EDSA IEC 909 SHORT CIRCUIT ANALYSIS

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1 1.0 Tutorial Exercise This tutorial exercise will serve as a validation and verification test for the EDSA IEC 909 short circuit program. The tutorial will be based on two examples documented in the IEC 909 Short circuit current calculation in three-phase AC systems. These examples are: Example 1: Example 2: Calculation of short-circuit currents in a low-voltage system. Appendix A, section A-1, page 109. EDSA File Name: IEC1.eds Calculation of balanced short-circuit currents in a medium-voltage system, including the influence of motors. Appendix A, section A-2, page 125. EDSA File Name: IEC2.eds Each example will first be solved by longhand calculations, and then the corresponding pre-created EDSA file will be used to re-calculate the short circuit results. Once both analyses have been completed, a table of comparison will be presented. It is assumed, for this exercise, that the user is familiar with building EDSA job files using the ECAD interface. If not, please refer to sections 1.0 and 2.0 of the EDSA User s Guide to review the process. The program options used in this tutorial are as follows: IEC 909 Methodology IEC Maximum Voltages Peak Method C 2.0 IEC 909 Example 1 / Longhand Calculations 250 MVA (cq = 1.1) X/R = j kv 15 kv T1 630 kva 15/0.4kV 4% P = 6.5 LOAD T2 400 kva 15/0.4 kv 4% P = 4.6 LOAD j j F1 C j F2 Un = 380 V C j C j14.85 F3 C j0.136 F j F2 Vpu = j j F j Figure 1. Single line diagram and equivalent impedance diagram for IEC Example 1. Page 1

2 Figure 1 shows the system under study for the IEC example 1. The diagram shown in this figure is given, both in ohms and per unit. The calculation, however, will be done in per unit. Original data for the example had the following cable info. C mm 2, 10 meter length and j ohms/meter/ph = j 0.79 milliohms. C mm 2, 4 meter length and j ohms/meter/ph = j milliohms. C3 2-70mm 2, 20 meter length and j ohms/meter/ph = j milliohms. C4 2-50mm 2, 50 meter length and j ohms/meter/ph = j milliohms. The PU bases selected are 10 MVA, 15 kv and 0.4 kv. Utility source - The source is <35 kv, therefore the X/R = 10 per IEC (X = 0.995*Z, R = 0.1 X). c (Z EQUIPMENT BASE ) (MVA COMMON BASE ) Z COMMON BASE = MVA EQUIPMENT BASE Transformers using equation below Z QT = 1.1 * 1.0 * 10/250 = PU R MULTIPLIER = , R PU = X MULTIPLIER = , X PU = (Z EQUIPMENT BASE ) (MVA COMMON BASE ) Z COMMON BASE = MVA EQUIPMENT BASE 630 kva transformer T1, impedances on 630 kva base is: R PU = P LOSS /KVA RATED = 6.5/630 = X PU = (Z 2 - R 2 ) ½ = ( ) ½ = , X PU /R PU = kva transformer T2, impedance on a 400-kVA base is: Convert to 10 MVA base R PU = P LOSS /KVA RATED = 4.6/400 = X PU = (Z 2 - R 2 ) ½ = ( ) ½ = , X PU /R PU = 3.33 Z PU = 10 * ( j0.0387)/0.63 = j for 630 kva Tx Z PU = 10 * ( j0.0383)/0.4 = j for 400 kva Tx Cables are converted from ohms to per-unit on 10 MVA and 0.4-kV bases Z PU = Z OHM * (MVA COMMON BASE )/kv 2 BASE C1= j0.395 mohms = j PU C2= j0.136 mohms = j PU C3= j1.740 mohms = j PU C4= mohms = j PU These per unit values are shown on Figure 1. Page 2

