iec 909 short circuit analysis

EDSA IEC 909 SHORT CIRCUIT ANALYSIS

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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: Calculation of short-circuit currents in a low-voltage system. Appendix A, section A-1, page 109. EDSA File Name: IEC1.eds Example 2: 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 Users 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

C35.420 + j 1.740

C418.52 + j14.85

F3

C10.385 + j 0.395

F2

T1 630 kVA15/0.4kV 4%P = 6.5

F1

LOAD

0.2875 + j0.9575

C20.416 + j0.136

Un = 380 V

T2 400 kVA15/0.4 kV 4%P = 4.6LOAD

15 kV

250 MVA(cq = 1.1)X/R = 10

0.02406 +j0.02469

1.1575 + j0.92813

0.33875 + j0.10875Vpu = 0.95

F3

F2

0.00438 + j0.04378

0.1638 + j0.6138

F1

0.0260 + j0.0085

15 kV

.

Figure 1. Single line diagram and equivalent impedance diagram for IEC Example 1.


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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. C1 2-240mm2, 10 meter length and 0.077 + j 0.079 ohms/meter/ph = 0.77 + j 0.79 milliohms. C2 2-150mm2, 4 meter length and 0.208 + j 0.068 ohms/meter/ph = 0.832 + j 0.272 milliohms. C3 2-70mm2, 20 meter length and 0.271 + j 0.087 ohms/meter/ph = 5.420 + j 1.740 milliohms. C4 2-50mm2, 50 meter length and 0.3704+ j 0.297 ohms/meter/ph = 18.52 + j 14.85 milliohms. The PU bases selected are 10 MVA, 15 kV and 0.4 kV. Utility source - The source is


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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 = 0.02406 + j0.02469 PU C2 = 0.02600 + j0.00850 PU T2 = 0.28750 + j0.95750 Sum 0.33756 + j0.99069 = 1.04662 @71.184 deg T1 = 0.16375 + j0.61375 = 0.63522 @ 75.061 deg Parallel of T1 and Sum = 0.11170 + j0.37940 Add source Z 0.00438 +j 0.04378 Total = 0.1161 +j0.4232 = 0.4388 @74.66, X/R = 3.645 VPREFAULT = (380/400V) = 0.95 PU V IPU400 =1/0.4388 = 2.2789 Ik= c *(VPREFAULT)(IPU)(MVACOMMON BASE)/(1.732*kV) Ik= 1.05*0.95 2.2789*10 /(1.732*0.4) = 32.81 kA sym @-74.66, X/R = 3.645. 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 = 0.00438 + j0.01751 T1 = 0.16375 + j0.24550 T2 = 0.2875 + j0.38300 C1= 0.02406 + j0.00988 PU C2= 0.02600 + j0.00340 PU C3= 0.33875 + j0.0435 PU C4=1.15750 + j0.37125 PU C1 = 0.02406 + j0.00988 PU C2 = 0.02600 + j0.00340 PU T2 = 0.28750 + j0.38300 Sum 0.33756 + j0.39628 = 0.52056 @49.5749 deg T1 = 0.16375 + j0.24550= 0.29510 @ 56.295 deg Parallel of T1 and Sum = 0.11123 + j0.15235 Add source Z 0.00438 + j0.01751 Total = 0.11561 +j0.16986 (X/R = 1.4693). X/RADJ = 2.5 * (X/R0.4) = 2.5*1.4693 = 3.67. 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 iPEAK = 2 Ik (1.02 + 0.98 e

