impedancetransformer
I have a power transformer with the transformer ratio 200kV/100kV, 100MVA rated power, and specified R = 0.001 p.u., X = 0.05 p.u. For a calculation I have to make, the literature I am using is referring to the "PI" model of the transformer:
How do I get the series impedance (R + jX) from the plate values? Or are R and X in the image the same as the plate values?
Dunno about the HAI Electric Strike, but yes, the inrush current may be 2 or 10 or 40 times the hold current. 4.5A rather than 450mA.
Not because the coil current is greater when the plunger is outside the coil, as is the case for AC solenoids.
Rather, because the hold-in current is artificially reduced, either by using a separate hold-in coil, or by switching a current limit into the circuit. You do this to get a good pull-in (what you are not getting) but not wasting power and burning out the coil when the door is left unlocked.
A simple tranformer plug pack had no effective limit to current supply: if you shorted it out for long enough, it burned out. A modern switching power supply does have an inherent current limit: if it doesn't burn out it will still only transfer the rated current.
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.
Best Answer
If you are doing short circuit calculations, you only care about the series portion R+jX. The parasitic capacitance to ground is negligible. Plus this would only be the positive and negative sequence branches. Zero sequence, of course, depends on the winding configurations and core design (core type, or shell type).
There is a pi model you would need if the set tap turns ratio used does not match the ratio of the terminal voltages (in a looped system).
Here are a few slides from a presentation i did many years back. This is based on a derivation that Neuenswander did in "Modern Power Systems", p. 251-254. That old text is $14 on Amazon right now. This is focused on load flow analysis but is easily extended to short circuit analysis by modifying negative sequence and zero sequence networks. Good short-circuit tools like CAPE can handle this as well.
Using admittance because it makes it easier here.
Modeling Transformer Bank with Off-nominal taps
Slide 1:
Slide 2:
Slide 3:
Slide 4:
Slide 5 (ZA is a parallel transformer):
Note: An easy way (really only slightly less accurate) to calculate the current that would circulate between paralleled transformer banks (where one has a slightly different turns ratio) is to take the voltage difference and divide by the series sum of the transformer impedances. For example, if you had a 69kV to 13kV bank in parallel with a 69kV to 13.8kV bank - you could calculate the current that would circulate in the bank as 800V / (XT1 + XT2) where XT1 is leakage reactance of bank 1 and XT2 of bank 2. This current circulates in the banks - as long as it does not exceed bank ratings it won't really hurt anything short term. There will be a var drop due to this circulating current (I squared X) and this will have to come from the system (69kV system in this example).