3 Fault at F1, the impedance network is reduced by having the impedance of T1 in parallel with sum of the impedances of T2 +C2 +C1. C1 = j PU C2 = j PU T2 = j Sum j = deg T1 = j = deg Parallel of T1 and Sum = j Add source Z j Total = j = X/R = V PREFAULT = (380/400V) = 0.95 PU V I PU 400 =1/ = I k= c *(V PREFAULT )(I PU )(MVA COMMON BASE )/(1.732*kV) I k= 1.05* *10 /(1.732*0.4) = ka X/R = To determine the circuit X/R ratio to be used to determine the peak current all reactances are multiplied 0.40 and the system then reduced. The new per-unit impedances are: Source = j T1 = j T2 = j C1= j PU C2= j PU C3= j PU C4= j PU C1 = j PU C2 = j PU T2 = j Sum j = deg T1 = j = deg Parallel of T1 and Sum = j Add source Z j Total = j (X/R = ). X/R ADJ = 2.5 * (X/R 0.4 ) = 2.5* = This compares with X/R = 3.64 for the 50 Hz system. The X/R = 3.67 is used to calculate the peak current from i PEAK = 2 I k ( e -3(R/X) ). i PEAK = 2*32.81*( e 3/3.67) ) = ka Page 3

4 Fault at F2, the impedance network is reduced by having the impedance of T1 and C1 in parallel with the impedances of T2 +C2. C1 = j PU T1 = j PU Sum j = @ deg C2 = j PU T2 = j pu Sum j = deg Parallel of branches = j Add source Z j Total = j = X/R = (V PREFAULT ) = (380/400V) = 0.95 PU V I PU 400 =1/ = I k= c *(V PREFAULT )(I PU )(MVA COMMON BASE )/(1.732*kV) I k= 1.05* *10 /(1.732*0.4) = ka X/R = For peak currents using 0.4 times the impedance. C1 = j PU T1 = j pu Sum j = deg C2 = j PU T2 = j Sum j = deg Parallel of branches = j Add source Z j Total = j , R/X = X/R ADJ = 2.5 * (X/R 0.4 ) = 2.5* = This compares with X/R = for the 50 Hz system. The X/R = 3.51 is used to calculate the peak current from i PEAK = 2 I k ( e -3(R/X) ). i PEAK = 2*32.333*( e 3/3.512) ) = ka IEC calculated using a rounded value for( e 3/3.512) )=1.44 instead of Fault at F3, the impedance network was reduced for fault F2 to the common point. Cables C3 and C4 are added to that impedance. From fault F2 = j Complex, j using 0.4*X C3 = j j C4 = j j Total = j j @ X/R = Page 4

Fault current I k is 6.595 ka @ -42.156 0, X/R = 0.9053. (V PREFAULT ) = (380/400V) = 0.95 PU V I PU 400 =1/ 2.183 = 0.4581 I k= c *(V PREFAULT )(I PU )(MVA COMMON BASE )/(1.732*kV) I k= 1.05*0.95 *0.

9055 is used to calculate the peak current from i PEAK = 2 I k (1.02 + 0.98 e -3(R/X) ). i PEAK = 2*6.595*(1.02 + 0.98e 3/0.9055) ) = 9.846 ka IEC-60909 calculated 9.

5 Fault current I k is , X/R = (V PREFAULT ) = (380/400V) = 0.95 PU V I PU 400 =1/ = I k= c *(V PREFAULT )(I PU )(MVA COMMON BASE )/(1.732*kV) I k= 1.05*0.95 *0.4581*10 /(1.732*0.4) = ka 0, X/R = system. X/R ADJ = 2.5 * (X/R 0.4 ) = 2.5* = This compares with X/R = for the 50 Hz The X/R = is used to calculate the peak current from i PEAK = 2 I k ( e -3(R/X) ). i PEAK = 2*6.595*( e 3/0.9055) ) = ka IEC calculated 9.89 using a rounded values I k and ( e -3(R/X) ). 3.0 IEC 909 Example 1 / EDSA Analysis Step 1. Select File. Step 2. Select Open. Step 3. Select IEC1.EDS. Step 4. Select Open. 3.1 Invoke the ECAD interface, and proceed to load the pre-formatted file for IEC example 1. The file is called IEC1.eds, and it can be loaded according to the procedure shown in the above screen capture. Page 5

6 Step 1. Click here to invoke the Short Circuit program. 3.2 Next, proceed to invoke the short circuit program, as indicated in the above screen capture. Page 6

3.3 Once the Short Circuit interface appears, proceed to specify the required short circuit components, the calculation

7 Step 6 Select OK. Step 1 Select the Options icon. Step 2 Fill out the Options screen exactly as indicated here. Step 3 Select IEC. Step 4 Fill out the IEC 909 Calculation Control screen exactly as indicated here. Step 5 Select OK. 3.3 Once the Short Circuit interface appears, proceed to specify the required short circuit components, the calculation methodology, and the specific IEC calculation controls. Follow the methodology outlined in the screen-capture shown above. Page 7

