Transformer short circuit fault calculation clarification

In summary, the transformer manufacturer provides specs for a 3 phase bolted short. If you use these specs, you can calculate the available current at the load if there is a 3 phase bolted short.
  • #1
lavalin
6
0
Hi,

I'm trying to perform a short circuit fault calculation on this transformer and load I have when a 3 phase bolted short (worse case) occurs.

The transformer specs are below:
Z(transformer)=5%, Z(cable)= .00306 + j.0027, Delta/Star 6.6kV/400V

I can understand calculating the short circuit available current at the load:
Ifault(load) = Vphase/Z(transformer)
where Z(transformer) is the impedance of one winding, and the voltage across it would be 400/1.73 ~=230V
Ifault(load) = 230/.05
= 4.6kA

Now calculating the fault down at the load is where I get a little confused:
The example I'm following states that the load current fault for a 3 phase bolted short is:
Ifault(load) = Vphase/(Z(transformer) + Z(cable))

My only guess/understanding of why they use Vphase is that Vphase represents the voltage drop from the neutral (mid point) of the WYE transformer to the end of 1 phase? I've drawn a diagram to help clarify what I mean, its hosted at:
http://img83.imageshack.us/my.php?image=61884112bb8.jpg

Can someone please explain to me why they use Vphase in the load fault calculation?
 
Engineering news on Phys.org
  • #2
If = Vph / Ztr(ohm)

Ztr(ohm) = %Z / Zb

Zb = V2/S

%Z : Transformer per unit impedance
V : Transformer Rated voltage (V)
S : Transformer Rated power (VA) :smile:


--------------------------------------
Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com
 
  • #3
Thanks for your reply. My question was however clarification on why they use phase voltage when they're calculating at the load end, total fault current that the transformer can deliver.
 
  • #4
Hi! I would say:

They use the phase voltage because this fits with the winding impedance, and the load is in star connection as well. This gives you the current in a secondary winding.

Within this logic, you could equally well compute everything with line voltages and equivalent delta connections. It would be consistent, but less direct and less usable.
 

1. What is a transformer short circuit fault?

A transformer short circuit fault occurs when there is a direct connection between the high voltage and low voltage sides of the transformer, bypassing the load. This can result in a high flow of current and can cause damage to the transformer and other equipment.

2. How is the fault current calculated in a transformer short circuit?

The fault current in a transformer short circuit is calculated by dividing the rated voltage of the transformer by the sum of the impedances on the primary and secondary sides of the transformer. This calculation takes into account the transformer's impedance as well as the impedance of any connected equipment or cables.

3. What factors can affect the fault current in a transformer short circuit?

The fault current in a transformer short circuit can be affected by factors such as the transformer's impedance, the impedance of connected equipment and cables, the source of the fault, and the available fault current from the power supply. Temperature and frequency can also impact the fault current.

4. Why is it important to calculate the fault current in a transformer short circuit?

Calculating the fault current in a transformer short circuit is important for determining the maximum amount of current that could flow through the transformer and other connected equipment. This information is crucial for selecting appropriate protective devices and ensuring the safe operation of the transformer and the electrical system as a whole.

5. Are there any limitations or assumptions in calculating the fault current in a transformer short circuit?

Yes, there are some limitations and assumptions in calculating the fault current in a transformer short circuit. These include assuming a balanced three-phase fault, neglecting the impedance of the transformer's windings, and assuming that the transformer is delivering its rated power. It is important to carefully consider these limitations and assumptions when performing fault current calculations.

Similar threads

  • Electrical Engineering
Replies
8
Views
1K
  • Electrical Engineering
Replies
4
Views
2K
  • Electrical Engineering
Replies
3
Views
2K
Replies
2
Views
945
  • Electrical Engineering
3
Replies
77
Views
5K
  • Electrical Engineering
Replies
16
Views
3K
  • Electrical Engineering
Replies
1
Views
4K
Replies
1
Views
2K
Replies
4
Views
6K
  • Electrical Engineering
Replies
11
Views
2K
Back
Top