IEC-909-SHORT-CIRCUIT-ANALYSIS.pdf

EDSA IEC 909 SHORT CIRCUIT ANALYSIS 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 cir cuit curr ent calcul ation in thr ee-phase ee-phase AC systems systems ”. ”. These examples are: Example 1:

“Calculation of short-circuit currents in a low-voltage system”. system ”. Appendix “A”, section 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”. motors” . Appendix “A”, section 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, 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 = 10

0.00438 + j0.04378

15 kV

T1 630 kVA 15/0.4kV 4% P LOAD = 6.5

15 kV

0.1638 + j0.6138

T2 400 kVA 15/0.4 kV 4% PLOAD = 4.6

F1

F1

C1 0.385 + j 0.395

F2

0.0260 + j0.0085

0.02406 +j0.02469

C2 0.416 + j0.136

F2

Un = 380 V

Vpu = 0.95 0.33875 + j0.10875

C3 5.420 + j 1.740 C4 18.52 + j14.85 F3

Figure 1.

0.2875 + j0.957

1.1575 + j0.92813 F3

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

Page 1

.

EDSA IEC 909 SHORT CIRCUIT ANALYSIS

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. 2

C1 – 2-240mm , 10 meter length and 0.077 + j 0.079 ohms/meter/ph = 0.77 + j 0.79 milliohms. 2 C2 – 2-150mm , 4 meter length and 0.208 + j 0.068 ohms/meter/ph = 0.832 + j 0.272 milliohms. 2 C3 – 2-70mm , 20 meter length and 0.271 + j 0.087 ohms/meter/ph = 5.420 + j 1.740 milliohms. 2 C4 – 2-50mm , 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 <35 kV, therefore the X/R = 10 per IEC-60909 ( X = 0.995* Z, R = 0.1 X ).

Z COMMON BASE =

c ( Z EQUIPMENT BASE) (MVACOMMON BASE) ______________________________________________ MVAEQUIPMENT BASE Z QT= 1.1 * 1.0 * 10/250 = 0.044 PU RMULTIPLIER = 0.00985, RPU = 0.00438 X MULTIPLIER = 0.09851, X PU = 0.04378

Transformers using equation below

Z COMMON BASE =

( Z EQUIPMENT BASE) (MVA COMMON BASE) ___________________________________________ MVA EQUIPMENT BASE

630 kVA transformer T1, i mpedances on 630 kVA base is: RPU = PLOSS/KVARATED = 6.5/630 = 0.010 2 2 ½ 2 2 ½ X PU = ( Z - R ) = (.04 - 0.0103 ) = 0.0387, X PU/ RPU = 3.76 400 kVA transformer T2, impedance on a 400-kVA base is: RPU = PLOSS/KVARATED = 4.6/400 = 0.0115 2 2 ½ 2 2 ½ X PU = ( Z - R ) = (.04 - 0.0115 ) = 0.0383, X PU/ RPU = 3.33 Convert to 10 MVA base Z PU = 10 * (0.0103 +j0.0387)/0.63 = 0.16375 + j0.61375 for 630 kVA Tx Z PU = 10 * (0.0115 + j0.0383)/0.4 = 0.2875 + j0.9575 for 400 kVA Tx Cables are converted from ohms to per-unit on 10 MVA and 0.4-kV bases 2 Z PU = Z OHM * (MVACOMMON BASE)/kV BASE

C1=0.385 + j0.395 mohms = 0.02406 + j0.02469 PU C2=0.416 + j0.136 mohms = 0.02600 + j0.00850 PU C3=5.420 + j1.740 mohms = 0.33875 + j0.10875 PU C4=18.52 + 14.85 mohms = 1.1575 + j0.92813 PU These per unit values are shown on Figure 1.

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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 I ”k= c *(VPREFAULT )(IPU)(MVACOMMON BASE)/(1.732*kV) I ”k= 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 i PEAK = iPEAK =

√2*32.81*(1.02 + 0.98 e –3/3.67)) = 67.426 kA

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√2 I ”k (1.02 + 0.98 e-3(R/X)).