8 Step 1 Select Update Answer File Step 2 Select Faults at all busses Step 3 Select OK. 3.4 Finally, run the analysis by following the procedure shown in the screen-capture above. Page 8

9 3.5 The IEC 909 report, showing the selected sections, is now presented in the output screen. At this point, the report can be printed out, copied to the clipboard or saved as a text file for third party software customisation. To exit, select Done from the menu. Page 9

10 4.0 IEC 909 Example 1 / Validation and Verification Table The following table shows a comparison between the results obtained using longhand calculations, EDSA and the results documented in the IEC 909 standard. IEC 909 example 1 (pp ) Location F1 F2 F3 Result Type Program Hand Calc Variance with IEC 909 Value Value Hand Calcs Example i peak % ½ cyc % I k % i peak % ½ cyc % I k % i peak % 9890 ½ cyc % I k % 6600 All variances with Hand Calculations and IEC 909 documents are attributed to round off on input data or results. 5.0 IEC 909 Example 2 / Longhand Calculations 750 MVA (cq = 1.1) X/R = j kv 33 kv C j ohm C j ohm C j C j T1 15 MVA 33/6.3 k V 15% X/R = 25 T2 15 MVA 33/6.3 k V 15% X/R = 25 T j T j Un = 6 kv F1 6 kv F1 M1 M2 M1 5 MW, U M = 6kV pf=0.86, eff = 0.97 I LR/ I M= 4, 4 pole X/R = 10 M3 M4 M2, M3, M4 1 MW, U M = 6kV pf=0.83, eff = 0.94 I LR/ I M= 5.5, 2 pole X/R = 10 M1 M2 M3 M j st Cy (IEC) j Int (IEC) M4 M2, M3, M j st Cy (IEC) j Int (IEC) Figure 2. Single line diagram and equivalent impedance diagram for IEC Example 2. Page 10

11 Figure 2 shows the system under study for the IEC example 2. The diagram shown in this figure is given both, in ohms and per unit. The calculation, however, will be done in per unit. The following table shows the data that corresponds to the motors. Size X/R Reactance >= 500 kw/pole (670 HP/pole) ZM < 500 kw/pole (670 HP/pole) ZM Grouped Low Voltage Motors ZM ZM = 100* IRATED/ILOCKED ROTOR IEC Motor X/R Ratios to be Used Example 2 - Impedance Calculations Source = MVA BASE * c/mva SOURCE = 10 * 1.1/750 = X/R = 10 = j PU Cable C1 & C2:Ohms (MVA BASE )/kv 2 = ( j0.485)10/33 2 = j PU Transformer T1 & T2: 25 MVA, X=15%, R = 0.6% MVA BASE (%Z)/(MVA TRAN * 100) = 0.1 X/R = 15/0.6 = 25 = j PU Equipment Data X/R Base kv PU R PU X Source 750 MVA Cable C j Cable C j Transf. T1 15 MVA,15%Z Transf T2 15 MVA,15%Z Data used in Short Circuit Calculations. Motor M1 at 6.0 kv, 5 MW and 4 Pole to 6.3 kv base 5MW/4pole = 1.25 MW/Pole has a X/R = 10 Motor MVA = MVA/PF/Eff = 5/0.86/0.97 = MVA 1st Cy Imp.= MVA BASE *kv MOT 2 /(kv BASE 2 *MVA MOT *I LR /I MOT ) = 10*6*6/(6.3*6.3*5.994/4)= PU = j PU Motors M2, M3, M4 at 6 kv 1 MW and 2 Pole to 6.3 kv base. 1MW/2pole = 0.5 MW/Pole has a X/R = 10 Motor MVA = MVA/PF/Eff = 1/0.83/0.94 = MVA 1st Cy Imp.=MVA BASE *kv MOT 2 /(kv BASE 2 *MVA MOT *I LR /I MOT ) =10*6*6/(6.3*6.3*1.282/5.5)= = j PU The standard provides both curves and equations to determine the currents from motors at breaking time. The interrupting time impedances are determined by using the factors : and q. Factor : accounts currents in both synchronous and asynchronous (induction motors) decaying from substransient to transient Page 11

