12 F Xg0XT10XL0XT20 XM20 F X1 X2 X0 Ia = 3 / (X1 + X2 + X0)
- 27. Fault Level Calculations27 Single Line to Ground Fault-Solved Example 3Ph 2850 MVA 8 Km 0.4 Ohms / Km 5 Km 0.4 Ohms / Km 50 MVA 10.21% 50 MVA 10.88 % 25 MVA 9.73 % 25 MVA 10.04 % 50 MVA 10.21% 50 MVA 10.88 % 25 MVA 9.73 % 25 MVA 10.04 % 50 MVA 10.03% 25 MVA 10.52 % 25 MVA 10.21 % 25 MVA 9.84 % 220/132kV Padegaon 132/33-11 kV Harsool 132/33-11 kV Chikalthana 132kV 33kV 11kV 132kV 33kV 11kV 1Ph 2230 MVA
- 28. Fault Level Calculations28 Single Line to Ground Fault-Solved Example 3Ph 2850 MVA 3Ph 2850 MVA 1Ph 2230 MVA 3*BaseMVA*100 ( XS0 + (XL10 || XL20) )2*( XS1 + (XL11 || XL21) )( )+ 1 Ph Short Circuit MVA at 132kV Harsool Bus
- 29. Fault Level Calculations29 Solved Example220/132kV Padegaon 132/33-11 kV Harsool 132/33-11 kV Chikalthana 132kV 33kV 11kV 132kV 33kV 11kV 1998 MVA 538 MVA 3Ph 2850 MVA 1Ph 2230 MVA 3Ph 2257 MVA 1Ph 1662 MVA 3Ph 668 MVA 1Ph 604 MVA
- 30.
Editor's Notes
- #6 Typical state level power system is shown here. This power system is consists of Generator, EHV lines, ICTs, Transformer, Distribution Feeders etc, Power generated get injected into the system either at 400kV or at 220kV. We can say that major portion of the power system is of 400kV Grid. (Shown violate) Some portion is of 220kV Grid (Shown brown) Generally most of engineers deals with small portion of the network shown inside. Which we may call as Sub-Transmission network. Majority portion of sub-transmission network is of radial nature. While dealing with such limited portion of the system from fault level calculation point of view we can replace the entire external grid system with a source and a thevinians equivalent impedance.
- #7 Here grid is replaced by thevinans equivalent impedance. Or to be more specific, while calculating fault levels we can neglect resistance of the elements Thus this impedance is considered as reactance Then next obvious question is how to determine this equivalent reactance. It is explained afterword. But before to proceed further let us have a closer look at Sub-Transmission network And it is necessary to recall our concepts about pu calculations
- #8 Here sub-transmission network enlarged ( and rotated clockwise 900) Though following points are un-important or known it is necessary to have common language while discussing some issues/problems regarding power system. Hence such terminology is reviewed here. S/S voltage levels means highest voltage level at that S/S Hence 400kV S/S – A 200kVS/S – B S/S – C 132kVS/S – D S/S – E S/S – F Sub-Stations are also recognized as 400/220kV or 220/33 kV or 220/132kV or 132/33kV or 132/33-11kV (S/S-F) etc. 400kV and one 220kV line of S/S-A is in grid. S/S-B and S/S-C get supplied through parallel lines. S/S-D and S/S-E are connected in local loop network (S/S-D is having LILO arrangement). S/S-F is of radial nature There may have standby lines too.
