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# Calculating Transformer Percent Impedance7

## Calculating Transformer Percent Impedance

(OP)
I know I'm missing something fundamental when understanding transformer impedances.

How is %Z calculated?  It seems to me that if you looked at one winding of a transformer secondary (on a typical 480delta to 208wye transformer) you would find less turns, but more conductor.  I am wondering if percent impedance is literally the ratio of impedance of the secondary (low) side as a percent of the impedance of the high side.  Is it just this simple?  I feel like I am missing something fundamental.

Also,
This information I understand to typically vary based on the transformer size/voltage and specs, which would make sense because this would impact the impedance on both the high and low side of a transformer.  It sounds like this is always something that should be verified from the manufacturer, but I've noticed that within SKM there is a calculator present to pop out values.

So the follow up is, for an isolation transformer, that is 480 high and also 480 low, lets say as an example, is the %Z then = 100%?

Please fill in the gaps.  I need to master this stuff.

### RE: Calculating Transformer Percent Impedance

(OP)
Also please explain how X, R, fit in with the original question.

Thanks!

### RE: Calculating Transformer Percent Impedance

2
Another way to understand what the %Z represents is that it is also the % of the primary voltage applied which would cause the FLA to flow through a shorted secondary winding.

### RE: Calculating Transformer Percent Impedance

%Z is also the % of secondary voltage drop from no load to full load on the transformer.

### RE: Calculating Transformer Percent Impedance

#### Quote:

%Z is also the % of secondary voltage drop from no load to full load on the transformer.
This is regulation, not impedance.  They are related, but not the same thing.

### RE: Calculating Transformer Percent Impedance

The transformer impedance is primarily the leakage reactance of the transformer and the winding resistance.  The leakage reactance is the reactance related to flux that does not fully link both windings.  If 100% of the flux in the core linked both windings, the transformer impedance would be only the winding resistance (and very low).  But real transformers do not have perfect flux linkage and there is always some leakage reactance.  This can be controlled to some degree to allow the impedance to be increased or decreased (within some range) by design changes in the coils and core of the transformer.

### RE: Calculating Transformer Percent Impedance

Hi.
Shortly and fully.
Best Regards.
Slava

### RE: Calculating Transformer Percent Impedance

%Z is calculated by a simple test.

Take a single-phase transformer and short out the secondary terminals.

Connect a variac to the primary and adjust the voltage until you measure rated current on the secondary.  Express that voltage as a % of rated primary voltage and the result is % impedance.

As an example, take a 7200 V 240 V tranformer.  Say you measure 144 V on the primary to give you rated secondary current.  144/7200 is 2% IZ where I is 1 per unit, so 2% is your Z.

You can put a wattmeter in the circuit to see how much of the impedance is resistive, and then work out %R and %X.

So it's not the %Z on one side of the transformer compared to the other.

Even an isolation tranformer will have some %Z and I wouldn't expect it to be as high as 100%.

### RE: Calculating Transformer Percent Impedance

I once sent a group of students to the shop to verify the % impedance of some small dry type transformers. They had significant errors compared to the nameplate value. The answer was that % imp is based on a transformer at operating temperature. If you try to verify it on a room temperature transformer you will get a low figure. (Higher current)

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Calculating Transformer Percent Impedance

(OP)
OK, this makes a bit more sense, I guess after reading these explanations.

The way that I am reading that is that essentially the high and low side of the transformer are not perfectly coupled, or that not all of the magnetic field is induced in the transformer's core (and the that conductor is not wrapped 'perfectly' perpendicular to the core).  This does make sense, however in a practical sense when calculating the short circuit current (for infinite bus), you use the formula FLA (based on XFMR size only) / %Z.    My only complaint in this is that it would seem that if there truly was an infinite supply, it would seem like the amount of energy lost due to this effect would (although not nessacarily a linear relationship), would keep increasing as your available fault current of your supply increased.

DanDel's:
This does help quite a bit, and jives with Magoo2's explanation of what to do experimentally to get the %Z of a transformer (this process I have encountered on these boards before).  I guess I'm trying to go a little deeper though into 'what' exactly this quantity/variable is.  I have no problem, for example of doing a linear regression and then coefficients pop out, and they don't really 'mean' anything.  They are just a part of a method for approximating fitting something to a linear model.  Maybe that's just what this %Z is, in that it doesn't really 'mean' anything.