12 impedance. Factor q is a second correction for asynchronous machines that accounts for different decay rates based on the motor size. The IEC example used 0.10 seconds for the breaking time. : = γ -0.26/X kg for t MIN = 0.02 seconds, Z kg = I kg/irg # : = γ -0.30/X kg for t MIN = 0.05 seconds : = γ -0.32/X kg for t MIN = 0.10 seconds : = γ -0.38/X kg for t MIN => 0.25 seconds, (not shown) # for a fault on generator terminals Z kg = c/[k G (Z G )] q = ln [m] for minimum time = 0.02 seconds q = ln [m] for minimum time = 0.05 seconds q = ln [m] for minimum time = 0.10 seconds m is the active power in MW per motor pole pair Motor M1, has 4 times inrush current at rated voltage. The inrush current is adjusted by the voltage factor c and is 4*1.1 = 4.4 and the motor has 2.5 MW per pole. Motors M2, M3, and M4 have 5.5 * 1.1 inrush current of 6.05 and 1 MW per pole. For motors => 1.0 MW per pole pair, the Standard specifies a X/R ratio of 10. The : and q values at 0.10 seconds are: Motor M1: : = 0.796, q = 0.680, :q = 0.541, Impedance multiplier = 1/:q = 1.85 Motor M2: : = 0.724, q = 0.570, :q = 0.413, Impedance multiplier = 1/:q = 2.42 The standard uses these multipliers to adjust the first cycle currents. The same total current will be calculated if the inverse multiplies are applied to the impedances. Motor MW Rating KV RPM Poles %X MVA HP per pole-pair X/R R PU Resistance X PU Reactance M1 M2 M3 M First Cycle Per-unit Motor Impedances on a 10-MVA base IEC First Cycle IEC Breaking Time (0.10 seconds) Motor MVA PU Resist. PU React. : Multiplier q Multiplier PU Resist. PU React. M1 M2 M3 M Motor Impedance for 1 st Cy. and Interrupting Time (10 MVA Base) Factors : and q for Rotating Equipment (Decay of Symmetrical Current) Following the procedure given in IEC-60909, the non-decaying ac fault current is first calculated on the 6 kv bus, then the motor contributions are added to it. Page 12

13 The equivalent impedance is determined by adding impedances C1 to T1 and C2 to T2 then paralleling the two and adding the remote source impedance. Transf T1 or T2 = j Cable C1 or C2 = j Total = Parallel transf and cables = j Source impedance = j Total = j = PU on 10 MVA, 6.3 kv base. I k of non-ac decay = 1.1*10 /0.067//3/6.3 = ka at X/R = Since the bus is operated at 6.0 kv, I k = *6.0/6.3 = = j ka Next, adding the decaying motor sources at 6.0 kv gives M1 = 1.1*10(6.0/6.3)/( j )//3/6.3 =2.553 = j 2.54 ka (X/R = 10) M2 = 1.1*10(6.0/6.3)/( j1.28)//3/6.3 = = j0.745ka (X/R =10) M3 and M4 are the same as M2. Total Sym. ka = * j( *0.745) = ka at X/R =11.2. The peak currents are added for each contribution separately. Using the equation for peak current I k PEAK =I k *[ , -3/(X/R) ]*/2 Transformer Source(X/R = 11.71) = *2.515 = ka = j ka peak Motor M1 (X/R = 10) = *2.469 = 6.3 ka = j ka peak Motor M1, M2, M3 (X/R = 10) = 3*0.748* = 5.540kA = j ka peak IbASYM = j = ka (Value in IEC Standard = ka due to rounding) Example 2 - Breaking Current Calculations at F1 To calculate the breaking time current at 0.10 seconds, the motor breaking currents are added to the nondecaying ac source current Ib ASYM = (I k 2 + I 2 DC ) ½ _(2Β f t /(X/R)), where I DC =I k */2, Transformers I k = j ka = ka I DC =I k*/2, _(2Β f t /(X/R)) = ( j )/2, _(2Β 50*0.1 /11.71) = j1.38 ka = Ib ASYM = [ ] 1/2 = ka M1 = 1.1*10 (6.0/6.3)/( j )//3/6.3 = ka at X/R = 10 (due to rounding of : and q Standard calculated 1.38 ka) Ik = j ka I DC =( j 2.54)/2, _(2Β 50*0.1 /10) = j ka = ka Ib ASYM = [ ] 1/2 = ka M2 = 1.1*10 (6.0/6.3)/( j )//3/6.3 = ka at X/R = 10 each motor (due to rounding of : and q Standard calculated ka) Page 13

