Energizing Power Transformer 1

[enggines]

Hi All!

What is the correct or better way of energizing a 15/20 Mva, ONAN/ONAF, 69-13.8 KV, DYn1 Power Transformer serving at least feeders? Without Load or at least one (1) feeder is initially loaded? My concern is the inrush current. One guy advice me to energize with load to damp the inrush while another is saying, should be without load to minimize. Which is which.

Any comments or alternative method would be highly appreciated.

Thanks.

 

[electricpete]

I would say that in comparing primary R to secondary R I made an error in assuming it is proportinal to R. That would holde true if the turns were of the same conductor size, but secondary turns are larger to handle higher current. I still think primary L/R is higher than secondary, but not by a whole factor of N=N1/N2 like I suggested above.

Now I'm wondering about the connection... probably delta on high side and grounded wye on low side. Will that make a difference? I'm not sure.

 

[electricpete]

Should be:"... I made an error in assuming R was proportional to number of turns N..."

 

[electricpete]

Still thinking about the impact of the transformer connections ie. delta/grounded wye or whatever (by the way what are they?)

The current has a shape that I showed in my link. Since there is a huge peak twice per cycle (different point in time and different magnitude for each phase), I would think that the excitation currents from three phases would not sum to zero. That also means the fluxes from excitatio currents would not sum to zero. But assuming 3-leg core form they pretty much have to. Leads me maybe to believe that if there is no path for current on the other (nonenergized) winding the peak excitation current would be limited. That in itself would seem to suggest that excitation current peak would be limited when energizing from the high side (no path for current on the secondary), but when energizing from the secondary (assuming there is a path for circulating current in a primary delta winding).

I'm not sure what to make of it.

 

[neps3]

Hallo electripete

I tried to give a rule of thumb figure. A 220/110kV transformer will have a lower inrush current if switched on the 220kV side. However the current magnitude depends on where the switch closes on the ac voltage wave.

Now

Initial peak value Imax

=1000*h*Ac(Bres +2Bmax-Bsat) / (3.2*n*As)
h=effective length of the magnetising winding
n=Number of turns in series in the winding.
Ac=Cross sectional area of core
As=Effective cross sectinal area of the air-core magnetic field within the excited winding,ie. between the core and the winding
Bxx are the corrosponding flux densities.

Merlin Gerin gives a value for I(rms)

=SQRT[0.125*Ie^2*te(1-e-(2t/te)) / t
te =current damping time constant in seconds.(Time after which the current falls to 37% of its intial value.
Ie=maximum peak current
t=Time in seconds after which the current is considered to have reached its final value.

In general approximation is;

t = 3*te


I have carried out onsite experiments of inrush current on a 220/110kV interconnecting transformer 300MVA of a switchyard. Some dumb operators close on the 110kV side which has tripped my thermal unit at times.

If you have access try and read the following:

Gravett KWE: "Magnetising Inrush currents in transformers",MSc(Eng) Thesis, London Oct 1951

Finzi LA & Mutschler WH Jr: "The inrush of magnetising currents in signle phase transformers" AIEE Transactions 1951, Vol70,Part II pp 1436-1438.

Unless we practice designing transformers we only need to know first pass only rule of thumb (accuracy of course is of concern) figures. Thats what Consultants do??.

 

[electricpete]

Hi Kantor. Good comments.

I think we all apply combine some degree of direct observation, thumbrules, theory in these things. Sometimes I lean heavily on the theory when trying to think through questions like this on the board. Hopefully I make it clear which things I know and which I am just thinking out loud on.

I will say that the simple single phase simulation in the link I provided seems to reflect inrush current very well (the saturation curve could use a little tweaking to get higher peaks). Maybe it is not well presented but it is just a simulation of ac voltage suddenly applied to a real (nonlinear) inductor in series with resistance.

Regarding the original question of effect of load on inrush:
Gord and others have suggested that the presence of load changes the L/R time constant and therefore duration of the inrush. This makes good sense to me. It is my understanding of your first post that you disagree. (do you?)

