transformer fault current calculations 1

[adapter3]

A bit embarrassed since i should know this...but....I have 1MVA transformer rated at 4160-600V z=6%(nameplate data)

What is the proper way of getting to the available fault current this transformer can generate? The units don't jive for me

available current is I=P/(1.73*E)..........lets assume pf=1

I = 1000000/(1.73*600) = 963Amps

I know if if divide by the impedance of the trabsformer I'l get the fault current avail....but the units are not correct in my thinking.....

I/z does not equal current? 963/.06 = approx 16kA

can someone set me straight

thanks

kck

 

[ahhie]

Hi,

I/z is equal to current for this case because the Z is in percentage. Transformer rated current divided by Z (percentage impedance) is the available fault current of the transformer at the HV voltage if the Z impedance reference voltage is HV.

 

[cuky2000]

Your calculation is OK (assuming infinite bus): 963Amp/0.06pu ~ 6 kA

Since 0.06 pu (per unit) is a non dimensional term, the result is the same unit as the numerator, Amp.

Check the enclose site for additional reference in this topic.

http://www.mathworks.com/access/helpdesk/help/toolbox/physmod/powersys/technic3.html#6759

 

[rbulsara]

16kA is the correct answer and so is your method.

The explanation for the units is: The actual formula is in per unit, that is
Isc in per unit=V per unit/Z per unit.

Since the transformer impedance, in % or per unit, uses its own rating as the base, V per unit=1.
Hence Isc=1/Z per unit. So actual current in amps will be
(Isc in per unit)x (Base current in amps). The base current is the rated FLA of the transformer.

Therefore the shortcut formula is Isc=FLA/Z. Where FLA is in amps and Z is in per unit and you get Isc in amps.

 

[mykh]

16kA above is the Maximum short circuit current the transformer will let through. In other words, transformer doesn't generate short circuit current in contrast to generators or motors.

Secondly, the actual fault current equals to 16kA only in case of unlimited source. In general it's lower and depends on the source short circuit level.

The nameplate data of z=6% represents the transformer 0.06 p.u. impedance for in the 1MVA base. For the calculation purpose, the transformer p.u. impedance should be converted to common base p.u. impedance using the following formula:

Xpu_trans = (Xtrans * MVAbase) / MVAtrans (1)

Source p.u impedance Xpu_source equals to MVAbase / SCMVAsystem.

Assuming equal x/r, the transfomer p.u. common base p.u. impedance found from (1) should be added to the source p.u impedance before short circuit current is calculated:

Xpu_total = Xpu_source + Xpu_trans

Total fault current at the transformer secondary terminals equals to I / Xpu_total. where I equals to MVAbase / 1.73*V

The short circuit current value can also be calculated using MVA method described and implemented on

http://www.arcadvisor.com

 

[robthecob]

Also, if you have a 100% motor load on the load side of the transformer, that will add another 3,900 amps due to motor contribution.

 

[weh3]

The definition of transformer impedance is

Percent of rated primary voltage to cause rated secondary current in a short-circuited secondary.

In your case, 250 V on the primary will cause 962 A in a short-circuited secondary. Therefore, the full primary voltage of 4160 V should cause I/Z or

962/0.06 A = 16 kA

in a short-circuited secondary.

It is a worst case until you consider the generative effects of motors and generators on your side.

William

 

[tommom]

16,040 amps would be the maximum available current at the 600 volt terminals.

However, make sure that if you have any motor loads (say 20 HP and over) that you include their fault contribution.

 

[fadihaj]

how do we size the ground ring bare conductor around the building?