14 Ik = j ka I DC =( j 0.745kA)/2, _(2Β 50*0.1 /10) = j ka = ka Ib ASYM = [ ] 1/2 = ka Total I SYM of transformer, M1, M2, M3, and M4 currents are: Transf. = j ka M1 = j ka M2(3) = j Total = j = ka Total I DC of transformer, M1, M2, M3, and M4 currents are: Transf. = j1.38 ka M1 = j ka M2(3) = j Total = j = ka (The Standard did a scale addition on the dc magnitude and left off the /2 in the IbASYM calcs.) IbASYM = [ ] 1/2 = ka (Standard gives ka) Since the sources having decaying ac current components are greater than 5% (>15% in this system) the fault currents are referred to as near to generator. This problem was redone using Method C for the first cycle peak. The motor impedances were included in the network reduction. Computer software was used reduce the network with reactances at 40% the 50Hz values. The final solution is given below with the Method C X/R AJD being used for both the peak and dc component. Total Sym. ka = ka at X/R =11.2. i PEAK = ka, X/R AJD = IbASYM = ka Comments on Calculation procedure The solution for Example 2 followed the procedure given in IEC To me it has several questionable items. 1. Why isn t method A, B, or C used in this example. It appears to present a 4 th method. Therefore, if this example is given to several engineers, a number of different correct answers can be obtained. Why not include the motors in the impedance reduction and let the math take care of the contribution? Including motor impedances would be more acceptable to computer programs. 2. From the IEC examples given, it is not clear how to handle a network in which the cable impedances to the motors are represented. If the fault currents at the terminal of the motors are to be calculated, and if the motors currents are added after the far from generator network is reduced, it appears that ohms law can be violated for some system configuration. Network action will affect the currents coming Page 14

15 from adjacent motors due to their cable impedances. To me, to motor current should not be added directly as if it does not make a difference. 3 The procedure shows that the first cycle network impedance from remote sources is assumed not to change for breaking time currents. While in example 2 this is correct, but in Example 3 ( not worked out here) the motor contribution from Busbar B and C have an influence on each other which was not taken into account during breaking time sample calculations. To me this again violates ohms law. 4 While I agree that a complex network reduction X/R ratio may not accurately represent the X/R ratio needed to obtain the peak current, IEC indicate that Method C is more accurate. But, the examples only use it on the first example. The Standard gives no references to support method C or several other procedures used in the Standard. 6.0 IEC 909 Example 2 / EDSA Analysis 6.1 Following the instructions outlined in steps 3.1 and 3.2, proceed to load the file IEC2.eds, and launch the short circuit program interface. Page 15

16 6.2 Following the same instructions outlined in step 3.3, proceed to select the options, and calculation control settings for this example. The above screen capture shows the what is needed. Notice that in IEC example 2, the 6 cycle X/R and AC component have been included in the calculation. Next, run the analysis according to the procedure explained in step 3.4. Page 16

17 6.3 Once again, the IEC 909 output screen, presents the selected output sections. At this point, the report can be printed out, copied to the clipboard or saved as a text file for third party software customisation. To exit, select Done from the menu. Page 17

18 7.0 IEC 909 Example 2 / Validation and Verification Table The following table shows a comparison between the results obtained using longhand calculations, EDSA and the results documented in the IEC 909 standard. IEC 909 example 2 (pp ) Location F1 Result Type Program Hand Calc Variance with IEC 909 Value Value Hand Calcs Example i peak % ½ cyc % I k % cyc Break % All variances with Hand Calculations and IEC 909 documents are attributed to round off on input data or results. Page 18

Fault Calculation Methods

ELEC9713 Industrial and Commercial Power Systems Fault Calculation Methods There are two major problems that can occur in electrical systems: these are open circuits and short circuits. Of the two, the