-3(R/X)). iPEAK = 2*32.81*(1.02 + 0.98 e

3/3.67)) = 67.426 kA


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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 = 0.02406 + j0.02469 PU T1 = 0.16375 + j0.24550 PU Sum 0.18781 + j0.27019 = [email protected] deg C2 = 0.02600 + j0.00340 PU T2 = 0.28750 + j0.38300PU Sum 0.31350 + j0.38640 = 0.49758 @50.946 deg Parallel of branches = 0.11769 + j0.38447 Add source Z 0.00438 +j 0.04378 Total = 0.12208 + j0.42825 = 0.44531 @74.09, X/R = 3.508 (VPREFAULT) = (380/400V) = 0.95 PU V IPU400 =1/0.44531 = 2.24563 Ik= c *(VPREFAULT)(IPU)(MVACOMMON BASE)/(1.732*kV) Ik= 1.05*0.95 2.24563*10 /(1.732*0.4) = 32.333 kA sym @-74.09, X/R = 3.51. For peak currents using 0.4 times the impedance. C1 = 0.02406 + j0.00988 PU T1 = 0.16375 + j0.24550PU Sum 0.18781 + j0.25538 = 0.31700 @53.6669 deg C2 = 0.02600 + j0.0034 PU T2 = 0.28750 + j0.38300 Sum 0.31350 + j0.38640 = 0.49758 @50.9464 deg Parallel of branches = 0.11762 + j0.15389 Add source Z 0.00438 + j0.01751 Total = 0.12200 +j 0.17140, R/X = 1.4049 X/RADJ = 2.5 * (X/R0.4) = 2.5*1.4049 = 3.5123. This compares with X/R = 3.508 for the 50 Hz system. The X/R = 3.51 is used to calculate the peak current from iPEAK = 2 Ik (1.02 + 0.98 e

-3(R/X)). iPEAK = 2*32.333*(1.02 + 0.98e

3/3.512)) = 65.71 kA IEC-60909 calculated 65.84 using a rounded value for(1.02 + 0.98e3/3.512))=1.44 instead of 1.437.

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 = 0.12208 + j0.42825 Complex, 0.12200 +j 0.17140 using 0.4*X

C3 = 0.33875 + j0.10875 0.33875+j 0.0435 C4 = 1.15750 + j0.92813 1.15750 + j0.37125

Total = 1.6183 + j 1.4651 1.6183 + j 0.58615 2.1830@ -42.1560 X/R = 0.3622


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Fault current Ik is 6.595 kA @ -42.1560, X/R = 0.9053. (VPREFAULT) = (380/400V) = 0.95 PU V IPU400 =1/ 2.183 = 0.4581 Ik= c *(VPREFAULT)(IPU)(MVACOMMON BASE)/(1.732*kV) Ik= 1.05*0.95 *0.4581*10 /(1.732*0.4) = 6.595 kA sym @-42.1560, X/R = 0.9053.

X/RADJ = 2.5 * (X/R0.4) = 2.5*0.3622 = 0.9055. This compares with X/R = 0.9053 for the 50 Hz system. The X/R = 0.9055 is used to calculate the peak current from iPEAK = 2 Ik (1.02 + 0.98 e

-3(R/X)). iPEAK = 2*6.595*(1.02 + 0.98e

3/0.9055)) = 9.846 kA IEC-60909 calculated 9.89 using a rounded values Ik and (1.02 + 0.98 e-3(R/X)). 3.0 IEC 909 Example 1 / EDSA Analysis

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.

Step 2. Select Open.

Step 1. Select File.

Step 3. Select IEC1.EDS.

Step 4. Select Open.


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3.2 Next, proceed to invoke the short circuit program, as indicated in the above screen capture.

Step 1. Click here to invoke the Short Circuit program.


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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.

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.

Step 6 Select OK.


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3.4 Finally, run the analysis by following the procedure shown in the screen-capture above.

Step 1 Select Update Answer File

Step 2 Select Faults at all busses

Step 3 Select OK.


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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.


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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 (pp109-119)

Location Result Type Program Value Hand Calc

Value Variance with

Hand Calcs IEC 909 Example

i peak 67597 67426 0.25% 67420 X/R @ cyc 3.645 3.645 0% F1

Ik 32898 32810 0.26% 32810 i peak 65883 65710 0.26% 65840

X/R @ cyc 3.507 3.51 0.08% F2 Ik 32418 32333 0.26% 32330

i peak 9871 9846 0.25% 9890 X/R @ cyc 0.905 0.9053 0.03% F3

Ik 6612 6595 0.257% 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

M2, M3, M41 MW , U = 6k Vpf=0.83 , e ff = 0 .9 4 I / I = 5 .5 , 2 p o le X/R = 1 0

M

LR M

M15 MW , U = 6k Vpf=0.86 , e ff = 0 .9 7 I / I = 4 , 4 po le X/R = 1 0

M

M

M1

LR

M2

U n = 6 k V

T1 15 MVA33/6 .3 k V 15% X/R = 2 5

C 10.485 + j 0.485 ohm

M4

M2, M3, M40.1920 + j1.9200 1st Cy (IEC)0.2001 + j2.0006 Int (IEC)

C20.00445 + j0.00445

0 .00146 + j0.01459 750 MVA(cq = 1 .1 )X/R = 1 0

M10.0374 + j 0.3742 1st Cy (IEC)0.0489 + j 0.4891 Int (IEC)

M4M3

C 20.485 + j 0 .485 ohm

3 3 k V

T2 15 MVA33/6 .3 k V 15% X/R = 2 5

F1

M1 M2 M3

C 10.00445 + j0 .00445

T10 .0040 + j0.0999

6 k V

3 3 k V

T20 .0040 + j0 .0999

F1

Figure 2. Single line diagram and equivalent impedance diagram for IEC Example 2.