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 I ”k= c *(VPREFAULT )(IPU)(MVACOMMON BASE)/(1.732*kV) I ”k= 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 i PEAK =

√2 I ”k (1.02 + 0.98 e-3(R/X)).

iPEAK = √2*32.333*(1.02 + 0.98e ) = 65.71 kA –3/3.512) IEC-60909 calculated 65.84 using a rounded value for(1.02 + 0.98e )=1.44 instead of 1.437. –3/3.512)

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, C3 = 0.33875 + j0.10875 C4 = 1.15750 + j0.92813 Total = 1.6183 + j 1.4651 2.1830@ -42.1560

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0.12200 +j 0.17140 using 0.4*X 0.33875+j 0.0435 1.15750 + j0.37125 1.6183 + j 0.58615 X/R = 0.3622


Fault current I ”k is 6.595 kA @ -42.1560, X/R = 0.9053. (VPREFAULT ) = (380/400V) = 0.95 PU V IPU400 =1/ 2.183 = 0.4581 I ”k= c *(VPREFAULT )(IPU)(MVACOMMON BASE)/(1.732*kV) 0

I ”k= 1.05*0.95 *0.4581*10 /(1.732*0.4) = 6.595 kA sym @-42.156 , 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 i PEAK = iPEAK =

√2 I ”k (1.02 + 0.98 e-3(R/X)).

√2*6.595*(1.02 + 0.98e –3/0.9055)) = 9.846 kA )

IEC-60909 calculated 9.89 using a rounded values I ”k and (1.02 + 0.98 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.

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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.

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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.

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.

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Step 1 Select “Update Answer File”

Step 2 Select “Faults at all busses”

3.4


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

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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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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

Program Value 67597 3.645 32898 65883 3.507 32418 9871 0.905 6612

Result Type

i peak X/R @ ½ cyc I”k i peak X/R @ ½ cyc I”k i peak X/R @ ½ cyc I”k

F1

F2

F3

Hand Calc Value 67426 3.645 32810 65710 3.51 32333 9846 0.9053 6595

Variance with Hand Calcs 0.25% 0% 0.26% 0.26% 0.08% 0.26% 0.25% 0.03% 0.257%

IEC 909 Example 67420

32810 65840 32330 9890 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 = 10

0.00146 + j0.01459

33 kV

33 kV

C2 0.485 + j 0.485 o hm

C1 0.485 + j 0.485 ohm

T1 15 MVA 33/6.3 kV 15% X/R = 25

M1

M2

M1 5 M W , UM = 6 k V pf=0.86 , eff = 0 .97 I LR/ I M= 4, 4 pole X/R = 10

Figure 2.

T1 0.0040 + j0.0999

T2 15 MVA 33/6.3 kV 15% X/R = 25

Un = 6 kV

M3

T2 0.0040 + j0.099

6kV

F1

M4

M2, M3, M4 1 M W , UM = 6 k V pf=0.83, eff = 0.94 I LR/ I M= 5.5, 2 pole X/R = 10

C2 0.00445 + j0.00445

C1 0.00445 + j0.0044 5

M1

M2

M1 0.0374 + j 0.3742 1st Cy (IEC) 0.0489 + j 0.4891 Int (IEC)

M3

M4

M2, M3, M4 0.1920 + j1.9200 1st Cy (IEC) 0.2001 + j2.0006 Int (IEC)

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

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F1


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 >= 500 kW/pole (670 HP/pole) < 500 kW/pole (670 HP/pole) Grouped Low Voltage Motors

Reactance 0.995 Z M 0.989 Z M 0.922 Z M

X/ R

10.0 6.67 2.38

Z M = 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 2

2

Cable C1 & C2:Ohms (MVABASE)/kV = (0.485 +j0.485)10/33 = 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 Source Cable C1 Cable C2 Transf. T1 Transf T2

Data X/ R Base kV 750 MVA 10 33 0.485 + j0.485 1 33 0.485 + j0.485 1 33 15 MVA,15% Z 25 6.3 15 MVA,15% Z 25 6.3 Data used in Short Circuit Calculations