- #10 We may define fault level as - Fault level at any location is that value of current which is expected to flow upon occurrence of fault considering all other elements in network are in service and all generations are in service and generating their maximum output. FOR FAULT LEVEL STUDY SELECTED LOCATION IS BUS IN THE SYSTEM Every year such calculations are done for the entire network of the company at corporate level using computer programs and results circulated to field offices. Generally these fault levels are expressed in MVA or KA. When expressed in MVA it is also called as Short Circuit MVA. Such a sample result is shown here. Generally while calculating fault levels, two type of faults are considered at every bus in the network (Grid). 3 Phase Fault Single Line to Ground Fault (SLG) Here 3 Ph fault is considered. Fault levels for S/S connected to generation are more (S/S-A). Also fault levels having more lines are more (S/S-B). Fault levels of S/S having higher rating transformers are also more
- #11 Let us have a closer look of S/S-B During fault level study not only bus fault levels are given but also contribution of this fault by various elements connected to that bus are also given. Not that arrow marking is not necessary. It is only shown here to get the contribution more clear. Such type of data can be given in SLD form as well as Table form too. However if given in SLD form it is more understandable
- #13 ================================================================================ Note :- we are using pu impedance and percentage impedance interchangeably throughout the discussion where Percentage Impedance = Per-Unit Impedance * 100 ================================================================================ While modeling a transmission network for fault level study following points are important. Transmission system (or network) is made up of lines and transformers. Line is characterized by inductance, resistance and line charging capacitance. For small length lines, charging capacitance can be neglected while modeling the transmission system. Small length lines are generally of lower voltage. From transmission point of view that is 132kV and 220kV. While modeling the 400kV line we have to consider line charging capacitance too. For fault level calculations only it is necessary to consider reactive impedance of the line. We can exclude line charging and resistance. Transformer impedances are always expressed in percentage (or PU ). ================================================================================ The per-unit impedance of the transformer is the same regardless of whether it is determined from ohmic values referred to high tension or low tension sides of the transformers. For example consider a 132/33 kV 25 MVA transformer under short circuit test. That means about 3 phase 440 V (i.e. 440/1.732 = 254 V Ph-N) supply is given from 132kV side and 33kV side kept short circuited. This transformer short circuit current at 440 V will be about 3.9 amps. That is its leakage impedance is 254/3.9 = 65.12 Ohms Its PU value (referred to 132kV side) is 65*25/(132*132) = 0.0933 ( or 9.33%) If the leakage reactance had been measured on the low tension side, the value would be 65/16 = 4.0625 ( Where 16 is square of transformation ratio (132/33)2 ) Its PU value (referred to 33kV side) is 4.0625*25/(33*33) = 0.0933 (or 9.33%) same as before Same case is about line modeling where base voltages for the lines connected on HV and LV sides of the transformers shall have same ratio as that of transformation ratio In short while calculating per-unit impedance of the line it’s rated voltage is selected as base voltage and base MVA will be common for entire system.
- #14 The per-unit impedance of the transformer is the same regardless of whether it is determined from ohmic values referred to high tension or low tension sides of the transformers. For example consider a 132/33 kV 25 MVA transformer under short circuit test. That means about 3 phase 440 V (i.e. 440/1.732 = 254 V Ph-N) supply is given from 132kV side and 33kV side kept short circuited. This transformer short circuit current at 440 V will be about 3.9 amps. That is its leakage impedance is 254/3.9 = 65.12 Ohms Its PU value (referred to 132kV side) is 65*25/(132*132) = 0.0933 ( or 9.33%) If the leakage reactance had been measured on the low tension side, the value would be 65/16 = 4.0625 ( Where 16 is square of transformation ratio (132/33)2 ) Its PU value (referred to 33kV side) is 4.0625*25/(33*33) = 0.0933 (or 9.33%) same as before Same case is about line modeling where base voltages for the lines connected on HV and LV sides of the transformers shall have same ratio as that of transformation ratio In short while calculating per-unit impedance of the line it’s rated voltage is selected as base voltage and base MVA will be common for entire system.
