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An Overview Of Short Circuit Current (part 3)

An Overview Of Short Circuit Current (part 3)

Continued from previous technical article: An Overview Of Short Circuit Current (part 2)

Sample calculation for small LT systemFault calculations are carried out to find the magnitude of fault current at various voltage levels of electrical system.

Short circuit calculations are actually just an elaborate version of Ohms Law. One of the key components in thecalculation process is to determine the total impedance of the circuit from the utility / source, through the transmissionsystem, transformers, and conductors, down to the point in question such as a panel or switchboard location. Theimpedances of the various circuit elements have both resistance and reactance and are often referred to as thecomplex impedance or polar notation.

Fault current values and time helps in deciding equipment short time withstand capacity and deriving settings ofprotection relays. Interrupting capacity of protection equipment should be high enough to open safely the maximumshort circuit current which the power system can cause to flow through that equipment.

One sample calculation for calculating the short circuit current at downstream of transformer is shown below.

Purpose and intent of this calculation is to calculate the short term current rating of a marshalling kiosk to be fed

by AC distribution board (ACDB). ACDB being fed by a source of 630kVA transformer .

Calculation basis1/ The busbar and switchgear of Marshalling kiosk is sized for short time rating as per contribution from MV sourcethrough LT transformer.

2/ For circuits connected by transformer PU system is particularly suitable. By selecting suitable base kV for circuitsthe per unit reactance and resistance remains same, referred to either side (HV or LV) of transformer.

3/ For circuits connected by transformer same base kVA is selected for both the circuits ( HV and LV) because powerremains constant throughout so same base kVA should be considered throughout.

4/ As a rule only two bases should be selected first and from these two the remaining bases should be calculated.This is so because kV, kVA, I and Z are interrelated. They must obey ohms law. If we select base kVA and base kVthan other base like base I and base Z are calculated from base kV and base kVA.

Vice-versa will be inconvenient that is selecting base I and Z and calculating other bases like kV and kVA will makecalculation difficult.

Input data to be collected: 1. Transformer Rating = 0.63 MVA2. Transformer Voltage ratio = 11/0.433 kV3. Frequency = 50Hz4. Transformer Impedance = 5% = 0.05 PU5. MV System fault level (Maximum) = 40 kA6. MV System fault MVA = 3 x 40 x 11 = 762 MVA

Calculation

Actual Fault Current available at AC distribution boardBase MVA = 0.63Base kV = 11Base Current in kA = Base MVA/(3 x Base kV) = 0.63/(3 x 11) = 0.033

Base Impedance = (Base kV)2 / Base MVA = 192.1

Source Impedance = MV System fault MVA / Base MVA = 0.0008LT Transformer impedance at 0.63MVA & 11kV Base = 0.05

Actual Fault Current available at marshalling kioskbusbar

Total MV system impedance (MV System + LTTransformer)

= 0.0508

Fault MVA contributed by Source through LT Transformer = Base MVA / Total Impedance= 0.63 / 0.0508 = 12.40

Fault current contribution in kiloAmpers from MV system at LV side through (Switchyard) LT Transformer:= Fault MVA x 1000 x 1000 / (3 x 0.433 x 1000 x

1000)= 12.40 x 1000 x 1000 / (3 x 0.433 x 1000 x 1000)= 17.245 kA

Actual Fault Current available at marshalling kioskbusbarBusbars and switchgear components of marshalling kiosk shall be braced for the peak value of the faultcurrentcontribution from the MV system through 630kVA rated source transformer .

Hereafter Marshalling kiosk to be referred as BMK and AC distribution board to be referred as ACDB.

