Single phase short circuit current calculation 1

[AggiesNM]

I have a single phase 75kVA transformer with 1.7 %Z. See attached file for oneline. Can someone please show me how to calculate the short circuit current on the secondary side (240V) of the tranformer?

Thank you for your time.

Regards.

 

[jghrist]

Treat the 75 kVA single phase transformer as a 225 kVA three phase wye-wye connected transformer with 1.7% impedance and calculate phase-to-ground current.

 

[cuky2000]

See if the simplified calcs could help.

 

[ScottyUK]

75kVA hanging off an 18MVA bar? The ratio is so large that you could reasonably treat the upstream network as an infinite bus and simplify the calc without worrying about what is upstream. That's unless this is for school and you have to show all the working.

FLC of the transformer is 312.5A.
Bolted S/C fault current is 312.5/0.017 = 18.4kA.

Upstream network impedance will reduce the result slightly but not to any extent worth noting.


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If we learn from our mistakes I'm getting a great education!

 

[AggiesNM]

Thanks for the response.

@Cuky: The mathcad implementation is pretty slick. I am assuming the 0.9 is the powerfactor.

@Scotty: Lol. Not for school.

 

[cuky2000]

AggieNM,

0.9 may be considered the voltage fluctuation @ -10% that provides the highest SC current. In other word, 216V < Vnom < 264V.

If the load of the system has low power factor, this also will be amplified accordingly.

 

[prc]

Cuky Is it +10% higher voltage?When Z remains same,current will go up at higher voltage.

 

[waross]

The available short circuit current (from which the PU impedance is derived) is measured with the transformer at normal operating temperature.
On small dry type transformers designed to run quite hot and with a fairly high X:R ratio, the actual current will be greater with a cold transformer.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[pie314]

Also if the 1.7% is mostly reactance and the short circuit occurs at a voltage zero, the the peak asymetrical current could be 18.4 X 2 X 1.41 = 51.9KA

 

[ijl]

If one wants to calculate the short circuit current with greater (apparent?) accuracy and take all impedances of the network into account, then one must be careful with the impedances and voltages. The current can be calculated either manually or using a program for the analysis of three phase networks.

When calculating manually, it may be easiest to convert all impedances to the 240V level. Note that the 240V voltage is a phase-neutral voltage, but the other voltages are phase-phase voltages. For example, the impedance of the source is about 5.1Ohms at the 115kV level. This can be converted to the 240V level as: 5.1*(240*sqrt(3)/115000)^2. The short circuit current at 240V is 16.9kA.

If a three-phase program is used, it may be simplest to think the single-phase transformer as one-third of a three-phase YY transformer, and to calculate the short circuit current for a three-phase fault. The programs generally use phase-phase voltages as inputs. This means that the single phase transformer must be entered as a 24.9kV/416V-transformer. The apparent power of the single phase transformer must also be multiplied by three. The short circuit current will again be 16.9kA.

 

[jghrist]

If a three-phase program is used, it may be simplest to think the single-phase transformer as one-third of a three-phase YY transformer, and to calculate the short circuit current for a three-phase fault.

I think you should calculate the short circuit current as a phase-to-ground fault. Only one of the phases of the assumed YY transformer is faulted.

 

[ijl]

Calculation of either three-phase or phase-ground fault is possible. Both give the same answer. But no need to think about zero or negative sequence impedances when calculating the three-phase fault current.

 

[jghrist]

Calculation of either three-phase or phase-ground fault is possible. Both give the same answer.

Both would give the same answer if the 115 kV source impedance and 24.9 kV line impedances are negligible.

 

[ijl]

Both give the same answer, when the positive, negative and zero sequence impedances are the same in each component separately. No need to be negligible. A simple network with only YY-transformers can (for practical purposes) be considered as three separate single phase networks. The networks do not interact. A three-phase fault is the same for one phase as a line-earth fault in such a network. Check with some program!

 

[AggiesNM]

IJL:

You would need a positive, negative, and zero sequence network (which connect in series) to calculate the single line to ground (SLG) fault current. Current for a three phase fault can be calculated using just a positive sequence network. Subsequently, higher impedance in a SLG network should give current magnitudes lower than say a three phase fault. I understand a three phase network is decoupled into three single phases in the symmetical component world but I dont understand how that equates to equivalent SLG and 3Ph fault currents.

Thanks.

 

[ijl]

Correct, the line-earth fault current is calculated by connecting the +,-,0 networks in series. But that calculation gives the sequence currents. The phase current is the sum of these.

 

[timm333]

I have a similar question. Let's say the available three phase short current at the three phase busbar is 50kA. So it is obvious that all the three phase breakers (L-L-L) at this busbar should be rated for at least 50kA.

Now if I want to connect a single phase 277V (L-N) breaker at one of the phases of the same busbar, what should be the rating of this single phase breaker? Should it also be at least 50KA?

 

[jghrist]

Should it also be at least 50KA?

Yes. 1Ø fault current could be more or less than 3Ø, but there's no need to complicate things.

 

[timm333]

Does anybody know the formula to convert three phase short circuit current into single phase short circuit current? In other words if three phase short circuit current (L-L-L) is 50kA, what will be single phase (L-N) short circuit current at the same point?

 

[davidbeach]

Tell me the ratio of the source Z1 to the source Z0 and I can answer your question. Oh, you don't have that value, well too bad, time to do a study to determine the SLG fault current. In other words, there is no one size fits any answer to your question.