- #15 This will be clear by an example consider a simple system as shown above of 400/220kV T/F 220/132kV Transformer and 132/33kV Transformer. Select base MVA of 200 MVA for entire system. Sectionalize the system as per voltage levels Section A is for 400kV Transformer T1 will get included in this section as pu values of transformer whether referred to primary or secondary are same Section B is for 220kV Transformer T1 and also transformer T2 will get included in this section due to same reason as said above. That means all transformers get overlapped by two sections Similarly section C is for 132kV and section D is for 33kV Select 400kV as base voltage in section A It will be converted into 220kV in section B, 132kV in section C and 33kV in section D =========================================================================== By this way line rated voltage will be base voltage for that line ===========================================================================
- #17 Now it is possible to calculate per-unit values of transformer and lines with above said base mva and base kv T1 Percentage Impedance = XT1% = 15*200/315 = 9.52% L1 Percentage Impedance = XL1% = (X/Km)*L*( Base MVA / ( kV2 ) )*100 =0.4*200*(200/(220*220))*100 = 33.05% T2 Percentage Impedance = XT2% = 14% L2 Percentage Impedance = XL2% = 0.4*50*(200/(132*132))*100 = 22.96% T3 Percentage Impedance = XT3% = 9.33 * 200/25 = 74.64% L3 Percentage Impedance = XL3% = 0.4*30*(200/(33*33))*100 = 220.38%
- #19 Total percentage impedance up to source 220 + 74.64 + 22.96 + 14 + 33.05 + 9.52 = 374.17% Short circuit MVA = 200*100/374.17 = 53.54 MVA Short Circuit Current = 17.5*53.54 = 936 Amp
- #20 Same results we get by considering ohmic values. How ever every time it is necessary to take care of transformation ratio hence pu method is always preferable. More ever when we works with large transmission grid extensive use of matrix is required and there is no alternative to pu values. Still we will calculate same results by ohmic method just to realize simplicity of pu method. 12 ohm referred to 132kV = 12*(4*4) = 192 Ohms ( Where 4 is transformation ratio 132/33kV) L3 + T3 + L2 65 + 192 + 20 = 277 ohms This referred to 220kV = 277*(1.6666*1.6666) = 769 Ohms ( Where 1.6666 is transformation ratio 220/132kV) (L3 + T3 + L2 + T2 + L1) Referred to 220kV = 769 + 34 + 80 = 883 Ohms This referred to 400kV = 883*(1.8181*1.8181) = 2918 Ohms ( Where 1.8181 is transformation ratio 400/220kV) (L3 + T3 + L2 + T2 + L1 + T1) = 2918 + 75 = 2993 Ohms Current at 400kV level = (400000/1.732)/2993 = 77.16 Amp Current at 33kV level = 77.16*400/33 = 935 Amp.
- #22 Let us again consider radial part of the part of the power system (Sub-Transmission Network) in which we are interested, Let we know the 3 phase short circuit MVA of the Root Bus. From this we can find out source impedance and from source MVA we can proceed further for calculating short circuit MVA of other busses in the radial Sub-Stations branched from Root Bus. As we have just calculated 3 Phase Fault Level at a bus in the grid = Base MVA / Equivalent Per-Unit Impedance up to source Hence Equivalent Per-Unit Impedance up to source from a bus in the system = Base MVA / 3 Phase Fault Level at that bus For the system under consideration its value is 2.16% which is considerable low as compared to other components pu impedances. Hence in such cases we may neglect it too without hampering further results much more
- #23 3 Ph Fault level calculations Calculate source impedance for 132kV Bus Padegaon from given fault level Consider Base MVA as 100 X1s = 100*100/2850 = 3.51% b) Calculate % impedance of Padegaon-Harsool single circuit XL1 = 0.4*8*(100*100)/(132*132) = 1.84% = XL2 c) Total impedance up to Harsool 132kV Bus = 3.51 + ( 1.84 || 1.84 ) = 3.51 + 1.84/2 = 4.43% d) 3 Phase Fault level at 132kV Harsool = 100*100/4.43 = 2257 MVA e) 132/33 kV 50 MVA Transformer percentage impedances at new base MVA XT1 = 10.21*100/50 = 20.42% XT2 = 10.88*100/50 = 21.76% f) Total impedance from 132kV Harsool S/S up to source X1s= 100*100/2257 = 4.43% g) Total impedance up to Harsool 33kV Bus = X1s + (XT1 || XT2) = 4.43 + ( 20.42 || 21.76) = 14.96% h) Fault level at 33kV Harsool Bus = 100*100/14.96 = 668 MVA Fault level at 132kV Chikalthana Bus XL1 = XL2 = 0.4*5*(100*100/(132*132)) = 1.15% Xeq = 4.43 + 1.15/2 = 5.005 Fault level at 132kV Chikalthana = 100*100/5.005 = 1998 MVA j) Fault level at 33 kV Chikalthana XT1new = 10.03 * 100/50 = 20.06% XT2new = 10.52*100/25 = 42.08 Xeq = 5.005 + ( 20.06 || 42.08 ) = 18.59% Fault Level = 100*100/18.59 = 538 MVA
- #24 Like wise fault levels of other busses also can be calculated
- #26 Without going into much details of the sequence network concept studied in academic years now we will recall some of them. Consider a SLD of stand alone system where a generator feeding 2 synchronous motor loads through a step up transformer, transmission line and a step down transformer. Let there is a SLG fault at F To find out the value of fault current first we will draw the equivalent sequence networks + Ve sequence network is with a source If all values are expressed in pu this source voltage shall be obviously 1.0 pu Subscript 1 in pu impedances of components stands for +Ve sequence impedances -Ve sequence network is exactly like +Ve sequence network Except there is no voltage source. As generated voltage (behind generator reactance Xg ) is always balance hence –Ve and zero sequence of supply voltage component is zero. Subscript 2 in pu impedances of components stands for -Ve sequence impedances NOTE THE REFERANCE BUS IN THE SEQUENCE NETWORK. It is referred to only that sequence network only. While drawing Zero sequence network special care for earthing provided and transformer winding connections has to be taken. Delta connected side get connected to Zero Bus directly. Where as star side get connected to component against it only if star point is earthed If earthing is done through reactance X , then in zero sequence network it is replaced by 3X For calculating single phase fault current we have to connect these three networks in series as shown in next slide.