Base kVA = same as above, since thisparameter remains constant throughout thecircuitBase kV = 0.415V Base kV at LV circuitDistance in meters of transformer from ACDB= 20Distance in meters of BMK from ACDB = 50Size of connecting cable in Sq mm fromtransformer to ACDB = 3.5C x 300 Sq mm Al,XLPEResistance in Ohms/kM of connectingcable from transformer to BMK = 0.128

Total resistance over route length = 200.128/1000 = 0.003PU resistance = Actual Resistance x BasekVA/ (BasekV2 x 1000)= 0.003 x 0.63 x 1000 / (0.415 x 0.415 x1000) = 0.009

Reactance in Ohms/kM of connecting cablefrom transformer to BMK = 0.0705Total reactance over route length = 0.070520/1000 = 0.001

PU reactance = Actual Reactance x BasekVA/(Base kV2x 1000)= 0.001 x 0.63 x 1000 / (0.415 x 0.415 x 1000)

= 0.0052

PU impedance of cable from LTtransformer to ACDB= ((PU resistance)2+(PU reactance)2)=(0.0092+ 0.07052) = 0.011

Size of conecting cable in Sq mm from ACDBto BMK = 3.5C x 35 Al, XLPEResistance in Ohms/kM of connecting cablefrom ACDB to BMK = 0.671Total resistance over route length = 0.671 x 50/1000 = 0.034PU resistance = Actual Resistance x BasekVA/(Base kV2 x 1000)= 0.034 x 0.63 x 1000 / (0.415 x 0.415 x1000) = 0.12

Reactance in Ohms/kM of connecting cablefrom ACDB to BMK = 0.0783Total reactance over route length = 0.0783 x50 /1000 = 0.004

PU reactance = Actual Reactance x BasekVA/(Base kV2x 1000)= 0.004 x 0.63 x 1000 / (0.415 x 0.415 x1000) = 0.14

PU impedance of cable from ACDB to BMK =((PU resistance)2+(PU reactance)2) =((0.12)2+(0.14)2) = 0.124

Total PU Impeadnce of connecting cable from LT transformer to BMK = 0.011 + 0.124 = 0.134Total PU Impedance from LT Transformer to BMK = PU Impedance of Transformer + Total PUImpedance of connecting cable from LT transformer to BMK = 0.05 + 0.134 = 0.1842

Fault MVA at BMK busbar = Base MVA/Total Impedance = 0.63 / 0.1842 = 3.42

Fault current in kiloAmps at BMK busbar = Fault MVA x 1000x 1000 / (3 x 0.415 x 1000 x 1000)= 3.42 x 1000 x 1000 / (3 x 0.415 x 1000 x 1000) = 4.757 kA

Hence selection of 10kA busbar and switchgear components like MCB is safe and appropriate as per the actualfault level existing at BMK main busbar.

Si. No Equipment CURRENT RATINGCALCULATED SHORT TERM CURRENT RATING IN kA

OPTIMUM SELECTION OF SHORT TIME CURRENT RATING IN kA

RMSSymmetrical

Assymmetricalpeak value =nxRMSSymmetrical

RMSSymmetrical

Assymmetricalpeak value =nxRMSSymmetrical

1 Main LT board 17.24 34.5 (n=2) 35 73.5 (n=2.1)2 Marshalling kiosk 4.75 7.1 (n=1.5) 10 17 (n= 1.7)

References:

1. Indian Standard 8623, part-1-SPECIFICATION FOR LOW-VOLTAGE SWITCHGEAR AND CONTROLGEARASSEMBLIES

2. Indian Standard 10118, part-2-CODE OF PRACTICE FOR THE SELECTION, INSTALLATION ANDMAINTENANCE OF SWITCHGEAR AND CONTROLGEAR

3. The Importance of the X/R Ratio in Low-Voltage Short Circuit Studies- Research paper DATE: November 17,1999 REVISION: 0 by AUTHOR: John Merrell

4. Short-circuit-current Calculating Procedures by Donald Beeman, Alan Graeme Darling, and R. H. Kaufmann5. Industrial Power Engineering and Applications Handbook by K.C. Agrawal

An Overview Of Short Circuit Current (part 3)Sample calculation for small LT systemCalculation basisCalculation

Actual Fault Current available at marshalling kioskbusbarReferences: