How to determine the short-circuit current across the terminals of a transformer?

1.Devide the Transformer MVA bu per unit impedance,per unit impedance is Z%/100.Then you will get short circuit MVA.

2.From short circuit MVA find out the short circuit current by diving with the transformer secondary voltage

Eg.

A1500 kVA,410 V transformer has5.75% impedance (Z). Determine the fault level, which could be produced by the transformer. Answer is1500 kVA ÷0.0575 =26086 kVA, or26 MVA. Fault current is26086 ÷ (415 ×1.73) =36 kA 

Dear All,

Inoder to calculate the short circuit current of a transformer, definitely we must know the Impedance of the transformer(%Z) which we can get from the name plate of the transformer. Different rated transformers has different %Z. Even same rated transformer but different manufacturers has different %Z.

Now in order to find out the short circuit current, follow the below steps

STEP-1: First find out the Fault(Short Circuit) MVA, from this we can easily derive short circuit current

MVASC =  (TRF MVA) / (% Z OF TRF)

STEP-2: Now as usual divide the short circuit MVA by the sytem voltage

I sc = MVASC  / (1.732*V)

Example will be attached as a image.

First to decide short circuit at which side? LV or MV ?

In general condition people do calculate the short circuit current at LV side cause it has a higher (KAmp) value.

To measure this first you should know the PER UNI IMPEDANCE of the transformer (its around0.05  -0.06 ), and then just calculate the full load current (FULL LOAD CURRENT = Transformer VA / ( SECONDARY VOLTAGE X1.732)).

FAULT CURRENT ACROSS THE SECONDARY TERMINAL will be = FULL LOAD CURRENT / PER UNIT IMPEDANCE.

remember this is not the SYSTEM FAULT CURRENT.

Short circuit current test is carried out to determine copper losses in transformer.Lets start with L,V. side is short circuited and now with the help of variac (variable a.c.voltage) applied voltage at rated frequeccy  is gradually increased untill the rated current in the primary &secmeterondary side reached.With the help of ampmeter,voltmeter & wattmeters thats are conneted in the H.V.Side apm,volt & watt readings can easily be found to determine copper losses in transformer.

By applying ohms law.

Using infinite bus method we can get the short ciruit  Isc by getting the Full Load Current (FLA) of Transformer over the % Impedance

Ex. Determine the available fault current at the secondary side of transformer?

Transformer Rating  =2000KVA

Voltage  =13.8kV /400V,3 Phase

% Impedance  =5.75%

FLA =2000 /1.73 x400 =2890Amps

Isc =2890 /0.0575 =50,260 Amps

We can say that the available short circuit/fault current on the secondary side is50,260Amperes, hence the size of kaic rating at secondary side is50kAIC, not included the short circuit capacity of the motors loads.

3 Phase,1500 KVA,410 Volt Trafo, Full load current =1500000/root3/410 =2112 Ampere at5.75 % Z.

So Short Ckt current should be =2112 X100/5.75 =36.3 KA.

THE SIMPLE METHODE IS( MVA/Z pu)

then I short circiut= MVApu/VOLT s

where Zpu-is impedanceof transformer in per unit

VOLTs is volt in secondery side of transformer

I think the polite answered is Z kapp or kapp impedance...

La Zcc=Ucc / Icc if we know Ucc and Icc from an essay we can obtain Zcc so we can deduce that the Icc depends directly from Zcc of eacg transformer...

Thank you

using infinite bus method., Get the secondary current  I from the transformer dividing by the % transformer Impedance..

I s.c =kva / (Ztr *V )