Regarding hi-side vs low-side transient: I defer to your experience if you say that hi-side is higher. I believe it should be possible to analyse this from standpoint that peak magnitude depnds upon per unit inductance (even though non-linear, the fact that one side has lower per unit inductance indicates it will go into saturation sooner), and the decay rate depends on L/R. Rethinking my previous analysis I come up thinking that the per-unit L is same on both high and low-side and L/R is similar. There is one other factor I've not considered is that the low side may actually have two parallel windings internally (even though it's not evident from looking at the bushings. I think this may push it towards a higher inrush. At any rate even if the in-rushes are comparable on per-unit basis, the hi-side system is generally equipped to handle the inrush so energizing from hi-side makes sense as you say. Still leaves us unable to respond to Adar.

Regarding your thumbrule of inrush duration... I know there is substantial variation since generator stepup transformers still have saturated inrush peaking pattern up to 30 seconds after energization. I believe this varation from your thrumbrule is due to L/R characteristics of either the transformer or the power system.

Thanks again the discussion. Looking forward to more.

 

[neps3]

electripete

Thanks for the info. I agree in most of what you have stated.

But I am unable to digest the fact that the presnece of load changes L/R the time constant. In that case a secondary open circuited transformer will have its R to infinity. The time constant then is: L/infinity.

I appreciate your comments on (your) approach because I also like to understand the inner part of an explanation, theorical & practical reasoning.

I am not a transformer designer, but I would like to research more into the inrush.

I like your approach.

I like an answer from gord on the L/R. Just want to understand how L/R infleunces the inrush.

Regards

 

[jbartos]

Suggestion to DanDel (Electrical) Feb 19, 2003 marked ///\\jbartos, it seems to me for the purposes of this question that v(t) and L are the constants and di(t)/dt is the quantity in question.
///Reference:
1. Gordon R. Slemon "Magnetoelectric Devices Transducers, Transformers, and Machines," John Wiley and Sons, Inc., 1966, Figure 3.4(a) Equivalent Circuit for Transformer on page 175 lends itself for the following relationships:
a) Eq.1 for the transformer with open secondary winding:
E1(t)=constant input voltage=R1 x i(t) + Lm x di(t)/dt
b) Eq.2 for the transformer with loaded secondary winding (sometimes the secondary may even be loaded with short circuit R2=0 Ohms)
E1(t)=constant input voltage=R1 x i(t) + [(Lm x LL)/(Lm + LL)] x di(t)/dt
Now, LL <<Lm<Constant, Henry; therefore, the term with di(t)/dt is much smaller in b) than in a). Consequently, the loaded transformer with R2=0 Ohms on its secondary, as the worst case, produces the smaller transient voltage than the unloaded transformer with R2 approaching to infinity.\\\

And I don't believe you can add a resistor in parallel with the inductor(L, in Henrys) in the formula as stated, since there are no values of impedance present.
///The formula stated was just and eng-tip. It is actually an essence of the inductive transient and a transformer equivalent circuit. Have you ever seen an equivalent circuit for a transformer? Also, I keep my belief for religeous places.\\\

 

[neps3]

Hello jbartos

I believe you are an engineer. Do you know the impedance voltage is the % voltage that you apply at the primary terminals with the secondary shorted. The secondary current is the full load current. So if Z% of the voltage produces FLC at the secondary (with secondary shorted), what do you think full voltage application at the primary will do to the transformer with the secondary shorted.

I have noticed your answers to some questions that harmonics produces vibration even in a generator.

If you think that transformers should be switched on with the secondary shorted, you are free to do so. But I am not.

 

[DanDel]

jbartos, I'm not sure what you are trying to say with your last post, but I will try to clarify mine. The original question asked about inrush current. Your original answer talked about using this equation:

v(t)=L x di(t)/dt

to find a change in voltage: &quot;the voltage v(t) cannot increase as much since the current will be flowing in the resistor in parallel thus reducing the inrush&quot;

My question to you was, the voltage(along with the impedance) is a known factor [as expressed in your reference equation: E1(t)=constant input voltage=R1 x i(t) + Lm x di(t)/dt], and the unknown was the inrush current.