- #27 After connecting sequence network in series ( with respect to fault point ). it is possible to reduce each sequence network into corresponding equivalent sequence impedance as shown above Here +Ve sequence network caries +Ve sequence current –Ve sequence network caries –Ve sequence current and Zero sequence network caries zero sequence current. Due to series configuration I1 = I2 = I0 And R Ph current involved in fault = I1 + I2 + I0 = 3I1 IR = 3/(X1+X2+X0) Or 1 Phase Short Circuit MVA = 3 * ( Base MVA) / (X1 + X2 + X0) if X is expressed in pu = 3*(BaseMVA)*100/(X1% + X2% + X0%) Note : Positive and negative sequence reactance of network components are equal. Zero sequence reactance for the transformer is same as that of +Ve sequence reactance Zero sequence reactance for line are given in data sheet and it is about 2.5 to 3 times its +Ve sequence reactance
- #28 As said previously 3 Phase and 1 Phase fault levels of busses in grid are given by studying the system using computer. With availability of this fault levels and formulae reviewed we can proceed further for calculation of single line to ground fault as below 1 Ph Fault level calculations Calculate positive sequence source impedance for 132kV Bus Padegaon from given fault level as previous Consider Base MVA as 100 X1s = 100*100/2850 = 3.51% b) Now 1 Ph fault level = 3*BaseMVA*100/(X1 + X2 + X0) But X1 = X2 = 3.51% Hence 2230 = 3*100*100/(3.51+3.51+X0) 7.02 + X0 = 30000/2230 X0 = 13.45 – 7.02 X0 = 6.43%
- #29 As said previously 3 Phase and 1 Phase fault levels of busses in grid are given by studying the system using computer. With availability of this fault levels and formulae reviewed we can proceed further for calculation of single line to ground fault as below b) Calculate % impedance of Padegaon-Harsool single circuit XL1 = 0.4*8*(100*100)/(132*132) = 1.84% = XL2 Let XL0 = 3*XL1 = 5.52% d) 1 Phase Fault level at 132kV Harsool = 3*100*100/(2*(XS1+(XL1||XL2)) + (XS0+XL0)) = 30000 / ( 2*(3.51 + ( 1.84 || 1.84 )) + ( 6.43 + ( 5.52 || 5.52 )) ) = 30000 / ( 8.86 + 9.19 ) = 30000 / 18.05 = 1662 MVA e) 132/33 kV 50 MVA Transformer percentage impedances at new base MVA XT1 = 10.21*100/50 = 20.42% XT2 = 10.88*100/50 = 21.76% h) 1 Phase Fault level at 33kV Harsool Bus = 30000 / ( 2*( Xs1 + (XL11 || XL21) + ( XT11 || XT21) ) + (Xs0 + (XL10 || XL20) + ( XT10 || XT20) ) ) = 30000 / ( 2*(3.51 + 0.92 + 10.53) + (6.43 + 2.76 + 10.53) ) = 30000 / ( 2*14.96 + 19.72 ) = 30000 / 49.64 = 604MVA Like wise you can proceed
- #30 Like wise fault levels of other busses also can be calculated