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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) 10.0 0.995 ZM < 500 kW/pole (670 HP/pole) 6.67 0.989 ZM Grouped Low Voltage Motors 2.38 0.922 ZM

ZM = 100* IRATED/ILOCKED ROTOR IEC Motor X/R Ratios to be Used

Example 2 - Impedance Calculations Source = MVABASE * c/MVASOURCE = 10 * 1.1/750 = 0.01467 PU @ X/R = 10 = 0.00146 + j 0.01459 PU

Cable C1 & C2:Ohms (MVABASE)/kV2 = (0.485 +j0.485)10/332

= 0.00445 +j0.00445 PU Transformer T1 & T2: 25 MVA, X=15%, R = 0.6%

MVABASE(%Z)/(MVATRAN * 100) = 0.1 PU @ X/R = 15/0.6 = 25 = 0.004 + j0.0999 PU

Equipment Data X/R Base kV PU R PU X

Source 750 MVA 10 33 0.00146 0.01459 Cable C1 0.485 + j0.485 1 33 0.00445 0.00445 Cable C2 0.485 + j0.485 1 33 0.00445 0.00445 Transf. T1 15 MVA,15%Z 25 6.3 0.0040 0.0999 Transf T2 15 MVA,15%Z 25 6.3 0.0040 0.0999

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 = 5.994 MVA

1st Cy Imp.= MVABASE*kVMOT 2/(kVBASE

2*MVAMOT*ILR /IMOT )

= 10*6*6/(6.3*6.3*5.994/4)= 0.3760 PU

= 0.0374 +j 0.3742 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 = 1.282 MVA

1st Cy Imp.=MVABASE*kVMOT 2/(kVBASE

2*MVAMOT*ILR /IMOT )

=10*6*6/(6.3*6.3*1.282/5.5)=1.2864 = 0.1280 +j 1.2800 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


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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.

: = 0.84 + 0.26 g-0.26/XkG for tMIN= 0.02 seconds, ZkG = IkG/IrG # : = 0.71 + 0.51 g-0.30/XkG for tMIN= 0.05 seconds : = 0.62 + 0.72 g-0.32/XkG for tMIN= 0.10 seconds : = 0.56 + 0.94 g-0.38/XkG for tMIN=> 0.25 seconds, (not shown) # for a fault on generator terminals ZkG = c/[KG(ZG)]

q = 1.03 + 0.12 ln [m] for minimum time = 0.02 seconds q = 0.79 + 0.12 ln [m] for minimum time = 0.05 seconds q = 0.57 + 0.12 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 M4

5.0 1.0 1.0 1.0

6.0 6.0 6.0 6.0

1500 3000 3000 3000

2 1 1 1

4.0 5.5 5.5 5.5

5.994 1.282 1.282 1.282

2.5 1.0 1.0 1.0

10 10 10 10

0.0374 0.12800 0.12800 0.12800

0.3742

1.2800 1.2800 1.2800

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 M4

5.0 1.0 1.0 1.0

0.0374 0.12800 0.12800 0.12800

0.3740 1.2800 1.2800 1.2800

0.796 0.724 0.724 0.724

0.680 0.570 0.570 0.570

0.0691 0.3102 0.3102 0.3102

0.6910 3.1017 3.1017 3.1017

Motor Impedance for 1st 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.


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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 = 0.0040 + j0.0999 Cable C1 or C2 = 0.00445 + j0.00445 Total = 0.00845 + 0.10435 Parallel transf and cables = 0.004225 +j 0.052175 Source impedance = 0.001459 +j 0.01459 Total = 0.00569 + j 0.06676 = 0.0670 PU 85.1280 on 10 MVA, 6.3 kV base. Ik of non-ac decay = 1.1*10 /0.067//3/6.3 =15.046 kA at X/R = 11.71 Since the bus is operated at 6.0 kV, Ik = 15.046*6.0/6.3 = 14.33 = 1.219 + j14.278 kA Next, adding the decaying motor sources at 6.0 kV gives M1 = 1.1*10(6.0/6.3)/(0.0374 +j 0.3742)//3/6.3 =2.553 = 0.254 + j 2.54 kA (X/R = 10)

M2 = 1.1*10(6.0/6.3)/(0.128 +j1.28)//3/6.3 = 0.748 = 0.0745 + j0.745kA (X/R =10)

M3 and M4 are the same as M2.