No one will argue that if you have less impedance in a circuit you would have less transient voltage(voltage drop). If you had zero resistance and zero inductance, you would have zero voltage drop. The question is, what is the current for a circuit with a known applied voltage and a known impedance. Use your equations to solve for i(t) with a known voltage, resistance, and reactance and perhaps we can arrive at the proper solution.

As far as your last paragraph, I understand that many people who use this forum are from different countries with perhaps different understandings for words used in common English. If you feel that religious issues were mentioned by me in this forum, rest assured that was certainly not my intention.
If, however, you feel that mentioning religious issues yourself(along with what appears to be an insult, excuse me if I'm incorrect), is appropriate here, then we do have a dissagreement.

 

[electricpete]

Good comments, Dan.

Some more thoughts on the general question:

My simulation linked was unloaded tranformer. The resistance included there was intended to represent resitance of transformer windings and power system losses.

Someone made a good point that if use a simple mode of he non-linear magnetizing inductor in parallel with a resistance representing the load, then the solution is the sum of the two currents, by kirchoff's law.

The only way the external resistance can influence the magnetizing current is by changing the voltage across that non-linear inductance, which can only be done if we include series leakage reactance and winding resistance.

I will try a simulation of that. Won't have time until next week though. Hope to hear more discussion from all in the mean time.

 

[Steve4444]

Impedance of a transformer is derived from both the resistance and reactance of the transformer, The resistance is derived from primarily the resistance of the wire and to some degree the silicone/steel laminate material. The reactance is a function of the turns of conductors around the core, and the size of the core material. As a Voltage is applied across the primary coil, the increasing EMF builds a magnetic field. Although current wants to flow through the Coil, the building magnetic field causes current to flow in the opposite direction from the excitation voltage with nearly the same force as that of the excitation voltage, effectively keeping the current at zero Amps through the first 90 degrees of the sine wave.

At peak voltage, the magnetic field has grown to it's maximum amplitude. As the excitation voltage falls off of peak, the magnetic field collapses, inducing a current into the coil, in the same direction as the excitation voltage. Current will flow in the same direction as the Excitation voltage intended, because of the collapsing magnetic field. As the Excitation voltage reverses direction, crossing zero, the magnetic field collapses faster, inducing peak current into the coil. If plotted, there would be zero current flow until the Sine wave reached peak voltage. As the voltage falls, the current will rise. As the voltage falls to zero, toward the opposite peak, the current peaks. As the Excitation voltage approaches peak, the magnetic field is nearly at it's maximum amplitude, the rate of change in the magnetic field causes the current to fall off, such that you have zero current at Peak Excitation Voltage again. As the Voltage falls off of the peak heading toward Zero, the Magnetic field collapses, again inducing current to flow in the direction that the Excitation Voltage intended. When the Excitation Voltage is at Zero, the Field collapses rapidly, causing current to peak again. As the Voltage rises toward the next peak, current falls off to zero.

This effectively causes the current to lag the voltage by ninety degrees, hence reactive. There is no tremendous inrush current because of the transformer. What current the transformer uses, is in the silicone steel laminated core of the transformer. In order for the magnetic field to build, the molecules of the laminate steel must align themselves magnetically north to south, as forced by the direction of the wire turns and by the Excitation Voltage. Saturation occurs when all the molecules that can align themselves have, a higher voltage will not increase the magnetic field. Heat is generated by the eddie currents of the silicone steel molecules as they move to align themselves, releasing their kinetic energy as they bump into one another. The steel is laminated to reduce this effect. Because the molecules align themselves and are nearly fluid with respect to their movement, the magnetic lines developing in the core are not necessarily straight and do in fact bend through the steel following the path of least reluctance created by those molecules that have or more nearly have aligned themselves.

For the purposes of this discussion, I'll go no deeper. E-mail me should reluctance be an issue in your design.