I also want to put a couple things I've noticed, expounding on your explanations thus far:

The 'X' and the 'R' quantities, which dpc is referring to as reactance and resistance of the transformer, respectively, form a ratio of X/R, which appears to be VERY CLOSE but not quite spot on the %Z of the transformer.  I'm wondering where this difference comes from, and how/why (if at all) the X and R variables are related to %Z.  Also, I am wondering how to experimentally verify (test) the values for X and R.  (We've already established how to test for %Z).  It sounds like in terms of power system studies, the %Z is the more important quantity because it allows for you to quickly get the short circuit current levels.

Thanks much guys as always.

### RE: Calculating Transformer Percent Impedance

(OP)
Bill did you do that to teach a lesson, or did you accidentally teach yourself a lesson in the process?

If there is such a substantial deviation based on operating temperature, does IEEE methods for calculating this stuff take that into account?  (In other words I'm wondering if my software is taking that into account)..

### RE: Calculating Transformer Percent Impedance

(OP)
Also magoo it looks like you gave details on how to get X and R of a transformer too, experimentally.  Makes sense, thanks

### RE: Calculating Transformer Percent Impedance

A word on the use of % impedance values.
The % Imp is on the nameplate of most transformers. This describes the steady state short circuit current.
The actual transient fault current may be much higher. It depends on the X/R ratio and the point on the sine wave that the fault occurred. Protection experts calculate the actual value based on the worst possible case of point on the wave.
The value of current based on % Imp is used to select equipment with adequate clearing ability. A switch, breaker \ or fuse rated for an available short circuit current of 10kA will safely interrupt the maximum instantaneous current available from a transformer with an available short circuit current of 10 kA or less.
Anyone who can divide the % Imp into the rated current of a transformer secondary winding can easily and confidently determine the steady state short circuit current even though the actual fault transient may be much higher.
The utilization equipment may then be selected based on this value. Equipment rated for the available short circuit current (also called the symmetrical current component) will have an adequate safety margin built in to safely handle the fault current or asymmetrical current.
The available short circuit convention and practice makes an important equipment selection very easy and simple.
By the way, this is not the only use for % Imp values, but it may be the most often used application by far.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Calculating Transformer Percent Impedance

Hi folks,

Philosophycally, the %Z is something like the percentage of "loss of voltage transformation" in a transformer when it is full loaded.

In a ideal transformer with two concentric windings, the leakage reactance is proportional to the volume of the space between the windings. Note that, in this case, this space will be the same does not matter if it is seen from the primary or the secondary side.

Best regards,

### RE: Calculating Transformer Percent Impedance

jimgineer,

I guess I don't understand your "complaint" regarding the system impedance versus the transformer impedance.

The %Z stamped on the nameplate IS the impedance voltage, determined by actual test after the transformer is built.  This will be temperature sensitive, due to the resistive component.

### RE: Calculating Transformer Percent Impedance

A word on temperature sensitivity of the impedance value. It all depends on what you are trying to calculate. For voltage drop under normal operating conditions, use the max operating temperature Z (highest). For available fault current, use the Z at the minimum expected ambient temperature (lowest).

### RE: Calculating Transformer Percent Impedance

(OP)
dpc let me rephrase my mis-understanding:

I don't see why/how an infinite source would ever be able to provide a finite short circuit current when put through an impedance of any sort.  There is something else going on in a transformer, no?

This would hold true in a circuit (although boring) that just consists of real impedence components.  (ie infinite source generates infinite available current in a fault)  I know there's not really such thing as an 'infinite' source, but what I'm not understanding, maybe, is the wording of calling something infinite if it isn't.  And if it is in the calculation, then why can't it supply infinite fault current?

I know AC gets a bit more interesting when you introduce your C or L in an RC circuit... that's where you have to look at your frequency response to really get a feel for what your output waveform magnitude (and phase shift) all come from.

From my experience, power engineers are often very far removed from these kinds of mathematical exactations and the understanding of the real underlying theory and math that goes into some of this stuff.  I'm not saying that's where you guys are here, only that I am trying to avoid falling into that trap myself.

### RE: Calculating Transformer Percent Impedance

An infinite bus has a source impedance of 0 (zero).  So, infinite bus on primary of transformer, fault on secondary, the only impedance to limit the fault current is the transformer impedance.

