Short Circuit Calculations for 2500A LV Switchgear Panel

If it is energised you can do it.Measure the current(say 200A) and voltage(400A).Switch off the load and read the voltmeter(402V) quickly before the AVR restore the voltage. Now calculate as under: For a drop of 2V current required is 200A Then for a drop of 400V(SC conditions) current required=200x400/2=40000A=40kA(SC)

My method given under item 1-is it not accurate?.

Thanks,,It is useful.....

i have calculated as below

FLA = 1500(kva)X1000 / 1.732X415(volt)X0.9 =2086A

Z=5%..... 100/5 = 20

KA = 2086X20

41720 Amps

Is it ok? actually I was searching a good format to send it one of my client.

Your switchgear panel is in word format???...

Why did client need it?...Normally consultants need it........short circuit calculation is not for a panel...it is for the system...in this case up to this panel....where are the other details such as cable, transformer, distance, ...etc

Poor employer!!...Poor Client....but you enjoy life

no thanks i dont want any help from u....cheers

http://www.east-med.schneider-electric.com/documents/electrical-distribution/en/shared/AMTED300014EN.pdf

http://www.bussmann.com/library/docs/spd02/SPDSection07.pdf

check the above links for formulas n notes. search cr4 for more info...

here are some examples.

1)The Transformer which is feeding the Panel is source of Short Circuit Current. To be in safer side the fault level is calculated by following formulae

Fault MVA = Base MVA / per unit impedance

Base MVA = Rating of Transformer feeding the Panel

Per unit impedance = Impedance of the transformer in % / 100

Note: All the impedance down line are ignored thereby adding to more safety.

Example :

Suppose the Transformer rating is 33/0.415 kV, 2000 kVA with Z = 7%

Fault MVA = 2/0.07 = 28.57 MVA

Fault Current = 28.57 / (1.732*0.415) = 40 kA (approx)

2)Short Circuit current is calculated depending on the percentage Impedance and line voltage. Line voltage( V _L ) is multiplied by 1.1 to take carte of the 10 % rise in voltage and Z .

As an example , In 3 phase 1000KVA Transformer ( single ), 22/ 0.415 kV,

the % Z = 5.25 and

Short Circuit Current

...... _{1.1 x V L}

I = -----------

......... ^{√ 3 x Z}

The Impedance Z should be calculated in ohms

........→ _{% Z x 10 x kV}²

Z = ----------------- say in this case 0.009 (calculated)

............→ ^kVA

..................................→ _{1.1 x 415}

Short circuit current I = ----------------- = 29.3 kA

. ................................^{√3 x 0.009}

The 0.8 given is a derating factor depending on the heat dispensing of the enclosure considering the panel is installed in indoor in well-ventilated room.

also check this link out.......

http://cr4.globalspec.com/thread/56152/Busbar-Design

Only by paying for it.

There is no facility in CR4 to send Word files electronically.

then I know how to make it in word

Lot of waste of space in your office!!!!!!

hello masseera...:P

Masseera di jai ho...........