Transformer Voltage Sag 2

[Mbrooke]

Does anyone have a table, or know how to calculate the voltage droop on the secondary output terminals of typical distribution transformers for a given over current or short circuit value above the nameplate rating?

I'm running some equations in regards to touch and step potential.

 

[Fischstabchen]

I don't think you can unless there is a special rule of thumb. You need the winding impedances from the manufacture.

 

[Mbrooke]

I have the data plate impedance, both for single and 3 phase units.

 

[waross]

The nameplate data typically gives you the Impedance voltage or the %Impedance.
That is the per unit voltage required to drive full load current into a short circuit.
For example, a single phase, 1000 KVA, 13,000 Volt to 480 volt transformer has a %impedance of 4%.
1000 KVA at 480 Volts is 2083 Amps full load current.
The Available Short Circuit Current will be 2083 Amps / 0.04 = 52 kA This is the figure used to size switchgear.
It is the symmetrical current, and the steady state current after the initial DC offset decays.
The amplitude of the DC offset depends on the point on wave, and on the X/R ratio.
At high values of X/R ratio and worst case point on wave, the peak offset may approach the peak to peak value of the RMS current or 2.828 times the RMS value or ASSC.
For voltage drop under load, you often need more information than is provided on the nameplate.
You need to know the percent regulation. That is the voltage drop under a full load at a reasonable power factor.
I don't recall the PF that regulation is specified at. You can Google it.
For overloads, you may get a reasonable estimate by extrapolating the PU regulation.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Mbrooke]

Thanks, epic reply!

To be open, Reason I'm asking is that I'm working out the touch voltage on various outdoor switchgear, panel boards and equipment rated 1000 volts and under.

Going by an infinite source with equally sized ungrounded and grounded/grounding conductors the touch voltage comes out as 1/2 the line to neutral voltage- ie a 139 volts to remote earth on a 277/480 volt Y supply.

However when secondary voltage droop is taken into account the touch voltage is greatly reduced- the closer the fault is to the supply transformer the lower the touch voltage.

 

[FacEngrPE]

The regulation calculator on this page (bottom of the page) might be helpful, but it stops at 100% load.

https://voltage-disturbance.com/power-engineering/transformer-voltage-regulation/

Inputs are X/R, Impedance, Power Factor, and Loading.
I think you might need some creative options, the following might provide some ideas.
[ul]
[li]evaluating-transferred-voltage-hazards

https://www.easypower.com/resources/article/evaluating-transferred-voltage-hazards

[/li]
[li]electrical-grounding Resources

https://www.easypower.com/resources/topic/electrical-grounding

[/li]
[li]touch-and-step-voltage-calculations

https://www.easypower.com/resources/article/touch-and-step-voltage-calculations

[/li]
[li]Calculation-of-the-Maximum-allowable-Touch-and-Step-Voltages

http://www.electrical-knowhow.com/2...aximum-allowable-Touch-and-Step-Voltages.html

[/li]
[/ul]

 

[7anoter4]

Generally speaking, you can overload a transformer for a while, but in order to limit ageing of insulation and so reducing the transformer life you have to limit the exposing time. So, you may consider it as a transient overheating and calculate the winding temperature according to IEC 60076-2,11,12 or or IEEE Std C57.96-1999 for dry type or IEEE Std C57.12.90-2010 for liquid immersed transformer.
The short-circuit reactance is not influenced by temperature-in my opinion-only the resistance and using the classic formula we can calculated the value.

 

[Mbrooke]

True, though I'm looking at short circuit conditions lasting up to 10-20 seconds and no more. Normal overloading will probably not take place.

 

[waross]

Does this help:
Transformer impedance is determined by the directed sum of Transformer resistance R_t and Transformer inductive reactance X_t
If there is a feeder adding impedance the impedance of the circuit becomes the directed sums of the line resistance, the the transformer resistance, the line inductive reactance (often ignored) and the transformer inductive reactance.
The formula: Total Impedance =√(R_t + R_l)²+(X_t)²
or
√(R_t + R_l)²+(X_t+X_l)²


Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Fischstabchen]

MBrooke,

I am not understanding why you are doing what you are doing. When you do grounding grid calculations in accordance with IEEE 80, you use the fault current that is available. The available fault current creates the voltage gradients as it passes through your grid and back to the source. When the available fault current is calculated, it includes the impedances of your transformer windings and the voltage droop caused by them. I would hesitate also in trying to play games to reduce the grounding grid due to it corroding with age and is a lot of work to redo if it ever needs to dissipate more current.

 

[Mbrooke]

@Waross: Yes that does help in that in confirms how I've been calculating the ground fault loop impedance thus far. Good to know that I'm on the right track. Thank you!


@Fischstabchen: Correct, but keep in mind things like light posts, score boards, kitchen equipment, portable equipment, panel boards ect do not come with a ground grid underneath. As such any fault within them must assume the person is referenced to remote earth IMO.

I'm not sure which codes your country uses, however NFPA-70 has 250.4 A 5 which mandates that an effective ground fault current path be present, and that path open an OCPD. The time is not specified however whereby I take that as meaning discretion is left up to the engineer or electrician.

 

[BlackWhite24]

The transformer model comprises the parameters are obtained from experimental measurements. The effect of the sag type, depth, duration and initial point-on-wave on the inrush current peak value is studied. The current summit has periodical dependence on sag length as well as the initial point-on-wave, and a linear dependence on thickness. Unsymmetrical sags can lead to present peaks as high as those of symmetrical sags.