### RE: Calculating Transformer Percent Impedance

Percentage impedance of a transformer is the percentage of voltage to be applied on primary to get rated current in the shorted secondary.It is also 1/Z  times of full load current that  will pass in secondary in the event of a shorting on secondary terminals with rated voltage on the primary.

It is the square root of X2 + R2,where X is the leakage reactance between primary and secondary winding, R is the total resistance of  both windings (or copper loss /rating in pu).X depends on the geometrical dimensions of the windings and number of turns( varies as square of turns) and when expressed as %pu on MVA of trf.When we say leakage flux please dont think it as a flux leaking out of core.It is the flux developed between two concentric windings( remember the current carrying solenoid with flux lines)It is zero at no-load and  rated %Z at full load ie %Z varies linearly with load ( approx) .For small  trfs R is significant compared to X and hence temperature sensitive.But for large trfs R is insignificant compared to X and hence Z is almost same as X and independent of temperature. % Z as per standards is always  to reference ambient temp of 75 C ( IEC) or 85 C (ANSI)

Coming to the physical explanation of it-it is the price ( in terms of voltage ie voltage regulation) that we pay to shift current from primary to secondary winding.This was said by that great Transformer engineer who lived in the first half of last centuary,L F Blume of GE.Voltage regulation ie secondary voltage dip with load is happening because of %Z.

For each kVA rating of a transformer,based on geometric dimensions there is an optimum % Z value.If you move from this value up or down, cost increases.When % Z goes up, we call it a copper machine as copper quantity goes up( load loss goes up) and core iron quantity comes down ( no-load loss comes down)When % Z comes down,it is called a iron machine and reverse is the truth.Many times the system operational requirements decide the %z rather than the optimum cost.

For parallel operation,the %z of transformers shall be same to the respective kVA of transformers  to avoid circulating currents between transfornmers.

### RE: Calculating Transformer Percent Impedance

#### Quote:

I don't see why/how an infinite source would ever be able to provide a finite short circuit current when put through an impedance of any sort.  There is something else going on in a transformer, no?
It doesn't appear that you are getting the concept of an infinite source.  This is the same as a source impedance of zero ohms.  Divide the source voltage by an impedance of zero ohms, and you get infinite amps of current.  Add a finite impedance and you now divide by a finite impedance and get a finite current.  The transformer has a finite impedance.

### RE: Calculating Transformer Percent Impedance

(OP)
I am falling short, grasping, and picking up some good pieces as I go.

In terms of the infinite source, I do now understand that it has more to do with the source impedance than it does with the capacity of the source to supply, as I was thinking of it before.  These two things are very much related,, but the abstraction of an infinite source calculation says basically that the source produces no I^2R losses, and is basically rated to provide an infinite amount of current.

Taken a step farther, if we connected an infinite source in parallel to a load bank (resistor), then we would find an infinite amount of current, right?

And then, let's say that we took a big slab of copper and on the load (downstream) side of the load bank, dropped it, inducing a fault.  Regardless of the size or resistance of the load bank, wouldn't we see an infinite amount of current passing through the current, with an infinite source?  And further, wouldn't an infinite amount of current pass through the load bank (resistor) in steady state operation, if we truly had an infinite source?

This is what I am getting at – something is fundamentally different in a transformer compared to the scenario I proposed before.  Because of this abstraction of the %Z, now we can say definitively, that even with an infinite source on the primary of the transformer, if we took a slab of copper and directly shorted Aphase and Bphase together, we can say that the fault current is limited to a certain value.

Is this because of the physical (getting damaged) limitations of the transformer based on I^2R losses (heat)?

### RE: Calculating Transformer Percent Impedance

Infinite source = ideal voltage generator (voltage V) in series with zero impedance.  Short circuit the infinite source and the current is V/0 = infinity.  I'm not sure what you mean by putting a load bank in parallel with an infinite source.  If you mean completing the circuit with a load bank of resistance R, then the current will be V/R, not infinity.  If this is what you mean, then putting a big slab of copper on the load side and dropping it doesn't make any sense.

A sketch would help.

### RE: Calculating Transformer Percent Impedance

#### Quote:

Taken a step farther, if we connected an infinite source in parallel to a load bank (resistor), then we would find an infinite amount of current, right?

I = V/R

An "infinite source" is just an idealized voltage source with no internal impedance, so the voltage provided is constant regardless of load.  But the VOLTAGE is not infinite, so once you connect this source to anything, the current is not infinite.