The mathematical models offered by another poster are just that, models. I particularly like his harmonics model, however his inrush model would only fit a switching power supply.

This day and age, most electronic office equipment, such as copiers, computers, UPS's and even the T-8 light fixtures all use switching power supplies. It is these connected devices that cause high inrush currents and outlandish harmonics effecting power quality.

The clear majority of these devices charge capacitors. A number 18 AWG Stranded TW conductor will see current spikes as high as 60 to 90 Amps of extremely short duration, considering power in the USA is 60 Hz. Many of these switching supplies operate around 5,000 Hz. To a power system, that can put unusual currents on the neutral. The National Electrical Code over the last few years has limited the System Neutral Reduction. Experience has dictated that in some instances the, Neutral is larger than the supply conductors. Now even though the switching currents are extremely high, they are normally of such short duration that the average heat generated within the conductors is well within range of the conductors and terminations. It is only the multiwire branch circuits that Neutral Conductor Harmonic currents become a concern. Steve4444

 

[Shortstub]

The &quot;observed&quot; effects to explain the variation of methods for energizing a transformer have no theoretical basis. That is, it doesn't matter whether energization occurs om the HV or LV side! Nor does it matter that the transformer is unloaded or lightly loaded!

The more likely reasons for &quot;observed&quot; variation involves the randomness of the following parameters:

o The instant the transformer is &quot;switched&quot; on, i.e, zero or maximum volts!

o The magnitude of the residual magnetism, i.e., zero or maximum magnitude!

o The polarity of the residual magnetism, i.e., equal or opposite to the flux build up under &quot;normal&quot; conditions!

o The robustness or &quot;stiffness&quot; or impedance from the source of supply to the transformer, i.e., up to the HV or LV terminals!

 

[electricpete]

Hi Steve - good comments. Why would the model apply to switching power supply but not power transformer?

Shortstub - no-one would argue with those effects. But I don't think the effect of L/R in determining the duration can be ignored. If we remove all R from the power supply circuit, transformer windings... also stray losses and hysteresis losses which can be modeled as R in equivalent circuit, then it seems to me inevitable that the dc component of the flux and dc component of magnetizing current would have no mechanism to decay and magnetizing inrush would last forever. (where would the energy go?... real power canot be carried by fundamental current which is 90 degrees from supply voltage, nor by dc or harmonic currents which are different frequency than supply voltage) Not a realistic situation, but it demonstrates that as we increase that series resistance associated with windings, power system, stray losses, the inrush duration must become shorter.

Resistances such as load resistance which are not in series are a little trickier to analyse... I'm not sure on those effects and will comment more next weekend.

I agree my initial comments on hi-side vs lo-side inrush were wrong.... based on an incorrect assumption about resistance as I discussed.

 

[Steve4444]

I had been up half the night researching an unrelated topic, when I came across this Forum. Unable to resist checking out anything to do with power quality, I jumped into the middle of an ongoing conversation. Little things like adding a current limiting resistors and inside humor set up a paradigm, I usually do not get caught in.

With a clearer head and a much greater respect, I'm eating a lot of &quot;Humble Pie.&quot;

Some years ago, I was investigating core hysteresis causes, permeability of materials and properties of the development of magnetic flux within the core materials in generators. I was also checking the validity of an expensive software product to predict these properties and map relative flux within the core. I did not realize that this too, was the topic of your discussion.

If I can contribute or have a question, I will not hesitate to respond. Until then I will follow your discussions as a casual observer, while I consume the pie on my face. Steve4444

 

[Steve4444]

To reduce the effects of residual magnetism, therefore reduce the inrush of Large Transformers, drop some load prior to de-energizing the primary.

When power is applied to the transformer primary again, it will not have to overcome the residual magnetism left from the last de-energizastion.

 

[Shortstub]

Steve4444,

As a matter of fact to ameliorate the inrush problem, series resistors and aux contacts to bypass them, were fitted to the supply breaker. While it was not a common US practice, it was used in Europe!

Electricpete,

Don't overlook the role of ther other two phases!