PU R 0.00146 0.00445 0.00445 0.0040 0.0040

PU X 0.01459 0.00445 0.00445 0.0999 0.0999

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 2

2

1st Cy Imp.=MVABASE*kVMOT /(kVBASE *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 γ -0.26/X”kG for tMIN= 0.02 seconds, Z ”kG = I ”kG/IrG # : = 0.71 + 0.51 γ -0.30/X”kG for tMIN= 0.05 seconds : = 0.62 + 0.72 γ -0.32/X”kG for tMIN= 0.10 seconds : = 0.56 + 0.94 γ -0.38/X”kG for tMIN=> 0.25 seconds, (not shown) # for a fault on generator terminals Z ”kG = c/[K G( Z G)] 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:

: = 0.796, q = 0.680, :q = 0.541, Impedance multiplier = 1/:q = 1.85 : = 0.724, q = 0.570, :q = 0.413, Impedance multiplier = 1/:q = 2.42

Motor M1: Motor M2:

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

X ”

PU

Resistance

Reactance

M1

5.0

6.0

1500

2

4.0

5.994

2.5

10

0.0374

0.3742

M2

1.0

6.0

3000

1

5.5

1.282

1.0

10

0.12800

1.2800

M3

1.0

6.0

3000

1

5.5

1.282

1.0

10

0.12800

1.2800

M4

1.0

6.0

3000

1

5.5

1.282

1.0

10

0.12800

1.2800

First Cycle Per-unit Motor Impedances on a 10-MVA base

Motor MVA

M1 M2 M3 M4

5.0 1.0 1.0 1.0

IEC First Cycle PU Resist. PU React.

0.0374 0.12800 0.12800 0.12800

0.3740 1.2800 1.2800 1.2800

IEC Breaking Time (0.10 seconds) : Multiplier

0.796 0.724 0.724 0.724

q Multiplier

0.680 0.570 0.570 0.570

PU Resist.

0.0691 0.3102 0.3102 0.3102

PU React.

0.6910 3.1017 3.1017 3.1017

st

Motor Impedance for 1 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 0 Total = 0.00569 + j 0.06676 = 0.0670 PU 85.128 on 10 MVA, 6.3 kV base. I ”k 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, I ”k = 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 I ”k PEAK -3/( X/R) = I ”k *[1.02 + 0.98 , ]*/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 I bASYM = 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 nondecaying ac source current _ (2 f t /( X/R)) 2 2 ½ I bASYM = ( I”k + I DC ) , where I DC = I ”k */2, Β

Transformers

I”k = 1.219 + j 14.273 kA = 14.33 kA

_ (2Β f t /(X/R))

I DC= I ”k */2,

2

_ (2 50*0.1 /11.71) = (1.219+ j 14.273)/2, Β = 0.1178 +j1.38 kA = 1.385

2 1/2

I bASYM = [1.385 + 14.33 ]

= 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) I k = 0.1389 + j 1.389 kA _ (2 50*0.1 /10) I DC =(0.254 + j 2.54)/2, Β = 0.0155 +j 0.1552 kA = 0.1560 kA

2

2 1/2

I bASYM = [0.156 + 1.396 ]

= 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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I k = 0.0308 + j 0.308 kA _ (2 50*0.1 /10) I DC =(0.0745 + j 0.745kA)/2, Β = 0.00455+j 0.0455 kA = 0.0457 kA

2

2 1/2

I bASYM = [0.0457 + 0.3111 ]

= 0.313 kA

Total I SYM 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 I DC 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 I bASYM calcs.) 2

2 1/2

I bASYM = [1.678 + 16.649 ]

= 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. i PEAK = 47.91 kA, X/ R AJD= 11.265 I bASYM = 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 isn’t 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 i n 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 i nfluence 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 interf ace.

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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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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

F1

Result Type

i peak X/R @ ½ cyc I”k 6 cyc Break

Program Value 47812 11.26 19092 16702

Hand Calc Value 47880 11.2 19130 16730

Variance with Hand Calcs 0.14% 0.53% 0.2% 0.167%

IEC 909 Example

19120 16650

All variances with Hand Calculations and IEC 909 documents are attributed to round off on input data or results.

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IEC-909-SHORT-CIRCUIT-ANALYSIS.pdf

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