### RE: Calculating Transformer Percent Impedance

If you drop that big slab of copper across the terminals of that infinite bus (without the load bank), you will get infinite current.

Don't forget that an infinite bus is a theoretical idea; I typically model this in a study as about 100kA at 13.8kV. During the calculation, the fault current on that bus will be 100kA.

### RE: Calculating Transformer Percent Impedance

(OP)
OK, forget the load bank.  I shouldn't have brought that up, it doesn't make sense.

Let's say you drop a big slab of copper on the secondary of the transformer, causing a L-L fault.

With an infinite source, what limits the fault current at this point, physically?  And why does %Z have any meaning for the transformer?

I guess I'm saying that in terms of a calculation, I understand you do FLA/%Z to generate it, but conceptually I am still not seeing why this calculation makes any sense.

### RE: Calculating Transformer Percent Impedance

The transformer has an impedance, it won't allow infinite current to pass through it. Period. This impedance is typically expressed in percent of the kVA rating of the transformer, that's where '%Z' comes from.

The model of a source is called a Thevenin or Norton equivalent.(You may want to Google these terms.) They are essential the same, but a Thevenin equivalent is a voltage source in series with a resistance, while a Norton is a current source in parallel with a resistance. You can see that no matter how large the voltage or current source, the resultant fault current in the external circuit will never be infinite if there is an impedance (transformer, load bank, cable, etc.) in that circuit.

### RE: Calculating Transformer Percent Impedance

Think of an infinite source as being capable of supplying enough current to any type of fault on the system you are analyzing.  For example if you are looking at a 1000W transformer, you will not be talking about 100kA of fault current.  If you are talking about higher kVA and MVA transformers you will be worried about these high currents.  An infinite bus, as many people have said above, is for theoretical discussions.

Another way to interpret %Z:

If you apply voltage to an unloaded transformer (open circuit secondary) it will require a certian exciting current which is related to the primary inductance.
Even though the primary is a low resistance circuit, the inductance of the primary circuit will limit the current draw from the line, you will not see a 'short circuit'.

If you now take this same transformer and short out the secondary terminals, the effective impedance of the transormer seen from the primary is now changed.  The short circuit impedance is now defined by the leakage inductance from the primary to secondary winding(s) and resistive losses of the transformer windings.  This is defined as %Z on the nameplate and the steady state short circuit current is determined by it.

This is what I picture in my head when I try to think about %Z.  I hope it helps...

Regards,
Jim

### RE: Calculating Transformer Percent Impedance

Infinite is relative.
The source has impedance.
The transmission lines have impedance.
The transmission transformers have impedance.
The distribution lines have impedance.
The distribution transformers have impedance.
The secondary conductors from the distribution transformers have impedance.
There are two reasons for assuming an infinite source. (Zero impedance.)
1> The upstream impedance is so low in relation to the impedance of sub-system or transformer under consideration, or so low that it has less effect on curent than normal voltage variation under normal operation that it may be safely ignored.
2> The upstream impedance is known but is subject to change without your control and/or knowledge. This may be due to utility upgrades or utility switching practice. Utility switching may change the source impedance several times a day. An engineer may decide to base his calculations on an infinite source so as to calculate the highest possible current in the event of a fault.
This approach may be erroneous for arc flash calculations but that is beyond the scope of this thread. Search this forum for arc flash comments. Arc flash may be a "Work in progress".
I hope this helps with your confusion.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Calculating Transformer Percent Impedance

(OP)
No source resistance  is hard to imagine, but thinking about it terms of an infinite current, but not an infinite voltage really helps.  It sounds like the whole reason it is considered infinite is because I=V/R where R is 0 or very close, therefore your kA available fault current = infinity.

I understand the reference to Thevenin/Norton equivalents and that also helps to visualize a voltage source with a series resistance that is, 0 ohms.

I think I get it ---

### RE: Calculating Transformer Percent Impedance

(OP)
Can anyone comment on the idea that the imaginary part of the impedance of the transformer has an impact on the available fault current?  I know we've covered X/R which is approximately equal to %Z but what is the reason for the slight deviation and why is the X so much bigger than the R, usually?

### RE: Calculating Transformer Percent Impedance

(OP)
Also (sorry for the separate posts) but I just thought of another: How do you experimentally verify the X and the R (well, the R I assume you can simply accmplish with an Ohmeter).  But after you get the real resistance across one winding, how do you get this to be the overall R of the transformer?

