Calculation of fault level at secondary side of transformer
Calculation of fault level at secondary side of transformer
(OP)
Hi,
I have little experience at HV side. Usually for my 800 kVA 11kV/400V 4% Ucc transformer I would calculate the fault at the LV side assuming an infinite bus at the primary, giving Isc = 800/3/230/0.04 = 29 kA
However in this particular instance we need to calculate the actual fault not this upper limit, and all we have as info is 3 phase fault level at primary is 7 kA.
How can I then calculate the fault level at the secondary, any additional info needed?
Thanks.
I have little experience at HV side. Usually for my 800 kVA 11kV/400V 4% Ucc transformer I would calculate the fault at the LV side assuming an infinite bus at the primary, giving Isc = 800/3/230/0.04 = 29 kA
However in this particular instance we need to calculate the actual fault not this upper limit, and all we have as info is 3 phase fault level at primary is 7 kA.
How can I then calculate the fault level at the secondary, any additional info needed?
Thanks.






RE: Calculation of fault level at secondary side of transformer
So, work out the upstream capability, Short circuit capability:
Ssc = (11000/ root 3)* 7000 * 3 = 133 MVA -> [phase to line voltage * current * 3 phases]
You now need to find the upstream impedance
HV = 11000 Ssc = 133 MVA
Zup = (HV^2)/Ssc = 0.907 -> the upstream impedance in ohms
Now need to convert upstream impedance to the 400 volt base:
Zup_LV = Zup * (400/11000)^2 = 0.0012
That gives an answer in ohms, so to continue we need to convert the % impedance of the transformer into an ohmic value.
Ztr = (4/100)*(400^2/800000) = 0.008 -> (percentage/100) * (Line_line voltage squared/transformer size)
So now to work out the short circuit current:
Isc = 400/[(root 3 * (0.0012 + 0.008)] = 25.1 kA -> Voltage L-L / [root 3 * (Zup_LV + Ztr)].
And I hope that all makes sense :)
RE: Calculation of fault level at secondary side of transformer
First: calculate the short circuit MVA of each piece of the system. (MVA at LV terminals if the equipment is connected to an infinite source)
Second- combine the MVA's. (MVA's combine like admittances, add when in parallel, useinvers in series).
Utility MVAsc= 11 kV x 7 kA x 1.732 = 133.4 MVA
Transformer MVA sc = MVA x 100%/%Z = 0.8 / 0.04 = 20 MVA
The two componenets are in series so we add them like admittances:
1/MVA system = 1/MVAutility + 1/MVAtx = 1/(1/133.4 + 1/20) = 1/ 17.4. MVA sc = 17.4 MVA
I sc = (17.4 MVA x 10^3) /(400V x 1.732) = 25.1 kA
RE: Calculation of fault level at secondary side of transformer
Transformer Impedance = 0.04 pu
System impedance = Transformer MVA/ System fault MVA = 0.8/133.4 = 0.006 pu
Total impedance for fault = 0.04+0.006 =0.046
Fault MVA =0.8/0.046 =17.4 MVA
Isc= 17.4 X 10000/ 1.732x400 =25.1 kA
RE: Calculation of fault level at secondary side of transformer
Once the actual tested value of impedance is available, a more accurate short circuit value can be derived.
RE: Calculation of fault level at secondary side of transformer
RE: Calculation of fault level at secondary side of transformer
RE: Calculation of fault level at secondary side of transformer
RE: Calculation of fault level at secondary side of transformer
RE: Calculation of fault level at secondary side of transformer
Note to self : check the numbers before posting
RE: Calculation of fault level at secondary side of transformer
To have that driven home can never be a bad thing, IMHO.
I must add, that is why I use the impedance method of calculation. As you progress down to the lower levels (boards) you just keep adding the impedances of the cabling, breakers, busburs etc, and get the resultant fault levels. This fault level can then be checked to see that distribution breakers aren't under-rated for the prospective fault current, but you can also check to make sure that they're able to trip in a timely manner.
As the above posts indicate, there is more then one way to skin a cat than the impedance method, but that's the one I learnt and is therefore the one I'm comfortable with.