transformer fault current calculations
transformer fault current calculations
(OP)
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
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






RE: transformer fault current calculations
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.
RE: transformer fault current calculations
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://ww
RE: transformer fault current calculations
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.
RE: transformer fault current calculations
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 www.arcadvisor.com
RE: transformer fault current calculations
RE: transformer fault current calculations
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
RE: transformer fault current calculations
However, make sure that if you have any motor loads (say 20 HP and over) that you include their fault contribution.
RE: transformer fault current calculations