### RE: Calculating Transformer Percent Impedance

The resistive portion can be calculated from the transformer losses, if known. Then, knowing Z and R, X can be computed.

### RE: Calculating Transformer Percent Impedance

The transformer impedance is not imaginary. The source is not infinite but may be assumed to be under certain circumstances.
The transformer impedance is a combination of transformer inductive reactance (X) and transformer resistance (R).
Vr = Rated voltage. Vt = Test voltage.
Impedance may be determined experimentally. Vr/Vt = % Imp.
Equivalent resistance. Use a watt meter.
Normally, X predominates Z to limit short circuit current.
With a resistive load added to the transformer,  the R of the transformer-load circuit becomes much higher than the X and so R often predominates in regulation calculations.
Stated another way.
Under short circuit conditions, X may be several times R and X predominates Z.
Under unity load conditions, the total R (transformer R in series with load R) may be many times the transformer X and so R may predominate the total Z of the transformer-load circuit. This is related to the regulation of the transformer. This is why regulation is a lower (better) percentage than impedance.
Start drawing triangles. X^2 + R^2 = Z^2.
Z, voltmeter and ammeter.
R, Wattmeter.
X calculate from X^2 = Z^2 - R^2.

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: Calculating Transformer Percent Impedance

#### Quote:

I know we've covered X/R which is approximately equal to %Z but what is the reason for the slight deviation and why is the X so much bigger than the R, usually?
Not to be snotty, but if you don't know the relation between R, X, and Z, I suggest you find a basic text on power systems before trying to understand infinite sources and fault currents through transformers.

### RE: Calculating Transformer Percent Impedance

(OP)
Thanks fellas.

Yeah, I definitely need to brush up on some of my power system math/theory, which I neither excelled at or flunked at the college level.  My goal has been to draw out some of your wisdom and intuition for what mathy/theoretical stuff is actually useful to know.

Bill, I wasn't referring to imaginary as in imaginary friend, but rather the imaginary component of a phasor quantity, when expressed in rectangular form.

IE:

Z = X + Rj

I know this relationship and I'm pretty familiar with phasors, complex math, etc, but the %Z is what is getting me, not Z or impedance of a load, etc, which is what I'm used to seeing.

It sounds like %Z is ~X/R , but not quite.

And it also sounds like we have a straightforward method for getting %Z (outlined by magoo among others), experimentally.  As well as R, (ohmmeter) experimentally.  Maybe what I'm hoping for is really not possible without getting down to the integration level and the core geometry, etc for something like this.  Which, I'm perfectly fine with this, but I just want to make sure that I'm going as far as I can with my understanding, and understanding when an advanced computer model would fill in the gaps.

### RE: Calculating Transformer Percent Impedance

It has been said so many times in this thread what Z is. Both methods are correct (the voltage drop and the X+R).

X is a lot larger simply because most transfromers are built that way. Having a certain Uk keeps short circuit current within limits. So, you need a certain Z. If most of that Z were resistive (large R and small X) then losses would be very high. Putting most of the Z in X avoids those losses.

And, please, %Z is NOT X/R. That quotient wold be tangent for short circuit current phase angle - a quantity that is not very often used.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

### RE: Calculating Transformer Percent Impedance

It is not that complicated.
The impedance is expressed in per-unit values or percent so that the impedance of the transformer primary and secondary can be added together more easily. The definition of % Imp is given in actual values under test conditions. There are some effects such as ski effect and core losses that may be difficult to calculate accurately but that are reflected or included in the impedance test. Skin effect is somewhat current dependent and the test includes the skin effect at actual short circuit current levels. The effect of eddy currents is also included. These may be difficult to calculate from first principles.
Take the rated KVA and rated voltage. Calculate the full load amps. Divide the % Imp into the full load amps. Now you have the steady state current under short circuit conditions.

% Imp came into use before any of us were born. Many small and medium sized systems were designed and installed without the aid of an electrical engineer. % Imp allowed electricians and technicians to safely select properly rated switchgear. It allowed uneducated linemen to safely parallel transformers. It allowed line supervisors with much less than a full engineering education to asses the performance and loading of transformers of unequal % Imp.
A few generations ago, % impedance was taught to third year apprentice electricians. These were students who panicked when faced with simple trigonometry. Sine, and cosine, often tangents were omitted.
Don't try to make it more difficult. Don't fight the problem.

Bill
--------------------
"Why not the best?"
Jimmy Carter

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