transformer
For a 3 phase open delta transformers:
When you are computing the short circuit current between the 208v high leg (point B) and neutral. Do you only use one 75kVA or do you add the other transformer to produce 75kVAx1.5 (or other value) = 112.5kVA?
Im thinking if you need to add because between point B and neutral, there are more than 1 winding (or 1 transformer) involved. What do you think?
The following is the nameplate of the transformers and the lever was set to 2nd:
I've never seen a seven-winding transformer but I have dealt with three-winding transformers.
The nameplate information for one particular three-winding transformer is as per below. (Graphic redrawn from photo to avoid showing the full extent of the very large nameplate.)
The impedances given are all on 60 MVA base. In this case tap no. 5 is the principal tapping, so consider the impedances as HV-LV 19.32%, HV-TV 33.08%, LV-TV 9.93%.
The impedance quoted for each pair of windings is with a voltage applied to the first winding, the second winding shorted, and all other windings open circuit. Refer IEC 60076.1:2005 Power Transformers - Part 1: General:
Therefore, assuming --
-- your fault current depends only on the impedance from the HV winding to the particular winding under fault.
In this example, a three-phase fault on the 33 kV winding would produce a maximum three-phase fault current of 60 MVA ÷ 19.32% ÷ 33000 V ÷ √3 = 5.43 kA. A three phase fault on the 11kV winding would produce 60 MVA ÷ 33.08% ÷ 11000 V ÷ √3 = 9.51 kA.
1) This is a typical delta to wye configuration. I might add that normally these 3 transformers would be built as one chunk of steel with 3 sections, one for each phase. The floating Vn wire is just to show that the primary is delta only, neutral is not used.
2) Because this is 3-phase the difference in phases is 120 degrees. The 30 degree phase shift is the same in each transformer so X1/X2/X3 are also 120 degrees out of phase.
3) A typical application for this would be 480vac main feed from a service entrance panel to risers (buss-bars) going to each floor of a 5 story building. Each floor has its own 3-phase step-down transformer to convert 480vac delta to 120vac wye (208vac delta) to power the outlets and lights on each floor from a 3-phase 42 breaker panel. There is plenty of heavy machinery that uses 208 single or 3-phase for power such as machine shops, ultrasonic welders, etc. This is typical of what I did as an electrician so it is easy to write about.
4) The common neutral (shown)for X1/X2/X3 is grounded at its sub-panel, and though it is not shown a ground runs back to the main service entrance panel. In large buildings it is mostly a distribution panel for all the sub-panels on each floor.
5) If Va/Vb/Vc is 480 vac delta then the phase to neutral voltage is 277vac, typically used for lighting and HVAC blower motors and heaters. It is not unusual to have a double-neutral feed to such a building, to account for imbalances in the loading of the phases.
EDIT: For more details about the 30 degree phase shift between phase windings, please see this article:Why does a delta/wye transformer make 30 degrees phase shift ? and this article:What are the Pros/Cons on the 30 Degree voltage phase shift between the primary and secondary of a Delta-Wye Transformer
Best Answer
To calculate the current in a short circuit between point B and the center tap of the A to C winding I would first calculate the winding impedance. The full 240 volt winding is rated 75 kVA. The impedance of a load that would draw 75 kVA is 240^2/75000 = 0.768 ohms. The transformer impedance is 2.59% or 0.0259 x 0.768 = 0.0199 ohms. The total impedance of a 120 V winding in series with a 240 V winding is 1.5 x 0.0199 = 0.0298 ohms. The short circuit current would then be 208/.0298 = 6971 amps.
I used the percent impedance to calculate impedance in ohms because the percent impedance is based on 75 kVA at 240 volts and the combination of windings results in 208 volts. It seems easier and more clear to used a fixed ohmic value when dealing with one single-phase transformer and part of another connected to three-phase power.
I believe the equivalent circuit is a shown below. Each transformer or part of a transformer is represented as an ideal voltage source and an impedance. The impedance of the 3-phase source is assumed to be zero.
Transformer resistive and reactive components
The transformer nameplate states the impedance as a percentage. Knowing only the percentage does not allow the resistive and reactive components, R and X, to be determined. Therefore the result of the above calculation is the combined resistance and reactance, Z. Often nameplates state X/R as well as percentage. That allows Z to be calculated when necessary. Another alternative is to find a listing of typical X/R for various transformer sizes. At this kVA level, X/R is likely more than 5 and therefore the impedance can be considered to be mostly inductive reactance.
If any added external impedance is mostly inductive, it can simply be added. If it is mostly resistive, it can be added using the square root of the sum of squares. In the case of adding the impedance of half of an additional winding, X/R is assumed to be the same for both components and can be simply added.