Total Sym. kA = 1.219 + 0.254 +3*0.0745 +j(14.27 + 2.54+ 3*0.745) = 19.13 kA at X/R =11.2.

The peak currents are added for each contribution separately. Using the equation for peak current Ik PEAK =Ik *[1.02 + 0.98 , -3/(X/R)]*/2 Transformer Source(X/R = 11.71) = 14.33 *2.515 = 36.04 kA = 3.0606 +j 35.910 kA peak Motor M1 (X/R = 10) = 2.553 *2.469 = 6.3 kA = 0.6269 + j 6.2685 kA peak Motor M1, M2, M3 (X/R = 10) = 3*0.748* 2.469 = 5.540kA = 0.5512 + j 5.5123 kA peak IbASYM = 4.2387 +j 47.6908 = 47.88 kA (Value in IEC Standard = 47.87 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 non-decaying ac source current IbASYM = (Ik

2 + IDC2), where IDC =Ik */2,

_(2B f t /(X/R)) Transformers Ik = 1.219 + j 14.273 kA = 14.33 kA

IDC=Ik*/2,_(2B f t /(X/R)) = (1.219+ j 14.273)/2,_(2B 50*0.1 /11.71) = 0.1178 +j1.38 kA = 1.385

IbASYM = [1.3852 + 14.332]1/2 = 14.397 kA

M1 = 1.1*10 (6.0/6.3)/(0.0691 +j 0.6910)//3/6.3 = 1.396 kA at X/R = 10

(due to rounding of : and q Standard calculated 1.38 kA)

Ik = 0.1389 + j 1.389 kA

IDC =(0.254 + j 2.54)/2,_(2B 50*0.1 /10) = 0.0155 +j 0.1552 kA = 0.1560 kA

IbASYM = [0.1562 + 1.3962]1/2 = 1.405 kA

M2 = 1.1*10 (6.0/6.3)/(0.3102 +j 3.1017)//3/6.3 = 0.3111 kA at X/R = 10 each motor

(due to rounding of : and q Standard calculated 0.307 kA)


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Ik = 0.0308 + j 0.308 kA

IDC =(0.0745 + j 0.745kA)/2,_(2B 50*0.1 /10) = 0.00455+j 0.0455 kA = 0.0457 kA

IbASYM = [0.04572 + 0.31112]1/2 = 0.313 kA

Total ISYM of transformer, M1, M2, M3, and M4 currents are:

Transf. = 1.219 + j 14.273 kA

M1 = 0.1389 + j 1.389 kA

M2(3) =0.0924 + j 0.924

Total = 1.4503+j 16.586 = 16.649 kA

Total IDC of transformer, M1, M2, M3, and M4 currents are:

Transf. = 0.1178 +j1.38 kA

M1 = 0.0155 +j 0.1552 kA

M2(3) =0.01365+j 0.1365

Total = 1.4695+j 1.6717 = 1.678 kA

(The Standard did a scale addition on the dc magnitude and left off the /2 in the IbASYM calcs.)

IbASYM = [1.6782 + 16.6492]1/2 = 16.73 kA (Standard gives 14.32 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/RAJD being used for both the peak and dc component. Total Sym. kA = 19.13 kA at X/R =11.2. iPEAK = 47.91 kA, X/RAJD= 11.265 IbASYM = 16.73 kA Comments on Calculation procedure The solution for Example 2 followed the procedure given in IEC-60909. To me it has several questionable items. 1. Why isnt method A, B, or C used in this example. It appears to present a 4th 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


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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-60909 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.


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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.


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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.


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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 (pp127-131)

Location Result Type Program Value Hand Calc

Value Variance with

Hand Calcs IEC 909 Example

i peak 47812 47880 0.14% X/R @ cyc 11.26 11.2 0.53%

Ik 19092 19130 0.2% 19120 F1

6 cyc Break 16702 16730 0.167% 16650 All variances with Hand Calculations and IEC 909 documents are attributed to round off on input data or results.

iec 909 short circuit analysis

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